Chapter 1: Introduction to Engineering Economy - ElCoM - HU Economy/Engineering... · Engineering Economy, Fourteenth Edition By William G. Sullivan, Elin M. Wicks, ... Engineering

Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling

Engineering Economy

Chapter 1: Introduction toEngineering Economy




The purpose of this book is todevelop and illustrate the principlesand methodology required to answerthe basic economic question of anydesign: Do its benefits exceed its

cost?




Engineering economy…

involves the systematicevaluation of the economicmerits of proposed solutionsto engineering problems.




Solutions to engineeringproblems must

• promote the well-being and survival of anorganization,

• embody creative and innovative technology andideas,

• permit identification and scrutiny of theirestimated outcomes, and

• translate profitability to the “bottom line” througha valid and acceptable measure of merit.




Engineering economic analysis canplay a role in many types of situations.• Choosing the best design for a high-efficiency gas

furnace.• Selecting the most suitable robot for a welding

operation on an automotive assembly line.• Making a recommendation about whether jet

airplanes for an overnight delivery service shouldbe purchased or leased.

• Determining the optimal staffing plan for acomputer help desk.




There are seven fundamental principlesof engineering economy.

• Develop the alternatives• Focus on the differences• Use a consistent viewpoint• Use a common unit of measure• Consider all relevant criteria• Make uncertainty explicit• Revisit your decisions




Engineering economic analysisprocedure

• Problem definition• Development of alternatives• Development of prospective outcomes• Selection of a decision criterion• Analysis and comparison of alternatives.• Selection of the preferred alternative.• Performance monitoring and postevaluation

of results.




Electronic spreadsheets are apowerful addition to the analysis

arsenal.• Most engineering economy problems can be

formulated and solved using a spreadsheet.• Large problems can be quickly solved.• Proper formulation allows key parameters

to be changed.• Graphical output is easily generated.




Engineering Economy

Chapter 2: Cost Concepts and DesignEconomics




The objective of Chapter 2 is toanalyze short-term alternatives

when the time value of money isnot a factor.




Costs can be categorized in severaldifferent ways.

• Fixed cost: unaffected by changes in activitylevel

• Variable cost: vary in total with the quantity ofoutput (or similar measure of activity)

• Incremental cost: additional cost resultingfrom increasing output of a system by one (ormore) units




More ways to categorize costs

• Direct: can be measured and allocated to aspecific work activity

• Indirect: difficult to attribute or allocate to aspecific output or work activity (alsooverhead or burden)

• Standard cost: cost per unit of output,established in advance of production orservice delivery




Some useful cost terminology

• Cash cost: a cost that involves a payment ofcash.

• Book cost: a cost that does not involve acash transaction but is reflected in theaccounting system.

• Sunk cost: a cost that has occurred in thepast and has no relevance to estimates offuture costs and revenues related to analternative course of action.




More useful cost terminology

• Opportunity cost: the monetary advantageforegone due to limited resources. The costof the best rejected opportunity.

• Life-cycle cost: the summation of all costsrelated to a product, structure, system, orservice during its life span.




The general price-demand relationship

The demand for aproduct or service isdirectly related to itsprice according to p=a-bD where p is price, D isdemand, and a and b areconstants that depend onthe particular product orservice.




Total revenue depends on price anddemand.

Total revenue is the product of the selling price perunit, p, and the number of units sold, D.




Calculus can help determine thedemand that maximizes revenue.

Solving, the optimal demandis




We can also find maximum profit…

Profit is revenue minus cost, so

for

Differentiating, we can find the value of D that maximizesprofit.




And we can find revenue/cost breakeven.

Breakeven is found when total revenue = total cost.Solving, we find the demand at which this occurs.




Engineers must consider cost in thedesign of products, processes and

services.

• “Cost-driven design optimization” is criticalin today’s competitive businessenvironment.

• In our brief examination we examinediscrete and continuous problems thatconsider a single primary cost driver.




Two main tasks are involved in cost-driven design optimization.

1. Determine the optimal value for a certainalternative’s design variable.

2. Select the best alternative, each with its ownunique value for the design variable.

Cost models are developed around the designvariable, X.




Optimizing a design with respect tocost is a four-step process.

• Identify the design variable that is the primary costdriver.

• Express the cost model in terms of the design variable.• For continuous cost functions, differentiate to find the

optimal value. For discrete functions, calculate costover a range of values of the design variable.

• Solve the equation in step 3 for a continuous function.For discrete, the optimum value has the minimum costvalue found in step 3.




Here is a simplified cost function.

where,

a is a parameter that represents the directly varying cost(s),

b is a parameter that represents the indirectly varying cost(s),

k is a parameter that represents the fixed cost(s), and

X represents the design variable in question.




“Present economy studies” can ignore thetime value of money.

• Alternatives are being compared over one year orless.

• When revenues and other economic benefits varyamong alternatives, choose the alternative thatmaximizes overall profitability of defect-freeoutput.

• When revenues and other economic benefits arenot present or are constant among alternatives,choose the alternative that minimizes total cost perdefect-free unit.




Engineering Economy

Chapter 4: The Time Value of Money




The objective of Chapter 4 is toexplain time value of moneycalculations and to illustrate

economic equivalence.




Money has a time value.• Capital refers to wealth in the form of

money or property that can be used toproduce more wealth.

• Engineering economy studies involve thecommitment of capital for extended periodsof time.

• A dollar today is worth more than a dollarone or more years from now (for severalreasons).




Return to capital in the form of interest andprofit is an essential ingredient of

engineering economy studies.• Interest and profit pay the providers of capital for

forgoing its use during the time the capital is beingused.

• Interest and profit are payments for the risk theinvestor takes in letting another use his or hercapital.

• Any project or venture must provide a sufficientreturn to be financially attractive to the suppliersof money or property.




