Economic Evaluation in the Petroleum Industry

This chapter covers the basic economic principles that govern the oil and gas industry. It can be

considered a microcosm of the topics covered in the remaining three chapters. This chapter is self-

contained in the sense that, by using this chapter alone, many oil and industry problems can be solved.

Many industry professionals use the principles described in this chapter to make daily economic

decisions. Through various case studies, we will illustrate how to apply these principles to real life field

examples.

This chapter contains five main sections. The first section covers the decision-making process in the

industry. In the second section, we cover the basic terminology used in the industry; this includes some

terminology specific to economic theories and some terminology specific to the industry only. In the

third section, we describeil and gas reserves. For any exploration and production company, the main

assets are oil and gas reserves. Understanding the basic definitions is important before we can estimate

those reserves. In the fourth section, we consider two economic methods which are commonly used in

the industry: payback period and profit-to-investment ratio. In the fifth section, we introduce the

concept of time value of money, which allows us to relate monies that are collected at different times.

Underlying all these principles is the common theme of assessing uncertainties. Without incorporating

’ uncertainties, no realistic economic decision can be made. Although we have an explicit chapter on

Economic Uncertainties, the idea of gaining a solid grasp on uncertainties is an important one.

Therefore, many examples will introduce the concept of uncertainties in an intuitive manner.

DECISION-MAKING PROCESS

Beginning with the decision to explore for oil and gas, economic decisions become an integral part of the

project. Do we need to sign an agreement to acquire a concession on a lease? Will a signing bonus be

paid? If so, how much? Also, what types of additional ’commitments, such as work programs and future

drilling activities, are needed? If a concession is acquired, does additional geological or geophysical data

need to be gathered before drilling? If sufficient data are available, where should we drill the first well?

If the well is successful, are the hydrocarbon reserves in sufficient economic quantities to justify

additional drilling and exploitation (Davis 1968)?

After finding hydrocarbons and knowing that they can be produced in economic quantities, we first

need to decide whether to develop or sell to another party. If we decide to develop, we need to

consider the decisions required during the drilling and production phases - drilling techniques, well

completion techniques, surface separation equipment, piping and tubing requirements, and the rate of

production. These factors are further compounded by the changing economics of oil and gas prices, over

which the producer may not have any control.

The economic decisions continue throughout the producing life of the project. Once the production

begins, we need to consider the possibilities of secondary and tertiary oil recovery techniques, drilling of

infill wells, and the implementation of artificial lift techniques. Also, the production scenario can be

significantly affected by changing government regulations related to production quotas or to

environmental and political concerns. At the same time, knowing that present reserves are going to last

a finite period, we have to make decisions related to future acquisition and leasing of onshore and offshore lands.

Economic Evaluation in the Petroleum Industry

Chapter 1 - Economic Principles

As discussed, the decision-making process is cyclical and continues throughout the perpetual existence

of the oil company. As old reservoirs are exhausted, new reservoirs need to be exploited.

Inherent in any economic decision process is the role of uncertainties. Without understanding

uncertainties, no economic decision can be made. In the oil and gas industry, three uncertainties play an

important role: technical, economic, and political.

Technical uncertainties are the uncertainties associated with applying a new technical concept to a

reservoir. This can be due to a new technology that is being implemented for the first time, or an old

technology that is being applied to a new reservoir. Technical uncertainty is highest when the

technology is implemented for the first time. As it becomes routine, uncertainty will be minimized. An

example of this is the drilling of a horizontal well in 1980’s. At that time, the technology was new and

the uncertainties were significant. As the technology matured, drilling a horizontal well became routine

and the technical uncertainties were reduced significantly.

Economic uncertainties relate to the uncertainties associated with benefits and costs. The biggest

uncertainty is the price of the commodity. For any economic evaluation, knowledge about the price of

oil and gas is important. Changes in the price of these commodities can have a first order impact on

economic decisions. Any operating company examining the economic viability of the project will be

remiss if it does not include the price uncertainties in its evaluation. Additionally, the uncertainties in

costs related to service activities are also important. For example, as commodity prices increase and

demand for services increases, service prices also increase. Having a good understanding of those

uncertainties is critical in economic evaluation.

Political uncertainties result from uncertainties from new regulations and laws that are anticipated but

not certain to be implemented. As unconventional resources become more important and exploitation

of those resources requires fracturing of the formation, regulations related to fracturing and their

impact on economic viability have become increasingly important. Other examples of political

uncertainties include changes in tax regime and changes in a country’s government after an election or

political turmoil. These changes can have significant impact on economic decisions. Although some of

these changes are difficult to anticipate, consideration should always be given before making any

economic decisions.

When considering uncertainties associated with any particular economic decision, it is not necessary for

all the uncertainties to be present in every decision. For example, running a routine gamma ray log in a

well does not require evaluation of technical or, for that matter, any other uncertainties. Conducting

multi-stage fracturing in a horizontal well has become routine, but could be impacted by new

environmental regulations related to fracturing fluids. The price of natural gas can also have a big impact

on the viability of producing gas from unconventional resources. Each situation requires identification of

uncertainties and ensuring that they are incorporated in the decision-making process.

Any decision-making process has to be a rational, well informed process. It should be consistent with the

available information and the constraints imposed. Unlike playing a card game or gambling, intuitive

deductions or "gut feeling" are rarely useful in making objective decisions related to economic

evaluation. The steps involved in making rational economic decisions are as follows (Newman 1991).

2 Mohan Kelkar, Ph.D J.D.

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RECOGNITION OF PROBLEM

Recognition of a problem includes defining the problem at hand, its importance, and associated

uncertainties and risks in dealing with the problem. For example, typical problems faced in the

petroleum industry are (DeGarmo, Sullivan and Bontadelli 1993):

� Should you buy a lease or concession?

� How many wells need to be drilled?

� What production scheme should be applied?

� What type of drilling method should be used?

If we consider the development and exploitation of a petroleum reservoir in chronological order, the

problems we will encounter can be stated as (Campbell 1987):

� Will geological/geophysical and surrounding reservoir data locate the reservoir?

� Will the wild cat well produce in commercial quantities to justify additional investigation?

� Will additional delineation be required before further development?

� Is full development of the field economically feasible?

� What is the optimum strategy for exploiting the reservoir?

In recognizing these problems, we also need to understand the associated risks and uncertainties.

Complete recognition of a problem should include the identification of a problem and the associated

risks. Without the inclusion of risks, the solutions obtained may be simplistic and are often

misleading.

IDENTIFICATION OF OBJECTIVE/GOAL

Once a problem is identified, the next step is to ascertain the objectives that need to be satisfied.

For the same problem, depending upon the objectives, economic analysis can be different. For

example, an oil company with a goal to optimize production and a lending institution with a goal to

provide a secured loan will analyze the same prospect differently. The oil company will consider the

economic analysis in the most favorable light to secure either a loan or the approval of its

shareholders. The lending institution, on the other hand, will analyze the prospect in the most

conservative fashion; its objective being to provide a secured loan. The lender may cut the potential

reserves of a prospect in half before granting a loan to assure that sufficient collateral exists on the

loan.

As another example of different objectives, a government-owned company may decide to opt for a

less efficient, less mechanized (and probably less profitable) operation to satisfy the objective of

retaining its employees, or an oil company may decide to operate an offshore facility at the expense

of additional safety to ensure the chances of environmental damage are minimized. Another

example is the decision of oil companies to maintain pre-war gasoline price during the first Gulf War

crisis. Although the crude price increased after the war started, by maintaining gasoline prices at

pre-war levels, the oil companies considered the objective of better public perception to be more

important than maximizing profit.

Economic Evaluation in the Petroleum Industry Chapter TI - Economic Principles 3

Briefly, typical objectives associated with any petroleum industry problem can be summarized as

follows (Campbell 1987) (lkoku 1985):

� Maximization of profit / minimization of loss

� Lending of capital using secured credit

� Diversification of activities

� Determining impact of government regulations

� Public perceptions

� Maximization of jobs

� Environmentally sound operations

� Selling and buying properties

� Capturing or increasing the market share

� Tax assessment

� Safety considerations

� Improving employee satisfaction

Not all of these objectives are monetary. The non-monetary objectives can result in choosing

different solutions rather than simply concentrating on the objective of maximizing profits or

minimizing losses. In this chapter and throughout this book, we will concentrate on this monetary

objective. Consideration of non-monetary objectives involves factors that are difficult to quantify

and, as a result, difficult to evaluate. If these factors have to be considered, every attempt should be

made to convert them in terms of monetary units, except when the non-monetary objectives have

ethical considerations.

ASSEMBLY OF RELEVANT DATA

Data collection is often tedious and sometimes the most time consuming process. Without careful

collection and analysis of available data, the resulting economic analysis may be worthless. The

assembly of data requires evaluation of available resources and feasibility of obtaining additional

data. This step itself involves decision making. For example, the design of a waterflood can be done

using some simplistic approximations and assuming analytical solutions. It may be quick, but may be

associated with a high degree of uncertainty. On the other hand, the design of a waterflood may

involve methodical collection of additional data through the drilling of additional wells or through

additional geological and geophysical information, extensive numerical simulation studies, and the

implementation of a pilot project. After assessment of the previous steps, field-wide waterflooding

will be implemented. Such an elaborate design will be associated with less uncertainty; but will cost

more to implement; therefore, collection of additional data before implementation of a waterflood

depends on the incremental benefits received by such collection. For small waterfloods, the

simplistic method may be appropriate; for large waterfloods, the detailed procedure will be needed.

Which route to follow will, in itself, involve making a decision. This decision will, essentially, depend

on one principle: the incremental benefits resulting from the collection of additional data should

outweigh the cost of collecting the data.

In oil and gas investment, additional data collection includes the quantification of uncertainties and

the risks associated with any project. More often than not, uncertainties are underestimated

4

Mohan Kelkar, Ph.D., J.D.

resulting in an improper evaluation of the project. For example, Table 1-1 summarizes the economic

evaluation of several Gulf of Mexico projects. As shown, the reserves are typically overestimated

(Brush and Marsden 1982), investments and time of completion are underestimated, and

profitability is significantly less than predicted. This table is a vivid illustration of what happens if

uncertainties are not properly accounted for in economic analysis. It is also an illustration of an

optimistic bias in petroleum economic analysis. As a rule, if in doubt with respect to quantification of

uncertainties, it is better to err on the conservative side.

Table 1-1: Economic Analysis of Gulf of Mexico Projects (Brush and Marsden 1982) Estimated Actual Overestimated! Optimistic!

Underestimated Pessimistic

Project Scale & Timing Measures Feet Drilled

Number of Wells

Days of Drilling

Platform Time, Months

Project Time, Months

Production Measures

Oil Produced, BOPD

Gas Produced, MMcf/D

Condensate Produced, BCPD

Reserves Measured Oil Reserves, bbl

Gas Reserves, Bcf

Condensate Reserves, Mbbl

Financial Measures Cost Per Footj’$/ft

Equipment Cost, Thousand $

Platform Cost, Thousand $

Well Cost, Thousand $

Total Cost, Thousand $

10,849 11,011 Underestimated Optimistic

6.2 7.8 Underestimated Optimistic

328 454 Underestimated Optimistic



3,612 3,297 Overestimated Optimistic

34.3 31.7 Overestimated Optimistic

368 287 Overestimated Optimistic

6,106 4,906 Overestimated Optimistic

73.7 70.3 Overestimated Optimistic

753 711 Overestimated Optimistic

113 129 Underestimated Optimistic


3,142 3,053 Overestimated Pessimistic



For a proper evaluation, it is critical that all relevant data be collected with a proper perspective. An

example would be an oil company making a decision to get all the core evaluations done by an

outside lab rather than doing it in-house. In considering the savings by "out-sourcing" the core work,

it is important to collect the information not only related to actual core measurements, but also the

cost of space requirements and the overhead costs associated with respect to that operation (total

cost of ownership). In other words, the perspective should be from the oil company rather than the

team responsible for measuring the core properties.

Also, if we are comparing alternatives lasting for long periods, relevant information with respect to

future cash flows should be gathered. For example, if we are interested in installing an electric

submersible pump on a well, the future benefits received as a result of installation of that pump

over the life of the pump should be collected.

Prediction of future benefits is an integral part of any economic analysis. We also need to quantify

the associated uncertainty. This can be done by analysis of historical data plus engineering judgment

derived from project-specific data. Over a long period of time, uncertainty in prediction will also

increase. Fortunately, the effect of uncertainty on economic evaluation diminishes with time;

therefore, the two effects will compensate for each other. Several techniques are used to forecast

various parameters (Campbell 1987). Most forecasting techniques rely on historical data to predict



future values. If sufficient historical data are available, future prediction over relatively short times

(times shorter than the available data) should be reasonable.

Prediction of oil and gas prices are also required as part of the collection of data. In general, it is

difficult to predict the price of oil and gas since the supply and demand of the commodities is

dependent on so many intertwined parameters, including political winds. However, the incremental

benefits of reducing price uncertainty are significantly greater than the cost of collecting such

information. As a result, operators always make an effort to come up with the best price prediction

and sometimes pay outside consultants to come up with future price predictions. It is, however,

important to remember that price predictions by different consultants are not truly independent

since they rely on essentially the same information. In the absence of additional data, assuming the

current price of the commodity to predict future prices is as good as any other possible prediction.

Overall, collection of relevant data is one of the most time consuming and the critical steps in

economic analysis. The goal of collecting data is always to reduce uncertainties. In the absence of

any uncertainty, there is no need to collect additional data; however, we always make a decision in

the presence of uncertainties. When data collection becomes expensive and does not significantly

reduce uncertainties, we stop collecting data and make an economic decision.

IDENTIFICATION OF FEASIBLE ALTERNATIVES

Analysis of a given problem may require considering several alternatives. Unless all alternatives are

identified, the overall analysis may result in a sub-optimal solution. Genuine creativity and

innovation are an integral part of this process. For example, during a primary depletion, initiating a

waterflood, drilling additional infill wells, or continuing the primary depletion may be some of the

alternatives to be considered. Sometimes, a combination of various alternatives will result in an

optimal solution. One alternative, which should not be ignored, is to continue the operation under

the existing conditions (the "do nothing" option).

SELECTION OF CRITERION

The selection of a criterion for evaluating alternatives should be consistent with the goals and

objectives of the problem. Using the selected criterion, we should be able to arrange the

alternatives for solving a problem in a way that allows us to select the most desirable alternative.

In selecting the criterion, only the differences in the alternatives are relevant to their comparison.

For example, if comparing two houses with the same price, we will only consider the differences in

location, type and annual maintenance costs. On the other hand, when buying a dining table, if two

tables are the same quality, built by different manufacturers, the differences in the prices would be

the only consideration. If we restrict ourselves to purely economic analysis, most of the criteria can

be grouped into three categories.


FIXED INPUT

This criterion is applicable where the amount of money or resources is fixed. The objective is to

effectively utilize the resources; that is, to maximize the benefits. An example is a fixed

exploration budget that needs to be spent on potentially attractive prospects. The number of

prospects and associated exploration costs are greater than the budget; therefore, prioritization

of the budget may be required depending upon the goals of the company.

FIXED OUTPUT

This criterion is used where a fixed task needs to be accomplished. The objective is to minimize

the resources required to accomplish the task; that is, to minimize the costs. An example would

be laying a pipeline from an offshore platform. We know the diameter and the length of the

pipeline, and the location where the pipeline will be laid. We need to secure the best method to

minimize the resources required for building the pipeline.

NEITHER INPUT NOR OUTPUT FIXED

This criterion is applied where both the input and output are flexible. The objective is to

maximize the output with a minimum of resources; that is, to maximize the difference between

the benefits and costs. An example would be oil production from a field. If we intend to increase

the production through drilling infill wells, we should select the number of infill wells in such a

way that the incremental cost of each new well justifies the incremental benefit received from

the additional production.

Using one of these three criteria, analysis of the various alternatives can be carried out.

CONSTRUCTION OF MODEL

In constructing a model for evaluation of the alternatives, we need to incorporate the goals of our

study, the criterion to be used, and the risks associated with the project. Typically, for economic

evaluations, the constructed model is comprised of a set of mathematical equations. Ideally, these

equations should reflect the time value of money, cash flow schedule, and quantitative evaluation of

the uncertainties. Depending upon the complexity of the problem, the complexity of the model will

also increase.

EVALUATION OF ALTERNATIVES

Using the proper mathematical model, each alternative to a given problem needs to be evaluated.

Application of the model may be computationally intensive and may require the use of computers.

Once the evaluation is complete, the alternatives can be ranked based upon the objectives of the

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles 7

project so that proper selection of the best alternative may be made. The best alternative is the one

that provides the most desirable solution under a given criterion.

IMPLEMENTATION OF THE BEST ALTERNATIVE

Once the proper selection of an alternative is made, the next step is to implement the best

alternative.

POST-EVALUATION OF THE ALTERNATIVE

After implementation of the best alternative, it is important to reevaluate the project after a

designated amount of time has passed. This evaluation involves comparison of the observed

performance with the predicted performance. If the comparison is acceptable, no changes need to

be made. If there is a difference, an assessment is needed to determine the reason(s) for this

difference. For example, the implementation of a waterflood may not produce an incremental

increase in oil rate as expected, resulting in reevaluation of underlying assumptions and the input

data. On the other hand, changes in the price of oil may result in an unexpected loss, forcing a

partial or total abandonment of a novel project to minimize the operating losses.

Example 1-1

As a petroleum engineer, you have been asked to evaluate the feasibility of installing a compressor for a gas well. Explain

all the necessary data you will collect. What are the alternatives you will consider? What criterion will you use to select

the best alternative?

Solution i-I

To consider the feasibility of installing a compressor, we need to investigate the additional production received by

installing the compressor versus the costs associated with the compressor. To assess this, the following information

needs to be collected:

1. The present capacity of the well.

2. The incremental capacity as a result of the compressor.

3. The price of gas.

4. The intake pressure of the pipeline.

5. The price of the compressor, if purchased.

6. The maintenance cost of the compressor.

7. The cost of leasing the compressor.

Alternatives

1. Produce under existing conditions.

2. Buy a compressor.

3. Lease a compressor.

Criterion

Maximizing the

N. Mohan Kelkar, Ph.D., J.D.

Sega

Sticky Note

Purchase or rent, Rent to reduce Risks (similar to hedging reasons). If the compressor proven to be working you can buy.

INTERACTION AMONG VARIOUS STEPS

Although the decision-making process is comprised of the steps discussed above, it is not necessary

for these steps to be taken in a sequence. In contrast, every step may require some feedback from

earlier, as well as later steps. The effects of one step on other steps in a decision-making process

have to be incorporated in the overall analysis. For example, definition of a problem may be

modified after collecting and assessing relevant data. New data may reveal that an important facet

of a problem has been ignored. In a waterflooding program, collection of additional data may

indicate that, without infill drilling, the program may not be successful; therefore, the problem may

need to be redefined. Identification of alternatives may reveal that we need to collect additional

data to analyze all alternatives. For example, when considering all possible alternatives for an

artificial lift process, if we realize that one of the feasible alternatives, the use of electrical

submersible pumps, to lift the liquids has been ignored, we may need to collect additional

information related to the submersible pump.

After evaluating the alternatives, a reality check may reveal that the evaluation is overly optimistic

or pessimistic and does not compare with prior experience. This result may require us to go back to

the data evaluation step to investigate the accuracy of collected data and the underlying

assumptions. For example if, when estimating the possible producible reserves in a prospect, after

analyzing all the data we realize that our estimated reserves are greater than nearby, depositionally

similar, reservoirs by an order of magnitude, we may have to re-evaluate our input data. After re-

evaluation, our analysis may reveal that some of the key assumptions we made are incorrect and

more data collection may be needed to properly evaluate the prospect.

Figure 1-1 shows the decision process in a schematic fashion. Although not precise, it is a fair

representation of the overall decision-making process. In the first stage, the problem and objectives

have to be defined. The two are independent of each other and do not require any interaction. For

example, drilling an offshore well with an objective of environmentally safe operation are two

separate facets of economic analysis. They are not necessarily dependent on each other; rather they

complement each other in carrying out the economic analysis.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

Figure 1-1: Decision-Making Process

The next phase in decision making includes data collection and identification of various alternatives.

These have to occur simultaneously for proper coordination. Unless the alternatives are selected,

the need for appropriate data collection may not be evident. On the other hand, the data gathering

phase may reveal some alternatives that were not previously evident.

Depending upon the complexity of the project and the alternatives identified, an appropriate

mathematical model needs to be developed. An application of a criterion consistent with the

objectives of the project should result in the ranking of various alternatives. The evaluation phase

identifies the best alternative among the various alternatives.

If the evaluation reveals that the solutions obtained do not satisfy the reality checks, or if none of

the alternatives are feasible, we may have to go back to the data collection phase. Additional data

may identify alternatives that were not analyzed before. The additional data may also identify

possible mistakes in the assumptions or approximations made during the data gathering phase.

Assuming that a feasible alternative is selected, it should be implemented, followed by the

evaluation of the performance after a reasonable period. If the results are not consistent with the

10 Mohan Kelkar, Ph.D., J.D.

estimations, or if the economic conditions have substantially changed after implementation, we may

have to redefine the problem and repeat the process of decision making.

As evident from this discussion, the feedback within the decision-making process is very crucial for a

rational, well informed answer. Figure 1-1 is just one way of incorporating the feedback into the

decision-making process. Suffice it to say, however, it is better to err toward using too much

feedback than none.

Example 1-2

As an engineer, you are investigating the feasibility of improving the productivity of a well by stimulating it. Based on the

calculations you have performed, you expect production to increase by 30 bbls/day. In reality, after the stimulation

treatment, the production has gone up by 5 bbls/day during the first week. What possible actions would you take? Why?

Solution 1-2

In this instance, the estimated performance does not match the observed performance. Based on Figure 1-1, some

feedback in the decision-making process will be needed. You may consider several options:

Option 1: Wait for a longer period to see if production improves after the treatment becomes more effective. In some

instances, stimulating the well will not result in an immediate increase in production. It is better to be patient. Check

with the service company to determine the typical time during which the effect should be observed.

Option 2: Check the consistency of the input data in the stimulation program. Several sources of error are possible:

a) The predicted damage, say, based on the well test data, may not be correct; therefore, the improvement is not

equal to the prediction. Re-analyze the estimated damage.

b) The rock properties may not be conducive to this particular stimulation. Investigate if this particular rock type

should be subjected to the prescribed stimulation program. Additional data collection may be necessary.

Option 3: If additional data analysis reveals that the stimulation treatment was not the right treatment for this well,

explore other possibilities of improving the production. These possibilities may include: fracturing the well, or increasing

the number of perforations, changing the tubing size, etc. Although these possibilities should have been investigated

before, it is still better to investigate them now rather than ignoring them completely. This requires re-defining the

problem. -

Problem 1-1

Suggest the economic criterion for the following situations:

1. A government is seeking bids for on offshore block. Several companies, equally competent, offer sealed bids. What should be the government’s criterion in selecting a successful bid?

2. An oil company is seeking a contractor to build a surface facility to process the hydrocarbons produced. The specifications of the surface facility are known. What should be the company’s criterion in selecting a contractor?

3. An engineer is evaluating the design of a compressor to compress gas at the surface. The higher the compressor power, the lower the required well head pressure; hence, more gas production. What should be the engineer’s criterion in selecting a compressor?

4. A drilling company has a spore drilling rig. The company con either lease it on a long-term basis at a discounted price

or lease it on a short-term basis at a higher price and keep it idle when not leased. In selecting the proper choice, what criterion should the company use?

5. A service company has observed that by reducing its price for acidizing wells, it can capture a bigger shore of the market. What criterion should the company use in deciding the price?

6. An oil company is evaluating several producing properties to determine which properties should be sold. What

criterion should the company select in ranking these properties so that the first ranking will be received by the


Chapter 1 - Economic Principles 11

property which needs to be sold first?

7. Refer to number 6. If on oil company receives several bids for the property on the market, what criterion should be

used to select a successful bidder?

8. After the initiation of a CO2 flood, which involved a substantial investment due to a drop in oil prices, an oil company

realized that it will lose money on any conceivable scenario. In selecting a possible scenario, what criterion should

the company select?

9. An oil company is interested in drilling a deep gas well at a known depth. In selecting a successful drilling company,

what criterion should be used?

10. After studying the production performance of an offshore well, an engineer realized that by injecting gas at the

bottom of the well, the production could be increased (gas-lift). The cost increases as more gas is injected. In deciding the injection rate, what criterion should the engineer use?

Problem 1-2

If you are assigned the responsibility of investigating the feasibility of drilling an infill well in a mature field, state the essential data you will collect. What are the criteria? What alternatives would you consider?

COMMON TERMINOLOGY

For any economic decision, an understanding of common terminology is important. In this section,

we introduce both the common economic terminology, as well as common oil industry terminology.

ECONOMIC TERMINOLOGY

BENEFITS

As the name indicates, benefits are the monetary awards received as a result of a given

investment. Typically, the benefits are accrued over a period of time as a result of the

present investment. For example, investing money in a newly implemented waterflooding

project is going to result in incremental production over several years. These benefits are

called future benefits. Most of the economic problems encountered are of this nature:

present investment followed by future benefits. However, in some instances, present benefits are relevant to a given decision making. In these instances, the influence of time on

money is not significant. If the benefits collected over a very short period of time after the

investment, then those benefits are considered present benefits. An example would be a

service company providing a service to the operating company at some cost and getting paid

within a few months for the services. In this case, the benefits are collected immediately

after the costs are incurred. Unless the service company provides a service which is risked to

the potential benefits from the project, in general, this is the key difference between the

service company and the operating company. An operating company will always make a

present investment to collect future benefits which will be accrued as a result of future

production. In contrast, the service company receives the benefits immediately after

incurring the costs of providing the service.

12

Mohan Kelkar, Ph. D-, 1. D.

FIXED COSTS

Fixed costs are not affected by the activity level (production changes) over a feasible range

of operating conditions. That is, they are not a function of production or output. These costs

remain fixed. These costs typically include insurance costs, management and administrative

salaries and interest paid on borrowed capital.

VARIABLE COSTS

Variable costs vary as a function of operating conditions. These costs are affected by the

rate of output. For example, utility costs (electricity, water, etc.) and labor costs are strongly

dependent on overall production.

INCREMENTAL COSTS

Incremental costs represent additional costs that will result from increasing the output of

the field. For example, if production from a field is to be improved by drilling two infill wells,

the incremental costs will include the cost of drilling the two wells as well as the additional

costs associated with the operation of the two wells.

NWIRWIMIN

Direct costs are the costs that can be directly allocated to a specific output. Costs of utilities

or labor are good examples, as well as the maintenance of pumping units.

INDIRECT COSTS

These costs are difficult to attribute to a specific output. These costs typically include costs

of administration, training programs, legal fees, etc.

SUNK COSTS

Sunk costs represent an important economic concept. These costs were incurred prior to

current and future to decision making. Since these costs have already been committed, they

bear no relevance to any future decision making. Any economic decision depends on future

costs versus future benefits. These are the things we can control based on our analysis.

An illustrative example would be an oil company obtaining two concessions. For one

concession the oil company paid a bonus of $2 million. If exploratory wells are drilled in

both of these concessions, whether to develop the concessions will depend on the relative

performance of the explored well as well as the future costs and benefits associated with

the development. Other than indirect effect on future tax liabilities, the initial paid bonus

will play no role in the development decision.


Chopter 1 - Economic Principles 13

OPPORTUNITY COST

An opportunity cost is the cost one has to pay by foregoing a potential investment that may

result in some benefit. A personal example is, if you lend money to your friend at a 0%

interest rate that you would have invested in the bank at a 10% interest rate, the

opportunity cost for you is 10%. On a commercial level if, due to a limited budget, you

cannot invest in all the projects you consider desirable, then you would rank them according

to some criterion. After selecting the projects in which you can invest, the best rejected

project represents the lost opportunity. The cost of this lost opportunity is called the

opportunity cost.

The opportunity cost should not be confused with the financial cost. For example, if we

need to borrow money at a 10% interest rate, then the financial cost is 10%. If you can

invest that money in a project which will provide you with a return of 20%, the project

provides you with an opportunity. If you forego this investment, then your opportunity cost

is 20%. In any economic analysis, the opportunity cost must exceed or be at least equal to

the financial cost. You would not borrow money at a 10% interest rate to invest in a project

that will provide you with a return of 6%.

PROFITS

Profit represents the difference between the benefits and the costs. If we define the

benefits as B and the costs as C , then profit can be written as:

P=B�C Equation 1-1

When a given project does not have either fixed benefits or fixed costs, an appropriate

criterion for economic decision making would be to maximize the profit or the difference

between the benefits and the costs.

INFLATION

Money has two faces: 1) It is capable of generating money through investment (earning

power of money). If you invest money in a productive project, it will earn additional money.

We are going to discuss the earning power of the money and its impact on economic

analysis in Chapter 2 - Economic Methods. 2) It is capable of buying goods (buying power of

money).

Inflation deals with the buying power of money (Steiner 1992). Inflation reflects the reduced

buying power of money as a function of time. In other words, the price of goods and

services increase over time - in some countries, very slowly (i.e., United States); in some

countries, at a very rapid rate (i.e., Brazil in the 1990’s). Deflation, although not very

common, is the opposite of inflation. It reflects reduction in prices as a function of time.

Inflation plays an indirect role in an economic decision-making process. If the inflation is

high, the decision maker will have to use a higher rate of return to counter the inflation. On

the other hand, if inflation is low, you can get by with a smaller rate of return.

14 Mo/ian Kelkar, Ph.D., J.D.

Example 1-3

I

An oil well requires pumping equipment. Two companies offer the pump for the well with the following

characteristics:

Company Company B

Initial Cost $50,000 $50,000

Maintenance Agreement $5,000/year $3,000/year

Pumping Costs $0.50/bbl $0.60/bbl

Using the information above, answer the following:

1. Which are the variable and fixed costs?

2. If the well produces 40 bbls/d, which pump is preferred?

3. What is the production from the well at which both pumps will deliver the same economic performance?

Solution 1-3

A. Fixed costs: Initial cost of the pump

Maintenance costs

Variable costs: Pumping costs

I I I I I

B. Since the initial costs for both pumps are the same, we need to compare only the annual costs to choose the

right pump.

Pump A:

’\ (

Annual costs = 5,000 + 40 bbl)

x 365( d

�) x 0.5 1 = $12,300 (

d yrl bbl)

I Pump B:

I

Annual costs = 3,000 + 40 x 365 x 0.6 = $11,760

II

Since the cost of operating pump B is smaller, it should be selected.

I

C. If we assume the production is x bbls at which both pumps will operate at the same cost, we can write

I 5,000 + x(365)(0.5) = 3,000 + x(365)(0,6)

lb Solving for x, x = 54.8 bbls/d.

That is, if the well produces 55 bbls/d, you can choose either of the two pum

Example l4

A proposed enhanced oil recovery project requires a supply of gas containing 90% CO 2 to achieve miscibility with the

remaining oil. Two possible sources of CO 2 are available. Source one contains 97% CO 2 and source two contains 70%

CO 2 . The price of source one gas at the delivery point is $0.35/MSCF; source two gas is $0.2/MSCF. Assume both

sources are abundant in nature and we can use a mixture of the two sources in any proportion. What is the

proportional mixture we should use to minimize the cost? What is the cost of the mixture per MSCF?

Solution 1-4

Considering that the cost of source two is small, we should use as much of source two gas as possible. Assume that X

is the fraction of source two gas used; therefore, (i -4 is the fraction of source one gas used.

We need a mixture which has a minimum CO 2 concentration of 0.9. Mathematica



15

x(O.7)+ (i - 40.97 = 0.9

Solving for x, x=0.26. Therefore, we will use a mixture of 26% of source two gas and 74% of source one gas. The price

of the mixture will be:

0.26($0.2/MSCF)+ 0.74($0.35 I MSCF) = $0,311/ MSCF

Example 1-5

The drilling of an oil well costs $250,000. Initial tests indicated a marginal well. The well will cost $40,000 to

complete and will produce 15 bbls/day for one year, declining at a rate of 20% per year. The operating costs are

expected to be $30,000 per year. Should this well be completed? If it is completed, in what year should it be

abandoned if the operating costs remain constant throughout the life of the project? Assume the revenue of oil

production to be $19/bbl. Neglect tax consequences.

Solution 1-5

In analyzing this problem, the cost of drilling should never be considered. This is a sunk cost. Irrespective of whether

we complete the well or not, we have already spent the money to drill the well. The decision as to whether to

complete it or not should be based on whether we can recover the cost of completion and the operating costs as a

result of the production. If the revenues generated are sufficient to cover these costs, the well should be completed.

This way, although we may not be able to recover the costs of drilling, we will minimize the overall loss.

The yearly profits from this well are shown below.

Year Production Revenue ($) Costs ($) Profit ($)

1 5,475 104,025 70,000 34,025

2 4,380 83,200 30,000 53,220

3 3,504 66,576 30,000 36,576

4 2,803 53,261 30,000 23,261

5 2,242 42,605 30,000 12,205

6 1,794 34,078 30,000 4,078

7 1,435 27,269 30,000 -2,731

In year 7, the costs exceed the revenues. Therefore, we should abandon production.

In year 1, the costs include the cost of completion plus the operating cost. The production in each year is calculated

by multiplying the previous year’s production by 0.8. For example, in year 1, the production is equal to,

= 1 5bbl I day x 365days = 5475bb1s

Therefore, production in year 2 is,

= 0.8x 5,475 = 4,380bbIs

As previously indicated, the decision whether to complete the well does not depend on the cost of drilling. Instead,

since we can recover the completion costs by producing the well, the well should be completed.

Although we recovered the cost of completion in the first year in this problem, it is not necessary to recover the

entire completion cost in the first year so long as the costs are recovered during the producing life of the well.

16 Mohan Kelkar, Ph.D., J. D.

Example 1-6

An enhanced oil recovery using CO 2 flooding is currently being investigated in a depleted oil field. The price of CO 2

including the transportation cost is $.75/MSCF. Based on the simulation studies, it will take anywhere between 3 to 8

MSCF of CO 2 to recover 1 incremental barrel of oil with the most likely value of 5 MSCF/bbl of oil. Due to extra

processing and separation costs, it is expected that the net revenue (excluding the cost of the CO 2 ) per barrel of oil is

25% of the sates price. The future price forecasts estimate the net revenue to be in the range of $16 to $22 per

barrel with the most likely value to be $20 per barrel. Estimate the economic feasibility of this project under the

worst, the best, and the most likely scenario. Should we invest in this project?

Solution 1-6

A. Most Likely

incremental profit = incremental revenue - incremental cost

= 0.25 x $20� 5 x $0.75

= $1.251bb1

This is a positive number indicating that the project is feasible based on the most likely estimate.

B. Worst Case

incremental profit = 0.25 x $16 � 8 x $0.75

=�$2/bbl

That is, we will lose $2/bbl of oil production.

C. Best Case

incremental profit = 0.25 x $22 �3 x $0.75

= $3.25Ibb1

That is, we will make a profit of $3.25fbbl of incremental production.

Examining these answers, we know that the best-case and thvIorst-case scenarios are highly unlikely, whereas, the

most likely scenario is the most likely estimate. That does not mean that we should select this project because it is

economically feasible under the most likely estimate. Instead, we need to know what is the likelihood that the most

likely estimate will be a reality, as well as the other two extreme cases will be a reality. In addition, we may also want

to know the likelihood of other possible in-between answers. Armed with this additional information, we may be

able to make an educated guess as to the feasibility of the project. Without it, our decision will most likely be based

on a gut feeling.

Case Study 1-1

Dac1ing Oil Field is one of the oldest fields in China. The field produces with a high water cut and several enhanced oil

recovery technologies have been used to improve the performance of the oil field. Starting in 1999, a colloidal dispersion

gel (CDG) pilot was conducted in a double five spot well pattern as shown in Case Study Figure 1-1.

Economic Evaluation in the Petroleum Industry rhnntpr I - Frnnmir Princinlps 17

cc tud3 h urr I-11- (IX p loti.n Da gang field ((hwy 200)

The well pattern consists of six injection wells and twelve production wells. Between 1999 and June 2003, three chemical slugs (about 0.53 pore volumes total) were injected followed by drive water. As shown in Case Study Figure 1-2, due to the CDC, slug, the pilot has shown reduction in water production and an increase in oil production.

Case Study F/gore 1-2: Oil production and water cut during Cl)G flood (Chan 200-I)

Based on base line decline curve analysis, incremental oil production from well 131-7.124, as well as the whole pilot, was gathered. The following table shows key production data.

Injected Well Polymer

(Ibs)

Incremental Oil ’bbls’ ’ /

Other Costs

Pilot 1173248 757484 84,535,793 Bi 7 P124 261140 171689 81,009,571

The information for an individual producer with respect to polymer injection and other costs (Bl-7-P124) is estimated. The amounts for pilot are actual numbers. The incremental oil is calculated based on projected decline for the pilot as well as individual wells. The other costs represent the cost of implementation including chemical plants, additional facilities, additional operating and facilities costs. Assume the cost of polymer is 81.30/11). Assume further that the average price of oil during the incremental production phase was S35/bbl. Using this information, calculate the Cost of Cl)G flood per barrel of oil produced. If the economic threshold is S10/bbl for the development cost of CIA; flood, is this project feasible? I low much is the profit generated from this project over its life? \X/hat is the ratio of profit to cost? If our CCOO( mic criterion for profit to cost ratio is greater than 2, does this project satisfy the economic cnrcrlon?


dw

r Sttzde .tOIU CiOsJ 1-i

’The following table shows the summary of all results. The explanation for the numbers is provided below the table.

Cost of

Total Cost Cost/bbl Revenue Well Polymer

" / / Profit ($) PIR

($) /

Pilot 1,525,223 6,061,016 800 26,511,940 20,450,924 3.4 B1-7-P124 339,482 1,349,053 7.86 6,009,115 4,660,062 3.3

The cost of polymer is calculated by multiplying the amount of polymer by $1 .30/lb. The total cost is cost of polymer plus other costs. By dividing the total costs by oil produced, we can calculate the cost/bbl. If the threshold requirement is less than 510/bbl, these costs satisfy that criterion. As explained in the section below, this cost represents the costs associated with accessing the oil in the ground. Knowing the value of the oil in the ground, we can determine what this cost has to be to ensure that it is profitable. The revenues are calculated by cumulative production by the price of oil. The difference between revenue and the cost is the profit. The last column represents profit-to-investment ratio or PIR. ’Uris is calculated by dividing the profit by the total costs. This number exceeds our threshold requirement of 2.0; therefore, the project is economically feasible. Profit-to-investment ratio can be important when the company has limited capital available. We will discuss this further in a future section.

OIL INDUSTRY TERMINOLOGY

Certain economic terminology is unique to the oil industry and understanding it is important in

making economic decisions.

E & P COMPANY

E & P stands for Exploration and Production. Sometimes this represents the "upstream" oil

industry. As the name indicates, E & P companies are mainly involved in exploring and

producing oil and gas. They are not concerned with refining, upgrading and marketing the

final products. In general, oil companies can be divided into two categories: E & P

companies and integrated oil companies. Integrated oil companies (e.g., ExxonMobil or

Shell) have an E & P division but they also own refineries and gas stations. E & P companies

only explore and produce oil; they do not refine or sell the products (e.g., Devon Energy or

Anadarko Petroleum).

MINERAL INTEREST

Mineral interest represents the ownership interest in oil and gas trapped in the sub-surface.

In most countries (with the exception of the U.S. and some parts of Canada), the minerals

are owned by the government. Even in the U.S., more than half of the on-shore minerals are

owned by either federal or state governments and all off-shore minerals are owned by

either federal or state governments. A mineral interest owner may or may not possess the

expertise to explore for or exploit the oil and gas from the sub-surface; therefore, the

mineral interest owner hires an operator (E & P company).

WORKING INTEREST

Working interest is owned by one or more parties. The parties owning the working interest

are responsible for paying for the expenses related to exploration and production of oil and

gas. Typically, there is only one operator and the working interest owned by the operator



IN

(responsible for day-to-day decision making) is called operated working interest. Other

parties, who own the working interest but are not involved in operation of the oil or gas

field, own non-operated working interest. It is possible, but not necessary, for a mineral

interest owner to also own a working interest.

ROYALTY INTEREST

Royalty interest represents cost-free interest in the producing property. Royalty interest

owner will receive a portion of the produced oil or gas (either as the product or the

proceeds from the product) cost free. That is, the royalty owner would not be responsible

for the costs associated with production. Typically, the mineral owner would receive royalty

interest in return for allowing the working interest owners to explore for and exploit the

hydrocarbons.

In some instances, royalty interest could be owned by a party who does not own the mineral

interest. For example, a geologist who develops a prospect (potential location for drilling a

well) sells that prospect to an E & P Company and, in return, receives a portion of the

production free of cost if the well is successful. This is called over-ride royalty interest. It is

also possible for one oil company to get over-ride royalty interest in return for assigning the

rights to explore for and produce hydrocarbons to another company.

NET REVENUE INTEREST

Net revenue interest (NRI) represents the portion of production owned by a party. It is

extremely rare that the working interest (WI) and NRI of a party would be the same. For a

mineral interest owner, typically, the NRI would be greater than the WI; whereas, for an E &

P Company, WI would be greater than NRI.

LEASE BONUS

As part of the enticement to sign an agreement, the operator will give the mineral interest

owner a signing bonus. This signing bonus is called a lease bonus. The amount is typically

determined per acre in the United States. In good areas, the lease bonus can be as high as

$20,000/acre.

SPACING

Spacing represents the surface area per well. For example, in a one square mile area

(equivalent to 640 acres), if we drill eight wells, the spacing of the well is 80 acres; that is

640 acres divided by S.

G & G COSTS

G & G costs represent geological and geophysical costs. These costs are normally incurred

during the exploration phase. These costs include, but are not limited to, seismic data

acquisition and processing, and geological mapping and modeling.


JOINT OPERATING AGREEMENT (iDA)

A joint operating agreement (JOA) is signed by the working interest owners to resolve any

disputes related to the operation of the field. JOA will also govern the rules related to

exploration of the project. Among other things, the JOA will include how to receive

permission from non-operated working interest owners to take a specific action (e.g., drilling a well), how to deal with non-consenting owners, and how to allow sell off the

ownerships.

EXPECTED ULTIMATE RECOVERY (EUR)

Expected ultimate recovery (EUR) represents the recovery expected from a well or from a

field based on current operating conditions. The units of EUR are either barrels or ft’

depending on whether it is mostly oil or gas production.

BARRELS EQUIVALENT

In reporting EUR or other methods of representing reserves, E & P companies will provide

barrels equivalent of the reserves. This is to combine oil and gas reserves into a single

number. The most common number used to convert gas into equivalent oil barrels is 6

MSCF equal to 1 barrel. This is based on an assumption that 6 MSCF of gas will generate the

same amount of energy as one barrel of oil. Please note that energy equivalence does not

translate into price equivalence. Historically, gas has always been sold at a discounted price

compared to oil because of the transportation costs associated with natural gas. It is

possible that the discrepancy between energy equivalence and price equivalence can be

large making it difficult to evaluate a company when only equivalent barrels of oil

production are provided.

HELD AND DEVELOPMENT COSTS

Field and development costs (F & D costs) hac,e the units of currency divided by units of

production (e.g., $/MSCF). These costs include all costs associated with securing the right to

drill a well, as well as the cost of drilling; in other words, all costs associated with securing

access to the resource. For example, if the operator had to spend $5 million to secure the

rights to drill a well, as well as costs of drilling and completion, and the operator expects to

get 5 BCF of gas from the well, then F & D costs would be $1/MSCF.

Example 1-7

A geologist proposes a new prospect (potential drilling location) to an operator in return for 3% over-ride royalty

(ORR). In addition, the operator needs to provide 3116th

royalty interest to the mineral owner to drill the well.

What is the working interest (WI) and net revenue interest (NRI) of the three parties?

If the operator, before drilling, assigns 40% of its working interest to another non-operator and, in return,

secures 1% ORR from the non-operator, what will be the WI and NRI for the four parties?

Solution 1-7

For the first scenario, neither the geologist, nor the mineral owner owns any working interest. Therefore, the

operator owns 100% of the working interest. The following table shows the proportion for each party:

Economic Evaluation in the petroleum Industry


21

WI NRI

Geologist 000% 3.00%

Mineral Owner 0.00% 1835%

Operator 1 100.00% 78.25%

In the second case, the WI will be split into two operators in the proportion 60 to 40% respectively. However, in

return, the NRI of the operator will not be split into the same proportion since Operator 1 will secure one

additional percentage of NRI. The following table shows the results.

WI NRI

Geologist 0.00% 3.00%

Mineral Owner 0.00% 18.75%

Operator 1 60.00% 47.95%

Operator 2 40.00% 30.30%

Example 1 -8

Based on decline curve analysis, an operator expects an EUR from a gas well to be 2.3 BCF. The well spacing is 80

acres. The cost of drilling and completion was $3.2 million. in addition, the operator paid $5,500 per acre in lease

bonus and the additional G & G costs were $300 per acre. What are F & 0 costs for this well? The royalty interest

is 20%. If operator requires a well to have F & D costs to be less than $1.50/MSCF, is this well economic?

Solution 1-8

Total costs to access the reserves = 5,500 x 80 + 300 x 80 + 3.2 x 106

Net EUR to the operator = 2.3 x 0.8 = 1.84 BCF

F & D Costs = 3664x10

= $2/MSCF 1.84x 10 6

Based on the F & D Costs for this well, it is not economic.

Example 1-9

In its July 2011 investor presentation (Chesapeake Energy 2011), Chesapeake Energy shows the results from

various plays in which they are involved (see Example Figure 1-1). Examining the data, what is the equivalence

Chesapeake is assuming in calculating boe/d for each well? Based on the prices of oil and gas, if we assume the

equivalence of 15 MSCF = 1 barrel of oil, how will the Boe/d change in each region?

DR

LJ.’ - P : -:-

EJ !I1.d

7,907 b~d

I t Wash - Grard te Was

h

i f 01 4,tjK-B 12P6

268 b�ld

i . J

22 Mohan Ke/kar, Ph.D., J.D.

Example Figure 1-1: Chesapeake results presented in Investor Presentation (Chesapeake Energy 2011)

Solution 1-9

Below are the results for all of the plays. If qg is gas production in MMSCF/D, q 0 is oil production in STB/D and q 0

is equivalent production, then the equivalence of gas corresponding to one barrel is calculated as:

equivalence = q9 x iO

(q0 - q 0 )

Play Gas Prod Oil Prod

Boe/d Equivalence Boe/d

MMSCF/D STB/d (MscF/bbl) (is MSCF = 1 Bbl)

Frontier 1.7 1,342 1,625 6.04 1,455

Tonkawa 1.2 1,151 1,351 6.00 1,231

Tx PH 20.5 4,496 7,907 6.01 5,863

valon Shale 2.3 1,153 1,528 6.13 1,306

Bone Spring 2.1 2,020 2,445 4.94 2,160

Eagle Ford 4.5 1,519 2,268 6.01 1,819

Wolfcamp 0.1 504 523 5.26 511

Granite Wash 12.0 3,273 5,279 5.98 4,073

Cleveland 7.1 1,404 2,586 6.01 1,877

Mississippian 1.1 1,427 1,609 6.04 1,500

Niobrara 0.3 655 705 6.00 675

As shown, with just a few discrepancies (due to rounding off), the equivalence assumed is 6 MSCF z 1 STB. If we

assume 15 MSCF z 1 STB, we will get lower values of BUE as shown in the last column.

Case Stud 1-2

Eagleford Shale is one of the largest unconventional discoveries in this decade. It is located in South Texas. Unlike other shale plays, which produce mostly gas, Eagleford Shale produces both oil and gas depending on where the well is located. According to EOG Resources, Eagleford Shale can be divided Into an oil window, condensate window and gas window as shown in Case Study Figure 1-3. The northern window is oil, the middle is condensate and the lower one is gas. Because oil is more valuable than gas in terms of prices, it is more valuable to have acreage in the oil or condensate window than in the gas window.

Case Study Figure 1-1, fiagle/i.’rdpiat in south 7.vas (EOG 2011)



The field is developed using horizontal wells with a lateral length ranging between 4,000 feet and 5,000 feet. The cost of dialling and completing these wells is between $5 million and $8 million. According to the 170C Resources wcbsite (EGG 2O11), some wells producing from the oil window are listed below. These rates represent the initial production (IF) of the well. On average, the CUR from these wells range between 400 and 460 MBOe (thousands of barrels equivalent - assuming 6 MSCF I STB). A typical spacing varies between 120 and 140 acres per well Assume that the cost of leasing the land is $6,000 per acre. Assume the average royalty interest to be 25 ° ’o

Oil Rate Gas Rate Well Name STB/d [j SCFD

Edwards 904 350

Sweet Unit 1182 1323

Hanson 311 1538 1512 Dullnig 511 1353 1224

Greenlow 607 386 Joseph 311 1317 1200

hoff 683 391

Wiatrek 857 682

Answer the following questions:

1. What is the range of field development cost per barrel if we assume 6 MS(:F z I STB. lithe requirement is that F& D cost should be less than S25/barrel, is that criterion satisfied?

2. EOG currently has 520,000 acres of leased acreage in the oil window. What will be the range of total estimated investment needed to drill and complete all of the wells?

3. If we assume that in ground gas reserves are worth S1/MSCF and in-ground oil reserves are worth S40/S1B, how much is EOG’s oil window acreage worth?

Casc Stssdy Solutfon 1-2

1. F&DCosts

We can calculate both the pessimistic and optimistic F & I) costs by using the following equations:

5 x iO + 120 x 6,000 Optimistic F & D Costs ’= = $16.61BBL

460,000 x 0.75 8 x 10 6 + 140 x 6,000

Pessimistic F & D Costs = ___________________ - $29.5/BBL 400,000 x 0.75

In calculating F & L) Costs, we have to consider both leasing and drilling and completions costs. We divided by net reserves to account for royal’ interest. The average cost is S23/bbl which is less than our threshold requirement of $25/bbl. I Iowever, the range of F & 1) costs also indicates that there is some risk involved in drilling these wells based on our Criterion.

2. Investment Needed

As with F & 1) Costs, we can also calculate the pessimistic and optimistic ranges. These costs do not include leasing costs since the land has already been leased. We consider both the van.itlollC in the spacing, as well as the uncertainties in the costs of drilling and completion. It is possible that, over time, these costs will come down as the technology of drilling and completion improves.

Optimistic Investment = 520,000

140 X 5 x 106 = $18.6 billion

Pessimistic Investment = 520,000

120 x 8 x 106 = $34.7 billion

The FOG wcbsite EOC 2011) uses a number of 10 to 15 billion dollars of investment. This probably assumes improvement in the drilling md COMPIC6011 costs.


3. Worth of In-Ground Oil and Gas Reserves

Using the well data provided, we can calculate the percent of gas present based on a 6 MSCF equivalence. The table below shows these calculations. We divided the gas rate by 6 and calculated the percent of gas present as the total equivalent oil production. The average of all these wells is 12%. that is, out of the total reserves produced from a well. 880/s will he produced as oil and 12% produced as gas. This number is consistent with the website.

Oil Rate Gas Rate % Gas Well Name STB/d MSCFD (6 MSCF eq)

Edwards 904 350 0.061

Sweet Unit 1,182 1,323 0.157

Hanson 311 1,538 1,512 0.141

Dullnig 511 1,353 1,224 0.131

Cireenlow 607 386 0.096

Joseph 311 1,317 1,200 0.132

Hoff 683 391 0.087

Wiatrek 857 682 0.117

Using this information, we can calculate the following values:

Optimistic Pessimistic

Gas/well, MSCF 248,400 216,000

Oil/well, S’FB 303,600 264,000

Total Gas (BCF) 1,076 802

Total Oil (SIMSTB) 1,316 981

S (billions) 53.70 40.03

In the table above, we assumed that 12i/ of the FUR is produced as gas and 88% of the FUR is produced as oil. For the optimistic scenario, we considered smaller spacing and higher FUR per well and for the pessimistic scenario, we considered larger spacing and lower EUR per well. By multiplying per well data with the total number of wells in 320,000 acres, we can calculate total gas and oil. The total FUR reported by FOG Resources is 900 MMSTB equivalent, which is slightly less than predicted in this table. This may be due to different spacing assumptions. Based on the optimistic and pessimistic scenarios, the in-ground oil and gas resources are worth anywhere from S40 billion to S54 billion. ’li’iesc values are calculated by multiplying gas FUR by S1/MSCF and oil FUR by $40/bbl.

Problem 1-3

Classify each of the following costs as fixed or variable:

� Materials required for maintenance � Direct labor � Supplies � Utilities � Property taxes � Administrative salaries � Insurance � Pumping costs � Interest on borrowed capital



Problem 14

An oil company is considering a routine core analysis to be done from on outside lab. The outside lab charges $20

per core plus $1 per core for shipping and handling. The manager of the core lab at the oil company contends

that costs associated with the equipment are approximately $101core the cast of material is $21care and the

cast of direct labor is $51care. Therefore, he contends that the care study should be done in-house. Why?

The company has conducted its economic study and has observed that, based on the overhead cast allocation,

the cast of lab space is $41core and the overhead costs are $31core. The company thinks that it can use the space

more efficiently.

Discuss why the core analysis should be done outside. What is the correct perspective in economic analysis?

Problem 1-5

An oil company is thinking of buying a computer. Because of the advancements in the technology, the computer

price depreciates at a rate of 25% per year. The maintenance and the upkeep costs increase gradually. The cast of

the computer is $5,000 and the maintenance and the upkeep costs in the first year are $700 with an increase in

the cost at a rate of 10% per year. What strategy should the company adopt to maximize the computer benefit?

Assume that due to advances in technology, it is worth buying a new computer when the yearly costs start rising

after reaching a minimum.

Problem 1-6

An operator is considering two options to improve the production from an oil well. Option 1 requires that the well

be stimulated at a cast of $10,000 with an increase in production of S bbls/day for a one-year period. Option 2 requires the well be fractured at a cast of $50,000 with an increase in production of 15 bbls/day for a one-year

period. Option 3 is to continue to produce under present conditions. If the oil price is assumed to be $201bbl,

which option should be chosen?

Based on prior experience, the operator knows that the incremental production after stimulation can be off by

–5% and the incremental production after fracturing can be off by –10%. How will this analysis affect the decision

in the previous paragraph?

77IThWA

An oil company is considering buying a compressor. Depending upon the horsepower, the cost of the compressor

will vary. The four options the company is looking at are:

Option A B C D

Initial Cost $100,000 $180,000 $250,000 $300,000

Annual Benefit $40,000 $60,000 $75,000 $90,000

Annual Operating Costs $20,000 $30,000 $30,000 $40,000

Assume that the unused portion of the money can be invested at a rate of 15%. That is, if we assume that the

company has $300,000 to invest, and if the company chooses Option A, then the remaining $200,000 can be

invested at a rate of 15%.

Which option should the company choose based an net annual benefit analysis?

26 Mohan Kelkar, Ph.D.. 1.0.

Problem 1-8 .

An oil company is interested in drilling in fill wells to accelerate production from a field. As more wells are drilled,

the production will increase; however, the incremental production will become smaller and smaller as more wells

are drilled. Based on the analysis of a nearby field, the company derives the following equations:

B = 400,0001 - 10,0001 2

C = 100,0001

where B is the total benefit (in dollars) received as a result of I infihl wells drilled and C is the cost (in dollars) of drilling I wells. To optimize the profit, how many wells should be drilled? What is the maximum profit the

company can receive?

Problem 1-9

A service company offers its employee two options. Either the company will provide him with a company car, or if

the employee chooses to use his own car (which he intends to buy), provide him with a mileage rote of $0.211mile. Assume that the employee drives the car 15,000 miles per year for business purposes. Assume further

that the new personal car the employee is going to buy will cost him $20,000. The employee will use his own car

for personal purposes for 10,000 miles. If the car depreciates per year at a rote equal to,

+ os)

where p is the price of the car at the beginning of the year and m is the mileage driven during a given year, which

option should the employee select in the first year? In which year should the employee switch to the other option?

Problem 1-10

An oil company hired a consultant to study the improvement of an oilfield. The consultant charged $10,000 and

recommended improvements which would cost the company $100,000 to be implemented. An in-house study,

costing two days of on engineer’s time at a rote of $3001day, revealed that the some improvements could be

done using on alternative scheme at a cost of $105,000. Which option should the company choose?

Problem 1-11

A drilling company is considering two alternative scenarios of drilling mud to drill a well. Alternative one will use on oil-based mud at a cost of $0.31gallon; however, due to environmental regulations and concerns, the cast of

disposal it is much higher. We can calculate the cost to be fixed at $10,000 plus $0.11gallon for each gallon of mud used. The second alternative is to use a water-based mud. Due to the addition of expensive components, the

cast of the water-based mud is $0.41gallon; however, the disposal casts are $5,000 plus $0.051gallon. Knowing

these costs, at what volume of drilling mud will both methods result in the some costs? Under what conditions

will you prefer one method over the other?

Problem 1-12

A service company wants to expand its business. An internal study indicates that one way to improve the business is to reduce the charges for a certain logging technique. For that technique, it costs the company $5,000 to log

the well and the company charges $10,000. Currently, the company has 200 customers. The study indicates that

for every 10% decrease in the charged costs, the number of customers can be increased by 150. What is the

optimum price the company should charge to maximize the benefits? What is the maximum benefit the company

will receive?

at Economic Evaluation in the Petroleum Industry

opt Chapter 1 - Economic Principles 27

Problem 1-13

An oil company is contemplating drilling a horizontal well at a cast of $400,000. The chance that the well will be successful is 30%. If successful the well will produce 500,000 bbls of oil. The company has an option of conducting

a detailed 3-D seismic survey at on additional cast of $400,000. By evaluating the seismic data a well can be

better located and, hence, the chance of success will be improved to 50%. If the net profit per barrel of oil is expected to be $3 in terms of present worth, should the company collect the seismic data?

An oil company implemented a waterflood on a marginal field. The cost of implementation was $1 million. Additionally, the injection costs were a fixed at $5,000 per month plus $0.41bbl of injected water. Currently, 6,000

bbls of water per day is being injected. Each barrel of injected water results in 0.1 bbl of incremental oil. After deducting all of the expenses excluding the injection costs, the net benefit received per barrel of oil is $6.50. If

after implementation, the all price in the market plunged resulting in the net benefit per barrel of oilfrom $6.50

to $5.50, should we continue to woterflood the reservoir? Assume that the change in the price is expected to be long-term. If we decide to continue, how much profit would we make at the new price?

Problem 1-15

Two oil companies owning equal operating interest in a property agree to sign a contract with the mineral owner to provide the mineral owner 15% royalty interest. Create a table showing the working and net revenue interests

of oil the properties.

Problem 1-16

Chesapeake Energy, owner of a substantial leasing interest in !-laynesville shale, agreed to sell some of its interest to Plains Exploration and Production Company. Chesapeake originally owned 100% working interest and 80% NRI. Chesapeake sold half of its interest to Plains during the transaction. However, the agreement had a clause which required that Plains pay for 50% of WI of Chesapeake until it paid out a certain amount of money (i.e., Plains paid

for its own WI plus 505v. of Chesapeake’s interest until a certain amount of cash was paid). What were the WI and NRlfor Chesapeake and Plains during that period?

Problem 1-17

An oil company (Company A) leased land with 100% WI and 75% 1VRI with the mineral owner receiving 15%

royalty and another oil company (Company B) receiving 106 of the override royalty interest. Company A decided

to sell Company B 40% of its working interest (and proportionate NRI). In return, it secured 2% override royalty from Company B. After the transaction, what were the WI and NRI of all the three parties?

Problem 1-18 -

An exploration and production company paid about $17,000 per acre in leasing costs. It expects that it can drill a gas well on 40 acre spacing with a drilling and completion cost of $5 million. The company expects to recover

about 4 BCF of gas (net to the company) in EUR. If the company’s threshold requirement is that it cannot exceed F

& D costs of more than $21MSCF, should it develop the ploy?

Problem 1-19

In the Woodford Cana ploy, the typical cost of drilling and completing a well is $4 million. The average leasing

costs are about $1,500 per acre. The standard spacing of a typical horizontal well is 80 acres. The well produces

30 STB/MMSCF of condensate throughout the depletion phase. If the gas EUR (net to the company) is 2.7 BCF,

what is the gas equivalent FUR if we assume 6 MSCF 1 STB? How much will it be if we assume 15 MSCF 1 STB?

How will the F & D Cost change depending on the equivalent assumption?


Problem 1-20

In an oil shale play, a company expects to have a success rote of 90% (the rest being unproductive) in completing the wells. It has acquired 100,000 net acres of land at a cost of $6,000 per acre. The cost of drilling and completing the well is $6 million. The well is expected to have an EUR of 220,000 barrels of oil with 1 BCF of gas. Assume the royalty interest to be 25%. The normal spacing of a well is 120 acres, If the company intends to develop the field, what will be the F & 0 cast per well? Assume 10 MSCF 1 STB to calculate equivalent oil production. The cost has to be less than $25 per barrel to be economical. Does the field satisfy this criterion?

Problem 1-21

An exploration and production company is producing a gas field by drilling one well per section (640 acres). Although there is some variation in the performance, the average EUR per well is about 2 BCF. The royalty interest is 25%. The total acreage the company has is 64,000 acres. The company wants to sell this property to another company for $400 million. In addition, it would like 10% override royalty. It is expected that the field can be developed on 80 acre spacing. In addition, an average of about 1 BCF per well still remains to be produced from the producing wells. The cost of drilling and completion is $3 million, If you are working for another company and your criterion is to buy property if F & 0 costs are less than $1.50/MSCF, would you buy this property?

OIL AND GAS RESERVES

Companies are valuable because of their assets. These assets typically constitute building, equipment,

manufacturing facilities, good will, people, long-term contracts, etc. For oil and gas companies, the main

asset is what is underground. It is not the buildings that provide value to a typical oil company; it is really

the future potential of oil and gas production that makes the company valuable. The common method

of defining this future potential is oil and gas reserves. By no means do oil and gas reserves represent

the precise value of an oil company. For one, the reserves do not tell us how quickly that oil and gas can

be produced. The reserves also do not tell us the cost of producing the oil and gas. However, oil and gas

reserves are an important measure of a company’s worth and understanding how those reserves are

reported is extremely important to anyone working in the oil industry.

It is important to distinguish between reserves and what is in place. Many times companies will report

oil in place (OIP) or gas in place (GIP). Sometimes, these numbers are reported as OOIP (original oil in

place) or OGIP (original gas in place). Typical units of QUIP and OGIP are barrels (or STB - standard

barrels) and SCF (standard cubic feet) respectively. These numbers represent the estimate of how much

oil and gas was originally present in the reservoir. Because of uncertainties, these numbers are only

estimates. However, the quantity that is produced is always smaller than what is present. Reserves

represent what can be economically produced from what is originally present. The recovery factor (RF)

can vary widely. Recovery factor represents the percentage of QUIP or OGIP that can be produced from

the reservoir. In extreme cases, it is possible that a reservoir containing a large amount of hydrocarbons

may have a recovery factor of zero, which means that the reserves are non-existent. The reasons for the

recovery factor being zero could be many, including technical difficulties (no current technology),

economic difficulty (price of commodity too low), or market difficulty (no current market available).

Nevertheless, it is important to understand that QUIP or OGIP may not tell us anything about the oil and

gas production potential of a company.

Many documents are available which define reserves. Individual countries have their own rules and

regulations regarding how the reserves are defined. Instead of providing alternate definitions of the

reserves, we will follow the document published by SPE (Society of Petroleum Engineers 2007). The

Petroleum Resources Management System (PRMS) (Society of Petroleum Engineers 2007) provides



definition of reserves under various conditions. These definitions are not tied to any specific country;

instead, they are based on opinions of people who have a wealth of experience in defining reserves. The

document has gone through many review processes and is the most robust available for defining the

reserves.

The goal of this section is not to make a person an expert in understanding the reserves. Instead, the

goal is to introduce the concept and provide the necessary rationale so that the reader can understand

why certain reserves are defined in a particular way.

PRMS defines petroleum resources in much broader terms than reserves. The Figure 1-2 shows these

categories.

!KODtCflON

conc_Eyr I T1 W

� 1

PROSPECMT

2

.-

..i n

Rartp ofUncmd1ny

Figure 1-2: Hydrocarbon Resources (Society of Petroleum Engineers 2007)

In Figure 1-2, hydrocarbon resources are divided into three categories: Reserves, Contingent Resources

and Prospective Resources. As shown in this graph, the y axis represents the commerciality; whereas,

the x axis represents the uncertainty associated with those resources. The top left corner (Proved

Reserves) represents the hydrocarbons we are most confident of producing and bottom right corner

(high estimate of prospective resources), hydrocarbons we are least confident of producing. Let us

examine the definitions of these categories as given in the PRMS.

RESERVES are those quantities of petroleum anticipated to be commercially recoverable by application of development projects to known accumulations from a given date forward under defined conditions. Reserves must further satisfy four criteria: they must be discovered, recoverable, commercial, and remaining (as of the evaluation dote) based on the development project(s) applied.


According to this definition, reserves represent what is commercially recoverable in the future based on

already discovered accumulations; for example, a gas field that is produced by ten wells under current

conditions. We can reasonably predict how much those wells can produce in the future based on their

past performance. Or we can consider an oil field which is producing under depletion. What that field

can produce under water flooding conditions is also reserves. However, a potential prospect, where not

a single well has been drilled, cannot be considered regarding reserves.

CONTINGENT RESOURCES are those quantities of petroleum estimated, as of a given date, to be potentially recoverable from known accumulations, but the applied project(s) are not yet considered mature enough for commercial development due to one or more contingencies. Contingent resources may include, for example, projects for which there are currently no viable markets, or where commercial recovery is dependent on technology under development, or where evaluation of the accumulation is insufficient to clearly assess commerciality.

According to this definition, like reserves, contingent resources also are discovered resources (i.e., well

or wells have already been drilled in the field). However, without overcoming a certain important

limitation, we cannot produce from it under current conditions. An example would be the discovery of a

large gas field on a remote island that cannot be produced without assurance that the LNG (liquefied

natural gas) produced from it can be sold to an end user. Another example would be the discovery of an

off-shore, tight oil, reservoir that cannot be produced without development of an efficient, economically

viable, fracturing technique.

PROSPECTIVE RESOURCES are those quantities of petroleum estimated, as of a given date, to be potentially recoverable from undiscovered accumulations by application of future development projects.

Prospective resources have both an associated chance of discovery and a chance of development.

According to this definition, prospective resources represent undiscovered accumulations. An example

would be a claim that in Alaska there are many fields containing as much as 10 billion barrels based on

geological assessment. Since no well has been drilled, there is an uncertainty associated with discovery.

In addition, even if it is discovered, there is no guarantee that it can be commercially produced

(uncertainty associated with development).

Examining these definitions, as we go from reserves to prospective resources, the uncertainty

associated with commerciality increases. However, within each of these categories, there is also an

uncertainty with respect to the percentage of hydrocarbons that can be produced (uncertainty with

respect to recovery factor). A certain percentage of hydrocarbons can be easy to produce within the

reservoir, whereas, additional recovery may require significant effort resulting in additional

uncertainties.

