ENG2000 Chapter 16 Evaluating and Comparing …...ENG2000: Eshrat Arjomandi (Based on R.I. Hornsey’s slides) CM: 4 Overview • So, we have seen that the act of investing is to sacrifice

ENG2000: Eshrat Arjomandi (Based on R.I. Hornsey’s slides) CM: 1

ENG2000 Chapter 16Evaluating and Comparing Projects:

The MARR


Exercise – Cash FlowYou decided to replace your old Toyota SUV. The dealer offers you a choice between buying andleasing. Buying the car, including all fees and taxescost you $36,534. Leasing the car cost you $470 permonth for the next 4 years and the value of the carafter 4 years would be $17681. Nominal interest rateis 5.9% compounded monthly. Which alternativewould you select?


Exercise – Cash Flow1. Calculate present value of $170002. Calculate present value for the uniform series

with payments of $4703. Add the above two to get the present value for

the lease option.

P1 = F * (P/F,i,N) = 17000 * 0.74137 = $13108P2 = A * (P/A,i,N) = 470 * 51.726 = $37,419Ptot = $13108 + $37419 = $37,419


Overview• So, we have seen that the act of investing is to

sacrifice something of present value in the expectation of receiving more value at a future time

• As an engineer making a decision about whether to go ahead with a project, your job is to determine whether the future returns merit the present investment

• In order to compare options, we must convert the transactions into a common value

present worthannual worthpayback period


Assumptions• Decisions are made on a purely financial basis

clearly this is not always the case, e.g. safety or environmentone can theoretically calculate the value of a safe workplace or the fines avoided by being environmentally sound, but it’s impossible in practise

• Future cash flows are known accuratelythere are techniques for dealing with this uncertainty

• We do not account for inflationcan also be handled in more complex treatments

• All options are equally feasiblein reality, some may be excluded because of insufficient resources etc


Assumptions continued• No taxes

yeah, right!

• All projects have an initial cash outflowcalled first costsand that all projects have subsequent cash inflows that equal or exceed the first costs


Relationships between projects• When there are several possible projects

available, we need to know what (if any) relationship there is between them

independentmutually exclusiverelated but not mutually exclusive


Independent projects• In this case, neither the expected costs nor the

expected benefits of the possible projects are in any way related

and so the numbers do not change if one or other (or both) projects go ahead

• In reality, there may not be enough funds to pursue two otherwise independent projects

but, in the current scenario, this violates one of the assumptions

• In the case of independent projects, each is treated on its own merits and the costs are compared directly (e.g. buying a microwave oven or a computer)


Mutually exclusive projects• In many cases, we will pursue either one option

or the other but not alle.g. we will build either an office block or a factory on the land we ownwe’ll buy Dell equipment rather than Sonywe will buy a BMW or a Lexus (mutually exclusive in most cases!)We will buy a car or lease it

• Again, each option can be considered on its own merits


Related but not mutually exclusive• Here, costs and benefits of one option depend on

whether or not a different option is chosene.g. opening two Starbucks franchises in close proximityclearly the returns will be different if one or both go aheadand costs may be reduced if both are constructed simultaneously

• In the case of Starbucks cafes, we can make four logical groups

open bothopen #1open #2open neither

• In general n related projects can be grouped into 2n sets which are mutually exclusive


Contingent projects• A special case of the “related but not exclusive”

is where one project is contingent on anothere.g. you can buy a PC and a printer, or a PC without a printerbut you’re not likely to buy a printer without a PChence the printer purchase is contingent on the purchase of a PC

• When there are several related projects 2n can be a large number

this can be reduced by determining feasibilities using a matrix technique


Contingent projectsAlternatives

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Garden x x x x x x x x

Old Lights

x x x x x x x x

Cobble Stones

x x x x x x x x

One way x x x x x x x x

Feasible N N Y Y N N Y Y N Y N N Y Y N


Summary – relations

Set of n projects

related but not mutually exclusive –

divide into 2n mutually exclusive sets

mutually exclusive: rank and pick the best

independent: evaluateeach one separately


Minimum acceptable rate of return (MARR)• The minimum acceptable rate of return (MARR) is

a lower limit for investment acceptabilitythe MARR is an effective interest rate that must be earned by a project in order to make the project acceptablein comparison to competing investment opportunities

