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John D. LeeIndustrial EngineeringEngineering Economy Review

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*Main conceptsModels are approximations of reality (THINK)

Time value of money, cash flow diagrams, and equivalence

Comparison of alternatives

Depreciation, inflation, and interest rates

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*Suggestions for solving problemsLookup unfamiliar terms in the indexDraw cash flow diagramsIdentify P, A, F, iBe flexible in using equations and tablesCheck with alternate methods

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*Cash flowsCash flows describe income and outflow of money over timeDisbursements =outflows -Receipts =inflows +Beginning of first year is traditionally defined as Time 0

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*EquivalenceTranslating cashflows over time into common units

Present values of future paymentsFuture value of present paymentsPresent value of continuous uniform paymentsContinuous payments equivalent to present payment

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*Single Payment Compound InterestP= (P)resent sum of moneyi= (i)nterest per time period (usually years)MARR=Minimal Acceptable Rate of Returnn= (n)umber of time periods (usually years)F= (F)uture sum of money that is equivalent to P given an interest rate i for n periods

F=P(1+i)nP=F(1+i)-nF=P(F/P,i,n)P=F(P/F,i,n)

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*Bank exampleYou 1000 deposit12% per year 5 yearsHow much do you have at end if compounded yearly?How much do you have at end if compounded monthly?

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*5.47 Income from savings $25,000 depositedAccount pays 5% compounded semiannuallyWithdrawals in one year and continuing foreverMaximum equal annual withdrawal equals?

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*5.47 Capitalized cost problemP=25,000A=?r=5%i=?

A=iP

01225,0003

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*Key points to rememberTime value of money$1000 today is not the same as $1000 one hundred years from nowMoney in the future is worth less than present sumsCash flow diagramsStarts at year zeroSuperposition to convert to standard formsEquivalenceFunctional notation, F=P(F/P,i,n)i and n must match units Capitalized cost, A=Pi, P=A/i

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*Comparison of alternativesPresent/Future worthCash flowRate of returnCost benefitPayback periodBreakeven analysis

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*Present/Future worthDetermine time period for analysis, least common multipleCalculate present value for each alternativeDraw cashflow diagramIdentify/calculate A, i, P, F, nUse present value equations to determine PCompare costs

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*Tomato peeling machinesMachine AMachine BPurchase cost=$52,000$63,000Annual cost=$15,000/year$9,000/yearAnnual benefit= $38,000/year $31,000 /yearSalvage value= $13,000$19,000Useful life= 4 years6 years

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*Present cost of A012352,00015,00013,000P4= -52000+(38,000-15,000)(P/A,12%,4)+13,000(P/F,12%,4)P12= P4+ P4(P/F,12%,4) + P4(P/F,12%,8)P12 =$53,25552,00052,00015,00015,00056715,00015,00015,00038,00038,00038,00038,00038,00038,00013,0009101115,00015,00015,00013,0001238,00038,00038,000P4P4P4A=38,000-15,000i= MARR=12%n=4F=13,000

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*Present cost of B012363,0009,000P6= -63000+(31,000-9,000)(P/A,12%,6)+19,000(P/F,12%,6)P12= P6+ P6(P/F,12%,6) P12 =$55,84663,0009,0009,0005679,0009,0009,00031,00031,00031,00031,000910119,0009,0009,00013,000129,00031,00019,00031,00031,00031,00031,00031,00049,000P6P6A=31,000-9,000i= MARR=12%n=6F=19,000

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*Cash flow analysis Determine time period for analysis: common multiple OR continuing operation then doesnt require least common multipleCalculate annual cost/benefit/profit for each alternativeDraw cashflow diagramIdentify/calculate A, S, i, P, F, nUse uniform payment equations to determine ACompare annual costs

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*Cash flow analysisProvides a shortcut for long/infinite analysis periods or when least common multiple might be a long time period with lots of calculationsCompare on the basis of annual cost if EITHERCommon multiple (e.g., 2 years and 8 years)OR Continuing operation (e.g., business will keep operating indefinitely with ability to replace equipment)

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*Rate of return analysisDraw cash flow diagram of each alternativeDraw combined cash flow diagram (higher initial cost- lower initial cost)Convert to Present worth and Present costs ORConvert to EUAB and EUACWrite equationSolve for iIf RORMARR, choose higher-cost alternative

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*7-52: Purchase vs. LeasePurchase machine: $12,000 initial cost$1,200 salvage value