Going from contingent resources to reserves requires an establishment of commerciality. According to

PRMS, the following factors are to be considered to establish the commerciality:

� Evidence to support a reasonable timetable for development.

� A reasonable assessment of the future economics of such development projects meeting defined

investment and operating criteria. � A reasonable expectation that there will be a market for all or at least the expected sales

quantities of production required to justify development.

Economic Evaluation in the petroleum Industry Chapter 1 Economic Principles 31

� Evidence that the necessary production and transportation facilities are available or can be

made available.

� Evidence that legal, contractual, environmental and other social and economic concerns will

allow for the actual implementation of the recovery project being evaluated.

Briefly, to categorize something as reserves requires that hydrocarbons will be produced within a

reasonable period after discovery. A reasonable period is, typically, less than five years; however, there

will be circumstances under which that period could be longer. For example, a company that has

discovered a giant field in deep water may require more than five years before beginning production

due to technical challenges; however, it has made a commitment to build the off-shore platform and has

started drilling the development wells. Then the company can claim the reserves associated with that

field.

DETERMINISTIC VS. PROBABILISTIC RESERVES

In defining reserves, we can use two methodologies. Although seemingly exclusive of each other,

these methods can be made consistent with each other. One underlying principle to remember

before defining the reserves is that there is always uncertainty with respect to reserves. Unless a

well or reservoir is abandoned, we truly do not know how much hydrocarbons are produced from it.

Until we reach abandonment, there is always some uncertainty with respect to how much

hydrocarbons will ultimately be produced from a reservoir. The uncertainty is much bigger in the

initial stages of development and, as more production data are collected and more is known about

the reservoir, the uncertainty will decrease. In defining the reserves, this uncertainty has to be

conveyed irrespective of the type of methodology used in calculating reserves.

Deterministic methods, as the name indicates, represent a single value to define the reserves.

However, in reality, to convey the uncertainties associated with reserves, deterministic methods will

report low, best and high estimates of the reserves. Low indicates a conservative estimate, best

indicates a best guess, and high indicates an optimistic estimate.

In probabilistic methods, instead of reporting a single value, a range of possible values are reported

for reserves along with the associated probabilities. Depending on how exhaustive the work is, it is

possible to have a continuous distribution of reserves. There is an implicit relationship between

probabilistic reserves and deterministic reserves. The low estimate corresponds to the 10th

percentile value in the distribution function. The 10th

percentile represents a value that is less than

at least 10% of the values. The best estimate corresponds to the 50 1h percentile which represents a

value which is greater than 50% of the values in the overall distribution. The high estimate

corresponds to the 901h

percentile value, which represents a value greater than 90% of the values in

the overall distribution. In other words, if we know the overall distribution of reserves, it is easy to

define deterministic reserves. Deterministic reserves are a sub-set of probabilistic reserves which

represent a much more exhaustive definition of the reserves. It is not always possible to create

probabilistic distribution of the reserves; however, unless the reservoir is very mature, we should

expect to see the low, best and high estimates. We also realize that the difference between low,

best and high estimates will decrease as the reservoir production maturity increases. This is also

true with probabilistic estimates. The difference in the possible range of reserves will decrease as

the reservoir matures.


CATEGORY OF RESERVES

Depending on the uncertainty associated with the reserves, we can define different categories of

the reserves.

PROVED RESERVES

According to PRMS, proved reserves are those quantities of petroleum, which by analysis of geoscience and engineering data can be estimated with reasonable certainty to be commercially recoverable, from a given date forward, from known reservoirs and under defined economic

conditions, operating methods, and government regulations.

Proved reserves represent the most important category of reserves. These are the numbers

reported by oil companies for public consumption. These are the numbers which are typically

accepted by government agencies. The key word in the definition of proved reserves is

"reasonable certainty." Although reasonable certainty is not defined explicitly, implicit in this

definition is high confidence. It is meant to be understood that the operator, with more than

90% confidence, can state that these are the reserves that can be produced. By nature, this is a

very conservative estimate. Also, these reserves need to be reported under defined economic

conditions, operating methods and government regulations. Typically what that means is that

no speculation is allowed in defining proved reserves. If we can only produce reserves if the oil

price reaches $150 per barrel, then those are not considered proved reserves if the current price

of oil is $100 per barrel. The same is true of the government regulations and operating methods

under which these reserves will need to be reported. If the regulations are expected to change

in the future or if new technology can change the way the field will be operated, without

proving it, we can only report the reserves based on existing government regulations and

existing operating methods.

If using determining methods, proved reserves represent the low estimate. If using probabilistic

methods, proved reserves represent the 10th

percentile (also called 1P) reserves. That is, there

should be a 90% probability that these reserves can be produced. An important theme in stating

proved reserves is high confidence and a conservative view. If a company reports proved

reserves correctly, it implicitly requires that those reserves will grow with time since the number

should represent the most conservative estimate.

According to PRMS, the area of the reservoir which is considered as proved includes (1) the area delineated by drilling and defined by fluid contacts, if any, and (2) adjacent undrilled portions of the reservoir that can reasonably be judged as continuous with it and commercially productive on the basis of available geoscience and engineering data. Again, the theme is conservative

estimation; if a well is drilled in a reservoir, the lowest known hydrocarbon (LKH) contact will

represent the lowest part up to which hydrocarbons are located unless proven otherwise. It is

possible that oil and gas may be below the known contact, but unless convincing proof exists,

the conservative estimate requires that the LKH is assumed to be the depth up to which

hydrocarbons are located. The same thing can be said about areal distribution of reservoir. If

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

33

few wells are drilled in the field, without substantial and convincing information, the area

around the existing wells, which is assumed to be commercially productive, is limited.

Proved reserves are further divided into sub-categories.

Proved, developed, producing (PDP) reserves represent the reserves with the most

confidence. These reserves are expected to be recovered from producing intervals within

wells that are currently open. These reserves continue to be produced without any

significant investment except for proper maintenance.

DEVELOPED, NON-PRODUCING, RESERVES (PDNP)

Proved, developed, non-producing (PDNP) reserves represent the reserves which can be

produced with very little investment compared to drilling a new well. For example, intervals

that are open but are not producing under current conditions, behind the pipe reserves that

are known but will require perforation of that interval, and pipeline connection to a newly

completed well where the available market is well established. The key to defining PDNP is

that the well has already been drilled and, with very little investment, the reserves can be

produced. If, by installing a compressor, we can increase the reserves and the cost of

compression is small compared to drilling a new well, it can be categorized as PDNP

reserves.

UNDEVELOPED RESERVES (PUD)

Proved, undeveloped reserves (PUD) represent the reserves known to exist but will require

substantial investment. These are the reserves that are expected to be recovered in the

future through large investments. Examples of PUD’s include those from (1) new wells on

undrilled acreage in known accumulations, (2) deepening existing wells to a different (but

known) reservoir, (3) infill wells that will increase recovery, or (4) where a relatively large

expenditure (e.g., when compared to the cost f drilling a new well) is required to (a)

recomplete an existing well or (b) install production or transportation facilities for primary

or improved recovery projects. Examining these examples, the key is the large expenditure.

For example, if a well is drilled and a shallower formation needs to be perforated, it is

cheaper and, hence, can be considered PDNP; whereas, a deeper formation needs to be

perforated; it would be considered PUD since the well will need to be deepened before the

new interval can be perforated. This will require significant investment compared to

completing a shallower interval.

PROBABLE RESERVES

Probable Reserves are those additional Reserves which analysis of geoscience and engineering data indicate are less likely to be recovered than Proved Reserves but more certain to be

recovered than Possible Reserves. As can be seen from this definition, the uncertainty is much

higher for probable reserves compared to proved reserves. A well which is currently producing

34 Mohan Kelkar.. Ph.D.. J.D.

can have both proved and probable reserves associated with it. Probable reserves represent

higher value than proved reserves but with less confidence or more uncertainty. If deterministic

definition is used, probable and proved reserves represent the best estimate of reserves. If

probabilistic definition is used, probable and proved reserves represent 50th

percentile reserves

(also called 2P). If we understand that proved reserves represent 10 th percentile reserves, then

probable reserves represent the difference between 50th

and 10th

percentile values. The key in

defining probable reserves is the degree of uncertainty. There is an equal likelihood that these

reserves can be produced or not produced. For example, if a new oil reservoir is discovered and

the recovery factor in all of the analogous reservoirs is at least 10%, then 10% of OOIP would

represent proved reserves. However, if fifty percent of the analogous reservoirs show 20%

recovery factor, then probable and proved reserves would be 20% of OOIP.

POSSIBLE RESERVES

Possible reserves are those additional reserves which analysis of geoscience and engineering data indicate are less likely to be recoverable than probable reserves. This definition indicates

even more uncertainty than probable reserves. If we are using the deterministic reserves

definition, possible plus probable plus proved reserves indicate a high estimate of the reserves.

If we are using the probabilistic reserves definition, possible plus probable plus proved (also

called 3P) represent goth percentile reserves. That is, there is only a 10% probability that these

reserves can be recovered. For a probabilistic definition, the difference between 90th and 50th

percentile represents possible reserves. An example of these reserves includes water flooding

on an existing oil reservoir when no analog for water flood exists. The reservoir properties of the

existing reservoir are substantially different from the water floods where data are available.

Without a pilot flood, significant laboratory data and any other evidence, the incremental

recovery from water flood is speculative and, therefore, should be categorized as possible

reserves. Another example is drilling spacing. Currently a field is being developed on 80 acre

spacing but there is a remote possibility that the field can be developed eventually on 20 acre

spacing, which means we might have an opportunity to drill more wells. However, without a

pilot project and government approval in the field, 20 acre spacing is speculative; hence,

reserves to 20 acre spacing can be considered as possible reserves.

CATEGORY OF CONTINGENT RESOURCES

Contingent resources have already been discovered; however, because of significant constraint

(economic, technical, etc.), these resources cannot be produced; therefore, the category of

contingent resources is very similar to reserves. Instead of 1P, 2P and 3P, we can consider 1C, 2C

and 3C, where 1C represents the 10th

percentile of contingent resource, 2C represents the 50th

percentile of contingent resource and 3C represents the 90th percentile of contingent resource. If

the contingency is removed, 1C, 2C and 3C should directly translate to 1P, 2P and 3P respectively.


Chapter 1 Economic Principles

35

ESTIMATION OF RESERVES

Estimation of reserves is beyond the scope of this book. However, we will briefly describe different

methodologies used in estimating reserves.

VOLUMETRIC ANALYSIS

During the appraisal and initial development phases, volumetric analysis is the most commonly

used method. The main advantage of the volumetric method is that it does not require any

production data. Volumetric analysis requires that we have an idea about the reservoir

dimensions. Knowing the dimensions we can calculate the volume of the reservoir and, by using

appropriate petrophysical properties such as porosity and saturation, we can calculate the

hydrocarbons in place. Then, using an analogous reservoir, we can estimate recovery factor.

Knowing the recovery factor and volume of hydrocarbons, we can calculate the reserves.

MATERIAL BALANCE

Material balance technique can only be used after the reservoir starts producing hydrocarbons.

Material balance is a simple technique and assumes a uniform reservoir without any variations

in reservoir properties, such as pressure. However, if the assumptions are satisfied, material

balance can provide hydrocarbons in place. This is similar to volumetric analysis, except that

since the method is based on hydrocarbons produced, it is more reliable than volumetric

analysis. Once hydrocarbons in place are calculated, by knowing the recovery factor, we can

calculate the reserves.

RESERVOIR SIMULATION

Reservoir simulation is a more sophisticated technique than material balance. Simulation

accounts for variations in physical properties of the rock as a function of space and time. It can

rigorously account for various scenarios and complex physical processes. It is an advanced

material balance technique. The technique can be used without any historical production data;

however, the technique becomes more reliable if historical production data are available. By

using reservoir simulation, we can match the historical production data and then predict the

future performance. Knowing the economic limit, simulation can predict how much

hydrocarbons will ultimately be produced. Unlike material balance, we do not need the recovery

factor since simulation is able to predict rate versus time until we reach abandonment. In

addition to predicting the remaining reserves, it can also predict rate versus time, which can be

valuable for economic evaluation. Reservoir simulation is more valuable than material balance

technique if properly used. It is important to understand that just because simulation

methodology is more complex, it does not necessarily mean that it reduces the uncertainties in

future prediction compared to material balance technique.

36

Mohan Kelkar, Ph.D., J. D.

RATE - TIME ANALYSIS

The rate-time analysis method also requires production data. It is a method by which production

data are plotted as a function of time and, using the historical trend, a curve is fitted to the past

data and is extrapolated to predict future performance. Assuming that the trend is correctly

predicted, we can assess future performance. Knowing the economic limit, we can calculate the

remaining reserves. Similar to reservoir simulation, rate-time analysis can also predict the future

rate as a function of time. As with any other method, when more production data are available,

this technique becomes more reliable. Although it is possible to use this method for the entire

field, normally, the method is used for individual wells to predict the reserves for that well.

There are other techniques that may also be used to calculate the reserves; however, the

techniques above are the most common.

RESERVES AND THE TRUE WORTH OF A COMPANY

Without question, assigning the right reserves to a particular project is important. Proved reserves

provide the most conservative estimate; therefore, in most countries, these are the reserves that

are reported for public consumption. It is assumed that an average person may not be able to

distinguish between different types of reserves that are assigned different levels of uncertainties.

However, when a company reports proved reserves, it implies more than 90% certainty that these

hydrocarbons can be produced. So an average investor can count on production of at least these

reserves with an expectation that, in most likelihood, more hydrocarbons will eventually be

produced.

A relevant question to ask is if this is the best estimate of company’s worth. We need to understand

two caveats in evaluating reserves. One, the reserves may not represent the net value of

hydrocarbons since reserves do not include the cost of producing them. When proved reserves are

reported, they represent the amount of hydrocarbons that can be produced in commercial quantity;

however, they do not tell us the cost of production. Clearly, if the production cost for producing a

barrel of oil is $10 for one operator versus $20 for another operator, although both operators are

reporting the same reserves, the one with the smaller cost of operation represents better worth of

the reserves since it will make more profit from selling the oil. Secondly is the timing. By examining

the reserves, we do not know when these reserves will be produced. The worth of these reserves is

clearly tied to when the reserves are produced. As we will see later in this chapter, the earlier we

produce these reserves, the more valuable they become. For example, if a company reports

reserves from a gas well equal to 2 BCF by assuming production from that well to be over sixty years

and the well is expected to produce 1.4 BCE over the first thirty years, economic evaluation indicates

that 0.6 BCF of gas produced after thirty years does not have any significant value. This means that

reserves by themselves may not tell us the whole story without knowing when they will be

produced.

Suffice it to say that reserves provide valuable information about the worth of the company;

however, it will be more valuable if, in addition to knowing the reserves, we also know the cost of

producing them, as well as the timing of producing those reserves.


Example 1-10

A company is producing from ten existing wells on 160 acre spacing. Using the rate-time analysis, the remaining reserves

are calculated. They add up to 8 BCF. Based on the analysis of all of these wells, the average EUR for each well is 1.5 BCF.

A local government entity has already approved 80 acre spacing in the same reservoir for other operators and they are

reporting the production from 80 acre spacing to be similar to 160 acre spacing wells. There is also a speculation that

government entity may allow 40 acre spacing for this reservoir. However, no one knows for sure. It is also rumored that

another operator in the same reservoir is using a different fracturing technique, which increases the EUR by 50% but, at

this time, no service company is providing this service and there is no reported information. For internal purposes, the

company needs to determine the reserves for this area. Provide the total reserves and assign an appropriate category for

these reserves.

Solution 1-10

We can divide the reserves into proved and contingent resources in this case.

Proved Developed and Producing (PDP) Reserves

Since production data from all ten wells is available, we are reasonably certain how much remaining gas can be

produced from these wells using rate-time analysis. Therefore, PDP reserves are 8 BCF. We are assuming that a

conservative estimate is used in determining the remaining reserves from these wells.

Proved and Undeveloped (PUD) Reserves

We are assuming that the reservoir is reasonably continuous around the 10 wells already drilled. That is, the properties

observed in the existing wells will be similar to wells that will be drilled on 80 acre spacing. If 80 acre spacing is already

approved in this reservoir, we may expect that similar spacing will also be approved in this field. If the operator is able to

drill 80 acre spacing wells, it will double the well count compared to 160 acre spacing. If the conservative estimate of

EUR for each of the ten wells is 1.5 BCF and if we know that the EUR from 80 acre spacing is not different from 160 acre

spacing (based on observations in other part of the reservoir), we can report 1.5 BCE times 10 as equal to 15 BCF as PUD

reserves.

Please note that the range of estimate observed in an individual well is not the same as the range observed in a group of

wells. That is, if for an individual well the range of possible EUR is from 1.0 BCF (low value) to 3.0 BCE (high value) with

1.5 BCE as the best estimate, it does not mean that the range of 10 wells drilled will be 10 BCE (low value), 15 BCE (best

estimate) and 30 BCF (high value). Instead, the range of uncertainty for the ten wells would be narrower than would be

calculated by just adding the numbers. Why? We will discuss this in more detail later in Chapter 4 - Economic Uncertainty.

Contingent Resources -

Contingent resources represent discovered resources which cannot be produced because of some contingency. If 40

acre spacing is approved, the well count would quadruple compared to the current number of wells. That is, additional

20 wells (compared to 80 acre spacing) can be drilled in the field. We do not know if those wells will produce the same

amount of gas as 80 acre spacing; however, assuming that is the case, 20 times 1.5 BCE would result is 30 BCF of most

likely contingent resources. If indeed a better technology is available and we can increase the EUR by 50%, on the high

side of contingent resources, we would be able to produce 45 BCF of gas from 40 acre spacing wells. - -

Example 1-11

An off-shore oil field is discovered and is currently approved for development. The development wells are being drilled

and production platform has been ordered. It will be another four years before the field will start producing. The OOIP is

estimated to be 800 million barrels. Based on the simulation results as well as comparison with analogs from similar oil

fields, the primary depletion recovery factor can be as low as 6% to as high as 20% with the best estimate being 12%.

Currently, additional slots for potential water injection wells are provided for on the production platform; however, no

wells are currently planned for. No laboratory results have been collected, nor has detailed simulation been done.

Examination of analog reservoirs indicates that it is possible to double the reserves if water flooding is properly


0 0 0 0 I I I

implemented and natural aquifer provides a weak support. The company would like to start the production and examine

the data before making a decision about when and if water injection should be implemented.

Explain different reserves categories based on this information.

Solution 1-11

Proved Reserves

Since the field is already under development, the proved reserves would be 6% of 800 million barrels which is equal to

48 million barrels. During the development phase, these reserves remain as proved undeveloped (PUD). However, once

the wells are drilled and if the company is waiting for pipeline connection which requires a small investment compared

to drilling the wells, these reserves would move into proved, developed, non-producing (PDNP). Once the production

starts, they would become proved, developed and producing (POP).

Probable Reserves

Probable plus proved reserves would be equal to 12% of 800 million barrels which is equal to 96 million barrels. By

subtracting proved reserves, probable reserves would 48 million barrels.

Possible Reserves

Possible plus probable plus proved would be 20% of 800 million barrels which is equal to 160 million barrels. Subtracting

96 million barrels would result in 64 million barrels of possible reserves.

Contingent Resources

Water flooding is a common process applied to many oil reservoirs. Therefore, the process is not new and technical

challenges are easy to overcome. However, we will need to add the additional reserves into contingent resources

because those resources depend on one important condition: support from an underlying aquifer which is not known

until production begins. If we assume that we can double the reserves using water flood, our low estimate of recovery

factor will be 12%, high estimate of recovery factor will be 40% and the best estimate will be 24%. Subtracting reserves

from our calculations, our low estimate of contingent resources will be 48 million barrels, high estimate will be 160

million barrels and the best estimate will be 96 million barrels.

If the company eventually determines that aquifer support is limited and based on detailed lab and simulation studies,

water flooding is economically feasible and approves the development, the contingent resources will be shifted into

reserves category. .- -

Case Srzdv 1-3

This case study is largely adopted from "Guidelines for Application of the Petroleum Resources Management System" (Sosie of Pr/ia/earn Engineers 2011).

An oil field was discovered about nineteen years ago. After two years of appraisal followed by three years of development, the field has been producing for fourteen years. The field has been under peripheral water flood. Based on a detailed reservoir study, a multi-disciplinary team built a reservoir model and matched the historical data using a detailed flow simulation study. Case Study Figure 1-4 shows the history match result for the most likely (best estimate) scenario. .\lthough only one figure is shown for the history match, two other scenarios were also run for conducting history matching which involves different amounts of oil in place. Case Study Table 1-1 shows some of the details about the history matched results from the three models. ’lhc field, to date, has produced 399 million barrels of oil. After history matching the results, the model was run in the future to understand a "do nothing" option (l’\Vl’); that is, continuing the water flood operations as they stand. Additionally, a peripheral water flood option with artificial lift using a submersible pump (P\y’F with ESP) was also run. The third option considered was conducting CO2 miscible displacement. All three scenarios were run and results were estimated to determine how much additional oil can be produced.



39

Caw Study .Inigure 1-4? Hi/tory match and future prediction fur the most likely scena rio

case Study Table 1-i: Swnnaary of oil recoveries under three scenarios

Bases and Estimates by Reserves Category Low Best High

Measured and Estimated Parameters Estimate Estimate Estimate

Original Oil in Place (MMSTB) 1,434 1,525 1,739

Original Gas in Place BSCF 367.7 434.6 545.0

Cumulative Production so far

Oil (MMSTB) 399 399 399

Gas (BSCF) 227.4 227.4 227.4

Recoverable hydrocarbons under PWF 5

Oil (MMSTB) 573.4 686.3 869.3

- Gas (BS(’F) 326.8 391.2 495.5

Recoverable I lydrocatbons under PWF with ESP*

Oil (MMSTB) 645.1 762.5 9562

Gas (BSCF) 367.7 434.6 545

Recoverable I -lydrocarbons under (102 Flood

Oil (MMSTB) 716.8 915.0 1217.1

Gas (BS(’F) 386.1 478.1 626.8

The 7% incremental oil recovery based on ESP is well supported by several nearby analog fields which showed incremental recoveries of as high as 9% of 00W. The economics of I SP is very attractive and the company has already made a commitment to implement I This project is expected to he implemented in four years and the investment is expected to be substantial.

The CO2 flooding project showed an incremental recovery of 18%. lhccc were two CO2 pilots running in analogous fields. lloth pilots indicated recovery efficiencies exceeding 21% of incremental recovery. The project economics of C() 2 flooding is also positive. Currently, delivery of the CO2 is somewhat in doubt since construction of the CO 2 pipeline is not finished and the pipeline company has yet to commit to building the pipelmc. ;\dditionally, the company suspects that the laws and rides regarding CO, sequestration may change resulting in potential positive impact. In the report describing the results of (02

AN

40 Mohcin Kelkar, Ph.D., J.D. a

p p p p flood study there was a note stating that "these estimates should be reviewed periodically to confirm whether these

impediments still exist and appropriate development decision should be made accordingly."

Using this information, categorize all of the reserves and explain your reasoning. Calculate equivalent oil recovery by Consorting gas production using 6 MSCF1S’1B. Report SIB and RIBI’. separately.

Case Stm tv Solution 1-3

The recovery of oil in this case can be categorized as:

Proved. Developed and Producing- QPD P) : These reserves represent the remaining reserves that can be produced using water injection. No significant investment is expected. These calculations are done only for a low (most conservative) estimate.

Proved and Undeveloped (PUD): These reserves represent the remaining reserves that can he produced through water injection and f’.SP combination. Once ESP’s are implemented, these reserves can he moved to the PDP category.

Probable: These reserves represent the difference between best case and low case scenario under water flood with FSP option.

Possible: These reserves represent the difference between the high estimate and the best case scenario under water flood with an ESP option.

Contingent Resources: Incremental recovery based on CO2 flood will be represented by contingent resources since these resources cannot be recovered by overcoming a significant contingency. The low, best and high estimates will be categorized as IC, 2C and 3C estimates respectively.

Case Study Table 1-2summariscs there results:

Case Study ’lSblc 12 AsnountufRemaining Resaut

Category Oil (MMSTB)

Proved

Proved, developed, producing 174.4

Proved, undeveloped 71.7

Probable 117.4

Possible 193.7

Contingent Resources

IC 71.7

2C 152.5

3C 260.9

car and tin i.e Respective Cats 50t!OS

Gas

Combined

(BCF) (MMBOE)

99.4

191.0 40.9

78.5

66.9 128.55

110.4 212.1

18.4 74.8

43.5 159.7

81.8 274.5

Note that, in calculating contingent resources, we calculated the difference between oil and Was recovered under CO2 flooding and under an US!2 option for each scenario. That is, the difference under the low estimate scenario represents IC reserves, the difference under the best ease scenario represents 2C reserves and the difference under the high ease scenario represents 3C reserves. ’l’his is because the entire CO2 flooding process is contingent upon satisfying certain conditions. If the conditions are satisfied, the IC should move into proved reserves, 2C should move Into probable reserves and 3C should move into possible reserves. By calculating the contingent resources as explained, this will happen logically.

Problem 1-22

After drilling a single well and testing it for production, a company predicts the low estimate of recoverable oil to be 30,000 barrels (with a GOR of 2,000 SCF/STB), the best estimate of recoverable oil to be 50,000 barrels (with a GOR of 2,500 SCF/STB) and high estimate of 100,000 barrels (with GOR of 3,000 SCF/STB). Categorize these reserves separately as oil and gas. If we define 6 MSCF ISTB, re-write the reserves in terms of oil equivalent



41

A gas field is being currently developed on 80 acre spacing. It began development on 640 acre spacing with an average

FUR of 2 BCF. However, over time, the government entity overseeing this particular area granted the spacing to be

reduced to 80 acre spacing A plot of FUR versus spacing indicates a declining trend That is every time the well is drilled

on a smaller spacing the FUR is slightly less than the wells drilled on a larger spacing There has been some dispute

among operators but a 15/ decline in FUR is observable every time the spacing is halved if the spacing goes from 640 to

320 acres the FUR of wells drilled on 320 acre spacing is about 15% lower than the EUR of wells drilled on 640 acre spacing. An operator has a lease interest over 3,200 acres which is producing at 80 acre spacing. Assume the royalty

interest to be 20%. If the government agency grants 40 acre spacing in this field, how much additional reserves can the

company book? How would the reserves be categorized? Assume that, for the wells producing on 80 acre spacing, an

average of 0.5 BCF of gas per well still remains to be produced.

Problem 1-24

Problem Figure 1-1 shows the progression of a reservoir. Initially, well A was drilled and tested. It produced at a high

enough rate to be economical. The lowest known oil (LKO) was established from the log data. The volumetric analysis

indicated 200 million barrels of oil in place based on LKO. The recovery factor in this type of reservoirs varies from 10% to

25%. Seismic analysis indicated that the oil contact could be much lower than shown in the well. If seismic analysis is to

be believed, the oil in place will be as much as 500 million barrels of oil. Assuming that field development is approved

based on this well, categorize all of the reserves.

Eventually, well B was drilled and entered the water zone. Based on the gradients in both oil and water, a new oil-water

contact was established at a different depth. This resulted in oil in place of 430 million barrels of oil. How would the

reserves change based on this information?

Well A Well B

Problem Figure 1-1: Cross section of on oilfield


Problem 125

A newly discovered oil field indicated original oil in place of 30 million barrels (with expected produced cumulative GOR of 800 SCF/STB). This was based on volumetric analysis. The recovery factor was expected to be in the range of 10% to 20% with the most likely value of 14%. How would the reserves be reported under these conditions?

The field started producing and, after one year of production, the material balance technique was applied. Based on the decline in reservoir pressure, the initial oil in place was revised to be 20 million barrels. The GOR remained unchanged. If the field has already produced I million barrels of oil by the end of one year, how will the reserves be reported at this point? Assume that the recovery factor has not changed.

The field can be subjected to water flood. There are some analog reservoirs which have indicated mixed success in indicating an incremental recovery between 8% and 15% of the original oil in place with the best estimate equal to 10%. However, the company has not done any detailed analysis of the cores, nor has it studied the water flood feasibility in detail. May be in another year, the company will start examining the feasibility of water flood and based on the evaluation, it may be approved. How would the reserves based on water flood characterized?

SIMPLE ECONOMIC METHODS

When economic projects are evaluated, we need to have the ability to decide which one is the best

project or the best alternative if several alternatives are available in one particular project. Many

sophisticated techniques are available to determine the best possible solution. Some of these

techniques are discussed in Chapter 2 - Economic Methods. Here we consider two simple methods

which are commonly used in the industry. We will also explain the rationale for using these methods.

Before we consider these two methods, let us consider two possibilities when decisions are made. We

can consider a possibility when the amount of investment needed for different alternatives is small

compared to the overall budget that we will select the best alternative without worrying if the company

has enough money; for example, if a company is considering the possibility of producing an oil well

under natural conditions, installing a rod pump, installing a submersible pump or a gas lift. If the cost of

all scenarios is insignificant compared to the capital commitment of the company, all four scenarios will

be evaluated and the best one will be selected. It is assumed that sufficient funds are available for

selecting the optimal solution. These alternatives are called mutually exclusive alternatives because if

the company selects one alternative, it will have to reject all others. That is, if the company selects a gas

lift option for an oil well, it has automatically eliminated the do nothing, rod pump and submersible

pump options.

On the other hand, if a company is considering multiple projects with different amounts of investment

needed and the investment amount is significant, it is possible that the company may have to pick and

choose between those projects based on the amount of money available. For example, if a company is

looking into development of three off-shore projects and has money to develop only two then, using an

appropriate economic criterion, it will have to decide which two should be selected and which one will

be rejected. These projects are considered independent since, by selecting one, the company does not

have to reject the other except for monetary constraints.



PAYBACK PERIOD

Payback (or payout) period method is one of the simplest methods used in economic evaluation. It is

the time it takes to recover the investment. Despite its simplicity, the method is extremely useful

under certain conditions.

Consider a small oil company with limited working capital. If a geologist enters the office and offers

a new prospect, the main consideration for the company’s CEO is how long the money will be tied

up before it can be re-invested. If the time is very long, even though the project is very attractive

and can make a significant profit, the CEO will reject it. For small companies, many times liquidity is

more important than profitability. If the initial investment can be recovered quickly, anything

beyond that is profit and the same money is put to other use. It is not the only criterion the

company will use, but will definitely be one of the important ones.

Even bigger companies will sometimes sell an asset that has long-term potential but has a very long

payback period. This type of monetizing is quite common so that the money can be deployed in

other assets that have faster turn-around time in terms of recovering investment. For example, in

2009, Devon Energy announced its intention to sell its offshore assets in the Gulf of Mexico and

Brazil so that the proceeds could be invested in North American assets (Devon Energy 2009). The

main reason for the sale of these assets was the long horizon over which those off-shore assets will

be developed. This means that a lot of money will be tied up for a long time without any potential

revenue. In contrast, North American assets are quickly developed and, therefore, quick revenue is

gained from it. This is an example where liquidity (quick payback) triumphs over potential

profitability.

In addition to working capital or liquidity problems, the payback period can also be valuable when

contractual obligations or political uncertainty makes it difficult to plan for a long-term, optimal

project. If a company is operating an oil field and the lease on the oil field expires in five years, the

payback period is an important consideration to ensure that the initial investment is recovered.

Payback period is definitely not a criterion if profitability ofthe project is important. A project that

recovers cash quickly but terminates after that would look more attractive using payback period

criterion compared to a project which generates cash over a long period of time but has a significant

lead time. Therefore, in evaluating a project, both payback period and some profitability criterion

should be considered in making an appropriate decision about the investment.

Example 1-12

An oil well is currently producing at a rate of 20 barrels per day. A service company claims that the well, after stimulation

at a cost of $50,000, will produce an incremental amount of 7 barrels/day. Using an internal program, the service

company provides you with the following table of old versus new (after stimulation) rates. What is the payback period

and what is the profit? Assume that net benefit from each barrel is $80. Assume that an average of 30.4 days exists in

each month.

44

Mohan Kelkar, Ph.D., ID.

Old rate New Rate Month

b/d b/d

0 20 27

1 19.2 25.2

2 18.4 23.6

3 17.7 22.0

4 17.0 20.5

5 16.3 19.2

6 15.7 17.9

7 15.0 16.7

8 14.4 15.6

9 13.9 14.6

10 13.3 13.6

11 12.8 12.8

12 12.3 12.3

13 11.8 11.8

14 11.3 11.3

15 10.8 10.8

Solution 1-12

Using the table above, we can construct the following table.

Old rate New Rate D oil per 0 revenue Cumulative

Month b/d b/d month per month Revenue

0 20 27 212.8 $ 17,024 $ 17,024

1 19.2 25.2 182.9 $ 14,636 $ 31,660

2 18.4 23.6 155.7 $ 12,456 $ 44,116

3 17.7 22.0 130.9 $ 10,468 $ 54,584

4 17.0 20.5 108.2 $ 8,658 $ 63,242

5 16.3 19.2 87.7 $ 7,013 $ 70,255

6 15.7 17.9 69.0 $ 5,519 $ 75,774

7 15.0 16.7 52.1 $ 4,165 $ 79,939

8 14.4 15.6 36.7 $ 2,940 $ 82,878

9 13.9 14.6 22.9 $ 1,833 $ 84,711

10 13.3 13.6 10.5 $ 836 $ 85,548

11 12.8 12.8 0 $ - $ 85,548

12 12.3 12.3 0 $ - $ 85,548

13 11.8 11.8 0 $ - $ 85,548

14 11.3 11.3 0 $ - $ 85,548

15 10.8 10.8 0 1 - $ 85,548

Note that after 10 months, the rates match; therefore, there is no incremental revenue from stimulation. The revenue

per month is calculated by multiplying incremental oil by $80/barrel. The last column, cumulative revenue, is just the

addition of revenue from the previous column. In month three, cumulative revenue exceeds the $50,000 investment;

therefore, the payback period is 4 months (starting with month 0). The profit from the project is $85,548 - $50,000 =

$35,548.



PROFIT-TO-INVESTMENT RATIO

Another important economic method extremely useful in evaluating independent projects is profit-

to-investment ratio (PIR or P1). This is simply a ratio of profit generated from a project to total

investment needed. PIR provides us with the information about which project gives us the "biggest

bang for the buck". If, for example, we are considering two projects, one requiring a $100

investment and the other requiring a $500 investment, and both projects generate a profit of $500,

the project with the $100 investment is much more valuable than the project requiring a $500

investment although both of them generate the same profit. This is because, if we can generate a

$500 profit with only a $100 investment, the remaining $400 (which would have been required had

we selected the other alternative) could be invested elsewhere.

When dealing with limited capital and multiple projects to consider, PIR is a method we can use to

rank these projects so that proper selection, which will maximize the profits, can be made.

Example 1-13

Consider an extension of Example 1-12. The same well can also be hydraulically fractured at a cost of $300,000. The

production profile after fracturing versus stimulation is provided in the table below.

Stimulation Fracturing Old rate

Month New Rate New Rate b/d

b/d b/d

0 20 27 50.0

1 19.2 25.2 45.0

2 18.4 23.6 40.5

3 17.7 22.0 36.5

4 17.0 20.5 32.8

5 16.3 19.2 29.5

6 15.7 17.9 26.6

7 15.0 16.7 23.9

8 14.4 15.6 21.5

9 13.9 14.6 19.4

10 133 13.6 , 17.4

11 12.8 12.8 15.7

12 12.3 12.3 14.1

13 11.8 11.8 12.7

14 11.3 11.3 11.4

15 10.8 10.8 10.8

Using this information, determine the profit from each of the options. If the company has enough money to either

fracture or stimulate every well, what is the best option? If the company can invest up to $600,000 and there are more

than 20 wells in the field, what is the best option? Assume the net oil revenue to be $80/barrel of oil and 30.4 days in

each month.

Solution 1-13

Using the information provided, we can construct the following table.

46


Old rate Stimulation

Fracturing Stimulation L Fracturing LI Month New Rate

b/d b/d

New Rate b/d revenue per month revenue per month

0 20 27 50.0 $ 17,024 $ 72,960

1 19. 25.2 45.0 $ 14,636 $ 62,746

2 18.4 23.6 40.5 $ 12,456 $ 53,669

3 17.7 22.0 36.5 $ 10,468 $ 45,613

4 17.0 20.5 32.8 $ 8,658 $ 38,470

5 16.3 19.2 29.5 $ 7,013 $ 32,144

6 15.7 17.9 26.6 $ 5,519 $ 26,550

7 15.0 16.7 23.9 $ 4,165 $ 21,610

8 14.4 15.6 21.5 $ 2,940 $ 17,256

9 13.9 14.6 19.4 $ 1,833 $ 13,425

10 13.3 13.6 17.4 $ 836 $ 10,062

11 12.8 12.8 15.7 $ - $ 7,115

12 12.3 12.3 14.1 $ - $ 4,541

13 11.8 11.8 12.7 $ - $ 2,299

14 11.3 11.3 11.4 $ - $ 352

15 10.8 10.8 10.8 $ - $ -

$ 85,548 $ 408,813

In this table, we have calculated the revenue from each of the options. Incidentally, the payback period for the fracturing

option is six months; slightly greater than stimulation option. However, four months versus six months shows a very

small difference and is not be an important consideration in making a decision about which method to choose.

The profit from stimulation is $35,548 per well; whereas, the profit from fracturing is $108,813 per well. If we assume

that the table above represents an average, typical well, then it is always better to fracture the well than stimulate the

well since the overall profit will be maximized by fracturing every well.

If the company has only limited funds, we will need to calculate the PIR for each option:

i PIR for stimulation = 35,548

= 0.71; whereas, PIR for fracturing s =108,813

300,000 = 0.36.

50,000

Therefore, we should select stimulation over fracturing.

Examining it differently, we can reach the same conclusion. If we have $600,000, we can either stimulate 12 wells or

fracture two wells. Therefore, the overall profit after stimulation will be 12 x 35,548 = $426,572; whereas, overall profit

after fracturing will be 2 x 108,813 = $217,626. If we want to maximize profit, stimulation is a better choice.

This example clearly shows how decisions can differ depending on a company’s availability of funds. PIR is an important

criterion when a company is strapped for cash.



case Study 1-4

In 2003, California Electric Utility (CET) decided to promote savings of electric energy to oil producers in the state of California. Working through P1TC (Petroleum Technology Transfer Council), C EU contacted several operators and promoted various strategies to reduce electric consumption. As an incentive to use new, more efficient technology, CPU also made a cash contribution toward implementation of this new technology. After a year-long effort, the following data were collected from various alternatives;

I. Pump-off Controller (POC)

Most of the wells in California use rod beam pump artificial lift equipment. Rod pumps are operated more efficiently if they operate with a full pump. POCs optimize the well run time so that the well operates periodically and with full pump. POCs sense when the well is full and optimize the pumping system. During the program, POCs were installed on 42 wells that were operated by 3 different operators with the following results;

Operator #1 Operator # 2 Operator # 3

Number of Wells 27 3 10

Gross Production b/d) N/A 180 623

� Oil Production (bid) N/A 26 104

Cost (5) 26,199 13,000 129,775

Incentive by CEU (5) 13,100 6,500 64,888

Annual Average Savings (KWh/yr) 391,207 74,720 822,915

Cost of Energy (S/Kwh) 0.080 0.115 0.080

2. Premium Ffficient and Downsizing Motors (PFDM)

PEDMs are 2% to 8% more efficient than standard motors. Replacing standard motors is a common practice in many industries. I lowever, in the oil industry, motors are typically run until they fall and are replaced with an existing motor on hand. By using state-of-the-art software, a motor could be optimized fora particular application at the oil well. It is also possible to optimize oil production at the same time. Two producers installed PEDMs on 21 wells with the following results;

Operator # 4 Operator # 5

Number of Wells 11 10

Gross Production (b/d) 661 17,500

Oil Production (b/d) N/A 400

Cost (S) 405.738 701,294

Incentive by CPU (S) 166,000 265,380

Annual Average Savings (KWh/yr) 3,194.047 5,019,790

Cost of Energy (S/Kwh) 0.080 0.080

3. Circuit Rider Controllers (CRC)

CRCs have three main components; surge suppression, capacitance, and line noise filtration. It isolates the motor or circuit from receiving or dispersing transient surges while reducing harmonics. In addition, CRCs filter out AC line noise problems, resulting in a much cleaner filtered AC current. Installing CRC allows for a better voltage/current balance, increases voltage at the motor, and improves power factor.

In addition, line loss is reduced. One producer installed CRC on 25 wells with the following result;

Operator # 6

Number of Wells 25

(,toss Production (b/d)

N /\

Oil Ploduction b/d)

N/A

Cost (5)

37.704 liicenrivc by (ldJ (S)

18.852

,\nno;d .\vcragc Savings (KWh/yr) 1.389.000

Cost of I inergy (S/ K’,vh) 0080

48


4. KOBE Systems

The KOBE system utilize ,-, production fluid as the power transfer medium instead of steel rod as used in traditional rod beam pumps. The working fluid is transmitted from a surface unit via surface piping to actuate a standard rod type pump in the well. The working fluid is cyclically applied and relieved at the surface unit to create pumping action in the well. A major drawback of the KOBE system is the large electrical requirements associated with the inefficiencies of the fluid piping system. The application of the KOBE system involved 43 wells with the following results. It should be noted that, in addition to reduced power requirements, the actual production increased and pump failures were reduced.

Operator # 7

Number of Wells 43

Gross Production (b/d) 960

Oil Production b/d) 530 � . Project Cost (S) 1,448,351

Reduction in annual Oper Fxp ($) 547,500

Incremental rev due to addl. Prod (S) 173,900

Incentive by CEU (5) 212,375

Annual average savings KWh/yr 2,644,046

Cost of energy (5/kWh) 0.073

Using this information, calculate the following for each of the four options. If multiple operators are involved, do the calculations for individual operators, as well as on an average basis.

1. Calculate the time it takes to recover the initial investment with and without CEU contributions. 2. Assuming that the annual benefit in each project will continue for 4 years, calculate the profit from each alternative with

and without CEU contributions. 1 Based on the evaluation of each of the four alternatives, which alternative is the best based on profit per unit investment?

Would your answer be different if we selected the best alternative as the one which takes the least amount of time to recover the investment? For part 3, assume that CPU contributions are zero.

(Sac Study Solution 1-4

1. Pump-off Controllers (PO(

Operator #1 Operator # 2 Operator #3 Total

Number of Wells 27 3 10 40 Gross Production (b/d) N/A 180 623

Oil Production (b/d) N/A 26 104

Cost (S) 26,199 13,000 129,775 168,974 Incentive by (i1U (S) 13,100 6,500 64,888 84,488 Annual Average Savings (KWIi/?i) 391,207 74,720 822,915 1,288,842 Cost of Energy (S/Kwh) 0.080 0.115 0.080 0.0917

Payout Period (w/ Incentive), mo. 5.0 9.1 11.8 8.6 Payout Period (w/o Incentive), urn. 10.0 18.2 23.7 17.2 Profit (w/ Incentive), $ 112,087 27,871 198,446 388,089 Profit (w/o Incentive), $ 98,987 21,371 133,558 303,601

Profit Per Unit Cost (win Incentive) 3.78 1.64 1.03 1.80

Sample calculations:

Payout period w/o incentive) = (26,199 )/(391,207x 0.080) x 12 = 10 months Payout period (w/ incentive) = (26199 - 13,100)/(391.207x 0.080) x 12 = 5 months Profit (w/o incentive) = 4091,207xO.03 26.199 = S98,987 Profit w/ incentive) = 4x391,207x0.08 26,199 + 13,100 = $112,087 Profit per unit cost (w/o incentive) - 98,987/261,199 = 3.7$

2.

PI C11141111 I f6cient and Downsizing Motors )l’S (Fl d)3l)

Economic Evaluation in the petroleum Industry Chapter 1 - Economic Principles

Operator # 4 Operator # 5 Total

Number of Wells 11 10 21 Gross Production (b/d) 661 17,500

� Oil Production (b/d) N/A 400 Cost (5) 405,738 701,294 107,032 Incentive by CI1J (5) 166,000 265,380 431,380 Annual Average Savings (K\Vh/yr) 3,194,047 5,019,790 8,213,837 Cost of Energy (S/Kwh) 0080 0.080 0.0800

Payout Period (w/ Incentive), mo. 11.3 13.0 12.3 Payout Period (w/o Incentive), mo. 19.1 21.0 20.2 Profit (wI Incentive), $ 782,357 1,170,419 1,952,776 Profit (w/o Incentive), $ 616,357 905,039 1,521,396 Profit Per Unit Cost (w/o Incentive) 1.52 1.29 1.37

3. Circuit Ride Controllers (CRC)

Operator # 6

Number of Wells 25 Gross Production (b/d) N/A Oil Production (b/d) N/A

� Cost (5) 37,704 Incentive by CEU (5) 18,852 Annual Average Savings KWh/vr 1,389,000 Cost of Energy (S/Kwh) 0.080

Payout Period (w/ Incentive), mo. 2.0 Payout Period (w/o Incentive), mo. 4.1 Profit (w/ Incentive), $ 425,628 Profit (w/o Incentive), $ 406,776

Profit Per Unit Cost (w/o Incentive) 10.79

4. KOBE Systems

Operator # 7 Number of Wells 43 Gross Production (b/d) 960 Oil Production (b/d) 530 Project Cost (5) 1,448,351 Reduction in Annual Oper Exp (S) 547,500 Incremental Rev Due to Addl. Prod (S) 173,900 Incentive byCEU (5) 212,375 Annual Average Savings (K\X’h/\ r) 2,644,046 Cost of Energy (S/kWh) 0.073

Payout Period (w/ Incentive), mo. 16.2 Payout Period (w/o Incentive), mo. 19.0 Profit (w/ Incentive), $ 2,421,685 Profit (w/o Incentive), $ 2,209,310 Profit Per Unit Cost (win Incentive) 1.53

Comparing the results of all four options, CRC is the best option in tcirns of both payback period and profit per unit Cost. The other three options are very similar if you compare the payback pc nod or the profit per unit cost (P1 R).


Problem 1-26

The initiation of o waterflood project will require an initial investment of $2 million. It is expected that the project will

generate additional revenues of $700,000 in the first year followed by a decline of 10% per year. Calculate the payback

period. --

Problem 1-27

You have $1,000 to invest and have the following potential projects in mind. To maximize your benefit, which projects

should be selected?

Project Cost Net Benefit

1 300 80

2 400 160

3 300 105

4 200 50

5 400 100

6 300 90

7 500 125

8 200 40

Problem 1-28

After investing $100,000 in drilling and development costs, it is expected that in the first year net income will be $50,000

declining at a rote of 10% per year over the next five years. Calculate the payback period. What is the profit-to-

investment (PIR)for this project?

Consider the following three alternatives for a project. Based on the payback period method, which alternative should be

selected?

Year A B C

0 -200 -100 -100

1 50 32 42

2 50 32 42

3 50 32 42

4 50 32 42

Problem 1-30

The following two alternatives are considered for a project. Based on a payback period method, which alternative would

be selected?

Year A B

0 -100 -300

1 -200 100

2 -100 150

3 50 100

4 150 50

5 300 0

6 300 0

7 300 0

8 200 0

Would the answer be different if the PIR method is used?



Problem 1-31

A company is considering two alternatives to improve production from a well. The cash flow of these two alternatives is

given below:

Year A B � 0 -10,000 -30,000

1 2,000 12,000

2 2,000 12,000

3 5,000 12,000 � : 4 5,000 8,000

5 5,000 6,000

6 5,000 0

Which alternative should be selected?

TIME VALUE OF MONEY

In any decision-making process, we have to account for the benefits and costs of a project. In a typical

project, the costs occur at the beginning of the project and the benefits accrue over a period of time. For

example, installation of a waterflood project results in benefits in terms of additional oil recovery over

several years. However, to install waterflood, we may have to incur significant costs at the present time.

This money has to come from the internal capital of a corporation or from a lending institute. Either

way, we will lose the opportunity to invest the money somewhere else or we will have to pay interest to

the lending institution. That is, instead of investing the money in a waterilooding project, we could have

earned interest from the bank by investing that money in the bank or, if we borrowed the money to

invest in this project, we have to pay interest on the borrowed amount. This lost opportunity

(opportunity cost) or the interest payment has to be accounted for in our cost benefit analysis. One way

to do this is through the understanding of the time value of money.

Money is a valuable commodity. People will pay you to use your money, whereas you will have to pay

someone to use their money. The cost of money is established and measured by an interest rate. An

interest rate is periodically applied and added to the amount of money borrowed (the principal) over a

specific period of time. For example, depositing $100 in the bank for one year may generate an interest

of $6 at the end of the year. That is, the bank has paid you 6% interest to use your money. In the same

vein, the bank will turn around and lend that money to another individual and charge 10% interest. The

borrower will have to pay back $110 at the end of one year. This means that your $100 today is

equivalent to $106 to you one year from now because of the interest earned on the principal. With the

same token, $100 lent by the bank is equivalent $110 to the bank one year from now. This principal is

called a theory of equivalence (Newman, Engineering Economic Analysis 1991) (Stermole and Stermole

1986) (Park 1993).

THEORY OF EQUIVALENCE

Money is a valuable commodity and it has earning power. We need to establish a method to

compare the money collected at different times. For example, if you have the option of receiving


$100 today or $110 one year from now, which would you prefer? The answer to this question

depends on what you would do with $100 if you received today. If you invest $100 in the bank at 5%

interest rate, you will only earn $5 in one year, leaving you with $105. Obviously, $105 is less than

the $110 you would receive if you had selected the other option; therefore, you should choose the

option to receive $110 in one year.

On the other hand, if you received a hot tip on a particular stock that is expected to grow at a rate of

25% in the first year, by investing in that stock, you can get $125 one year from now. Therefore, you

would prefer to get $100 today rather than $110 one year from now.

Only if you receive a 10% interest rate on your $100 investment will you earn $110 one year from

now. Whether you receive $100 today or $110 in one year, it will not make any difference to you. In

other words, $100 today is equivalent to $110 one year from now if you can earn a 10% interest rate

on your investment.

This example illustrates the theory of economic equivalence. When cash flows can be traded for one

another in the financial world, those cash flows are considered equivalent to each other. In the

example above, at a 10% interest rate, $100 today is equivalent to $110 one year from now. By the

same token, at a 25% interest rate, $100 today is equivalent to $125 one year from now. Studying

these examples illustrates one fundamental aspect of the economic equivalence. It depends on the

interest rate earned. The equivalent amounts will be different for different interest rates.

The economic equivalence is also affected by time. Consider the extension of the previous example.

If someone offers you $100 today, $110 one year from now, or $121 two years from now, which

would you prefer? As before, it depends on what you can do with the $100 you will have today. If

you invest it at a 5% interest rate, after one year you will have:

$ioo+sioo(os)= $105

After two years, you will have:

$ios+sios(os)=$i 10.25

Since $110.25 is less than $121, you should choose to receive $121 two years from today. On the

other hand, if you can earn a 25% interest rate, you will have, after one year:

$100+$100(0.25)=$125

After two years:

$125+ $125(0.25)= $156.25

This amount is greater than $121; therefore, you should choose to receive $100 today rather than

$121 two years from now.

Only if you earn a 10% interest rate, after one year you will have:

I Economic Evaluation in the Petroleum Industry


$ioo+sioo(i)= $110

After two years, you will have,

s110+s110(1)= $121

At a 10% interest rate, whether you choose to receive $100 today, $110 one year from now or $121 two years from now, it will not make a difference. That is, $121 two years from now is equivalent to $100 today at 10% interest rate. This enforces another factor which affects the principle of equivalence - time. $100 today is equivalent to $110 one year from now and $121 two years from now at 10% interest rate; $100 today is equivalent to $125 one year from now, and $156.25 two years from now at 25% interest rate. Notice that equivalence is affected by both the time and the interest rate.

Think about why this principle is so important. When conducting the cost benefit analysis of any project, if the benefits are received in the future, we cannot directly compare the present costs to the future benefits unless we can convert the future benefits to equivalent present benefits; or, alternatively, we will have to convert the present cost to equivalent future costs.

EQUIVALENCE RELATIONSHIPS

In typical economic analysis, cash flows occur in many different ways. The benefit may occur at the end of the useful life of a project (i.e., withdrawing the principal and all accumulated interest at the end of five years from a bank), or the benefits may be received periodically during the life of the project (i.e., opening a restaurant and receiving profit each day). Before we can consider whether the investment made in a given project is justified in light of future benefits, we need to establish systematic procedures for relating monies collected at different times to some common frame of reference. Only then can we compare them correctly. We use the theory of equivalence for this purpose.

To understand the theory of equivalence in a more rigorous way, we need to establish certain relationships starting with some basic parameters. We can define these parameters as:

P = present sum of money F = future sum of money A = end of period cash payment or receipt

= interest rate per period n = number of periods

In establishing the relationships between the various types of cash flows, we will begin with the simplest type of relationship; a relationship between the present sum of money and the future sum of money. Then, we will follow it with much more involved and complex relationships.

54

Mohan Kelkar, PhD., iD,

The cash flow diagram for establishing the relationship between the present sum and the future

sum can be drawn as:

Let us assume that the present sum P is invested at an interest rate of i per period for one

period. The amount collected after one period will be:

P+Pi=P(1+i)

If we continue to invest the new principal, P(1 + i), for another period, we will have, after the

end of two periods,

P(1 + i) + iP(1 + i) = P(1 + i)(1 + i) = P(1 + i) 2

Notice that the exponent of (1 + i) term is equal to the number of periods the principal is

invested.

Extending this analysis, after periods, the sum colletted is equal to P(1 + i)’. We can write,

therefore, the future sum after periods as:

F = P(1 + i)- Equation 1-2

Since, in this equation, we increased the principal for each period (compounded) by the

accumulated interest, this method is called the compounding interest method. Equation 1-2 establishes a relationship between the future sum of money and the present sum of money.

Equation 1-2 can also be written as:

P = ( 1+)11 Equation 13

This relationship allows us to calculate the present sum if the future sum is known. The

following examples illustrate the application of these equations.



Example 1-14

If you need $10,000 after 5 years, how much should you invest today at an interest rate of 10%?

Solution 1-14

Given: F = $10,000, n = 5 years, t = 10%

Find: P

Using Equation 1-3,

F $10,000

= (1 + j)fl = Ti + 0.1)5= $6,209

You need to invest $6,209 today.

Example 1-15

If you invest $4,000 in a bank at an interest rate of 6.25% per year, how much money will you have at the end of

three years?

Solution 1-15

Given: P = $4,000, n = 3 years, i = 6.25%

Find: F F

Using Equation 1-2,

F = P(1 + i) = 4,000(1 + 0.0625) = $4,798

You will have $4,798 after three years.


Example 1-16

If you want to invest $3,000 at an interest rate of 7%, how long will it take to double the initial investment?

Solution 1-16

Given: P = $3,000, F = $6,000, I = 7%

Find: n

Using Equation 1-2,

6,000 = 3,000(1 +.07)-

(1 + .07 1 = 2

Taking log on both sides

nlog(1 + .07) = log(2) - log(2)

- Iog(1.07)

= 10.2 years

The amount will double in 10.2 years. It is interesting to note that in economics, there is a famous rule called the

Rule of 72. It says that if you divide 72 by the interest rate in percentage, you can calculate the number of years

needed to double the money. In our example, dividing 72 by 7 will result in 10.3 years which is close to the actual

answer.

RELATIONSHIP BETWEEN A AND F

Let us extend the previous relationships to a case where a payment is made at the end of each

period. We want to calculate the future value of these payments at the end of the total period.

A cash flow diagram for this arrangement is shown in Figure 1-4.

F

1 "1 Figure 1-4: Periodic payments

This case is similar to the periodic investment in a bank at a fixed interest rate for a certain

period followed by a total withdrawal at the end of that period.



F

_ �___

(n-2) 4

(n-i)

Figure 1-5: Accumulation of interest for periodic payments

Considering that for the first payment, we earned interest for (n - 1) periods (see Equation 1-

2), and for the last payment we earned no interest, using Equation 1-2, we can write,

F = A(1 + j)fl_l + A(1 + Ofl_2 + + A Equation 1-4

Multiplying Equation 1-4 by (1 + i),

F(1 + i) = A(1 + j)fl + A(1 + i) 1 + + A(1 + i) Equation 1-5

Subtracting Equation 1-4 from Equation 1-5 and rearranging, we obtain,

F(1+i)�F=Fi=A[(1+i)-1]

Therefore,

F = A [ (1h1 Equation 1-6

We can rewrite Equation 1-6 as,

A = F I Equation 1-7

Equation 1-6 and Equation 1-7 establish the relationships between A and F.

58

Mohan Kelkcir, Ph, D., ID,

Example 1-17

If you deposit $10,000 at the end of each year, how much would you accumulate at the end of five years at an

interest rate of 6%?

Solution 1-17

Given: A = $10,000, n = 5 years, i = 6%

Find: F

Using Equation 1-6,

F = A [(1 + in -

I j $10,000 [(1 + 0.06) s- ii =1 0.06

= $56,371

You would have $56,371 at the end of five years.

Example 1-18

If you need $100,000 at the end of ten years for a college education, how much should you invest at the end of each

year at an interest rate of 8%?

Solution 1-18

Given: F = $100,000, n = 10 years, i = 8%

Find: A

Using Equation 1-7,

A = F 1(1 + M.0

= $100,000 1(1 + 0.08)10 - 11 = $6,903

You should invest $6,903 at the end of each year to receive $100,000 at the end of ten years.

Example 1-19

You intend to invest $1,000 per year in mutual funds. If the average annual yield from this fund is expected to be

12%, how long will it take before you will accumulate $15,000 in your account?

Solution 1-19

Given: A = $1,000, F = $15,000, i = 12%

Find: n

Using Equation 1-6,

P - (1 + j)n 1

i



(1 + i) Th = 1 +

Substituting,

(1+i)=1+.12x 15,000 =2.8 1,000

nlog(1 +.12) = Iog(2,8)

Therefore,

n = 9.1 years

It will take approximately 9.1 years before $15,000 will be accumulated.

Example 1-20

After graduating from college, Betty plans, in 5 years, to buy a house worth $150,000 with a 20% down payment.

Betty’s father gives her $10,000 as a graduation gift. If Betty invests that money at a 6% interest rate, how much

additional annual savings will she have to invest at the same interest rate to accumulate the desired 20% down

payment at the end of five years?

Solution 1-20

Given: F = 20% of $150,000, P = $10,000, i = 6%, n = 5 years

Find: A

In this example, Betty is investing $10,000 at the beginning of year 1 plus additional annual investments to get

$30,000 at the end of five years. The cash flow diagram can be drawn as shown in Example Figure 1-2.

F = $30,000

IF Ilr

AAAAA

$10,000 Example Figure 1-2: Cash flow diagram for Example 1-20

The first step is to can calculate the future value (F1 ) of $10,000 after five years.

Using Equation 1-2,

F1 = P(1 + i)’- = 10,000(1 + .06)

= $13,382

The remaining future value has to be the result of annual investments. We can calculate the remaining future value.

F2 = 30,000 + 13,382 = $16,618

Using Equation 1-7,

60 Mohon Kelkar, Ph.D., J.D.

I I

� A = F t(l+iflh1 _ 0.06

� - 16,618

� = $2,948

I Therefore, Betty needs to invest $2,948 at the end of each year.

I

I

I RELATIONSHIP BETWEEN A AND P

Let us extend the relationship one step further by relating the present value to the periodic

payments. As shown in Figure 1-6, we want to calculate the present value of future periodic

payments.

Figure 1-6: Relationship between P and A

From Equation 1-6, we know that

FP(1+i)

From Equation 1-6, we know that

F - A (1+i)-1 �

Substituting Equation 1-2 in Equation 1-6, we can write,

P(1+i)=A (1+i)’-1

Simplifying,

P - A n Equation 1-8

Equation 1-8 can also be written as:


i(i+i) I A = P

{(1+)fl_1i Equation 1-9

Equation 1-8 and Equation 1-9 establish the relationship between the periodic payment (A) and

the present worth (P).

Example 1-21

If you take a home improvement loan of $10,000 to be paid over a five year period, what will be the yearly payment

if the interest rate is 12% per year?

Solution 1-21

Given: P = $10,000, n = 5 years, i = 12%

Find: A

Using Equation 1-9,

[ i(1+i) A = P

[(1 + j)fl - 1

.12(1 +.12) 5

= (1 + .12) - 1

= $2,774

The yearly payment will be $2,774.

Example 1-22

If you want to invest sufficient money in the bank so that, at an interest rate of 8%, you will receive $20,000 per year

for the next 10 years, how much should you invest in the bank?

Solution 1-22

Given: A = $20,000, n = 10 years, i = 8%

Find: P

Using Equation 1-8,

P = A [(1 + j)fl

- 1]

i(1+i)’ J (1 +.08)’0 1

= $20,000 [.08(1 +.08)10

= $134,202

You will have to invest $134,202 today to earn $20,000 per year for the next ten years.

Example 1-23 -

Able borrows $1,000 from a loan shark. In return, the loan shark demands that Able pay $100 per month for one

year. What is the monthly interest rate the loan shark is charging?

Solution 1-23

Given: P = $1,000, A = $100 per month, n = 12 months


Find:

Notice that the periodic payment and the period are given in terms of monthly units.

Using Equation 1-8,

[(1 P A

+ jY’ - ii =I

j i(1+i)’1

Substituting

1,000(1 + 012 - 1

i(1+i)12 =10

There is no explicit solution for i. By trial-and-error,

For

P 1=1%

P

P = 2.9% -10.01

I Therefore, the interest rate charged is 2.9% per month.

Example 1-24

Betty buys a new computer at a price of $10,000. Betty expects that the use of the computer should result in an

annual income of $2,500. If Betty wants to earn at least a 15% return on her investment, at what price will the

computer have to be sold after 4 years?

Solution 1-24 -

Given: P = $10,000, A = $2,500, i = 15%, n = 4 years

Find: Salvage value (resale price) of the computer

We can draw a cash flow diagram for this example as shown in Example Figure 1-3.



F=? A = $2,500

: : :

P = $10,000 Example Figure 1-3: Cash flow diagram for Example 1-24

Using Equation 1-8, we can calculate the present value of periodic payments as:

(1 + j)’ - 1 P1=A

i(l+i)

Substituting

(1 + .15) - 1 P1

= 2,500 .15(1 + .15)

= $7,137

The remaining present investment will have to be recovered by the future resale price.

The remaining present investment is

= 10,000 - 7,137 = $2,863

Using Equation 1-2,

F = P(1 + j)" = 2,863(1 +.15)4

= $5,007

The computer will need to be sold at a price of $5,007 at the end of 4 years.

In solving these and other examples and problems, please remember that the units of periodic

payment, interest rate per period and the number of periods have to be consistent. For

example, if the payment is per month, then the number of periods has to be in months and the

interest rate has to be defined per month. Similarly, if the payment is per year, the interest rate

has to be defined per year and the number of periods has to be in years.

GEOMETRIC GRADIENT PAYMENT

In the previous section, we developed the relationships when the periodic payment is constant.

However, it is possible that instead of being constant, the payment can change as a function of time.

For arbitrary changes in payment, there is no analytical solution. However, for some simplified

cases, an analytical solution is available. One case that has practical implications in the oil industry is

64

Mohan Kelkar, PhD., 1.0.

geometric gradient assumption. Geometric gradient assumes that, instead of being constant, the

payment is changing by a constant percentage. This is useful in the petroleum industry because, in

many instances, we assume that production from a field or from a well is declining by a constant

percentage. If we want to capture future production, we can use a geometric gradient series. As

with production, while doing an economic evaluation, we may assume that the price of the

commodity (either oil or gas) is increasing by a constant percentage. This assumption can also be

incorporated using a geometric gradient series. The relationship between the changing payment and

the future sum is shown in Figure 1-7.

F

12 n�i n

A A(l+g)

A(1 +g)’

u-I A(1 +g)

Figure 1-7: Geometric gradient series

In this case, we assume that the periodic payment changes by a constant percentage every period. If

payment A is made at the end of the first period, it will change to A(1 + g) at the end of the second

period where .g is a constant fraction. For the last period,.n, the payment is A(1 + g)fl_l. Knowing

that the interest earned on the first payment is for (n - 1) years, and no interest is earned on the

last payment, we can write the equation for the future sum as,

F = A(1 + 0_’ + A(1 + g)(l + ofl_2 + + A(1 + g)--1 Equation 1-10

(1+1) Multiplying the Equation 1-10 by -----3, we get,

F� = A (1+O

+ A(1 + 0- + A(1 + 0(1 + ) n2 Equation i-Il (1+9) (1+g)

Subtracting Equation 1-10 from Equation 1-11, we obtain,

(1+i) (1+i’ F �F=A �A(1+i)

(1+g) (1+g)

Simplifying,



F Equation 1-12 (i-g)

This equation is applicable only if i g. If i = g, Equation 1-10 simplifies to,

Adding,

F = nA(1 + j 1 Equation 1-13

Knowing the relationship between F and P, Equation 1-2, we can write:

F = P(1 + j)fl Equation 1-2

Substituting Equation 1-2 in Equation 1-12 and rearranging,

P = -4- Ft - (1+9)1 Equation 1-14 (i�g) I (i+0Th1

Similarly, for i = g, Equation 1-2 can be substituted in Equation 1-13. Therefore,

P(1 + j)fl = nA(1 + j)fl_l

and,

P = nA-- Equation 1-15 (1+i)

Example 1-25

If you invest $10,000 in the first year with a 10% increase in each subsequent year, how much money would be

accumulated at the end of ten years at an interest rate of 8%?

Solution 1-25

Given: A = $10,000, p = 10%, 1 = 8%, n = 10 years

Find: F

Using Equation 1-12,

F=---f(i+i)-(1+g)} (’ - 9)

10,000

= (1.08-1) [(1 +.08)’0 - (1 + .1) 10 1 = $217,410

After ten years, you would collect $217,410.

1-26

If, as an employer, you guarantee one of your employees an initial salary of $20,000 per year and an annual increase of

M. Mohan Kelkar, Ph.D., J.D.

at least 6% for the next five years, what is the minimum amount of money you need to set aside at an interest rate of 7%

to cover the cost of the employee’s salary? If the interest rate is 6%, how much more needs to be set aside?

Solution 1-26

Given: A = $20,000, p = 6%, j = 7%, n = 5 years.

Find: P


A (1+g)

P=(j_g) 1(l+i)’ 20,000 (1+.06) 1

= 1 - (.07 .06) (1+.07) = $91,727

At a 7% interest rate, $91,727 has to be set aside to cover the cost of the employee’s salary.

If the interest rate is 6%, i = p. Therefore, using Equation 1-15,

nA P =

(1+i) 5 x 20,000

= (1 = $94,340

+ .06)

Therefore, at an interest rate of 6%, an additional $2,621 ($94,340 - $91,719) has to be set aside. -

Example 1-27

Able pays $500,000 for a producing oil field that generates a profit of $10,000 in the first month. The field profit is

expected to decline at a rate of 1.0% per month. If Able expects to get a rate of return of 1.25% per month on his investment, at what minimum price will he have to sell the field after 10 years of production?