Simple Interest: infrequently used

When the total interest earned or charged is linearlyproportional to the initial amount of the loan(principal), the interest rate, and the number ofinterest periods, the interest and interest rate are saidto be simple.




Computation of simple interestThe total interest, I, earned or paid may be computedusing the formula below.

P = principal amount lent or borrowed

N = number of interest periods (e.g., years)

i = interest rate per interest period

The total amount repaid at the end of N interestperiods is P + I.




If $5,000 were loaned for five years at asimple interest rate of 7% per year, theinterest earned would be

So, the total amount repaid at the endof five years would be the originalamount ($5,000) plus the interest($1,750), or $6,750.




Compound interest reflects both the remaining principaland any accumulated interest. For $1,000 at 10%…

Period

(1)Amount owedat beginning of

period

(2)=(1)x10%Interest

amount forperiod

(3)=(1)+(2)Amount

owed at endof period

1 $1,000 $100 $1,100

2 $1,100 $110 $1,210

3 $1,210 $121 $1,331

Compound interest is commonly used in personal andprofessional financial transactions.




Economic equivalence allows us tocompare alternatives on a common basis.• Each alternative can be reduced to an

equivalent basis dependent on– interest rate,– amount of money involved, and– timing of monetary receipts or expenses.

• Using these elements we can “move” cashflows so that we can compare them atparticular points in time.




We need some tools to find economicequivalence.

• Notation used in formulas for compound interestcalculations.– i = effective interest rate per interest period– N = number of compounding (interest) periods– P = present sum of money; equivalent value of one or

more cash flows at a reference point in time; the present– F = future sum of money; equivalent value of one or

more cash flows at a reference point in time; the future– A = end-of-period cash flows in a uniform series

continuing for a certain number of periods, starting atthe end of the first period and continuing through thelast




A cash flow diagram is an indispensabletool for clarifying and visualizing a

series of cash flows.




Cash flow tables are essential tomodeling engineering economy

problems in a spreadsheet




We can apply compound interestformulas to “move” cash flows along the

cash flow diagram.Using the standard notation, we find that apresent amount, P, can grow into a futureamount, F, in N time periods at interest ratei according to the formula below.

In a similar way we can find P given F by




It is common to use standard notation forinterest factors.

This is also known as the single paymentcompound amount factor. The term on theright is read “F given P at i% interest perperiod for N interest periods.”

is called the single payment present worthfactor.




We can use these to find economicallyequivalent values at different points in time.

$2,500 at time zero is equivalent to how much after sixyears if the interest rate is 8% per year?

$3,000 at the end of year seven is equivalent to howmuch today (time zero) if the interest rate is 6% peryear?




There are interest factors for a series ofend-of-period cash flows.

How much will you have in 40 years if yousave $3,000 each year and your accountearns 8% interest each year?




Finding the present amount from a seriesof end-of-period cash flows.

How much would is needed today to providean annual amount of $50,000 each year for 20years, at 9% interest each year?




Finding A when given F.

How much would you need to set aside eachyear for 25 years, at 10% interest, to haveaccumulated $1,000,000 at the end of the 25years?




Finding A when given P.

If you had $500,000 today in an accountearning 10% each year, how much could youwithdraw each year for 25 years?




It can be challenging to solve for N or i.

• We may know P, A, and i and want to findN.

• We may know P, A, and N and want to findi.

• These problems present special challengesthat are best handled on a spreadsheet.




Finding NAcme borrowed $100,000 from a local bank, whichcharges them an interest rate of 7% per year. If Acmepays the bank $8,000 per year, now many years will ittake to pay off the loan?

So,

This can be solved by using the interest tables andinterpolation, but we generally resort to a computersolution.




Finding iJill invested $1,000 each year for five years in a localcompany and sold her interest after five years for$8,000. What annual rate of return did Jill earn?

So,

Again, this can be solved using the interest tablesand interpolation, but we generally resort to acomputer solution.




There are specific spreadsheet functionsto find N and i.

The Excel function used to solve for N is

NPER(rate, pmt, pv), which will compute thenumber of payments of magnitude pmt required topay off a present amount (pv) at a fixed interestrate (rate).One Excel function used to solve for i is

RATE(nper, pmt, pv, fv), which returns a fixedinterest rate for an annuity of pmt that lasts for nperperiods to either its present value (pv) or future value(fv).




We need to be able to handlecash flows that do not occur until

some time in the future.• Deferred annuities are uniform series that

do not begin until some time in the future.• If the annuity is deferred J periods then the

first payment (cash flow) begins at the endof period J+1.




Finding the value at time 0 of adeferred annuity is a two-step

process.1. Use (P/A, i%, N-J) find the value of the

deferred annuity at the end of period J(where there are N-J cash flows in theannuity).

2. Use (P/F, i%, J) to find the value of thedeferred annuity at time zero.




Sometimes cash flows change by aconstant amount each period.

We can model these situations as a uniformgradient of cash flows. The table belowshows such a gradient.

End of Period Cash Flows1 02 G3 2G: :N (N-1)G




It is easy to find the present valueof a uniform gradient series.

Similar to the other types of cash flows, there is aformula (albeit quite complicated) we can use to findthe present value, and a set of factors developed forinterest tables.




We can also find A or Fequivalent to a uniform gradient

series.




End of Year Cash Flows ($)1 2,0002 3,0003 4,0004 5,000

The annual equivalent ofthis series of cash flows canbe found by considering anannuity portion of the cashflows and a gradientportion.

End of Year Annuity ($) Gradient ($)1 2,000 02 2,000 1,0003 2,000 2,0004 2,000 3,000




Sometimes cash flows change bya constant rate, ,each period--this

is a geometric gradient series.

End of Year Cash Flows ($)1 1,0002 1,2003 1,4404 1,728

This table presents ageometric gradient series. Itbegins at the end of year 1and has a rate of growth, ,of 20%.