Lease machine$3,000 annual payment

15% MARR, 8 year useful life

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*n=87-52: Purchase vs. LeasePurchase-Lease30004200PW of Benefits-PW of Costs=0 3000(P/A,i,7)+4200(P/F,i,8)-12,000= 0

i=17% 3000(3.922)+4200(0.2848) -12,000= 962

i=18% 3000(3.812)+4200(0.2660) -12,000= 553

i=20% 3000(3.605)+4200(0.2326) -12,000= -208

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*7-52: Purchase vs. LeaseInterest rate (%)i *NPW+- 55318%20%-208i=18%+2%(553/761)

i=19.45%

13,200-12,000

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*7-52: Purchase vs. Lease

Internal rate of return =17.6%

17.6%>15% therefore choose purchase option

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*Evaluation of multiple alternativesIdentify all alternativesCompute rate of return of all alternativesDelete alternatives with a return< MARRArrange remaining alternatives in order of increasing investment (find alternative where investing component dominates)Evaluate first two alternativesRepeat evaluation until all alternatives have been evaluated

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*Repeated evaluation of alternativesABCDMultiple comparisons of return on incremental investment

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*General suggestionsThink about alternativesi

*Payback period analysisApproximate rather than exact calculationAll costs and profits are included without considering their timingEconomic consequence beyond payback period are ignored (salvage value, gradient cash flow)May select a different alternative than other methodsFocus is speed versus efficiency

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*Benefit cost ratioBenefit cost ratio analysis (PW of benefit/PW of cost 1)Compare incremental investment, similar to rate of return analysis

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*9.9 Three alternatives ABCInitial cost50150110AB first 28.839.639.6Useful life 264Rate of Return 10%15%16.4%Compare using MARR=12%Future worthBenefit costPayback period

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*Future worth: Option C39.6110FF=-110(F/P,12,4) -110(F/P,12,8) -110(F/P,12,12) +39.6(F/A,12,12)F=81.61

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*Future worth analysis ABCInitial cost50150110AB first 28.839.639.6Useful life 264Rate of Return10%15%16.4%Future worth-18.9475.1781.61Benefit costsPayback period

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*Benefit-cost ratio analysis YearCAC-A0-110 -50-60139.6 28.810.8239.628.8-5060.8339.628.810.8439.628.810.8Present worth of Cost=60Present work of benefit=10.8(P/A,12,4)+50(P/F,12,2)B/C=72.66/60>1Reject A

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*Benefit-cost ratio analysis YearBCB-C0-150 -110-601-439.6 39.6040-1101105-639.639.606-1500-1507-839.639.6080-1101109-1239.639.60

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*Benefit-cost ratio analysisPW of cost=40+150(P/F,12%,6)PW of cost=115.99PW of benefits= 110(P/F,12%,4)+110(P/F,12%,8)PW of benefits=114.33

B/C=114.33/115.99

*Payback periodA 50/28.8 = 1.74 yearsB150/39.6=3.79 yearsC110/39.6=2.78 yearsSelect A

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*Summary ABCInitial cost50150110AB first 28.839.639.6Useful life 264Rate of Return 10%15%16.4%Future worth -18.9475.1781.61Benefit costC-A=1.21B-C=0.98Payback period1.743.792.78

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*Motor comparisonGraybarBlueballInitial cost$7,000$6,000Efficiency89%85%Maintenance300/year300/yearElectricity cost $0.072/kW-hour200 hp20 year useful life, No salvage valueInterest rate =10%Hours used to justify expense

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*Motor comparisonGraybar-Blueball>0NPC of Graybar-Blueball=1000+(300-300)+ (P/A,10%,20)200*0.746kW/hp*0.072$/kWhr*HRS(1/0.89)-(P/A,10%,20)200*0.746kW/hp*0.072$/kWhr*HRS(1/0.85)

1000= 8.514*0.568*HRS206.7 hrs

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*Key points to rememberPresent/Future worthUse least common multipleCash flowUseful for infinite analysis periodsRate of returnDo not use rate of return, but incremental rate of return as criterionSet up cash flow as investmentCost benefitsUse incremental comparison similar to rate of return analysisPayback periodApproximate method that makes huge assumptionsBreakeven analysis

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*Interest rates, depreciation, and inflationConcepts that allow more precise modeling of economic decisionsNominal vs effectiveDepreciationStraight lineMACRS (Modified Accelerated Cost Recovery System)Book valueInflation moderates value of rate of returns

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*Nominal and effective interest ratesEffective interest rate, iP, (period of compounding=period of interest) is used in formulas:i=iP=(1+ is)m-1i=iP=(1+rP/m)m-1is=interest per subperiodm=number of subperiods in period P rP=nominal interest per period PNominal interest rate, rP=m X isContinuous compounding: ia=er -1F = P(1+ ia ) n = P* ern

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*DepreciationDepreciation basis= Initial cost(C)- Salvage value (S)Book value = C-Accumulated depreciation

Straight line depreciationDi=(C-S)/nn= service lifeMACRSDi =C X Factor from table

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*Methods for depreciationBook value=cost-depreciation chargesStraight line (SL)Same amount each yearProportional to years of useful life and (initial cost-salvage)Sum-of-years (SOYD)Initial rate is faster than SLProportional to sum of digits in useful life and (initial cost-salvage)

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*Methods for depreciationDeclining balance, double declining balance (DDB)Double declining = 200% of straight lineProportional to years of useful life and book valueSalvage value not consideredDeclining balance/conversion to straight line (DDB/SL)Optimal switch year CANNOT be determined from a simple comparison of depreciation schedulesUnit of production (UOP)Modified Accelerated Cost Recovery System (MARCS)

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*Depreciation calculationsMethodAnnual Book value (year J)Straight line (P-S)/NP- (P-S) J/N

SOYD(P-S)[(N-J+1)/(N(N+1)/2)]P-sum of dep.