Solution 1-27

Given: A = $10,000/month, p= -1%, i = 1.25%, n = 120 months.

Find: P, salvage value

First calculate the present sum of all future profits for a ten year period.


A F (1+y)l Ii� I

(i�g) (1+i)’j

10,000 1 (1 - .01) 120 1

C0125+.01)[1 1+0125 j=$414478

Since Able paid $500,000 for the property, the balance has to come out of the salvage value after 10 years.

balance at present = 500,000 414,478

= $85,522

However, we need to calculate the future value of this balance. Using Equation 1-2,



F = P(1 + j)n

= 85,522(1 +.0125) 120

= $379,736

Able will have to sell the oil field for $379,736 after 10 years.

Example 1-28

In the oil industry, a rule commonly used to evaluate oil property is to multiply daily production (in STB/d) by one

thousand times the price of oil to determine the present worth of the field. How can this rule be justified based on the

geometric gradient series?

Solution 1-28

Equation 1-12 can be simplified for a large value of n. We can write P as:

A P =

The reason for this simplification is that the second term in the bracket goes to zero when a fraction less than one is

taken to a higher power. As is common, because of the risk involved in the oil business, if we assume that the rate of

return you want to earn on an investment is 25% (i = 25%), and a typical decline of a field is 10% (p = �10%), we

can write the equation above as:

daily production x 365 x $/STB

(0.25+0.1) daily production x

$x 1,000

STB

Since the interest rate is per year, we also need to use production per year. That is why we multiply by 365. We can use

this equation to calculate worth of an oil field. For example, if the field is producing 50 bbl/d and the oil price is $30/bbl,

then the field is worth 50 x 1000 x 80 = $4,000,000. Please note that this is a simplified form and not every field

depletes at a rate of 10%. It is also true that the net revenue from a barrel of oil is not the price of oil. We have to

account for production costs as well as royalty interest. However, this is a good approximation that works well in many

situations.

Example 1-29

Following is a newspaper article about Anadarko Corporation selling its Louisiana assets to EXCO Corporation in 2006.

The oil price was trading around $50/barrel at that time. Assume 6MScF 1STB:

"Anadarko Petroleum Corp. has confirmed it has agreed to sell two gas fields in Louisiana to EXCO Resources. Inc. for $1.6

billion in cash. .... The company said it agreed to sell its Vernon and Ansley fields, located in Jackson Parish, Louisiana, to

EXCO, which is focused on acquiring and developing onshore North American oil and natural gas properties. The fields

produced 192 million cubic feet equivalent per day from about 350 wells on 66,000 net acres as of November 1, according

to Anadorko."

How does our rule compare to the actual purchase price?

Solution 1-29

Using the equivalence, we can calculate 192 million SCF = 32,000 STB. Using our rule, the present worth of property

would be 32,000 x 50 x 1000 = $1.6 billion. This is consistent with the price EXCO paid for the property.

(Size Study 1-5

Gas wells start loading when the gas flow rate is not high enough to remove the liquids accumulated in the tubing Several solutions are used to minimize liquid loading. lhesc solutions include: (1) injection of chemicals (surfactants) to create a stable foam so that the foam can he produced to the surface; (2) changing the tubing size; (3) gas lift by injecting additional gas, and (4) plunger lift which does not allow the liquid fall back while liquid is being lifted to the surface. For marginal wells, gas lift is

Mohan Kelkar, P/iD., J.D.

not economically viable; therefore, techniques commonly used include plunger lift or use of capillary strings to inject chemical downhole to lift the liquid.

Traditional plunger lift requires shut in time for the plunger to fall hack to bottom and to allow for bottomhole pressure to build. This thut-in time results in lost production and forces liquids back into the formation. Chevron is testing a new plunger (called Pacemaker) which has several advantages over conventional plunger lift. The Pacemaker plunger operates as two interdependent pieces a cylinder and a ball (Case Study Figure 1-5) - each of which fall separately and are designed to do so against a significant gas flow rate. Once the cylinder reaches the bottom, it encounters the ball, which seals off the cavity in the cylinder. Gas is now forced to travel around the cylinder. The gas velocity around the cylinder results in a drag force, causing the cylinder and ball to return to surface, thus acting like a piston. Once on surface, a rod in the lubricator separates the ball from the cylinder, allowing the ball to fall back to the bottom.

� .

-

a C

ç -

cam Study - 1 çu re 1-5 Different sacs’s anJ (ESteeM/s of two-piece f/eu’-through plungers

Only 5-10 seconds of shut-in time per cycle is required. This minimizes production fluctuations and means less interference for wells sharing the same facilities and/or compression. The longer flow period means more production, and liquids are not forced back into the formation. Stabilized production allows liquids to be produced as opposed to accumulating in the near-weilbore area, reducing the relative gas permeability.

Pacemaker has some limitations. Since this plunger system is driven by gas velocity, it works best at low wellhead pressures. With 80 to 100 psi of flowing tubing pressure, the plunger system performs best when unloaded gas rates exceed 150 to 200 Mcfd. Very high liquid rates impede the ball’s fall and can result in the cylinder catching the ball prior to reaching bottom. Wellborc restrictions, tubing set too high or excessive sand production will prevent optimum performance. Finally, those involved must understand that the cit/ica/Jartor is gas velosatj, ii 01 pressure as Will) 0 csni’entionalp/uqger 1fi’ ys/ern. Training as to how to set the controller, troubleshoot and optimize is critical.

An example of application of Pacemaker plunger lift is shown in a South Texas well that originally was operated by capillary string through which chemicals were injected. A capillary string subsequently began producing more condensate and became more difficult and expensive to operate (Case Study Figure 1-6). The Pacemaker plunger system was installed in October 2002. Production increased and become more consistent, producing an incremental 40 Mcfd. Based on the evaluation, the incremental production is declining at a rate of 1 1/o per month. Prior chemical costs of $1,740 per month have now been eliminated. The cost of Pacemaker is $6,000. Assume the net price of gas to be S3/MSCF. Assume the life of the well to be 5 years after installation.



500 � ---

. t i ��- � -

ji", QU

M r , t 111 FTP

L i1 Auqll2 27-Auçj02 22-5tp-02 0L02 13-l&v-02 cc-02

Case

St nlj Pagure 1-6: South Texas example: capithay srrlsag vs. Pacemaker

Case Study Figure 1-7 compares the results of standard plunger lift with Pacemaker. The results are from West Texas gas field where the production increased by 1,200 MSCFD for 10 wells after installing Pacemaker plunger lift. Again the cost is S6,000 per installation and the net price of gas is assumed to be S3/MSCF. Assume the average life of wells to be 5 years.

&litJO

13,500

3,000

2.500

2.000

1.500

1.000 0

U)tiys Case Study FIgure 1- 7 Wear Texas field test: normaliaedproductkan fi,im all wells

As an engineer working for Chevron, you are asked to evaluate the performance of Pacemaker against capillary string as well as conventional plunger lift. Using the information above:

1. Calculate the present value of incremental revenue by installing Pacemaker compared to capillary string. What is the net profit? Assume interest rate to be 15% per year.

2. Continuing the first part, if the Pacemaker will have to be replaced after 3 years, how would the economics change? Assume that capillary string will not have to he replaced and continued to be used for 5 years.

3. Compare the effectiveness of Pacemaker compared to conventional plunger lift on an average well basis using the West Texas information.

4. 1 low will the economics in Part 3 change if the Pacemaker has to be replaced after 2 years

Case Studj’ Solution 1-5

Capillary String Replacement

For the first part, we can use the equation for geometric gradient series. Since we know the decline of a well to he 1%/month, we also need to define the interest rate per month. It would he 15/12 = 1.25% per month. Incremental production is 40 MSCIO). We will assume 30.4 days in each month. Using this assumption, we can calculate the present value of future o.venues as:

70


40 x 30.4 x 3 (1 -0.01)60 PVprod (00125 + 001) 1 - (1 + 0.0125)60 = $120,033

Note that is equal to -0.01. ’I’hc number of years over which this is valid is 5 years or 60 months. In addition, we also save the Cost of chemicals. ’Ihe cost savings can be written as:

I (l+0.0125) ° -1 PVc6em savings = 1,740 0.0125(1 + 0.0125)60 = $73,140

I Combining the two and subtracting $6,000 which is the plunger cost, we can calculate the profit as 120,033 + 73,140� 6,000 = $187,173.

Even if we have to replace the plunger after three years, the cost will he 6,0001(1+0.0125) 36 $3,836. ’Ibis is the present value of the costs after three years. Subtracting this from the profit, we still make $183,337 profit.

Conventional Plunger

We can calculate the present value of incremental gas as:

(1+.012 5)60 1 PVprod = 1,200 X 30.4 X 3 0.0125(1 + .0125)60 = $4,600,265

I Subtracting the cost of 10 plungers, profit = 4,600,265 - 60,000 - $4,540,265. Even if we have to replace the plunger after two )ears, the additional cost will be 60,000/(1+0.0125) 24 . Subtracting it, we still make S4,495,733 profit.

I Overall, Pacemaker is a good investment for this project.

I I ---. - .. ....................................................

NOMINAL AND EFFECTIVE INTEREST RATES

I In the previous section, we were careful to define the periodic payments that were based on certain

periods, and the interest rate that is also defined based on per period. Although many examples we

considered were based on annual payments (yearly payments), we also considered some examples

in which payments were made at shorter intervals (months). As long as we defined our interest rate

consistently with the payment period, we do not have.to make any special adjustments to our

economic analysis. The equations developed in the previous sections can easily be adapted to

shorter durations as long as the periodic payment and the interest rate per period is defined for the

some period.

I In practice, however, the interest rates are rarely defined for a period shorter than one year. If you

invest money in the bank or if you borrow money from a lending institution, the interest is, typically,

- defined on an annual basis. However, depending on how frequently the payments are paid, the

effective interest rate could be different than the annual interest rate charged. Consider a simple

example to illustrate this concept.

Esampie 1-30

$1,000 is invested in the bank at a rate of 12% per year. The bank statement declares, "the interest is compounded

monthly based on the average daily balance during that month." How much money will you accumulate at the end of

one year?

Solution 1-30



The interest rate is equal to 12% per year. Since the interest is compounded monthly, we need to calculate the monthly

interest rate.

-. 12%/year - 1%/month

- 12 months/year -

The average daily balance (in the absence of any withdrawals or deposits) will be the principal plus the accumulated

interest from the previous months. For example, at the end of the first month, the interest will be:

1,000 x .01 = $10

Therefore, we will accumulate $1000 + $10 = $1010 in the account. In the second month, the interest will be:

1,010 X.01 = $10.10

Therefore, at the end of the second month, the total accumulated will be:

1,010 + 10.1 = $1,020.10

In general, we are simply compounding the principal each month by the designated interest rate. We use Equation 1-2,

F = P(1 + j)n

In this example, P = $1,000, i = 1%/month, n = 1 year = 12 months.

Substituting,

F = 1,000(1 +.01)12 = $1,126.80

Therefore, we will accumulate $1,126.80 by the end of the one-year period.

Instead of accumulating the interest monthly, if the interest is accumulated at the end of the year, we will receive,

F = 1,000(1 +.12) 1 = $ 1,120.00

The two amounts are not equal. By compounding the interest more frequently, we accumulate more money. We can

define an equivalent yearly rote that will give us the same amount of money as the monthly compounding. For example,

if we assume that our interest rate is 12.68%, and the interest is compounded at the end of each year, at the end of one

year, we will accumulate,

F = 1,000(1 + .01268)1 = $1,126.80

This is exactly the same amount we will receive if the interest is 12% per year, but is compounded monthly.

12.68% in this example, therefore, is the effective interest rate, and 12% is the nominal interest rate.

This brings us to the definitions of nominal and effective interest rates. The nominal interest rate is

the interest rate advertised, typically, on an annual basis. Most of the banks or lending institutions

will use the annual interest rate to advertise the attractiveness of the loan or the interest

accumulated on the principal. The nominal interest rate need not be defined on an annual basis,

although it is the most common method. Nominal interest rate for any period can be calculated

using the following equation,

nominal annual interest rate Equation 4 nominal interest rate per period=

number of periods in one year

72


For example, if the nominal interest rate per year is 6%, we can calculate the nominal interest rate

per quarter as,

6% - - - 1.5% per quarter

4

In the example above, 4 in the denominator represents four quarters per year. The nominal interest

rate per month will be calculated as,

6% =

-j--1

2 = 0.5% per month

I The effective interest rate is the actual interest rate paid based on the number of compounding sub-

periods as well as any other hidden charges. More often than not, the effective interest rate will be

higher than the nominal interest rate. However, in some cases, the reverse may be true. Consider

that if you pay off a credit card balance as soon as it is accumulated, you pay no interest. That is, the

effective interest rate is zero, whereas the nominal interest rate charged may be higher. The

number of compounding sub-periods (we define them as M) represent the number of times the

interest is compounded within the period for which the nominal interest rate is defined. Note that

the nominal interest rate is equal to the effective interest rate if the number of compounding sub-

periods is equal to one.

I I I I I I I I I

Let us consider the development of the relationship between the nominal and the effective interest

rate. If we define,

j = nominal interest rate per period

M = number of compounding sub-periods

Using Equation 1-14, we can calculate the nominal interest rate per sub-period as,

I M

If we invest principal P for one period (M sub-periods), we can calculate the future value of the

principal as, using Equation 1-2,

F = P (i + j Equation 1-15

If we want to define the effective interest rate, i, we need to accumulate the same future value at

this interest rate by compounding it only once during that period. That is,

F = P(1 + 0 1 Equation 1-16

Equating Equation 1-15 and Equation 1-16,

1 + = (i +)M

Economic Evaluation in the petroleum Industry


or

= (1+ - 1 Equation 1-17

Equation 1-17 defines the relationship between the effective interest rate per period, i, and the

nominal interest rate per period,]. Note that if compounding sub-periods, M, is equal to one, then,

= (1 + j)1 - 1 = j Equation 1-18

Let us consider some examples illustrating the applications of these equations.

Example 1-31

A bank advertises in a newspaper: Invest a minimum of $5,000 today in a CD (certificate of Deposit) account at an

interest rate of 8.5% per year and receive an effective yield of 8.87% per year by compounding the interest daily." Do

you believe that this statement is accurate?

Solution 1-31

Given: j = 8.5% per year, M = 365 (365 sub-periods in one year)

Find: i


/ .085 I = (1

+)365

= 8.87%

The effective interest rate is 8.87%. Therefore, the statement is accurate.

Example 1-32

A credit card agreement states, "The finance charge will be calculated on a monthly basis based on the Average Daily

Balance." Further, it states that the finance charge is "135% which is an ANNUAL PERCENTAGE RATE OF 21%." Is this an

accurate statement?

Solution 1-32

The finance charge is calculated on a monthly basis. If we assume that the nominal interest rate per year is 21%, then the

monthly nominal interest rate is, using Equation 1-14,

21 = = 1.75%

Therefore, the statement in the credit card is accurate to the extent that 21% is the nominal annual percentage rate.

However, the interest is compounded monthly; therefore, M is equal to 12. Using Equation 1-17,

= (i +)M

+ 0.21) ’

= 23.14%

74 Mohon Kelkcxr, Ph.D., J.D.

The effective interest rate is 23.14%, which is much greater than the 21% advertised.

We, therefore, can say that the statement in the agreement is misleading because it does not specify whether the

interest is nominal or effective.

In defining the time value of money, we should use the effective interest rate rather than the

nominal interest rate in our computations. The effective interest rate is the actual interest rate

charged to the principal. In applying any of the equations we learned in the previous section, it

should be remembered that the units for many of the terms should be consistent. The determining

factor in most instances is the periodic payment. The interest rate used in any calculation should be

the effective interest rate per period (period defined based on the periodic payment) and the

number of periods should be defined in a consistent unit as the periodic payment. For example, if

the periodic payment is per month, then the effective interest rate should be per month, and the

number of periods should be in months.

Example 1-33

Able borrows $50,000 for a lending institution for building a house. The lending institution charges 8% per year nominal

interest rate compounded daily. If Able intends to pay the loan back in 10 years, what is his monthly payment?

Solution 1-33

Given: P = $50,000, j = 8% per year compounded daily, n = 10 years

Find: A per month

In this example, the payment needs to be calculated per month. Therefore, we need to define the effective interest rate

per month as well as the number of periods in months. Using Equation 1-14,

8% nominal interest rate per month = - j-- = 0.6667%

The number of sub-periods per month = 30.4 days/month. Using Equation 1-17,

=(

.006667 304

i+ 30.4 ) �1

= .6688% per month

The number of periods - 10 years - 10 x 12 - 120 months. Using Equation 1-9,

F i(1+i)" 1 A=Pl

[(1 +O’ -1 j 006688(1 + .006688) 120 1

= ]so 00 [ (1 + .006688) - 1 j

= $607.30

The monthly payment will be $607.30.

Example 1-34

A department store advertises on September 1a,

"Buy now and don’t make the first payment until January 31 at 0% interest.’ At the bottom of the advertisement, in small print, the advertisement states, ’After January 31a,

the interest will be charged at 18% annual rate compounded daily."



75

If you buy a computer for $2,500 on September 1’, and pay it off on January 31st

what effective interest rate have you

paid?

Instead of paying on January 31, you make the first payment on February 281h

You make twelve consecutive monthly

payments; how much is your monthly payment? What is the effective interest rate charged to you?

Solution 1-34

Part I: Payment on January 3l

Since the interest charged until January 31t is 0%, the payment will be $2,500 on January

31st The payment and the

borrowed amount is the same (by purchasing the computer and not paying for it, you are borrowing $2,500 from the

store!), therefore, the effective interest rate is zero.

Part II: Constant monthly payments starting at the end of February

In this case, the interest will not start accumulating until the first day of February. In other words, whether we bought

the computer on January 31st or September 1st, the loan amount will be the same: $2,500. We, therefore, can calculate

the monthly payment on $2,500 using the appropriate monthly effective interest rate. Using Equation 1 - 14,

18% nominal interest rate per month

= �j-- = 1.5%

For daily compounding, M = 30.4 days. Using Equation 1-17,

01S )m =(i~-) �1

= .0151% per month

Using Equation 1 -9, for n = 12,

i(1 + i)’ A = P

(1 +i)-1 [.0151(1 + .01S1) 12

= 2500[ (1 +.0151) 12 - 1

= $229.21 -,

You will have to make a monthly payment of $229.21 for 12 months to pay off the $2,500 loan.

To calculate the effective rate, note that for five months (September 1 - January 31), you did not pay any interest. After

that, over a 12 month period, you paid an interest rate of 1.51% per month. The effective rate should be somewhere

between.

If we define the effective interest rate = e per month, we can write that the interest on the $2,500 loan should have

accumulated at that rate. Therefore, on January 31st

the future value of the loan should be, using Equation 1-2,

F = P(1 + O = 2,500(1 + te) 5

This will be the new principal that should be paid off in twelve months. So, using Equation 1-9, we can write our monthly

payment as,

No 76

Mohan Kelkar, Ph.D., J . D. No

[_i(l+i)1 A

=P (1 + n - 1

j(1 + ie) 12 1 A = 2,500(1 + e) [( 1 + je)12 il

We already know that our monthly payment is $229.21. Therefore, we need to find the value of i, so that the right side

of the above equation is equal to $229.21. By trial and error,

for ie=1% A=$233.40

i,=.85% A$229.50

That is, the effective interest rate is 0.85% per month. This is less than the nominal interest rate.

In some instances, we can simplify our relationship between the nominal and the effective interest

rate if the compounding is continuous. Recall that,

i = (i +)M �1 Equation l-17

represents the relationship between the effective and the nominal interest rate. If M is very large

(continuous compounding), we can write,

= 11m 7n .large (i +� j )m - 1 = e 1 - 1 Equation 1-19

where i = the effective interest rate per period and is the nominal interest rate per period. This

equation represents the relationship between the nominal and the effective interest rate for

continuous compounding.

Example 1-35

If the nominal interest rate is 16% per year, calculate the effective interest rate if,

a. The interest is compounded quarterly.

b. The interest is compounded monthly.

C. The interest is compounded daily.

d. The interest is compounded continuously.

Solution 1-35

Given: j= 16% per year

Find: i for various compounding periods


= (i +)M 1

a. Quarterly compounding,

0.16 �1=16.98%



b. Monthly compounding,

/ 0.16\12 �1=17.23%

C. Daily compounding,

0.16 365 �1=17.35%

d. Continuous compounding, using Equation 1-19,

= e’ - 1

= e 0 ’ 16 - 1 = 17.35%

Notice that the effective interest rate increases as the compounding sub-periods becomes decrease. However, there is

practically no difference between daily compounding and continuous compounding. (The difference is at the third

decimal point). Unless the nominal interest rate is very large (in practice, not very likely), there is no difference between

the daily compounding and continuous compounding.

Case Study 1-6

The Wail Street Journal (Wall Street Journal 2009) reports that during Britain’s boom, easy credit helped Eduardo lreneo pile up £1,750 (52,848) in bank debt. When the boom turned to bust in 2007, the banks cut him off and he turned to a loan shark, an illicit credit source that is gaining popularity.

According to Mr. Ireneo’s testimony in a recent court case, Mr. Irenco went to a south London loan shark in April 2007, seeking money to send to family members in the Philippines after his bank refused to lend him any more money. A friend introduced him to Greg Dc Guzman, a fellow Filipino immigrant to the U.K. who sat him down at his kitchen table and scribbled an agreement on a sheet emblazoned with his logo - a mouse in a hat next to the words "General Speedy." He would give Mr. Ireneo £1,500 in exchange for £1,950 to be paid over six months. Mr. lreneo signed the paper but never received a copy. After making three monthly payments of £325, he called Mr. Dc Guzman in October 2007 to say he was having difficulty keeping up and wanted to reduce his monthly payment, he told the court. Mr. Dc Guzman agreed, saying Mr. Irenco could pay £100 a month for an additional 13 months. Mr. lreneo kept paying, and in June 2008 (after 9 payments) phoned to say lie wanted to pay the remaining balance. But Mr. Dc Guzman told him he still owed £1,950, saying that what he had paid until then was interest only.

Calculate:

1. What is the initial effective monthly interest rate Mr. Irenco agreed to pay in April 2007? 2. After making three payments of £ 325, Mc lreneo paid £ 100 for additional nine months. Assuming that the loan

- would be retired at that point, what is effective monthly interest rate Mr. lrcneo would have paid? 3. After making all the payments, if Mr. Dc Guzman believes that Mr. Ireneo still owes him additional £ 1,950, what is

the monthly effective interest rate he is expecting to charge to Mr. lrcneo?

Case Study Solution 1-6

Initial Agreement

We need to solve the following equation

(1 + 06 - 1 1,500 = 325

i(1 + 0 6

ihrough trial-and-error, the effective interest rate is 8% per month

Second \crccmcnt

lhis would include three payments of 325 and nine payments of 100. Notice that in the equation below, we consider the fact that the £ 100 payments started after three months.

78 Mohan Ke/kor, Ph.D., i.D.

( 1 +O -1

1,500 = 325 [

____

+ i)3 I +100 i(1 + 012 I Using a trial and error, the effective interest rate i 5% per month.

Third irccicit

\X’e need to solve the following equation.

1,500=325l� [(1 + ) 3 - 11 +1001 1(1 +

i) 9 - 11 1,950

- I [ i(1+i)3 J i(1+i)12 j+(1+.)12

Through trial-and-error, the effective interest rate is 14% per month.

Problem 1-32

If you invest $1,000 in a five year band paying 8.5% interest annually, how much will you get back after five years?

Problem 1-33

If $5,000 is invested in a bank today followed by $3,000 after three years, how much money will be accumulated at the

end of six years if the interest rate is 6%?

Problem 1-34

To receive $2,000 three years from now, how much should you invest in the bank at a rate of 7%?

Problem 1-35

You need to have $7,500 seven years from today. If you currently have $5,000, at what rate should you invest to receive

the needed amount?

Problem 1-36

Able has invested $2, 000 in the bank at a rate of 6%. If Able needs $4,500 after five years, how much money does he need

to save and deposit after three years to receive the needed amount?

Problem 1-37

If you want to double your investment at an interest rate of 10%, how long will it take?

Problem 1-38

If you invest $1,000 today at an interest rate of 8%, how long will it take before the amount becomes $1,500?

Problem 1-39

Betty purchased 50 shores of stock at a price of $30 each. After five years, she sold the stock, after deducting the

commission, at a price of $65 each. What was the interest rate Betty earned on her investment?

Problem 1-40

If you need $5,000 five years from now, how much in yearly deposits should you make at an interest rote of 5%?

Problem 1-41

How much money will be accumulated in the bank if $1,000 is deposited each year for ten years at on interest rote of 8%?



Problem 1-42

Able invested $5,000 per year in a bank for three years at an interest rate of 6% At the end of three years the interest

rate changed from 6% to 8%. If Able continues to invest the money for another six years, how much money is

accumulated at the end of that period?

Problem 143

If a Betty invests $2,000 per year in an IRA (Individual Retirement Account) and, at the end of 30 years, receives $250,000

in cash, what annual interest rate did Betty receive?

Problem 1-44

Baker wants to buy an expensive piano that costs $20,000. If he invests $3,000 per year at on interest rate of 7%, how

long will it take before he can purchase the piano?

Problem 1-45

After investing $1,000 per year for the first five years, a person decided to invest $2,000 per year for the next five years. If

the interest rate is 8%, how much money will be accumulated at the end often years?

Problem 1-46

To buy a house, Able took a loon of $50,000 at an interest rate of 12% for a period of 15 years. What will be the yearly

payment?

Problem 1-47

A car dealer advertises a down payment of $250 with a monthly payment of $250 for five years (60 months) for a new

car. If the interest rate charged is 1% per month, what is the price of the car?

Problem 1-48

If you have $100,000 and would like to invest it so that you will receive $25,000 per year over the next ten years, at what

interest rate should the money be invested?

Problem 1-49

A magazine provides two subscription options. You can pay $250 now for a five-year subscription, or you can make an

annual payment of $65 at the end of each year for the next five years. If you can invest the money at an interest rote of 10%forfive years, which option should you choose?

Problem 1-50

When buying a car for $10,000, Baker agreed to pay a yearly payment of $2,000. At an interest rate of 8%, how long will

it take to pay off the loon?

Problem 1-51

A TV worth $500 is purchased on a monthly installment plan. After making a cash payment of $50, what will be the

monthly installment if the interest rote is 1% per month and the loon is to be paid off in one year?

80 Mahan Kelkar, Ph.D., J.D.

Problem 1-52

Carson purchases a stereo from a distributor. The stereo is worth $2,000. The distributor asks Carson not to make a

payment for the first 12 months. If Carson will have to make twelve equal payments during the second year to pay off the

stereo what will be the payment? Assume the interest rate to be 1% per month and the interest is accumulated during

the first year when no payment is mode.

MWOM Dawn buys a new boat for $20,000. With a 20% down payment, she is going to pay for the boot with monthly payments

over a five year period. The interest rote is 0.8% per month. After making payments for 2 years (24 payments), Able offers

Down $11,500 for the boot. If Down can pay off the remaining loan balance without penalty, will she make a profit by

selling the boot? If so, how much?

Hint: Calculate the remaining principal balance at the end of 2 years based an the remaining payments.

Problem 1-54

A person invests $3,000 in 0401K mutualfund account in the first year followed by a 5% increase per year over the next thirty years. If on overage yield on the mutual fund account is 11%, how much money will she collect at the end of thirty

years?

Problem 1-55

You are interested in saving money for your daughter’s education. After twelve years, when your daughter turns 18, the college tuition plus other expenses are expected to be $100,000 in the first year, rising at an 8% rate during each of the

remaining three years. If you can invest a fixed sum per year over the next twelve years at an interest rote of 8%, how

much money should you be investing to cover your daughter’s college costs?

Problem 1-56

Concerned about medicol insurance costs, you wont to purchase a life insurance policy that will cover the medical

insurance costs of your family in case of your death. The current family medical insurance is $500 per year and is

expected to rise at a rote of 13% per year. If you wont to pay for the next twenty years of medical insurance and the life

insurance money con be safely invested at a rate of 6.8%, how much life insurance should you purchase?

Problem 1-57

An oilfield is currently producing at a rate of 100,000 bbls per year. The net revenue received from the field is $111bbl.

The field is expected to decline at a rate of 10% per year over the next eight years before becoming uneconomical. If the

revenues earned are invested at a rote of 8%, how much money will the oil company have at the end of eight years?

Problem 1-58

If a Betty invests $2,000 per year with on 8% increase during each subsequent year for the next ten years at an interest

rate of 6%, how much will she accumulate? How much of a fixed sum of money will she have to invest per year over the

next ten years to receive the some future sum? How will the fixed sum of money per year change if the interest rate is 8%

instead of 6%?

Problem 1-59

A gas field is currently producing 1.2 BCE per year. The net revenue earned from the field is $0.61MSCF. The field is

expected to decline at a rate of 12% over the next eight years before production will cease. If the interest rote is 15%, how

much will you be willing to pay to buy this gas field?



Problem 1-60

Four years from now, Carson would like to buy a nice diamond ring for $10,000 for his wife for their 25th wedding

anniversary. He can save $2,000 this year with a 5% increase per year over the next three years What is the minimum

interest rote he needs to earn to have $10,000 at the end of four years?

Problem 1-61

Able has $200,000 to invest at the age of 65. He wants to earn $20,000 at the end of the first year followed by a 4%

increase in each year to account for inflation over the next 20 years. What is the minimum interest rate Able is required

to earn on his investment?

Problem 1-62

If the effective interest rate is 20% and the money is compounded continuously, what is the nominal interest rate?

If a loan company charges you $10 per month for every $100 you borrow, what is the nominal yearly interest rate? What

is the effective interest rate per year?

A petroleum engineer wants to buy a house. If she can afford to pay $10,000 as a down payment and can afford to pay

$1,000 as a monthly payment at a 12% interest rate compounded monthly for a 30-year period, what is the maximum

price of house she can afford?

Problem 1-65

A small company wants to borrow $100,000 from the bank. The bank charges a 12% nominal interest rate, compounded

monthly, and 2% of the loan as a loan origination fee to be paid in cash. What is the effective interest rate? Assume the

loan is paid over a five-year period. Calculate the part of the payments that go toward the principal and interest for the

first and the last payments.

If you invest $100 per month over a five-year period, how much money will be accumulated at the end of the five years if the nominal interest rate is 8% compounded continuously? -

Problem 1-67

If you require $30,000 per year for living expenses for the next twenty years, how much should you invest today? Assume

that the interest earned would be 9% to be compounded continuously.

Problem 1-68

A credit card company offers an instant credit scheme. if you withdraw money on the credit card, the company will

charge you a minimum $5 processing charge if the withdrawal is less than $200, and charge 2.5% of your total

withdrawal if the withdrawal is greater than $200. in addition, the company will charge a 19% interest rote, compounded

daily, on payment of the cash withdrawal plus any charges.

� If you withdraw $100 and pay it over a three-month period, what is the monthly payment? What is the effective

interest rate?

� If you withdraw $1,000 and pay it over a twelve-month period, what is the monthly payment? What is the

effective interest rate?

82 Mohan Kelkor, Ph.D., J.D.

Problem 1-69

A new credit card advertises a low annual interest rate of 12% to be compounded continuously. The card charges an annual membership of $100. If you wish to purchase a sofa worth $2,000 to be paid over a one-year period, what will be your monthly payment? What will be the effective interest rate on the payment? If another credit card offers an 18% annual interest rate compounded monthly with no membership fee, would you prefer this credit card over the previous one? Why?

WORKS CITED

Brush, R. M., and S. S. Marsden. Journal of Petroleum Technology, February 1982: 433-439.

Campbell, J. M. et al. Analysis and Management of Petroleum Investments: Risk, Taxes and Time. Norman,

Oklahoma: CPS Publishing Company, 1987.

Chang, et al. ’Successful Field Pilot of In-Depth Colloidal Dispersion Gel (CDG) Technology in Daqing Oil Field." SPE

Improved Oil Recovery Symposium. Tulsa: Society of Petroleum Engineers, 2004.

Chesapeake Energy. Investor Presentation. Chesapeake Energy. July 2011. www.chk.com .

Davis, L. F. Journal of Petroleum Technology, May 1968: 467-474.

DeGarmo, E. P., W. G. Sullivan, and J.A. Bontadelli. Engineering Economy. 9th Edition. New York, New York:

MacMillan Publishing Company, 1993.

Devon Energy. Summary Annual Report. 2009. www.dvn.com .

EOG. Investor Presentation. EOG Resources. June 2011. www.E0GResources.com .

Ikoku, C. U. Economic Analysis and Investment Decisions. New York, New York: John Wiley, 1985.

Newman, D. G. Engineering Economic Analysis. San Jose, California: Engineering Press, 1991.

-. Engineering Economic Analysis. San Jose, California: Engineering Press, 1991.

Park, C. S. Contemporary Engineering Economics. Reading, Massachusetts: Addison-Wesley Publishing Company,

1993.

Society of Petroleum Engineers. Guidelines for Application of the Petroleum Resources Management System. 2011.

www.spe.org .

-. Petroleum Resources Management System. 2007. www.spe.org .

Steiner, H. M. Engineering Economic Principles. New York, New York: McGraw-Hill, 1992.

Stermole, F. J., and J. M. Stermole. Economic Evaluation and Investment Decision Methods. Golden, Colorado:

Investment Evaluation Corporation, 1986.

Wall Street Journal. September 1, 2009.



R-VMZ 11 s

In this chapter, we will discuss a variety of methods used for evaluating the economic feasibility of

petroleum projects. The foundation for these methods was laid in Chapter 1 - Economic Principles when

we discussed the importance of the time value of money and its impact on cash flow. In this chapter, we

will formalize those concepts through various methods and illustrate the application of these methods

by numerous examples.

As discussed in Chapter 1, in decision making, we typically encounter two types of projects; those where

we have to make a decision among mutually exclusive alternatives, and those where we have to make a

decision about selecting the appropriate independent projects.

In this chapter we will concentrate on the selection of an alternative among mutually exclusive

alternatives. We will only cursorily examine independent projects.

In simple terms, mutually exclusive alternatives are exclusive of each other. By selecting one alternative,

we will automatically eliminate the other alternatives. In the petroleum industry, in many instances you

will deal with projects where only one alternative will be selected after evaluating several alternatives.

For example,

� Selection of a contractor to conduct a 3-D seismic survey for an exploration venture.

� Selection of a drilling contractor to drill a well.

� Selection of a service company to conduct log surveys.

� Selection of a pumping unit to improve production.

� Evaluation of an infill drilling option to increase production.

� Selection of a compressor to increase the gas production.

In all of these, we can select only one of the alternatives being considered. Once a particular alternative

is selected, the other alternatives are automatically eliminated from further consideration.

Throughout this chapter, we will use a minimum rate of return (MROR) to evaluate the attractiveness of

various alternatives. The choice of MROR is very critical in evaluating the alternatives. In simple terms,

the MROR is the minimum rate required by a corporation or an individual to make the project attractive.

For example, if you borrow money from a bank at an interest rate of 10% per year to invest in a drilling

venture, the MROR is 10%. This is because, if the project does not yield at least a 10% return on the

investment, you will be a net loser. If the project earns a 15% return, then after paying 10% interest to

the bank, you make money. If the project only earns a 5% return, you will have to pay out of your own

pocket to cover the interest payment to the bank. If the project earns 10%, then there is no net gain or

loss. Therefore, 10% becomes the minimum acceptable rate. tl

MROR is sometimes called the cost of capital. The computation of the cost of capital for a corporation is

much more complex than the previous example due to the varied sources of capital. The discussion

about MROR calculation is beyond the scope of this book. We will assume that this information is

provided to us. Any alternative that does not satisfy the MROR criteria will be automatically rejected.


Chapter 2 - Economic Methods

1

Ideally, when evaluating mutually exclusive alternatives, we prefer the economic criterion possess the

following characteristics:

� It should be suitable for ranking various alternatives.

� It should reflect the cost of capital.

� It should incorporate uncertainties in our assumptions.

� It should reflect the goals and objectives of the corporation.

In practice, no single criterion will be able to satisfy these characteristics. Sometimes, multiple criteria

may be appropriate. We will defer the discussion related to uncertainties to Chapter 4. Regarding the

fourth bullet, we will assume that our goal will always be to minimize cost or to maximize profit or

benefit. We will not consider other objectives in our analysis.

Many economic criteria are available in the literature. We will only concentrate on the ones which are

most commonly used. These include present value (PV) analysis and its cousin, annual value (AV)

analysis, rate of return (ROR) method and profit to investment (PIR) ratio. We have already discussed

payback period and PIR. The only difference in this chapter is that we will consider the impact of time

value of money on PIR. 1

L) frff’’ Ji4.L

PRESENT VALUE ANALYSIS

Present value (PV) or worth analysis evaluates projects based on the financial position of various

alternatives at the present time. This technique accounts for the time value of money and provides a

way of comparing various alternatives with the same frame of reference. If the project has a fixed

output, the objective should be to minimize the present worth of costs. As discussed in the previous

chapter, if the project has a fixed input, the objective should be to maximize the present worth of

benefits and, if the project has neither fixed input nor output, the objective should be to maximize the

difference between the present worth of benefits and present worth of costs. Let us illustrate these

three alternatives through various examples.

Example 2-1

A company is considering two alternatives to satisfy its photocopying requirements. The cost of each machine is shown below:

(a) (b)

Initial cost $10,000 $8,000

Annual Cost $1,000 $1,400

The annual cost includes the replacement and maintenance costs. If both machines have a life of five years and the minimum

rate of return (MROR) is 10%, which project should be selected?

Solution 2-1

Both alternatives will perform the same functions, i.e., they offer the same output. Therefore, our objective should be to

minimize the cost.

For alternative (a),

1(1 +

PV = $10,000 + $1,000 = $13,791 [.1(1~1) 5 ]

2 Mohan Kelkar, Ph.D., J . D.

For alternative (b),

PV, ,),, = $8,000 + $1,400 = $13,307

Comparing the present worth costs of the two alternatives, alternative (b) should be selected.

Example 2-2

A proposal calls for an investment of $100,000 in drilling a new well. It is expected that the production will generate revenue

of $30,000 per year for six years. At the end of six years, the production equipment can be sold for $10,000. Another proposal

requires a $100,000 investment. It will generate $50,000 in the first year followed by 8% decline in each year. The life of the

project will be six years. There is no salvage (remaining) value associated with the proposal. If the minimum rate of return is

12%, which project should be selected?

Solution 2-2

We have a fixed amount of investment. Our objective should be to maximize our output or to maximize the present value of

benefits received after the fixed investment.

For the first proposal,

(1 +.12) 6 - 1 10,000 PVbenefl

= 30,000 .12(1 + .12)6 + (1 + .12)6 $128,408

For the second proposal,

50,000 (1 -.08)6

PVbenefl = (.12 + .08) 1 - (1 + .12)6 = $173,200

Comparing the PVb ene fits for both proposals, the second proposal should be selected.

Example 2-3

A company is considering two alternative computer models to satisfy its computer needs. Due to their differing powers, the

benefits received by both the computers are different. Based on the work projections, the company thinks that either of the

computers can be used to their fullest potential. If the minimum rate of return is 10%, which computer should the company

select?

(a) (b)

Initial Cost $4,500 $3,200 Annual Benefit $1,500 $1,000

Life, Years 5 5

Salvage Value $600 $300

Solution 2-3

The salvage value indicates the price received if the computer is sold after five years. Since neither the costs, nor the benefits

are fixed, we need to select an alternative that maximizes the difference between the benefits and costs. We can define the

difference as,

NPV = PVh,fit - PV 0 Equation 2-1

where NPV is the net present value of the alternative.




3

NPV=$1,500I + [(1. + .i) - lj 600

.1(1 + .1) (1 + .1) �4,500 = $1,559


+ .i) NPV = $1,000 I

[(1. - lj 300

[.1(1 + .1) + (1 + .1) 3,200 = $777

Based on the comparison of the two alternatives, (a) should be selected.

(se .Study 2-1

You work for JLF Production Company as a production engineer and are responsible for 16 gas wells in western Oklahoma. Currently the wells are on plunger lift and are managed by a pumper on a daily basis. They have an expected life of three years at which time the wells will be abandoned. The average daily production for the past month for each well is 136 MSCF.

Autoplunger, Inc. has approached you with a method of remotely monitoring the wells. The automation device consists of a special flow meter that is able to connect to an Autoplunger, Inc. server remotely using a cellular phone. The data can then be monitored by you using the internet.

Autoplunger claims that the systcm can reduce down time associated with well problems and, therefore, prevent lost production due to down time. From previous wells’ production history, they claim that by installing the system they can increase production initially by 10% over the base and it will decline at 11% per month. The base decline is 10% per mouth. The installation of the flow meter and transmitter equipment costs $3,000 per well for the first 4 wells and $1,000 for each additional well thereafter and the monthly maintenance fee is $80 per month per well. Currently you pay the pumper $125 per well per month to monitor the wells- If the automation system is installed the pumper will no longer be needed on a daily basis, but he will be needed as problems occur. lie requires a $25 fee per visit. It is expected that initially the required visits will be 0.5 per well per month, but will increase exponentially to 2 visits per well per month at the end of the well’s life, that is in month 0 there will be 0.5 visits required and in month 36 there will be 2 visits required.

Perform an economic analysis using a MROR of 1 5%/year over the life of the wells to determine the minimum number of automation units to install for gas prices of S1.50/mcf, S2.50/mtf and $3.00/mef. We need to make the project viable. Assume that the initial (year 0) average monthly production is last month’s average production and the production decline is 10%/month.

Consider the problem starting with 4 installations and go up to 16 using an increment of four.

case Swdr Solution 24

Many variations exist in this case study. We will show the detailed calculations for one scenario and then summarize the other scenarios in the table following these calculations.

Scenario: 16 wells with ps price of SI .50/MSCF

Total Cost of Equipment = 3,000 x 4 + 1,000 x 12 = $24,000 (1 + 0.0125)36 - 1

Cost of Maintenance = 80 x 16 x [0. 0125(1 + 0.0125)361 = $36,924

To calculate the visiting Ice, we realize that the number of visits Start at 0.5 per month and reach 2 visits per month at the end of 36 months period. Therefore, we can calculate the rate at which the visits arc increasing by knowing that.

2 = 0.5(1 + rate)36

Solving for the rate, the visits are increasing at a rate of 3.9% per month.

25 x Cost of Visits = 1 $11,583

0.5 x 16 [

(1+0.03 9)36

0.0125 - 0.039 - (1 + 0.0125)3 61

lhc current cost for using the pumper is:

to


(1+0.0125) 36 - 1 Current Cost of pumpers = 125 x 16 x 00125>< (1 + 0.0125)361 = $56,694

These costs will bc saved

lherefore, the net increase in present costs = 24,000+36,924+11,58356,694$14,813

As a result of autoplunger, the production will improve by I0 0/s from the original value but it will decline at a higher rate compared to the previous case- Production revenue can he explained by geometric gradient. Therefore, net present value of benefits is given by:

PV Benefits

136x30.4x 16x 1.50x(1+0.1) 136x30.4x16x1.50[ 1-0.1 36

(0.0125 + 0.11) 1 - (1 + 0.0125) - (0.0125 + 0.1) [ - (125) $13,121

NPV = PVbenetts - PV rosi = 13,121 - 14,813 = -$1,692

This means that the project is not feasible under these conditions. We have done similar calculations for other scenarios in the following tables.

Gas Price = $1.50 per MSCF

Wells Equip. Cost Monthly Fee Visiting Fee Pumper Costs PV Costs PV Benefit NPV 4 $ 12,000 S 9,231 S 2,896 S 14,424 $ 9,703 S 3,280 S (6,423) 8 $ 16,000 $ 18,462 S 5,792 S 28,847 $ 11,407 S 6,561 S (4,846)

12 S 20,000 S 27,693 $ 8,687 S 43,271 $ 13,110 S 9,841 S (3,269) 16 $ 24,000 S 36,925 S 11,583 S 57,695 S 14,813 S 13,121 S (1,692)


Wells Equip. Cost Monthly Fee Visiting Fee Pumper Costs PV Costs PV Benefit NPV 4 S 12,000 S 9,231 S 2,896 $ 14,424 S 9,703 S 4,374 S (5,330) 8 S 16,000 S 18,462 S 5,792 S 28,847 S 11,407 5 8,747 S (2,659)

12 S 20,000 S 27,693 S 8,687 S 43,271 S 13,110 S 13,121 S ii 16 S 24,000 5 36,925 S 11,583 S 57,695 S 14,813 S 17,495 S 2,682


Wells Equip. Cost Monthly Fee Visiting Fee Pumper Costs PV Costs PV Benefit NPV

- 4 S 12,000 S 9,231 $ 2,896 $ 14,424 S 9,703 S 5,467 S (4,236) 8 S 16,000 S 18,462 S 5,792 S 28,847 S 11,407 S 10,934 S (472)

12 S 20,000 S 27,693 $ 8,687 S 43,271 S 13,110 $ 16,401 $ 3,291 16 $ 24,000 S 36,925 S 11,583 S 57.695 S 14,813 $ 21,868 S 7,055

Fxamining these numbers, we can see that for a gas price of S1.50/MSCF, the project is not feasible under any scenario. For the gas Price of $2.00/$ISCF, the project breaks even if we adapt it for 12 wells and for gas price of S2.50/MS( -F, the project is profitable if we adapt it for at least 12 wells.

Case Study 2-2

Adapted from: http://ww\vbpcom/genericarrielc.dQc43cgnild=201 2968&contcnt1ri7046356)

lit ) America acquired approximately 90,000 net acres of leasehold and producing properties in the Woodford shale in ( )khihoina’s Arkoma basin from Chesapeake Fncrgy Corporation. BP paid $1.75 billion dollars for this acquisition. The current gross production from the area is approxiniatclv 50 lMSCF/day. ’lie relevant information about Woodford Shale is provided below:


Chapter 2-Economic Methods 5

Statistical Data about Woodford Shale

Gas in Place/640 acre 40 to 120 BCF Recovery Factor SO% Initial Production (MMSCFD) 3 to 6 Average Well Cost (Drill, Complete, and Facilities) $6 to $8 million Typical Well Spacing 80 acres Royalty Interest 25% No. of days to complete and begin well production 90 days

A. Assuming S6,000/MSCFID to $10,000/MSCFD, how much did BP America pay for the right to drill the wells? Assume that only 80,000 net acres are available for drilling.

B. Using 80 acre spacing and assuming that only 80,000 net acres are available for drilling (other parts are already drilled), using the average well cost range, what is the possible range of field and development (F& D) cost?

C. Assuming that for a project of this kind to be feasible, the cost of in place reserves has to be less than $3/MSCl, do you think that BP America made a good deal? Note that BP America has only access to new reserves from 80,000 acres.

D. A typical production profile from the best and the worst well in Woodford Shale is available. The profiles are provided below:

Year Worst Best

� (MMSCF/Year)

1,095 2,190 2 383 767

� 3 249 498 � 4 212 423 �

5 191 381 6 172 343 : 7 154 309 8 139 278

� 9 125 250 10 113 225 11 101 203

� 12 91 182 13 82 164 14 74 148 15 66 133 16 60 120

� : 17 54 108 18 48 97

Using these production profiles, calculate the N1’V at 15% (assume monthly compounding) using the worst of the scenario and the best of the scenario. Note that for the worst case, you should consider everything (e.g, drilling costs) to be the worst. Assume the range of gas price to be S6/MSCF to S9/MSCF held constant. Assume that the operating costs per mouth per well to be In the range of 54,000 to $6,000. Do not forget to subtract royalty interest in your calculations.

E. Assume that the NPV you have calculated per well represents the present value of that well at the time the well was started to drill. BP America believes that it will secure 10 to 20 drilling rigs to initiate the development of the field. If we assume that 10 drilling rigs are running, then we will be able to complete 40 wells per year assuming a 90 day period for beginning production for each well. Again, consider the worst and the best case scenarios based on renting 10 to 20 drilling rigs. Using the NIV values from the previous section, calculate the NIT for the whole project using a 15 0/s MROR (continuous compounding. For this part, remember that you will accumulate 10 to 20 wells’ NPV every 3 mouths until all the wells are drilled.

F. Based on your evaluation, is this a good deal for LIP? Why?

(7 ,cc .’olution 2-2

\. \mount paid per acre for right to drill wells We can examine both a pessimistic and an optimistic case to determine the amount paid for the right to drill. To achieve this, first we need to calculate how much lIly paid fur the existing production.

Net Production = 50 MMSCFD x 0.75 = 37.5 MMSCI’D


This is based on 25°/s royalty interest. Depending on the amount paid for existing production, BP paid between 37,500 x S6,000/MSCFD -. $225 million and 37,500 x 510,000/1NTSC D = $375 million.

The balance will be paid for the right to drill at undrilled locations within 80,000 net acres. Since the total amount paid is 1,750 million dollars, the balance will be between $1,375 million and $1525 million. Dividing that amount by 80,000 acres, BP paid between $17,188 and $19,063 per acre for the right to drill the wells.

B. F and D Costs Assuming 80 acre spacing, the amount of gas in place is between 5 to 15 BCF per well. Assuming 50% recovery of the gas, the amount of gas that can be recovered per well is between 2.5 and 7.5 13CR Knowing the cost of drilling and completion is between 6 and 8 million dollars per well, we can calculate F & U costs.

8 x 10 6 + 19,063 x 80 $5.08 Pessimistic F&D Costs =

2.5 < < 0.75 = MSCF

6 x 106 x 17,188 x 80 Optimistic F&D Costs=

75 > 106 0.75 = $1.31/MSCF

C. Is it a good deal? Under the optimistic scenario for F & 1) costs, it is a good deal; whereas, under the pessimistic scenario it is a bad deal. The average of the t-wo (assuming equal likelihood) is S3.19/MSCF. This number is slightly greater than 53/MSCF; therefore, the project is marginal.

D. NPV of Individual well The effective interest rate is -e 0 ’ 5 - 1 = 16.2%

Using the production data, the following table shows yearly revenues for the worst case and best ease scenarios. In calculating, we assumed that, in the worst ease scenario, we had the highest cost of drilling and production as well as operations and the worst production and lower price. For the best case scenario, we assume the smallest cost of drilling and completion and operations, and the best production profile and the best price scenario. Using the net revenue at the end of each year, we calculated the NPV for each scenario.

Worst Case Scenario Year Prod Revenue Op Costs Net Rev

MMSCF (MM $) (MM $) (MM $)

0 -8.00 -8.00 1 1095 5.75 0.072 5.68 2 383 2.01 0.072 1.94 3 249 1.31 0.072 124 4 212 1.11 0.072 1.04 5 191 1.00 0.072 0.93 6 172 0.90 0.072 0.83 7 154 081 0.072 0.74 8 139 0.73 0.072 0.66 9 125 0.66 0.072 0.58 10 113 0.59 0.072 0.52 11 101 0.53 0.072 046 12 91 0.48 0.072 0.41 13 82 0.43 0.072 0.36 14 74 0.39 0.072 0.32 15 66 0.35 0.072 0.27 16 60 0.32 0.072 0.24 17 54 0.28 0.072 0.21 18 48 0.25 0.072 018

Best Case Scenario Year Prod Revenue Op Costs Net Rev

MMSCF (MM $) (MM $) (MM $)

0 -6.00 6.00 1 2190 14.78 0.048 14.73 2 767 5.18 0.048 5.13 3 498 3.36 0.048 3.31 4 423 2.86 0.048 2.81 5 381 2.57 0.048 232 6 343 2.32 0.048 2.27 7 309 2.09 0.048 2.04 8 278 1.88 0.048 1.83 9 250 1.69 0.048 1.64 10 225 1.52 0.048 1.47 11 203 1.37 0.048 1.32 12 182 123 0.048 1.18 13 164 1.11 0.048 1.06 14 148 1.00 0.048 0.95 15 133 0.90 0.048 085 16 120 0.81 0.048 076 17 108 0.73 0.048 0.68 18 97 0.65 0.048 0.61

NPV = 1.51 MM $ NPV = 19.24 MM $

S I ’nilur both the scenarios, the NPV is positive-

S


S Chapter 2 - Economic Methods

7

S S

V. NPV of the Overall Pro ject In this calculation, we assume that we paid the upfront costs for leasing, and then, at the end of every 3 months, we started collecting revenue based on the NOV of all the wells drilled. As a worst case scenario, we will start collecting $15.1 million at the end of ever’, 3 months. This is based on an assumption of 10 wells in each period (1.51 MM a 10). We will need to drill a total of 80,000/80 - 1,000 wells. That is, it will be 100 periods of 10 wells each before all the wells are drilled. For the best case, we will drill 20 wells in each period. It will take only 50 periods before all the wells are drilled. Each pet id is 3 months- The interest rate per period is 15/4 = 3.75%. Since the interest is compounded continuously, the effective interest rate per period is = e 00375 � 1 = 3.82%/period.

NPV under the worst case scenario

(1 + 0.0382)100 - 1 = �1,525 + 15.1 X

0.0382 x (1 + 0.0382)100 = �1,139 million

NPV under the best case scenario

(1 + 0.0382) ° - 1 = �1,375 + 384.8 X

0.0382 x (1 + 0.0382)° = 7,153 million

F. Is this a good investment? Depending on which scenario is selected, the project can have positive or negative NPV. Knowing the uncertainties in these calculations, we can examine the average of the two and find that the average value is 3,007 million. This is clearly a viable project based on the assumptions we made.

It 15 worth remembering that, in addition to the uncertainties we considered, we should realize that there are other uncertainties present in this project that we did not consider. Examples of these uncertainties include:

� Drilling and completion efficiencies: BP may be able to improve the drilling and completion techniques over time resulting in better completion at smaller cost and over a smaller period. This will allow to them to drill and complete the wells over a shorter period using a smaller amount of money.

� The rules and regulations of fracturing and completing wells can change over time making it difficult to dispose of produced water and emit produced gas. This can increase the cost of drilling and completion.

� The surplus gas from the shale play can depress the gas prices making it less attractive from a revenue perspective. � The actual spacing can become smaller based on the evaluation of the wells. This can result in higher recovery of gas as

well as more revenue over time.

Problem 2-1

A company is considering the following two alternatives for installing a compressor. The annual benefits of the compressor depend upon the horse-power.

A B

Initial Investment $30,000 $70,000 Annual Benefit $8,000 $20,000 Life, Years 10 10

Salvage Value $5,000 $10,000

Which alternative would you select lithe MROR is 12%? Use present worth analysis.

Problem 2-2

An oil company is considering building an off shore pipeline. Depending upon the pipeline, the cost of maintenance and the installation varies. Which of the three alternatives should the company choose?

A B C

Installation Cost $30 $20 $25 Annual Cost 2 4.5 3 Life, Years 20 20 20

The numbers are in millions. Assume the MROR to be 10% and the salvage value to be zero.


0 0

Problem 2-3

A prospect is expected to fetch $100,000 from oil revenues per year over the next eight years. The annual operating costs ore

estimated to be $10,000. The salvage value can be assumed to be zero. If the MROR is 15%, what is the maximum price you are willing to pay for this property?

0 Problem 2-4

An oil company paid a consultant $10,000 to analyze various alternatives with respect to a producing property. Based on the analysis, the consultant suggested the following scenarios. Which one should be selected?

Alternative Investment Annual Benefit Salvage Value

Do Nothing $0 $20,000 $10,000 5 In-Fill Wells 200,000 60,000 25,000 10 In-Fill Wells 400,000 100,000 40,000

If the MROR is 15%, which option should the company select? Assume the life of the project to be six years

Problem 2-5

A periodical having a cover price of $5.50 per magazine, published monthly, advertises 020% discount off the cover price if subscribed on a yearly basis and a 30% discount off the cover price if subscribed for a two-year period.

Assuming that you buy the magazine every month, which option is best? Assume the minimum rate of return to be 8%.

92POTTO" "M.

Three alternatives are considered for installing a pipeline. The initial cost is less for a smaller diameter pipe; however, due to a higher pressure drop, the pumping cost is higher for the some pipe. The costs are shown below:

Pipe Size/Inches Cost Per BBL of Pumping Construction Cost

2 $.031bbl $30,000 3 $.0251bbl $50,000 4 $.021bbl $75,000

We can assume the useful life of the pipe to be 12 years with no salvage value. Assume also that the flow rate per year is

constant. Calculate the range of flow rates for which different pipe sizes will be the most beneficial. Assume the minimum rote

of return to be 10%.

Problem 2-7

An investment proposal calls for an initial investment of $427,000. The estimated annual benefits are $120,000 and the

estimated annual casts are $30,000. The costs are decreasing at a rate of $2,500 per year. If the life of the investment is 8 years, and the MROR is 15%, should the money be invested? Assume the salvage value to be zero.

Problem 2-8

Consider the following four alternatives for a project. Calculate the NPV for each project. Which ones are acceptable if the MROFI is 12%?

Period Alternative 1 Alternative 2 Alternative 3 Alternative 4

0 -4,000 0 4,000 -6,000 I 10,800 -4,500 -16,000 -1,500 2 28,800 1,500 4,000 22,500 3 14,400 3,000 8,000 19, 500


Chapter 2 - Economic Methods 9

Two producing properties are considered for a potential investment of $100,000. The first property will generate net revenue

of $3,000 in the first month with a 10% decline per year. The property is expected to be sold for $30,000 after S years. The

second property will generate revenue of $6,000 in the first month with an operating cost of $1,000. The revenues are

expected to decline at a rate of 15% per year and the operating costs are expected to increase at a rate of 5% per year. The

property can be disposed of at the remaining value of the property. Using five years as a basis, which property should be

purchased?

David and his wife decide to set a sum of money each year such that they can withdraw $30,000 per year perpetually after

retirement. If the interest rate is 10%, and they start saving money each year for 20 years before retirement, how much money

do they need to save every year?

Hint: First calculate the PV of periodic payment received in perpetuity.

Problem 2-I1

Betty bought a producing property at a price of $190,000. The revenue received in the first year is $90,000 declining at 6% per

year. The operating costs are fixed at $15,000 per year. If after 5 years of operation. Carson offers Betty $85,000 for the

property, should Betty sell it? Assume the MROR to be 14%.

ANNUAL VALUE ANALYSIS

The difference between the present value analysis and the annual value (AV) analysis is that in the case

of PV analysis we compare the alternatives based on present value, whereas in the case of AV analysis,

we compare the alternatives based on the annual value. All the problems that can be solved by AV

analysis can be solved by PV analysis. However, AV analysis does possess certain advantages; the biggest

advantage being it is easier to explain to someone not familiar with economic analysis principles than

the PV analysis.

Specifically, some instances in which AV analysis is useful are (Park 1993) (Steiner 1992):

� Financial Reporting: Most corporations report to their shareholders the company’s performance

on an annual basis. It is much easier to present the annual cost or benefit of a project than the

net present value.

� Cost/Unit Calculation: In comparing alternatives, it is easier to illustrate the benefit of one

method over the other by showing that a particular method costs less to produce one unit than

the other.

Example 2-4

Two alternatives are being considered for leasing a copier machine. The costs associated with both machines are given below:

v l

Cost Alternative I Alternative II

Installation $2,000 $1,500

Annual Maintenance $800 $900

Life, Years 5 5

Salvage Value 0 0

Which is the better alternative? Assume that the operating cost to make one copy is the same for both machines. The MROR =

10%.


Solution 2-4

We can solve this example using PV analysis; however, it is more convenient to compare the annual costs associated with

each machine. The installation cost or the initial cost is called the capital cost. The annual equivalent of capital cost is called

capital recovery cost.

Alternative I

Using the equation,

i(1 + O A=P (1+i)-1

capital recovery cost,A 2,000 [(�l+ .15 - 1

= $527 total annual cost = 800 + 527 = $1,327

. Alternative II

capital recovery cost = 1,500 1

[(1 + � .1),- ij = $395

Total annual cost = 900 + 395 = $1,295

Based on the annual costs, Alternative II is superior.

If we use present value analysis,

Alternative I

(PV)0 1(1

= 2,000 + 800 - 11 +

= $5,033

. Alternative II

10.1( 1 (PV) 0 . = 1,500 + 900 (1

+ 0.1) 11 = $4,911

I Since Alternative II has smaller costs than Alternative I, the second alternative should be selected.

Exom pIe 2-5

Two submersible pumps are being evaluated for an oil well. Both have the same capacity to produce liquid. The costs are

shown below:

Pump I Pump If

Installation $15,000 $20,000

Maintenance Per Year $5,000 $6,000 Operating Costs (Utilities) Per Barrel $0.04 $0.03

The well is expected to produce 2,000 barrels of fluid per day with a water oil ratio of 95. Which pump should be purchased?



11

Assume the life of both pumps to be 10 years. MROR = 15%.

Solution 2-5

In this example, it is much more illustrative to compare the cost of lifting one barrel of fluid using both the methods.

Total production per year,

= 2,000 x 365 = 730,000 bbls

Pump I

annual costs = capital recovery cost + maintenance costs + operating costs

(1 +.15)b0(. 152 = 15,000 (1 + .15)10 - 1 + 5,000 + 730,000 x 0.04

= $37,189

cost/bbl = 37,189 730,000 = $0.05 1/bbl

. Pump II

annual costs = 20,000 [(1 + .15)10 - 1 j + 6,000 + 730,000 x 0.03

= $31,885

cost/bbl = 31,885 = $0.044/bbl 730,000

The cost per barrel of lifting is less for Pump II than Pump I. Therefore, Pump II should be selected. As stated before, this

comparison is much more illustrative than the (PV) c ,, , t, comparison.

O.sscStudv2-3

As a new petroleum engineer working for Stetson Petroleum Corporation, you are in charge of evaluating the newly installed rod-pump controllers (RPC’s) which replaced the time clock in its East Linden field in East Texas. Fifteen wells are currently producing from the field. The wells are about 10,000 ft. deep, producing 42 to 46 API gravity oil, with an average water cut of 30%.

In the past, Stetson had always relied on contract pumpers. Timers were utilized for intermittent operation and, unfortunately, the pumpers could not evaluate the cyclical status of the wells during the single visit every 24 hours. lbev had to set the timer manually based on a series of trial-and-error adjustments. During the trial-and-error period, sometimes some of the wells would pound the fluid fora considerable period of time resulting in damage to the rod and pump. Another problem was wider-pumping the wells. The pump efficiency at this depth was considerably impacted due to rod stretch. On certain wells, the times were set for as low as 3 hours per day. During the well’s idle time, gas would break out of the fluid in the tubing and extra pump time would be required to return the liquid level in the tubing to surface where production could be "put in the tank." ’l’his extra pump-up time was not consistent and, combined with decreased pump efficiency due to pump wear over time, resulted in additional complication while trying to estimate the time required to produce the well efficiently with a time clock.

l\vo years ago, Stetson hired eProduction (eP) Solutions to install RPC’s. ’Ihe version installed in the held is shown in Case Study Figure 2-1. ’Ihe installation cost for RPC for the whole field was $25,000. In addition, el’ charged S1.99 per well per day

12 Mohan Kelkor, Ph.D., J. D.

maintenance costs. These controllers were supposed to eliminate a lot of guessing. Rather than using trial-and-error for adjusting the time clocks, RPC immediately shut down the pump and adjusted the idle time based on buffered data in the controller from past Cycle times .

The to \car survey revealed that rod failure frequency and tubing repairs were reduced by installing RP( In addition to reducing the rod repair costs another savings -was elimination of the belts burn offs The belts would burn off the electric motor when the well had a shallow rod part The RPC shut the well down on a low load limit which resulted in saving the belts For over two years about 6 Per year repairs were eliminated savirigabout S3,500 per repair,

A major cost of production for Stetson was lifting cost. The average production from the well is 30 barrels of oil per day in addition to water production. The cost per barrel of lifting was about $1.0/bbl of liquid production. By optimizing the well performance using RFC’s, the cost of lifting was reduced by about 30 0/s.

Using this information over the last two years and assuming that this will last for about 5 years total, do the following evaluations. Assume the costs and saving on a monthly basis. The MROR is 15% per year compounded monthly.

a) Using annual value (AV) analysis, calculate the monthly savings. Flow much are the savings per barrel of oil produced?

b) In reality, the oil production rate is not constant. Instead, the initial rate of oil was 30 barrels per day. However, it is declining at a rate of 10/0 per month. What would be the current monthly savings and saving per barrel of oil produced?

e) Stetson also operates the extension of Fast Linden field containing approximately the same number of wells and similar production profiles. The only exception is that the rod pumps are better optimized in the extension of the field. Based on talking to the field supervisor, you conclude that the rod repair savings will be three per year with an average saving of about S2,700 per repair. In addition, the reduction in the lifting cost will be about 20% instead of 30%. All the other factors are assumed to be same as in part (a). Will you recommend installation of RPC in the extension of the field?

Case Study Solution 2-3

A. Constant oil production at 30 hhl/day

The initial cost is S25,000. We can convert it to capital recovery cost per month. The interest rate is 15% per year or 1.25% per month.

0.0125 x 1.0125 60 capital recovery costs = 25,000(1.012560 - 1) j =

monthly maintenance costs = 1.99 x 30.4 x 15 = $907 3,500 x 6

rod pump savings per month = 12

= $1,750

30 electrical savings per month = (1 - 0 3) X 30.4 x 1 x 15 x 0.3 = $5,863

net savings per month = 5,863 + 1,750 - 907� 595 = $6,111 6,111

net savings per barrel = 30 > s

< 30.4 = $0.45/barrel

This is a positive value and, hence, economically feasible project.

B. Oil production dcclming 1 °/o per month

In this case, we can calculate equivalent oil production per month by comparing geometric gradient series with constant payment series.

F=Aeq (1+i)-1 = __[(1+i)n(1+gyi] (t �g)

Comparing the future values for both constant payment and geometric gradient series, we can calculate a value of A which will provide us svith the same future value as geometric gradient series. tJsing this equation, we can calculate equivalent oil production per niuiiih.

30x30.4 0.0125 Aeq = (0.0125 + 0.01) (1.01250 - 0.9960) (1.012560 - 1) = 714 bbl/month


Chapter 2- Economic Methods 13

All the other costs will remain the same except the electric savings costs:

714 electrical savings per month = (1 - 0,3) x 1 x 15 x 0.3 = $4,589

net savings per month = 4,589 + 1,750 - 907 - 595 = $4,837 4,837

net savings per barrel= 1----j- = $0.45/barrel

It is still an economically feasible project The reason that savings per barrel is the same as case (a) is because electrical savings dominate the savings in our analysis. Since it is based on per barrel, savings do not change when calculated on a per barrel basis.

C. Extension of East Linden Field

All the other costs and savings would remain the same except the following two items

2,700x3 � rod pump savings per month=

12 = $675

30 � electrical savings per month = - 0 3)

X 30.4 x 1 x 15 x 0.2 = $3,909

net savings per month = 3,909 + 675 - 907 - 595 = $3,081 3 , 081

net savings per barrel = 30 15 < 30.4 = $0.23/barrel

The savings are reduced by half compared to the previous case, but it is still an economically feasible project.

Problem 2-12

A company is considering two options for using a computer. It can buy o computer for $15,000 and use itforfive years with a salvage value of $3,000, or it can lease the computer for an annual cost of $4,200. If the MROR is 12%, which is the better

option? Use the annual value method.