We can find the present value of ageometric series by using the appropriate

formula below.

Where is the initial cash flow in the series.




When interest rates vary withtime different procedures are

necessary.• Interest rates often change with time (e.g., a

variable rate mortgage).• We often must resort to moving cash flows

one period at a time, reflecting the interestrate for that single period.




The present equivalent of a cash flow occurring atthe end of period N can be computed with theequation below, where ik is the interest rate for thekth period.

If F4 = $2,500 and i1=8%, i2=10%, and i3=11%, then




Nominal and effective interest rates.• More often than not, the time between successive

compounding, or the interest period, is less thanone year (e.g., daily, monthly, quarterly).

• The annual rate is known as a nominal rate.• A nominal rate of 12%, compounded monthly,

means an interest of 1% (12%/12) would accrueeach month, and the annual rate would beeffectively somewhat greater than 12%.

• The more frequent the compounding the greaterthe effective interest.




The effect of more frequentcompounding can be easily

determined.Let r be the nominal, annual interest rate and M thenumber of compounding periods per year. We canfind, i, the effective interest by using the formulabelow.




Finding effective interest rates.For an 18% nominal rate, compounded quarterly, theeffective interest is.

For a 7% nominal rate, compounded monthly, theeffective interest is.




Interest can be compoundedcontinuously.

• Interest is typically compounded at the endof discrete periods.

• In most companies cash is always flowing,and should be immediately put to use.

• We can allow compounding to occurcontinuously throughout the period.

• The effect of this compared to discretecompounding is small in most cases.




We can use the effective interestformula to derive the interest

factors.

As the number of compounding periods getslarger (M gets larger), we find that




Continuous compounding interestfactors.

The other factors can be found from these.




Engineering Economy

Chapter 5: Evaluating a SingleProject




The objective of Chapter 5 is todiscuss and critique

contemporary methods fordetermining project

profitability.




Proposed capital projects can beevaluated in several ways.

• Present worth (PW)• Future worth (FW)• Annual worth (AW)• Internal rate of return (IRR)• External rate of return (ERR)• Payback period (generally not appropriate

as a primary decision rule)




To be attractive, a capital projectmust provide a return that exceeds

a minimum level established bythe organization. This minimum

level is reflected in a firm’sMinimum Attractive Rate of

Return (MARR).




Many elements contribute todetermining the MARR.

• Amount, source, and cost of money available• Number and purpose of good projects

available• Perceived risk of investment opportunities• Type of organization




The most-used method is thepresent worth method.

The present worth (PW) is found bydiscounting all cash inflows and outflows tothe present time at an interest rate that isgenerally the MARR.

A positive PW for an investment projectmeans that the project is acceptable (itsatisfies the MARR).




Present Worth ExampleConsider a project that has an initialinvestment of $50,000 and that returns$18,000 per year for the next four years. Ifthe MARR is 12%, is this a goodinvestment?PW = -50,000 + 18,000 (P/A, 12%, 4)

PW = -50,000 + 18,000 (3.0373)

PW = $4,671.40 This is a good investment!




Bond value is a good example ofpresent worth.

The commercial value of a bond is the PW ofall future net cash flows expected to bereceived--the period dividend [face value (Z)times the bond rate (r)], and the redemptionprice (C), all discounted to the present at thebond’s yield rate, i%.

VN=C (P/F, i%, N) + rZ (P/A, i%, N)




Bond exampleWhat is the value of a 6%, 10-year bond with apar (and redemption) value of $20,000 that paysdividends semi-annually, if the purchaserwishes to earn an 8% return?

VN = $20,000 (P/F, 4%, 20) + (0.03)$20,000 (P/A, 4%, 20)

VN = $17,282.18

VN = $20,000 (0.4564) + (0.03)$20,000 (13.5903)




Capitalized worth is a specialvariation of present worth.

• Capitalized worth is the present worth of allrevenues or expenses over an infinite lengthof time.

• If only expenses are considered this issometimes referred to as capitalized cost.

• The capitalized worth method is especiallyuseful in problems involving endowmentsand public projects with indefinite lives.




The application of CW concepts.

The CW of a series of end-of-perioduniform payments A, with interest at i%per period, is A(P/A, i%, N). As Nbecomes very large (if the A are perpetualpayments), the (P/A) term approaches 1/i.So, CW = A(1/i).




Future Worth (FW) method is analternative to the PW method.

• Looking at FW is appropriate since theprimary objective is to maximize the futurewealth of owners of the firm.

• FW is based on the equivalent worth of allcash inflows and outflows at the end of thestudy period at an interest rate that isgenerally the MARR.

• Decisions made using FW and PW will bethe same.




Future worth example.A $45,000 investment in a new conveyorsystem is projected to improve throughput andincreasing revenue by $14,000 per year for fiveyears. The conveyor will have an estimatedmarket value of $4,000 at the end of five years.Using FW and a MARR of 12%, is this a goodinvestment?FW = -$45,000(F/P, 12%, 5)+$14,000(F/A, 12%, 5)+$4,000

FW = $13,635.70 This is a good investment!

FW = -$45,000(1.7623)+$14,000(6.3528)+$4,000




Annual Worth (AW) is anotherway to assess projects.

• Annual worth is an equal periodic series of dollaramounts that is equivalent to the cash inflows andoutflows, at an interest rate that is generally theMARR.

• The AW of a project is annual equivalent revenueor savings minus annual equivalent expenses, lessits annual capital recovery (CR) amount.




Capital recovery reflects the capital costof the asset.

• CR is the annual equivalent cost of thecapital invested.

• The CR covers the following items.– Loss in value of the asset.– Interest on invested capital (at the MARR).

• The CR distributes the initial cost (I) andthe salvage value (S) across the life of theasset.




