Chapter 1: Introduction to Engineering Economy - ElCoM - HU Economy/Engineering... · Engineering Economy, Fourteenth Edition By William G. Sullivan, Elin M. Wicks, ... Engineering
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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 Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
“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, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.”
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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).
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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)
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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%.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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?
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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).
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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).
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.)
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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…
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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)
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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?
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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.
Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
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