Problem 2-13

An oil company is considering buying a compressor at a cost of $60,000. The maintenance cost per year is estimated to be

$3,000. As a result of the compressor, on incremental production in the first month is estimated to be 30 MSCFD declining at a

rate of 10% per year. The operating cost is estimated to be $0.2 0/MSCF of gas. If the price of the gas is $2. 0/MSCF, should be compressor be bought? What is the cost of producing the incremental gas per MSCF? Assume the life of the compressor to be

10 years and the MROR to be 12%. The salvage value is zero.

Problem 2-14

Able has the option of using either his car and claiming the cost on a mileage basis or to use a company owned car. He purchased a new car at a cost of $15,000. He is expected to drive approximately 13,000 miles per year for business related

travel. The car, based on the personal mile and business miles estimate, will depreciate at orate of 25% per year.

The cost of scheduled maintenance per year is $150. The cost of registration is $100 per year. Other non-scheduled repairs and maintenance will cost $50 in the first year increasing at a rate of $20 per year. The car is expected to give an average of 25

miles per gallon. Assume the gas price to be $1.00 per gallon. The parking and toll costs per year are $125. If the car is kept for 5 years, should Able claim the mileage cost? The mileage costs are $0.301mile. Assume the MROR to be 10%. The salvage value

at the end of 5 years is the un-depreciated value.

Problem 2-15

An oil company is considering two alternatives for routine core analysis. The cost of the machine is $100,000 and the annual

maintenance cost is $8,000. The cost of labor is $30,000 per year. The cost of supplies is $2.00 per core analyzed. The allocated

space will cost approximately $250 per month. If an outside lab charges $25 per care for routine analysis, how many cores

does the company need to analyze in-house so that it will make no difference whether the cores are analyzed using outside


I

0 services or an in-house facility. The life of the equipment is 15 years with a salvage value of zero. The MROR is 8%.

Problem 2-16

An independent oil producer wants to buy well testing equipment to test individual wells. The cost of the equipment is $30,000

with a maintenance cost per year of $4,000. The life of the equipment is 7 years. An outside service using the same equipment

costs $2,500 per well test. The service also includes analysis of the well. If the data are analyzed in-house, the cost is an

additional $200 per well. If the producer estimates that it will conduct 15 well tests per year, should it buy the equipment?

Assume the salvage value to be zero. The MROR is 15%.

Problem 2-17

to 14

A cable company manufactures a particular type of cable for supporting tools in a well. It currently produces 20,000 units of

cable per year. The costs if produced in house, are:

Direct Materials $120,000

I Direct Labor 360,000

Variable Overhead (Power) 270,000 Fixed Overhead (Light, Space) 140,000

I I It is anticipated that the production will last for at least 5 years. The cost of material is expected to rise at a rate of 5% per

year, the direct labor is expected to increase at a rate of 6% per year, and the variable costs will rise at a rate of 3% per year. The fixed overhead will remain fixed at the present level. Another cable company offers to sell the some cable at a rote of $50 per unit. If this offer is accepted, the space used currently can be used for manufacturing other types of cable saving the

�

company $70,000 per year. In addition, the variable costs will be reduced by $10 per unit. If the MROR is 15%, should the cable

be manufactured in-house or outside?

RATE OF RETURN ANALYSIS

� The rate of return analysis is probably the most popular criterion in economic analysis. Its popularity

� stems from the ease with which a common person can understand the meaning of rate of return. Most

investment brochures will use rate of return on your investment as a criterion to show how good a given

investment opportunity is. It is much easier to understand that ’a project will provide a 20% return on

� your investment" than "the project will result in a NPV of $5,000." Unfortunately, although simple to

understand, the technique has some major drawbacks. In this section, in addition to explaining how to

calculate the rate of return (ROR), we will discuss the advantages and disadvantages of this technique � (Park 1993) (Steiner 1992).

� Rate of return has two definitions. One definition can be stated as "the interest rate earned on the

unpaid balance of a loan such that the payment schedule makes the unpaid balance equal to zero when

the final payment is made." Consider a simple example to illustrate this definition. Assume that you take

a loan of $1,000 from a bank at an interest rate of 10% for a period of four years. Every year, including

� last year, you pay an interest of $100 to the bank. At the end of four years, you pay the principal amount

of $1,000. Therefore, at the end of four years the unpaid balance is zero. The rate of return for the bank

is (1,000/100=) 10%. Schematically, the cash flow is shown in Figure 2-1.



This definition can be turned around to state that the "rate of return is the interest rate earned on the

unrecovered investment such that the payment schedule makes the unrecovered investment equal to

zero at the end of the life of the investment. Using a similar example as before, let us assume that you

have invested $10,000 in the bank at an interest rate of 6% for five years. At the end of each year, you

withdraw $600 in interest and at the end of five years, you withdraw $10,000. The investment in the

bank at the end of five years is, therefore, zero. You can consider that the rate of return on the

investment is (600/10,000=) 6%. Schematically, the cash flow profile is shown in Figure 2-2.

Figure 2-2: Investment of .10,000 with a uniform receipt of interest

Mathematically, the rate of return (ROR) is defined as the rate at which net present worth (NPV) for a

given investment is equal to zero. In equation form, the rate at which,

"costs - "benefit s= 0

Equation 2-2

is the rate of return.

In other words, the rate at which:


NPV = 0 Equation 2-3

If we assume that the cash flow for a particular project is given by Aj where A1 represents the cash flow

in yearj, we can write the equation for NPV as,

A 1 A 2 A NPV �A O+(l + ) + (l + .) 2 ++(l + .)n

A = J° (1+01

iu.i4IflPr

If we define the rate R corresponding to the rate at which NPV is zero, we can write the equation for i as,

A 2 A =0 0 +

(1–i) + (i+ i)2 + + (i+.)" Equation 2-5

Observing Equation 2-5, we notice that the equation represents a polynomial in R which may result in n possible solutions for tR which will satisfy Equation 2-5. In economic analysis, we are only interested in

real solutions. Although negative rate of return is a real value, we may not be interested in an

investment of negative rate of return. As a practical matter, we are searching for positive, real solutions

of this equation. In most instances, we will obtain only one positive, real solution which represents the

rate of return. This is shown in the following examples.

Example 2-6

Calculate the rate of return for the following cash flow.

Year Cash Flow

0 -4,000

1 2,500

2 1,800

3 1,300

4 900

Solution 2-6

Using the cash flows, we can write the equation for NPV as,

2,500 1,800 1,300 900 NPV = �4,000 + (1 + .) + (

1 + ) 2 + (1 + i)3 + (

1 + 0

Since this is a polynomial equation in i, we will have to solve it by trial-and-error,

i=15% NPV=904 = 35% NPV = � 361

Since the value of NPV changes a sign between i = 15% and i = 35%, the rate of return should fall in between the two values. By

linear interpolation, we can write an approximate equation for the rate of return (ROR) as,

- f ""

1 ROR i, + ( t - t+) tNPV-NPVJ

quatiot1 2-6

Where i and L respectively represent the trial values which resulted in positive and negative NPV values, and NPV+ and

NPV_ represent the positive and the negative NPV values respectively. In our example, -- -



NPV = 904 NPV_ = � 361

= 15% L = 35%

Therefore,

r 904 ROR 15 + (35-15) 904

+ 3611 29.3%

We can calculate the NPV at 29.3%.

NPV = � 66.5

Although close to zero, we can try one more interpolation between 15% and 29.3%.

904 ROR 15 + (29.3 - 15) [904

+ 66.51 28.3%

NPV at233.% = �10.2

We assume this value to be close enough to zero. You may note that the higher the difference between the i and i_, the

bigger the deviation between the true ROR and the interpolated value. Therefore, the interpolation may have to be carried out

- more than once to obtain a correct value of the ROR.

Example 2-7

By investing $10,000 in a project, you are promised that you will earn $2,700 per year for a period of six years. What is the ROR

for this investment?

Solution 2-7

Using Equation 2-2,

PVcosts - PV eneits = 0

I That is,

[(1 + 06 - 11 10,000-2,7001

i(1-I-i) j

I For i=10%,

F(1 11 �10,000 + 2700[01(1 + 0.1)6] = 1,759

I For i=20%,

[(1 + 0.2)6 - 11 �10,000 + 2700[o2(1 �+0.2) 61 = �

1,021

I Using Equation 2-6,

r 1

ROR 1,759

10 + (20� 10) 1 I 1,021 - 1,759]

For 16.3%, NPV = -129

18 Mohan Kelkar, Ph.D., lU

Doing interpolation one more time,

1,759 1

ROR 10 + (16.3 - 10) [1,759 + 1291

15.8%

At

= 15.8% NPV = 1.7 0

Therefore, the rate of return is 15.8%.

The examples above show that the ROR calculation has to be done by trial-and-error. With the advent of

software programs, this problem can be easily solved by using standard functions available in

spreadsheet programs. One easy way to guess the initial value of ROR is by knowing that if the initial

investment is equal to the salvage value, the ROR can be calculated as,

ROR = periodic payment

initial investment Equation 2-7

Using Equation 2-7, if the salvage value is less than the initial investment,

ROR < periodic payment


On the other hand, if the salvage value is greater than the initial investment,

ROR > periodic payment


Equations 2-7 through 2-9 are applicable only if the investment is made at the beginning of the project

and the periodic payments are equal to each other.

Example 2-8

As an investment, you bought a house for $50,000. If you can rent the house for $800 per month, and can sell the house for

$70,000 at the end often years, what is the ROR on your investment?

Solution 2-8

In this problem, using Equation 2-9, the salvage value ($70,000) is greater than the initial cost ($50,000). Therefore,

800 ROR> = 0.016/month

50,000

Let us assume the ROR to be .017/month. Using Equation 2-2,

PVcosts - P17 115 = 0

1(1 + .017)120_lI (1 ) 120

70,000

50,000-800 � 017(1 + .017)120 - + .017

= �$93 0

where 120 is the number of months in which the rent is collected. Therefore, the ROR is 1.7%/month, or 20.4%/year



As shown in the example above, by using the correct initial guess, we did not have to use too many trial-

and-error methods to find a solution. A similar equation can be developed for geometric series as

explained in the example below.

Example 2-9

A proposal calls for an investment of $25,000 in an oil property that will result in an initial income of $6,000 per year declining

at a rate of 8% per year over the next twenty years. What is the rate of return? Assume the salvage value to be zero.

In this example we have a geometric series.

Given: A= $6,000, n= 20 years, g= -0.08

Using the equation for geometric series,

A f1+’1 NPV= �l----1 1-25,000= 0

(i�g) 1

\1+

9\

tJ j

For a geometric series with negativeg, with long life, as a rule, the initial guess of ROR can be used as equal to,

first periodic payment initial guess = + g

initial investment

In this problem,

initial guess = 6,000- 0.08 = 0.16 25,000

Using the initial guess,

6,000 [ 1 - NPV=(16+08)[1_(l+16) j_25000=$242

After one additional trial-and-error, the ROR = 15.7%.

Equation 2-10

As stated before, the ROR technique is probably the most used technique in economic analysis. It is easy

to understand. Since everyone understands the interest rate, rate of return is equated to return on

investment in terms of an interest rate that would be earned. Intuitively, when comparing two

investments, one fetching a higher ROR is always more attractive.

In a corporate structure, to evaluate the feasibility of a project, we need to compare the ROR to the

minimum rate of return (MROR). If the ROR>MROR, the project is selected; if the ROR<MROR, the

project will be rejected.

Example 2-10

An oil property is on sale for a value of $100,000. It was observed that the property will result in an annual net income of

$30,000 in the first year followed by a decline of 10% per year. If the property will have to be abandoned after 10 years with a

salvage value of $10,000, should property be purchased if the MROR is 20%?

Solution 2-18

A= $30,000 with a decline of 10% (p = -0.1), initial investment = $100,000, salvage value = $10,000, n= 10 years

20 Mohan Kelkar, PhD., J.D.

Using the expression for geometric gradient series,

NPV= Ii �100,000= 30,000 1 - (1+0)n} 10,000

(i�g)[ 1+i (1+i)

Using Equation 2-10, we can start with an initial guess of

30,000 =0.1 = 0.2

For,

= 0.2 NPV = �4,016

= 0.15 NPV = 12,129

By linear interpolation,

12,129 = 0.15 + (.2 + .15) (12,129 + 4,016)

ROR = 0.1875

Assuming that the ROB is correct, it is compared with the MROR. Since 18.75% < 20%, the project should be

The procedure is slightly more involved when two alternatives are being considered. In addition to

comparing the ROR to the MROR for individual alternatives, we also have to carry out an incremental

analysis. This procedure is illustrated in the following example.

The following two alternatives are considered for a project,

(a) (b)

Initial Investment $50,000 $500,000

Annual Benefit $25,000 $125,000

Life, Years 5 5

Salvage Value $50,000 $500,000

If the MROR is 20%, which alternative should be selected?

Solution 2-11

The first step is to estimate the ROR of the individual alternatives and compare the ROR with the MROR. If the ROB is less than

the MROR, the alternative(s) should be rejected.

In this example, since the salvage value is equal to the initial investment, using Equation 2-7,

25,000

RORa =S0,000 = 50%

125,000 RORb

= = 25%

500,000

Since the ROR ( > MROR and ROR 1, > MROR, both alternatives satisfy the feasibility criterion.

Intuitively, since ROR 1 > R0R 5 , one may be inclined to select (a) over (b), but notice that the initial investment for both

alternatives is not the same One of the drawbacks of the ROB analysis isits inability to account for the investment amount. In



considering the two alternatives with a different investment amount, our implicit assumption is that we have $500,000 to

invest; otherwise, we would not consider alternative (b). Therefore, if we choose alternative (a), we have to assume that the

remaining $450,000 will have to be invested at MROR. To properly account for the investment, we need to conduct

incremental analysis. That is, to find out by investing an additional (incremental) $450,000 in alternative (b), what incremental

benefits are received? Subtracting values related to alternative (a) from alternative (b), we obtain,

(b)-(a) Investment $450,000

Annual Benefit $100,000

Life, years 5

Salvage Value $450,000

For incremental investment, we can calculate the ROR by Equation 2-7 (since investment = salvage value),

AROR b_ a = 100,000

= 22.2%

450,000

This number indicates that the ROR on incremental investment is 22.2% that is greater than the MROR. In other words, by

investing an additional $450,000, we will earn a ROR of 22.2%. On the other hand, if we do not invest an additional $450,000 in

alternative (b), we will earn only MROR on that additional amount. Therefore, it is more attractive to invest the additional

$450,000 in alternative (b). That is, to select alternative (b) over (a).

This analysis can be easily confirmed by calculating the NPV for both the alternatives at MROR.


(NPV) a = 25,000 [(1 + .2) - 11 50,000

.2(1 + .2) + (1 +.2) - 50,000

= $44,859


[(1 +.2) 5 - 11, 50,000

(NPV)b = 125,000 [ .2(1+.2)5 j +

(1 + .2)1 - 500,000

= $74,765

Since (NPV)b > (NPV) a, alternative (b) should be chosen. This is consistent with the answer we obtained from the

incremental analysis.

To generalize, if two alternatives requiring different amounts of investment need to be compared, we

should carry out an incremental analysis. If AROR>MROR, we should select an alternative requiring a

larger investment. If tROR:~ MROR, we should select an alternative requiring a smaller investment.

The procedure can be easily extended when considering more than two alternatives. Briefly, the

stepwise procedure for incremental analysis can be stated as:

a. Calculate the ROR for each alternative. If ROR>MROR, assume that the alternative is feasible and

retain it for further incremental analysis. If the ROR<MROR, remove tle alternative from further

analysis.

b. Take two alternatives requiring the smallest investments. Calculate the ROR on the incremental

investment by subtracting the smaller investment from the larger investment. We denote the

ROR on incremental analysis as LXROR. If AROR~!MROR, select the alternative requiring the

22 Mohan Kelkcir, Ph.D., J . D.

larger investment; if AROR<MROR, select the alternative requiring the smaller investment.

Remove the rejected alternative from further analysis.

c. Take the remaining alternative and compare it with the alternative requiring the next largest

investment. Calculate the incremental ROR. If AROR ~: MROR, select the alternative requiring the

larger investment; if LROR<MROR, select the alternative requiring the smaller investment.

Remove the rejected alternative from further analysis.

Repeat step (c) until only one alternative remains.

Example 2-12

The following three alternatives are considered for a project. If MROR is 15%, select the appropriate alternative.

(a) (b) (c)

Initial Investment $1,000 $3,000 $6,000

Annual Benefit $300 $1,000 $1,800

Life, Years 10 10 10

Salvage Value $1,000 $3,000 $6,000

p 0 p p

Solution 2-12

Using Equation 2 -7,

300 ROR a = = 30%

1,000 1,000

ROR b = = 33.33% 3,000

ROR C = 1,800

= 30% 6,000

I I I I I I I

Since ROR for all of the alternatives is greater than the MROR, all are feasible. In step (b) take the two alternatives requiring the

smallest investments.

In this example, we will consider alternatives (a) and (b) for incremental analysis,

(b)-(a)

Investment $2,000

Annual Benefit $700

Life, Years 10


Therefore,

700 1\ROR ba = = 35%

2,000

Since ARORb_ a > MROR, select (b) over (a). Eliminate alternative (a) from further analysis.

In the next step (step c), compare (b) with the remaining alternative (c). For incremental analysis,

(c)-(b)

Investment $3,000

Annual Benefit $800

Life, years 10



Chapter 2- Economic Methods

23

Therefore,

800 LIROR C_ b =

3, -000 = 26.67%

Since AROR C _b > MROR, select (c) over (b). After eliminating (b), we are left with only alternative (c). This will be our choice.

To summarize the economic criterion applied for the rate of return analysis for a single project: if the

ROR>MROR, the project is selected; if the ROR<MROR, the project is rejected. For a project having

multiple alternatives, an incremental analysis needs to be conducted so long as there is a difference in

the cash flow profiles of two projects. Only after applying the incremental analysis is the solution

consistent with the NPV analysis.

In addition to the requirement of incremental analysis, the ROR analysis method has another drawback.

This method works well when a given alternative requires an initial investment that is followed by future

benefits. For this type of alternative, the cash flow profile can be shown as negative cash flow in the first

year followed by positive cash flow in future years. For example, if we consider an investment of $1,000

that will result in a $300 annual benefit for the next six years with a $500 salvage value at the end of six

years, the cash profile can be written as,

Year Cash Flow

0 -1,000

1 300

2 300

3 300

4 300

5 300

6 300+500

In this profile, there is only one sign change in cash profile between years 0 and 1. Such a profile is

amenable to conventional ROR analysis.

Note that the ROR calculation requires solving a polynomial of i. We calculate the value of i for which

the NPV is zero. For economic analysis, we are only interested in obtaining positive, real values of i for

which the NPV is equal to zero. When there is only one sign change in the cash flow profile, as shown

above, we can only obtain one or zero positive solutions.

In some instances, however, the sign changes more than once in a cash flow profile. Under these

circumstances, we may obtain more than one real ROR. The rule of signs for polynomial solution states

that the number of real solutions between -1 and Oo is never greater than the number of sign changes.

That is, if we have two sign changes, we may obtain a maximum of two rates of return values between -

100% and oo. In practice, a typical cash profile includes initial investment followed by positive cash flow.

Since this type of cash flow has only one sign change, only one rational value of ROR can be obtained.

However, there are practical situations where it is possible to obtain multiple solutions. Under those

conditions, the ROR technique is not very useful.

24 Mohan Kelkar, PhD., J. D,

Example 2-13

An in-fill drilling project is being considered for an existing oil field to accelerate oil recovery. The following two scenarios,

based on two alternatives (no in-fill drilling versus in-fill drilling) are predicted. Which alternative would you select?

Alternative A Alternative B Year (no drilling) (in-fill drilling)

0 0 -20

1 30 60

2 20 40

3 18 6

4 14 4

5 10 2

6 6 0

The numbers are in millions. Assume that MROR is 20%.

Solution 2-13

The first step in ROR analysis is to compare individual ROR’s for each alternative with the MROR. For alternative A, there is no

sign change in the cash flow profile. Therefore, the ROR for alternative A is . For alternative B, ROR can be shown to be

greater than 20% (the ROR for alternative B is 260%). Therefore, both alternatives satisfy the requirement that the ROR be

greater than the MROR.

The next step is to conduct the incremental analysis. The cash flow profile for incremental values can be written as,

Year B-A

0 -20

1 30

2 20

3 -12

4 -10

5 -8

6 -6

The cash flow profile shows more than one sign change. This indicates the possibility of more than one positive ROR solution.

NPV for any interest rate can be calculated as,

30 20 12 10 8 6 NPV= _20+ (l+) + _ _ _ (l+)2(l + .)a(l + .)4(l + )s(l + ) 6

Example Figure 2-1 shows a plot of NPV as a function oft,



25

$3

$2

U)

> $0 EL C0.3 0.6 0.7 08

($3)

Example Figure 2-1: Plot of NPV vs. ifor Example 2-13

As stated before, the ROB is the rate at which the NPV is equal to zero. Based on Example Figure 2-1, two L\ ROR’s are possible;

11% and 72%. If we assume /X ROR to be 11%, then alternative A (the alternative requiring a smaller investment) should be

selected (i ROR<MROR). If we assume A ROR to be 72%, then alternative B (an alternative requiring a larger investment)

should be selected (\ ROR>MROR). Obviously, our answer changes depending upon the selected value of L ROR. When faced

with this issue, ROR cannot resolve the correct answer. We have to use NPV analysis to come up with a correct answer. As

Example Figure 2-1 shows, the NPV value at 20% MROR is positive. This means that drilling wells is a better option than not

drilling since the NIPV represents the difference of the two.

One easy way to confirm this analysis is to calculate the NIPV at the MROR (=20%) for incremental cash flow,

iT � � 0 30 20 12 10 8 6

NP

�2 + (1 +.2) + (1 +.2)2 (1+.2)3 (1+.2)4 (1+.2)s (1+.2)6 = 1.9

Since NPV is positive, alternative B should be selected. This is the same answer predicted in the previous paragraph.

Case Study 2-4

Cimarex is one of the major operators in the Woodford play in Oklahoma. Cimarcx has net 94,000 acres in western Oklahoma for which they paid an average of $700 per acre for the leasing fee. The infrastructure is excellent and with better fracturing and completion technology, Cimarex has improved the performance of the well significantly. The map of well locations is shown below.

Eiornt � �., a, ,,a �

* a. �,

a, a *1 4 4

� a c s,,, n . a, * *

* a, 4,

jc


All the wells drilled are horizontal, typically on 160 acre spacing, and are about 4,300 feet long with multistage fracturing. Fhcre is a potential for 80 acre spacing in the future which will increase the production significantly. The cost of drilling and completing a typical wcll is about S7.2 million.

Cimarex has approximately 50% Net Revenue Interest (34R1) in each well (after subtracting royalty costs). The operating costs are assumed to be about 85,000 per month for first year followed by about $’s,OOO per month for the rest of the life of the well. In addition to producing gas, the wells typically also produce some condensate. Although the condensate rate varies widely, about 20 S’FB/MMSCF is a good approximation.

A typical rate profile ofawell is shown below.

Year Gp, MMSCF 1 1,046,0 2 6147 3 471.1 4 393 5 3423 6 306.1 7 2787 8 257.1 9 239.5 10 224.8 11 218.2 12 205.1 13 192.8 14 181.2 15 170.4 16 160.1 17 150.5 18 141.5 19 133 20 125 21 117.5 22 110.5 23 103.8 24 97.6 25 91.8 26 86.2 27 81.1 28 76.2 29 71.6 30 67.3

Assume, as a baseline, the price of gas to be S5/MSCF and the price of oil to be S70/bbl. We have arbitrarily cut off the production after 30 years, but it is possible that, based on the economic cut-off, the well can produce for 60 years. Assume that, starting after 30 years, yearly production reduces by 5.8% every year until one reaches 60 years.

A. Assuming the life of the well to be 30 vs. 60 years, what is the EUR expected from this well? how much of a percentage increase will you get by assuming 60 years of life?

B. Assuming 30 vs. 60 years, what is the field and development cost per MSCIi? Do not forget to include the leasing costs as part of the initial investment. Is there a big difference between the two? iiow much difference will it make if we assume 160 acre vs. 80 acre spacing?

C. What is the RfR for 30 vs. 60 years of life? Assume 160 acre spacing. Assume the MR(.)R for Cimarex is 10%. D According to the \’ice President of Cimacx, Field and Development Cost per MSCP is one of ’the’ most worthless pieces of

information an investor can get. She ROR is much more valuable information according to him. Do you agree with his statement? Why or why not?

Ii. Keeping the oil price flat at $70/hhl, if we assume that price of gas is 53/1\ISCEi, would the well still remain economical based on R( )R analysis? Use 30 year life only.

E,\twhat price of gas, the ROR would exceed 100%_1 Do you believe that this price of gas is reasonable or iicliiev’al,le? Use 30 year life only.


Chapter - Economic Methods 27

Case Study 5Iution 24

1. EUR and F & 1) Costs

To calculate F & D Costs, we first need to determine EUR for 30 years production and 60 years production. Since the well produces condensate, we also will need to convert STB into 5CC by using I SIB z 6 MSCF.

By summing the total gas production over thirty years, we obtain 6,755 MMSCF of cumulative production. Using 20 6,755xZOX6

bbl/MMSCU, we can calculate equivalent gas production as =1,000

= 811 MMSCF. I hcrefore, net equivalent production

is 0.8 (6,755 + 811) = 6,052 MMSCF. We can calculate F & D Costs as:

700 x 160 + 7.2 x 106

= 6,052 x 10 $1.21/MSCF

If the spacing is 80 acre spacing instead of 160 acre spacing, the only cost that would change in the above equation is leasing costs. It would be 700 x 80 instead of 700 x 160, The F & 1) Costs for 80 acre spacing would he 51.20/MSCF.

If the production continues for more than sixty years, based on a 5.85’s decline every year beyond 30 years, we can calculate the production profile for the remaining thirty years as follows. Starting at 30 years, we multiplied production from the previous year by 0.942 to get the result.

Year Gp, MMSCF

31 63.4 32 59.7

� . . ��. . 33 56.3

34 53.0 35 49.9 36 47.0

. .. 37

. . . .

44.3 � .. ,

� 38 41.7 39 39.3 40 37.0 41 34.9 42 32.9 43 31.0 44 29.2 45 27.5 46 25.9 47 24.4 48 23.0 49 21.6 50 20.4 51 19.2 52

18.1 53 17.0 54 16.0 55 15.1 56 14.2 57 13.4 58 12.6

59 11.9 60 11.2

Additional production for the remaining thirty years is 1,020 MMSCF gas ciiuivalcnt assuming the same condensate ratio. The net production is 0.8( 7 ,565 – I,020) = 6,868 i\IMSCI’. Therefore, F & 1) Costs based on sixty years production For 160 acre spacing is:

700 x 160 + 7.2 x 10 6

6,868x103 = $1.061MSCF

-

By assuming I\tinstead of thirty years, we reduced the I & I) Costs by about 14%. The FUR has changed from 6,052 MMSCI to 6,868 1 \1S( 1’. ’I Ins represents 14% increase In the I UR value based on 60 years production compared to 30 years


production.

5 2. Rate of Return Calculations

0

The following table is constructed based on 30 years production. The net revenue is calculated for both the gas price of

5 S5/SISCP as wdll as S3/MSCE The net revenue is determined by assuming S70/bbl of condensate, and 0.8 NR1. We subtracted the operating costs from the gross revenue to calculate the net revenue.

S , Gas Price

$5/MSCF $3/MSCF

0 � Year Gp, MMSCF STB Op Costs Net Revenue Net Revenue

0 0 (7,312,000) (7,312,000

1,046.00 20,920 60,000 5,295,520 3,621,920 � . .� . . 2 614.7 12,294 36,000 3,111,264 2,127,744

0 . 3 471.1 9,422 36,000 2,376,032 1,622,272

4 393 7,860 36,000 1,976,160 1,347,360 $ 5 342.3 6,846 36,000 1,716,576 1,168,896

e 6 306.1 6,122 36,000 1,521,232 1,041,472

7 278.7 5,574 36,000 1,390,944 945,024 � 8 257.1 5,142 36,000 1,280,352 868,992

O 9 239.5 4,790 36,000 1,190,240 807,040

10 224.8 4,496 36,000 1,114,976 755,296

5 . . 11 218.2 4,364 36,000 . 1,081,184 732,064 . 12 205.1 4,102 36,000 1,014,112 685,952

13 192.8 3,856 36,000 951,136 642,656 � 14 181.2 3,624 36,000 891,744 601,824

S. 15 170.4 3,408 36,000 836.448 563,808

16 160,1 3,202 36,000 783,712 527,552 $ 17 1503 3,310 36,000 734,560 493,760

S 18 141.5 2,830 36,000 688,480 462,080

19 133 2,660 36,000 644,960 432,160

5 . 20 125 2,500 36,000 604,000 404,000

S. 21 117.5 2,350 36,000 565,600 377,600

22 110.5 2,210 36,000 529,760 352,960

5 23 103.8 2,076 36,000 395.456 329,376

S.. 24 97.6 1,952 36,000 462,712 307,552

25 91.8 1,836 36,000 434,016 287,136

5 . 26 86.2 1,724 36,000 405,344 267,424

S. 27 81.1 1,622 36,000 379,232 249,472

28 76.2 1,524 36,000 354,144 232,224

29 71.6 1,432 36,000 330.592 236,032

30 67.3 1,346 36,000 308,576 200,896

The initial investment in year zero includes both the cost of drilling and completion is well as leasing costs. It could also be argued that the cost of leasing isa"sunk cost" and should not be included in the costs for evaluating individual wells. The cash flow profile has only one sign change and, hence, only one feasible value of R( )R. The R( )R cari’bc calculated by any trial-and-error method by determining the rate at which NPV is 7cm. In RXCI 1., a function IRR can be used to calculate the R( )R. The rate of return for S5/MSCI" price is 43% whereas the ROR for S3/1NISL" price is 24°/s. based on 6IROR of 10%, under both scenarios, the project is feasible.

If, instead of a 30 year life, one assumes a 60 year life, a similar table its contructcd above can be constructed for 60 years. The net rcvcilue would continue for another 30 yeats; however, the ROR of the pluject for a gas pi ice of S5/MSCF is still 43% and the ROR for a gas price of S3/SISCI’ is still 24%.


Chapter 2 Economic Methods 29

3. ROR versus F & D Costs

’l’here is an important difference between ROR and F & I) Costs, as is illustrated in this Case Study. ROR accounts for future discounting of revenue; hence, even though the life of the well is extended beyond thirty years, the revenues generated beyond thirty years do not contribute to the ROR value. A similar observation can be made if NPV is calculated. ’ftc NPV will not be different if we assume 30 years versus 60 years life since the production at later stages of life of the well is highly discounted and does not contribute to the present value of the well. In contrast, RUR or F & D Costs do not account for the discounting of the reserves or revenues. ’l’herefore, the longer the well is expected to produce, the larger the EUR will he and the F & I) Costs will be smaller. By extending the life of the well based on certain, arbitrary, economic limits, it is easy to manipulate the F& I) Costs as well as CUR; however, ROR and NPV of the well will not be affected by the extension of the life of the well. hence, the ROR is a much more robust measure of the well’s economic feasibility than CUR or F & D Costs. ’This is especially true for shale wells, which tend to have quick, early decline followed by potentially a long life with a slow decline at the end.

4. 100% Rate of Return

Using the table above, through trial-and error we can determine the price of gas, which will provide us with I00% ROR. ’l’he price of the gas has to be 510/tsISCF. Historically, the price of gas has reached S10/MSCF; therefore, it is possible that the gas price can reach $I0/MSCF in the future. however, based on current conditions, it appears unlikely.

Before we close the discussion about rate of return analysis, it is worth repeating an important

distinction between NPV and ROR. NPV is an absolute measure of the profitability of the project;

whereas, ROR is a relative measure of the profitability of the project. A large NPV means more profit

from the project. A large ROR means relative to investment, the project is economically attractive. This

is one of the reasons NIPV is a much more appropriate measure when comparing mutually exclusive

alternatives. If our goat is to maximize profit, then the alternative that provides the maximum value of

NPV is the most appropriate. In contrast, as we have seen in this section, the largest ROR may not

indicate the best mutually exclusive alternative without conducting incremental analysis. However, the

disadvantage of ROR is also a big benefit when we are comparing alternatives that are independent.

Since ROR is a measure of profitability relative to investment, the highest ROR among many

independent projects tells us which project will provide the biggest bang for the buck. By ranking various

projects according to ROR, we can make a decision about the order in which various projects should be

selected. This can be important when the budget is tight and we do not have enough money to invest in

all of the projects. The following example illustrates the usefulness of ROR under these conditions.

Example 2-14

A company is considering multiple independent projects for potential investments. The company currently has $70,000 to

invest and the MROR of the company is 10%. Which projects should the company consider for investment?

Project Initial Investment Annual Benefit Life

A $50,000 $14,500 5

B $20,000 $5,500 5

C $30,000 $9,500 5

0 $70,000 $21,000 5

E $40,000 $12,500 5

Solution 2-14

Knowing the life of the project, the annual benefit, and initial investment, we can calculate both the ROR and NPV of the

project. The following table shows the ranking of the projects based on both the ROR and NPV.


Project ROR Ranking NPV Ranking

A 14% 4 $4,966 4

B 12% 5 $849 5

C 18% 1 $6,012 3

D 15% 3 $9,607 1

E 17% 2 $7,385 2

The ranking based on ROR (the highest being the best) and the ranking based on NPV are different. If we only have a $70,000

budget, based on the ranking of NPV, we will select project D and that will use up the $70,000 and, hence, we cannot select any

other project. Based on ROR, we will select C and E and that will use up the $70,000. The selection of C and E is the correct

choice since the summation of NPV for C and E adds to $13,397 which is better than $9,607 for project D. Since ROR is based on

relative benefit, it favors projects that bring in more profit relative to investment. When limited capital is an important

consideration, ROR provides us the correct ranking because it tells us the order in which we need to select projects that will

maximize the profit. -- ---- --

This topic is not as simple as illustrated in this example. We will add more detail in the next section.

Problem 2-18

An investment proposal promises to pay $1,000 per year for the next ten years if you invest $5,000 today. Assume the salvage value to be zero. What is the rate of return on your investment?

Problem 2-19

A proposalfor drilling on oil well states that if you invest $25,000 in the project, you will own 33% of the working interest (pay 33% of the operating costs) and will receive 25% of the revenue interest (receive 25% of the revenues). If the operating costs in the first year are expected to be $20,000, rising at a rote of 5%, and the revenues are expected to be $70,000, declining at a rate of 8%, should you invest in this project? Considering the risks involved in the project, you want to receive at least 25% MROR. Use the ROR analysis. Assume the life of the project to be five years.

Problem 2-20

An oil company is considering a proposal to automate a plant to reduce the labor costs and increase the efficiency of the producing field. It has an option of completely automating or partially automating the field. The cash flows from both the

alternatives are shown below.

Alternative Investment Annual Benefit

Full Automation $4,000,000 $1,150,000 Partial Automation $1,500,000 $500,000

Assume the life of the project to be seven years and the salvage value to be zero. If the MROR is 20%, which alternative should

be selected? If the MROR is 15%, does the answer change? Use the ROR analysis.

Problem 2-21

A company is considering four alternatives for copying machines. The costs and the annual benefits are shown below:

A B C D

Cost $6,000 $5,000 $10,000 $8,000 Annual Benefit $900 $500 $1,800 $1,300

Assume the salvage value to be zero, life of each machine to be ten years and the MROR to be 6%. Which alternative should the

company choose? Use the ROR analysis.



Problem 2-22

Able bought 50 shares of stock in an oil company at a price of $20 per shore. After holding the stock for 10 years, he sold it for

$45 per share. For the first 5 years, he received an annual dividend of $1.50 per share; for the last five years, he received a dividend of $0.50 per share per year. What is the rate of return on his investment?

Problem 2-23

Consider an investment in a gas producing property. After investing $70,000, annual revenue of $36,600 is expected with annual

expenses of $22,000 per year.

� If the life of the project is 8 years with a salvage value of zero, what is the rate of return?

� If the annual expenses increase at a rote of 7% per year, but the revenues are constant, what is the rate of return?

At what rate in the above option would the annual revenues have to increase so that the rate of return is the same as in the first

option?

Problem 2-24

Consider the following two mutually exclusive investments.

Period I II

0 -30,000 -40,000 1 15,000 16,000

2 15,000 30,000 3 15,000 10,000

Using the ROR analysis, select the appropriate alternative if the MROR is 10%.

Problem 2-25

A company is considering buying workstation computers for internal use. Model I costs $27,000 and Model I! costs $39,000. Bath models are expected to provide the some service. Due to the increasing power of computers, these machines will be traded

in after four years of service. Model I can be traded in for $13,500 versus Model II which can be traded in for $17,000. If the MROR is 1091o, which model should be selected?

Problem 2-26

Consider the following four, mutually exclusive projects.

Period A B C 0

0 -2,000

1 1,800

-1,500

900

-4,000

1,800 -2,400

800 2 1,000 750 1,800 800 3 200 750 1,800 800 4 100 150 1,800 800

If the MROR is 12%, which project should be selected? OW


Problem 2-27

The following three mutually exclusive alternatives are presented for a project. If the MROR is 13%, select the appropriate alternative.

Period I II Ill

0 -150 -450 -300 1 45 150 1230 2 75 150 -1674 3 120 150 756 4 300

Problem 2-28

An oil field is currently under consideration for in-fill drilling. The net revenues for two alternatives are shown below. Which option will you select?

Year No Drilling In-Fill Drilling

0 0 -1,170,000 1 600,000 1,900,000 2 475,000 900,000 3 400,000 440,000 4 325,000 100,000 5 260,000 0 6 130,000 0

Assume the MROR to be 18%.

Problem 2-29

An oil producing property was expected to make a profit throughout its lifetime. Unfortunately, due to fluctuations in oil prices the property resulted in the following cash flow. What is the rate of return on the investment?

Year Cash Flow

0 -$100,000 1 70,000 2 55,000 3 -35,000 4 -10,000 5 30,000 6 20,000

If the MROR is 1096, based on NPV analysis, was this a profitable investment?

Problem 2-30

An oil field is under consideration for a potential in-fill drilling project. The feasibility study indicates that without in-fill drilling,

the field will generate net revenues of $200,000 over the next year and will continue to generate additional revenues over twenty-five years declining at a rate of 8%. If we invest $1 million in in-fill drilling, the field will generate net revenue of $580,000 over the next year and will continue to generate revenues over an additional six years declining at a rate of 10%. For what range of ROR’s is the in-fill drilling a suitable alternative? Assume the salvage value to be negligible. Assume the MROR to be 15%.

Problem 2-31

An oil well is currently producing 10 bbl/day under natural flow. The analysis of the inflow performance curve reveals that installing an electrical submersible pump at a cost of $50,000 will increase the production to 18 bbl/day. It is expected that,



under natural flow (active water drive), production will continue at the some rate over a period of 15 years. If we use an electrical submersible pump, production will continue only for eight years. If we assume that the net revenue under natural flow conditions is $131bbl and the net revenue with electrical submersible pump is $12.301bbl, which option is preferable? Assume the MROR to be 15%.

Problem 2-32

A producing gas property is currently receiving total revenues of $2,000 per month. The revenues are expected to decline at a rate of 6% per year. The operating costs are $600 per month and will increase at a rate of 4% per year. If you offer $60,000 for

this property, what is the rote of return on your investment? Assume the salvage value to be zero.

Problem 2-33

You are considering two properties for a potential investment. The first property requires on investment of $1 million and will result in $60,000 in net revenues in the first month followed by a decline of 9% per year over the next ten years. The second

property requires an investment of $3 million and will result in $220,000 in net revenues in the first month followed by a decline

of 40% per year over the next ten years. Assume continuous compounding. If the MROR is 15%, which investment should you select? Use incremental analysis.

PROFIT TO INVESTMENT RATIO

We discussed the PIR in Chapter 1. The only difference in this chapter is that we will account for time

value of money in defining it. Profit to investment ratio (PIR) is the ratio of the NPV at MROR to the

present value of out of pocket investment. We can write it as,

NPV PIR Equation 2-11

Pvcosts

This number is an indication of the efficiency of the investment. In other words, PIR is the amount of

money earned per dollar invested. Similar to ROR, PIR is also a measure of the profit relative to

investment. If we are interested in evaluating projects that are mutually exclusive, PIR does not add any

value compared to NPV. This is because once we determine the NPV of all the mutually exclusive

alternatives, we already know which alternative is the best alternative. Also, similar to ROR, to make an

appropriate selection we will have to conduct incremental analysis. For a project to be feasible, the PIR

has to be greater than zero. The following examples illustrate the application.

Example 2-I5

The following two alternatives are considered for a project. Based on the PIR analysis, select the best alternative.

(a) (b)

Initial Investment $10,000 $20,000 Annual Benefit $3,000 $7,000 Life, Years 5 5 Salvage Value $8,000 $3,000

Assume the MROR to be 10%.


I 0 S

Solution 2-15 + .i) - ii 8,000

NPVa 3000 [(1

= ’ .1(1+.1) b 0 339 1+(1+1)5O0O_$6

+ .1) s- ii 3,000 NPVb =

’ 7 000 .1(1+.1) J+(1+1)s_2O O O O_$8 398

NPVa 6,339 PIRa

= Pvcos = 10,000 = 0.6339 > 0

NPVb 3,398 PIRb = PV COStSb = 20,000

= 0.4199> 0

Although PIRa > PlR, we need not select (a) unless we carry out one additional calculation. As shown in the NPV calculation, alternative (b) is a better alternative; however, we will have to do incremental analysis to arrive at the correct conclusion.

NPVa NPVb PIR

PVinVeStMeDtb PVinvestment

8,398 - 6,339

20,000 - 10,000 = 0.2059> 0

Since LXPIR > 0, an alternative with a higher investment should be selected (i.e., we should select (b) over (a)). If LXPIR < 0, an alternative requiring a smaller investment should be selected.

Like ROR analysis, the criteria for the selection of an alternative depend on the value of APIR. The above

example illustrates the application of incremental analysis when only two alternatives are involved. The

analysis can be easily extended when more than two alternatives are involved. Rather than discussing

the method, what is important to understand is that the PIR does not add any value compared to NPV

analysis when selecting mutually exclusive alternatives. However, just like ROR, when capital constraints

are important, the PIR can provide a much better approach in making the correct selection of

independent projects. See the Example 2-16.

Example 2-15

We have $50,000 to invest. We are considering four possible projects in which to invest. The information with respect to each project is provided below. Which projects should be selected? Assume the MROR to be 10%.

Project Initial Cost Annual Benefit Life, Years Salvage Value

A $10,000 $3,000 5 $1,000 B $20,000 $5,500 5 $6,000 C $30,000 $8,500 5 $10,000 D $50,000 $13,000 5 $15,000

Solution 2-15

We will have to calculate PIR for each of the projects.

Project A

[(1 +.1) 5 - U 1,000 NPV = 3000[

.1(1 +.1), j + - 10,000 = 1,993

NPV 1,993 PIRA == 10,000 0.1993


Chapter 2 Economic Methods 35

Project B

NPV = 5,500 [(1 +.1)5 - ii 6,000

00O= .1(1+.1) 1+(1+1)5_204575

4,575 PIRH = 20,000 = 0.2287

Project C

NPV = 8,500 [(1 + .i)5 - 11+ 1(1 T.155 ] + (1 T.15 5

�30,000 = 12,222

12,222 PIRc = 30,000 = 0.4074

Project 0

f(i +.1) 5- 11 15,000 NPV = 13,0001

.1(1 + .1) j + (1 + .i) - 50,000 = 8,594 8,594

PIRD = 50,000 = 0.1719

All of the projects satisfy the minimum criterion that PIR>0. Based on the analysis, we can rank the projects as,

Cumulative

Project PIR Investment Investment

C .4074 30,00 30,000

B .2287 20,000 50,000 V

A .1993 10,000 60,000

0 .1719 50,000 110,000

The table above shows the investment required for each alternative and the cumulative investment required if we select the

projects in the same order. It is obvious that after selecting C and B, we have exhausted our limited budget. Therefore, we will

select only B and C projects. If we had $60,000, we should have selected A, B and C. Our selection is only limited by the

budgetary constraints; not by the selection of a particular project. In principal, we can select any combination of projects so

long as the budget constraints are satisfied. In this particular case, by combining projects C and B our NPV is $12,222 + $4,575 =

$16,797. This is the maximum value we can achieve with the existing budget constraint.

For this particular problem we selected the projects so that our cumulative investment matched precisely with the budget

requirements of the projects. There was no residual budget remaining after the selection. In some instances, where the

projects are ranked so that if we select i projects, our cumulative investment is less than the total budget, and if we select

(i + 1) projects, the cumulative investment is greater than the total budget, we may require a trial-and-error procedure to

select the appropriate combination.

Let us consider the above example, except we will assume that the salvage value for project B is $4,000 instead of $6,000. Then

for project B,

[(1 + .1) 5- 11 4,000 NPV = 5,500 [ .11 + .1) + (1 + .1) - 20,000 = 3,333

3,333 PIR0 = 20,000 = 0.1667

If we rank the projects,


Project PIR Investment Cumulative Investment

C .4074 30,000 30,000

A .1993 10,000 40,000

0 .1719 50,000 90,000

B .1667 20,000 110,000

we realize that we can select projects C and A for a cumulative investment of $40,000. This investment is less than the total

budget, $50,000, we currently have. On the other hand, if we add project 0, it will result in $90,000, which exceeds our

budgetary limits.

If such a case arises, we do not have a systematic way of obtaining a solution to maximize our net benefit - NPV.

If we consider only projects C and A, our total net benefit is,

NPVtotai = NPVc + NPVA = 12,222 + 1,993 = $14,215

However, with this combination, the remaining $10,000 ($50,000 -$40,000), un-invested budget can only earn the MROR. That

is, the NPV of the remaining $10,000 in budget is 0. To utilize $50,000, if we select only project 0, the NPVD is $8,594. This amount is smaller than the combination of C and A. Therefore, we will have to reject the selection of D. Another alternative is

to select C and B, which satisfies the budgetary requirement as well. If we select C and B, the net present value is,

NPVtotai = NPVc + NPVa 12,222 + 3,333 = $15,555

This is better than selecting C and A, which results in a NPV of $14,215.

The only way we can obtain the correct combination of C and B is through trial-and-error. Other sophisticated methods such as

linear programming may be used to achieve this result. These methods allow for maximization of a given function (NPV) with

external constraints. The discussion of these methods is beyond the scope of this book.

In practice, operating companies will use a combination of the methods to determine the feasibility of a project. For example, it

may use a criterion that the payback period is less than three years and the PIR should exceed 1.5. If both conditions are

satisfied, then it accounts for liquidity criterion as well as limited budget criterion.

Problem 2-34

A project has three possible alternatives:

Alternative Investment Annual Benefit Salvage Value

A $3,000 $800 $1,000 B $2,500 $700 $600 C $1,200 $600 $100

If the MROR is 10%, use PIR analysis to choose the correct alternative. Assume the useful life to be five years.

Problem 2-35

Consider four alternatives for the project. Assume the salvage value to be zero and the MROR td be 10%. Using PIR analysis, which alternative will you select? Assume the life of the project to be five years.

Alternative Investment Annual Benefit

A $15,000 $3,750 B $10,000 $2,750 C $3,000 $900 D $18,000 $4,750


Chap ter 2 - Economic Methods 37

Problem 2-36

A newly discovered oil field needs to be developed. The expected costs ore: $20 million in the first year, $80 million each in

second and the third years and $150 million in the fourth year. The net income will be $110 million in the fifth year declining at a

rate of 8/ per year. If the producing life of the project is 15 years is the project feasible? Assume the MROR to be 20% Use the

PIR method.

Problem 2-37

The following seven projects are being considered for possible investment. If $360,000 is available for investment, which

projects should be selected? All numbers given are in thousands.

Project A B C D E F G

Investment 150 140 100 70 120 50 140

Pljb ene fi ts 210 190 126 110 152 64 200

Problem 2-38

The following alternatives are considered for potential investment. If the available investment is $2,600, which projects should

be selected? Assume the MROR to be 10%.

Project Cost Annual Benefit Useful Life, Years Salvage Value

1 400 96 10 0

2 800 160 10 0

3 200 140 2 0

4 400 90 5 400

5 400 80 10 400

6 400 72 10 400

7 1,200 340 5 0

8 1,200 192 10 400

9 200 28 10 200

Use the method of a common denominator to adjust for varying useful lives. Use the PIR technique.

WORKS CITED

DeGarmo, E. P., W. G. Sullivan, and J. A. Bontadelli. Engineering Economy. New York, New York:

MacMillan Publishing Company, 1993. -

Park, C. S. Contemporary Engineering Economics. Reading, Massachusetts: Addison-Wesley Publishing

Company, 1993.

Steiner, H. M. Engineering Economic Principles. New York, New York: McGraw-Hill, 1992.

ADDITIONAL CASE STUDIES

Case Ssdv 2-5

IJassi R’mel field in Algeria is developed using mostly horizontalwells. The field is at a depth of approximately 5,000 ft., and is overlain by gas cap at the top and underlain by aquifer at the bottom. The gross thickness of oil rim varies between 25 to 40 feet. To prevent gas or water coning, the best solution is to develop the field using horizontal wells.

As a reservoir engineer working on the field, you are undertaking a performance of a new horizontal well - well I IRZ09 - which is going to he drilled, ’the wcil has been drilled using inverted high angle technique as shown in Case Study Figure 2-2. this technique involves drilling the entire oil column at a high angle and then turning the wellbore hack up through the reservoir at an angle exceeding 90 ° . ’l’he actual path of the well is shown in Case Study Figure 2-2 along with oil-water and nil-gas contacts. We need to complete this well. ’the company is going to case this hole, and asarcservoir engineer, your job is to figure out the optimal well

38 Mohan Kelkar, Ph.D., J. 0.

HRZO9 Path

1 01 550 653 750 000 000

Dsiancs is the Fast tract the Well head, m

Case Study Figuse 2-2’ Actual well path

The three perforation strategies you can use are: 1) perforating only slanted zone, 2) perforating both slanted and horizontal zones, and 3) perforating only horizontal zone A possible completion with completion in both slanted and horizontal zones is shown in Case Study Figure 2-3.

Case Study J’i’ure 23: Possible corupletton in both sdantcd and horizoneilpomotis

The cost of drilling the well is S million dollars, and the cost of perforation for all the three options is $1.3 million. The cost of transportation is S1.2 per barrel of oil. Assume that the net revenue per b"Irrel of oil is S25. Assume further that the expected rate of return on the investment is 15". . \isumc that the gas produced has to be flared. The cost of lost revenue is $0.2/idS( :i’. ’lie following data ate provided for all the three options:



Np (Mbbls) GOR (SCF/STB)

Year Horizontal Slanted Both Horizontal Slanted Both

1 275.06 171.92 275.06 1,685 2,808 2,246

2 275.06 171.92 275.06 5,054 5,054 5,054

3 275.06 171.92 275.06 6,177 6,177 8,423

4 242.57 172.39 268.58 14,599 8,984 17,407

5 132.65 134.52 148.92 21,337 12,353 23,583

6 78.93 92.11 77.21 25,829 14,599 25,829

7 50.64 66.88 49.49 29,198 16,845 26,391

8 35.53 50.64 33.43 32,567 19,091 27,514 9 26.38 40.32 22.17 34,252 20,776 30,321 10 20.11 33.33 13.6 36,498 21,899 35,375 11 15.1 27.32 8.6 38,744 23,022 40,428 12 11.47 22.74 5.73 42,113 23,583 47,166

13 8.03 19.49 4.39 47,728 24,145 52,781

Choose the best option. Will your answer be different if the previous wells have been completed in horizontal portion

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Cast Swdv 2-4

Cobra Oil Company operates five wells on the \7ogtsberger lease in Texas. Vogtsbergcr field, originally drilled in the 1940s, comprises numerous wells completed in several different zones. One of the wells produces from a depth of 5,200 ft and averages about 8.2 bopd and 35 bwpd, with 4 Mcfd of natural gas.

The well was experiencing downhole gas locking, which resulted in an inefficient lifting operation. To cure this problem, Cobra Oil Company hired Ecometer Inc., a company that specializes in well diagnostic problems. An acoustic depth measurement indicated that the liquid level was 148 ft. over the pump when the well was producing at stabilized conditions (continuous pumping). Casing pressure increased by about 0.1 psi per minute when the casing valves were closed, indicating that free gas was being produced and flowing up through the annular liquid. To determine stabilized conditions, a dynamometer test was run without shutting down the pumping unit. It showed that pump fihlage was 27% (Case Study Figure 2-4). Traveling and standing valve tests verified that the pump was in good condition.

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Poor downhole gas separation occurs when a high liquid level is present above the pump and the pump does not fill with liquid on the upstroke. In this case, the pump was set at 5,173 ft., which is above the formation. The completion is open hole (4-3/4 in) below the 5-1/2-in. casing from 5,235 to 5,247 ft. 1 he well was shut down for 10 minutes, then restarted with the same result �low pump fil]age. liven after a 20 minute shut down, poor fillage continued. As is often the case, the "poor boy" gas separator was not working effectively.

The producing bottomhole pressure was about 94 psi, but the reservoir pressure was unknown. To evaluate reservoir characteristics, Cobra performed an acoustic liquid-level pressure buildup test, allowing the well to be shut in for five days. Pressure buildup tests using computerized acoustic liquid level equipment are cost efficient, since the rods and pumps do not need to be pulled � a primary impediment to running a conventional pressure bomb. The result of this shut-in was loss of production for 5 days, plus an additional $1,000 for well test evaluation and data collection.

Although the well pressure did not stabilize, reservoir pressure (p5) was in excess of 1,000 psi, according to pressure buildup data. This data also showed that the well did not have skin damage, meaning re-stimulation was not necessary. permeability also was low. Furthermore, the data showed that calculated liquid flow into the wdllbore after shut in was only about 10 bpd of liquid, considerably less than the well’s production (Case Study Figure 2-5). Low liquid flow into the wdllbore may indicate that some cross flow occurs in the formation. This suggests that the pump should probably be set as low as possible in the well.

WO-11: 1WOCTA8 -1’0

6 40

0 25 W " icc 12575

Tirn5hr 6hr Stczdr J’Yguxr’ 2-5: Llllcoiatcdl7cjssld aJie.r-ulow

The high reservoir pressure (pt) indicated that, even after the gas interference problem was solved, very little additionalproduction would be obtained. Therefore, before doing any further work, Cobra knew that the primary benefits of the project would come from reduced electrical and maintenance costs resulting from less run time and better equipment loading. .\ substantial increase in production was not expected.

The rods and tubing were pulled and the pump serviced. In addition, a worn pull tube was replaced. It had probably deteriorated from the continuous "pounding" of the plunger when the pump was 20% to 50 0/s filled with liquid. The pump also had two standing valves, but only one was used when the pump was run back into the well. A tubing anchor was in the well when it was pulled, but it had a broken spring and was not run back into the well. As rerun, the tubing string consisted of 167 joints of 2-3/8-in, tubing, a seating nipple and a 2-3/8-in. collar-size gas separator that was six feet long. The bottom of the gas separator was placed 3 ft. from the bottom of the well in the middle of the producing formation.

A solid-state percentage timer, low voltage transformer and relay were installed in the motor panel. The goal was to reduce the amount of run time, since the 24-hour pump capacity exceeded the well’s producing capacity. The three items were purchased for Si 10. In addition the work over rig for one day cost S1,000.

After pumping the well overnight, the liquid level was at the seating nipple, 166.9 joints from the surface. Without shutting down the pumping unit, a dvnamoinctcr test was conducted. with the pump card showing about 30% fillage. With the liquid level at the pump and partial punip tillage occurring, the gas separator was operating efficiently. The well was shut-down for time minutes, then restarted and run for about ten minutes.

The first 36 strokes indicated full pump fillage (Case Study Figure 2-6, stroke 2 of 60). ’l’he next seven strokes showed that the well was ix’imig pumped down and, from stroke 43 to the final stroke 60, pump fillagc was relatively stablest around 35%. The collar-size go separator was working king effectively. A percentage 11111C r ’a as installed and set to run one third of the time.


Chapter 2 Economic Methods

41

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Case StuJy I’sorc 2-6: dtSus cc and dow nhole cards showthg pump d!Js,gc attic Sc ’ke 2

Liquid level and dynamometer tests were important for diagnosing well problems, but it was necessary to review the downhole equipment to get the full picture. The pressure buildup analysis helped determine whether the well needed to be re-stimulated, and what production level to expect.

The beam pump now operates five minutes on and ten minutes off, with a full pump until near the end of the on cycle. While there w efficiency has been only a slight increase in oil, ater and natural gas production, the overall electrical eciency was improved from 35% to

59O/ Power costs are approximately $108 per month, about half of the earlier level ($203 per month). With smoother operation and less stress on equipment, maintenance Costs are expected to decrease by about S25 per month. Assume the oil price to be S25 per barrel and the price of gas to be S5/MSCF.

As a newly hired production engineer for Cobra, you are asked to evaluate the cost effectiveness of the diagnosis and implementation. You are asked to find the answers to the following questions:

a) How long did it take to recover the initial investment? Assume that period has to be less than 3 years to make it viable. Ignore interest rate.

b) If we expect that in the subsequent wells, we do not have to conduct well tests, how long will it take to recover the initial investment? Ignore the interest rate.

C) If we expect to see the savings Over five year period for this well, what is the present value of future benefits assuming an interest rate of 15%? What is the profit at present conditions? Use tile original conditions as in (1).

d) If, in addition to electricity savings, we also can obtain an incremental production of 5% over the original production, what will the present value of future benefits and profit at present conditions? Use 15% interest rate.

Case Study 5,7

Renco Energy, Inc., a small independent in Tulsa, OK, operates several stripper well leases throughout northeastern Oklahoma. As is typical with stripper well leases, profit margins are thin and often nonexistent if leases contain problem wells. In many cases, high costs can be attributed to high water production with its associated operating costs. In managing its leases, Renco identified five problem wells that needed something different done to restore or improve profitability.

While looking for solutions, Renco began exploring whether the balanced oil recovery system BORS Lift, manufactured by ’loops ’l’echnohigy Iiccnsing, would be helpful. Fquipnient is quite different from a conventional rod pumping unit. From a mechanicalstandpoint, oil is lifted to the surface in a flexible tube (having appropriate valves and bottom weight) with a nylon strap (Case Study Figure 2-7).

Surf aes card

GPM

42 Mohan Kelkar, Ph.D., i.D.

At the surface, the strap is wound/unwound on a reel. As the full tube arrives at the surface, a sensor tells the tube valve to open, dumping the fluids into a small holding tank. The Current design does not capture casing head gas, thus losing potential gas revenues. � small transfer pump with a level controller periodically transfers fluids from the holding tank into the well flowline.

� wellsite computer controls both the depth from which fluid is lifted and the number of cycles per hour. These parameters, determined in the pre-installation analysis, can (and often need to) be adjusted at the wellhead - a simple matter. By achieving a balance, the system removes oil at a rate that corresponds to the natural migration of oil through the formation.

Not all wells are amenable to this system. To determine if a well is a candidate, Toups requires information about porosity, water saturation and permeability, in addition to standard well data (depth, perforations, oil / water / gas production, drive mechanism, oil and water gravities). Typically, well logs are analyzed and permeability estimated from the nearest available core or regional experience. As a final screening test, Toups logs the wellbore (preferably after a 3-day shut-in period) to determine fluid level and oil-water contact within the weilbore. If adequate oil entry is not occurring, wells are not good candidates.

Renco had several problem wells. Renco drilled and completed the Keefer 8 in the Bartlesville sand at about 1,400 ft. and Peru sand . at about 950 ft. There had been an old watcrulood, but not in the area where Keefer 8 was drilled. For more than a year (except for a

few instances after well work when it would produce 9 bopd for several days), the well produced essentially all water at around 80

0 bwpd, if one chose to pump that much.

The Ogelsby #1 is on a single-well lease and produces from the Peru sand at about 950 ft. The well could produce at 80%, or higher,

O water cut. But since the lease does not have a saltwater disposal well, water must he hauled at a cost of $600 per 80-bbl tank of oil sold ($7.50 per bbl oil sold). Operations were uneconomic, but to hold the lease, the well was operated on a time clock, producing for two, 15-mm periods a day. Production averaged about 0.5 bopd.

Renco operates four wells on the Comstock lease. ftc Comstock I produces from the Wayside sand at about 500 ft. Sand production, with its associated mechanical and well service needs, was a real problem in the Comstock 1. Pulling frequency was excessive, often being only days (or weeks apart). Well servicing costs were excessive - averaging over $1,000 per month. With production of only 3.5 bopd, the lease was unprofitable if the problems associated with sand production could not he solved. In fact, the well had been inactive for a couple of months when the BORS system was installed.

The operator installed B( )RS units on two Bartlesville completions on the Cominonwcalth lease. Prior to installation, one of the wells produced about 3 bopd plus 70 l.wpd, which is marginally economic. The second well produced a trace of oil plus 70 bwpd, but they could not "keep ahead" of the water production. Like the Keefer 8, the phlcni for the second well was obvious - try something or plug the well.

Renco installed I3ORS unit on several wells. Results from 5 wells are available in Case Study Table 2-1.



Case Study Table’ 2-1: Composite results ofaltemthve ardfi’clal life system

Production bopd/bwpd

Time system has

been Reduction in Reduction in installed, power cost, operating cost,

Well months Before After S/month s/month General Comments

Keefer 8 Unprofitable "water’ well 9 0/80 �

Bartlesville & 6.0/3.0 65 700 converted to profitable

� Peru sands producer

Essentially eliminated water Oglesby 1 Peru 9 0.5/3.0 2.7/trace 60 0 hauling costs and increased oil sand production

Comstock 1 4 3.5/trace 3.5/trace 60 1,000 Unit has operated trouble-free

çrylcr A) with no well service cots

Wayside sand

Commonwealth 4 little/70 7/trace 40 500 Essentially eliminated water lease production, converting too

unprofitable wells to profitable

Too Bartlesville 4 trace/70 4/trace 40 500 producers

comolctions

The improved artificial system is still work in progress. Equipment frequently breaks down and some of the parts have to be replaced. On an average, the cost of maintenance per unit is $2,400 per year spread over the year. Also, not every well has been successful. Based on Toup’s experience with other wells, about 80% of the wells will benefit from such installation.

As a petroleum engineer working for Renco, you are assigned to investigate the feasibility of BORS so that the same type of system can be installed in other wells. Assume that the average response from a successful well can be determined from averaging the responses from the table above. Assume that the price of oil is S25/barrel and the life of a well operating under given conditions to be three years. Answer the following questions:

a) What is the payback period for this new installation? Calculate discounted payback. b) What is the NPV if the ?s[ROR is 15% per year? c) Should Renco purchase the unit or lease it?

Cc’ Study 2-8

In most reservoirs, after the oil is produced through primary depletion, the next stage is to use water flooding to produce more oil. Significant quantities of oil still remain in the reservoir after water flood. Therefore, other tertiary oil recovery processes can be used to receiver additional oil from such reservoirs.

One of such experiment was conducted in David Pool in L]oydminstcr, Alberta, Canada. David Pool was first depleted using primary recovery, followed by water injection, and then ultimately a tertiary recovery process, called alkalinc-surfactant polymer (ASP) flooding. Between 1987 and 1997, David Pool was subjected to ASP, and was closely monitored for oil recovery from tertiary oil recovery. Case Study Figure 2-8 shows the size of the Pool. Notice that ASP flood was only applied to part of the field. The North, South and West parts of the pool were only water flooded.


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-South

Case Study Fi’ure 2-8; Pa aid Pool oetpav isopa oh

\Ve are going to evaluate the ASP flood in the middle part of the reservoir. The pore volume (P\’) of the ASP flooded area is 12.5 million barrels. The oil in place (OOIP) in that area is 9.4 million barrels at standard conditions. The chemical flood - ASP flood - required that we design a mixture of chemicals - to be injected in the reservoir, so that we can displace additional oil from the reservoir. After spending in 1986 about $150,000 for laboratory and reservoir studies, the oil company determined that the best combination of the chemicals is 1% by weight Na2CO3 sJkah), 0.2% by weight Petrostep B-100 (surfactant) and 600 mg/Litre Alcoflood Pusher 1000 F (polymer. The company also determined that 0.3 1’V of chemical (ASP) slug needs to be injected in the reservoir for succcssful,\SP flood, and that slug will be followed by 0.2 PV of 400 mg/Litre Alcoflood Pusher polymer slug. The cost Of Na2CO3 was S0.06/lb, the cost of surfactant was $1.00/lb and the cost of polymer was $0.90/lb. Assume that the density of ASP and polymer slug is 62.4 lb/ft. Assume that 0.1 PV of ASP slug was injected in 1987, 0.2 P N I of ASP slug was injected in 1988 and 0.2 PV of post-flush, polymer only, slug was injected in 1989. The operating cost was $3.50 per barrel of oil produced. To initiate the ASP flood, additional 12 wells were drilled and completed in 1986 at a cost of $300,000 per well. Since the chemicals in ASP plants are sensitive to salinity of injected water, a special softening plant was built in 1986 to soften the water. In addition, special separation facility to account for emulsion of oil-water had to be built. The total incremental cost of facilities was S1.1 million. The incremental production and the price per barrel of oil is provided below. The field was assumed to reach economic limit at the end of 1997. The price reflects an average price received for the oil during that year - which can be used to calculate the gross revenue during that year.

Oil Price Year Mbbl ($/hbl) 1986 0 1987 113 18.1 1988 259 12.65 1989 290 15.94 1990 275 1788 1991 249 13.72 1992 216 15.69 1993 193 13.98 1994 169 12.94 1995 127 14.35 1996 69 18.12 1997 16 16.91



Using the numbers above, calculate the payback period and the NPV of the project if the MROR is 15%. Also calculate ROR for the project. is the project economical?

Fast forward to current conditions. As a reservoir engineer, working for an oil company, you are investigating ASP flood to be implemented in present year. Assume that the production profile starting next year, will be similar to above table, except that year 1987 and onwards will change to next year and onwards. The cost structure has changed though. You have already spent $280,000 on the lab study which has determined that 035 1/V of chemical slug needs to be injected in the reservoir for successful ASP flood, and that slug will be followed by 0.2 PV of 400 mg/Litre Alcoflood Pusher polymer slug. The cost of NO2CO3 is $0.08/lb, the cost of surfactant is Si.50/lb and the cost of polymer is S1.05/lb. Assume that the density of ASP and polymer slug is 62.4 lb/ft 5 . Assume that 01 PV of ASP slug will be injected next year, 0.25 PV of ASP slug will be injected in year after and 0.2 PV of post.-flush, polymer only, slug will be injected in the third year. The operating cost will he S5.00 per barrel. To initiate the ASP flood, an additional 12 wells will have to he drilled and completed at a cost of $600,000 per well. That can be accomplished this year. The plant for softening the water and the cost of separation combined will be $1.6 million which will have to be spent this year. For economic calculation purposes, we will assume that the price of oil next year will he $40 per barrel and will remain steady throughout the life of the project.