A project requires an initial investment of $45,000,has a salvage value of $12,000 after six years, incursannual expenses of $6,000, and provides an annualrevenue of $18,000. Using a MARR of 10%,determine the AW of this project.

Since the AW is positive, it’s a good investment.




Internal Rate of Return• The internal rate of return (IRR) method is

the most widely used rate of return methodfor performing engineering economicanalysis.

• It is also called the investor’s method, thediscounted cash flow method, and theprofitability index.

• If the IRR for a project is greater than theMARR, then the project is acceptable.




How the IRR works• The IRR is the interest rate that equates the

equivalent worth of an alternative’s cashinflows (revenue, R) to the equivalent worthof cash outflows (expenses, E).

• The IRR is sometimes referred to as thebreakeven interest rate.

The IRR is the interest i'% at which




Solving for the IRR is a bit morecomplicated than PW, FW, or AW

• The method of solving for the i'% thatequates revenues and expenses normallyinvolves trial-and-error calculations, orsolving numerically using mathematicalsoftware.

• The use of spreadsheet software can greatlyassist in solving for the IRR. Excel uses theIRR(range, guess) or RATE(nper, pmt, pv)functions.




Challenges in applying the IRRmethod.

• It is computationally difficult withoutproper tools.

• In rare instances multiple rates of return canbe found. (See Appendix 5-A.)

• The IRR method must be carefully appliedand interpreted when comparing two moremutually exclusive alternatives (e.g., do notdirectly compare internal rates of return).




Reinvesting revenue—theExternal Rate of Return (ERR)

• The IRR assumes revenues generated are reinvested at theIRR—which may not be an accurate situation.

• The ERR takes into account the interest rate, ε, external toa project at which net cash flows generated (or required)by a project over its life can be reinvested (or borrowed).This is usually the MARR.

• If the ERR happens to equal the project’s IRR, then usingthe ERR and IRR produce identical results.




The ERR procedure• Discount all the net cash outflows to time 0

at ε% per compounding period.• Compound all the net cash inflows to period

N at at ε%.• Solve for the ERR, the interest rate that

establishes equivalence between the twoquantities.




ERR is the i'% at which

where

Rk = excess of receipts over expenses in period k,Ek = excess of expenses over receipts in period k,N = project life or number of periods, andε = external reinvestment rate per period.




Applying the ERR method

Year 0 1 2 3 4Cash Flow -$15,000 -$7,000 $10,000 $10,000 $10,000

For the cash flows given below, find the ERR when theexternal reinvestment rate is ε = 12% (equal to the MARR).

Expenses

Revenue

Solving, we find




The payback period method issimple, but possibly misleading.

• The simple payback period is the number ofyears required for cash inflows to just equalcash outflows.

• It is a measure of liquidity rather than ameasure of profitability.




Payback is simple to calculate.The payback period is the smallest value of θ (θ ≤ N) forwhich the relationship below is satisfied.

For discounted payback future cash flows arediscounted back to the present, so the relationship tosatisfy becomes




Problems with the payback periodmethod.

• It doesn’t reflect any cash flows occurringafter θ, or θ'.

• It doesn’t indicate anything about projectdesirability except the speed with which theinitial investment is recovered.

• Recommendation: use the payback periodonly as supplemental information inconjunction with one or more of the othermethods in this chapter.




Finding the simple anddiscounted paybackperiod for a set of cashflows.

End ofYear

Net CashFlow

CumulativePW at 0%

CumulativePW at 6%

0 -$42,000 -$42,000 -$42,000

1 $12,000 -$30,000 -$30,679

2 $11,000 -$19,000 -$20,889

3 $10,000 -$9,000 -$12,493

4 $10,000 $1,000 -$4,572

5 $9,000 $2,153

The cumulative cashflows in the table werecalculated using theformulas for simpleand discountedpayback.

From the calculationsθ = 4 years and θ' = 5years.




Engineering Economy

Chapter 6: Comparison and SelectionAmong Alternatives




The objective of chapter 6 is toevaluate correctly capital

investment alternatives when thetime value of money is a key

influence.




Making decisions meanscomparing alternatives.

• In this chapter we examine feasible designalternatives.

• The decisions considered are those selectingfrom among a set of mutually exclusivealternatives—when selecting one excludesthe choice of any of the others.




Mutually exclusive alternatives(MEAs)

• We examine these on the basis of economicconsiderations alone.

• The alternatives may have different initialinvestments and their annual revenues and costsmay vary.

• The alternatives must provide comparable“usefulness”: performance, quality, etc.

• The basic methods from chapter 5 provide thebasis for economic comparison of the alternatives.




Apply this rule, based onPrinciple 2 from Chapter 1.

The alternative that requires the minimuminvestment of capital and producessatisfactory functional results will be chosenunless the incremental capital associated withan alternative having a larger investment canbe justified with respect to its incrementalbenefits. This alternative is the basealternative.




For alternatives that have a largerinvestment than the base…

If the extra benefits obtained by investingadditional capital are better than those thatcould be obtained from investment of thesame capital elsewhere in the company at theMARR, the investment should be made.

(Please note that there are some cautions when consideringmore than two alternatives, which will be examined later.)




There are two basic types ofalternatives.

Investment AlternativesThose with initial (or front-end) capital investment

that produces positive cash flows from increasedrevenue, savings through reduced costs, or both.

Cost AlternativesThose with all negative cash flows, except for a

possible positive cash flow from disposal of assetsat the end of the project’s useful life.




Select the alternative that givesyou the most money!

• For investment alternatives the PW of allcash flows must be positive, at the MARR,to be attractive. Select the alternative withthe largest PW.

• For cost alternatives the PW of all cashflows will be negative. Select thealternative with the largest (smallest inabsolute value) PW.




Investment alternative exampleUse a MARR of 10% and useful life of 5 years to selectbetween the investment alternatives below.

