36636052-Engineering-Economy

8/7/2019 36636052-Engineering-Economy

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Engineering Economics

November 3, 2004


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Engineering Economy

It deals with the concepts and techniquesof analysis useful in evaluating the worthof systems, products, and services in

relation to their costs


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Engineering Economy

It is used to answer many differentquestions Which engineering projects are worthwhile?

Has the mining or petroleum engineer shown that

the mineral or oil deposits is worth developing? Which engineering projects should have ahigher priority?

Has the industrial engineer shown which factoryimprovement projects should be funded with the

available dollars?

How should the engineering project bedesigned?

Has civil or mechanical engineer chosen the best

thickness for insulation?


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Basic Concepts

Cash flow

Interest Rate and Time value of money

Equivalence technique


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Cash Flow Engineering projects generally have economic

consequences that occur over an extendedperiod of time

For example, if an expensive piece of machinery isinstalled in a plant were brought on credit, the simpleprocess of paying for it may take several years

The resulting favorable consequences may last aslong as the equipment performs its useful function

Each project is described as cash receipts ordisbursements (expenses) at different points intime


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Categories of Cash Flows The expenses and receipts due to

engineering projects usually fall into one ofthe following categories: First cost: expense to build or to buy and install

Operations and maintenance (O&M): annual

expense, such as electricity, labor, and minorrepairs

Salvage value: receipt at project termination forsale or transfer of the equipment (can be asalvage cost)

Revenues: annual receipts due to sale of productsor services

Overhaul: major capital expenditure that occursduring the assets life


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Cash Flow diagrams

The costs and benefits of engineeringprojects over time are summarized on a cashflow diagram (CFD). Specifically, CFDillustrates the size, sign, and timing ofindividual cash flows, and forms the basis forengineering economic analysis

A CFD is created by first drawing asegmented time-based horizontal line,divided into appropriate time unit. Each time

when there is a cash flow, a vertical arrow isadded pointing down for costs and up forrevenues or benefits. The cost flows aredrawn to relative scale


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Drawing a Cash Flow Diagram

In a cash flow diagram (CFD) the end of period t isthe same as the beginning of period (t+1)

Beginning of period cash flows are: rent, lease, andinsurance payments

End-of-period cash flows are: O&M, salvages,

revenues, overhauls The choice of time 0 is arbitrary. It can be when a

project is analyzed, when funding is approved, orwhen construction begins

One persons cash outflow (represented as a

negative value) is another persons inflow(represented as a positive value) It is better to show two or more cash flows occurring

in the same year individually so that there is a clearconnection from the problem statement to each cashflow in the diagram


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An Example of Cash Flow Diagram A man borrowed $1,000 from a bank at 8%

interest. Two end-of-year payments: at theend of the first year, he will repay half of the$1000 principal plus the interest that is due.

At the end of the second year, he will repay

the remaining half plus the interest for thesecond year.

Cash flow for this problem is:

End of year Cash flow

0 +$1000

1 -$580 (-$500 - $80)

2 -$540 (-$500 - $40)


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Cash Flow Diagram

$1,000

0

1 2

$580 $540


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Time Value of Money

Money has value Money can be leased or rented

The payment is called interest

If you put $100 in a bank at 9% interest for one timeperiod you will receive back your original $100 plus$9

Original amount to be returned = $100

Interest to be returned = $100 x .09 = $9


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Compound Interest

Interest that is computed on the originalunpaid debt and the unpaid interest

Compound interest is most commonlyused in practice

Total interest earned = In = P (1+i)n - P

Where,

P present sum of money

i interest rate n number of periods (years)

I2 = $100 x (1+.09)2 - $100 = $18.81


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Future Value of a Loan With

Compound Interest Amount of money due at the end of a loan

F = P(1+i)1(1+i)2..(1+i)n or F = P (1 + i)n

Where, F = future value and P = present value Referring to slide #10, i = 9%, P = $100 and say

n= 2. Determine the value of F.

F = $100 (1 + .09)2 = $118.81


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Notation for

Calculating a Future Value Formula:

F=P(1+i)n is the

single payment compound amount factor.

Functional notation:

F=P(F/P,i,n) F=5000(F/P,6%,10)

F =P(F/P) which is dimensionallycorrect.


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Notation for

Calculating a Present Value P=F(1/(1+i))n=F(1+i)-n is the

single payment present worth factor.

Functional notation:P=F(P/F,i,n) P=5000(P/F,6%,10)

Interpretation of (P/F, i, n): a present sum P,

given a future sum, F, n interest periodshence at an interest rate i per interestperiod


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Spreadsheet Function

P = PV(i,N,A,F,Type)

F = FV(i,N,A,P,Type)

i = RATE(N,A,P,F,Type,guess)

Where, i = interest rate, N = number of interestperiods, A = uniform amount, P = present sumof money, F = future sum of money, Type = 0means end-of-period cash payments, Type =

1 means beginning-of-period payments,guess is a guess value of the interest rate


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Equivalence Relative attractiveness of different

alternatives can be judged by using thetechnique of equivalence

We use comparable equivalent values of

alternatives to judge the relativeattractiveness of the given alternatives

Equivalence is dependent on interest rate

Compound Interest formulas can beused to facilitate equivalencecomputations


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Technique of Equivalence

Determine a single equivalent value at apoint in time for plan 1.