Calculate the payout period. Calculate NPV if MROR is 20°/s per year. What is the cost of chemical per incremental barrel of oil? Is the project feasible? If to account for uncertainty, we assume that only half of the projected incremental oil is produced every year, how will the economics be affected?

Case Stud 2-9

Granite Wash is a formation at a depth of more than 12,000 ft. It is located in the Western part of Oklahoma. The permeability is less than 0.001 md and the well will not typically produce without fracturing it. After the success of horizontal wells in Barnett Shale (another very tight formation), operators are exploring the possibility of using horizontal wells to exploit Granite Wash formation, in the last few years, some operators have drilled the horizontal wells with some success. As a production engineer, your job is to evaluate the efficacy of both the vertical and horizontal wells and recommend your boss which type of well is a better option. The vertical wells have been used in this formation for a very long time and typical production data from the best vertical well is shown below. Assume that the production from a vertical well can be as low as 33% of the given data. With limited production data available from horizontal wells, the production profile is estimated based on decline curve analysis. This is the data from the worst horizontal well. Assume that the production data from a horizontal well can be as high as 100% above the worst horizontal well. Production data for water is not provided; however, most wells will produce significant quantities of water over the life of the well. The condensate production for both the vertical and horizontal wells can vary between 30 STB/I’VIMSCF to about 70 STB/M1\ISCF. Most wells produce naturally for the first yean lloevever, starting in the second year, compressor will have to be installed to reduce the well head pressure, and starting third year, gas lift need to be installed to lift both the condensate and water. The economic parameters are provided as below:

� \’ertical well drilling and completion costs: S2.2 million to S2.8 million � horizontal well drilling and completion costs: S6.2 million to $8.8 million � Cost of operating vertical well: In year 1, S1,000/month; starting year 2, the well will have to be fitted with a compressor

increasing the operating cost to S6,000/month. In year 3, to lift the liquids, we will have to install gas lift ($70,000. In addition, the cost of operation will increase to Si 1,000/month (due to compression required for artificial lift plus compression at the well site

� Cost of operating horizontal well: In year I, S 1,300/month; starting year 2, the well will have to be fitted with a compressor increasing the operating cost to S7,000/month. In year 3, to lift the liquids, we will have to install gas lift S70,000. In addition, the cost of operation will increase to S12,500/month (due to compression required for artificial lift plus compression at the well site)

� Price of gas: $5/MSCF (constant over the entire period) to S5/MSCI 5 in the first year escalating at 2%/year � Price of condensate: S60/S’I’B (constant over the entire period) to S60/STB in the first year escalating at 2%/year � Assume that the abandonment cost is negligible. The MROR is 15% per year compounded daily.

Evaluate the data using this information. Use various economic parameters (NPV, bR, PIR, payback, etc.) to determine which one is the best option. Calculate the pessimistic and optimistic development costs for both the options. Evaluate the risks involved in choosing one or the other option. I Iltunatcly, what do you think is the best option and why?

Vertical Well Horizontal Well Year

MSCF MSCF 800,199 1324,946

2 319,642 445,137 3 186,869 262,388

127,682 175,438 3 95,014 137.830 6 74,645 109.383


7 60,866 84,100 8 51,003 69,519 9 43,639 63,520 10 17,958 57,992 11 33,458 49,054 12 29,817 39,978 13 26,819 34,705 14 24,313 29,088 15 22,191 26,776 16 20,373 22,952 17 18,802 21,161 18 17,432 22,992 19 16,228 21,936 20 15,162 21,093 21 14,214 21,882 22 13,365 24,029 23 12,601 15,638 24 11,911 19,756 25 11,284 18,611

Cjse Study 2-JO

CamWest, Inc., an independent based in Denver, Colorado, and McKinney, Texas, operates more than 120 oil wells in Lander, Wyoming. These wells are located in a remote region that is difficult to reach due to the rough terrain and harsh winter climate. To better manage its waterflood and oil production activities, CamWest installed web-based pump-off controllers.

For CamWest, like many producers, electrical and well pulling costs are a significant part of the total lease operating expense. In remote areas, operations with manual data retrieval and site inspection complicate things further. historically, the company’s pump- off controls were activated based on pre-set timing mechanisms not linked to changing well fluid levels. Furthermore, unit failures

0 were primarily discovered by inspecting each individual well throug

m h field visits. Many of these breakdowns were the direct result of

over-pumping’ (fluid pound), leading to premature pump or rod failures. As a result, the company experienced higher expenditures and lost production revenue.

To minimize electricity costs, lower repair expenditures and optimize production, Cam West implemented a wireless, electronic pump-off controller (POC) that could be monitored and controlled via the Internet. Aunion Technologies, Inc., a Dallas-based provider of turnkey, web-based telemetry services for oil and gas Producers and gas gatherers, provided and installed pump-off control R1’Us and a spread-spectrum wireless communication network. The company provides Internet monitoring and support services that include POC software applications within an application service provider (ASP) format. The first 10 POCs were installed in November 2000.

O

By March 2001,49 POCs had been installed with 51 units being in place at the end of May 2001. That is, out of the total of 59 units installed, 51 were operational by May 2001.

CamWcst utilizes Aurion’s vebite to monitor fluid level, production and pump-off controller activity, and to control load limits, stroke and load references to activate the pump-off Control. The wehsitc (Case Study Figure 29) also allows both management and field operations personnel to access various pump-off control reports, such as morning, event and runtime reports; analyze trending and dynagraph card graphics; and allow multiple users to comment on-line about well maintenance issues. Additionally, the website allows users to respond iriOre quickly to equipment problems and production interruption through the receipt of POC data about four times per day.



Wait 7/02 Avomga Data 1tna Status Notes Runbrn,u Runtime CST

Wigitaryl 111 53 0

NvtfmARtit NtwrnJRn .a Uoetnelflzn li Nniial Run

11.0 7/03 12:1S iSA ISA 7/03 2l4 J. 5 53 7103 12:23 I 15.0 7/03 i4 100 100 7103 321

5$ 7t 2d4 5A 5. 14 iI03 12- O 55 &S 7103 1818

" rrPhplsoda-$lf-S 24.0 24.0 1103 12t22 N5wm41 Burt

A if Vie ceita 1rif Stun i s jean easanq lof a aii tit.s *, zSO urttB th asS liSa 001 boat painS twSa, ii ir,i 7 aiurnt b t, Ft I iSbrnJ 1nn1is Edit

Case Study Figure 2-9: An on-line "Morning Report" provides operator persoiatiel the most recent opera dog di to from all weh-basedpurnp -off controllers. Personnel can access the report fivm any loco iion with Internet access.

By using the pump-off control RTV unit and web software, the operator, CamWest, reduced actual electricity usage by maximizing pumping activity during periods of high fluid levels, while eliminating pumping during periods of low fluid levels. Additionally, peak electricity demand usage was reduced by decreasing the rate of pumping activity during peak period.

As a new engineer just hired by CamWest, your job is to present the cost effectiveness of these l’OC’s to upper management. The cost of installation of each POC is S2,500. In addition, Aurion Technology charges $5.99 per month fee for service and monitoring per well. The production at the time of installation per well was 20 bbl/day, and it is declining at a rate of 15% per year. Based on the evaluation of the data, electricity costs have dropped from an average of $1.20/bbl of oil to about S1.04/bbl of oil. In addition, CamWest expects that additional economic value can be received through production optimization, and by reducing rod and well repair expenditures. Although no firm data is available to you, you can assume an average additional savings of S500 per year per well through reduction in rod and repair expenses and production optimization. Assuming MROR of 20%, calculate:

a) Payback period in months by discounting the cash flow b) NP\ if we assume that the well will produce for 4 years.

What will be the NPV if the $500 savings per year does not materialize’ ,

case Smdy 2-fl

Silver Pines Energy Corporation operates the Sun T1 I and Stuart City Fields in LaSalle Counts, Texas. The leases have 32 wells producing either sour gas from Edwards lame at about 11,000 ft. deep or Sweet gas from Olmos gas sand at a depth of 7,500 ft. Both formations arc tight and, as production matured, the wells began loading water. Silver Pines started using soap sticks to keep the wells unloaded. The soap sticks reduced the interfacial tension between water and gas and, hence, reduce the loading velocity required to lift the water. The soap sticks were launched by a gauger each day. I lowever, Silver Pines discovered that the soap sticks launched by the gauger would not keep the wells unloaded for a full 24 hours. Production was being lost, and more frequent visits by a contract gauger is not an economic option.

&J Oilfield and Electric Service (J&J) approached Silver Pines to propose a solution. J&J had’-just invented in automatic soap launcher as shown in Case Study Figure 2-10.

’a

’a ’a


r

H

Sen dyFi,c uromdlie soap Ia all chr

The cost of installing the soap launcher is about S5,000 per well- Once the launcher is installed, one soap stick could be launched by gauger during his daily rounds, and two additional soap sticks will be automatically launched at 8 hour intervals. The cost of additional soap sticks is about $6 per day, per well. Installation of these launchers takes just a few hours, and if not successful, can be re-installed at other well at an additional cost of SI ,000 per well. j&J claims that the launchers are reliable. If any of the launcher breaks down, it can be fixed at an average cost of $500 per repair. J&J claims 90i/o reliability for the launcher per year. That is, 90% of the installed launchers will not breakdown during a year.

J &J also claims that the launchers increase the production for 90% of the wells. If the launcher is unsuccessful on a well, we can assume that we will not get any incremental production. The same launcher can be used for another well at a cost of S -1,000.

As an example of the success enjoyed by J&) Launehers,J&J provided the following information:

Gas Production MSCFD

Well Before After

1 18 70

2 83 120

3 17 135

4 120 185

These installations took place about four months ago, and J&J claims that the production increase has sustained over this period.

As an engineer working for Silver Pines, you are in charge of evaluating the efficacy of installing launchers at the well Site. You assume that the average performance of a well will be consistent with what has been reported by J&J in their field experience. however, you also know that the production from the wells is declining at a rate of about 15% per year. You can assume that the incremental production will also decline at about the same rate.

Assume that the price of gas is S2/MS( l and the MROR to he 15% per year. Assume that you will install the launcher on 10 gas wells at a time, followed by another ten wells after 3 months, till the installation is complete on all the thirty-two wells. If the installation is not successful in one of the wells, we will reinstall the launcher on another well, thus saving the cost of bu Ing a new launcher and spending only SI ,000 for reinstallation.

Assume that the last two wells, where the installation will be done after 9 months, will be successful. l)o your calculations on a monthly basis.

a) Assuming the life of the wells to be live years, is It worth iiistalllng the launcher? b) What is the cost of thc launcher (installation plus operation) per \lS( F of gas produced? e If the gas price changes to SI /i\ISCP, how will the N l’V change?


Chapter 2 Economic Methods

Case Mudt 22

Citation Oil and Gas Corporation operates \Xasioger Unit of Bemis Scbutts Oil Field in the State of Kansas, The location of the wells is shown in Case Study Figure 2-11.

CITATION OIL AND GAS CORP. 0 Commingled Arbuckle Producers

Wasinger Unit 31111990 Bemis Shutts Field , TopekaILkC Producers

Ellis Co.. Kansas 31111990 Unit Boundary

EI2Sec.21 r

TIIS

RI8W I 17SWD

14 + 12

0

8 2 18

4 0

: 1 o

7 18 5

6 + 0 19

0

Case Stvclv Figure 211: Well locations

Several wells in the Unit were either temporarily abandoned, or were about to be abandoned because of large water production. The cost of water disposal is about S0.12 per barrel.

’ITORCO, Inc., a service company specialized in polymer gel treatment, suggested to Citation that by treating the well with polymer gel, the overall water production could he reduced. This will increase the oil production and hence extend the life of the well. The cost of polymer treatment is roughly S135,000 per well. This includes the cost of polymer gel treatment, rig costs, stimulation, down time, frac trucks, prep time and change in artificial lift equipment.

Depending on the type of job, the cost could be as low as S60,000 and as high as S200,000. Citation gave contract to TIORCO, Inc to treat well # I in 1997. This well is shown in Case Study Figure 2-11. Prior to November 1997, when well I was treated, the Unit was producing about 42 barrels of oil declining at 6.5% per year. The WOR was 70 and increasing at a rate of 9.5% per year. Economic cut-off calculations indicated that when the ’tX’OR reaches a value of 100, it is uneconomical to produce oil from the Unit.

Well I was treated in November of 1997. The well was not producing at that time. The production increased from 42 to 100 barrels from the Unit. The unit \VOR decreased to 55. Since that time WOR has increased at a rate of 9.5 0/a per year, however, the oil production has declined at a rate of 31% per year.

a) I low much are the remaining reserves without treatment? h I low much incremental production will you expect from Well # 1? One measure of success is the cost per incremental

barrel. You prefer to keep it below S3/bbl. c) At what point you will abandon the Unit based on W( )R value? d) What is the rate of return on your investment? c) If wells 2, 3, 4, 5 and l2 arc candidates for gel treatment, will you recommend gel treatment for those wells? t) If you want to be conservative, and assume that the incremental production from a well will go to 75 b/d, will you

recommend it? Assume that \X( )R xvill decrease to 55.


Do the calculations on a monthly basis. Assume MROR to he 20%.

Assume the net revenue per barrel of oil to he S11; however, assume that this does not include the Cost of water disposal. For oil production declining at a fixed percentage, we can assume exponential decline. For this tvpc of decline the oil rate at any given time is calculated as:

q0 q0te_t

where D is the declining rate and q01 is the initial rate. Integrating rate w.r.t, time, we can calculate hydrocarbons produced when the rate has declined to q by using the equation:

N _q0gq0

D

(.rssc Stride 2-13

Parallel Petroleum Corporation (NASDAQ: PLLL) on September 151, 2009 announced that it has entered into a definitive agreement for the Company to be acquired by an affiliate of Apollo Global Management, LLC, a leading global alternative asset manager, in a transaction valued at approximately S483 million, including the assumption or repayment of approximately $351 million of net indebtedness. The agreement was unanimously approved by Parallel’s Board of Directors.

Commenting on the transaction, Sam Oh, Partner at Apollo, said, ’AVc believe Parallel’s high-quality assets and its outstanding management team will he a positive addition to our investment portfolio and we look forward to working with the Company."

Larry Oldham, Parallel’s President and Chief Executive Officer commented, "Apollo’s interest in the Company is a clear recognition of the attractiveness of Parallel, its business plan and the success that has been achieved Apollo has a strong track record of growing businesses. Under its ownership, Parallel will be better capitalized to execute its current business plan and develop new opportunities for growth."

Your job is to evaluate this transaction to see who got a good deal out of it. Most of the information I am providing is taken from a second quarter 2009 presentation provided by Parallel Petroleum. Here we summarize all the properties operated by Parallel. In the table below, we provide the name, net production either in MSCFD (1000’s of cubic feet per day) or STB/D (barrels per day), the remaining proved reserves (oil and gas remaining in the ground but vet to be produced) in BCF (lxIO 9 Cubic feet) or MMSTBO (millions of barrels), and the upside potential in each of the field.

Parallel’s Properties

Net Daily Reserves Upside Property

Production

Gas

Barnett Shale Project 10,344 MSCI’I) 19.5 BCF Additional infill wells potential

31) seismic being collected - additional well locations New Mexico Wolfcamp 10,152 MS(FD 22.4 BCF otcntial

AR

Permian Basin of West Texas

Diamond NI Canyon Reef 606 STB/D 4.1 M1\lBO 30 MMBOE gross potential based on analog; additional CO2 flood potential

Carm-ann San Andres 303 SIB/I) 5.4 MMBO lnfill and step out drilling

I larris San Andres 412 SIB/I) 5.5 ’tiNillO Infill and step out drilling; \vaterflrioding potential

Fullerton San Andres 1,421 5Th/I) 3.2 ’tIMI-IO Immature warcrflood

Diamond NI Shallow 104 SIB/I) 1.0 NIMBO Infill Potential

Oilier Pcrmian Basin 220 SlB/l) 1.6 NINIBO lnfill potential

On Shore Gulf Coast 262 S. ..I -I/I) 2.2 MMBO Infill well potential

P specific interest ainnng all these prnperiics is Bai ileir Shale production. In February of 2009, Bir:i]ld entered into an agreement with Chesapeake. Parallel currently has pi )tcnfi:Il 2-I0 li ciii ins where the wells can he drilled. tin flu nina xlv. due to linv gas prices and



prior commitments, Parallel did not have money to contribute its portion of the drilling and operating costs. Parallel has 35% working interest in these wells and Chesapeake has 65%. That is, Parallel has to pay for 35% of the drilling, completion and the operating costs. Instead, Chesapeake will "carry" Parallel’s interest in these wells. That is, Chesapeake will drill these wells by paying for all the costs for drilling, completion and production. As a penalty, Chesapeake will recover ISOb/o of the drilling and completion costs from the net revenue before Parallel is allowed to "back in" with 17.5% of the working interest. For example, if the cost of drilling and completion is $1 million, and each year the net revenue is S250,000, then Chesapeake will have to be paid $1.5 million before Parallel would receive anything. That is, it will take six years before Parallel can back in with 17.5% interest. At that point, Parallel will start receiving 17.5% of the net revenue.

The cost of drilling and completing a typical Barnett well is about S3.2 million. The operating cost of a well is about S4,000 per month. Assume a 20 year life for a well. According to Chesapeake’s website, the production profile for a t)pical Barnett Shale well is provided and it is reproduced below: The production provided is for each year.

Year BCF 1 0.650

� 2 0.228 � 3 0.164

4 0.140 5 0.129 6 0.119 7 0.109 :

� 8 0.100 9 0.092 10 0.085

� 11 0.078 12 0.072 13 0.066

� 14 0.061 15 0.056 16 0.052 17 0.047 18 0.044 19 0.040 20 0.037

Assume that royalty interest is 25%. For gas prices, assume S3.50 for the first year and S4.50 for the rest of the years, held flat.

Using this information, answer the following questions:

a) Using a purchase price of S483 million, and using S70/barrcl oil price, what was the conversion factor for price equivalence between MSCF1) to barrels per day? Use the standard rule to calculate equivalent barrels of oil.

b) Using the same conversion factor, convert all the gas reserves into equivalent oil reserves and assume S20/barrel in the ground price and calculate the worth of the reserves. Is this price higher or lower than what Apollo paid? Knowing

� what Apollo paid, how much did Apollo pay for an equivalent barrel in the ground? c Using the information about Barnett Shale well, determine the NP\T for each well from both Chesapeake and Parallel’s

perspectives. Assume that the nominal interest rate is 15% per year compounded continuously. d) Assume that Chesapeake is planning to drill 30 wells each year so that it will take eight years to drill all of the well ,,,-.

Knowing the N 1 3 \7 for each well, how much is the present worth to Parallel for all those horizontal wells which will be drilled in the future? For convenience, assume that the NP\T for the first set of 30 wells will be collected at the

� beginning of year 2011. You need to know present worth of these wells for Parallel at the beginning of year 2010. � Using Part 1, do you think that Parallel received any compensation from Apollo for these horizontal \VClls?

e) If, on a long-term basis, you believe that the gas price on a MSCF basis is going to be 1/15 of the price of barrel of oil, �

hew much will you pay for the property based on current production if you expect the (ill price to remain steady at S70/barrel? I low much will you pay based on reserves if you par S20/barrel in the ground?

1) In making this deal, under what conditions would Apollo get the worst end of the’bargain? I/oiler what conditions would Parallel get the worst end of the bargain?

1)o you believe that Apollo paid anything to Parallel for upside potential?


lq

This chapter covers the importance of taxes and contract agreements on decision making in the oil and

gas industry. Although taxes, in general, are not unique to the oil and gas industry, certain rules and

regulations related to taxes are uniquely applicable to the upstream industry. Complicating the impact

of taxes is the type of contract signed between the operator and mineral owner.

In the Basic Tax Structure section, basic tax concepts are explained in a simplified manner. These

concepts are common to any industry and are not necessarily restricted to the oil and gas industry.

Understanding of these basic concepts is important before other complicated tax consequences are

considered. The two concepts that are discussed in this section are depreciation and income taxes.

Depreciation relates to deduction of part of the durable equipment costs from potential taxes; whereas,

income taxes relate to part of the income received by the government. Both concepts are further

explained through multiple examples.

In the Domestic Agreements section, the effect of domestic (U.S.) agreements on overall cash flow

analysis is discussed in detail. These agreements are signed between the mineral owner and oil

company. Due to the terms of the agreements and the overriding tax laws, the cash flow analysis may

be significantly affected. In the Chapter 1, we discussed basic terminology in the oil and gas industry.

Additional terms are explained in this chapter. In the second part, the concept of depletion is discussed.

Similar to depreciation, depletion is also a tax benefit provided to oil companies. Depletion relates to

deduction of costs associated with locating hydrocarbons. Although used in the United States, it is rarely

used in other countries. In the third part, the effect of income taxes is discussed for domestic contracts.

In the International Contracts section, international agreements, signed between governments or their

national oil companies and international oil companies, are discussed. International agreements are

becoming more common. After a brief history and discussion of the purpose of these international

agreements, three common types of agreements - concession, production sharing, and service - are

explained. The host country government exerts a progressively larger amount control as we move from

the concession agreement to the service agreement. The tax consequences of these agreements are

presented in general terms. Although the specific agreements signed can vary from country to country,

we do not cover the impact of specific agreements on economic analysis. Instead, we cover only the

basic concepts of the different types of agreements. The sources cited in this chapter provide greater

detail.

The Summary section, we will review the concepts discussed in this chapter. It also highlights the

similarities and differences between the various agreements.

BASIC TAX STRUCTURE

In this section, we discuss two important concepts common to most types of tax analysis. The only

requirement is that the business invests in durable (lasting more than one year) goods. For an industry

exclusively service, requiring no investment in durable goods, the concept of depreciation may not be

relevant. Otherwise, an understanding of depreciation and income taxes is, predominately, a universal

requirement for proper economic evaluation of any project.


Chapter 3- Economic Analysis

DEPRECIATION

Depreciation, in the literal sense, means a decrease in value. For example, if you buy a car today,

one year from today, the same car cannot be sold at the same price. The value of the car is

depreciated. As time goes by, the price of the car will depreciate further. In economic analysis,

depreciation has a different meaning. Realizing that the existence of a corporation is perpetual,

depreciation is treated as a replacement cost. That is, if you buy a piece of equipment that has a

useful life greater than one year, the assumption is that you will have to replace that piece of the

equipment at the end of its useful life. Therefore, some money has to be set aside during the useful

life so that the equipment can be replaced. This yearly replacement cost is called depreciation; it is

an allocation of the cost of the equipment spread over its useful life. The government allows the

corporation to set aside this money for future replacement without taxing it.

Although the depreciation is most commonly used for durable equipment, it is also used for other

properties. To generalize, a property can be depreciated if it satisfies the following characteristics

(Newman, 1991):

� It must be used in business or production of income.

� It must have a life longer than one year.

� It must be something that wears out, decays, becomes obsolete or loses value from natural

causes.

These characteristics are most appropriate for equipment; however, they may also be applied to

buildings, patents and copyrights.

Investment in such property is called capitalized cost. In production operations, this cost will include

the cost of casing, piping, tank batteries, pumps, compressors, buildings, etc. Considering that these

costs are incurred over tangible items, they are sometimes called tangible costs. Costs related to

patents and copyrights are intangible costs; however, they still need to be capitalized. In contrast,

the costs incurred for regular maintenance, upkeep, utilities, and labor costs, are considered

operating costs and can be deducted from the operating revenue in the same year for tax purposes.

Although the distinction between capitalized costs and operating costs is clear in most instances,

sometimes the distinction is difficult to achieve. In the context of the petroleum industry, a good

example would be the costs associated with drilling a well. These include (Campbell, 1987):

� Costs related to agreements with drilling contractors.

� Surveys related to the location of a well.

� Road costs and dirt work related to well location.

� Rig transportation and set up costs.

� Fuel, water, and drilling mud costs.

� Labor costs including supervisors, well site geologists, and the testing of well.

� Stimulation costs.

� Cementing costs (does not include casing).

� Surface damage costs.


Sagdi

Highlight

Since the drilling costs are incurred before production begins, they cannot be considered operating

costs. On the other hand, the costs for labor, utilities, drilling mud, and rig rental cannot be

considered tangible costs. How to treat these costs is indeed a legislative decision. The industry calls

these costs intangible drilling costs (IDC). Different countries have different rules with respect to

categorizing IDC.

Before we can understand the different methods for calculating the depreciation, we need to

understand some basic definitions.

COST BASIS

Cost basis represents the cost of acquiring a property. This is the total cost that will be

capitalized over the life of the property. This cost includes the cost of the property and all other

incidental expenses such as installation, freight, and site preparation.

BOOK VALUE (BK)

Book value is cumulative depreciation subtracted from the cost basis of the property. In year k, the book value can be defined as,

Bk = cost basis - D Equation 31

where Dj is the depreciation in yearj. The amount depreciated each year will depend upon the

method used.

SALVAGE VALUE

Salvage value is the price obtained from the sale of a property at the end of its useful life. In

many instances, for tax purposes, this is assumed to be zero.

REPAIR COSTS

Repair costs are the costs related to keeping the property in ordinary, efficient operating

condition.

CAPITALIZED COSTS

Capitalized costs pertaining to the existing property are costs related to alterations, additions, or

improvements that increase property’s useful life or value or make it adaptable for a different

use.

The distinction between repair costs and capitalized costs for an existing property is useful

because repair costs can be deducted as operating costs; whereas, capitalized costs will have to

be depreciated. This difference can have significant tax consequences, as we will examine in the

Domestic Agreements section.


Chapter 3- Economicflnalysis 3

USEFUL LIFE

Useful life is the time over which the property is kept in productive use.

TAX LIFE

Tax life is the period of time over which depreciation is calculated to offset taxable income. This

period is usually shorter than the useful life of the property. It is determined by the tax code of a

particular country depending on the type of property. For example, in the United States the tax

life can be as short as three years for taxi cabs and race horses, and as long as 31.5 years for

non-residential real estate.

Some popular methods of estimating the cost of depreciation are discussed below.

Example 3-1

An oil company ordered a new submersible pump for improving production from an old well. The pump costs $20,000. The

company will have to pay 7% sales tax on the pump. Additionally, the shipping cost is $300, and the installation cost is $1,000.

What is the cost basis for the pump?

Solution 3-1

In calculating the cost basis (any cost associated with the purchase), shipping, installation and site preparation should be

included. Therefore, the cost basis for the pump is,

Cost of Pump $20,000

Sales Tax (7%) $1,400

Shipping Cost $300

Installation Cost $1,000

Cost Basis $22,700

In calculating cost basis, it is critical that all costs, including some non-monetary transactions, are included. One example of a

non-monetary transaction is the trade-in benefit. If you trade in an oil pump for a new pump so that the cost of the new pump is

$18,000, the cost basis will not be reduced by $2,000. Since the trade-in value is considered to be $2,000, it will be added in the

cost basis as a separate item.

STRAIGHT-LINE DEPRECIATION

Straight-line depreciation is one of the easiest methods of estimating depreciation.

Mathematically, the depreciation is calculated as:

Dk (B0 - S)

Equation 3-2

where Dk is the depreciation in year k, B0 is the cost basis, S is the salvage value at the end of

tax life, and n is the number of years of tax life.

Alternately, depreciation can also be calculated as,

Bk_ 1 � S

n-k+1 Equation 3-3

Mohan Ke/kar, Ph.D., J.D.

where Bk_i is the book value at the end of period k - 1. Both equations will give the same result. The book value is calculated using Equation 3-1.

A common term, S. in Equation 3-2 and Equation 3-3, is difficult to estimate at the beginning of

taxable life for some properties. Many tax codes simplify the calculations by assuming that the

salvage value is zero at the end of tax life.

Example 3-2

Consider a piece of equipment purchased for $5,000. If the tax life is five years, and the salvage value is zero, calculate the depreciation schedule over the five year period.

Solution 3-2

Using Equation 3-2,

B0 �S 1 Dk = = (5,000) = $1,000

Therefore, the depreciation schedule can be written as,

Book Value at the

Year Depreciation Beginning of the Year

1 1,000 5,000 2 1,000 4,000 3 1,000 3,000 4 1,000 2,000

5 1,000 1,000

Using Equation 3-3, we could reach the same result. For example, in year 2, the book value at the end of year 1 is $4,000; therefore, depreciation is,

B1 �S 4,000 - DECLINING BALANCE DEPRECIATION

The declining balance method allows depreciation at a faster rate than the straight-line method.

The amount of depreciation in each year is a certain percentage of the book value at the end of

the previous year. This percentage is constant throughout the tax life of the property and is

usually defined as the ratio of declining balance rate to tax life. For example, if the declining

balance rate is 200% and the tax life of the property is five years, then the percentage of the

book value used as depreciation is 200%/5=40% per year. If we denote this percentage change

in fraction as R, we can calculate the depreciation in year k as,

if Dk = RB k _ l Equation 3-4

where the book value at the end of year k can be calculated as, � � BkBO_>IDk e k=1

B0(1 - R)k Equation 3-5 0


Chapter - Economic Analysis 5

Substituting Equation 3-5 in Equation 3-4, we can calculate the depreciation in year k as,

Dk = RB 0 (1 - R)’

Lciuii4ii

where B0 is the cost basis.

The declining balance rate can vary between 100% and 200% (double declining balance); the

most common value being 200%.

Example 3-3 Using the same information as in Equation 3-2, and using the double declining balance method, estimate the depreciation

schedule.

Solution 3-3 For year one, depreciation is calculated as (using Equation 3-6),

where R is 200%/5 = 40%, and B0 = $5,000.

Substituting,

= 0.4 x 5,000 = $2,000

The book value at the end of year 1 can be calculated using Equation 3-5,

131 =B0 (1 - R) 1 = 5,000(1 - 0.4) = $3,000

Similarly, for year two,

D2 = RB0 (1 .- R) 2 ’ = 0.4(5,000)(1 - 0.4)

= $1,200

Or, using Equation 3-4,

D2 = RBe_ 1 = 0.4(3,000)

= $1,200

The book value is,

13 2 = (5,000)(1 - 0.4)

= $1,800

I Calculations can be repeated for other years as well, as shown in the table below:

Year Book Value Deprecation

1 5,000 2,000

2 3,000 1,200

3 1,800 720

4 1,080 432

5 648 259

Total $4,611

After five years, the total amount of depreciation allowed is $4,611, which is not equal to the cost of the equipment. This is one

6 Mohan Kelkar, Ph 0., J.D.

Sagdi

Highlight

of the major disadvantages of the declining balance method. That is, the cost of the equipment cannot be recovered fully over

the tax life. Only over long periods of time can the total cost be fully recovered. J

SUM-OF-YEARS DIGITS DEPRECIATION

The sum-of-years digits depreciation method also results in a larger depreciation at the

beginning of a useful life than the straight-line depreciation method. It is not a very common

method. To calculate the depreciation using this method, we first need to know the sum-of-

years digits. This sum represents the sum of all the years in a tax life. For example, if n is the tax

life of a property, the sum is calculated as,

sum = 1 +2+3+...+n Equation 3-7

Alternately, we can also calculate the sum as,

n (n+ 1) sum = Equation 3-8

2

Depreciation in year k can be calculated as,

remaining tax life at the beginning of period k (B - s) Equation 3-9 Dk = sum

We can write Equation 3-9 as,

n- k+1 - 2(n-k+1) Dk n(n+1) (B0 - S)

n(n+1) (B0 - S) Equation 3-10

2

Book value can be calculated using Equation 3-1.

Example 3-4

Using the same information as in Equation 3-2, calculate the depreciation schedule using sum-of-years digits depletion.

Solution 3-4

Using Equation 3-10, depreciation in year 1 can be calculated as,

2(5 - 1+1)

= (5)(6) (5,000) = $1,667

For year two,

2(5-2+ 1)

D2 = (5)(6)

(5,000) = $1,333

The book value after year two is,

B2 = B0 - Dk

= 5,000 - (1,667 + 1,333) = 2,000


Chapter 3- Economic Analysis 7

Similar calculations can be done for other years.

Year Book Value

1 1,667 3,333

2 1,333 2,000

3 1,000 1,000

4 667 333

5 333 0

Total Depreciation $5,000

Unlike the declining balance method, the sum-of-years digits method allows the depreciation of the entire cost of an asset.

DECLINING BALANCE AND STRAIGHT-LINE DEPRECIATION

To remove the major disadvantage of the declining balance method (i.e., not being able to

recover the entire cost of an asset) and, at the same time, allow for rapid depreciation as

calculated by the declining balance method, a combination of declining balance and a straight-

line method is used. This method allows the calculation of depreciation using both the declining

balance and the straight-line methods and allows for switching from one method to another

when one method predicts a higher depreciation.

Recall that straight-line depreciation is calculated as,

D - Bk_I � S

k - n�k+1 Equation 3-3

and the declining balance depreciation is calculated as,

Dk = RBk_ l Equation 3-4

For each year, we compute the depreciation using both methods and choose the value that is

higher. If

RB k1 > i-:i

Equation 3-11

we select the declining balance method. If

RB k _ l < Equation 3-12

we select the straight-line method.

If we assume that the salvage value is zero, we can simplify Equation3-12 to write,

1 R< ork>n+1----

1 n�k+1 I? Equation 3-13

Equation 3-13 provides a condition for switching to a straight-line method. For example, if

n = 5, and R = 0.4, k > 3.5 to switch to a straight-line method. That is, we will switch in year 4

to a straight-line method. The following example illustrates the technique.


Example 3-5

Using the same information as in Example 3-2, calculate the depreciation schedule using the combination (double declining and

straight-line) method. Assume double declining balance (DDB).

Solution 3-5

Depreciation Depreciation Book Value Selected

Year (Declining Balance) (Straight-line) At The End Depreciation

1 2,000 1,000 3,000 2,000

2 1,200 750 1,800 1,200

3 720 600 1,080 720

4 432 540 540 540

5 216 540 0 540

For example, for year one, for the declining balance method,R = = 0.4. Using Equation 3-4,

D1 = RB0 = 0.4 x 5,000 = $2,000

For straight-line method, using Equation 3-3,

B0 5,000

n-1+1 -= $1,000

Since 2,000>1,000, we should select the declining balance method; therefore, the book value at the end of year 1 will be,

5,000 + 2,000 = $3,000

For year two, depreciation using the declining balance method,

D2 = RB 1 = 0.4 x 3,000 = $1,200

and depreciation using a straight-line method,

B 1 3,000 11 3 = =�=$ 750

n-2+1 4

As before, we select the declining balance method.

Using Equation 3-13, we expect to switch after 3 years. In year 4, depreciation using the declining balance method,

/14 = RB3 = 0.4 x 1,080 = $432

and depreciation using the straight-line method,

B3 1,080 114

= 4 + 1 = = $540

Depreciation using the straight-line method is higher; therefore, the straight-line method should be selected.

Problem 3-1

A pump for injecting water costs $10,000. If the salvage value is assumed to be zero, calculate the depreciation schedule using

the following methods if the useful life is five years.

� Straight-line depreciation

� Sum-of-years digits depreciation

� Double declining balance (DDB) depreciation



DDB depreciation with switch to a straight-line depreciation

Problem 3-2

A hauling truck costs $12,500. If the useful life is three years and the salvage value of the truck is $2,000, calculate the

depreciation schedule using:

� Straight-line depreciation

� DDB depreciation

� DDB depreciation with switch to a straight-line depreciation.

Problem 3-3

Tubing for a well costs $20,000. If the life of the tubing is seven years and the salvage value is zero, calculate the depreciation

schedule using:

� Straight-line depreciation.

� DDB depreciation.

� DDB depreciation with switch to a straight-line depreciation.

Calculate the year in which you will switch to a straight-line method and verify your answer with Equation 3-x.

21Y1I

A drilling company purchased a new drilling rig for $5 million. The tax life is seven years, and the depreciation is to be calculated

using the double declining balance method. After three years, due to relatively limited use, the company sold the rig for $2

million. Based on the depreciation schedule, did the company earn any taxable income after three years? If so, how much?

Problem 3-5

An oil company bought a new workstation for $100,000. The workstation is to be depreciated over a five year period using a DDB/straight-line combination method. After four years, because of advances in computing, the company traded in the old

machine for a new machine. With trade in, the new workstation costs $100,000; without a trade in, the value is $115,000.

Calculate the depreciation for a new machine assuming five years useful tax life. Calculate the taxable income, if any, for the oil company after four years due to trade-in.

Problem 3-6

A computer was purchased for $10,000 at the beginning of 1991. It was depreciated using the straight-line method assuming a

tax life of 5 years. In early 1994, $5,000 was spent to significantly upgrade and improve the performance of the computer. Assuming that depreciation begins in the same year as capital expenditure, and the tax life of the upgraded expenditure is also five years, calculate the depreciation in years 1994, 1995, and 1996.

Problem 3-7

An oil company bought a gas oil separator for $80,000. The sales tax is 85v., the delivery cost is $500, and the site preparation

required another $1,000. The maintenance insurance for the compressor was offered at $1,000 per year. If the tax life of the compressor is seven years, and is to be depreciated using a combination DDB and straight-line method, calculate the

depreciation schedule.

Problem 3-8

Tubing for a new oil well was purchased for $50,000. Assuming a salvage value of zero and a tax life of seven years, calculate the depreciation schedule using the sum-of-years digits method. If, after five years, the well needs to be shut in and the tubing

can be sold for $10,000, what is the taxable income for the company, if any?


INCOME TAXES (PARK, 1993) (DEGARMO, SULLIVAN, & BONTADELU, 1993)

In general terms, taxes are a way by which authorities share part of the income of an individual or a

corporation. These authorities may be the federal government, individual states, local

municipalities, counties and school districts. Taxing income can take several forms; the most

common being income tax that is levied on the difference between the gross revenue and allowed

deductions for tax purposes. Other types of taxes include property taxes that are based on the value

of real estate; sales taxes based on the price of goods; luxury taxes based on items that are not

considered necessities; entertainment taxes based on entertainment expenses; and value added

taxes based on the difference in the value of goods as they pass from one party to another. There

may be other types of taxes not included here.

In this section, we will concentrate on income taxes which, in most instances, are levied by either

federal or central government and local or state governments. The laws governing the calculations

of income tax are extremely complex and replete with exceptions. Therefore, rather than

considering the details of the tax laws in various countries, we will focus on the basic principles of

tax laws common in most instances.

The first principle of income tax is that it is assessed based on taxable income. Taxable income is the

difference between gross income or gross revenue and allowable deductions according to tax laws.

Logical tax deductions are operating costs or other expenses associated with operating a business.

Other tax deductions typically included are depreciation and interest payments.

The second principle common to most income tax laws is that the rates at which the tax applied

tends to be progressive. That is, the tax rates are lowest when the taxable income is the least. As the

taxable income increases, the tax rate also increases.

In most economic evaluations, we need to use the marginal tax rate for calculations. Marginal tax

rate represents the tax rate applied to the last dollar earned in the taxable income. For example, if

the corporation has earned $30,000 last year and expects to add another $10,000 in taxable income

with the addition of a new project, it is reasonable to assume a marginal tax rate of 15% (which is,

for example, based on the tax rate paid by a corporation for $30,000 of income); however, if the

corporation earned $1,000,000 last year and is considering a new project with a potential additional

taxable income of $500,000, it is better to use a marginal tax rate as the tax rate applicable beyond

the $1,000,000 income.

In the Basic Tax Structure section, we only need to understand these two principles. The following

examples use these concepts.

TAX COMPUTATIONS

Taxable income is calculated by subtracting the allowable deductions from gross revenue. In

mathematical terms,

taxable income = gross revenue tax deductions Equation 3.-14


Chapter Economic Analysis 11

Most common deductions include operating costs, depreciation, overhead costs and other

related expenses; therefore, we can write,

taxable income = gross revenue - operating costs

�related expenses - depreciation

FARMITWIMMy

Once the taxable income is calculated, the income tax is calculated using the applicable marginal

tax rate. That is,

income tax = taxable income x tax rate

To calculate net revenue, we need to subtract the actual expenses from the gross revenue.

These expenses do not include depreciation or other deductions that are only deducted for tax

purposes. Net revenue is calculated as,

net revenue = gross revenue - operating costs - related expenses - tax Equation 3-17

If we use the following notations:

A = gross revenue in year]

Oj = operating costs in year]

D1 = depreciation (or tax deductions) in year]

T = marginal tax rate

E = other related expenses in year]

we can write taxable income as,

= A1 - - - Dj

we can write the income tax as,

=T(A�0 1 �E 1 �Dj)

and we can write the equation for net revenue as,

_�A�0 1 �E�T(A 1 �O�E 1 �Di) net revenue = (1 - T)(A - - E) + TD 1

Equation 3-20 can be repeated for all years in which a project is in operation.

Example 3-6

�7IptI’j

A company wishes to purchase a workstation for $15,000. This will save the company $5,000 in outside computing costs. The

maintenance agreement calls for a payment of $800 per year for the machine. Assume that the useful life of the machine is five

years. Assume further that a straight-line depreciation is allowed over a five year period. If the marginal income tax rate is 28%,

calculate the NPV of this investment with a MROR of 15%.

Solution 3-6

12 Mohan Kelkc,r, Ph.D., J.D.

Sample calculation for year one,

gross revenue (savings) = 5,000


taxable income = gross revenue - operating costs - depreciation = 5,000 - 800 - 3,000 = $1,200


tax = taxable income x tax rate 1,200 x 0.28

= $336


net revenue = gross revenue - operating costs - taxes = 5,000 - 800- 336 = $3,864

In these calculations, we assumed a straight-line depreciation. Therefore,

15000

depreciation =5

= $3,000 per year

The following table lists the calculations for other years.

Gross Taxable Net

Year Income Income Tax Revnue 0 -15,000 -15,000 1 5,000 1,200 336 3,864 2 5,000 1,200 336 3,864 3 5,000 1,200 336 3,864 4 5,000 1,200 336 3,864 5 5,000 1,200 336 3,864

We are assuming that all of the income is collected at the end of the year,

3,864 3,864 3,864 3,864 3,864 NPV = -15,000 + (1 + .15) + (1 + .15)2 + Ti - + .15)3+

(1+.15) + = -$2,047

Since the NIPV is negative, the project is not feasible.

Alternately, if we use Equation 3-20, we can write the net revenue equation as,

net revenue = (1 - T)(A 1 - - E) + TD

Substituting,

= (1 - 0.28)(5,000 - 800) + 0.28 x 3,000 = $3,864

Since the revenue is constant each year, we can write,



NPV = PVbenefits - PVosts [(1 + .15)

- .3 864{ 15(1 + .15) 5

15,000

= $2,047

We get the same answer as before.

Example 37

A gas well is expected to produce at an exponential decline rate based on the adjacent producing wells. Per initial testing, the following production profile is predicted over the next five years. The total capitalized cost for this well was $2,000,000.

Operating costs per year are $600,000 and are expected to increase at a rate of 5% per year. Assume that gas is sold at a rate of $1.75/MSCF, and all money is collected at the end of each year. Assume that the local tax is 7% of the gross revenue. Assume that the capitalized cost is depreciated over a five year period using a combination method (declining balance and straight-line). Assume that depletion costs are negligible. Using this information, calculate the ROR for this investment. If the minimum ROR is 20%, is this investment feasible?

Production Profile Year Production, MSCF

1 1,561,515

2 1,134,922

3 823,531

4 597,688

5 434,578

Solution 3-7

The following table illustrates the results over a five year period. Sample calculations are shown below.

Production, Gross Operating Local*

Year MSCF Revenue ($) Costs ($) Tax ($) 1 1,561,515 2,732,651 600,000 191,286 2 1,134,922 1,986,114 630,000 139,028 3 823,531 1,441,179 661,500 100,883 4 597,688 1,045,954 694,575 73,217 5 434,578 760,512 729,304 53,236

*Loca l tax may be treated as other expenses since these taxes are assessed based on the gross revenue, and not taxable income.

For year one,

revenue = production x price = 1,561,515 x $1.75 = $2,732,651

local tax = revenue x .07 = $2,732,651 x.07 = $191,286

Depreciation Taxable Net Year ($) Income ($) Tax ($) Revenue ($)

1 800,000 1,141,365 319,582 1,621,783 2 480,000 737,086 206,384 1,010,702 3 288,000 390,796 109,423 569,373 4 216,000 62,162 17,405 260,757 5 216,000 -238,028 -66,648 44,620

14 Mohon Kelkor, Ph.D., J,D.

For year one,

Using Equation 3-4,

2 2

depreciation = x 2,000,000

= $800,000


taxable income = revenue - operating costs - local tax - depreciation

= 2,732,651 - 600,000 - 191,286 - 800,000

= $1,141,365


tax = taxable income x tax rate

= 1,141,365 x .28

= $319,582


net revenue = revenue - operatingcosts - localtax - incometax

= 2,732,651-600.000-191.286-319582

= $1,621,783

Note that for the last year, the taxable income is negative. As a result, the tax is also negative. Therefore,

net revenue = 760,512 - 729,304 - 53,236 - (-66,648)

= $44,620

The reason we use negative tax is because of the assumption that this is one of several projects the company is involved in and,

as a result, the loss in this project will result in a reduction in profit in other ventures and, thus, result in an overall lower tax

payment by the negative amount. To calculate the ROR,

0 = PVcos - PVbenefits 1,621,783 1,010,702 569,373 260,757 44,620

0=-2,000,000+(1j)

+ (1+i)2 +(1.)3+(li)4+(1)s

Through trial-and-error, i = ROR = 38.3%. Since ROR>MROR, the project is feasible.

EFFECT OF DEPRECIATION SCHEDULE

As shown in the previous section, depreciation plays a major role in net revenue calculations. In

this section, we will investigate the impact of the depreciation schedule on the after-tax analysis

of a project. Recall Equation 3-20 which states that the net revenue is,

Equation 3-28

That is, we receive an "additional" net revenue of TD1 as a result of the depreciation deduction.

This additional revenue is collected over the life of the project. If we assume that the tax life of

the asset is n, then we can calculate the present value of the additional future revenues as,

�TV ___D j

- ’ J=1 (1+1)1 Equation 3-21


Chapter - Economic Analysis

15

Where i is the effective interest rate. This is the benefit we will receive as a result of the capital

expenditure B0 . Therefore, after tax consideration, the net cost associated with that asset is,

net cost = B0 - TYi (i+t)J WEEMSEM

Obviously, the smaller the net cost, the better off the corporation. This net cost will depend on

the depreciation schedule. Examining Equation 3-22, one can see that more accelerated the

depreciation; the smaller the net cost. Time value of money of earlier depreciation is much

more valuable than later depreciation. This means an accelerated depreciation is more valuable

than straight-line depreciation. It also means that an item depreciated over a shorter period will

reduce more of the net cost than an item depreciated over a longer period. If the item is

depreciated over one year, there is no difference between capitalized cost and operating cost

since both costs will be deducted in the same year in which they are incurred.

Example 3-8

A water pump at a cost of $90,000 is to be used in a waterflooding operation. The additional water injected, because of the

pump, is expected to result in incremental production of 8,000 bbls of oil in year 1, decreasing at 8% per year. Additionally,

initial costs include reperforation and acidization of some of the wells which will cost an additional $70,000. The operating

costs, as a result of the new pump, are expected to be $7.50 per additional barrel of oil. If the oil price is assumed to be $18/bbl

and the life of the project is expected to be 10 years, calculate the NPV of this project. Assume the MROR to be 15%. Use both

the straight-line depreciation and DDB depreciation methods. The marginal tax rate is 34%. Assume the salvage value to be

zero. Compare the results between straight-line and DDB depreciation methods.

Solution 3-8

Straight-line Depreciation

The following table shows all of the computations. In performing these calculations, we assumed that $90,000 was expended at

the beginning and all the revenues are collected at the end of the year. $70,000 acidization and re-perforation costs were

considered operating costs; hence, they were deducted at the end of the first year along with the operating costs. It is possible

that a different convention may be used to define the expenses and revenues. For example, one of the most common

conventions used is the half-year convention where all numbers are defined in the middle of the year.

Gross Operating Taxable Net Year Production Revenue Costs Depreciation Income Tax Revenue

0 -90,000 -90,000 1 8,000 144,000 130,000 12,857 1,143 389 13,611 2 7,360 132,480 55,200 12,857 64,423 21,904 55,376 3 6,771 121,882 50,784 12,857 58,241 19,802 51,296 4 6,230 112,131 46,721 12,857 52,553 17,868 47,542 5 5,731 103,161 42,984 12,857 47,320 16,089 44,088 6 5,273 94,908 39,545 12,857 42,506 14,452 40,911 7 4,851 87,315 36,381 12,857 38,077 12,946 37,988 8 4,463 80,330 33,471 12,857 46,859 15,932 30,927 9 4,106 73,904 30,793 43,111 14,658 28,453 10 3,777 67,991 28,330 39,661 13,485 26,176

Calculations are carried out using previously described methods. The NPV for this operation can be calculated as,

13,611 26,176 NPV

= -90,000 + (1 + .1s + + (1 + .15) 0 = $103,175


I I I I I I I I I

I I I I I

Since the NPV is positive, the project is feasible.

Double Declining Balance Depreciation (DDB)

The following table shows all the computations.

Gross Operating Taxable Net

Year Production Revenue Costs Depreciation Income Tax Revenue

0 -90,000 -90,000

1 8,000 144,000 130,000 25,714 -11,714 3,983 17,983

2 7,360 132,480 55,200 18,367 58,913 20,030 57,250

3 6,771 121,882 50,784 13,120 57,978 19,713 51,385

4 6,230 112,131 46,721 9,371 56,039 19,053 46,357

5 5,731 103,161 42,984 6,694 53,483 18,184 41,993

6 5,273 94,908 39,545 4,781 50,582 17,198 38,165

7 4,851 87,315 36,381 3,415 47,519 16,156 34,778

8 4,463 80,330 33,471 46,859 15,932 30,927

9 4,106 73,904 30,793 43,111 14,658 28,453

10 3,777 67,991 28,330 39,661 13,485 26,176

The only difference between this table and the previous table is the amount of depreciation per year. The NPV can be

calculated as,

NPV = - - 90,000 + 17,983 26,177+ +

(1 + .15) (1 + .15) 10

= $104,338

The difference between the two NPV values is equal to

= 104,338 - 103,175 = $1,163

This additional NPV using the 0D13 method is a result of accelerated depreciation. Instead of depreciating, if the cost of the

pump is allowed to be expensed in the first year, the NPV will be $142,686. This improvement is due to recovering all of the tax

benefit of depreciation in the first year.

Many companies will maintain separate accounting books; one for tax purposes and one for

balance sheets. For tax purposes, the company will use the most favorable depreciation

schedule available that will maximize the tax benefit. For balance sheets, the company will use a

depreciation schedule that allows for the smallest depreciation possible each year. This

apparent dichotomy serves two purposes. By using the quicker depreciation schedule for tax

purposes, the company can improve the net revenue as shown in the previous example. At the

same time, by using a slower depreciation schedule for balance sheet purposes, the company

can show that the asset base of the company is not reduced significantly over the year. The

asset base is the cumulative book value in the previous year minus the depreciation. If the

depreciation is smaller, the asset base will be larger providing the shareholder with a larger

equity base. To account for the differences between the two methods, a company will report

differed taxes in its balance sheet. These differed taxes represent the difference between what

the company had would have to pay if it had used a slower depreciation versus what it is paying

due to accelerated depreciation. In mathematical terms, the differed taxes are:

T (Daccj - D st tine.) Equation


Chapter 3 - Economic Analysis

17

Where Daccj represents the accelerated depreciation amount used for tax purposes versus

Ds t.iinej represents a slower depreciation used for asset base purposes. In Equation 3-8,

depreciation in year 1 using DDB is $25,714, whereas, depreciation using the straight-line

method is $12,857. For a marginal tax rate of 0.34, the differed taxes will be,

0.34(25,714 - 12,857) = $4,371

This is the exact difference in the net revenue in the first year using different depreciation

schedules. Most corporations, in their annual and quarterly financial reports, include a line item

for differed taxes. These are the taxes resulting from different depreciation schemes.

EFFECT OF BORROWED MONEY

Most tax laws allow a deduction for an interest payment on borrowed money. The interest

payment is considered the cost of operating a business; hence, it is considered an operating

cost. Because of the deduction of interest rate, the effective interest paid by the corporation is

equal to (1 - T)I where T is the marginal tax rate and I is the interest rate on the loan. As long

as (1 - T)I is less than the MROR, a corporation can borrow money and "leverage" its own

investment and make additional NPV. The following example illustrates the benefit.

Example 3-9

A potential prospect has been shown to an oil company. It will require the oil company to invest $2,000,000 to install a

compressor in a gas field. The compressor can be depreciated over a seven-year period using a combination of DDB and the

straight-line method. After installing, the gas field is expected to initially produce an incremental gas of 600 MSCFD declining at

10% per year. Beyond ten years, there will not be any incremental production. Assume the gas price to be $4/MSCF with 12%

of the revenue for the royalty payment and about $0.50/MSCF in operating costs. The company uses 15% as a hurdle rate

(MROR). The company has enough cash to invest in this project; however, using its other assets, it can borrow the entire

amount at a 6% interest rate and redeploy the money elsewhere. The marginal tax rate is 35%. Which is the better choice?

Assume that the money is borrowed over a ten year period. At what interest rate will it not make a difference if the company

borrows money or invests its own capital?

Solution 3-9

We can solve this problem by first assuming that the oil company used its own money. The table below shows the summary of

the calculations.

Production Gross Operating Taxable Net

Year (MMSCF) Revenue Costs Depreciation Income Tax Revenue

0 -2,000,000 -2,000,000

1 219 876,000 214,620 571,429 89,951 31,483 629,897

2 197 788,400 193,158 408,163 187,079 65,478 529,764

3 177 709,560 173,842 291,545 244,173 85,460 450,257

4 160 638,604 156,458 208,247 273,899 95,865 386,181

5 144 574,744 140,812 173,539 260,393 91,137 342,794

6 129 517,269 126,731 173,539 216,999 75,950 314,588

7 116 465,542 114,058 173,539 177,946 62,281 289,203

8 105 418,988 102,652 316,336 110,718 205,618

9 94 377,089 92,387 284,702 99,646 185,057

10 85 339,380 83,148 256,232 89,681 166,551

Production is calculated by multiplying the daily rate by 365 and diving by 1,000. Operating costs include both royalty payment

and operating costs. The net present value at 15% is $41,371. The project is barely economical.


If the oil company borrows $2,000,000 at 6% interest, no money from its own pocket is invested. Instead, the company will

have to make a payment on the borrowed amount. The yearly payment is

[0.06(1 + 0.06) 10 1 A

= 2,000,000 [(1 + 0.06)10 - i.j = $271,736

This payment is split into an interest part and a principal part. The portion going toward principal is calculated as

Principalk+l = (l+-k� This represents the principal payment for (k+l)t period. The balance would go toward interest. For

example, in year 2, we can state Principa12 = (1 2101 = $ 160,840. Therefore, the interest payment in year 2 will be

=271,376 - 160,840 = $110,896. These amounts change over time as the interest payment decreases and principal payment

increases. The functions to calculate the interest payment and principal payment are available in standard spreadsheet

program. The following table shows the details of the calculations.

Production Gross Operating Interest Taxable Net

Year (MMSCF) Revenue Costs Depreciation Payment Income Tax Revenue

0 0 0

1 219 876,000 214,620 571,429 120,000 -30,049 -10,517 400,161

2 197 788,400 193,158 408,163 110,896 76,183 26,664 296,842

3 177 709,560 173,842 291,545 101,245 142,927 50,025 213,957

4 160 638,604 156,458 208,247 91,016 182,883 64,009 146,401

5 144 574,744 140,812 173,539 80,173 180,220 63,077 99,119

6 129 517,269 126,731 173,539 68,679 148,320 51,912 66,890

7 116 465,542 114,058 173,539 56,496 121,450 42,508 37,241

8 105 418,988 102,652 43,581 272,755 95,464 -50,864

9 94 377,089 92,387 29,892 254,810 89,184 -76,217

10 85 339,380 83,148 15,381 240,851 84,298 -99,802

The NPV for this scenario is $826,043. In effect, by borrowing money, a marginal project became very attractive.

For example, for year two,

taxable income = gross revenue - operating costs - depreciation - interest payment

= 788,400 - 193,158-408,163 - 110,896 = 76,183

net revenue = gross revenue - operating costs - interest payment - tax - principal payment

= 788,400 - 193,158- 110,896 - 26,664- 160,840

= $296,842

Note that the principal payment has to be deducted from the net revenue since it is an actual expense. The only difference is

that it is deducted after taxes are calculated. To calculate the interest rate at which it will not make any difference, we can use

trial-and-error. In the table below, an interest rate of 23.1% was paid on the borrowed amount. The NPV for this project is

$41,371; the same as when no money was borrowed. This interest is also calculated by noting that (1-0.35) x 0.231 = 0.15. 0.35

is the marginal tax rate and 15% is the hurdle rate. The higher the marginal interest rate the company pays, the higher the

interest rate it can borrow and improve the NPV of the project.

Production Gross Operating Interest Taxable Net

Year (MMSCF) Revenue Costs Depreciation Payment Income Tax Revenue

0 0 0

1 219 876,000 214,620 571,429 461,540 -371,589 -130,056 263,732

2 197 788,400 193,158 408,163 446,271 -259,193 -90,717 158,255

3 177 709,560 173,842 291,545 427,479 -183,307 -64,157 72,171

4 160 638,604 156,458 208,247 404,350 -130,451 -45,658 100

5 144 574,744 140,812 173,539 375,881 -115,491 -40,422 -53,351

6 129 517,269 126,731 173,539 340,848 -123,849 -43,347 -93,819

7 116 465,542 114,058 173,539 297,728 -119,782 -41,924 -134,296

8 105 418,988 102,652 244,656 71,680 25,088 -236,456

9 94 377,089 92,387 179,337 105,365 26,878 -279,879

10 85 339,380 83,148 98,945 157,287 55,051 -326,522



Borrowing is useful and leverages the project and improves the NPV as this example illustrates.

Case Study 3

An article appeared in the paper (Wall Street Journal, 2009) about a federal program to stimulate the purchase of electric vehicles. Golf carts which run on clectncity, qualify as electric vehicles. The federal credit provides from 54,200 to $5,500 for the purchase of an electric vehicle, and when it is combined with similar incentive plan in many states the tax credits can almost pay the entire cost of a golf cart Even in states that do not have their own tax rebate plans the federal credit is generous enough to pay for half or even two-thirds of the as erage sticker price of i golf cart which typically ranges in price from $8,000 to $1 0,000. "The purchase of some models could be absolutely free says Roger Caddis of Ada Electric Cars in Oklahoma Is that about the coolest thing you’ve ever heard?"

\ golf cart boom followed an IRS (Internal Revenue Service which governs taxes in the 12 S) ruling that golf carts qualify for the electric car credit is long as they are also road worthy. These qualifying golf carts are essentially the same as normal golf carts with the addition of extra safety features, such as side and rearview mirrors and three-point seat belts. They typically go 15 to 25 miles per hour.

In South Carolina, sales of these carts have been soaring as dealerships alert customers to the government tax rebate. ’The Golf Cart Man" in the Villages of Lady Lake, Florida is running a banner online ad that declares: "GET A FREE GOLF CART or make S2,000 doing absolutely nothing!"

The Golf Cart Man is referring to his offer in which you can purchase the cart for $8,000, get a $5,300 tax credit off your year’s income tax, lease it back for $100 a month for 27 months, at which point The Golf Cart Man will buy back the cart for $2,000. "This means you own a free golf cart or made 82,000 cash doing absolutely nothing!!!"

Let us assume that you are interested in purchasing a golf cart for $8,000. You lease it to The Golf Cart Man for $100/month-Assume that The Golf Cart Man only pays you at the end of the year. Let us assume that you bought the car on January 1st, Since the golf cart is car that you are using for investment, you can depreciate it over a three year period. Use a combination of double declining/straight-line method. At the end of the first year (December 311, you -will be able to get a $5,300 tax credit. The tax credit is not ordinary income. First calculate the tax owed in that year and then add $5,300 as a tax credit. For example, let us assume that you calculate the tax to be paid as $2,000. If you are owed a tax credit of $5,000, then the tax liability is $3,000. That is, the government will pay you 53,000 instead of you paying the government $2,000. If you return the car at the end of 27 months and collect S2,000, what is the return on your investment (NPV) based on a 3 year project? Do your calculations based on the assumption that the transaction occurs at the end of the year. That is, assume an income of S2,000 at the end of the third year. Instead of taking $2,000. you decide to keep the cart. Every month, starting with the 27th month, you believe that you can save about $100/month by utilizing the cart for your own use. However, The Golf Cart Man charges you a $50/month storage fee to store your cart at the golf course. You intend to use the golf cart until it is six years old. At that point, you expect to sell it for about $500. Is it better to collect S2,000 from The Golf Cart Man or keep the car for six years? Assume a marginal tax rate of 31%. Assume the MROR to be 15%.

Consider another scenario. You want to buy the cart but you do not have the cash. You approach The Golf Cart Man and he agrees to lend you S8,000 at an interest rate of 15% per year. You will have to pay back the entire cost over a three year period on an annual basis. If you decide to take on that offer, what will be your NPV under the first scenario described above?

Case Stud Solution 3.1

Option 1: Sell the cart after 27 months for 52,000

Here are the details:

Year Gross Revenue Depreciation Taxable Income Taxes Net Revenue

0 (8,000) (8,000) -3,000 1,200 5,333 (4,133) (6,581) 7 1781

2 1,200 1,778 (578) (179) ,. 1,379 3 2,300 889 1,411 437 1,863

Notice that in year 1 taxes include a tax credit of 85,300. so the amount stated (6,581) is calculated as -4,133 x 0.31 - 5,300. The NP\T for this option is $1,034.

)ptn in 2: Sell the car after six years

I Ieee are the details:

20 Mohan Kelkur, Ph.D., J.D.

Year Gross Revenue Depreciation Taxable Income Taxes Net Revenue o (8,000) (8,000) 8,000 1 1,200 5,333 (4,133) (6,581) 7,781 2 1,200 1,778 (578) (179) 1,379 3 750 889 (139) 43 793 4 600 600 186 414 5 600 600 186 414 6 1,100 1,100 341 759

S

The NPV for this scenario is $1,101, which is slightly better than Option I

Option 3: Similar to option 1 except that the money is borrowed at 15% interest

Here are the details:

Gross Interest Principal Taxable Net Year Revenue Depreciation Payment Payment Income Taxes Revenue

0 0 1 1,200 5,333 1,200 2.304 (8,333) (6,953) 4,650 2 1,200 1,778 854 2 1649 (1,432) (444) 1,860 3 2,300 889 457 3,047 954 296 -1,500

In this case, S8,000 is borrowed; therefore, there is no initial investment. The borrowed money is returned over a three year period using an interest rate of 15%. The interest payment is deducted for tax purposes. The NP’ for this option is $1,651. This is probably the best of the available options.

Problem 3-9

An oil company is considering installing gas lift equipment on a well. The equipment will cost $60,000 which can be depreciated

using sum-of-years digits depreciation. The equipment has five years of useful life and a zero salvage value. it is expected that the annual net benefit, as a result of installment will be $1 7,000 for the five year period. If the company’s MROR is 12% after taxes and the tax rate is 34%, should the company invest in this equipment?

Problem 3-10

An oil company is considering buying a compressor that costs $500,000. An alternative is to lease the compressor for $200,000

per year. The maintenance cost is $25,000 per year if the compressor is purchased. Assume that the compressor can be depreciated using a DDB and straight-line combination depreciation method over a five year period. Further assume that the

income tax rate is 34%. At a MROR of 15%, which option is preferable? Would the answer be different if the MROR is 25%? Use

a five year period to complete the calculations.

Problem 3-11

A small operator has a total revenue of $1.5 million from gas production. Following are expenses for the company:

� Labors $150,000

� Materials = $70,000

� Depreciation = $20Q,000

� Operating expenses = $300,000

� Royalty income from other investments = $32,000

� Interest = $61,000

� Proceeds from sales of old equipment with a book value of $30,000 = $65,000

What is the taxable income for the operator? If the tax rate is 34%, what is the net revenue received after income taxes? If the

state levies a 6% grass revenue tax on gas production, what is the net revenue? If the operator lost $100,000 in the previous year and is allowed to carry forward the loss into the next year, what is the net revenue?


Chapter 3 Economic Analysis 21

Problem 3-12

Clean Environmental Company started its operation with an initial investment of $2.5 million. Its primary target is oil well site

cleanups Out of $2.5 million $1.5 million is invested in equipment with a seven year life (DDB depreciation) The remaining $1

million are invested in start-up costs that can be depreciated over a five year period using the straight-line method The

company is expected to gross $1 million in revenues in year I with a projected increase of 10% per year over the next 10 years

The operating costs will be $150,000 in year one increasing at 10% per year. If the tax rate is 34% calculate the NPV of this

venture if the MROR is 15% If the company can deduct the start-up costs in the first year of operation and is allowed to carry

forward losses into future years, how will the NPV change?

Problem 3-13

An oil company is considering two options for the development of a project. It can continue to produce under existing conditions

or it can stimulate all of the wells to accelerate the production from the field. Proceeding under existing conditions will result in

the production of 150 bbl/d with a decline of 6% per yeorfor the next 10 years. The cost of production is $100,000 in fixed costs

plus $61bbl. If the stimulation project is undertaken, it will cost $20,000 per well. A total of 20 wells are operating in the field.

Production is expected to increase to 230 bbl/d, but is expected to decline at 10% per year over the next 10 years. The operating

cost structure will not change. The costs associated with stimulation can be considered as operating casts and can be deducted in the same year. Assume that the company has other profitable operations such that the company can have "negative taxes"

from this project. If the marginal tax rote is 39% and the MROR is 12%, which option should the company choose? The price of

oil is $1 71bbl.

Problem 3-14

A promoter wants to drill a well in a yet-to-be explored area. The cost of drilling is expected to be $1 million. Other start-up

costs are expected to be $250,000. The equipment costs related to the well, if successful, will be $200,000. Based on the wells in

the nearby area, the well is expected to produce 5,000 MS CF/B. Production will decline at a rate of 10% per year. The well is

expected to produce at least for 15 years. Assume that the equipment cost con be depreciated over 7 years using a combination

of DDB and a straight-line method. A total cost of drilling of 70% can be deducted in year one and the rest over the next five

years with a straight-line depreciation. The other start-up costs are deducted over the life of the well proportional to the fractional gas produced each year. Assume that the total gas produced from the well is equal to 15 years of total production.

The operating cost of the field is $10,000 per year plus $0.15/Ms CF of gas production. If the marginal tax rate is 34% and if all

the money is invested by the promoter, what is the rate of return on this investment? If the promoter only invests 10% of the

required initial investment and 90% of the money is borrowed at 14% interest per year, what is the rate of return?Assume that

the interest payment is tax deductible. The price of gas is $1.75/MSCF1n the first year and is expected to increase at a rate of 3%

per year. Assume zero salvage value.