• Hence, projects making a higher return than MARR are desirable

those projects earning a lower return are undesirable since the money can get better returns elsewhere

• In order to attract investors, a company must participate in projects returning more than the MARR

the minimum return required by investors is called the company’s cost of capital


• The MARR represents an opportunity costinvestors have a choice of investing in your company or elsewhereif the company invests in one project, it gives up the opportunity to invest in another project which would pay at least at the rate of MARR

• The ‘A’ in MARR is sometimes ‘attractive’• The point is clearer if one imagines that the

money to be spent on the project is being borrowed at a specified interest rate

so the project must give a return in excess of that interest ratehowever, investors or management may set a different level for the MARR


Example 4.2• Alex Student notices that students frequently

have to wait for computer terminalshe proposes to offer an alternative location in a nearby mallit will cost Alex $70 000 to set up the servicethe annual net cash flow (after expenses) is expected to be $30,000 for five years (after which the new university terminal room will be available)the salvage value of the 5-year-old equipment is negligible

• If investors in such a hi-tech venture expect an annual return of 20%, is Alex’s proposal a good investment?


Example 4.3• A mechanical engineer is considering building

automated materials handling equipment for a production line

on one hand the equipment would substantially reduce the manual labour required for the materials handlingon the other hand, it will consume energy, require insurance, and maintenance

• Option 1continue with present method; labour costs $9 200/year

• Option 2build new equipment with a life of 10 yearsFirst cost – $15 000; labour – $3 300/year; power –$400/year; maintenance – $2 400/year; tax & insurance –$300/year

• Which is preferable if the MARR is 9%?


Present worth (PW) comparison – independent projects

• PW analysis compares the present worth of all the cash flows using the MARR as the interest rate

the highest PW is preferableand indeed the value of a company can be assessed as the sum of the PWs of all the current projects

• The PW of a project exactly at the MARR is zerosince the future receipts exactly balance the initial disbursementand the project is deemed marginally acceptable

• In order to be acceptable we need our project to have a PW > 0

otherwise the ‘do nothing’ option (= invest elsewhere) is preferable


Present and annual cost• While we have been talking about the worth of a

project, we can also compare costs• This assumes that the major benefits of the

projects are identicaland can therefore be neglected in the calculations, since they are the same for each option

• This approach also assumes that the value of each project clearly exceeds its costs

i.e. they are all viable projects and we are looking for the lowest cost viable option


Example 4.2• Alex Student notices that students frequently

have to wait for computer terminalshe proposes to offer an alternative location in a nearby mallit will cost Alex $70 000 to set up the servicethe annual net cash flow (after expenses) is expected to be $30 000 for five years (after which the new university terminal room will be available)the salvage value of the 5-year-old equipment is negligible

• If investors in such a hi-tech venture expect an annual return of 20%, is Alex’s proposal a good investment?


4.2 Solution• PW = -$70,000 + $30,000(P/A,20%,5)

= -$70,000 + $30,000 * 2.9906= -$70,000 + $89,718 = -$70000 + $90,000= $19,718

Namely, investors are prepared to offer Alex$90,000 for his venture.


Example 4.3• A mechanical engineer is considering building

automated materials handling equipment for a production line

on one hand the equipment would substantially reduce the manual labour required for the materials handlingon the other hand, it will consume energy, require insurance, and maintenance

• Option 1continue with present method; labour costs $9200/year

• Option 2build new equipment with a life of 10 yearsFirst cost – $15,000; labour – $3,300/year; power –$400/year; maintenance – $2,400/year; tax & insurance –$300/year

• Which is preferable if the MARR is 9%?