AlternativeA B

Capital investment -$100,000 -$125,000Annual revenues less expenses $34,000 $41,000

Both alternatives are attractive, but Alternative B providesa greater present worth, so is better economically.




Cost alternative exampleUse a MARR of 12% and useful life of 4 years to selectbetween the cost alternatives below.

AlternativeC D

Capital investment -$80,000 -$60,000Annual expenses -$25,000 -$30,000

Alternative D costs less than Alternative C, it has a greaterPW, so is better economically.




Determining the study period.• A study period (or planning horizon) is the time

period over which MEAs are compared, and itmust be appropriate for the decision situation.

• MEAs can have equal lives (in which case thestudy period used is these equal lives), or they canhave unequal lives, and at least one does notmatch the study period.

• The equal life case is straightforward, and wasused in the previous two examples.




Unequal lives are handled in oneof two ways.

• Repeatability assumption– The study period is either indefinitely long or equal to a

common multiple of the lives of the MEAs.– The economic consequences expected during the

MEAs’ life spans will also happen in succeeding lifespans (replacements).

• Coterminated assumption: uses a finite andidentical study period for all MEAs. Cash flowadjustments may be made to satisfy alternativeperformance needs over the study period.




Comparing MEAs with equal lives.When lives are equal adjustments to cash flows are notrequired. The MEAs can be compared by directly comparingtheir equivalent worth (PW, FW, or AW) calculated using theMARR. The decision will be the same regardless of theequivalent worth method you use. For a MARR of 12%, selectfrom among the MEAs below.

AlternativesA B C D

Capital investment -$150,000 -$85,000 -$75,000 -$120,000Annual revenues $28,000 $16,000 $15,000 $22,000Annual expenses -$1,000 -$550 -$500 -$700Market Value (EOL) $20,000 $10,000 $6,000 $11,000Life (years) 10 10 10 10




Selecting the best alternative.Present worth analysis select Alternative A (but C is close).

Annual worth analysis—the decision is the same.




Using rates of return is anotherway to compare alternatives.

• The return on investment (rate of return) is a popularmeasure of investment performance.

• Selecting the alternative with the largest rate of return canlead to incorrect decisions—do not compare the IRR ofone alternative to the IRR of another alternative. The onlylegitimate comparison is the IRR to the MARR.

• Remember, the base alternative must be attractive (rate ofreturn greater than the MARR), and the additionalinvestment in other alternatives must itself make asatisfactory rate of return on that increment.




Use the incremental investmentanalysis procedure.

• Arrange (rank order) the feasible alternativesbased on increasing capital investment.

• Establish a base alternative.– Cost alternatives—the first alternative is the base.– Investment alternatives—the first acceptable alternative

(IRR>MARR) is the base.• Iteratively evaluate differences (incremental cash

flows) between alternatives until all have beenconsidered.




Evaluating incremental cash flows• Work up the order of ranked alternatives smallest to

largest.• Subtract cash flows of the lower ranked alternative from

the higher ranked.• Determine if the incremental initial investment in the

higher ranked alternative is attractive (e.g., IRR>MARR,PW, FW, AW all >0). If it is attractive, it is the “winner.”If not, the lower ranked alternative is the “winner.” The“loser” from this comparison is removed fromconsideration. Continue until all alternatives have beenconsidered.

• This works for both cost and investment alternatives.




Incremental analysisAlt. A Alt. B Alt. B-Alt. A

Initial cost -$25,000 -$35,000 -$10,000Net annual income $7,500 $10,200 $3,200IRR on total cash flow 15% 14% 11%

Which is preferred using a 5 year study period and MARR=10%?

Both alternatives A and B are acceptable—each one has a rate of returnthat exceeds the MARR. Choosing Alternative A because of its largerIRR would be an incorrect decision. By examining the incremental cashflows we see that the extra amount invested in Alternative B earns areturn that exceeds the IRR—so B is preferred to A. Also note…




Comparing MEAs with unequal lives.• The repeatability assumption, when

applicable, simplified comparison ofalternatives.

• If repeatability cannot be used, anappropriate study period must be selected(the coterminated assumption). This is mostoften used in engineering practice becauseproduct life cycles are becoming shorter.




The useful life of an alternative is lessthan the study period.

• Cost alternatives– Contracting or leasing for remaining years may be

appropriate– Repeat part of the useful life and use an estimated

market value to truncate• Investment alternatives

– Cash flows reinvested at the MARR at the end of thestudy period

– Replace with another asset, with possibly different cashflows, after the study period




The useful life of an alternative isgreater than the study period.

• Truncate the alternative at the end of thestudy period, using an estimated marketvalue.

• The underlying principle in all such analysisis to compare the MEAs in a decisionsituation over the same study (analysis)period.




Equivalent worth methods can beused for MEAs with unequal lives.

• If repeatability can be assumed, the MEAsare most easily compared by finding theannual worth (AW) of each alternative overits own useful life, and recommending theone having the most economical value.

• For cotermination, use any equivalent worthmethod using the cash flows available forthe study period.




We can use incremental rate of returnanalysis on MEAs with unequal lives.Equate the MEAs annual worths (AW) over theirrespective lives.

A BCapital Investment $3,500 $5,000Annual Cash Flow $1,255 $1,480Useful Live (years) 4 6

Solving, we find i*=26%, so Alt B is preferred.




Engineering Economy

Chapter 7: Decpreciation and IncomeTaxes




The objective of Chapter 7 is toexplain how depreciation affectsincome taxes, and how incometaxes affect economic decision

making.




Income taxes usually represent asignificant cash outflow. In this

chapter we describe how after taxliabilities and after-tax cash flows

result in the after-tax cash flow(ATCF) procedure. Depreciation

is an important element infinding after-tax cash flows.




Depreciation is the decrease in value ofphysical properties with the passage of

time.