Problem 3-15

As part of restructuring, an oil company is proposing to automate port of the field operation. The cost of automation is $230,000 and the equipment con be depreciated over a five year period using a combination of DDB and straight-line

depreciation. Automation is expected to eliminate annual payroll expenses of $120,000 per year over the next six years. Additionally, due to automation, production problems are expected to be detected earlier saving another $25,000 per year in

maintenance expenses. If the marginal tax rate is 39 01., and the MROR is 15%, should the company install the automation? If the

proposed field is expected to be sold to a potential buyer in three years, and the addition of automation can bring in an

additional $50,000 for the price of the field, will the decision be different?

Problem 3-16

A consulting company sends a newsletter to its clients updating them about recent events in the oil industry. Currently, the

newsletter is printed by an outside company at a cost of $15,500 per year. The cost is expected to increase over the next five

years at a rate of 6% per year. An alternative is to purchase a desktop publishing software package, a color laser printer, and a

high quality scanner. The cost of the software package is $3,000, a laser printer is $6,000, and a scanner is $4,000. All of this equipment can be depreciated over five years using the straight-line method. The salvage value of the equipment is 15% of the

original equipment cost. An additional $3,000 per year will be spent on labor and material costs. If the marginal tax rote is 34%,

and the MROR is 12%, is it worth printing the newsletter in-house?


Problem 3-17

An oil company is offered $3 million for a producing property based on an assessment of the property. The fair market value of the surface property is $500,000. The fair market value of the equipment is $1 million. The remaining price goes toward the

hydrocarbons that will be produced in the future The company originally paid $200 000 for the surface property and $2 million

for the equipment. The equipment is already depreciated over ci five year period using the double declining balance method

assuming a seven year useful life The gain due to selling the surface property is taxable immediately as is the gain due to selling

the depreciated equipment five years ago. The field is currently producing at a rote of 60,000 bbls in the existing year with on expected decline of 10% per year over the next ten years. The operating costs are $101bbl of oil. The tax rote is 34%. Assume the salvage value of the equipment after 10 years is zero. If the oil price is expected to be $201bbl over the life of the field, should the oil company sell the property at the expected rate of return of 20%? What is the minimum price at which the oil company

should expect to sell this property?

The buyer of the property can depreciate the equipment over a seven year period using a combination of DDB and straight-line method. Further, by cutting some of the operating costs, the buyer can reduce the operating costs to $8.51bbl of oil. At the end

of 10 years, the buyer can sell the surface property for $1 million. The salvage value of the equipment will be zero. What is the

rate of return the buyer should expect from purchasing the property?

DOMESTIC AGREEMENTS

In the Basic Tax Structure section, we discussed the importance of depreciation and taxes on cash flow

analysis. In this section, we will discuss the importance of domestic operating agreements on cash flow

analysis. These agreements are specific to the U.S. Some of the details were provided in Chapter 1 and

will not be repeated here. To briefly review, a contract (called a tease) has to be signed between the

mineral owner and operators so that the operator will have the right to explore and produce from the

property (L. G. Mosburg, 1989). In return, the mineral owner receives royalty interest (cost-free

production or proceeds from cost-free production), and the operator receives the remaining interest

and has to pay all expenses (working interest). The operator and mineral owner typically pay the state a

percentage of production in the form of severance taxes. The lease, in most cases, can be held in

perpetuity if the operator can produce hydrocarbon in economic quantities (the operation is

’ economical), The operator will also pay a lease bonus to the mineral owner to entice him into signing

the lease.

Royalty payment and severance taxes are treated as operating expenses; therefore, from a tax

perspective, no unique treatment is needed. However, certain costs associated with oil and gas

exploration and production do require unique treatment. For example, leasehold costs and intangible

drilling costs (IDC). lDCs were explained in the Depreciation section. The rules regarding lDCs are also

complex and subject to many exceptions beyond the scope of this book. In general, for small,

independent operators, IDCs are may be deducted in the same year they are incurred; for large

operators,70% of the IDCs may be deducted in the same year they are incurred, and 30% of the costs

are depreciated over a five year period using the straight-line method. Capitalized costs are depreciated

using a combination of DDB and straight-line method. The specific rules are complex but, in general, in

the first year of operation, only a half year of depreciation is allowed and, in the last year, another half

of a year of depreciation is permitted. For example, if the taxable life is 3 years, the capitalized cost is

depreciated over 4 years with a half year on both ends and 2 years in between. Leasehold costs are

explained if the following section.


Chapter 3 - Economic Analysis 23

LEASEHOLD COSTS

Leasehold costs include the lease bonus, costs related to the lease and property acquisitions such as

legal and administrative costs, assessment costs and most of the geological and geophysical costs.

Since these costs occur prior to production, they cannot be considered operating costs. Since these

costs do not result in tangible equipment, they cannot be considered capitalized costs. However, the

closest analog to leasehold costs is capitalized costs, because leasehold costs result in the

acquisition of minerals (instead of equipment). As with equipment, the value of minerals will

deplete over time until it reaches zero when the field is abandoned. Therefore, as with depreciation,

leasehold costs are deducted over the life of the field. The process of deduction is called depletion.

Only if the lease is terminated (e.g., dry hole), are the costs deducted in the same year. The

leasehold costs are depleted either through percentage depletion or cost depletion with few

restrictions. Both methods have a logical explanation as to how they work. The rules regarding the

specific application and eligibility for depletion are complex and beyond the scope of this book.

Remember that the percentage depletion option is normally afforded only to small, independent,

operators and they can choose between the two types of depletion. For large companies, only cost

depletion option is available.

DEPLETION

The costs (leasehold) associated with acquiring the right to drill a well, as well as the initial

assessment of feasibility to drill in a productive reservoir, is subjected to depletion. Depletion spans

the life of the reservoir.

The two most commonly used methods to recover the leasehold costs are cost depletion and

percentage depletion. Both methods have an underlying logic. If the discovery is greater than

expected, percentage depletion is the better option. The cost depletion method is strictly a recovery

of leasehold costs over the useful life of the reservoir. In contrast, percentage depletion is a tax

incentive, where part of the gross income is waived irrespective of the leasehold costs. Both

methods are discussed below.

COST DEPLETION

Cost depletion is based on the number of units produced in a given year. It can be written as,

_ \ depletion - (L - D,)

) --) Equation 3-24

where L is the leasehold cost, DC is the cumulative depletion up to a given year, u is the number

of units produced in a given year and R is the remaining reserves at the end of a given year.

Considering the uncertainties involved in estimating the remaining reserves, the value of the

remaining reserves, R, can change reflecting any new knowledge gained over the period of

production.

Example 310

Leasehold costs associated with finding a new field are $1 million, It is estimated that the field contains 200,000 bbls of

recoverable oil. Assuming that the field has produced 30,000 bbls in the first year, 27,000 bbls in the second year and

20,000 bbls in the third year, calculate the amount of depletion this period usingthe cost depletion method.

24 Mohcin Kelkcir, Ph.D., J. D.

0 0 0 0 0 0

Solution 3-10


depletion = (L - D) ( U

T1 i)

For the first year,

depletion = (1,000,000) 30,000 00000)

= $150,000

where L = $1,000,000, u = 30,000 bbls and R = 200,000 - 30,000 = 170,000 bbls.

For the second year,

/ 27, 000 \ depletion = (1,000,000 - 150,000) (170000)

= $135,000

For the third year,

(143,000

)

depletion = (1,000,000 - 150,000� 135,000) 1

20,000

= $100,000

The amount will change over the production period. Eventually, the leasehold costs will be recovered by the end of the

production period. However, the method has the flexibility to account for reduction or increase in the remaining reserves.

For example, after the third year, based on evaluation of the production data, if we observe that the remaining reserves are

150,000 barrels instead of 123,000 barrels, the depletion in the third year can be calculated as:

\ depletion = (1,000,000 - 150,000 - 135,000)

20,000

+ 20,000) = $84,117

PERCENTAGE DEPLETION

Unlike the cost depletion method, which allows recovery of the leasehold costs only, the

percentage depletion is calculated based on a certain percentage of gross income per year. The

idea of percentage depletion is analogous to depreciation of tangible property. Depreciation is a

certain percentage of the cost of the asset each year over the useful life of the asset. Percentage

depletion is a certain percentage of the diminished value of the producing property over the

producing life of a property. In the United States, for oil and gas properties, percentage

depletion is restricted to 15% of the gross revenues with the constraint that the amount should

not exceed 50% of the taxable income.

Exam ple 3-11

An oil property has a gross income of $500,000 per year. The deductible expenses excluding depletion are $380,000 per year.

Calculate the amount of depletion using the percentage depletion method.

Solution 3-11



25

Wo

percentage depletion = 0.15 x 500,000 = $75,000

However, we need to check if it exceeds 50% of the taxable income,

taxable income before depletion = gross income - expenses = 500,000 - 380,000 = $120,000

50% of taxable income = $60,000

Since $75,000 exceeds $60,000, the allowed is

For a prospective oilfield, the leasehold costs ore estimated to be $60,000. The field is expected to produce 100,000 bbls? in the

first year with a decline of 12% per year. If the initial reserves are estimated to be 1 million bbls?, calculate the allowed cost

depletion for the first five years.

Problem 3-19

For a gas field currently in development leasehold costs are estimated to be $1 million. The field is estimated to produce 5

billion SCF of gas over its life. The field produced 700 million SCF in the first year and 650 million SCF the following year. After

two years, based on the field data, it was determined that the remaining reserves beyond two years of production are 3 billion

SCF as against 3.65 billion SCF as originally anticipated. If the field declines in production at a rate of 791. per year calculate the

allowed cost depletion for the first seven years of the production.

Problem 3-20

An oil company has a gross income of $500,000 for the year. Operating expenses, excluding depletion, are $420,000. If the

percentage depletion allowed is 15%, calculate the allowable percentage depletion.

Problem 3-21

A producing property is expected to generate $200,000 in the first year with operating expenses of $90,000. If revenues decline

at a rate of 10% and operating expenses increase at a rate of 4%, develop a percentage depletion schedule for the first five

years.

The leasehold costs for this property are $200,000. If the original reserves are estimated to be 120,000 bbls and the well

produces 20,000 bbls in the first year with a decline of 10% per year, calculate the cost depletion schedule for the first 5 years.

Compare it with the percentage depletion schedule.

ITAX CONSIDERATIONS

In calculating the taxable income for the purpose of tax assessment, we first need to calculate gross

revenue (or income). This income represents the total production multiplied by the sales price per

unit of production. Once the royalty interest is paid (percentage of gross revenue), the remaining

gross revenue is subject to production-related taxes.

Production related taxes are assessed by state and local governments. These taxes include

severance and ad valorem taxes. Severance taxes are assessed as a percentage of the remaining

gross revenue. The percentage varies from 0% to 35% for different states. Ad valorem taxes are

levied against the assessed value of the tangible equipment. The equipment is appraised and the tax

is assessed as a certain percentage of the appraised value.


Mathematically, the taxable income can be calculated as,

taxable income before depletion = gross income - royalty payments

�production related taxes

�direct operating cost

�amortization (of IDCs)

�depreciation

Using the taxable income before depletion, for small producers, depletion is calculated using both

the methods of cost depletion and percentage depletion, and a higher of the two values is chosen.

For integrated oil companies, depletion is calculated using the method of cost depletion,

taxable income = taxable income before depletion

�depletion

fl14DII1

Once the taxable income is calculated, the applicable federal tax is calculated using the appropriate

tax rate. Using the tax rate, the federal tax is calculated as,

federal tax = taxable income x tax rate Equation 3-27

Once the federal tax is calculated, the net revenue after tax is calculated using,

net revenue = gross revenue - royalty payments - operating costs

�production related taxes - federal taxes

WRT-THERIM

The following two examples illustrate the economic evaluation of a property, one owned by a small,

independent, producer and the other owned by a large oil company.

Example 3-12

A large oil company is interested in bidding on a prospect. The leasehold costs are expected to be $500,000. During the same

year, $1 million will be spent on intangible drilling costs and $1 million will be spent on tangible equipment. During the second

year, an additional $1 million will be spent on tangible equipment. Assume a useful equipment life of seven years. Depreciation

is carried using a combination of DDB and straight-line method. Production is expected to begin in the third year. Based on the

analysis of nearby wells and geophysical data, the initial production will be equivalent to 25,000 barrels in the first year and will

decline at a rate of 8% per year. Production is expected to last at least 12 years. Assume that the recoverable reserves are

estimated to be approximately 200,000 barrels. The oil price can be assumed to be $80/bbl during the entire production phase.

The royalty payment will be 12.5%; the severance tax rate is 7% and the income tax rate is 34%. The operating costs, including

the ad valorem tax, are expected to be $200,000, increasing at a rate of 4% per year. Assume the salvage value at the end of 12

years to be zero, calculate the net present value if the MROR is 15%. What is the RUR on this property?

Solution 3-12

The following table shows a summary of the calculations. Sample calculations follow.

p Economic Evaluation in the Petroleum Industry


I

Gross Operating Severance Taxable Net

Year Production Income Costs Royalty Tax Depreciation Depletion Amortization Income Tax Revenue

0 (2,500,000) (2,500,000)

1 (1,000,000) 285,714 700,000 (985,714) (335,143) (664,857)

2 25,000 2,000,000 200,000 250,000 122,500 489,796 63,258 60,000 814,446 276,912 1,150,588

3 23,000 1,840,000 208,000 230,000 112,300 349,854 58,197 60,000 821,249 279,225 1,010,075

4 21,160 1,692,800 216,320 211,600 103,684 249,896 53,541 60,000 797,759 271,238 889,958

5 19,467 1,557,376 224,973 194,672 95,389 190,893 49,258 60,000 742,191 252,345 789,997

6 17,910 1,432,786 233,972 179,098 87,758 173,539 45,317 60,000 653,102 222,055 709,903

7 16,477 1,318,163 243,331 164,770 80,737 173,539 41,692 614,094 208,792 620,533

8 15,159 1,212,710 253,064 151,589 74,278 86,769 38,357 608,653 206,942 526,837

9 13,946 1,115,693 263,186 139,462 68,336 35,288 609,421 207,203 437,506

10 12,830 1,026,438 273,714 128,305 62,869 32,465 529,085 179,889 381,661

11 11,804 944,323 284,662 118,040 57,840 29,868 453,912 154,330 329,450

12 10,860 868,777 296,049 108,597 53,213 27,478 383,440 130,370 280,549

13 9,991 799,275 307,891 99,909 48,956 25,280 317,239 107,861 234,658

Sample Calculations

For year one, there is no production. The gross income reflects $500,000 in leasehold costs, $1 million in tangible costs and $1

million in intangible costs (IDCs). 70% of the lDCs are deducted in the first year and the rest is amortized over five years using

straight-line depreciation.

0.3($1,000,000) amortization = = $60,000

Depreciation is calculated using a combination of DDB and straight-line method. The first million starts depreciating in the first

year, and the second million dollar starts depreciating in the second year. According to the accelerated cost recovery system

(ACRS), the depreciation will begin in the same year during which the equipment is placed in service. Since $1 million has been

spent on tangible items during this year, assuming a seven year useful life for equipment,

depreciation = ($1,000,000) = $285,714

taxable income = -amortization - depreciation

= -0.7(1,000,000) - 285,714

= $985,714

During this year, the company makes no revenue; therefore, taxable income includes only the costs that are allowed to be

deducted. Assuming that the company has several ongoing projects resulting in possible tax savings in other profitable projects

due to negative income for this property, we can calculate the tax paid as,

tax paid = taxable income x tax rate = 0.34 x -985,714

= -$335,143

Therefore,

net revenue = gross revenue - royalty payment - operating costs - severance tax - tax paid

= -1000,000 + 335,143

- -$664,857

For year three,


WO

bbl 80 gross income = 25,000�x $-

year bbl

= $2,000,000 royalty = 0.125 x 2,000,000

= $250,000 severance tax = (gross income - royalty)state tax rate

= (2,000,000 - 250,000)(. 07)

= $122,500

For a large oil company, only the cost depletion option is allowed. Assuming that the cumulative oil production before

abandonment is 197,604 barrels (summation of the production over 12 years), using Equation 3-24,

"

cost depletion = (L - D) (fi 25,000 \

= (500,000) (25,000 + 172,604) = $63,258

Using Equation 3-25 and Equation 3-26,

taxable income = gross revenue - royalty payments - production related taxes - operating costs - depreciation - amortization - depletion

= 2,000,000� 250,000 - 122,500 - 2000,000 - 489,796� 60,000 - 63,258

= $814,446


federal tax = taxable income x tax rate

= 814,446 x 0.34

= $276,912


net revenue = gross revenue - royalty - production related taxes - operating costs - federal tax

net revenue = 2,000,000� 250,000 - 122,500 - 200,000 - 276,912

= $1,150,588

Similar calculations are carried out for other years. Production in each year is 0.92 times the production in the previous year;

whereas, the operating costs are 1.04 times the costs in the previous year.

The net present value at a 15% interest rate is calculated to be $450,130 and the rate of return is 19%. This indicates that the

project isfeasible. -

Example 3-13

A small producer has spent $200,000 in leasehold costs on a potential gas prospect in year one. The producer needs to spend

$300,000 in intangible development costs and $500,000 in tangible costs in the second year before the well can be brought to

production. Based on geological, geophysical, and reservoir information, the well will produce 200,000 MSCF of gas in the first

year followed by a 15% decline per year over eight years before abandonment. In the lease, 12.5% royalty interest is awarded

to the mineral owner and a 2.5% override royalty is given to the geologist. All tangible items are assumed to have a useful life

of seven years. Assume that the total production from the well is expected to be approximately 1 BCE. Assume the gas price to

be $4.25/MSCF. The severance tax is levied at 7% and the federal tax is levied at 34%. The operating costs are expected to be

$80,000 during the producing life of the well. If the MROR is 20%, calculate the net present value of the project. What is the

ROR for this project?

Solution 3-13

The following table shows the summary of the calculations. The sample calculations follow.



29

Production Gross Operating Severance Depletion Taxable Net

Year (MMSF) Income Costs Royalty Tax Depreciation Cost Percentage Selected Income Tax Revenue

0 (1,000,000) (11 0001 000)

1 200 850,000 380,000 127,500 50,575 142,857 39,043 74,534 74,534 74,534 25,342 566,583

2 170 722,500 80,000 108,375 42,989 102,041 21,447 108,375 108,375 280,720 95,445 395,691

3 145 614,125 80,000 92,119 36,540 72,886 3,091 92,119 92,119 240,461 81,757 323,709

4 123 522,006 80,000 78,301 31,059 52,062 - 78,301 78,301 202,283 68,776 263,870

5 104 443,705 80,000 66,556 26,400 43,385 - 66,556 66,556 160,809 54,675 216,074

6 89 377,150 80,000 56,572 22,440 43,385 - 56,572 56,572 118,180 40,181 177,956

7 75 320,577 80,000 48,087 19,074 43,385 - 48,087 48,087 81,945 27,861 145,555

8 64 272,491 80,000 40,874 16,213 - 40,874 40,874 94,530 32,140 103,263

9 54 231,617 80,000 34,743 13,781 - 34,743 34,743 68,351 23,239 79,854

Sample Calculations

The sample calculations are not repeated here since they are very similar to the previous example with two important

exceptions. The lDCs are deducted as operating costs in the first year; therefore, only in the first year are the operating costs

$380,000, which includes $300,000 in IDC5. The depletion is calculated using percentage depletion and cost depletion and the

highest value is selected. In our case, percentage depletion is selected in each year. The cost depletion is calculated in each

year by examining cumulative depletion in previous years. Therefore, after 3 years, since all the leasehold cost is depleted using

percentage depletion, no cost depletion exists. The NPV of the project is $451,131 and the rate of return is 33%. The project is

feasible. - -

Casc Study 3-2

In June 2008, Chesapeake (CHK) signed an agreement with Plains Exploration & Production Company (Plains) and formed a joint venture. Your job is to evaluate from the Plains perspective. Here are the essential facts. Plains agreed to pay CHK $1.65 billion dollars in cash for 20% working interest in CHK’s leasehold in J-Taynesvillc shale. CHK has 80% NM in these leases. CHK had a net leasehold of 550,000 net acres, which means Plains will hold about 110,000 net acres. In addition to $1.65 billion dollars in cash, Plains agreed to pay 50% of CI-IK’s 80% share of drilling costs until an additional $1.65 billion dollars is paid (i . e., for initial wells, Plains will pay 60 1/6 of the drilling and completion costs until, based on CHK’s share of 40%, all the funds are exhausted. For example, if the cost of drilling is $1 million, Plains will pay $600,000 and CI-IK will pay S400,000. Plains’ CI-IK share is S400,000. Therefore, Plains has to pay $400,000 for every well until the $1.65 billion is exhausted. After that, for every well, Plains’ share will be only $200,000 (20%). The spacing of the wells is expected to be 80 acres.

It is assumed that it will take approximately 2 months to drill and complete a well. CIIK is expected to have 40 drilling rigs ready to drill starting at the beginning of 2009. The royalty interest is 20% and the severance tax is 7%. The operating cost for these wells is assumed to be $12,000 per month and will remain constant.

The cost of drilling and completion is from S6 million to $10 million. Assume that 30% of these costs are IDC and 70% are capitalized expenses, which can be depreciated using a 7 year DDB/straight-line combination. The IDC costs are split so that 70% of the lDCs are deducted in the first year and the remaining 30% are amortized over a five year period- Please note that Plains is eligible to take the depreciation and IDC deduction for the expenses it paid. For the initial investment of S1.65 billion, assume that, based on 80 acres spacing, the leasehold bonus will be calculated as leasehold costs and, using cost depletion, deducted over the life of the well. Assume that the gas price can vary from S5/MSCF to S7/MSCF held constant over the life of the well. The income tax rate is assumed to be 35% k typical production profile for the best and worst well is provided below.

Production (MMSCF/Yeai) Year Best Worst

1 4745 2555 2 902 485 3 595 320

464 250 5 385 207 6 335 180

298 161 5 271 146 9 250 134 10 232 125


I. Assuming that Plains paid $1.65 billion up front, how much did Plains pay per acre? 2. Assuming the ten year cumulative production represents the EUR, what is the 1° & U Coot for Plains? Assume 80 acre sparing.

If the cost needs to be less than S2/i4S( li, do you think that Plains made a good deal? 3. Assuming that the MROR is 15%, what is the NPV for Plains for the entire project based on the best and the worst ease

scenarios? Is it a good deal?

(’se Study So/ursosa 32

1. Leasehold Costs per Acre $ $1.65 x iO

110,000 = $15,000/Acre

Acre2. F&DCosts

The cumulative production for the best and the worst case scenarios is between 8.477 BCF and 4.56 BCF. Only 80% of the production belongs to the operator; therefore, under the worst case scenario, F & D costs are:

80 x 15,000 + 10 x 10 6 F & D Costs= 4.56 x 10 6 .a

= $3.1/MSCF

Under the best case scenario, F & D Costs are:

80 x 15,000 +6x 10 6 F & D Costs

8.477 x 10’ x 0.8 = $1.06/MSCF

The costs vary between S1.06/MSCF and S3.1/MSCF with an average of $2.1/MSCF. This number is slightly greater than S2/MSCF and there is a lot of uncertainty in F & D costs making this project uneconomical.

3. Net Present Value per Well

To calculate the net present value of the well, we need to consider four scenarios. In one scenario, Plains will pay 60 1/0 of the drilling costs but will only receive 20% of the production, and in another scenario, Plains will pay for 20% of the drilling costs and get 20% of the production. We need to consider these scenarios under the best and worst cases.

Assuming 550,000 acre spacing, the total number of wells that can be drilled using 80 acre spacing is equal to 6,875. Using an estimate of 40 drilling rigs, and assuming a period of 2 months to drill and complete each well, we can calculate how many 2 month periods it will take to drill the wells. It will be 172 periods or 28.7 years. Plains will have to contribute SO°/s of CIIK’s drilling and completion costs until it exhausts S1.65 billion. SO% C1IK’s contribution is equivalent to 40% of the drilling and completion costs. Under the worst case scenario, Plains will contribute 40% of SlO million for each well’s drilling and completion costs. That will be S4 million. Under the best ease scenario, Plains will contribute S2.4 million toward CIIK’s costs. Please note that in addition to contributing toward Cl IK’s 50% of drilling and completion costs, Plains will also be contributing 20% of its own drilling and completion costs. Under the worst ease scenario, we can calculate the number of periods required to exhaust S1.65 billion.

1.65 x io No. of periods for 60% contribution = 40 4 <

= 10 periods

For the best case scenario,

1.65 x 109

No. of periods for 60% contribution= 40 2.4 < 106 = 17 periods

That means under the worst case scenario, Plains will start contributing only 20% of drilling and completion costs starting in the 1 1’ period; whereas, for the best case secnaiii. Plains will start contributing 20% of the costs starting in the 18I period.



NPV - Optimistic Case - 60% Contribution

Gross Operating Taxable Net Prod Revenue Cost Severance IDC Deprec. Depletion Income Taxes Revenue

Year MMSCF (8MM) ($MM) (8MM) (8MM) (8MM) ($MM) ($MM) ($MM) (8MM) 0 -3.60 -3.60

4,745 5.31 0.0288 0,372 0.76 0.72 0.67 2.77 0.97 3.95 2 902 1701 0.0288 0.071 0.065 0.51 0.13 020 007 0.84 3 595 0,67 0.0288 0.047 0.065 0.37 0.08 007 0.03 0.56 4 464 0.52 0.0288 0.036 0.065 0.26 0.07 0.06 0.02 0.43 5 385 0.43 0.0288 0.030 0.065 0.22 0.05 0.03 001 0.36 6 335 0.38 0.0288 0.026 0.065 0.22 0.05 (0.01) (0.00) 0.32 7 298 0.33 0.0288 0.023 0.22 0.04 0.02 0.01 0.27 8 271 0.30 00288 0.021 0.04 0.22 0.08 0.18 9 250 0.28 0.0288 0.020 0.04 0.20 0.07 0.16 10 232 0.26 0.0288 0.018 0.03 0.18 0.06 0.15

In the table above, royalty interest calculations are implicitly included in the gross revenue. The gross revenue - production x 0.2 x 0.8 x gas price. This includes the 20% share of Plains and 80% NRI. It is assumed that Plains will be allowed to depreciate its own contribution to tangible costs. Depiction costs are the leasehold costs per well, which is 515,000 x 80 for 80 acre spacing. The price of the gas is assumed to be S7/MS(-1. The NPV for this situation is S1.65 million.

NPV Pessimistic Case - 60°/s Contribution


Year MMSCF (8MM) (8MM) (8MM) (8MM) ($MM) (8MM) (8MM) (8MM) (8MM) 0 -6.00 -6.00 1 2555 2.04 0.0288 0.143 1.26 1.2)) 0.67 (1.26) (0.44) 2.31

� 2 485 0.39 0.0288 0.027 0.108 0.86 0.13 (0.76) (0.27) 0.60 3 320 026 0.0288 0.018 0.108 0.61 0.08 (0.60) (0.21) 0.42

� 4 25)) 0.20 00238 01014 0.108 0.44 0.07 (0.45) (0.16) 0.32 5 207 0.17 0.0288 0.012 0.108 0.36 0.05 (0.40) (0.14) 0.27 6 180 0.14 0.0288 0.010 0.108 0.36 0705 (0.41) (0.15) 0.25 7 161 0.13 0.0288 0.009 0.36 0.04 (0.32) (0.11) 0.20 8 146 0.12 0.0288 0.008 0.04 0.04 0.01 0.07

� 9 134 0.11 0.0288 0.008 0.04 0.04 0.01 0.06 10 125 0.10 0.0288 0.007 0.03 0.03 0.01 0.05

This procedure is similar to the previous case. The gas price is S5/MSCF. ’l’he NPV for this option is 5-2.71 million.

NPV - Optimistic Case - 20% Contribution


Year MMSCF (8MM) ($MM) ($MM) ($MM) (8MM) ($MM) (8MM) (8MM) (8MM) 0 -1.20 -1.20 1 4,745 5.31 0.0288 0.372 0.25 0.24 0.67 3.75 1.31 3.60 2 902 1.01 0.0288 0.071 0.022 0.17 0.13 0.59 0.21 0.70 3 595 0.67 0.0288 0.047 0.022 0.12 0.08 0.36 0.13 0.46 4 464 11.52 0.0288 0.036 07)22 0.09 007 028 010 0.36 5 385 0.43 0.0288 0.03)) 0022 (11)7 0705 0.22 0.08 0.29 6 335 0.38 0.0288 07026 0022 0.07 0.05 0.18 0.06 0.26 7 298 0.33 0.0288 0.023 0.07 0.04 0.17 0106 0.22 8 271 0.30 0.0288 0.021 0.04 0.22 0.08 0.18 9 251) 0.28 11.0288 0.020 11.04 0.20 0.07 0.16 10 232 11.26 0.0288 0.018 ((03 0.18 01)6 0.15

The NP\T for this situation is S3.46 million.


?’.PV Pessimistic Ca se 20% Contribution

Gross Operating Taxable Net Prod Revenue Cost Severance IDC Deprec. Depiction Income Taxes Revenue

Year MMSCF ($MM) ($MM) ($MM) ($MM) ($MM) ($MM) ($MM) ($MM) ($MM) 0 2.00 2.00

255 2.04 0.0288 0,143 0.42 0.40 0.67 0.38 0.13 1.74 2 485 0.39 0.0288 0.027 0.036 0.29 0.13 (0.12) (0.04) 0.37 3 320 0.26 0,0288 0.018 0.036 0.20 0.08 (0.11) (0.04) 0.25 4 250 0.20 0.0288 0.014 0.036 0.15 0.07 (0.09) (003) 0.19 5 207 0.17 0.0288 0.012 0.036 0.12 005 (0.09 (0.03) 016 6 180 0.14 0.0288 0010 0036 0.12 0.05 (0.10) (0.()3) 0.14 7 161 0.13 0.0288 0.009 0.12 0.04 (0.07) (0.03) 0.12 8 146 0.12 0.0288 0.008 0.04 0.04 0.01 0.07 9 134 0.11 0.0288 0.008 0.04 0.04 0.01 006

10 125 0.10 0.0288 0.007 0.03 0.03 0.01 0.05

The NPV for this situation is S0.30 million.

Using this information for individual wells, we can calculate the NP\T of the overall project by assuming that 40 drilling rigs are available and it takes 2 months to drill and complete each well. We will do these calculations based on a 2 month period. The interest rate used will be 15/6 = 2.5% per period in our calculations. Recall that, in the pessimistic case scenario, Plains will have to contribute 60% for 10 periods; whereas, it will have to contribute 60% for 17 periods under the optimistic scenario. Knowing the NP\ under all situations, we can calculate NPV for the overall project.

NPV Overall Project Pessimistic Scenario

(1 +.025) 10 - 1 1 [(1 + 025)162 - 1 NPV = -1,650 + 40 x (-2.71) + 40 x (03) = -$2,128 million

0.025(1 + .025)101 t0.025(1 + .025)162

NP\ - Overall Project - Optimistic Scenario

NPV (1 +.025) 17 - 1 (1 +.02 S)155 - 1

= -1,650 + 40 x (1.65) [0025(1 + .025)17] + 40

x (3.46) [0.025(1 + .025)155] = $4,671 million

Under the pessimistic scenario, Plains will lose 52.13 billion; whereas, under the optimistic scenario, Plains will make S4.67 billion. The average of the two is S1.271 billion, making the project economically feasible.

Considering that the price of gas plummeted in 2009 and continues to be low to this date makes this project less attractive. It is possible that, with enhancement in drilling and completion technk1ucs, the cost of drilling and completion will go down and the amount of gas produced per well may be increased. l’hat is the only possible scenario that would make this project more attractive.

Problem 3-22

An oil company purchased a property for $2 million. Operating expenses are $1 million per year with an additional $31bbl

pumping cost. The well is estimated to produce 400,000 bbls in the first year with on annual decrease of 10% over the next five years. In addition, the company has purchased oil field and pumping equipment for $3 million that is subject to a DDB and

straight-line combination depreciation method over a five year period. The oil percentage depletion allowance is 15% and the local tax is expected to be 7%. If the income tax is levied at a rote of 34%, calculate the present worth of the project if the MROR is 15%. Assume the price of the oil to be $201bbl. The life of the project is five years.

Problem 3-23

An oil company wants to dispose of a producing property. The property is currently producing 200,000 bbls/year and is expected to decline at a rate of 10% per year. The oil price over the life of the project is assumed to be $20/bbl. The existing equipment on

b the property is worth $300,000, which is subject to the DDB/straight-line method over a seven year period. Operating costs are $51bbl. The royalty interest in the property is 118. Cost depletion is al/owed on the leasehold costs (purchase price-equipment

worth). If the remaining reserves are estimated to be $1 million and the tax rote is 34%, what is the minimum price you will be willing to pay for the property? Assume the MROR to be 20% and the life of the property to be seven years.

If you are the seller and will have to pay a 28% tax on the proceeds and wont to recover the taxes in your soles price, at what

price will you be willing to sell the property if the MROR is 15%? Assume that the seller has $1 million remaining in the leasehold



cost basis Also the seller purchased the tangible property four years ago at price of $800,000. It has already been depreciated

over the last four years using the DDB/straight-line method

Problem 3-24

What is the maximum price you (a large oil company) are willing to bid on an offshore property (excluding the lease bonus) if it is estimated to cost $3 million to drill an exploratory well at the end of the first year and, if successful, another $4.5 million to

drill delineation wells at the end of the second year. The platform construction should be complete by the end of the third year

with a cost of $15 million. Drilling the remaining wells and installing production facilities will be done at the end of the fourth

year at a cost of $35 million. The production should commence at the beginning of the fifth year with initial production

estimated to be 2 million bbls in the fifth year. Based on nearby fields production will decline at a rate of 10% per year for the

next ten years. The operating costs are $71bbl and the sale price is $211bbl. Assume that 30% of the intangible drilling costs

need to be considered amortized costs and 705Y. can be expensed in the year production begins. The amortized costs are

amortized over a five year period using a straight-line method starting in the year IDC5 are incurred. The capitalized cost can be depreciated over a ten year period using the ACRS system. Assume that intangible drilling costs in year one are $1.5 million, in

year two are $2.5 million, and in year four are $10 million. Assume that the recoverable oil is 13 million barrels. The income tax

rate is 3016, the royalty interest is 3116, and the MROR is 18%.

INTERNATIONAL CONTRACTS

This section discusses the implications of international contracts on economic evaluation. We introduce

the basic concepts of international contracts; however, each country adds its own criteria to these

contracts. We will not discuss the details of those criteria.

The oil business has always been a global business. Most international oil companies have searched for

oil wherever there are attractive targets. The key parameter is the cost of finding the oil. A country,

where that cost is small, attracts many suitors (Bertagne, 1992). In addition to the low cost of finding oil,

the international oil companies are also interested in favorable terms in the contract.

This section will discuss various contracts used between the host country and the international oil

companies to proceed with the exploration and development of potential hydrocarbon reserves. It is

important to remember that the objectives of the host country and the international oil company can

differ significantly. Most countries are afraid of exploitation, pollution, loss of national pride, and the

repetition of recent history at the hands of the western civilization (Bertagne, 1992). The host countries

like to be treated as equal and be part of the development of their own mineral resources so that it may

benefit the entire population of the host country. On the otherhand, the international oil companies are

afraid of varying tax rules, expropriation of oil and other assets, nationalization of a private companies,

and political uncertainties. The oil companies’ main interest is economical. They prefer to produce the

hydrocarbons in the most optimal way so they can maximize the benefits. To structure a contract

between these two parties that will create a win-win situation for both parties is a challenging task. The

solution is the various types of contracts that have evolved over the last thirty years that try to balance

the interest of both the host country and the international oil company.

In the first part of this section, we present the purpose of each of the parties involved in the contract so

that the understanding of the terms in the contracts becomes easy. In the next three sections we

illustrate the three types of contracts most commonly used: concession agreements, production sharing

contracts and service contracts. For host countries, concession agreements require the least

involvement; whereas, the service contracts require the most involvement. We will discuss the

advantages and disadvantages of these contracts and illustrate their application with several examples.


OBJECTIVES

When a contract between a host country and an international oil company is signed, both parties

want to achieve certain objectives. Independent of the type of contract that is signed, both parties

try to satisfy most of their objectives. To understand the terms included in the contract, it is

important to consider the objectives of the parties involved.

OBJECTIVES OF THE HOST COUNTRY LMLi ske ll, 1984

The objectives of the host country can be divided into three broad categories; financial, political

and technical. Each of these objectives is described below.

FINANCIAL

The host country is ultimately interested in building a foreign exchange and capital base so

that the money can be invested in other capital-intensive projects. Unfortunately, to

develop this capital base requires an initial investment. Most exploration efforts involve

significant risk and capital investment. The host country may have only limited capital

resources that it may not want to tie in limited projects. By inviting an international oil

company to explore for and produce hydrocarbons, the host country can eliminate, or at

least minimize, the initial capital investment. It is also true that international oil companies

can get loans with more favorable terms than the developing country. If the project is

successful, by ensuring the appropriate terms in the contract, the host country can secure

significant financial benefits from the production of hydrocarbons. These benefits may be

reinvested in other projects of national interest.

Most developing countries have been previously colonized by Western countries. Awarding

contracts to international oil companies is a sensitive issue and involves national pride. To

overcome foreign dependence, the host countries prefer to maintain control over the

operations of the project, as well as the hydrocarbons produced as a result of the contract.

Instead of being passive beneficiaries, the host countries prefer to play an ctive role in the

field development so that the natural resources are optimally exploited to the country’s

benefit. Further, by controlling the hydrocarbon production, the country may be able to

influence desired foreign policy goals. It can also reduce imports and gain economic

efficiency.

TECHNICAL

Exploration of new areas requires increasingly sophisticated technologies which the host

country may not possess. By signing a contract with an international oil company, the host

country can benefit from the technical expertise within that company. Not only will this

reduce the risk in the exploration efforts, it will utilize the latest technology in the

development of the oil field.



Additionally, the host country wants to gain technological independence so that, eventually,

future fields can be discovered using local talent and local companies.

Therefore, the host country prefers that the international oil companies hire a domestic

workforce, provide education grants and local R&D effort, and transfer technology to the

local talent. The companies also prefer local companies for the outside contracts.

The objectives of the host countries should be balanced by the objectives of the international

companies.

(OBJECTIVES OF INTERNATIONAL OIL COMPANIES

As with the host countries, the objectives of international oil companies can be described in

three broad categories: financial, operational and political.

FINANCIAL

International oil companies are incurring significant risks exploring for and producing

hydrocarbons from yet-to-be-explored areas. In return, the companies expect that they are

provided with a reasonable return commensurate with their risk. The companies want to

maximize their benefits with a minimum amount of cost. They also prefer that the initial

investment be recovered as quickly as possible (short payback period) in order to minimize

the impact of any political uncertainties. The companies also prefer that the revenue

generated be repatriated and their share of crude sold in the open world market.

International oil companies prefer to have operational control over the project. Rather than

sharing operational responsibilities with some bureaucratic agency, they prefer to operate

the field so as to preserve the economics. The international oil companies fear that by

sharing control with a host country or its wholly owned subsidiary, production decisions may

be made based on domestic political considerations rather than sound economics.

POLITICAL

International oil companies realize that their existence depends on the ability to explore for

and produce hydrocarbons from host country’s sovereign land. It is, therefore, important

that good working relationship be maintained with the host country and the local

community so that the company can secure additional concessions when awarded. The

company, therefore, is willing to spend resources on the development of local communities

as well as local talent.

On the other hand, the companies want to be careful as to what terms are agreed to in signing

the contract. The companies prefer not to set a precedent in one contract that can be used

against them in either the same country or in other countries. If the company gives in too much,

it will be subjected to similar demands in the future.


As shown, the objectives of the two parties differ significantly. The types of contracts that are

signed between the parties reflect some objectives from both parties. We will discuss these

contracts in detail below.

TYPES OF CONTRACTS (BLINN, DUVAL, LELENCH, & PERTUZIO, 1986) (KATZ, 1992)

The three types of contracts most commonly used between a host country and an international oil

company are concession agreements, production sharing agreements and service contracts. In this

section, we will briefly describe each of these contracts highlighting their common features and the

fiscal impact of each of these contracts. Detailed economic evaluation of each of these contracts is

differed until the next section. It is important to note that none of these contracts is economically

inherently superior to the other types. By changing the individual terms of a particular contract, any

contract can be made more or less beneficial to the parties involved.

Although we will be discussing the contracts between the host country and an international oil

company, as a practical matter, a contract is typically signed between a nationalized oil company

and an international oil company. Creation of national oil company is helpful in separating the

financial benefits recovered from oil revenues. Also, by creating a company, it is easier to develop

technical expertise among the citizens of the country. It is also believed that a separate company

will be able to run more efficiently without the burden of imposing bureaucracy. Although separate,

these national companies have the same objectives as the host countries. Table 3-1 lists the

countries and their national oil companies. The most notable exceptions in this list are the United

States and the United Kingdom who do not have nationalized oil companies.

CONCESSION AGREEMENTS

A concession agreement is similar to the standard oil and gas lease signed between a mineral

owner and producer. In this agreement, the host country assigns to the international oil

company the rights to explore for and produce from a concession area. In return, the host

country receives royalty payments and income taxes from the proceeds. The host country may

also receive a signing bonus and other incentives, such as training of the domestic work force. In

this type of agreement, the host country does not usually participate in the day-to-day

operations of the field; however, the agreement may be modified to change that. This modified

contract is called a participation agreement.

PRODUCTION SHARING CONTRACTS

In this type of contract, the international oil company bears the cost and risk during the

exploration phase. However, if successful, the host country participates during the production

phase by paying for part of the costs and receiving part of the production based on the agreed

upon formula. In addition, the host country also receives income taxes on the international

company’s profits.

SERVICE CONTRACTS

In service contracts, the international oil company acts as a service contractor. The host country

will invite the international oil company to provide certain services. This includes exploration of



a new area as well as operation of the field. The international oil company in return will receive

fixed fees as well as fees based on barrels of oil produced. The contractor does not hold any

interest in the producing field. In that sense, the service contracts have the least control by the

international oil companies.

Table 34 National Oil Company Names in Various Countries (Allen & Seba, 1993)

Company Country

ADNOC Abu Dhabi

SONANGOL Angola

SONATRACH Algeria

YPF Argentina

VPFB Bolivia

Petrobras Brazil

Petro-Canada Canada

ENAP Chile

EcoPetrol Columbia

CEPE Ecuador

EGPC Egypt

CPGTotaI France

TOTAL France

Pertamina Indonesia

ONGC India

NIOC Iran

INOC Iraq

ENI Italy

JNOC Japan

KazMunaiGas Kazakhstan

KOC Kuwait

Petronas Malaysia

Pemex Mexico

NNPC Nigeria

Statoil Norway

OGDC Pakistan

PetroPeru Peru

Trintoc Trinidad & Tobago

ANCAP Ururguay

PDVSA Venezuela

Table 3-2 lists the countries that have adopted a particular type of contract. The particular terms

of the contract vary significantly. In general, a country with less geologically promising regions is

more likely to set lenient terms to attract international investment. On the other hand, a

country with geologically promising regions can set stringent terms to leverage the benefit from

a successful venture.


Table 32: Current Distribution of Exploration and Production Contract Types (Allen &

Seba, 1993)

Concession

(Royalty/Income Production

Region Tax) Sharing Risk Service

North Canada

America United States

Latin Argentina Bolivia Brazil

America Bahamas Guatemala Chile

Barbados Guyana Columbia

Belize Honduras Ecuador

Costa Rica Panama Peru

Paraguay Uruguay

Surinam

Trinidad

Europe Austria Albania

Bulgaria CIS

Denmark Romania

France Yugoslavia

Germany Hungary

Greece

Greenland

Ireland

Italy

Malta

Netherlands

Norway

Poland

Portugal

Spain

Sweden

Turkey

United Kingdom

Africa Cameroon Algeria Madagascar

Chad Angola

Congo Benin

Gambia Burundi

Ghana Cote D’lvoire

Guinea Bissau Egypt

Mali Eq. Guinea

Morocco Ethiopia

Namibia Gabon

Niger Guinea

Nigeria Mauritania

Senegal Kenya

Seychelles Liberia

Sierra Leone Libya

Somalia Mozambique


Chapter 3 Economic Analysis 39

Concession

(Royalty/Income Production

Region Tax) Sharing Risk Service

South Africa Sudan

Tunisia Tanzania

Zaire Togo

Zambia

Near Israel Bahrain

East United Arab Jordan

EmirateS Oman

Qatar

Syria

Yemen

Far East Australia Bangladesh Philippines

and Brunei China

Australia Cambodia India

Fiji Indonesia

New Zealand Laos

Pakistan Malaysia

Papua New Mongolia

Guinea Myanmar

South Korea Nepal

Thailand Sri Lanka

Vietnam

Irrespective of the type of contract signed, there is enough flexibility within individual terms of

the contract to make it more or less attractive. In evaluating the economic impact of the

contract, the following economic considerations are very important.

HOST COUNTRY’S NET SHARE

Depending on the terms of the contract, the net share of the host country can vary significantly

We can define the host country’s share as,

host country’s share = cash flow after costs - net revenue of contractor

where cash flow after costs is defined as,

cash flow after costs = gross revenue - project capital and operating costs

The net revenue of the contractor is the revenue received by the international oil company after

paying all royalties, taxes, and production to the host country; the net profit the contractor (the

company) will receive. As a percentage, we can calculate the host country’s share as,

host country’s share = cash flow after costs � net revenue of contractor

cash flow after costs

Equation 3-31

I.

40 Mohan Kelkar, Ph.D., ID.

Equation 3-31 presents the percentage of the revenue after costs received by the host country.

The larger the share, the harder it is to make the project more economically favorable. Figure 3-

1 presents the typical shares received by various countries. What is remarkable is that

approximately 90% of the countries receive more than 50% of the share. Also, another

noteworthy aspect of this graph is that the share of a host country does not depend on the type

of the contract. Ireland with the smallest share (25%) and Norway with the largest share (88%)

both have concession types of agreements. On the other hand, Guyana with a share of 51% and

Benin with a share of 88% both have production sharing types of agreements. These shares are

average shares calculated based on various scenarios. It is important that, before signing a

contract, this type of analysis is conducted to understand what share the host country will

receive under different scenarios.

Concession Agreement

rtnTh n-rrm U(("fl enjiway

uTunisia

Senegal Netherlands

Nigeria rm Papua New Guinea

Cameroon Poland

Somalia Trinidad

___-1 Thailand Morocco Denmark

u Pakistan Namibia Greece

Greenland Italy

Australia Germany

U.S.A. ,,rnrrrrr Portugal

New Zealand -rn France

___________________ Spain Argentina Turkey

Paraguay U.K.

0 10 20 30 40 50 60 70 80 90 100

Country’s Share



Production Sharing Contract

to Divoire

n

0

Panama

0 10 20 30 40 50 60 70

60 90 100

Country’s Share

Service Contract

Ecuador

Chile

Peru

Philippines

0 10 20 30 40 50 60 70 80 90

Country’s Share

Figure 3-1: Country’s share as afunction of the type of contract (Allinson, 1995)

ECONOMIC EFFICIENCY OF THE CONTRACT

An economically efficient contract will tax production from marginal fields very lightly, whereas,

it will heavily tax the most profitable fields. This efficiency is especially relevant when the

international oil company discovers marginal oil fields. If, for marginal oil fields, the host

country’s share does not change and the tax structure remains the same, these fields will be

very difficult to produce and cannot be economically justified. It is becoming increasingly

difficult to find giant oil fields. The focus is becoming more concentrated on small or marginal


fields. The host country will have to show some flexibility in contractual terms so that the small

fields may be developed in an economic efficient manner.

Efficient/"

0- /

Inefficient

/ --

NPV before ’ country’s share

Figure 3-2 shows a plot of an economically efficient contract and an economically inefficient

contract. For economically efficient contracts, the NPV of cash flow before the host country’s

share is closely tied with the NPV of the host country’s share, the line passing through the origin.

For inefficient contracts, the NPV of cash flow before the host country’s share is positive;

however, the NPV after the host country’s share is negative. That is, by participating in this

contract, the host country prevented the development of the field, which would have been

developed without their participation. The inefficiency results from the terms of the contract

that relate to gross revenue or a certain level of production rather than the profitability of the

project. A common example is a royalty payment that is a percentage of the gross revenue and

ignores costs. Another example is a local tax that is levied based on the gross revenue. Marginal

projects are particularly affected by such terms of the contract.

TAX STRUCTURE

The overall tax structure can be progressive or regressive. Progressive tax structure involves

increasing share of the host country as the profitability of the project increases. When the

project is less profitable, the proportionate share of the host country is small as well. This type

of progressive structure also encourages development of marginal or small fields. Figure 3-3

shows schematically both progressive and regressive tax structures. In regressive structures, the

host country’s share increases as the profitability of the project gets smaller. A neutral fiscal

region is one where the host country’s share is not affected by the profitability of the project.



100%

75°h

a) cc

to (O 50%

0 0

25%

0 0/c

Project Profitability

Figure 3-3: Progressive vs. regressive tax regime

In analyzing the economic impact of any project, the three sections above are important.

Analysis of these considerations will allow a proper evaluation of the projects under various

scenarios before reaching a final decision as to whether to invest money in a given project. In

the following sections, we will discuss these contracts in additional detail and evaluate their

economic impact.

CONCESSION AGREEMENTS

As stated before, in modern concession agreements, the host country has no direct involvement in

the management of day-to-day operations. In most instances, the host country will open up certain

areas for a sealed bidding process. The contracting companies compete for that area with incentives

including a signing bonus (up front) payment, royalties, and tax arrangements. If successful, the host

country will assign rights to the contracting company to explore for and develop certain areas in

return for a share of proceeds (royalty) and taxes. Although individual contracts vary, some common

features included in the concession agreement are:

� The international oil company, solely at risk, has the exclusive right to explore for and

exploit petroleum reserves in the concession area. The company is also responsible for costs

associated with these rights.

� The international oil company has a contractual obligation to supply part of the production

to the domestic market. Beyond that, the company can dispose of the remaining production

on the open market.

� During the exploration phase, the company has to pay surface rentals to the host country.

� The international oil company has to pay part of the gross revenue in kind or in cash.

� The international oil company has to pay income tax on the taxable income.

� The international oil company may have to pay a signing bonus and production bonus tied

to attaining a certain level of production.

� The international oil company owns the equipment and installations used in the operations.

� The contract will be in effect for a limited duration depending upon whether the exploration

is successful or not. For successful exploration, the contract may be in effect for 30 to 50

years.


One main disadvantage of the concession agreement is that the host country has no control over

the operation of the field. Also, the citizens of the country may not get a chance to learn the

technical skills so that they can pursue an independent energy course. One way to overcome some

of these difficulties is to require in the contract that part of the proceeds should be obligated

towards technical training of the domestic workforce. Also, the host country can enforce a

requirement that the number of foreign nationals be restricted to a certain quantity that can be

reduced as the operation proceeds.

A modification of the concession agreement is a joint venture agreement. In a joint venture

agreement, the host country (or its fully owned national oil company) participates with the

international oil company in the exploration and development phases. In general, the international

oil company carries the host country’s interest through the exploration phase. If the exploration

effort is successful, the host country backs in with the pre-determined interest. The percentage of

the host country can vary greatly; as little as 5% or it can be as high as more than 50%. The

percentage may also be different during the exploration phase and the production phase.

Another modification of a joint venture agreement is the formation of a new operating company in a

host country that is co-owned by the international oil company and the host country. To retain

management control, the host country may own more than a 50% share. Even under this

modification, the international oil company has an obligation to carry the interest of the host

country during the exploration phase and, in some instances, the development phase.

The advantage of a joint venture agreement from the host country’s perspective is the management

control or, at least, the influence. Also, by participating in the day-to-day operations, the domestic

labor force can learn the practical side of the operation. This should help in the future when the

domestic oil company ventures out on its own. The schematic of the overall cash flow calculations

are shown in Figure 3-4 for standard concession agreements and joint venture agreements. The flow

diagram also shows sample calculations starting with 1,000 units of gross revenue. The shaded

portion indicates the host country’s share. At the bottom of the figure, the host country’s share is

calculated as a percentage of the revenue after deducting costs. As shown, depending on the tax

rate, royalty rate and participation percentage, the host country’s share may remain the same

irrespective of the type of contract.

The following example illustrates the application of royalty, as well as joint venture agreements.

I I I



45

I

Concason Agreement Joint VaitureAgriiait

Gross

;

Less

Gross Revenue

$1,000

Less

TaxabI

$650

I Less

Equal To

F NCflow

$325

Host Country’s Share

Host Country’s She

2 59 4% 800

Figure 3-4: Schematic of cash flow profile for concession agreements

Example 3-14

A royalty agreement is signed between an international oil company and a national oil company. The agreement calls for 20%

of the royalty interest on the gross income. The capitalized expenditure can be depreciated over a seven year period using a

double declining balance method with the remaining balance to be depreciated in year seven. Depreciation can only start after

production has started. IDCs and leasehold costs are recovered as soon as production has commenced (this is called cost

recovery). If the taxable income is negative, it is zeroed out and the balance is carried forward into the next year. The income

tax rate is 55% per year. The following data for the expected costs and production rates have been gathered:

Signing Bonus $5 million in year zero

Capital expenditure: $30 million in year one

$50 million in year two

$60 million in year three

$80 million in year four


DC $5 million in year 1

$8 million in year 2

$10 million in year 3

Initial production: 3 million barrels in the fifth year

5 million barrels from year 6 to 10

Start decline in year 11

Production decline rate: 18% per year

Years of production: 15 years

Operating costs: $18 million in year five

Operating cost decline rate: 4% per year

Oil price: $80 per barrel assumed constant

MROR: 20%/year

Examine the economic feasibility of the project.

Solution 3-14

The following table shows the detailed calculations:

Gross Operating 0 � Cost Taxable � Net

Production Revenue Costs U U

Royalty1 Depreciation Recovery Income I Taxe Revenue

Year (MMSTB) ($MM) ($MM) (SliM) (5MM) ($MM) (5MM) MM) (5MM)

0 -500 -5.00

1 -35.00 I U I -35.00

2 -58.00 -58.00

3 -60.00 U I -60.00

4 -80.00 -80.00

5 3.00 240.00 18.001 48*1 62.86 28.00 83.14 41.73 128.27

6 5.00 400.00 17.28. �

80.0 44.90 257.82 1.80 160.92

7 5.00 400.00 16.59 � 80.00 32.07 271,34 U 149.24 154.17

8 5.00 400.00 15.931 � 80.0 22.91 281.17 � 1164 149.43

9 5.00 400.00 15.290 80.10 16.36 288.35 1.59 146.12

10 5.00 400.00 14.68k �

80.0 11.69 293.64 16.50 143.82

11 4.10 328.00 14.09 � 65.60 29.22 219.09 I 120.50 127.81

12 3.36 268.96 13.531 � 53,9 201.64 � 11.90 90.74

13 2.76 220.55 12.99� 44*1 163.45 89.90 73.55

14 2.26 180,85 12.471 � 36,7 132.21 7.72 59.50

15 1.85 14830 11.97 � 29,66 106.67 � 58.67 48.00

16 1.52 121.60 11.49 24.2 � 85.79 4.19 � 38.61

17 1.25 9931 11.03� 19*4 68.74 35.81 30.93

18 1.02 81.77 10.59 �

16.5 54,82 � 3.15 24.67

19 0.84 67.05 10.16 U 13.41 43.47 N 23.91 19.56


Years zero to four, the gross revenue is the expended amount and includes lease bonus, lDCs and tangible costs. Tangible costs

are depreciated starting year 5 with a double declining balance (DDB) method with a twist in that in year 7 the remaining

balance is depreciated. This is needed; otherwise, the operator will not be able to depreciate the total amount of capitalized

costs. IDCs and leasehold costs are treated differently than in domestic contracts. The total IDC and leasehold costs is equal to

$28 million. This amount can be deducted as soon as production begins with a caveat that the taxable income cannot be less

than zero. This means that if the taxable income is less than zero, only the cost recovery will be allowed up to the point that the

taxable income becomes zero. The balance for cost recovery is carried forward in subsequent years until all of the lDCs and

leasehold costs are deducted. This is a significant advantage over domestic contracts. Instead of spreading it out over the life of

the project, the leasehold costs are immediately recovered after production begins.

In year five,



gross revenue = $80 x 3 x 106 = $240 million

royalty = 0.2 x 240

= $48.0 million

depreciation = x (60 + 50 + 60 + 50)

= $62.86 million taxable income = 240.0 - 18.0 - 48.0 - 62.86 - 28

= $83.64 million tax = 83.64 x 0.55

= $45.73 million net income = gross revenue - royalty - operating costs - tax

= 240.0 - 18.0- 48.0 - 62.86- 28

= $128.27 million

Using the last column of net income, the NPV of the company can be calculated to be $134.05 million. The shaded portion in the table represents the contribution made to the host country. The host country collects the money in two different ways,

through taxes and through royalties. If we calculate the NPV of the royalty payment, it is equal to $142.41 million and if we calculate the NPV of taxes, it is equal to $252.97 million. Combining the total amount received by the host country divided by the total amount received by the operator and the host country, the host country’s share is 75%. For a highly profitable project, we expect the royalty payment (20%) and taxes (55%) will roughly add up to the host country’s share. For marginal projects, this can make a difference and host country’s proportionate share would increase significantly.

Let us consider the previous data except we will assume that, under a pessimistic case, the production is half the original case

(i.e., 1.5 million barrels instead of 3 million barrels, etc.) Using those numbers, we can construct the following table:

Gross Operating Cost Taxable Net

Production Revenue Costs � Royalty. Depreciation Recovery Income Taxe1 Revenue

Year (MMSTB) ($MM) ($MM) ($kIM) ($MM) ($MM) ($MM) MM) ($MM)

0 -5.00 � 0.10 � .00 -5.00

1 -35.00 N 0*0 *00 -35.00

2 -58.00 0.1 � .(t) -58.00

3 -6000 N 0.00 � 0.00 -60.00

4 -80.00 � 0.0 � ,00 -80.00

5 1.50 120.00 18.00� 24*0 62.86 15.14 - N - 78.00

6 2.50 200.00 17.28w � 40.0 44.90 12.86 84.96 4.73 95.99

7 2.50 200.00 16.59 N 40.00 32.07 111.34 � 61.24 82.17

8 2.50 20000 15.93 � 40.0 22.91 121.17 � 6.64 77.43

9 2.50 200.00 15.290 40*) 16.36 128.35 7*59 74.12

10 2.50 200.00 14.68 � 40.0 11.69 133.64 7.50 71.82

11 2.05 164.00 14.09 � 32.80 29.22 87.89 N 48.34 68.77

12 1.68 134.48 13.53 � 261 94.06 � 5.73 42.33

13 1.38 110.27 12.990 22.*) 75.23 4*.38 33.86

14 1.13 90.42 12.47_ � 18.8 59.87 � 3.93 26.94

15 0.93 74.15 11.97 N 14.83 47.35 N 26.04 21.31

16 0.76 60.80 1149 � 12.16 37.15 � 2.43 16.72

17 0.62 49.86 11.030 9*7 28.86 18.87 12.99

18 0.51 40.88 10.59 0 � 8.8 22.12 � 1.17 9.95

19 0.42 33.52 10.16 N 6.70 16.66 U 9.16 7.49

In this case, the cost recovery had to be split over two years because we could not subtract the entire $28 million in the first year and make the taxable income not less than zero. The NPV of operator is $5.4 million, the PV of the royalty payment is $71.21 million, the PV of taxes is $96.79 million, and the host country’s share is 97%. As you can see, it is conceivable that the NJPV of the operator is negative, but the host country’s share can be positive because royalty payment is no matter whether the

taxable income is positive or negative. The more marginal the project, the more likely it is that the host country’s share will be

larger. In other words, this contract is both regressive and inefficient.

48 Mohan Kelkar, Ph.D., 1. D.

Example 3-15

A participation agreement is signed between the host country and an international oil company. The agreement calls for a 10%

royalty payment based on the gross revenue, 45% income tax rate and 30% participation by the host country. The agreement

further requires that the host country will not participate during the exploration and exploitation phases, and will be "carried"

by the contractor until production begins. The contractor will be allowed to recover 30% of the host country’s contribution to

all initial costs in the cost recovery phase. This is also true with lDCs and leasehold costs. Cost recovery is limited by the

requirement that taxable income cannot be less than zero. The capitalized expenditure can be depreciated using the double

declining balance method with the remaining balance depreciated in year seven. Assume the field data is the same as Example 3-14.

Solution 3-I5

The calculations are summarized in the following table.

Gross Operating Cost Taxable � Host Contry Contractor

Production Revenue Costs � Royalty � Depreciatio Recovery Income � U

Taxes Net Renuei Net Revenue

Year (MMSTP) (5MM) (5MM) ($rÆ\4) n (5MM) ($MM) (5MM) ($Vl) 15MM) � (5MM)

0 -5.00 � 0.1 � 0. � � 0.0Q, -5.00

1 -3500 � 0.41 U 8.41 .(100 -35.00

2 -58.00 0

&T ac � 0.0 -58.00

3 -60.00 � 8.00 0.00 � 0.001 -60.00

4 -80.00 � 01 U � � �0.0C -80.00

5 3.00 240.00 18.00 � 24.1* 62.86 91.00 44.14 � 19.1* *3.44 124.70

6 5.00 400.00 17.28 ’ 40.(Q 44.90 297.82 � 134.4 2.6r 146.09

7 5.00 400.00 16.59 �40.00 32.07 311.34 *40.10 � 60.9* 142.32

8 5.00 400.00 15.93 � 40.( 22.91 321.17 44.S � 139.68

9 5.00 400.00 15.29 � 40.1* 16.36 328.35 � 147.1* *9.89 137.87

30 5.00 400.00 14.68 w 4041 11.69 333.64 501 � 58.5e 136.63

11 4.10 328.00 14.09 �32.80 29.22 251.89 113.35 U 503* 117.43

12 3.36 268.96 13.53 � 26.E 228.54 � 377

87.99

13 2.76 220.55 12.99 U 22.41 18S.51 U 8141 *0.51 71.42

14 2.26 180.85 12.47 � 18.0 150.30 � 67.E 4.81 17.86

15 1.85 148.30 13.97 014.83 121.50 054.57 U 20.0* 46.78

16 1.52 121.60 11.49 _121 97.95 U 16.1% 37.71

17 1.25 99.71 11.03 U 9.41 78.71 � 35.41 *2.99 30.30

18 1.02 81.77 10.59 �

63.00 28.3 0.4t 2426

19 0.84 67.05 10.16 U 6.70 50.18 U22.58 U 8.2* 19.32


Years 0 through 4, the contractor will have to pick up all costs and carry the host country’s participation share (30%). Once

production begins, the contractor can recover the host country’s contribution through cost recovery. Unlike in the previous

case, the amount that can be recovered through cost recovery is the 30% contribution of the host country plus the contractor’s

contribution (70%) to (DC and leasehold costs. This can be calculated as:

cost recovery = 0.3 x (5 + 35 + 58 + 60 + 80) + 0.7 x (S + 23) = $91 million

As in the previous case, the contractor is only allowed to recover the costs until taxable income reaches zero.

For year five,

gross revenue = $80 x 3 x 106 = $ 240 million

royalty = 0.1 x 240 = $24 million

depreciation = x (60 + 50 + 60 + 50) = $62.86 million

taxable income = 240 - 18- 62.86- 91 - 24 + 0.7 x 50 = $44.14 million

tax = 44.14 x 0.45 = $19.86 million

net income(host country) = (240 - 18.0- 24.0 - 18.0 - 19.86)0.3 = $53.44 million net income (contractor) = (240- 18.0- 24.0- 18.0-19.86)0.7_= $124.70 million



The key difference between a participation agreement and concession agreement is that host country receives a percentage of

the net revenue after taxes. The host country has three sources of income in this contract: royalty, taxes and after tax net

revenue.

We can calculate the PV of royalty ($71.21 million), PV of taxes ($227.62 million) and PV of after tax revenue ($113.50 million).

The contractor’s NPV is $117.09 million. The total revenue is $529.43 million.

We can calculate the host country’s share as,

412.33

= =78% 529.43

The contractor’s share is 22%.

Similar to Example 3-14, let us assume that under a pessimistic scenario, production is 50% of the optimistic case. The following

table shows the calculations under that scenario.

Gross Operating U U Cost Taxable U U Host ContryI Contractor

. U U

Production Revenue Costs � Royalty � Oepreciatio Recovery Income Taxes Net Rev1nue Net Revenue

Year (MMSTB) ($MM) ($MM) ($MJv1) n I$MM) ($MM) ($MM) ($âJJ) $MM) i (5MM)

0 -5.00 � o.c � � 0.0f -5.00

1 -35.00 � 0.111 � 0.111 11110.00 -35.00

2 _S8.00 o.q O.c �0.01 -58.00

3 -60.00 U 0.00 U 0.00 U 0.0(1 -60.00

4 -80.00 � 01 � 0.1 � �0.0Q -80.00

5 1.50 120.00 18.00 U 1200 62.86 27.14 - U U U7.00 63.00

6 2.50 200.00 17.28 �

20.1 44.90 63.86 53.96 � �24. 138.44

7 2.50 200.00 16.59 E20.00 32.07 131.34 U59.10 U 31.291 104.31

8 2.50 200.00 15.93 �20. 22.91 141.17 .63.51 111

30.11 100.55

9 2.50 200.00 15.29 � 20.1* 16.36 148.35 � 56.111 *9.39 97.95

10 2.50 200.00 14.68 �

U20.q 11.63 153.64 � 69.3k 8.86 96.19

11 2.05 164.00 14.09 U16.40 29.22 104.29 04593 U 25.9JI 86.58

12 1.68 334.48 13.53 � 13.’ 307.51 111 �48. � IT71 59.13

33 1.38 110.27 12.99 U 11.111 86.26 U 38.1* *4.23 47.44

14 1.13 90.42 12.47 �

U 9.1w 68.92 U

31.q 37.90

15 0.93 74.15 11.97 � 7.41 54.77 �24.64 U 9.041 30.12

16 036 60.80 11.49 � 6.01 43.23 � 19. �

�73 23.78

17 0.62 49.86 11.03 U 4.1* 3184 U 15.31 �5.58 18.61

18 0,51 40.88 10.59 �

4.1w 26.21 �

U 117 U 14.41

19 0.42 3152 10.16 U 3.35 20.01 � 9.00 U 3311 1100

The cost recovery was spread over two years due to limited cash flow in the first year. The PV of royalty is $35-6 million, PV of

taxes is $85.35 million and PV of the host country’s net revenue share is $60.05 million. The contractor’s NPV is $41.58 million.

The total revenue is $222.60 million. We can calculate the host country’s share as:

181.01

= 2 = 81%

22.6

Notice an important difference between the concession agreement and participation agreement. In the concession agreement,

as the project became marginal, the host country’s share increased from 75% to 97%. In the participation agreement, the host

country’s share increased from 78% to 81% as the project became marginal. The key difference is how the host country is

rewarded. In the concession agreement, the royalty payment forms a big part of the payment irrespective of the profitability of

the project. In the case of the participation agreement, the host country receives a small percentage of royalty but most of the

income comes from taxes and a percentage of net revenue. This means that if the project is not economically successful, the

host country will not receive significant benefit. This makes the participation agreement more efficient and less regressive than

the concession agreement.