4.3 solution• The 2nd option has an annual cost of: $3300 +

$400 + $2400 + $300 = $6400• The operating cost is less for the life of the new

equipment which is 10 years. However, we have to take into account a capital investment of $15,000.

• The task is to evaluate the present worth of investing $15,000 which gives us a saving of $9200 - $6400 = $2800 per year for 10 years at the rate of 9%.


4.3 solution continued• PW = -$15,000 + $2800 * (P/A, 9%, 10)

= -$15,000 + 2800 * 6.4176= $2969.44

Hence the investment is worthwhile.


Present Worth for Mutually Exclusive Projects

• As long as all projects have the same service life, we do as we did in the case of independent projects. For example if you are buying a car and deciding between Toyota and Mazda. As long as they have the same life, we use PW.


Another ExampleBarb was let go by her company due to cutbacks.She was offered two choices for her settlement:

1. $70,000 in cash today2. An annual payment of $17,500 per year for the

next 5 years.Such choices are given to save employees taxes.However, we do not take taxes into account.Therefore if the nominal interest rate of return is 6%compounded monthly, which option is moreattractive for Barb?


Annual worth (AW) comparisons• These are essentially the same as the PW

calculations except that the PW is converted to an uniform series at the MARR

using the capital recovery factor, (A/P,i,N)

• One gets the same result with PW and AW provided the lives are the same

• The difference between AW and PW is often perceptual

it’s sometimes easier to grasp an annual cost, rather than a single present worthSometimes may make sense to compare future values


Revised 4.3 (MARR = 9%)• Option 1: build equipment

first cost – $15 000labour – $3 300/yearpower – $400/yearmaintenance – $2 400/yeartax & insurance – $300/yeara life of 10 years

• Option 2: buy off-the-shelf equipmentfirst cost – $25,000labour – $1 450/yearpower – $600/yearmaintenance – $3 075/yeartax & insurance – $500/yeara life of 15 years


Unequal lives• It is important to cover the same time periods for

even the PW calculationsso the full advantages and costs are included in both

• There are two approaches to unequal lives• Use a repeated lowest common multiple from the

choicesknown as repeated service liveseach repetition has the same cash flowsmay not be valid for equipment which quickly becomes obsolete

• Use a defined study perioddefine a time period for the calculationsaffects the salvage value that can be assumed


Revised 4.3 (MARR = 9%)• Option 1: build equipment

first cost – $15 000labour – $3 300/yearpower – $400/yearmaintenance – $2 400/yeartax & insurance – $300/yeara life of 10 years

• Option 2: buy off-the-shelf equipmentfirst cost – $25,000labour – $1 450/yearpower – $600/yearmaintenance – $3 075/yeartax & insurance – $500/yeara life of 15 years


Revised 4.3 solutionOption 1 has to be repeated twice and option 2repeated once. Note we assume every time thesituation is the same. For example, 10 years fromnow and 20 years from now in option 1 we stillspend $15000

• So in option 1, for the second and third repetitions, we have to calculate present value for $15000 after 10 years and after 20 years. Our uniform series now runs for 30 years. Hence:

PW1 = -$15,000 -$15000 * (P/F, 9%, 10) - $15,000 * (P/F, 9%, 20) - $6400 * (P/A, 9%, 30)

= -$15,000 - $15000 * 0.42241 - $15000 * 0.17843 - 6400 * 10.273

= -$89760• Similarly for option 2 we get: -$89649So very little difference between the two options.


Revised 4.3 solutionIn this example it may be more convenient to usethe annual worth (cost) method, assuming costs donot change no matter how many times we repeatthe project.

AW1 = -$15000 * (A/P, 9%, 10) - $6400= -$15000 * 0.15582 = -$8737

AW2 = -$25000 * (A/P, 9%, 15) - $5625= -$25000 * 0.12406 - $5625= -$8726

Once again no significant difference.