• It is an accounting concept, a non-cash cost,that establishes an annual deduction againstbefore-tax income.

• It is intended to approximate the yearlyfraction of an asset’s value used in theproduction of income.




Property is depreciable if• it is used in business or held to produce

income.• it has a determinable useful life, longer than

one year.• it is something that wears out, decays, gets

used up, becomes obsolete, or loses valuefrom natural causes.

• it is not inventory, stock in trade, orinvestment property.




Depreciable property is

• tangible (can be seen or touched; personalor real) or intangible (such as copyrights,patents, or franchises).

• depreciated, according to a depreciationschedule, when it is put in service (when itis ready and available for its specific use).




Straight line (SL): constant amount ofdepreciation each year over the

depreciable life of the asset.

• N = depreciable life• B = cost basis• dk = depreciaton in k

• BVk = book value atend of k

• SVN = salvage value




Declining-balance (DB): a constant-percentage of the remaining BV is

depreciated each year.

The constant percentage is determined by R,where R = 2/N when 200% declining balance isbeing used, R = 1.5/N when 150% decliningbalance is being used.




The units-of-production method can beused when the decrease in value of the

assset is mostly a function of use, insteadof time. The cost basis is allocated

equally over the number of unitsproduced over the asset’s life. The

depreciation per unit of production isfound from the formula below.




The Modified Accelerated CostRecovery System (MACRS) is the

principle method for computingdepreciation for property in engineeringprojects. It consists of two systems, the

main system called the GeneralDepreciation System (GDS) and the

Alternative Depreciation System (ADS).




When an asset is depreciated usingMACRS, the following information is

needed to calculate deductions.• Cost basis, B• Date the property was placed into service• The property class and recovery period• The MACRS depreciation method (GDS or

ADS).• The time convention that applies (half year)




Using MACRS is easy!

1. Determine the asset’s recovery period (Table 7-2).

2. Use the appropriate column from Table 7-3 thatmatches the recovery period to find the recoveryrate, rk, and compute the depreciation for eachyear as




There are many different types of taxes.

• Income taxes are assessed as a function of grossrevenues minus allowable expenses.

• Property taxes are assessed as a function of thevalue of property owned.

• Sales taxes are assessed on the basis of purchaseof goods or services.

• Excise taxes are federal taxes assessed as afunction of the sale of certain goods or servicesoften considered nonnecessities.

We will focus on income taxes.




Taking taxes into account changesour expectations of returns on

projects, so our MARR (after-tax) islower.




The after-tax MARR should be at leastthe tax-adjusted weighted average cost of

capital (WACC).

= fraction of a firm’s pool of capital borrowedfrom lenders

t = effective income tax rate as a decimalib = before-tax interest paid on borrowed capitalea = after-tax cost of equity capital




Depreciation is not a cash flow, but itaffects a corporation’s taxable income, and

therefore the taxes a corporation pays.

Taxable income = gross income

– all expenses except capital invest.

– depreciation deductions.




Federal taxes are calculated using a setof income brackets. each applying a

different tax rate on the marginal valueof income. State taxes vary widely.

• Tax rates are found in Table 7-5.• Corporations need to know their effective tax rate,

which is a combination of federal and state taxesaccording to either formula below.




The disposal of a depreciable asset canresult in a gain or loss based on the sale

price (market value) and the currentbook value

A gain is often referred to as depreciation recapture,and it is generally taxed as the same as ordinaryincome. A loss is a capital loss. An asset sold formore than it’s cost basis results in a capital gain.




After-tax economic analysis isgenerally the same as before-taxanalysis, just using after-tax cashflows (ATCF) instead of before-

tax cash flows (BTCF). Theanalysis is conducted using the

after-tax MARR.




Cash flows are typically determined foreach year using the notation below.

Rk = revenues (and savings) from the projectduring period k

Ek = cash outflows during k for deductibleexpenses

dk = sum of all noncash, or book, costsduring k, such as depreciation

t = effective income tax rate on ordinaryincome

Tk = income tax consequence during year kATCFk = ATCF from the project during year k




Some important cash flow formulas.Taxable income

Ordinary income tax consequences







Acme purchased a pump for $250,000 andexpended $20,000 for shipping and

installation. The addition of this pump willresult in an increase in revenue of $80,000,

with associated increased expenses of$10,000, each year. The pump has a GDSrecovery period of five years, and Acme’s

effective tax rate is 41%. What is the ATCFfor this project for the fourth year of service

of the asset?







Economic value added, EVA, is anestimate of the profit-earning potential of

proposed capital investments inengineering projects. It is the difference

between a company’s adjusted netoperating profit after taxes (NOPAT) ina particular year and its after-tax cost of

capital during that year.




where,

and




For Acme, what is the EVA for year 4 iftheir after-tax MARR is 8%?




Engineering Economy

Chapter 9: Replacement Analysis




The objective of Chapter 9 is toaddress the question of whether acurrently owned asset should bekept in service or immediately

replaced.




What to do with an existing asset?

• Keep it• Abandon it (do not replace)• Replace it, but keep it for backup purposes• Augment the capacity of the asset• Dispose of it, and replace it with another




Three reasons to consider achange.

• Physical impairment (deterioration)• Altered requirements• New and improved technology is now

available.

The second and third reasons are sometimesreferred to as different categories of obsolescence.




Some important terms forreplacement analysis

• Economic life: the period of time (years) thatyields the minimum equivalent uniform annualcost (EUAC) of owning and operating as asset.

• Ownership life: the period between acquisitionand disposal by a specific owner.

• Physical life: period between original acquisitionand final disposal over the entire life of an asset.

• Useful life: the time period an asset is kept inproductive service (primary or backup).




Replacement: past estimation errors

• Any study today is about the future—pastestimation “errors” related to the defenderare irrelevant.

• The only exception to the above is if thereare income tax implications forthcomingthat were not foreseen.