Few countries in the world will go "straight up" in a participation agreement. That is, the host country will participate during

the exploration phase and pay for its share from the beginning. This makes the project more attractive for the contractor.


Below we show the calculation under a "straight up" contract for the original case (optimistic). The only difference is that, from

year 0, the contractor only has to contribute 70% of the total costs and is able to recover 70% of IDC and leasehold costs after

production begins. The other numbers remain the same. Since the host country contributes from the beginning, the economics

are improved for the contractor.

Cross Operating 0 Cost Taxable - Host CountryU Contractor

V U Production Revenue Costs � Royalty.. Deprectatio Recovery Income z Taxes Ne

Ut Rev1nue Net Revenue

Year (MMSTB) ($MM) ($MM) ($Pj(\n) n ($MM) ($MM) ($MM) ($MJv) I$MM) � ($MM)

0 -3.50 0.t 0.0 0.0Q -350 � � 1 -24.50 � 0., � o �0.00 -24.50

2 -40.60 0.4 � 0.4 � -40.60 ,o.cjd ’ �

3 -42.00 U 0.00 U 0.00 � 0.001 -4200

4 -56.00 0-cm 1 0.0q -56.00 � � �

5 3.00 240.00 18.00 0 2459 62.86 19.60 115.54 U SI-99 13.80 102.20

6 5.00 400.00 17.28 44.90 297.82 2.6f 146.09

7 5.00 400.00 16.59 040.00 32.07 311.34 140.10 U 60.991 142.32

8 5.00 400.00 15.93 22.91 � 40d 321.17 44.S’ 139.68 � 9 5.00 400.00 15.29 U 40.(,* 16.36 328.35 U 147.A 199.09 137,87

10 5.00 400.00 14.68 40.Q 11.69 333.64 � 150.4 � 136.63

11 4.10 328.00 14.09 �32.80 29.22 251.89 113.35 � 50.331 117.43

12 3.36 268.96 13.53 _26.91 228.54 �37,7

87.99 � 13 2.76 220.55 12.99 U 22.S 185.51 U 83.41 10.61 71,42

14 2.26 180.85 12.47 � �18.Q 150.30 � 676 42 57.86

15 1.85 148.30 11.97 114.83 121.50 154.67 � 20.041 46.78

16 1.52 121.60 11.49 12.:M . 97.95 �44.1 6.16 37.71 � 37 1.25 99.71 11.03 U 9.99 78.71 � 35.41 12.99 30.30

18 1.02 8137 10.59 8.1k 63.00 � ’28.31 �

0.4c? 24.26

19 0.84 67.05 10.16 U 6.70 50.38 �22.58 U 8.231 19.32

In this case, The PV of royalty is $71.21 million, PV of taxes is $240.54 million and PV of the host country’s net revenue share is

$109.63 million. The contractor’s NPV is $152.38 million. The total revenue is $573.75 million. We can calculate the host

country’s share as:

421.37 = =73% 573.75

The contractor’s share has increased from 22% to 27%. This makes the contract more efficient than a "carried" contract.

............................................................ PRODUCTION SHARING CONTRACT

The production sharing contract was first signed between Pertamina (the national oil company of

Indonesia) and Mobil in 1966. Subsequently, production sharing contracts became extremely

popular among the non-OPEC countries. Today, it is the most widely used contract between host

countries and an international oil companies.

The basic difference between the production sharing contract and the concession agreement is the

extent of control exercised by the host country in the day-to-day operations. As a partner in the

operation of a field, the host country can monitor and participate in the decision making process. As

with the concession agreement, the host country will open up certain concession areas for bidding

purposes. The international oil company will bid on the concession and offer several incentives,

including the signing bonus and work program, to receive a contract. If successful, the host country

will sign a production sharing contract with the international oil company.

Some key features of a production sharing contract are:



� The international oil company, solely at risk, has the exclusive right to explore for and

exploit petroleum reserves in a concession area. The company is responsible for the

exploration costs associated with these rights.

� If the exploration effort is successful, the international oil company is allowed to recover the

exploration costs from future production of the contract area.

� Any production belongs to the host country.

� After recovering the cost from the production proceeds, the balance of production is shared

on a pre-determined percentage split between the host country and the international oil

company.

� The income of the international oil company is subject to income taxes.

� Equipment and installations are the property of the host country. This can happen at the

onset of production or progressively in accordance with the approved schedule.

The main advantage of the production sharing contract is that the host country can develop its own

natural resources at no cost without losing managerial control of the overall operation. Further, by

acquiring equipment and installations, it can develop the necessary infrastructure for future

development. The host country can also benefit by having domestic personnel working directly with

personnel from the international oil company and gaining valuable experience.

The schematic of the overall cash flow calculations is shown in Figure 3-5 for a standard production

sharing contract. As shown, the contractor can recover the operating, capitalized (through

depreciation), and exploration costs from the gross revenue. The remaining revenue, called profit

oil, is split between the host country and the contractor based on the approved formula. This

formula may involve a volume-related sliding scale. The contractor’s share of profit oil is subject to

income taxes.

Some variations of this standard production sharing contract include assessment of royalty based on

the gross revenue before cost recovery, or the contractor’s share of profit oil is not subject to

income tax. These variations will change some of the calculations. Overall, though, the host

country’s share in a production sharing contract is rarely less than 50%.


Gross Revenue 1000

Less

Operating, Capitalized and Exploration Costs (Cost Recovery)

400

I Equal To

Profit Oil 600

Equal To

Contractor’s Share 300

Less

Lqnl to

Net Cash Flow 150

Host Country’s Share I

450 I --=75% I

600 I

The following examples illustrate the application of a production sharing contract on the economic

evaluation of a project.

Example 3-16

A production sharing contract is signed between the host country and an international oil company. The contract calls for a profit oil share of 60% with an income tax rate of 50% on taxable income. The capitalized costs are to be depreciated over a seven year period using the double declining balance method with the balance to be depreciated in the seventh year. The expected costs and other relevant parameters are stated below.

Signing Bonus $5 million in year zero Capital expenditure: $30 million in year one

$50 million in year two $60 million in year three $80 million in year four

IDC $5 million in year 1 $8 million in year 2 $10 million in year 3

Initial production: 3 million barrels in the fifth year 5 million barrels from year 6 to 10 Start decline in year 11



53

Production decline rate: 18% per year

Years of production: 15 years

Operating costs: $18 million in year five

Operating cost decline rate: 4% per year

Oil price: $80 per barrel assumed constant

MROR: 20%/year

Estimate the feasibility of the project.

Solution 3-16

The calculations are shown in the following table:

Year

Production

(MMSTB)

Gross

Revenue

($MM)

Operating

Costs

($MM)

Depreciation

($MM)

Cost

Recovery

($MM)

� Host M IS

Country’s.

Profit Oil Shire

($MM) � ($MM) � Taxable

Income � ($MM)

E

� Taxes � ($re1 )

Contractor

Net

Revenue

($MM)

0 -5.00 � - 0(1 - � 0.0 -5.00

1 -35.00 M 0.00 0.( -35.00 � U U 2 -58.00 � 0.00 � 0(1 -58.00

3 -60.00 acm 0.00 -60.00 � . U 4 -80.00 � 0(1 � 0.00 -80.00

5 3.00 240.00 18.00 62.86 28.00 131.14 078 . 69 82.46 N 26.I 117.09 � U a 6 5.00 400.00 17.28 44.90 337.82 IJ02.69 135.13 � 60.71 119.22

7 5.00 400.00 16.59 32.07 351.34 0 210-EM 140.54 063.24 109.37 � U U 8 5.00 400.00 15.93 22.91 361.17 � 216.71 144.47 �65.01 102.36

9 5.00 400.00 15.29 16.36 36835 21.01 147.34 � 66.3 97.40 � U U 10 5.00 400.00 14.68 11.69 373.64 Q24.18 149.45 � 67.21 93.89

11 4.10 328.00 14.09 29.22 284.59 170.01 113.88 051.24 91.85 � a U 12 3.36 268.96 13.53 255.43 � 153.71 102.17 045.98 56.20

13 2.76 220.55 12.99 207.56 24.54 83.02 37JI 45.66 � U U 14 2.26 180,85 12.47 16838 IJ01.03 67.35 � 30.71 37.04

15 1.85 148.30 11.97 136.33 M 81.b 5453 024. 54 29.99 � . . 16 1.52 121.60 11.49 110.11 � 66(1 44.05 �19.82 24.23

17 1.25 99.71 11.03 88.69 53.21 35.47 15.919 19.51 � . U 18 1.02 81.77 10.59 71.18 .42.71 28.47 � 12.71 15.66

19 0.84 67.05 10.16 56.88 34. A 22.75 M 10.24 12.51


The calculations are carried out with the constraint that the profit oil can never be less than zero. Similar to the concession

agreement, lDCs and leasehold costs are recovered through the cost recovery method. The total amount is $28 million.

For year five,

gross revenue = 3 x 106 x 80 = $240 million

DC and leasehold costs can be deducted in the first year of production as long as the profit oil is not less than zero.

profit oil = gross revenue from oil - operating costs - depreciation - cost recovery

= 240- 18- 62.86- 28

= $131.14 million

The depreciation is calculated by using the double declining balance method.


host country’s share of profit oil = 0.6 x 131.14 = $78.69 million

taxable income = 0.4 x profit oil = $52.46 million tax = 52.46 x 0.5 = $26.23 million

net income = gross income - operating costs - 0.6 x profit oil - taxes 240.0 - 18.0 - 78.69- 26.23

= $117.09 million

Net income represents revenue minus actual outlays. These outlays include 60% of the profit oil given to the host country. The rest of the calculations are carried out in a similar fashion. The host country receives money in two ways in a production sharing contract: a portion of the profit oil and taxes. In this example, the PV of the profit oil share is $361.41 million, PV of taxes is $109.48 million and NPV of the contractor is $55.54 million. The host country’s share is,

470.89 - - 55.54 + 470.89 =89%

If we consider a pessimistic case where production is reduced by half, we see the following results:

� Host Contractor

Gross Operating Cost out?try’s Taxable � Net

Production Revenue Costs Depreciation Recovery Profit Oil Shire Income � Taxes � Revenue

Year (MMSTB) ($MM) ($MM) ($MM) ($MM) l$MMI � ($MM) � ($MM) ($r*1) ($MM)

0 -5.00 � o. U U � oe -500

1 -3500 000 0. cffi -3500 � U U 2 -58.00 U 0.00 U 0.0 -58.00

3 -60.00 U adg � U 0.00 -60.00

4 -80.00 � 0.( U � 0.00 -80.00

5 1.50 120.00 38.00 62.85 28.00 11.14 6.69 4.46 2.I 93.09 � . U 6 2.50 200.00 17.28 44.90 137.82 U8269 55.13 U 27.4 72.46

7 2.50 200.00 16.59 32.07 151.35 gam 50.54 03027 62.34 U U U

8 2.50 200.00 15.93 22.91 161.17 � 96. 64.47 .3123 55.14

9 2.50 200.00 15.29 16.36 168.35 � 1101.01 . 67.34 � 33.ER U

50.03

10 2.50 200.00 14.68 11.69 173.64 104.18 69.45 U 34.lj 46.41

11 2.05 164.00 14.09 29.22 120.69 M 72.1 48.28 024.14 53.36 U

M K

12 1.68 134.48 13.53 120.95 � 72.1k 48.38 .24.19 24.19

13 1.38 110.27 12.99 97.29 05837 38.92 M 19.’! 19.46 U U U

14 1.13 90.42 12.47 77.96 .46.78 33.18 U 15.4 15.59

15 0.93 74.15 11.97 62.18 0 37.’! 24.87 � 12.44 12.44 � U U 36 0.76 60.80 11.49 49.31 U 29.8k 19.73 U 9.86 9.86

17 0.62 49.86 1103 38.83 m 23.30 15.53 M 7.’! 7.77 U � U

18 0.51 4088 10,59 30.30 �18.18 32.12 � 6.4 6.06

19 0.42 33.52 10.16 23.36 34.’! 9.34 � 4.67 4.67

In this case, the PV of the host country’s share is $147.79 million and PV of taxes is $49.26 million, but the NPV of the contractor is -$23.66 million. Although the host country will make money on the project, the contractor will lose. This happens because of the initial investment made by the contractor. Although the contractor is allowed to recover those costs, the time value of money makes the project uneconomical. Production sharing contracts, in general, are more efficient than concession agreements because they do not contain royalty payments (payments based on gross revenue). However, for marginal projects, it is possible for contractor to lose money while the host country makes money.

This type of marginal project discourages companies from developing fields and, hence, the loss will be to the contractor and the host country. To avoid this problem, many production sharing contracts have clauses that change the percentage of profit oil as a function of production rate. The higher the production rate, the more oil that is shared by the host country. In some instances, the profit oil share is tied to the rate of return on investment. If the oil prices increase and, hence, the profit, a higher percentage will be taken by the host country. Other variations in production sharing contracts include a domestic market obligation. In essence, the contractor is required to provide a portion of the production at a reduced price to the host country. This is another form of royalty payment except that it is subsidized rather than free.



[SERVICE CONTRACTS

As we move from the concession agreement to the service contract, the involvement of the host

country increases. In essence, under a service contract, the international oil company provides the

host country with services and information to help the country develop its own resources. In return,

the company receives a fee or share of production at a reduced price. Most significant, the oil

company has no equity position in the production. The company may receive production at discount

prices, a production bonus tied to reaching a certain level of production, and the host country may

also pay the company’s taxes so that the proceeds are tax free. On the other hand, the host country

retains control and ownership of the minerals.

The service contracts can be further divided into three types of agreements: (i) pure service

contract, (ii) technical assistance agreement, and (iii) risk service contract.

PURE SERVICE CONTRACT

In a pure service contract, the host country or its national oil company will contract with an

international oil company to perform a specified service for a fixed fee. These service contracts

do not provide for any right to production. To make the contracts more attractive, the host

country may include a buy back arrangement where the international oil company can obtain

crude oil instead of a fixed fee. This allows the oil company (the contractor) to receive a fixed

share of production (similar to overriding royalty) as a fee.

These contracts are common in the Middle East. In Saudi Arabia, for example, international

companies received a fixed fee per barrel of production. In Qatar, an oil company is paid back

with a fixed percentage of crude produced while providing the service.

TECHNICAL ASSISTANCE AGREEMENT

In a technical assistance agreement, the contractor provides technical assistance related to

exploration, development, production and the refining of oil. The contractor’s services may

include providing equipment as well as training employees to operate the facilities. In return,

the company agrees to pay for the expenses plus a fee tied to production.

Venezuela entered into such an agreement with several international oil companies, where the

oil companies agreed to operate the facilities they built in return for a fixed per barrel fee.

RISK SERVICE CONTRACT

In the risk service contract, the risk of exploration is borne by the contractor (the international

oil company). The contractor agrees to explore a specific area and evaluate its potential. If a

commercial discovery is made, the contractor is obligated to develop the reservoir. Once the

field is developed, depending on the terms of the contract, the field is operated by either the

contractor or the national oil company. The contractor’s investment is paid back with interest

and a fee. In addition, the contractor may receive a portion of the production at a reduced price.

The countries that employ this type of contract are Argentina and Brazil.

56


Unlike the concession and production sharing contracts, we have not included any examples to

illustrate the applications of the service contracts. The main reason is the limited use of service

contracts compared to the other two types of contracts. Further, due to wide variations in the

terms of service contracts, it is very difficult to consider a typical service contract. Additionally,

the service contract can be analyzed as a reverse of concession agreement with some

modification, where the contractor becomes a royalty owner and the host country becomes the

operator.

An agreement is signed between an international oil company and a national oil company to develop a concession. The contract

calls for 020% royalty interest and 55% income tax rate. The oil production and cost profile follows.

Capitalized Costs Operating Costs Production

Year ($ Million) ($ Million) (10 6 bbls/year)

1 180

2 400

3 520

4 600 24 80

5 26 9.0

6 28 11.0

7 28 15.0

8 28 15.0

9 27 14.0

10 26 13.0

11 26 12.2

12 25 11.5

13 24 10.4

14 23 9.9

15 22 9.3

16 22 8.2

17 20 6.0

18 1] 4.5

19 12 2.6

20 10 1.3

Assume that capitalized costs can be depreciated over a 10 year period using the double declining balance method once

- production begins. The oil price is expected to be $801bbl in year four, increasing at a rate of 2.5% per year.

A. Calculate the ROR of the project using the conditions above. If the MROR is 15%, is the project economically feasible?

B. If the oil price is assumed to be $75/bbl at the present time, how will the economics of the project change?

C. If the predicted production is reduced by half each year, how will the economics of the project be affected?

D. Calculate the NPV of the contractor for A, B and C. Based on these calculations, calculate the NPV per barrel of oil. If the finding cast in the United States today is $241bbl, will you select the proposed project or the one in the United States?

E. If the national oil company is willing to negotiate the royalty percentage, as well as the tax rate, to what percentage are you willing to increase the royalty interest if the tax rate can be decreased by 4%?

(Hint: See the Works Cited section (Moran, 1992) for details)

Problem 3-26

Assume the production data and other information given in Problem 3-25.

A. If the host country requires that its participation interest is 30%, how will the economics change? Assume that the initial

investment is carried by the oil company.

B. If the capitalized costs are higher by 50%, how will the economics of the project be affected?



57

C If the operating costs are higher by 50% how will the economics of the project be affected?

D. What is more critical in terms of costs capitalized costs or operating costs? Why?

E. Repeat ports B, C and Dfor a "straight-up"contract.

F. If the exploration casts for the given data are $10 million in year zero, how will the project economics change? Assume that

the exploration costs can be deducted in the first year of production as long as the taxable income is not less than zero

otherwise it is carriedforwardi Repeat the problem for both carried and straight up options

Problem 3-27

If, in year eight, the oil company is required to drill two additional wells costing $6 million, will the net revenue in that year be

affected with and without drilling the wells?

Repeat calculations for,

A. Royalty interest only.

B. Royalty and participation interest.

Assume the royalty interest to be 2016 and the participation interest to be 20%. Use all the other relevant data from Problem 3-25. Assume that exploration casts can be deducted in the some year.

1. For parts A and B, is your net revenue reduced by $6 million? Why?

2. In terms of incremental investment (such as drilling an in-fill well), are these contracts efficient or inefficient?

3. If the royalty interest is increased to 40%, how will the net revenue be affected with and without incremental costs?

4. If the participation interest is increased to 40%, how will the net revenue be affected with and without incremental casts?

5. Is it more beneficial to have increased royalty interest over increased participation interest? Why?

Problem 3-28

Use the production and cost data from Problem 3-25. A host country has signed a production sharing contract with an

international oil company with the following conditions.

The capitalized costs are depreciated over a seven year period using a combination of the double declining balance and the straight-line method. The capitalized costs are to be depreciated starting in the year production begins. The host country’s share

of profit oil is 50%. The host country also receives a royalty interest on gross revenue of 10%. The tax rate is 40% on taxable income.

If the MROR is 15916, and the price of oil is expected to be $801bbl over the life of the project, is the project economically feasible? Assume that $30 million is spent in IDCs in year zero, which can be deducted once production begins.

Consider the variations in the following parameters and assess their impact on the profitability of the project.

� –20% variation in capitalized costs.

� –20% variation in production rote

� –20% variation in operating costs.

� –20% variation in exploration costs.

A. Which input parameter has the most impact on the profitability of the project? Based on this sensitivity analysis, discuss the progressive or regressive nature of the project. -

B. If we assume that the royalty interest is reduced to zero and repeat port A, how will the nature of the project be affected? C. If the host country wants to increase the profit oil share by 5%, how much does the tax rate need to be reduced not to

affect the profitability of the project? (Hint: See the Works Cited section (Moran, 1992) for details.)

D. If another oil company is willing to buy your interest at a price of $41bbl of oil in the ground, will you sell it?

Problem 3-29

Use the production data given in Problem 3-25.

In year six, if the oil company is required to spend an additional $10 million in exploration costs, how will the revenue be


0 affected? Assume that there is no royalty interest, the host country takes 40% of the profit oil, and the tax rote is 55%. The exploration costs can be deducted in the some year.

In terms of incremental costs, is this contract efficient or inefficient?

SUMMARY

In this chapter, we considered the effect of income taxes on the economic analysis of a project. The

effect can be significant depending on the tax structure. Although various nuances of the tax structures

are not considered, the basic principles were illustrated through several examples.

In this chapter, we also concentrated on the tax implications on the development of oil and gas

properties domestically and internationally. As the landscape of oil production changes and oil fields in

the United States become mature, international production becomes more relevant to the major oil

companies and large independents. International agreements can vary significantly in terms of the

control of the operation and type of compensation the oil companies receive. Hwever, there is one

* significant feature that is common to most of them - the fractional share of production received by the

host country. Unlike typical royalty contracts signed in the United States, the majority of production

goes back to the host country. As shown in the examples provided in this chapter, this type of

arrangement significantly affects the cash flow analysis of the project and may make marginal or small

oil fields difficult to develop. Understanding the implications of the type of contract a company signs is,

__ therefore, very crucial in making the project economically profitable.

WORKS CITED

Wall Street Journal. (2009, October 17). Wall Street Journal.

Allen, F. H., & Seba, a. R. (1993). Economics of Worldwide Petroleum Production. Tulsa, Oklahoma, USA:

OGCI Publishing Company.

Allinson, G. (1995). Economics of Petroleum Exploration and Production. Australia: PetroConsultants.

Bertagne, R. G. (1992). International Exploration by Independents. (R. Steinmetz, Ed.) The Business of Petroleum Exploration, 319-330.

Blinn, K., Duval, C., LeLench, H., & Pertuzio, a. A. (1986). International Petroleum Exploration and

Exploitation Agreements: Legal, Economic and Policy Aspects. New York: Barrows Company.

Campbell, e. a. (1987). In Analysis and Management of Petroleum Investments: Risk, Taxes and Time.

Norman, Oklahoma: CPS Publishing Company.

DeGarmo, E., Sullivan, W. G., & Bontadelli, J. A. (1993). Engineering Economy (9th Edition ed.). New York,

New York: MacMillan.

Katz, S. B. (1992). Types of International Petroleum Contracts: Their History and Development. (R.

Steinmetz, Ed.) The Business of Petroleum Exploration, 297-306.

L. G. Mosburg, J. (1989). Tax Consequences of Oil and Gas Exploration and Development under Tax

Reform. Tulsa, Oklahoma: K.S. Enterprises/ElI.

Miskell, R. F. (1984). Petroleum Company Operations and Agreements in the Developing Countries (Vol.

Resources for the Future). Washington, D.C., USA: Johns Hopkins Press.

Newman, D. G. (1991). In Engineering Economic Analysis. San Jose, California: Engineering Press.

Park, C. S. (1993). Contemporary Engineering Economics. Reading, Massachusetts: Addison-Wesley.

Economic Evaluation in the Petroleum Industry Chapter 3 - Economic Analysis 59

ADDITIONAL CASE STUDIES.

Casa Stu(v 3-3

Among stripper gas wells, data reveals that brine accumulation is the primary reason for declining natural gas production and often subsequent well abandonment. Just a few tenths to a few barrels per day can and does kill gas production. Conventional fluid removal techniques using beam pumps, periodic swabbing of fluids with workover rigs, siphon strings and tubing plungers or rabbits have both physical limitations and significant capital, operating and maintenance cost considerations. Regular, rhythmic removal of fluids as they accumulate In the well increases gas production. Averting a brine column and a brine-wet zone in the production horizon fosters the ease of movement of gas into the well.

A new pump, called the Gas Operated Automatic Lift (GOAL) PetroPump, has a unique "on-tool’ actuator-valve assembly. The tool is designed to freely operate within the well easing, descending downhole when wellhead pressure and production drop. Upon descending into the fluid, afforded by a through-tool passageway, a fixed column or weight of fluid accumulates atop the tool that offsets the preset actuator pressure, and the valve automatically closes.

Flex wall cups surround the tool and actuator body and make a circular seal with the well. Additional pressure and gas accumulating in the well below the tool build and lift the tool and fluid load to the surface. On reaching the surface, the fluids are pushed ahead of the tool and processed off the topside of a wellhead lubricator. Follow-on gas production is produced from below the tool on a bottom side port in the lubricator.

The tool is "smart" in both directions, In that the tool, upon ascending the wellhore to the wellhead lubricator, does not have to latch into a wellhead catcher to await physical release by man, clock or other mechanical device. The on tool actuator valve assembly senses tool pressure, and at such time as wellhead pressure declines to within a preset limit of sales-line pressure, the actuator opens. The tool then again descends for another load of fluid.

The tool (Case Study Figure 3-1) has evolved significantly through a multi-year development and field-testing process. The original tool was more than 6-ft long, weighed over 100 lbs and had more than 60 components. The current fourth-generation convertible-tool design, which can accommodate 4-in, or 3-in. II) casing and/or tubing, has lust 14 components, is only 38-in, long and weighs less than 45 tbs. Metallurgical upgrades from carbon steel to stainless steel and I lastelloy (a proprietary alloy) largely eliminated corrosion problems, and elastomer modifications have proven effective in reducing wear. The convertible tool now being tested will broaden the tool’s application environment, since tubing wells with new 4-in, and 3-in. materials will not present the irregularities of older cased wells. BED - 0 is researching the use of non-metallic, continuously spooled casing and tubing to further reduce lift lost.


(nsc Study Figure 3-1: G(1:tL I otosop tool c 1010 tIon (most recent pump Is on

Required wellhead modifications include a full-port ball valve to allow passage of the tool and, atop the full-port valve, a lubricator to allow top end (above tool) takeoff of fluids, and bottom end (below tool) production of post-run gas. A top-of-lubricator catcher for periodic tool retrieval for cup replacement and/or actuator adjustment has also been developed. Actuator operation can be changed or reset by the removal of only one tool component: a top-end catcher rod. Cups can be removed and changed by the removal of just two components: a lock nut and washer. Periodic tool service and adjustment can be achieved using two IS-in, adjustable wrenches. Field testing and pumper input have led to a physical configuration comprised of standard field threads, wide flats and easily accessible key components.

The gas/fluid ratio for the current tool is about 3 Mef/bbl per tool cycle. Pressure differential to move the tool within the well is about 12 psi in conventional steel casing. Data demonstrate the tool is capable of lifting 0.1 barrels to as much as 40 bbl of fluid per cycle, but normal operating range is 0.25 - 6.0 bbl per tool cycle. Cup life on the tool has proven to be greatly extended over conventional tools and cups due to the regular prcssurelift cycling and exertion of uniform forces on the cups. Cup life on some wells now exceeds six months.

Teoape Resources used the initial and successor tools in Well 52, a 3,200-ft Medina formation completion in New York drilled in 1986. Well 52 has 4.5-in. J-55 casing. It produces through a horizontal wellhead separator into a gas gathering system, with pressure ranging 50 - 100 psi. Recent wellhead shut-in pressure is about 120 psi. By late 1996, the producing method had evolved from unassisted flowing, to a 1-1/2-in, tubed well that was periodically soaped, shut-in and vented to a standard 4.0-in, casing swab that was mcchanically/ physically dropped and operated.

An example of tool perforniaoce is shown in Case Study Figure 3-2. Annual production before tool deployment was about 1,100 Mef/ycar. Production since the tool installation has averaged 3,416 Mcf/vcar. Case Study Table 3-1 shows results for other tight Medina wells. Production multiples in the other wells are also economically attractive. The average success rate of these wells is about 0.7; that is, 70% of the wells, where the tool is installed, have been successful.



4

2

Wolf 52 docilne cume annual production

6 .6 . q 2 a T 91 i S 7 ’i Dl rQ 0.1

PIDUCtcn ’POF

Care Stu dv laure 3 2 4naala s proc/ac tion from L flap e 52 we /I before and rlterdi s taiia tion

C as c., S tu cip 1 51- 1: 1 e norm a ace ian pro vone ii ts on pa r wells u s1ag the ile w p amp Wells

(all wells are in Pre-pump Queenstonc/ installation Fluid Production,

Medina sandstone) Depth, ft flow, Mcfd Post-pump installation flow, Mcfd bwpd Notes L-52 3,200 6 13-19 05-1.0 2-4 tool cycles!

day (rhythmic) L-54 3,300 3-5 8-10 < 0.5 1 cycle every 2-3

days (possible casing leak)

L-274 3,400 < 2 6-8 025-0.5 1-2 cycles/ day (well in adjustment)

L-332 3,200 5-6 18-20 0.5-075 2-3 cycles! day (regular)

L-29 2,400 < 7 13-16 < 0.25 2-3 cycles! wk S-2023 2,700 0-10 28-35 0.5-0.75 3-4 cycles! day

(well was swabbed,

As a production engineer, you want like to evaluate the effectiveness of this new technology. Assume that the cost of pump is S1 1,500. The surface equipment and the installation cost is S9,700 per pump. These costs can be depreciated over five years using the DDB!straight -line methods. Assume that Case Study Table 3-1 represents an average response from a typical field for a similar environment with a 70% success rate. The royalty payment is 15%, the severance tax is 8%, and the incremental operating cost for each well is S500 per year. Assume that the price of gas is $4.50!MSCF and assume that the production in Case Study Table 3-1 represents the incremental production in the first year declining at 10% per year. The marginal tax rate is 34%. If the MROR is 15% , is the project feasible? What is the ROR on the project? If the price is assumed to be $3.50/MSCF, will the feasibility of the project be affected?

Case Study 3-4

Coal bed methane (CBM) production has been receiving a lot of attention lately. Methane is primarily adsorbed to the organic matter in coal. Coal contains large number of fissures and fractures which are filled with water. These are the primary conduits of gas. As the reservoir is depleted, the wells initially produce significant quantities of water. As the water from the fissures is depleted, the natural gas is slowly desorbed into the fractures and is produced. Over time, the water gas ratio decreases as less and less water is produced and more and more gas is produced. A typical profile of gas production includes a slow increase at the beginning followed by a steady decline in gas production.

CBM production first started in San Juan basin in New Mexico. Because of economic success of that basin, eventually operators started exploring other basins for CR61 production. Figure Case 3-4-1 below shows the locations nfCBM production in Oklahoma.


C2 se Study Figure 3-3: CBM production placs

One of the basins which have been extensively explored is the Arkoma basin in Oklahoma As a production engineer, your job is to understand the feasibility of drilling vertical vs. horizontal wells. Currently, you have collected data from 153 vertical wells and 53 horizontal wells. Based on a cursory analysis, Out of 153 wells, roughly 51 vertical wells produce from a poor quality coal with distinctive different rate behavior. The other 102 wells exhibit a distinct behavior. For 53 horizontal wells, roughly 15 wells exhibit a distinctive behavior because of low quality rcsei’oir. The other 38 wells exhibit distinct behavior indicating better coal quality. The typical production profiles of vertical wells (poor and good) and horizontal wells are provided below: Note that we have only provided the production profiles for wells for 20 years. These productions represent yearly production from a single well in MSCF. In principle, the wells will continue to produce for a longer period of time; however, from present value analysis perspective, the production beyond 20 years is not going to contribute significantly; therefore, we can ignore the production beyond 20 years.

Gas Production (MSCF)

Vertical Well Horizontal Well

Poor Good Poor Good

9,600 24,000 50,400 120,000

14,400 44,400 39,600 78,000

12,000 39,600 32,400 57,600

10,200 32,400 27,600 45,600

8,400 25,200 22,800 36,000

7,200 20.400 20,400 30,000

6,600 17,400 18,000 25,800

5,760 15,000 16,200 21,600

5,160 12,600 14,400 19,800

4,56() 11,400 13,200 16,800

4,32() 10,200 12,000 15,000

3,84)) 9,000 11,400 13,800

3,36)) 8,160 10,560 12,60()



14 3,240 7,200 96,000 11,400

15 3,000 6,600 9,120 10,200

16 2,880 6,120 8,400 9,600

17 2,580 5,320 7,920 9,000

18 2,400 5,400 7,320 8,160

19 2,280 4,800 7,200 7,800

20 2,160 4,560 6,960 7,200

Assume that the cumulative production over 20 years for each of the scenarios represents the reserves from that well. For vertical wells, typical spacing is 80 acres (i.e., we need to drill 8 wells per 640 acre spacing). For horizontal wells, typical spacing is 160 acres; (i.e., we need to drill 4 wells per 640 acre spacing). Do your calculations based on the development per section (640 acres). The coat of drilling and completing a vertical well is 595,000; whereas the cost of drilling and completing a horizontal well is $400,000. These costs can be divided as 50% IDC and 50% tangible costs. These costs are incurred in year 0. Assume that depreciation and depletion begins in year 1. ID( s are accounted for starting in year 1. The leasehold costs include a lease bonus, calculated at $50 per acre, plus an additional $2,000 in legal costs per section and S4,000 in G & G costs per section. The operating costs for a vertical well average about S650 per month; whereas, the operating costs for a horizontal well average about S1,000 per month. The operator has a working interest of iOO°/s with a net revenue interest of 85%. The severance tax is 7°/s. Assume an independent operator. Use the ACRS method for depreciation using a 7 year taxable life. The marginal tax rate is 31%. Assume the gas price for the first 3 years to he $6/N,-IS( - I-, and then, starting year 4, it will hold steady at S5/MSCF until the cud of its life. The MROR is 15%.

Calculate the payback period for good and poor, vertical and horizontal wells. Also calculate the NP\T for all four options. Based on this information, will you drill horizontal wells or vertical wells? Why?

Case Stuc/4 3-5

Downhole Oil/Water Separator (DOWS) operates on the principle of disposing of water in another formation below the producing formation. The produced water is separated downhole and part of it is disposed of in a formation below the producing formation. The reduced amount of water is produced at the surface, which is eventually disposed. For wells producing significant amounts of water, this technology can prove to be effective for the following reasons:

� Since much of the produced water is not pumped to the surface, treated, and pumped back into deep formation, the cost of handling the produced water is greatly reduced.

� The production of oil can be increased by reducing the requirement of surface processing and disposal facilities. � By reducing the water/oil ratio, the back pressure on the formation is reduced; thereby, increasing the productivity of the

well.

The two main components of DOWS are an oil/water separation system and dowohole pump. Separation can be achieved using either hydrocyclones or gravity separators. The downhole pump can be either an electric submersible pump, progressive cavity pump or beam pump. A typical downhole installation is shown in Figure. In this configuration, there is one hydroeyclone for separating oil and water and two pumps, one for disposal and one for lifting. The separator is designed such that under flow of the hydrocyclone produces water clean enough for injection into disposal zone. The over flow typically contains a water/oil ratio of 50%. The reason that high water/oil ratio exists at the surface is because of the constraint that no oil be accidentally injected in the disposal zone. Case Study Figure 3-4 shows schematically how the process works.


Motor shroud

Production zone

Injection zone

[ Tubing to surface

Concentrate pump

Motor upper protector

Motor

Motor lower protector

Pump intake

Case Study lCgurc3-4: DOWo operation

You are working for a small independent in Oklahoma, who has recently discovered a new oil field. This oil field produces from a carbonate reservoir that appears to be fractured. The reservoir characteristics are not clearly understood; however, one important factor common to all the new producing wells is the high water-oil ratio. The oil is currently produced from a zone that is approximately 5,000 ft deep, and the produced water is disposed of in a zone that is approximately 6,700 ft deep.

A typical producing oil well has the following characteristics:

� Oil production rate = 80 hbl/d � Water production rate = 3,000 bbi/d � OCR = 3,000 SCF/STB � Operating costs = 86,000 per month � Expected life of a well = 8 years with 30% decline in production � Cost of drilling a vertical well = S350,000

You read all the material available on DOWS and come tip with the following numbcis in terms of improvement in production if DOWS is installed on a well:

� New WOR = I � Operating costs with reduction in water handling = S4,480 per month � % increase in oil production = � Expected life of it well = 8 years with 330/s decline in production � in addition, following information is also available: � Cost of implementation of 1)1 lOWS = S250,000 � Cost of drillmg a producing well deeper by 1,700 ft = S40,000 � Price of oil = $25/bbl held constant � Price of gas = 83/ MSCF hckl constant � Working interest = 1 00h � Net revcnite ititcrest = � Scvcreticc tax in Oklahoma = 7Wo



� Marginal income tax rate = 31%

DOWS can he depreciated over 7 years using a combination of DDB and straight-line methods. The drilling costs can be deducted in the some year as IDCs. Determine the feasibility of Dl lOWs operation by assuming MROR of 15%.

Cisc Studt’ .3-6

As gas and oil prices rise, the oil fields in remote locations are becoming increasingly attractive. I lere is an example of a condensate reservoir located in North Slope, Alaska. The condensate field contains a large amount of gas along with liquid condensate. No gas pipeline exists; therefore, gas cannot be sold and will have to be re injected. The cost of a gas pipeline from Alaska to lower 48 states is pegged at $25 billion, a cost that is difficult to overcome by a single project. However, the condensate produced from an oil field can be sold through the Trans-Alaska Pipeline. Here are some economic parameters to consider:

� Working interest = 9% � Net Revenue Interest = 7.88% � Income tax rate - 35% � 31140R = 20% � Number of wells in the field after it is developed = 8 producers and 6 injectors

The following costs are anticipated after the project begins:

Intagible Tangible Intangible � Year Facilities Development Drilling Drilling

(M $) (M $) (M $) (M $) 0 0 10,000 0 0

� 1 0 20,000 0 0 2 0 70,000 13,800 32,200

� 3 400,000 50,000 0 0 4 400,000 40,000 77,280 180,320

5 220,000 10,000 77,280 180,320

6 200,000 168,360 392,840

The costs are provided in 1000’s. The facilities costs are tangible costs to be depreciated using a 7 year DDB/straight-line combination method. The intangible development Costs include preliminary engineering studies, geological and geophysical costs and other leasehold costs. These costs are to be depicted using the cost depletion method.

The tangible drilling costs are depreciated over 7 years using the DDB/straight-line method. \lreoth eed this in pa igtaph ,ib we. The intangible drilling costs are lD(2s. 70 0/s of IDCs and are deducted in the first year of production with the remaining 30% depreciated over five years using the straight-line method.

Production is expected to begin in year 6 and, for the first 7 years, will be 16,000 Mbbis per year at a constant rate. After 7 years, production will decline at a rate of 2.5% per year for an additional 23 years. The severance tax is 6%. Assume, for simplicity, that depreciation of all equipment will only begin after production starts. The operating cost in the first year is expected to be $21,000 million, increasing at a rate of 2.5% per year. The transportation cost per barrel of oil is Si. At the end of field life, the abandonment cost is expected to be $50,000 million. The price of condensate in the first year can be assumed to he S25 per barrel escalating at 4% per year. -

Is the project feasible under the given conditions? If not, what is the initial break-even price needed for oil? Do you think that the price you estimated is reasonable to achieve? What do you think is the maximum initial oil price you can assume to evaluate this project? \Xhy?

Let us assume that, in year 10, a pipeline connecting the Alaskan gas fields to the lower 48 states will be built and can carry the gas to the lower 48 states. Let us assume that, beginning in vcar 10, the gas from the field can be sold into the pipeline. The cost of building the futilities for processing and piping the gas to the main pipeline is S100,000 million. This -amount is spent in year 9 and will be olepicciated as tangible costs starting in year 10, over 7 \’ears using a DDB/straight-line combination method. Assume that the gas-oil ratio will be maintained at 10,000 S(,] ,’/S’I’B throughout the remaining life of the project. The price of the gas in year 10 is assumed to he SI .50/3lSCli increasing at a rate of 3% per yese Note that the low gas puce results from the Let that markets are so far away and transportation costs are very high.


Assuming the baseline oil price, will selling the gas result in making the project feasible?

If our high estimates of the gas price are $2.50 increasing at a rate of Si/s per year, what will the economics look like?

If you prefer not to rely on the gas pipeline, another alternative of disposing of the gas is to use GTI. (gas to liquid) technology. This technology allows to covert gas into liquids that can be sold like crude. Do research and find out the cost of building a (i/IL plant by assuming the producing gas liquid ratio is 10,000 SCF/SIB. Is it worth using (’l’[, plant to convert gas?

Case Srucly 3-7

Westbrook (Clearfork) is the Permian basin’s discovery field. It was discovered in 1920 on the Eastern Shelf in northwestern Mitchell County. The first well pumped 10 bopd from an openholc completion in the Upper Clearfork after shooting with nitroglycerine. Main pay in the 3,000 ft Lower Clearfork was found when this well was deepened in 1921. Three large secondary units now account for most Clearfork production.

By 1964, 99 wells were on leases that would later be incorporated into the Southwest Westbrook Unit. The unit was formed in 1969 and operated by Union Texas. In its final form, the unit covers 3.833 acres, with calculated original oil in place of 61.6 million bbl. Union Texas immediately drilled injectors and converted producers to injectors to initiate injection. There were 26 injectors on 160-acre spacing.

Beginning in 1971, the unit’s south end was infill drilled on a 40 acre five-spot pattern. ’l’his continued until 1986, when the 40-acre five-spot pattern in the unit’s north end was implemented in areas thought to be productive. In 1991. Meridian (Burlington) took over as operator and drilled, on 10-acre spacing, selected areas that its studies indicated to be the most productive. The injection pattern then became an east west line drive, due to an east-west preferential permeability trend that was noted in several areas of the unit as it was down-spaced.

Gruy purchased the unit from Burlington in 1997 and continued to drill the 10-acre locations in what were thought to be the more productive, less risky areas. Currently, Cimarex is operating the field.

During the Meridian and pre 2003 Gruy tenures, stimulation of the Middle Clearfork zone included small acid breakdowns, followed by cross linked sand fracture treatments diverted with graded rock salt and some surged acid treatments. Before Gray became operator, producers and injectors were sand fracture-treated. However, Gruy was concerned about the ability to control water placement in the injection wells by using hydraulic fracture treatments.

Cimarex started investigating various options to improve performance and learned about BI Service’s LiteProp fracturing technique. T.itcProp is a proppant used in hydraulic fracturing that has a specific gravity of 1.25 and a closure range tolerance up to 5,000 psi. When it is pumped in a near saturated brine water with a specific gravity of 1.20, the result is a near buoyant proppant. This buoyancy allows the propparit to be placed farther out in the induced hydraulic fracture without viscosity-enhanced fluids for proppant transport.

These non-gelled or slick brine water fluids are essentially non-damaging compared to polymer-laden fluids that have been pumped historically at Westbrook. The lack of viscosity also works favorably in controlling fracture height growth development, increasing the potential to stay in zone (and out of water at Westbrook), thus creating a longer effective fracture length. It was also believed that a proppant partial monolayer could be created in this low-closure reservoir to further enhance conductivity in the induced hydraulic fractures. The average cost of fracturing was $140,000, and some relevant tangible costs associated with re-completion were $20,000, which would be depreciated over a five year period using a DDB/straight-line method. If the life of the project is less than five years, assume that you will not be able to depreciate the entire cost. All factors considered, the team believed the Clearfork formation to be an excellent candidate. This, combined with having no other ideas for new or innovative stimulation treatments, led to the implementation of slick water hydraulic fracture treatments using lightweight proppants.

Treatments yielded 80 to 100 bopd during flush production, and declines decreased significantly after six months so that, after a year-and-a half in production, the rate was still 20 bopd (Case Study Figure 3-4). It should he noted that most of the wells drilled or recompleted in 2004 are in areas not supported by injection. In 2005, no drilling was scheduled, although 20 recomplctions were performed, with results similar to those of 2004. For calculation lliu, assume that initial incremental production is 100 b/d, and the first yearly production varies betsvccn 21,900 hbls (worst) and 25,550 bbls (best) and the decline in production per year is anywhere between 55% (best) and 70% (worst).



iso

�’ -. o 114 l,;, .-,--44 --. -.

iF) s 512O

113 !�.-4-2 10- 29

90

i] 30 40

-

-

20

(’0 (’0 CSI (".1 C’) C’) C") C") C’) C") ’a Z; ca c, c es a ’ � :i , :, ;-:: ’ I, x3 -j -, - us .- ,,

C’se Study Figure 3-4: SWWU stimulAtIon effect on well produetionAs a Cimarex engineer, you are interested in evaluating this fracturing treatment. Assume that the operating cost per well is about SI ,500 per month, royalty interest is 20°/, severance tax is 6%, and the income tax rate is 34%. Stop production when the operating revenues equate to the operating benefits (before taxes). Assume the oil price to be between S40/bbl and 565/bbl held constant.

Based on the data of the existing wells, for both pessimistic and optimistic cases, has the project been successful? Assume the costs of fracturing and tangible costs are incurred in year 0 and then revenue from production starts in year I. Begin depreciation and depletion in year 1.

Looking into the future, the cost of fracturing increases to $180,000 and tangible costs increased to $40,000. Since the best wells are already stimulated, you expect that the first year’s production will vary between 17,600 bbls to 20,500 bbls (pessimistic and optimistic) with decline rates the same as in the past. The operating Cost per month has increased to about S2,500. Is the project feasible in the future if you expect similar oil prices? I valuate both pessimistic and optimistic cases.

For calculation purposes, assume that the fracturing costs are depleted over the life of the project. Use Cost depletion only. The reserves are to be determined based on the economic cut-off Assume MROR to be 10°/.

Cisc Study 3-8

On January 26, 2010, Chesapeake Energy Corporation (NYSE: ClIK) announced the closing of its $2.25 billion Barnett Shale joint venture transaction with Total E&P USA, Inc., a wholly-owned subsidiary of Total S.A. (NYSE: TOT, FP:FP) ("Total"), whereby Total acquired a 25% interest in Chesapeake’s upstream Barnett Shale assets. The assets in the joint venture include approximately 270,000 net acres of leasehold in the Core and Tier 1 areas of the Barnett, approximately 700 million cubic feet of natural gas equivalent per day of current net production and approximately 3.0 trillion cubic feet of natural gas equivalent (tcfe) of proved reserves (0.75 tcfe net to Total). Chesapeake also believes that this leasehold position will support the drilling of approximately 3,100 additional net locations with approximately 6.3 tcfe of unrisked, unproved reserves (1.6 tefe net to Total).

Total paid Chesapeake approximately $800 million in cash at closing and will pay an additional $1.45 billion over time by funding 60% of Chesapeake’s share of drilling and completion expenditures until the $1 45 billion obligation has been paid, which Chesapeake expects to occur by year-end 2012.

In the framework of the joint venture, Chesapeake plans to continue acquiring leasehold in the Barnett and Total will acquire its 25% share of the new acreage on promoted terms until December 31, 2015. After this date, Total’s right to acquire its 25% proportionate share of Chesapeake’s leasehold will he on an unpromotcd basis and ’loud will also begin paying 25% of Chesapeake’s support costs related to the joint venture’s corporate development activities.

Aubrey K. i\lcClendon, Chesapeake’s Chief lxccntivc Officer, commented. ’VSc ale pleased to close our joint venture transaction with Total and look forward to cicating substantial value for both companies in the years ahead. We are honored to partner with one of the largest and most respected industrial enterprises in the world to further develop the Barnett Shale. This transaction allows

68 Mohan Kelkcxr, Ph.D., J.D.

S S S S

S S I S I S

I

I

I I I I I I I

Chesapeake to reduce its financial leverage and future capital expenditures and further positions us to deliver industry-leading finding and development costs and returns on capital for years to come."

Your job is to evaluate this transaction from total’s perspective. Chesapeake’s wehsite provides a typical decline curve for the Core and 11cr 1 areas in Barnett Shale. Assume that decline curve to be representative of what to expect in this area. 1he joint venture will drill about 260 wells per scar. For convenience, assume that all the wells start at the beginning of the year. ’1he cost of drilling and completing a well is $2.6 million. Starting Jan 2010, lotal will contribute 70% (why?) of the drilling costs and Chesapeake will contribute 30% of the drilling costs until the end of 2012 when approximately $1.45 billion of lotal’s contribution will be exhausted. lotal will still receive only 25% of gross production, although it is paying 70% of the drilling costs until that time. Beyond 2012, Chesapeake will contribute 75% of the drilling costs and Total will contribute 25%. Drilling will continue at a pace of 260 wells per year until all of the wells are dnllcd. The operating costs of S4,000 per month per well (assume constant) will always be split 75% and 25% between Cl 1K and Total respectively. Assume the average royalty interest to be 25%. llcre is a typical production profile:

Gp Year (MMSCF)

0 � 1 480.4

2 260.1 � 3 196.3

� - 4 162.7 5 141.1 6 125.9 7 114.5

� 8 105.4 9 98.1

� 10 92.0 11 89.3 12 83.9 13 78.9 14 74.2

� 15 69.7 16 65.5 17 61.6 18 57.9 19 54.4 20 51.2 21 48.1 22 45.2 23 42.5 24 39.9 25 37.5

Neglect depletion expenses and assume that drilling and completion costs can be subtracted in the same year in which they are incurred. Assume the severance tax to be 6.0% and federal income tax to he 35%. MROR is assumed to be 12%.

Using this information, answer the following ucStiOfl5:

1. how much did Total pay for its share of current production per MSCFD? is this price reasonable based on an assumed oil price of S80/barrel at the time of transaction? Assume 6 MSCFD is equivalent to 1 S113/D.

2. flow much did ’1 oral pay for the reserves (in ground) per MSCI’? Calculate this number based on both proved reserves and total of proved plus probable plus Possible reserves. What do you think about this price?

1 The biggest uncertainty in this transaction is gas price; therefore, we need to do the economic evaluation using two gas prices, S5/MSCF and S8/MSCl, held constant. Using these prices, calculate the NPV of a typical well when lotal is carrying Chesapeake’s interest and when there is no promote (the split is based on working interest).

1 By using individual well evaluations for both gas prices, determine the NP\’ for ’lotal for the whole project. Assume that, for the first 3 years, Total carried CI 1K and then for the next 9 years, the cost was divided according to working interests.

S Using your evaluation, do you think Intal gist a grind deal for these assets?


Chapter 3- Economic Analysis

rJ

wi�o-’s"11

In this last chapter, we cover the impact of economic uncertainties on economic evaluation. The topic of

uncertainties is extremely broad and the purpose of this chapter is to simply introduce the topic to first-

time readers. We will cover all principles related to this topic, but will not delve into great detail. Please

refer to the Works Cited section at the end of this chapter. (Mian, 2002) (Newendrop & Schuyler, 2000)

(Megill, 1984)

In the previous chapters, we incorporated economic uncertainties into some of the case studies. Most

solutions dealing with uncertainties require a better understanding of statistics; therefore, in this

chapter, we will formally introduce statistical principles and show some basic applications of statistics.

We will also discuss the two important extensions of statistical principles, the decision tree analysis, and

the application of Monte Carlo simulation.

PROBABILITY

Probability is the likelihood of a certain outcome if we repeat an experiment that can result in multiple

outcomes (a random experiment). If a random experiment is repeated many times, N, and an outcome, A, is observed n times, the probability of event A is defined as,

p(A) = Equation 4.4

For example, if we toss a coin 1,000 times and observed an outcome of heads 500 times, the probability

of heads is 0.5 (as it should be for a fair coin).

Similarly, if we repeat the experiment of rolling a die many times, eventually, we will observe that the

probability of each of the six outcomes of rolling a die experiment is 1/6.

For controlled experiments like tossing a coin, it is easy to repeat the experiment numerous times to

determine the probability of a particular outcome. In the petroleum industry, when drilling a well, which

is a random experiment (why?), the act cannot be repeated numerous times for the sake of correctly

determining the probability. Further, we cannot conduct the experiment under controlled conditions.

Each experiment is unique because of geological variations. As a result, when we describe the

probability of a certain outcome, in many cases, it is a subjective estimate of likelihood rather than an

objective assessment of an outcome. For example, when a geologist predicts the probability of the

success of drilling an exploration well to be 0.3, it implies that there is a 30% chance that particular

location will result in a successful producer. It does not mean that, if we drill ten geologically similar

prospects, we will have three successful wells. Statistics do not distinguish between the subjective or

objective estimates of probabilities. However, the distinction is important because subjective

probabilities have associated uncertainties (i.e., the number itself is not guaranteed and is subject to change); whereas, objective probabilities are not subject to change (e.g., the probability of heads in tossing a fair coin is always going to be 0.5).

LAWS OF PROBABILITY

a) For any outcome, A, the probability is not less than 0 and not greater than 1. Mathematically,


Chapter 4 Economic Uncertainties

0 :!~ p(A) < 1 .iwiJi71’

b) If A i represents all of the mutually exclusive outcomes that constitute the entire sample space,

then the addition of the probabilities of all instances of A 1 should add up to one,

1 p(A) = 1 Equation 4-3

For example, when rolling a die, the probability of any of the six outcomes is 1/6. Adding the

probabilities of all the outcomes will result in a value equal to one. If the probability of drilling a

successful well is 0.3, the probability of failure has to be 0.7.

c) If A and B are two mutually exclusive outcomes (i.e., these two outcomes cannot occur at the

same time), then,

p(AUB) =p(A)+p(B)

Equation 4-4

and,

p(A fl B) = 0

LiffDHt;7;1 Di1

where p(A U B) is the probability that either of the two outcomes will occur, and p(A fl B)

represents a probability that both A and B will occur simultaneously. Knowing that drilling a well

can result in two mutually exclusive events - a producer or a dry hole - the probability that the

well can be both a producer and a dry hole is zero.

d) If A and B are two independent outcomes (each outcome has no bearing on the other), then,

p(A r B) = p(A)p(B) Equation 4-6

where p(A n B) represents the probability that A and B will occur together. For example, if we

drill two wells in a basin and assume that the drilling of these two wells are independent events,

assuming that the probability of finding a producer is 0.2, the probability that both wells will be

producers is 0.2 x 0.2 = 0.04, or the probability that both will be dry holes is 0.64. Why?

e) Conditional probability represents the probability of an outcome given another outcome has

occurred. The notation used is p(AB). This represents the probability that outcome A will occur

given that B has already occurred. One of the most famous equations representing conditional

probability is Bayes’ rule. It is given by:

p(AJB) = Equation 4-7

Equation 4-7 can take many convenient forms, some of which are provided below.

p(A n B) = p(B)p(AB) = p(A)p(BIA) Equation 4-8

2 Mohcin Kelkor, Ph.D., J.D.

If we assume that A 1 represents mutually exclusive events that cover all the outcome space

[Z p(A 1 ) = 1], then we can write:

p(B) = X p(Ai fl B) = Z p(A)p(BA 1 ) Equation 4-9

Combining Equations 4-8 and 4-9, we can write:

p(A)p(BlA1) Equation 4-10 p(A1IB) = X p(Aj)ptBIAj

Example 4-1

The probability for the success of an exploration well is dependent on many factors. Assume all these factors to be

independent. They include the probability of geological structure (pg), reservoir rock (pr), sealed trap (pt) and presence of

hydrocarbons (ph). For a particular area, assume that the following values are known:

pg = 0.8,pr = 0.7,pt = 0.6,ph = 0.8

Calculate the probability of success for this well. Assume that another operator drilled a well in the same basin and discovered

the presence of source rock as well as hydrocarbons. You believe that uncertainty related to the presence of hydrocarbons has

been removed. What is the new probability of success?

Solution 4-1

Since the four factors are independent, using Equation 4-6, we can calculate the probability of success (P(S)) as:

Pg X PT X Pt X Pb = 0.8 X 0.7 X 0.6 X 0.8 = 0.269

If we assume the outcome of hydrocarbons presence is H, we can use Bayes’ rule:

p(SIH) =p(SnH) 0.269

- - 0.336

p(H) 0.8

Alternately, this value can also be calculated by multiplying pg, pr, and pt together.

Note that being successful means having all four factors satisfied; therefore, p(S 11 H) = p(S) since being successful is just a

subset of having discovered hydrocarbons. Without hydrocarbons, there is no possibility of a successful well.

Example 4-2 -

Based on the geological basin modeling, an oil company believes that the probability of a favorable structure for oil and gas

accumulation is 0.4. Based on the seismic data collected in a nearby region, if a favorable structure exists, the probability that

seismic data will indicate it is 0.85. If the structure does not exist, the probability that seismic data will incorrectly indicate a

structure is 0.15. Seismic data is collected on the site and it indicates the presence of a structure. What is the probability that

the structure exists?

Solution 4-2

Let us define an outcome that the structure exists as E. We can then write p(E) = 0.4. This also means that (Ec) = 0.6, where Ec represents an outcome that the structure does not exist. Let us define the outcome that seismic data are favorable as S. We

can then state that p(SE) = 0.85 and p(S(Ec) = 0.15. We are interested in calculating p(EIS). Using Bayes’ rule, we can

write:

E - p(S(E)p(E) - 0.85 >< 0.4

- � (5()() + p(SIEC)p(Ec) - 0.85 x 0.4 + 0.15 x 0.6 - 0.7


Chapter 4- Economic Uncertainties 3

If seismic data are favorable, the probability of a structure will increase to 0.79. This also means that p(ECIS) = 0.21. The

probability that the structure does not exist, even in the presence of favorable seismic data, is 0.21.

Problem 4-1

A proposed exploration well will be drilled through two independent prospects. Prospect A has a 0,2 probability of success and Prospect B has a 0.4 probability of success. If a well is drilled, what is the probability that at least one prospect will be successful?

71I1I

An exploration well is proposed. The well is expected to test two potential prospects. Both prospects are expected to have similar structures. The probability of a favorable structure for both prospects is 0.6. The probability that the shallower structure is a

potential reservoir is 0.3 and the probability that the deeper structure is a potential reservoir is 0.2. As the well is being drilled

through the shallower prospect, it is realized that the structure is favorable. Based on the geological modeling, it is believed that

the probability of a favorable structure at a deeper formation is 0.95 if the structure is favorable in the shallower formation. What is the revised probability of success at the lower prospect?

Problem 43

A well is drilled with a prior probability of success equal to 0.3. After the well is drilled, a resistivity log is run. The log indicates the presence of hydrocarbons. Based on other wells in the field, when the well is a successful producer, the resistivity log is 80% accurate in identifying the presence of hydrocarbons. As a compliment, 20% of the time, the log incorrectly identifies the presence of hydrocarbons. Knowing that the log has indicated the presence of hydrocarbons, what is the likelihood that the well will be a producer?

CUMULATIVE DISTRIBUTION AND PROBABILITY DENSITY FUNCTIONS

Before we discuss these two functions in detail, let us define a random variable. A random variable is

defined as the outcome of a random experiment. If it is discrete, it will only have a finite number of

values; if it is continuous, it will have numerous values. An example of discrete random variable is that

the outcome of drilling a well can result in either a producer or a dry hole. An example of a continuous

random variable is the distribution of porosity values in a reservoir. Porosity can take a large number of

values from minimum to maximum.

PROBABILITY DENSITY FUNCTION f(x)

The probability density function is sometimes abbreviated as pdf. This is a pdf of a random variable.

If a random variable is discrete, then we can write,

5 f(x) = 1

–Mtlirm*

where the probability density function over the sample space should add up to one. For example,

when rolling a die, if we define the probability density function for each outcome as 1/6 (i.e., f(1) = f(2) = = f(6) = 1/6,) then all of the numbers should add up to one. A plot of

probability density function for rolling a die is shown in Figure 4-1.


f(x)

1/6

1 2 3 4 5 6

x

Figure 4-1: pdf for rolling a die

If a random variable is continuous, then we can write,

f f(x)dx = 1 Equation 4-12

where the integration of pdf over the sample space is equal to one. Refer to Figure 4-2. In this

figure, f(d) is plotted as a function of the variable value. The only requirement that this be pdf is

that the area under the curve has to be equal to one.

0.45

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0

0 2 4 6 a

Value of Variable

Figure 4-2: pdf of continuous distribution


Chapter 4 - Economic Uncertainties 5

The probability of any given outcome is related to the pdf. For example, the probability that

outcome A will occur for discrete pdf can be written as,

p[X E Al = >Af(X) Equation 4-13

For example, when rolling a die, the probability that the outcome will be an even number (i.e., either 2, 4 or 6) will be the addition of pdf for all the even number outcomes, or:

p[xE 2,4, 6]=

For a continuous variable, a similar equation can be written as,

p[a :!~ X < b] = ff(x)dx Equation 4-14

where Equation 4-14 defines the probability that the random variable will be between an interval

(a, b). As shown in Figure 4-3, Equation 4-14 represents the area under the curve within the interval

(a, b).

r iI

0.4

0.35

0.3

- 0.25

0.2

0.15

0.1

0.05

0 Value of Variable

Figure 4-3: Probability value that random variable will fall within an interval [a,b]

CUMULATIVE DISTRIBUTION FUNCTIONJ(x)

Cumulative distribution function is abbreviated as cdf and is defined as,

F(x) = p[X :5 x] Equation 4-15

It is the probability that the value of a random variable is less than or equal to a given outcome.

Knowing how we calculate the probability, for discrete variables, we can write cdf as,

F(x) f(t) Eatio 4-16

6 Mohan Ke/kar, PhD., ID.

Equation 4-16 states that if we sum all the values of pdf up to and including the value x, we will

obtain cdf. For example, when rolling a die, if we want the cdf for all the values less than or equal to

3, we can calculate it as,

1111 F(3)=f(1)+f(2)-i-f(3)=�+�+�=-

A plot of cdf for rolling a die is shown in Figure 4-4. As is evident in the figure, the probability that

the outcome is less than any of the six outcomes can be calculated using this figure. For example,

the probability that the outcome will be less than or equal to 4 is 4/6.

A similar equation can also be written for a continuous random variable. Mathematically,

F(x) = p[X x] = ff(t)dt

Graphically, refer to Figure 4-2. Using the pdf shown in Figure 4-2, we can calculate the cdf as shown in Figure 4-5. For any given value, the area under the curve up to that point is equal to F(x). The entire plot for F(x) can be constructed this way. For a discrete variable, cdf appears as steps,

whereas, for a continuous random variable cdf increases gradually (and continuously) until it

reaches a value of one.

0.83

0.67

0.50 LL

0.33

0.17

0.00

0 1 2 3 4

5 6

Value of Variable

Figure 4-4: cdf for rolling a die



1

0.9

0.8

0.7

0.6

U-

0.4

0.3

0.2

0.1

0

0 2 4 6 8

Value of Variable Figure 4-5: cdf for continuous variable

We can use cdf to determine the probability that the value of a random variable is less than a

certain threshold. The two important characteristics worth remembering about cdl are that it is a

non-decreasing function and its limiting values are always zero and one.

EXPECTED VALUE

Expected value is defined as the average outcome of a random experiment if the experiment is

conducted numerous times. Expected value has no meaning if we conduct the experiment a limited

number of times. The following example illustrates the concept of expected value more clearly.

Example 43

In a casino, the wheel of fortune contains 80 slots. Each time you bet you have to place $1. Of the 80 slots, if you hit one of

seven designated slots, you will win $10. If you hit any of the other 73 slots, you will lose your $1. What is the expected value of this experiment?

Solution 43

Knowing the total number of slots and the number of slots that will allow you to win, the probability of winning is (7/80) and

the probability of losing is (73/80). The expected value is calculated as,

EV = 1� ($10) + - (-1) = $00375 f7) (73)

80 80

In the equation above, we multiplied the probability by the associated amount. The expected value for this experiment is -

$0375. Obviously, we know that, for a given bet, we will either win $10 or lose $1; therefore, we will never lose $00375. This

number is, however, telling you that if you play the game many times, on average, you will lose $0375 per game. That is, if you

play the game 1,000 times, you will lose 1,000(0.0375) z $37 over 1,000 games.


As illustrated, the expected value only achieves meaning if an experiment is conducted numerous times.

Let us consider another example. If we roll a die, what is the expected value of the outcome? We know

that the probability of any of the six outcomes is 1/6. We can calculate the expected value as,

EV �3.5

Obviously, we know that none of the experiments will result in an outcome of 3.5. What this number

indicates is that if we roll the die 1,000 times, and calculate the average of all our outcomes, it will be

close to 3.5.

Mathematically, the expected value is defined as:

E[u(X)} 5 u(x)f(x) Equation 4-13

for a discrete variable, and:

E[u(X)] = fu(x)f(x)dx

where u(x) is any function of a random variable X.

For our purposes, we are most interested in two expected values of a random variable; mean and

variance. The mean for a discrete variable is defined as,

E[X] = xf(x) =

A similar equation can be written for a continuous random variable. A variance is defined as,

E[(X - ii)2] = - Lt)2f(x) =

Traditionally, ji denotes the population mean and a 2 denotes the population variance. In the two

equations above, f(x) represents the probability density function. For discrete variables, this function

represents the probability of a certain discrete outcome. We can, therefore, simplify the two equations

above as:

[1 =

where p1 IS the individual probability value for an outcome x. The variance is calculated as:

a2 = 1=1 Pt (x

Equation 4-23

where i is the mean given in Equation 4-22 and Pt IS the probability of a discrete outcome x.

In economic calculations, instead of using mean, we use expected monetary value (EMV). The definition

is the same as shown in Equation 4-22, except that we use our outcome as money. The following

example illustrates the concept.



Example 4-4

Based on the exploratory analysis of an oil field, the following probability distribution for possible outcomes is expected.

Calculate the expected monetary value of this venture.

Probability, p- Barrels of Oil(bbls) NPV ($)

0.4 0 -400,000

0.3 100,000 200,000

0.2 300,000 600,000

0.1 500,000 900,000

Solution 4-4

As evident from the table above, there is a 40% probability that the well will be dry and we will lose $400,000, the cost of

drilling. On the other hand, there is a 10% probability that the well will make $900,000. Using Equation 4-22,

EMV = p1(NPV)

where n is the number of possible outcomes, (NPV)1 is the net present value of the outcome i, and pi is the probability of the

outcome i,

= 0.4(-400,000) + 0.3 (200,000) + 0.2(600,000) + 0.1(900,000)

= $110,000

The expected monetary value (EMV) of this venture is $110,000. If we participate in a similar, large number of ventures, on

average, we will make $110,000 per venture. Unfortunately, no two ventures are exactly alike. Nevertheless, the EMV concept

is still attractive from the decision-making perspective. In a more general sense, assuming that our probability distribution

represents the true nature of our outcomes, we can say that, if we select the projects that have positive EMV’s, on average, we will be profitable if we participate in a relatively large number of similar adventures.

EMV can be a very useful tool in making economic decisions. As discussed in the Probability section, in

the petroleum industry, the probability determination is often made using subjective evaluation. As a

result, there is always some uncertainty associated with the estimation of probability itself. EMV can be

used to assess the impact of uncertainty with respect to probability. The following examples illustrate

this application.

Example 4-5

It is expected that drilling a well will cost $100,000. Based on the analysis of wells nearby in similar areas, there is a 50%

probability that the well will be successful. If successful, the annual revenues are expected to be $40,000 over the next seven

years with a salvage value of $10,000. Calculate the EMV of this venture. What will be the probability of success for the venture

to break even? Assume the MROR = 15%.

Solution 4-5

Based on the values given, the probability of success is:

Ps = 0.5

The probability of failure (dry hole) = 1 �p = 0.5.