Revised 4.3 solutionNow let’s use the specified period method. Weuse this approach because after 10 years the costsmay not be the same. So we decide on a study

period of 10 years, and for comparison reasons wecome up with a salvage value for option 2. Assumea salvage of $5000 after 10 years.

PW1 = -$15000 - $6400 * (P/A, 9%, 10) = -$56,073

PW2 = -$25000 -$5625 * (P/A, 9%, 10) + $5000 * (P/F, %9, 10)

= -$58,978 In this case option 1 is more attractive.


Payback method• The payback period is the number of years taken

for an investment to be recouped with 0% interest rate

• Useful for quick assessment of the economic viability of projects

roughly giving the amount of time needed to recover an initial investment (i.e. pay for itself)

e.g. an initial investment of $20,000 saves $8,000 per year, gives a payback period of 2.5 years

• Normally it is followed up by PW or Aw assessments as well

Payback period = first costannual savings


• If the annual returns are not equal, simply subtract each return from the first cost until it vanishes

e.g. $20 000 investment saves $5 000 in year 1, $6 000 in year 2, $7 000 in year 3,etc.now $5 000 + $6 000 + $7 000 +$8 000 = $26 000so the payback period is either 4 years (if the $8 000 is not received until the end of the year)or 3.25 years if the $8 000 is spread out throughout the year

• Now, the project with the shortest payback period is the preferred option

typically a payback between 2 and 4 years is considered reasonable, although it depends a lot on the circumstances


Advantages of payback approach• Straightforward and easy to understand• Easy to calculate• Identifies period of cash-flow and speed of

recovery of capital• Given uncertainties about future returns on

investments, a more precise calculation may not make sense.


Disadvantages of payback approach• Tends to make long-term projects seem

unattractive• Timings of cash flows within the payback period

are ignored. Interest rates are ignored.no account taken of the time value of money

• Ignores service life and benefits that occur outside the payback period


Discounted payback period• A better estimate can be obtained (at the expense of more

complex calculations) if the PW of each year’s savings are deducted from the first cost

• If you buy a computer at $5 000 and receive $2000 per year in benefits for the next 4 years and the interest rate is 10%, your benefits after the payback period continues for another 1.5 years.

compare with standard technique, which gives 5000/2000 = 2.5 years

Year Present worth CumulativeYear 1 $2 000(P/F,10%,1) = $1 818 $1 818Year 2 $2 000(P/F,10%,2) = $1 653 $3 471Year 3 $2 000(P/F,10%,3) = $1 503 $4 974Year 4 $2 000(P/F,10%,4) = $1 366 $6 340


Example 4.10• Self Defence Systems is going to upgrade their

paper shredding facility and have a choice between two models

• Model 007first cost: $50 000life: 7 yearssaving: $10 000/year

• Model MXfirst cost: $10 000life: 20 yearssaving: $1 500/year

• MARR is 8%


Example 4.10 solutionUsing payback period:

• Model 007: Payback = $50,000/$10,000 = 5 years• Model MX: Payback = $10000/$1500 = 6.6 years

Using Annual Worth (AW):

• AW Model 007 = -$50000 * (A/P, 8%, 7) + 10,000 = $396.50

• AW Model MX = -$10,000 * (A/P, 8%, 20) + $1500= $481.50


Example 4.10 Solution• In each of the cases,

after the payback period, Model 007 provides benefits for 2 more years and Model MX for almost 14 more years.

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Summary• Here we considered how to choose between

different projectsindependentmutually exclusiverelated nut not mutually exclusive

• For these situations we primarily examined the present worth method

where all cash flows are converted to the present worth for comparisonalthough this can be converted to an annual worth as well

• We also looked at the payback perioda rough way of determining how long it takes to recover an initial investment

Documents

ENG2000 Chapter 16 Evaluating and Comparing …...ENG2000: Eshrat Arjomandi (Based on R.I. Hornsey’s slides) CM: 4 Overview • So, we have seen that the act of investing is to sacrifice