DDB2(Book value)/N 2P/N(1-2/N)j-1 P-P(1-(1-2/N) j )

UOP(P-S)Prod. in year/Total prod. P-sum of dep.

MARCSTable lookup (Property class, year) P-sum of dep.

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*Depreciation of machineInitial cost of $50,000Salvage value of $10,000Service life of 10 yearsStraight line depreciation=dn=(P-S)/Ndn =(50,000-10,000)/10dn =4,000/year

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*10.3 Capsulating machineInitial cost= $76,000Five year useful lifeNo salvage valueFind depreciation scheduleStraight lineSum of years digitsDouble declining balanceDDB with conversion

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*10.3 Straight lineYearDep/yearCumulative DepP-S/N000176,000/5=15,200 15,200215,20030,400315,20045,600415,20060,8005 15,20076,000

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*10.3 Sum of year digits YearDep/yearCumulative Dep (P-S)[(N-J+1)/(N(N+1)/2)] 000176,000(5)/15=25,333 25,333220,26745,600315,20060,800410,13370,9335 5,06776,000

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*10.3 Double declining balance YearDep/yearCumulative Dep 2P/N(1-2/N)j-1 OR 2/N(Cost-cumulative dep)000176,000(2/5)=30,400 30,4002(76,000-30,400)(2/5)=18,24048,640310,94459,58446,56666,1505 3,94070,090

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*10.3 Summary of depreciation schedulesYearSLSOYD DDB1 15,20025,33330,400215,20020,26718,240315,200 15,20010,944415,200 10,133 6,5665 15,200 5,067 3,940

What is best year to switch from DDB to SL depreciation?

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*Straight line depreciation if DDB has been used in previous yearsBook value in year three for DDB =76,000 - 30,400 - 18,240 = 27,360SL depreciation = Book value/ remaining useful lifeSwitch year BVSL dep3 27,36027,360/3= 9,120 6,5665 9,8509,850

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*InflationInterest rate adjusted for computing present worth and other valuesIncreases the value of the MARR to account for loss in value of future dollars due to inflation

Inflation adjusted interest rate = i + f + iff= rate of inflation

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*13.33 Value of a 10,000 investmentInterest rate 10%General price inflation is projected to be:3% for next 5 years5% for five years after that8% for following five yearsCalculate future worth of investment:in terms of actual dollarsin terms of real dollars at that timereal growth in purchasing power

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*A) Future value of actual $ =10,000 (F/P,10%,15)= $41,770

B) Future value in real $, constant value =41,770 (P/F,8%,5)(P/F,5%,5)(P/F,3%,5)0.6806 0.7835 0.8626=19,213C) Real growth rate of investment=19213=10,000(1+i)15 =4.45%13.33 Value of a 10,000 investment

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*Alternate solution solving for real dollarsUse real rather than market interest rateReal interest rates; i=(i - f)/(1+f)First five years: 6.796%Second five years: 4.762%Third five years: 1.9608%Real dollar value in 15 years10,000*(1.06796)5 *(1.04762)5 *(1.019608)519,318

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*13.30 Comparison of alternatives with inflation3 year lives with no salvage valueInflation = 5%Income tax rate of 25%Straight line depreciationMARR=7%Using rate of return analysis which is preferable?

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*13.30 Cash flowYearAB0-420-300120015022001503200150

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*Cash flow for option AYearAActual DepTax IncTaxATCFATCF Y0$0-420-420-420-420120021014070-17.5192.5183.32200220.514080.5-20.1200.4181.83200231.514091.5-22.9208.6180.2

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*Cash flow for option BYearAActual DepTax IncTaxATCFATCF Y0$0-300-300-300-3001150157.510057.5-14.4143.1136.32150165.410065.4-16.4149.0135.13150173.610073.6-18.4155.2134.1

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*Incremental ROR analysis A-BYearAB Y0$A-B0-420-300-1201183.3136.3472181.8135.146.73180.2134.146.1

Guessing 7%NPW = -120 + 47(P/F,7%,1) +46.7(P/F,7%,2) + 46.1(P/F,7%,3)= 2.3, therefor ROR >7% choose more expensive alternative

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