Determine a single equivalent value at apoint in time for plan 2.

Both at the same interest rate and at the same time point.

Judge the relative attractiveness of thetwo alternatives from the comparable

equivalent values.


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Engineering Economic AnalysisCalculation

Generally involves compound interestformulas (factors)

Compound interest formulas (factors) can

be evaluated by using one of the threemethods

Interest factor tables

Calculator

Spreadsheet


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Given the choice of these two planswhich would you choose?

Year Plan 1 Plan 2

0 $5,000

1 $1,000

2 $1,000

3 $1,000

4 $1,000

5 $1,000Total $5,000 $5,000

To make a choice the cash flows must be altered

so a comparison may be made.


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Resolving Cash Flows to Equivalent PresentValues

P = $1,000(PA,10%,5)

P = $1,000(3.791) =$3,791

P = $5,000

Alternative 2 is better

than alternative 1 sincealternative 2 has agreater present value


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An Example of Future Value

Example: If $500 were deposited in a

bank savings account, how much wouldbe in the account three years hence ifthe bank paid 6% interest compounded

annually? Given P = 500, i = 6%, n = 3, use F =

FV(6%,3,,500,0) = -595.91

Note that the spreadsheet gives anegative number to find equivalent of P.If we find P using F = -$595.91, we getP = 500.


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An Example of Present Value

Example 3-5: If you wished to have

$800 in a savings account at the end offour years, and 5% interest we paidannually, how much should you put into

the savings account? n = 4, F = $800, i = 5%, P = ?

P = PV(5%,4,,800,0) = -$658.16

You should use P = $658.16


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Economic Analysis Methods

Three commonly used economic analysismethods are

Present Worth Analysis

Annual Worth Analysis Rate of Return Analysis


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Steps to do present worth analysis for asingle alternative (investment)

Select a desired value of the return oninvestment (i)

Using the compound interest formulas bringall benefits and costs to present worth

Select the alternative if its net present worth(Present worth of benefits Present worth of

costs) 0


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Steps to do present worth analysis forselecting a single alternative (investment)from among multiple alternatives

Step 1: Select a desired value of the return on

investment (i) Step 2: Using the compound interest formulas

bring all benefits and costs to present worthfor each alternative

Step 3: Select the alternative with the largestnet present worth (Present worth of benefits Present worth of costs)


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A construction enterprise is investigating the

purchase of a new dump truck. Interest rate is9%. The cash flow for the dump truck are asfollows:

First cost = $50,000, annual operating cost =$2000, annual income = $9,000, salvage valueis $10,000, life = 10 years. Is this investmentworth undertaking?

P = $50,000, A = annual net income = $

9,000 -$2,000 = $7,000, S = 10,000, n = 10.

Evaluate net present worth = present worth ofbenefits present worth of costs


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Present worth of benefits = $9,000(PA,9%,10) =$9,000(6.418) = $57,762

Present worth of costs = $50,000 +$2,000(PA,9%,10) - $10,000(P`F,9%,10)=

$50,000 + $2,000(

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8) - $10,000(.4224) =$58,612

Net present worth = $57,762 - $58,612 < 0 donot invest

What should be the minimum annual benefit formaking it a worthy of investment at 9% rate ofreturn?


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Present worth of benefits = A(PA,9%,10)= A(6.418)

Present worth of costs = $50,000 +$2,000(PA,9%,10) - $10,000(P`F,9%,10)=

$50,000 + $2,000(6..418) - $10,000(.4224)= $58,612

A(6.418) = $58,612 A = $58,612/6.418

= $9,312.44


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Cost and Benefit Estimates

Present and future benefits (income) andcosts need to be estimated to determinethe attractiveness (worthiness) of a newproduct investment alternative


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Annual costs and Income for a Product

Annual product total cost is the sum ofannual material, labor, and overhead(salaries, taxes, marketing expenses,office costs, and related costs), annual

operating costs (power, maintenance,repairs, space costs, and relatedexpenses), and annual first cost minus theannual salvage value.

Annual income generated through thesales of a product = number of units soldannuallyxunit price


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Rate of Return Analysis

Single alternative case

In this method all revenues and costs ofthe alternative are reduced to a singlepercentage number

This percentage number can be comparedto other investment returns and interestrates inside and outside the organization


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Steps to determine rate of return for a

single stand-alone investment Step 1: Take the dollar amounts to the same

point in time using the compound interest

formulas Step 2: Equate the sum of the revenues to the

sum of the costs at that point in time andsolve for i


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An initial investment of $500 is being

considered. The revenues from thisinvestment are $300 at the end of the firstyear, $300 at the end of the second, and

$200 at the end of the third. If the desiredreturn on investment is 15%, is the projectacceptable?

In this example we will take benefits andcosts to the present time and their presentvalues are then equated


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$500 = $300(P`F, i, n=1) + 300(P`F, i, n=2) +

$200(P`F, i, n=3) Now solve for i using trial and error method

Try 10%: $500 = ? $272 + $247 + $156 = $669(not equal)

Try 20%: $500 = ? $250 + $208 + $116 = $574(not equal)

Try 30%: $500 = ? $231 + $178 + $91 = $500(equal) i = 30%

The desired return on investment is 15%, theproject returns 30%, so it should beimplemented

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36636052-Engineering-Economy