Replacement: sunk-cost trap• Only present and future cash flows are

considered in replacement studies.• Past decisions are relevant only to the extent

that they resulted in the current situation.• Sunk costs—used here as the difference

between an asset’s BV and MV at aparticular point in time—have no relevanceexcept to the extent they affect incometaxes.




Replacement: the outsider viewpoint• The outsider viewpoint is the perspective taken by

an impartial third party to establish the fair MV ofthe defender. Also called the opportunity costapproach.

• The opportunity cost is the opportunity foregoneby deciding to keep an asset.

• If an upgrade of the defender is required to have acompetitive service level with the challenger, thisshould be added to the present realizable MV.




Replacement: economic lives of thechallenger and defender

• The economic life of the challenger minimizes theEUAC.

• The economic life of the defender is often oneyear, so a proper analysis may be betweendifferent-lived alternatives.

• The defender may be kept longer than it’sapparent economic life as long as it’s marginalcost is less than the minimum EUAC of thechallenger over it’s economic life.




Replacement: income taxes

• Replacement often results in gains or lossesfrom the sale of depreciable property.

• Studies must be made on an after-tax basisfor an accurate economic analysis since thiscan have a considerable effect on theresulting decision.




Before-tax PW exampleAcme owns a CNC machine that it is consideringreplacing. Its current market value is $25,000, but it canbe productively used for four more years at which timeits market value will be zero. Operating and maintenanceexpenses are $50,000 per year

Acme can purchase a new CNC machine, with the samefunctionality as the current machine, for $90,000. In fouryears the market value of the new machine is estimated tobe $45,000. Annual operating and maintenance costs willbe $35,000 per year.

Should the old CNC machine be replaced using a before-tax MARR of 15% and a study period of four years?




Example solutionDefender

Challenger

PW of the challenger is greater than PW of the defender(but it is close).




Proper analysis requires knowing theeconomic life (minimum EUAC) of the

alternatives.• The EUAC of a new asset can be computed

if the capital investment, annual expenses,and year-by-year market values are knownor can be estimated.

• The difficulties in estimating these valuesare encountered in most engineeringeconomy studies, and can be overcome inmost cases.




Finding the EUAC of the challengerrequires finding the total marginal costof the challenger, for each year. Theminimum such value identifies the

economic life.This equation represents the present worth, through year k,of total costs. (Although the sign is positive, it is a cost.)




Total marginal cost formulaThe total marginal cost is the equivalent worth, at theend of year k, of the increase in PW of total cost fromyear k-1 to year k.

This can be simplified to




Finding the economic life of the newCNC machine.Year 1 Year 2 Year 3 Year 4

O&M costs $35,000 $35,000 $35,000 $35,000Market value $75,000 $60,000 $50,000 $45,000

Marginal costs:Year 1 Year 2 Year 3 Year 4

O&M $35,000 $35,000 $35,000 $35,000Depreciation $15,000 $15,000 $10,000 $5,000Int. on capital $13,500 $11,250 $9,000 $7,500TC $63,500 $61,250 $54,000 $47,500




The economic life of the defender• If a major overhaul is needed, the life

yielding the minimum EUAC is likely thetime to the next major overhaul.

• If the MV is zero (and will be so later), andoperating expenses are expected to increase,the economic life will the one year.

• The defender should be kept as long as itsmarginal cost is less than the minimumEUAC of the best challenger.




Finding the economic life of thedefender CNC machine.

Year 1 Year 2 Year 3 Year 4O&M costs $50,000 $50,000 $50,000 $50,000Market value $15,000 $10,000 $5,000 $0

Year 1 Year 2 Year 3 Year 4O&M $50,000 $50,000 $50,000 $50,000Depreciation $10,000 $5,000 $5,000 $5,000Int. on capital $3,750 $2,250 $1,500 $750TC $63,750 $57,250 $56,500 $55,750




Replacement cautions.

• In general, if a defender is kept beyond where the TCexceeds the minimum EUAC for the challenger, thereplacement becomes more urgent.

• Rapidly changing technology, bringing about significantimprovement in performance, can lead to postponingreplacement decisions.

• When the defender and challenger have different usefullives, often the analysis is really to determine if now is thetime to replace the defender.

• Repeatability or cotermination can be used whereappropriate.




Abandonment--retirement withoutreplacement

• For projects having positive net cash flows(following an initial investment) and a finiteperiod of required service.

• Should the project be undertaken? If so, and givenmarket (abandonment) values for each year, whatis the best year to abandon the project? What is itseconomic life?

• These are similar to determining the economic lifeof an asset, but where benefits instead of costsdominate.

• Abandon the year PW is a maximum.




Abandonment exampleA machine lathe has a current market value of$60,000 and can be kept in service for 4 moreyears. With an MARR of 12%/year, whenshould it be abandoned? The following data areprojected for future years.

Year 1 Year 2 Year 3 Year 4

Net receipts $50,000 $40,000 $15,000 $10,000

Market value $35,000 $20,000 $15,000 $5,000




Abandonment solution

Keep for two years

Keep for one year

Keep for three years (BEST!) Keep for four years




Taxes can affect replacement decisions.

• Most replacement analyses should considertaxes.

• Taxes must be considered not only for eachyear of operation of an asset, but also inrelation to the sale of an asset.

• Since depreciation amounts generallychange each year, spreadsheets are anespecially important tool to use.




The effect of taxes.The economic life of an asset becomes, after taxes,

which reflects not only annual taxes but also tax effectsof the sale of the asset. The total marginal cost, foreach year, is




We must also consider the possible taxeffects of the sale of the defender.

The MV of the asset must be compared to the BV toassess the possible tax implications, and this should bereflected in the opportunity cost of keeping thedefender. The net ATCF, if the defender is kept, aftertaxes, is




Engineering Economy

Chapter 10: Evaluating Projects withthe Benefit-Cost Ratio Method




The objective of Chapter 10 is todemonstrate the use of thebenefit-cost ratio for the

evaluation of public projects.