We can calculate the EMV as:


EMV = p1 (NPV)

where n represents the number of outcomes. In this case, we have only two outcomes. If the well is successful:

(1+.15)-1 10,000 NPV = 40,000 .15(1 +.15)7 1 + (

1 +.15) - 100,000

= $70,176

If the well is failure:

NPV = �$100,000 = cost of drilling

Using the EMV equation,

EMV = p(NPV) 1

= 0.5(70,179) + 0.5(-100,000) - -$14,912

Knowing that there is always some uncertainty with respect to probability value, we can extend this analysis by plotting EMV as

a function of the probability. Figure 4-6 shows this plot where EMV is calculated for probability values between 0 and 1. As

expected, [MV is -100,000 for probability equal to zero and EMV is 70,176 for probability equal to one. For probability of 0.5,

EMV is negative. However, probability greater than 0.58, [MV becomes positive. In other words, if we know that the probability

of success is at least 0.58, we can say that the project will be successful. We do not need to know the exact value of the

probability of success; we only need to know the minimum value of probability for success and be confident that that value can

be achieved.

80000

�� I ....

LU

120000 _ _

Probability Figure 4-6: EMV versus probability

Farming out is another application where this type of graphical representation is useful in investigating the feasibility of a

project. Farming out is a procedure whereby a company that owns the working interest in a lease gives away the working

interest in return for override royalty or some other interest. A company owning a leasehold interest may not have enough

capital to drill a well, or it may have interest in other areas where it is more confident about success. Rather than lose the lease,

it may opt to farm out the interest to another company. In return, it will receive an override royalty or a "back in" interest.

Back in interest represents a working interest after it pays a penalty for not participating in the initial drilling of the well



0 I a a S S a, S S 0

Example 4-6

A company owns 100% working interest in a lease. If a development well is drilled on the lease, based on the geological

knowledge, there is more than a 65% chance that a successful gas well will be drilled. The cost of drilling is expected to be

$1,200,000. If successful, an initial income of $700,000 is expected, which will decline at a rate of 20% per year for the next 20

years. The operating costs are assumed to remain constant at $50,000 per year. Assume the salvage value to be negligible. The

company is also considering a farm out option where, after assigning 100% working interest, the company can acquire 1/16 of

the override royalty interest. Which option should be selected? Under what circumstances will the other option be the better

option? Assume the MROR = 10%.

Solution 4-6

The NPV, if the well is successful, can be calculated as:

700,000 1 0.8 20 1.120 - I

NPV = (0.1+0.2) [ - - 50,000 0.1 >

1.120 - 1,200,000 = $703,656

If the operator opts to override royalty interest, the NPV will be:

700,000 x() 0.8 20 NPV = (

OT +O.2) 0.2) 1 - ---j- = $ 145,583

Although the NPV using an override option is smaller, there is an advantage because there is no cost for failure since the

operator does not have to participate in the drilling. So the EMV for maintaining the working interest can be written as:

EMV = p(703,656) + (1 - p)(-1,200,000)

The EMV for only override royalty interest can be written as:

EMV = p(145,583)

The two equations are plotted in Figure 4-7. As shown, for probability of success greater than 0.68, drilling a well always results

in better EMV than securing an override royalty interest. If the operator believes that there is at least a 65% chance of success,

it is better to drill the well rather than secure an override royalty interest.

1100000

600000

1/16 Ri

100000 ....................

Uj

z::::i::: .... :1 ....................

100..........................................

-1400000 ______ -------- -------- - ______

0 0.2 0.4 0.6 0.8 1

Probability

Figure 4-7: The impact of probability on drilling well vs. override royalty

We can add a little more complexity to this analysis by considering a "back in" option. Unlike the farm

out option, where working interest is traded in for an override royalty interest, the "back in" option


allows trading in interest with the option of backing in with the same or smaller working interest after

paying a penalty for not taking any risk. For example, a company with 50% working interest may assign

the working interest to another company with the option of gaining back the 50% working interest after

the other company recovers the cost of drilling plus the penalty cost. Typically, the penalty cost is a

percentage of the drilling cost. If the penalty is 100%, the other company will be allowed to recover

twice the cost of drilling as a penalty before the first company can back in with a working interest. The

following example illustrates this option.

Company A owns 50% working interest in a new prospect; Company B owns the other 50%. Company B wants to drill a well in

the prospect. It is expected that the cost of drilling the well will be $2 million. The revenue, after subtracting the royalty

interest, is expected to be $1,200,000 in the first year, declining at a rate of 8% per year for ten years. The operating costs are

$60,000 per year, held constant. Company A is considering the following options:

� Participate in the well with 50% working interest.

� Do not participate in the drilling with the option of backing in with 50% working interest after 100% penalty.

� Farm out the interest with 3/16 royalty interest.

If p is the probability of success, develop equations for EMV for all three options and calculate the probability range for which

each of the options will be successful. Assume the MROR to be 10% per year.

Option (1)

Since company A owns 50% working interest, it will have to pay $1 million for drilling the well. In return, we can calculate the

NPV as:

600,000 1 0.92101 F 1.110 - 11

NPV = (0.1 + 0.08) [i - 1110 j - 1,000,000 - 30,000 [1.110 x o.4 = $1,590,744

The EMV can be written as:

EMV = Ps X 1,590,744 + (1 - jj 5 ) X 1,000,000

Option (2)

Company A wants to back in with 50% working interest after paying a 100% penalty. That is, Company A will have to pay

$2,000,000 in drilling costs (instead of $1,000,000) before it can "back in" with 50% working interest. Company B will have to

recover $2,000,000 from the net revenue of Company A before Company A can "back in." That is, for the first year, Company

A’s revenue ($600,000 - $30,000 "$570,000) will go to Company B. Based on an 8% decline each year, for the first four years,

Company A will not recover any money. NPV under this option can be written as:

429,836 1 0.92 1 1 [1.1 6 11

NPV = (0.1+0.08) [ - --jj----30,000 [ 1.110 j = $983,528

EMV can be written as

EMV = Ps X 983,528

Option (3)

In this case, Company A will get 3/16 royalty interest if the well is successful. NPV can be written as:



600,000 x 3/16 0.92 10

NPV = (0.1+0.08) 1 - 1:110 = $520,328

EMV can be written as:

EMV = Ps ( 520,328

The three options are graphically presented in Figure 4-8. Examining Figure 4-8, Option 2 is always better than Option 3 for any

probability value. In both cases, there is no cost for failure but the benefit increases more rapidly under Option 2 than Option 3.

Comparing Option 1 with Option 2, if the probability of success can be assumed to be greater than 0.6, Option 1 is always going

to be better than Option 2. On the other hand, if we are confident that the probability of success is less than 0.6, it is better to

choose Option 2 rather than Option 1.

-Option I - - Option II - -Option III

2100000

F ... .................... ..... ............................................

LU

-1400000 - T I1l Probability

Figure 4-8: Comparison of WI versus "back in" us. override royalty

From this analysis, it is clear that if prolific production is expected from the well, the option of paying the penalty is a better

option. On the other hand, if the well is expected to be marginal, securing override royalty is a better choice. For example,

instead of $1,200,000 revenue in the first year, if we assume that the revenue is $800,000 in the first year, it will take about

seven years to recover the drilling penalty before Company A can "back in". The graph will be similar to Figure 4-9. Under this

scenario, Option 3 is always better than Option 2. That is, it is better to accept the override royalty than pay a drilling cost

penalty. Further, the probability of success has to be greater than 80% before Option I becomes better than Option 3. As the

well becomes more marginal, the option of accepting an override royalty looks increasingly better.

-Option I - - Option II - Option III

1100000

600000 - ......, ..

Uj

100 _

Probability

Figure 49: Impact of less revenue on decision making


Another interesting application of probability analysis is to estimate the expected monetary value for a

multi-zone prospect (Rose, 1992). Recall that Equation 4-6 states that the probability of success can be

calculated as a product of four factors: reservoir rock, favorable structure, presence of hydrocarbons

(source rock), and the presence of a trap.

Pi - Pr X Pst X Ph X Pt

If we are considering two prospects, one underneath the other, some of the probabilities on the right

side will be common to both the prospects. For example, if both zones have the same source and similar

structures, Ph and Pst may be identical for both; whereas, the probabilities Pr and Pt are different. Also,

remember that when investigating two zones, four possibilities exist: both are successful, both are

unsuccessful, or one of them is successful. We need to estimate the probabilities of all four possibilities

before we can estimate the expected value.

Example 48

A company is investigating the potential of a two zone prospect: zones x and Y. Both are expected to have the same source

rock with Ph equal to 0.85. The probability of structure is 0.60 for both the zones. For zone x, p1 = 0.7 and Pt = 0.8 ; for zone

y, Pr = 0.4 and Pt = 0.7. Drilling and completing a well in multiple zones will cost $1 million. If production from zone x is a

success, the NPV will be $4 million. If production from zone y is a success, the NPV will be $10 million. Calculate the EMV for

this project.

Solution 48

To calculate the probabilities of all four possibilities, we need to separate dependent and independent factors. Dependent

factors are Ph and Pst� These factors have to co-exist for both at the same time.

For both zones x and y, the probability of dependent factors:

Pde = Ph X Pst

= 0.85 x 0.6 = 0.51

For zone x, the probability of independent factors, if successful:

Pm = Pr X Pt

= 0.7 x 0.8 = 0.56

For zone y:

Pin = Pr X Pt = 0.4 x 0.7 = 0.28

The probability of independent factors represents the probability of success.

Using these numbers, we can calculate the probability that both x and y are successful,

= (Pin)x(Pin)yPde = 0.56 X 0.28 X 0.51 = 0.08

the probability that x is successful and is unsuccessful,

= (Pin)x( 1 - Pin) yPde = 0.56(1 - .28)(. 51) = 0.2056

and the probability that x is unsuccessful and is successful,



= (1 - Pin)x(Pin)yPde = (1 -.S6) 28(. 51) = 0.0628

Knowing that probabilities of all four possibilities should equal one, we can calculate the probability that both x and y are

unsuccessful:

= 1 - 0.08 - 0.2056 - 0.0628 = 0.6516

Using the equation:

EMV = pL(NPV)I

= 0.08(14) + 0.2056(4) + 0.0628(10) + 0.6516(-1) = $1.92 million

Since the expected monetary value is positive, the project is feasible.

Problem 4-4

A new well is going to cost $2,000,000 to drill. There is a 50% probability that the well will be a dry hole, a 30% probability that the well will produce 40 bbls/day, a 15% probability that the well will produce 70 bbls/day and oS% probability that the well will produce 150 bbls/day. If production is expected to decline at a rote of 12% per year over the next seven years, what is the EMV of this property? Assume that the MROR is 20%. Based on the EMV, how much lease bonus are you willing to pay? The price of oil is expected to be $901bbl and the operating costs are $101bbl. The salvage value is negligible.

Problem 4-5

A drilling program involves drilling two wells at two potential targets. Based on the geological analysis, the probability of success is 0.2 at location A and 0.4 at location B. The cost of drilling wells at either location is $1,000,000. If successful, a producer will result in a NPV of $8,000,000 at location A and $5,000,000 at location B. What is the EMV of this venture? Assume that the drilling at these two locations are independent events.

Problem 4-6

An oil company is considering two alternatives for exploring a new area. Alternative A requires drilling a vertical well at a cost of $1,000,000 and obtaining production of 30 bbls/day with a probability of 0.6. There is a 40% chance that the well is a dry hole. Alternative B requires drilling a horizontal well at a cost of $3 million and obtaining production of 150 bbls/doy with a probability of 0.4, obtaining a production of 450 bbls/day with a probability of 0.2, or hitting a dry hole with a probability of 0.4. Production is expected to decline at a rote of 11% /year. If production is expected to last for a six years, the oil price is $801bbl with an operating cost of $100,000 for a vertical well and $300,000 for a horizontal well, select the correct alternative. Assume that the MROR is 15% and the salvage value is negligible.

Problem 4-7

An oil company is considering a new enhanced oil recovery method for increasing production from existing reservoirs. The

research effort is going to cost $1 million per year over the next four years. The probability that the process will be successful in

the laboratory is 0.4. If, after four years, it proves to be successful, further testing will be needed in the field. This testing will cost $15 million in the fifth year. The probability of success in the field, once successful in the lob, is 0.7. If successful, the project will result in on additional $25 million, after tax, income over the next ten years. From the FMV perspective, is the research effort worth investing? Assume the MROR to be 15%.

Problem 4-8

An independent producer has 100% working interest in a lease. Due to deep drilling, the cost of drilling a well is expected to be $3,000,000. If successful, production will result in a NPV of $12 million. Another option for the producer is to form out the working interest and retain 118 royalty interest. Plot the FMV for both options as a function of the probability of success. At what value of probability will drilling the well by the producer be more profitable?


allowt An oil company has a 40% working interest in a lease. There is a 40% probability that drilling a well will result in a successful producer with an oil production of 100 bbls/day, declining at a rote of 15% per year over the next six years. There is a 60% probability that the well will be a dry hole. The price of oil can be assumed to be $801bbl with an operating cost of $121bbl. The cost of drilling the well is going to be $1,180,000. The company is also considering farming out the working interest and receiving, in return, 1116 royalty interest. Which option should be selected? At what probability value will the other option be

more suitable? Assume the salvage value to be zero and the MROR to be 15%.

Problem 4-10

Oil Company A has on overriding royalty interest of 3116 in a potential prospect. Oil Company B wishes to trade its 60% working

interest for the overriding royalty. If the cast of drilling is $700,000 and the NPV of a successful well is $5,000,000, for what range of values of the probability of success should Oil Company A trade its interest to Oil Company B?

Problem 4-11

An oil company, which currently owns 100% working interest in a potential prospect, has two possible options of trading in its

working interest. Under option one, it can trade 100% of its working interest for 3116 royalty interest. Under option two, if the well is successful, the company can back in the well with 80% working interest after paying a 150% penalty. If the cost of drilling

is $1,600,000 and the expected NPV of a successful project is $5.5 million, state the probability of success ranges for which either of the options (including retaining the 100% working interest) will be the most profitable.

Problem 4-12

A company owns 40% working interest in a potential oil prospect. The cast of drilling a horizontal well is expected to be $6,000,000. If a well is successful, it is expected to produce 30,000 bbls of oil in the first year, declining at a rate of 15% per year. The price of oil is $851bbl and the operating casts are $300,000 fixed cost plus $81bbl lifting cost. Assuming five years of production and negligible salvage value, if the company can trade in its working interest with the option of backing in at 30% working interest with 150% penalty, which option should the company choose if the probability of success of a horizontal well is 0.4?At what range of probability of success values will the other option be vioble?Assume the MROR to be 10%.

Problem 4-13

A two zone prospect is currently being investigated by an oil company. The probability of structure is 0.5 for both zones, L and M. The other probabilities are given below.

Probability L M

Pr 0.7 0.4

Ph 0.9 0.7

Pt 0.8 09

Pc 0.9 0.6

Pc represents the probability of completion success. If drilling the well costs $2 million, a successful completion in L will result in a NPV of $3.5 million and a successful completion in M will result in a NPV of $12 million. Calculate the EMV of the prospect. Is the project feasible?

DISTRIBUTION FUNCTIONS (Mian, 2002) (Newendrop & Schuyler, 2000)

A distribution function is a way by which probability density function can be defined with a minimum

number of parameters. The functions are defined so that, once we define some key parameters, the

entire probability function is defined. Most of these functions are applicable for a continuous random

variable. One important function, which is applicable for discrete random variable, is a binomial

distribution function. We will discuss it first.



B INOMIA L DISTRIBUTION FUNCTION

As indicated, this function describes a discrete random variable. The function has several

characteristics.

� The function can have only two outcomes, typically denoted as success (S) or failure (F) � The two outcomes are mutually exclusive.

� The probability of success is p in every trial.

� Subsequent experiments are independent of each other.

A simple example, which satisfies these criteria, is a coin flipping experiment. A coin flip has only

two possible outcomes, heads or tails. For a given experiment, only one of the two outcomes can be

observed (mutually exclusive), the probability that we will observe heads or tails is always 0.5, and

what happened in the previous toss has no reflection on the outcome of the next toss (subsequent

experiments are independent).

Example 4-9

Calculate the probability that if we flip three coins simultaneously, we will observe the outcome of heads on all the three coins.

Solution 4-9

Since the three coin flips are independent of each other, extending Equation 4-6 for three independent events, p(A n B n C) =

p(A)p(B)p(C). Knowing that the probability of observing heads is 0.5,

p(A fl B 11 C) = 0.5 x 0.5 x 0.5 = 0.125

The probability that we observe three heads simultaneously is 0.125.

Example 4-10

A company is planning to drill three wells in a new area. Based on past experience, the probability of finding a producer is 0.2.

Calculate the probability that 1) all the wells are producers; 2) two wells are producers and one is a dry hole; 3) one well is a

producer and the other-two holes are dry; and, 4) all three wells are dry holes.

Solution 4-10

We know that drilling three wells are three independent events, we also know that the probability of success is 0.2 and the

probability of failure is 0.8.

Let us write all the possible combinations in which the four events listed above can occur and calculate the probabilities. Assign

the letter P for a producer and the letter D for a dry hole.

1) All three producers

p(PPP) = 0.2 x 0.2 x 0.2 = 0.008

2) Two producers and one dry hole

p(PPD) = 0.2 x 0.2 x 0.8 = 0.032 p(PDP) = 0.2 x 0.8 x 0.2 = 0.032 p(DPP) = 0.8 x 0.2 x 0.2 = 0.032

0.096


3) One producer and two dry holes

p(DDP) = 0.8 x 0.8 x 0.2 = 0.128 p(DPD) = 0.8 x 0.2 x 0.8 = 0.128 p(PDD) = 0.2 x 0.8 x 0.8 = 0.128

0.384

4) Three dry holes

p(DDD) = 0.8 x 0.8 x 0.8 = 0.512

In parts (2) and (3), since the order in which the given event occurs is not important, we have added the probabilities of

individual combinations to calculate the probability of that event. If we add the probabilities of all four events,

= 0.008 + 0.096 + 0.384 + 0.512 = 1.0

will add up to one indicating that these four events cover the entire samples

Although we can solve every problem exhibiting binomial distribution this way, for a large number

of wells this can become cumbersome. Fortunately, we can write a simple formula to calculate the

probability of any combination,

p(S =x) = cpx(1_p)n_x r..thii.Z’J

where x is the number of successes, n is the number of trials (experiments) and c is the number of

combinations in which x successes can occur and is defined as,

n� - x!(n�x)!

Example 4-11

In the initial drilling program, a company wants to drill six exploratory wells. Based on the analysis of preliminary data, chances

of finding a producer are 0.3. calculate the probabilities of all the possible combinations in which these trials will result.

Solution 4-11

n = 6 = the total number of trials and p = 0.3. If we define a producer by a letter P and a dry hole by a letter D, we can calculate the probabilities of all the possible events using Equation 4-24,

1)

p(6P) = C(0.3) 6 (0.7) ° = 7.29 x iO

=-j= 1 (note that 0! is one) 6 6!

2)

p(5P, 1D) = C(0.3) 5 (0.7) 1 = 0.01021

3)

p(4P,2D) = C(0.3) 4 (0.7) 2 = 0.05954

4) p(3P, 3D) = C.(0.3) 3 (0.7) 3 = 0.1852



p(2P,4D) = C(0.3) 2 (0.7) 4 = 0.3241

6)

p(1P,5D) = C , (03)’(0.7)’ = 0.3025

7)

P(6D) = C6 (0.3) 1 (0.7) 6 = 0.1176

The probabilities of all seven combinations add up to 1.0.

A related issue to binomial distribution is what is commonly referred to as gambler’s ruin. Gambler’s

ruin is a normal run of bad luck; a dry spell. You keep on betting and you keep on losing over a long

period. If you don’t stop, you may lose everything. We can easily equate the gambler’s ruin to

drilling wells in exploratory areas. With a normal run of bad luck, you may exhaust your resources

and hit all dry wells. To estimate the chances of exhausting all the resources, we can use the

binominal distribution function.

Assuming that drilling of subsequent wells is an independent event, to avoid gambler’s ruin we need

to hit at least one producer. That is, we need to find probabilities of all events where the number of

producers is greater than or equal to one. Knowing that the probabilities of all the combinations will

add up to one, we can calculate the probability of at least one success by:

p(at least 1P) 1 - p(all dry holes)

= 1 - C(p)°(1 - p’ Equation 4-26

If we subtract the probability of all the holes being dry, we can calculate the joint probability that at

least one of the wells will be a producer.

Example 4-12

A company has an exploration budget of $500,000. Each wildcat well is expected to cost $100,000. If the probability of success

is 0.1, what is the probability of hitting a producer using the exploration budget?

Solution 4-12

The number of wells that can be drilled:

$500,000

$100,000

Therefore, n = 5. Using Equation 4-26,

p(at least 1P) = 1 -

= 0.4095

There is a 40.95% chance that at least one of the wells will be a producer. Considering that the company is exhausting its entire

budget, it is not a very good chance.

One alternative to reduce the risk of gambler’s ruin is to participate in more wells. Instead of drilling the

wells on your own, join with other oil companies so that the company can participate in more

exploratory wells. This has been a common practice in the oil industry especially in frontier areas. Let us

consider the above example and assume that the company has decided to jointly explore the wildcat


area by reducing its interest to 50%. Therefore, instead of drilling only five wells, the joint operation can

drill 10 wells in the same area. We can calculate the probability of at least one success by,

p(at least 1P) = 1 - CO(0.1)0(0.9)b0

= 0.6513

By reducing the interest and participating in more wells, the probability of success has improved by 25%.

A company can further reduce risk by reducing the interest in individual wells and participating in more

wells.

Example 4-13

A company is anticipating drilling a well in a wildcat area. The cost of drilling is expected to be $1,000,000. If successful, the well

is expected to generate $10 million in net present revenues. The probability of success is calculated to be 0.2. The two options considered are:

1. Drill a well by owning 100% interest.

2. Drill two wells jointly owning 50% interest.

Solution 4-13

1) Under option one, the probability of success is 0.2.

EMV = >Pj(NPV)j

For the possible alternatives,

EMV = 0.2($10,000,000) + 0.8(�$1,000,000) = $1,200,000

2) Under option two, three possibilities exist:

p(2P) = 0.2 x 0.2 = 0.04 p(1P, 1D) = (0.2 x 0.8)2 = 0.32

p(2D) = 0.8 x 0.8 = 0.64

The probability of at-least one success is 0.32 + .04 = 0.36, which has increased substantially from 0.2. It can easily be shown

that the expected value does not change by participating in two wells; however, the probability of at least one success increases substantially.

In using binomial distribution, we made an assumption that each event is independent; that is, what

happens in the previous event has no bearing on the next. In a basin where exploration is carried

out, we have a limited number of prospects. If we assume a certain success probability, what has

happened in the past can influence what will happen next. This is because only a certain number of

prospects are productive. If we discovered a productive prospect, we will need to eliminate the

number of productive prospects by one and recalculate the probability of success for the remaining

prospects. A separate distribution function exists to describe these characteristics called

hypergeometric distribution. In practice, since it is very difficult to determine the total number of

prospects as well as number of productive prospects, the hypergeometric function is rarely used.

Problem 4-14

If you flip three coins, calculate the probability that we will observe two heads and one tails. Will the result be different if we



want to calculate the probability that we observe one heads and two tails?

Problem 4-15

An oil company wishes to drill five wells in on exploratory drilling program. If the probability of finding a producer is 0.2 and the drilling of wells is an independent event, calculate the probabilities of all possible events.

Problem 4-16

A company wishes to drill four development wells. The probability of success is 0.4. Based on the analysis of other wells in the area, a successful well will result in a $5 million NPV. The cost of drilling an individual well is $2,000,000. Calculate the EMV of this venture. Assume the events to be independent.

Problem 4-17

An independent oil producer anticipates that, using the existing exploration budget, he can drill five wells on his own. The probability of success in the area is 0.15. What is the probability that at least one well will be successful? If he wants to increase the probability of success to 0.9, in how many wells does he need to participate?

Problem 4-18

Based an geological analysis of nearby fields, one in ten drilled we/Is will be a successful producer. How many we/Is should an oil company participate in to increase the probability of success to 0.8? If the cost of drilling on individual well is $1 million and the exploratory budget of an oil company is $4 million, what is the working interest owned by the oil company in an individual well? If a successful well will result in a NPV of $15 million, what is the EMV of this venture?

UNIFORM DISTRIBUTION

Uniform distribution is one of the simplest distributions for a continuous random variable. The pdf for uniform distribution function is shown in Figure 4-10 along with its cdf.

f(x) F(x)

a

x

Figure 4-10: Uniform distribution function

As shown, the uniform distribution function assumes that there is equal probability that a random

variable is selected from the interval (a, b). The only two parameters that needed are the minimum

and the maximum values. Knowing that the area under the pdf curve should be one, we can write:

f(x)=O x<a 1

axb


f(x)=O x>b

MIUMMIMM

Knowing that

F(x) = f f (x)dx

we can write:

F(x) = Equation 4-28

An advantage of using uniform distribution is the minimal requirement of data. For variables that

are not very well sampled, if the minimum and maximum are known, a uniform distribution may be assumed.

TRIANGULAR DISTRIBUTION

Triangular distribution is arguably the most popular distribution function used in economic analysis.

This distribution requires three pieces of information of a random variable: the minimum, the

maximum, and the most likely value. The pdf and cdf of the triangular distribution are shown in Figure 4-11.

I

>( U.

-II a c b a b

X x

Figure 4-11: Triangular distribution

In this figure, (a) represents the minimum value, (c) represents the most likely value and (b) represents the maximum value. Knowing that the area (a) under the curve for pdf function should be one, we can write:

f(x)=O x<a 2(x�a)

f(X ) = (b a)(ca)

2(b�x) f(X)=(ba)(b_c) cx~b



AX) = 0 x > b Equation 4-29

Knowing the relationship between cdf and pdf, we can write,

(x

ax~c

F(x) = 1 - (b-x)2 c :!~ x b Equation 4-30 (b-a)(b-c)

Example 4-14

If payzone thickness for a reservoir can be represented by a triangular distribution where the minimum is thirty feet, the most

likely value is fifty feet and the maximum is ninety feet, calculate the probability that the thickness is less than sixty feet for this

distribution.

Solution 4-14

In this example, a =3Oft.,b=90ft., and c=50ft.

To calculate the p (thickness :5 60 feet), we have to use Equation 4-30 because c < 60 < b. Recall that F(x) = p[X < x].

Equation 4-30,

(b �X)2

p[thickness :560] = 1 (Li - a)(b - c)

(90 - 60) 2 = 1

- (90 - 30)(90 - 50) = 0.625

The popularity of triangular distribution is a result of the triangular distribution’s simplicity as well as

flexibility. It can be constructed with relatively limited information. By adjusting the most likely value,

the triangular distribution can approximate the normal and log-normal distributions, which will be

discussed later.

NORMAL DISTRIBUTION

Normal distribution is the most commonly used distribution in conventional statistics. It is also

called a Gaussian distribution. The pdf and cdf for normal distribution are shown in Figure 4 -12. As

shown, the distribution is bell shaped and is symmetric with respect to the maximum pdf value.

f(x) F(x)

0 x

x

Figure 4-12: Normal distribution

24 Mohan Kelkar, Ph.D., J . D.

The pdf for normal distribution is defined as,

f 1 e_(x_ 2/2Z - -: < x <

Knowing the relationship between pdf and cdf, we can write an equation for cdl. Before we define cdf, we can simplify Equation 4-31 further. It can be shown that the mean of the normal distribution

function is it and the variance is cr 2 . If we define a standardized variable:

Table 4-1: z values for normal distribution (MegilI, 1984) (Davis, 1986) Standard Deviations

from the Mean Cumulative Probability

Standard Deviations from the Mean

Cumulative Probability

-3.0 0.0014 +0.0 0.5000 -3.0 0.0014 +0.0 0.5000 -2.9 0.0019 +0.1 0.5398 -2.8 0.0026 +0.2 0.5793 -2.7 0.0035 +03 0.6179 -2.6 0.0047 +0.4 0.6554 -2.5 0.0062 +0.5 0.6915 -2.4 0.0082 +0.6 0.7257 -2.3 0.0107 +0.7 0.7580 -2.2 0.0139 +0.8 0.7881 -2.1 0.0179 +0.9 0.8159 -2.0 0.0228 +1.0 0.8413 -1.9 0.0287 +1.1 0.8643 -1.8 0.0359 +1.2 0.8849 -1.7 0.0446 +1.3 0.9032 -1.6 0.0548 +1.4 0.9192 -1.5 0.0668 +1.5 0.9332 -1.4 0.0808 +1.6 0.9452 -1.3 0.0968 +1.7 0.9554 -1.2 0.1151 +1.8 0.9641 -1.1 0.1357 +1.9 0.9713 -1.0 0.1587 +2.0 0.9773 -0.9 0.1841 +2.1 0.9821 -0.8 0.2119 +2.2 0.9861 -0.7 0.2420 +23 0.9893 -0.6 0.2743 +2.4 0.9918 -0.5 0.3085 +2.5 0.9938 -0.4 0.3346 +2.6 0.9953 -0.3 0.3821 +2.7 0.9965 -0.2 0.4207 +2.8 0.9974 -0.1 0.5602 +2.9 0.9981 -0.0 0.5000 +3.0 0.9987

We can write Equation 4-31 as,


Chapter 4 - Economic Uncertainties

25

AZ) = =e_z2/2 - 00 <z <cx)

Integrating, cdf can be written as,

I z F(z) =-f e_t2,72dt

The right side of the equation is called an error function. For all practical purposes,

F(-3) 0 F(3) 1

–I!Ii1t;17wr4

WIZZIRM

The right side of Equation 4-34 has to be integrated numerically. Standard tables (see Table 4-1) are

available to calculate F(z) for a given value of z.

The two parameters that describe the normal distribution completely are the mean (ii) and the

variance (a 2 ). Once these two values are known, we know the normal distribution function.

Example 4-15

Based on the analysis of the sample porosity data, it was concluded that porosity was normally distributed. If p = 0.2 and

a 2 = 0.0004, calculate the probability that porosity values will be 1) less than 0.18; 2) between 0.18 and 0.22; and 3) between

0.2 and 0.24.

Solution 4-I5

To solve this problem, we need to standardize all the values. Using Equation 4-32 and knowing that p = 0.2 and a = 0.02,

x�l2 z=�

a

1) When

0.18 - 0.2

x1 = 0.18,z1 = .02

= �1

we know that

p[Z <zi ] = F(z1 )

Looking at Table 4-1, F(-1) = 0.1587. Therefore, p{porosity < 0.18] = 0.1587.

2) When

0.22 - 0.2 x2 = 0.22,z2 = 0.2

= 1,F(z2 ) = 0.8413

we know that

p1z1 <Z <z2 1 = F(z2 ) - F(z1 )

= 0.8413 - 0.1584 = 0.6826

Therefore, [0.18 < porosity < 0.22 = 0.6826.

3) When


0.2 - 0.2 X3 = 0.2,z3 =02 = 0,F(z3 = 0.5)

0.24 - 0.2 = 0.24,z4 =02

= 2,F(z4 = 0.9773)

p[z3 <Z < z] = F(z4) - F(z3 ) = 0.9773 - 0.5 = 0.4773

Therefore, p[0.2 <porosity < 0.24]= 0.4773.

As shown, once the mean and the variance are known, the probability value for any interval can be calculated.

Normal distribution can be approximated by a triangular distribution if the difference between the

most likely and the minimum is equal to the difference between the maximum and the most likely

values.

LOG NORMAL DISTRIBUTION

A log normal distribution is closely related to the normal distribution. If a log of random variable is

normally distributed, then the variable is assumed to be log normally distributed. This relationship is

shown in Figure 4-13. In this figure, the random variable is log-normally distributed because log x is

a normally distributed variable.

f(x) f(Iog x)

/ ’I

log

Figure 4-13: Transformation of log-normal to normal distribution

Several variables exhibit log-normal behavior. These include permeability values in the reservoirs,

the size of sand grains in the sediment, the size of oil and gas fields resulting from past discoveries,

EUR distributions in a shale gas reservoir, and economic variables such as well costs and completion

times. We can use log probability plot to verify the behavior of a variable to see if indeed the

variable is log normally distributed. If we observe a straight line, it is an indication of log normal

behavior. Figure 4-14 shows a permeability distribution indicating a log-normal behavior in a

probability plot.

Economic Evaluotion in the Petroleum Industry


9 9 9

> 9

I 4 3

:5 9 E

C)

001 - 0.010

o 100 too 100 100 1000.

PerrnebOty. md

Similar to the normal distribution, the log-normal distribution can be described using the

parameters mean and variance. The probability density function for a log normal variable can be

written as,

1 [Inxa] f(x) = exp

xf3f 213 2 J

where x is log normally distributed, ci is the arithmetic mean of natural logarithm of x values and /32

is the variance of natural logarithm of x values.

In addition to knowing the probability density function, we can also define some additional useful

relationships for a log normal distribution. For example, if i and a 2 are the arithmetic mean and variance of the variable x, we can relateji and cr 2 to ci and p 2 by:

p 2 = In (i + Equation 4-37

and

a =


Example 4-16

A survey of oil and gas reservoirs in a given basin indicates that the arithmetic mean of all the reserves is 200,000 bb!s and

the standard deviation is 400,000. If the distribution of the reserves is log-normal, calculate the probability that the

prospect will produce less than 400,000 bbls. Also calculate the value of reserves so that 10% of the reserves are less than

that value.

Solution 4-16

In most instances, the mean and variance are easy to calculate based on the sample data. If the distribution is log-normal,

however, then we need to convert the arithmetic mean and variance to the mean and variance of log of the values.


= ln(1 +_-)

(400,000) 2 1 = In Fi + ( 200,000)2j

= 1.609


j2 a =

ln(200,000) - 1.609 -----

= 11.4

Since distribution of the log of the variable is normal, we can use Table 4-1 to estimate the probability that reserves are

less than 400,000 bbls,

ln(400,000) = 12.90

To define a standardized variable,

- In(x) - a - 12.9-11.4 Z-

= 1.1825

F(1.1825) = 0.88 from Table 4-1.

Therefore, there is an 88% probability that a future successful prospect will produce less than 400,000 bbls. The z value corresponding to the 10th percentile is 1.28 (Table 4-1). By manipulating Equation 4-32,

X10 = ea-1.28fl = ehl 4_l 28 X 1268

= 17,613 bbls

Problem 4-19

If the area of the reservoir is assumed to be uniformly distributed, with the minimum area equal to 60 acres and the maximum area to be 200 acres, what is the most likely value of the reservoir area? What is the probability that the area will be less than 160 acres?



Problem 4-20

A reservoir recovery factor is assumed to be triangularly distributed with a minimum of 0.15, a most likely value of 0.23 and a

maximum of 04 What is the probability that the value of the recovery factor is between 0.2 and 0.3? What is the value of recovery at which you are 90/b sure that the recovery factor will be less than that value?

Problem 4-21

The saturation in the reservoir is assumed to be normally distributed. If the mean saturation is 0.3 and the standard deviation is

0.05, calculate the probability that the saturation lies between 0.27 and 0.33.

Problem 4-22

lithe formation porosity is normally distributed with a mean of 0.15 and a variance of 0.0016, and the reservoir quality rock is

assumed to have a porosity of at least 0.10, calculate the probability that the porosity in this reservoir is less than 0.10. Does this

value represent the fraction of the reservoir that is unproductive?

Based on the survey of various economic analysts, the oil price forecast for the next year is normally distributed with a mean of $951bbl and a standard deviation of $101bbl. If this analysis is assumed to be true, what is the probability that oil prices will be

lower than $701bbl? What is the probability that the price of oil will be greater than $1101bbl?

Problem 4-24

Permeability of a reservoir is assumed to be log-normally distributed. If a sample of well bore data indicates a mean of 30 md

and a standard deviation of 50 md, what is the probability that the permeability of a given sample is less than I md? What is the

probability that the permeability is greater than 100 md?

Problem 4-25

The following data represents average porosities in Canadian fields. Verify whether or not the data represent log normal

distribution (Capen, 1992). Use a log probability plot.

0.03 0.04 0.05 0.05 0.06 0.06 0.06 0.06

0.07 0.07 0.07 0.07 0.07 0.09 0.09 0.1 0.1

0.1 0.1 0.12 0.13 0.14 0.14 0.15 0.15 0.15 0.18 0.18

0.19 0.19 0.19 0.19 0.19 0.2 0.22 0.27 -

DECISION TREE ANALYSIS

Decision tree analysis is a Logical extension of probability analysis. In the Expected Value section, we

discussed how to make an economic decision based on EMV analysis. That is, for a given decision, if we

find that the EMV is positive, we consider that decision to be feasible. It does not mean that we are

always going to be successful in making the decision. It just means that if we make similar decisions for a

large number of problems, on average, we will be successful. Decision tree analysis allows us to extend

EMV analysis when we are examining more complex problems that require more than one decision and

may depend on earlier decisions. Decision analysis provides us with a structure so that understanding

the subsequent decisions that are dependent on earlier decisions can help us make an optimal decision.

Similar to [MV analysis, we can conduct sensitivity analysis to understand the impact of uncertainty with

respect to probability on decision tree analysis.


In a typical decision tree plot, we always denote decision by a square node; whereas, the probability (or

chance) node is described by a circle. A payoff node provides the value of payoff for a particular

decision. This is the final value at the end of the tree and it is defined for every branch of the tree. The

best way to illustrate the decision tree analysis is through a simple example.

Example 4-17

An operator is thinking about drilling a well at a cost of $2 million. The probability of a dry hole is 0.6 and, if successful, there is

a 0.3 probability that the P Vb ene jts will be $4 million, 0.4 probability that the PVb,,,,Fu will be $8 million and 0.3 probability that

the PVbenefits will be $13 million. Should the operator drill the well? Row will the decision affected by the probability of success?

Solution 4-17

We can draw the decision tree diagram as:

NPV

P = 0.12 $2m

p = 0.16 $6 m

Drill p = 0.12 $11 m

111111 No Drill

$Om

Figure 4-15: Decision tree for Example 4-16

The square represents the decision whether to drill the well or not. The circle represents the probability or trial node. The

probability associated with each of the successful options is calculated by multiplying the probability of success (0.4) with

likelihood of each of the successful outcomes. If the probability of "benefits is 4 million (03), then 0.3 x 0.4 gives us the

probability 0.12 that the NPV of that branch will be $4 -$2 = $2 million. This requires subtracting the cost of drilling the well.

We can do the calculation at the probability node to determine the EMV if the well is successful.

EMV - 0.12 X 2 + 0.16 x 6 + 0.12 x 11 + 0.6 x �2 = $1.32 m

Therefore, at the decision node, the EMV is calculated as:

EMV = 2.52 - 1.2 = $1.32 m

Since the EMV is positive, the decision should be to drill the well. The final decision tree is shown in Figure 4-16.



N PV

P = 0.12 - --------.- $2m

p=0.16 $6m

EMV Drill N p = 0.12

=$1.32m7 $llm

o EMV=$1.32m =06

EMV$Om

No Drill $-2m

re 4-16: Finel answer from decision tree in Example 4-16

As with the [MV analysis, we can examine the sensitivity of the [MV at the decision node for different

probability of success values. This is shown in Figure 4-17.

8

6

4

> LU

0

-2

-4

0 0.2 0.4 0.6 0.8 1

Probability of Success Figure 4-17: Sensitivity of EMV to probability of success in Example 4-16

As Figure 4-17 shows, unless the probability of success is less than 0.23, the [MV will be positive. This

means that, as long as the operator believes that the probability of success is at least 0.23, this project is

feasible.

Example 4-18

A company is considering buying a new prospect. The lease hold bonus is expected to be $100,000. Based on the current

analysis, there is a 20% chance that a successful well can be drilled resulting in a present worth of $1.2 million. Once bought,

another alternative is to conduct a 3-0 seismic to increase the chance of recovery. Based on the survey of nearby areas, there is

a 0.7 probability that seismic survey will give a favorable result. If favorable, the probability of a successful well will increase to

0.8. If seismic survey is unfavorable, the probability of success will decrease to 0.05. The cost of conducting the seismic survey is

estimated to be $300,000. The cost of drilling is estimated to be $100,000. Using decision tree analysis, what actions should the

company take?


Solution 4-13

This problem requires decision making at several occasions. The three major decisions are 1) whether to buy a lease, 2)

whether to conduct a seismic survey, and 3) whether to drill a well. Figure 4-18 shows the decision tree for this problem.

Drill

Ps° 8

Favorable No Drill

Ps’0 7

Drill

p 5 = 0 05

Seismic 2

Unfavorable No Drill

pS=O. 3

4

No

Buy Seismic Drill

Lease Ps° 2

5 3

Don’t Buy No Drill

Lease


In this decision tree, there are five decision nodes indicating the point at which a decision is needed. Treat the decision at each

decision node to be an isolated one where, depending upon the EMV for each cAernative, we can select the alternative that has

the highest EMV.

Consider node 1 where, after a favorable seismic survey, we need to decide whether to drill or not. Using Equation 4-22,

EMV = p(NPV) 5 - (NPV) f ] + (NPV)J.

If we drill the well,

(NPV) = 1.2 x 10 6 - 100,000 - 100,000 - 300,000

= $700,000

In the calculation above, 1.2 x 106 represents the present worth if the well is successful, $100,000 represents the cost of

drilling, the second $100,000 represents the cost of leasing and $300,000 represents the cost of the seismic survey.

(NPV) 1 = �100,000� 100,000-300,000

= � $500,000

Therefore,

Economic Evaluation in the Petroleum Industry Chapter 4 - Economic Uncertainties 33

EMV = 0.8[700,000 - (-500,000)1 + (-500,000) = $460,000

If we do not drill the well,

EMV = �$400,000

where -$400,000 represents the costs of leasing and of conducting seismic survey.

Comparing the two expected values, we should drill the well if the seismic survey is favorable.

The (EMV) corresponding to the selected decision is carried forward for further analysis.

At decision node 2, we can carry out a similar analysis. This node represents the decision of drilling a well if the seismic survey is

unsuccessful.

If we drill the well,

EMV = 0.5700,000 + 500,0001 - 500,000 - -$440,000

If we do not drill the well,

EMV = �$400,000

where -$400,0 00 represents the costs of leasing and of conducting a seismic survey.

Since the EMV drilling option will result in a bigger loss, we should opt to not drill the well.

At decision node 3, where we have to consider drilling a well without conducting a seismic survey,

(NPV) S = 1.2 X 106 - 100,000� 100,000

= $1,000,000

1.2 x 106 represents the present worth if the well is successful, the two $100,000 values represent the cost of leasing and the

cost of drilling respectively. There is no cost of seismic for this option. Using Equation 4-22, if we decide to drill a well,

EMV = 0.2[1,000,000 - (-200,000)] - 200,000 = $40,000

If the option of not drilling the well is chosen,

EMV = �$100,000

where -$100, 000 represents the cost of leasing the land.

Comparing the two options, we need to select the option of drilling the well.

After we have made the decisions at the three nodes, we can redraw our decision tree as shown in Figure 4-19.

We have a probability node after deciding to conduct a seismic survey. We can calculate the [MV at the probability node by:

EM = Equation 4-39

where pi represents the probability of the event i, and (EMV) 1 represents the expected monetary value of the event i.


EMV = 460,000

Favorable

Ps° .7

a

Seismic

Unfavorable

EMV =-400,000

4 Buy No

Lease Seismic

5

Don’t Buy Lease

EMV = 40,000

EMV=0

Figure 4-19: Partial decision tree for Example 4-17

Using Equation 4-22, the EMV at the probability node "a" is,

EMV = 0.7 x 460,000 + 0.3 x (-400,000) = $202,000

Consider decision node 4. The EMV, if we conduct a seismic survey, is $202,000. If we do not conduct a seismic survey, it is

$40,000; therefore, we should select the option to conduct the seismic survey. The decision tree now can be drawn as:



EMV = $202,000

Don’t Buy Lease

I \EMv=$0

Figure 4-20: Partial decision tree for Example 4-17

Lo o k ing at the decision tree for Figure 4-21, the option to select is to purchase the lease.

Example 4-19

A company is considering a waterflooding project in a depleted solution-gas drive reservoir. The current estimates of decline

curve analysis indicate that, if produced under existing conditions, the present worth of the reserves is $20 million. One option

under consideration is to use the existing wells and convert half of the welts to the injection wells. It is estimated that this

conversion will cost $1 million and, due to additional reserves, with a 40% probability, will result in a present worth of $32

million and, with a 60% probability, will result in a present worth of $26 million. Another option is to drill injector wells and then

initiate a water flood. Drilling an additional 10 injectors and converting the current operation to watertlood will cost $5 million.

It is estimated that, due to additional reserves, with a 60% probability, it will result in a present worth of $35 million and, with a

40% probability, it will result in a present worth of $28 million. Based on the EMV analysis, which option should be selected?

Solution 4-19

I The decision tree can be drawn as follows,

Buy Lease

5


P P P

Based on the decision tree, we need to make two decisions: 1) whether to conduct waterflood or not, and 2) if we decide to

conduct, whether to drill additional wells for waterflooding. We have two probability nodes in the decision tree. The first step

will be to calculate the EMV at each probability node.

At probability node ’ct’, for p = 0.6,

EMV = $35m - $5m = $30m

where $35m is the present worth and $5m is the cost of drilling additional wells.

For p = 0.4,

EMV = $28m - $5m = $23m


EMV = p, (EMV) 1

= 0.6(30) + 0.4(23) = $27.Om

At probability node ’b’, for p = 0.4,

EMV = $32m - $lm = $31m

For p = 0.6,

EMV = $26m - $lm = $25m


EMV = 0.4 x 31 + 0.6 x 25 = $27.4m

At decision node 1, the option to drill wells results in an EMV of $27.2m versus $27.4m for the option not to drill wells;

therefore, the option not to drill wells should be selected.

At decision node 2, the option for watertlooding will result in an EMV of $27.4m versus $20m with no waterflooding; therefore,

we should select the decision to initiate a waterflood.

In Example 4-18, we saw an application of decision tree analysis for understanding value of information;

-to determine if it is worth conducting seismic survey or not. In many instances evaluating the value of

information requires the application of Bayes’s rule. We will illustrate this application in the following

example.

Example 4-20

An oil company has acquired a new concession in offshore Gulf of Mexico. Based on an analog study and an old 3-D seismic

shoot done 10 years ago, the probability of finding successful hydrocarbons is determined to be 0.2. The cost of drilling a well is

estimated to be $50 million. If the first well is successful, there is a 0.2 probability that the present value of benefits (after

subsequent costs of drilling are accounted) will be $200 million, a 0.5 probability that the present value of benefits will be $400

million, and a 0.3 probability that the present value of benefits will be $800 million. There is a proposal by an exploration group

that an additional 3-D seismic survey be performed at a cost of $10 million. Based on other exploration plays, when a 3-D

seismic survey was performed, the survey was accurate in predicting the successful location of hydrocarbons 80% of the time;

whereas, successful seismic surveys have incorrectly predicted the correct location for drilling a well 20% of the time. The

probability of the present value of benefits does not change with seismic data. Should the seismic data be collected? We will

assume that, if seismic data indicates an unfavorable result, we will not drill the well.



Solution 4-20

The decision tree for this example can be drawn as shown in Figure 4-22. The probability values shown in the decision tree

require some explanation. As shown in Figure 4-22, there are three decision nodes. Let us assume that event S represents

successful drilling (finding hydrocarbons) and that the event of successful seismic data is D. We already know that p(S) = 0.2). We also know that p(DIS) = 0.8 and p(DISC) = 0.2, where SC represents a failure event, which is complimentary to success.

Using these values, we need to calculate p(SID). This represents the probability of success if seismic data are successful. We

also need to calculate p(SC D), which represents the probability of failure if seismic data are successful. This requires the use of

the Bayes rule. Using Equation 4-10, we can calculate

SD) - p(DIS)xp(S)

p(DIS)xp(S)+p(DS c)xp(Sc )

0.8 x 0.2

p(SD) =

0.5 0.8 x 0.2 + 0.2 x 0.8 =

Using a similar equation, we can calculate p(5C ID) = 0.5.

P = 0.1 $ 140 m

p = 0.25 ..- --... ..................................................$ 340 m

Drill (p, = 0.5)

/ N p=O.15

740 m

\.. p=0.5 -$ 60 m

Seismic No Drill

- ---. -$ 10 m

I I No Drill

____ ..--- ................------- $Om

No Seismic

p = 0.04 $150 m

’ Drill (p,=0.2)

$350m

N p=0.06 $ 750 m

P =0.8 -$50 m


The probabilities of three successful possibilities are calculated by multiplying the probability of particular outcome with the

probability of success. For example, if the probability of PVb enefl ts is 0.2, then multiplying 0.2 by 0.2, we obtain the value of 0.04.

In reality, after collecting seismic data, we should have a probability (chance) node indicating the probability of favorable

seismic versus the probability of unfavorable seismic. If the seismic is unfavorable (we can define the event as DC), we will need

to calculate p(SIDC) and p(SdIDT). There will be another decision node associated with unfavorable seismic data

corresponding with the decision to drill or not to drill well; however, in this example, we made asimplifying assumption that we


will not drill a well if the seismic data are unfavorable. Therefore, that decision is already known and has been eliminated from

decision tree analysis.

The NPV for each of the options is calculated by subtracting the cost of drilling of a well from the present value of benefits. The

reason for slightly different results when seismic data are collected is due to the additional $10 million cost of conducting

seismic data. We can calculate the EMV at the first chance node (after seismic data are collected) as:

EMV = 0.1 x 140 + 0.25 x 340 + 0.15 x 740 + 0.5 x �60 = $180m

We can compare this EMV with EMV of not drilling the well after collecting seismic data, which is -$10 m. Since 180 is greater

than -10, we will choose the option to drill the well.

If we calculate the EMV at the probability node without seismic, we can calculate it as:

EMV = 0.04 x 150 + 0.1 x 350 + 0.06 x 750 + 0.8 x �50 = $46m

Comparing this value with the EMV of the not drilling option (zero), even without seismic data, we should choose the option to

drill the well.

Comparing the two options, $180 m is greater than $46 m; therefore, we should choose the option of collecting seismic data.

Problem 4-26

An oil company is considering leasing a highly faulted prospect for potential gas reserves. The cost of leasing is estimated to be

$7501acre. The company has the option of leasing 640 acres and drilling a well by itself or leasing 160 acres and unitizing the

leased area with other companies to receive 25% working interest in 640 acre unit. If the 640 acre land is leased, the company

has the option to conduct a 3-D seismic survey at a cost of $1.5 million. If the seismic survey is successful (the probability of the seismic survey being successful is 0.6), the probability of a successful well will increase from 0.2 (before seismic) to 0.6. If the

seismic survey is unfavorable, the probability of a successful well will reduce to 0.03. If the company only leases the 160 acres, it will have to participate in the seismic survey by sharing 25% of the seismic survey cost. The cost of drilling a well is $3 million. If successful, the well will produce at a rote of 2 MMSCF/D during the first year declining at a rote of 10% per year. If the life of the gas well is seven years and the average price of gas received is $2. 70/MSCF, which option should the company choose? Assume that the operating costs for the lease are $200,000 per year, the MROR is 12%, and the salvage value is negligible. Use the

decision tree analysis.

Problem 4-27

An oil company is considering drilling an infill well to reach reserves otherwise not contacted by existing wells. The company has

several options. Under option 1, the company can drill a vertical well without collecting any additional data. The chances of hitting a successful target are 0.4. If successful, it is expected that, after drilling the well at a cost of $800,000, tin well will

produce 7,000 bbls/yeor with an operating cost of $101bbl over the next six years. Under option 2, the company can collect additional geological, well testing, and cross-well seismic data at a cast of $1,000,000. The chances that the additional data

collection will indicate favorable results are 0.7. Based on other available data, the additional data is accurate 70% of the time if the results are favorable. If unfavorable, the additional data is accurate in indicating unfavorable drilling prospect 80% of the time. With option 3, the company can choose to drill a horizontal well after collecting additional data and receiving favorable

results. Drilling a horizontal well will cost $3,000,000. The chance of hitting a successful target is 0.6. If successful, the well will produce 15,000 bbls/year for the next five years with an operating cost of $121bbl. If the price of oil is $801bbl the MROR is 18%, and the salvage value is negligible, which option should the company choose? Use the decision tree analysis.

Problem 4-28

An oil company, which owns a 100% working interest, is considering drilling a potential prospect. The drilling casts are expected

to be $2,000,000. The well has a 60% probability of being a thy hole, 030% probability of producing 10,000 bbls in the first year, and a 10% probability of producing 20,000 bbls in the first year. Based on nearby wells, the well will decline at a rate of 18% per year. Economic analysis indicates that there is a 60% probability that the oil price will remain at $801bbl over the production

period and a 40% probability that the oil price will remain at $1001bbl. Assume the operating costs will remain steady at

50,000 per year. Assume the income tax rate to be 28% and negligible royalty interest. With option 4, the company will receive

3116 overriding royalty interest in the some well. If the life of the well is assumed to be six years, which option should the



company choose? The MROR is 15%. Use the decision tree analysis. . . . : �.� .....

Oil Company A owns a lease on a £40 acre wild cot area. Based on geological and geophysical analysis, the company has a 20%

chance of drilling a successful gas well. The drilling of a well will cast $500,000 and will result in $4 million PVb, fl itS. Company B,

owning the adjoining land, is proposing to drill a well on their land. Both lands indicate similar geological and geophysical

features. Company B offers two options to Company A:

1) Company A can receive the welibore data from the drilled well at a cost of $60,000. If the results of the welibore data are

favorable (a 50% chance), company A’s chances on its own lease will improve to 0.5. If unfavorable, the chances of a

successful well will be reduced to 0.05.

2) Company A can participate in the well Company B is drilling at a 50% working interest with some override to Company B.

The chances of a successful well on company B’s lease are 20% with the cost of drilling being $600,000 and if successful,

resulting in a PV to Company A of $1.8 million. If the well is successful, the chances of Company A drilling a successful well on its own lease will improve to 50%. If unsuccessful, then Company A’s chances will reduce to 5%.

Which option should Company A choose? Use the decision tree analysis.

MONTE-CARLO SIMULATION (Megi!!, 1984), (Bratvold & Begg, 2010)

Monte-Carlo simulation is an extension of the probability analysis. It is the method by which a

probability distribution of an output variable can be determined if the probability distributions of input

variables are known. Schematically, Figure 4-23 represents the concept of Monte-Carlo simulation. On

the left side, all the input variables are shown. We assume that the distribution functions of all the input

variables are known. We also know the relationship between the output variable and the input variable,

y=f(x 1 ,x2 ...,x)

ytadl7’DJ

where y is the output variable and x 1 ,",x are input variables. As a simplistic case of Monte-Carlo

simulation, if we assume that only one value of all the input variables is known, then we will obtain only

one output of value of y. Unfortunately, due to uncertainties with respect to input variables, we need to

estimate the uncertainty with respect to output variables. Monte-Carlo simulation allows us to generate

the probability distribution function of an output variable knowing the distribution of the input variables

and the relationship between the output variables and the input variables.

In principle, if we know the distribution of input variables, it is possible to generate the distribution of

output variables under some limiting circumstances. This can be obtained analytically. If we assume that

all the input variables have Gaussian distribution and we know the mean and variance of those

distributions and, if we further assume that those input variables are independent of each other, it is

possible to determine the distribution of the output variables (including mean and variance) under

certain conditions. In practice though, the relationship between the input variables and output variables

can be complex and non-linear. Additionally, it is possible for input variables to be dependent on each

other. That dependency needs to be accounted for in our analysis. Monte-Carlo simulation can allow for

such complex relationships.


INPUT RELATIONSHIP OUTPUT

Figure 4-23., Concept of Monte-Carlo simulation

The Monte-Carlo simulation method requires numerous computations; therefore, it can be used only

with the help of computers. A logical flow diagram of the Monte-Carlo simulation is represented in

Figure 4-24. Each of the steps in the flow diagram is explained in more detail below.

Step 1 requires input of all relevant data. This includes the distribution functions of all input variables.

These functions include uniform, triangular, normal, log normal or any other type of function. These

functions are specified by inputting the appropriate parameters for each distribution function. For

example, for a uniform distribution function, we need to input the minimum and maximum or, for

normal distribution function, we need to specify the mean and variance. We also need to specify the

functional relationship between input and output variables. The input should also include the number of

passes the computer has to loop through before the program is terminated. This number should be

sufficiently large in order to obtain a representative distribution of the output variables. With the recent

speed of computers, it is very easy to generate the output variables by using a large number of passes

unless the problem is extremely complex. Ten-thousand passes is not uncommon in Monte Carlo

simulation.



Read In Data

y=f(x1,x,. X 11)

Number of Passes, N

Select Values for Input V ariables

_ x n i xli

[

Determine for Pa,i

y i=f (x1i,x, xj)

No

SucientPas:s?

yes

Distiibution of y

V

E STOP

Figure 4-24: Flaw diagram of Monte-Carlo simulation

Step 2 of the Monte-Carlo simulation requires selecting values for input variables for a given pass. This

can be accomplished using a uniform random number generator and converting the probability density

function into a cumulative distribution function. A uniform random number generator generates

random numbers between 0 and 1 with equal probability. That is, any number between the interval 0

and 1 can be selected with an equal chance. Knowing that the cumulative distribution function also


varies between 0 and 1, the value of a random number is assigned to the cumulative distribution

function and the corresponding value of the variable is selected.

Let us illustrate this procedure through a simple example. Assume that we have defined a variable by a

uniform distribution between an interval [a, b]. The cumulative distribution function for this interval can

be defined as:

F(x) = x-a- Equation 4-41

b-a

This cumulative distribution function is shown in Figure 4-25.

_:____

As evident in Equation 4-27, it is a linear relationship. A random number is selected within a uniform

random number generator; any number between [0, 1] interval. We assign this random number RN to

the cumulative distribution function. Substituting RN for cumulative distribution function F(x) in

Equation 4-27, results in,

RN = Equation 4-42

Rearranging Equation 4-40, we can write,

x i = a + RN(b - a) Equation 4-43

We assign the value xi to this particular variable.

Although the equations illustrated above show how to select a value from a uniform distribution

function, a similar technique can be applied to any other distribution function. Once we identify a

relationship between F(x) and x for any distribution function, selecting a value for the variable is easy.

From the discussion on distribution functions, we know that such a relationship exists for every

distribution function including the discrete distribution functions.

Note that the value of x1 in Equation 4-41 will change depending upon the value of a random number.

We repeat a similar procedure for every input variable. For selecting a variable, every time, a new

random number is drawn within the interval [0, 11.

Economic Evaluation in the Petroleum Industry Chapter 4 Economic Uncertainties 43

Once the values for all the input variables are selected, in Step 3, we can use the relationship between

the output and input variables. This will result in one value of the output variable y. This completes the

calculation for one pass.

Step 4 in the Monte-Carlo simulation is to check to see if the number of passes for which the

calculations are done exceeds the maximum number of passes. If not, the procedure in steps 2 and 3

(selecting the values of the input variables using uniform random number generator and calculating the

value of y using the relationship between the output and the input variables) is repeated.

Depending upon the number of passes, we will generate an equal number of output values. Since, in

each pass, different values of the random numbers are selected, we will have different values of the

output variable. Therefore, if we have gone through 1,000 passes, we will have 1,000 values of the

output variable. Using these 1,000 values, we can construct a frequency distribution function of the

output variable. This function should be able to quantify uncertainties with respect to the prediction of

the output variable.

As explained above, Monte-Carlo simulation is a procedure that describes and quantifies uncertainties

through a distribution function of an output variable. Instead of a conventional procedure, which may

allow an estimation of output variables based on the most likely estimates of the input variables,

Monte-Carlo simulation describes the alternative values the output variables can take and the relative

frequency with which it can take those values.

In making economic decisions, understanding of uncertainties with respect to output variables is

immensely important. By using the range of estimates as a basis, we can quantitatively evaluate the

risks involved in a project. This type of evaluation is much more realistic than simply assigning a single

value to the output variable.

A note of caution is warranted about the use of the Monte-Carlo simulation. The technique is as good as

the knowledge about the distribution of the input variables. If the distribution of the input variables is

not adequate, the distribution of the output variables will not reflect the extent of uncertainties.

Hopefully, even in the presence of sparse information, we will at least be able to describe the input

variables by quantifying the minimum and the maximum. If such information is not available, we will not

be able to use the Monte-Carlo simulation to correctly describe the uncertainties of the output

variables.

The Monte-Carlo simulation may be extended to a situation where two input variables are related to

each other. If we know the type of relationship available between the two variables, we can incorporate

such a relationship very easily. We can use a correlation coefficient to define the relationship and,

hence, sample the value of a second input variable knowing the value of the first input variable. If the

relationship is not linear, we can use conditional distribution functions to sample the value of a second

variable knowing the value of the first variable.

Example 4-21

We are interested in estimating potential reserves under a new prospect. Based on the geological and petrophysical analysis of

nearby wells, it is estimated that following the distribution of variables should be observed.

OW

OW

ON

44 . Mohan Kelkar, Ph.D., J.D.

Most

Variable Distribution Minimum Likely Maximum

Area (acres) Triangular 20 40 120

Thickness (feet) Triangular 20 30 50

Porosity (fraction) Triangular 0.2 0.25 0.3

Water saturation (fraction) Uniform 0.3 - 0.35

Recovery Factor (fraction) Uniform 0.2 - 0.3

Formation Volume Factor (bbl/STB) Single - 1.2 -

The equation for calculating the reserves is given by,

7758Ahł(l - S) N=

B01 RF

where,

A = area in acres

h = payzone thickness in feet

0 = porosity in fraction

S,,1 = water saturation in fraction

RF = recovery factor in fraction

B0 , = formation volume factor in bbl/STB

N = recoverable oil in STB

Estimate the distribution function for the oil that can be recovered.

Solution 4-21

Following the procedure discussed in the previous paragraphs, we input the distribution functions of all the variables, the

functional relationship between the output variables and the input variables and the number of passes in the Monte-Carlo

simulation program.

Using a Monte-Carlo simulation program with 2,000 passes, we have created the distribution of the amount of oil that can be

recovered in Figure 4-26. As shown in the histogram plot, this distribution appears close to log-normal distribution. As

previously discussed, this is a result of a central limits theorem that states that the product of a large number of independent,

random numbers tends to be log-normally distributed. Considering that recoverable oil is a product of several independent

random variables, it should be log-normally distributed.

In practice, when you evaluate the distribution of oil reserves in a given basin, you will notice a similar trend - a log-normal

distribution. This can be attributed to the fact that the reserves recovered are a product of various independent variables.

Knowing the distribution of the recoverable reserves, we can calculate the expected value of the reserves. If we identify a class

mark for each class, then

EV =Y p 1 Np 1

where P1 is the probability of class i, and Np1 is the class mark. Defining the class mark as the center of each class, the expected

value in this case is 4.24 x 10 s STB. This value represents the average amount of oil that can be recovered if several, similar

reservoirs are explored. This number can be used as a ’single outcome for further economic analysis if such analysis is needed.



0.16

0.14

0.12

0,1

Prnt,abiffty 008

0.06

0.04

0.02

0

1.06 lUS 2.46 345 4.24 5.03 5.83 8.62 7.4* 8.2 S.0 LB 10.6 11.4 12.2 13 13.1 14.5

Od Reserves. SIB x 10 -5

Figure 4-26: Distribution of recoverable oil

Example 4-21 illustrates the utility of the Monte-Carlo simulation for generating the distribution of

recoverable reserves. The method, however, can be extended to other estimations where uncertainties

with respect to input variables are known. For example, by coupling the economic evaluation and its

related uncertainties, we can generate NPV distribution for different scenarios. The distribution of NPV

values presents us with the economic uncertainties distribution. By studying this figure, we can make a

decision as to the feasibility of the project.

To summarize, the Monte-Carlo simulation is a very powerful technique for assessing and quantifying

the uncertainties of important variables. Knowing the uncertainties with respect to input parameters,

the Monte-Carlo simulation can generate the distribution of output variables. Unlike the other

techniques we studied in this chapter, the Monte-Carlo simulation gives a range of output values. This

range of values should allow us to objectively evaluate the feasibility of a given project in light of

uncertainties. Such evaluation is much more realistic than the evaluations that are based on single-value

analysis.

Problem 4-30

Estimate the reserves distribution for a reservoir with the following property distribution:

Variable Minimum Most Likely Maximum

Porosity (%) 13 20 26

Area (acres) 1,500 - 3,000

Payzone (feet) 50 100 130

Water Saturation (%) 15 - 25

Formation Volume Factor (bbl/STB) 1.1 - 13

Recovery Factor .3 .4 .45

Generate the distribution of reserves using the Monte-Carlo simulation method. Make two runs using 100 passes and 2,000

passes. Do you observe any difference? Which one is more representative of uncertainties? Calculate the expected value of

46 Mahan Kelkcir, Ph.D., J.D.

reserves. If the cost of drilling in this reserve is $250,000 and the PV/bbl in this reservoir is $10 (including operating Costs),

should we invest the money in this project?

Problem 431

Estimate the reserves distribution for a reservoir with the following properties distribution

Variable Minimum Most Likely Maximum

Porosity (%) 20 30 36

Area (acres) 100 640 750

Payzone (feet) 30 - 70

Water Saturation (%) 15 18 25

Formation Volume Factor (bbls/STB) - 1.25 -

Recovery Factor .25 - .4

Calculate the distribution of the possible reserves using the Monte-Carlo simulation method. Calculate the expected value of the

reserves. How does it compare with the short-cut method?

9711471=111114M

The following data are given for a particular reservoir:

1. Porosity: Histogram Type Distribution

Range Probability

.14-16 .1

.16-.18 .15

,18-.20 .30

.2-.22 .20

.22-24 .15

.24-26 .07

.20-28 .03

2. Area: Uniform Distribution

Minimum 80 acres

Maximum 160 acres

3. Thickness: Triangular Distribution

Minimum 60 feet

Most Probable 90 feet

Maximum 50 feet

4. Saturation: Histogram Distribution



Range Probability

.14-.18 .1

.18-.2 .2

.2-.24 .30

.24-27 .2

.27-30 .12

.30-.35 .08

5. Formation Volume Factor: Triangular

Minimum

Most Probable

Maximum

Minimum .20

Maximum .50

6. Recovery Factor: Uniform

Calculate and predict the producible oil distribution after making 1,000 passes. Estimate the expected value of the producible reserves. If the NPV per bbl of oil is $8, and the cost of drilling is $200,000, what is the maximum lease hold costs you are willing

to pay?

WORKS CITED

Bratvold, R. B., & Begg. S. H. (2010). Making Good Decisions. In R. B. Begg. Richardson, TX: SPE Publications.

Capen, E. (1992). Dealing with Uncertainties. (R. Steinmetz, The Business of Petroleum Exploration, 29-62.

Davis, J. (1986). Statistics and Data Analysis Geology. New York: John Wiley and Sons.

Megill, R. E. (1984). In An Introduction to Risk Analysis (pp. Chapters Eight, Nineteen and Fourteen). Tulsa, OK: PennWell Publishing Company.

Mian, M. A. (2002). Project Economics and Decision Analysis; Volume II: Probabilistic Models, Tulsa: PennWell Publications.

Newendrop, P., & Schuyler, J. (2000). Decision Analysis for Petroleum Exploration. Aurora, CO: Planning Press.

Rose, P. R. (1992). Chance of Success and Its Use in Petroleum Exploration. (R. Steinmetz, Ed.) The Business of Petroleum Exploration, 7 1-86.


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Economic Evaluation in the Petroleum Industry