Public projects are unique in many ways.• Frequently much larger than private ventures• They may have multiple, varied purposes that

sometimes conflict• Often very long project lives• Capital source is ultimately tax payers• Decisions made are often politically influenced• Benefits are often nonmonetary and are difficult to

measure• more...




These elements make engineeringeconomy studies more challenging.

• The Flood Control Act of 1936 requires thatbenefits must exceed costs to justifyfederally funded projects, this is a criterionnow used in most public projects.

• There can be difficulty defining benefits,and even in establishing costs.




For any project, the proper perspective isto consider the net benefits to the owners

of the enterprise.• For government projects, the owners are

ultimately the taxpayers.• Benefits are favorable consequences of the project

to the public (owners).• Costs represent monetary disbursements required

of the government.• Disbenefits represent negative consequences of a

project to the public (owners).




Self-liquidating projects are expected torepay their costs.

• These projects generally provide utility services(power, water, toll roads, etc.).

• They earn direct revenue that offset their costs, butthey are not expected to earn profits or pay taxes.

• In some cases in-lieu payments are made togovernments in place of taxes and fees that wouldhave been paid had it been under privateownership.




Cost allocations in multiple-purpose,public-sector projects tend to be arbitrary.

• Some projects naturally have multiplepurposes—e.g., construction of a dam.

• Some of the costs incurred cannot properlybe assigned to only one purpose.

• Purposes may be in conflict.• Often support for a public project, and its

many purposes, is politically sensitive.




Difficulties inherent in engineeringeconomy studies in the public sector.

• Profit standard not used to measure effectiveness• Monetary effect of many benefits is difficult to

quantify• May be little or no connection between the project

and the public (owners).• Often strong political influence whenever public

funds are used, with little consideration to long-term consequences.




Difficulties inherent in engineeringeconomy studies in the public sector.

• Public projects are more subject to legalrestrictions than private projects

• The ability of governmental bodies to obtaincapital is more restricted than that of privateenterprise

• The appropriate interest rate for discountingbenefits and costs is often controversially andpolitically sensitive.




Selecting the interest rate to use in publicprojects is challenging.

• Main considerations are– the rate on borrowed capital,– the opportunity cost of capital to the

governmental agency, and– the opportunity cost of capital to the taxpayers.

• If money is borrowed specifically for aproject, the interest rate on the borrowedcapital is appropriate to use as the rate.




More interest rate considerations…

• The 1997 Office of Management and Budgetdirective states that a 7% rate should be used, asan approximation of the return tax payers couldearn from private investments.

• Another idea is to use a market-determined risk-free rate, about 3-4% per year.

• Bottom line: there is no simple formula, and it isan important policy decision at the discretion ofthe governmental agency.




Applying the benefit-cost ratio method• The consideration of the time value of

money means this is really a ratio ofdiscounted benefits to discounted costs.

• Recommendations using the B-C ratiomethod will result in identicalrecommendations to those methodspreviously presented.

• B-C ratio is the ratio of the equivalent worthof benefits to the equivalent worth of costs.




Two B-C ratiosConventional B-C ratio with PW

Modified B-C ratio with PW

A project is acceptable when the B-C ratio is greaterthan or equal to one.




B-C ratios for annual worth.Conventional B-C ratio with AW

Modified B-C ratio with AW




Disbenefits (D) can be included in the B-Cratio in either the numerator or denominator,

as shown with AW below.

or




Added benefits vs. reduced cost• As with the different types of ratios, the

question arises if classifying certain cashflows as either added benefits or reducedcosts.

• As before, while the numerical value of theratio may change, there is no impact onproject acceptability regardless of how thecash flows are handled.




Selecting projects• If projects are independent, all projects that

have a B-C great than or equal to one maybe selected.

• For projects that are mutually exclusive, aB-C greater than one is required, butselecting the project that maximizes the B-Cratio does not guarantee that the best projectis selected.




Incremental B-C analysis for mutuallyexclusive projects.

• Incremental analysis must be used in the case of B-C and mutually exclusive projects.

• Rank alternatives in order of increasing totalequivalent worth of costs.

• With “do nothing” as a baseline, begin with thelowest equivalent cost alternative and determine theincremental B-C ratio (B/C), selecting thealternative with the higher equivalent cost if theratio is greater than one.




Which, if any, of the MEA projects below shouldbe selected using B-C analysis? Assume a 20

year study period and MARR=10%.A B C

Investment $125,000 $160,000 $180,000Annual O&M 10,000 10,000 9,500MV (20 yrs.) 40,000 50,000 50,000Benefit/yr. 35,000 42,000 44,000PW(10%)-costs 204,190 237,703 253,447PW(10%)-benefits 297,975 357,570 374,597B-C ratio 1.46 1.50 1.47

Each alternative is attractive.




Incremental analysis∆ (Β-A) ∆(C-B)

Investment $35,000 $20,000Annual O&M 0 -500MV (20 yrs.) 10,000 0Benefits/yr. 7,000 2,000PW(10%)-costs 33,514 15,743PW(10%)-benefits 59,595 17,027B-C ratio 1.78 1.08Conclusion B is better C is better

Choose alternative C.




Criticisms and shortcomings• B-C is often used as an “after-the-fact”

justification tool.• Distributional inequities (one group

benefits, another pays the cost) may not beaccounted for.

• Qualitative information is often ignored.• Bottom line: these are largely reflective of

the inherent difficulties in evaluating publicprojects rather than the B-C method itself.

Documents

Chapter 1: Introduction to Engineering Economy - ElCoM - HU Economy/Engineering... · Engineering Economy, Fourteenth Edition By William G. Sullivan, Elin M. Wicks, ... Engineering