工程經(jīng)濟學第2講Sean含答案工程經(jīng)濟學第2講Sean含答案工程經(jīng)濟學第2講Sean含答案工程經(jīng)濟學第2講Sean含答案工程經(jīng)濟學第2講Sean含答案Timevalueofmoney1.Interest:thecostofmoney2.Economicequivalent3.Interestformulasforsinglecashflows4.Uneven-paymentseries5.Equal-paymentseries6.Dealingwithgradientseries7.Compositecashflows8.Today’stakeaway2Timevalueofmoney1.InterestMoneyisacommodity,moneycostsmoneyThecostofmoneyisestablishedandmeasuredbyaninterestrateWhatisinterestrate?3Moneyisacommodity,moneyco1.Interest:thecostofmoney(1).Thetimevalueofmoney(2).Elementsoftransactionsinvolvinginterest(3).Methodofcalculatinginterest41.Interest:thecostofmoney(1).ThetimevalueofmoneyAnexampleWhatdoesthisexampleillustrate?WemustconnectearningpowerandpurchasingpowertotheconceptoftimeItalsoreflectsthatmoneyhasatimevaluePrincipleofthetimevalueofmoneyTheeconomicvalueofasumdependsonwhenthesumisreceivedAdollarreceivedtodayhasagreatervaluethanadollarreceivedatsomefuturetimeMarketinterestrate(usedinthetext)Considerstheearningpowerofmoneyaswellastheeffectofinflationperceivedinthemarketplace5(1).ThetimevalueofmoneyAnGainsachievedorlossesincurredbydelayingconsumption$100$106$108$104$106Case1:InflationExceedsearingpowerCase2:EaringpowerExceedsInflationYourmoneyYourmoneyPeriod1Period0CostofrefrigeratorCostofrefrigeratorNetloss$2Netgain$26Gainsachievedorlossesincur

(2).Elementsoftransactionsinvolvinginterest

ElementsarecommontoallthetypesoftransactionsCashflowdiagramsEnd-of-periodconventionInpractice,cashflowscanoccuratthebeginningorinthemiddleofaninterestperiodoratpracticallyanypointintimeOneofthesimplifyingassumptionswemakeinengineeringeconomicanalysisistheend-of-periodconventionend-of-periodconventionisthepracticeofplacingallcashflowtransactionsattheendofaninterestperiod7

(2).ElementsoftransactionsElementsPrincipal(P)Interestrate(i)Interestperiod(n)Numberofinterestperiods(N)Aplanforreceiptsordisbursements(An)Afutureamountofmoney(F)8ElementsPrincipal(P)8Cashflowdiagrams$22,123$6215$6720$883510,324$10,815i=7%0123459Cashflowdiagrams$22,123$62151010AcashflowdiagramforSabathia’scontractwiththeNewYorkYankees.11AcashflowdiagramforSabath(3).MethodofcalculatinginterestSimpleinterestF=P(1+iN)CompoundinterestF=P(1+i)N12(3).Methodofcalculatinginte2.EconomicequivalentHowtomeasuretimevalueofmoneyWhetherweshouldprefertohave$20,000today,and$50,000tenyearsfromnowor$8,000eachyearforthenext10years?Method:economicequivalenceAcashflowthatcanbeconvertedtoanequivalentcashflowatanypointintimeAnexample132.EconomicequivalentHowtomAnexampleSupposeyouareofferedthealternativeofreceivingeither$3,000attheendoffiveyearsorPdollarstoday.Becauseyouhavenocurrentneedforthemoney,youwoulddepositthePdollarsinanaccountthatpays8%interest.WhatvalueofPwouldmakeyouindifferenttoyourchoicebetweenPdollarstodayandthepromiseof$3,000attheendoffiveyears?Solution:GivenF=$3,000,N=5years,i=8%peryearFindPP=F/(1+i)NP=3,000/(1+0.08)5=$2,04214AnexampleSupposeyouareoff3.Interestformulasforsinglecashflows(1).Compound-amountfactor(復利系數(shù))(2).Present-worthfactor(現(xiàn)值系數(shù))(3).Solvingfortime(4).Solvingforinterest(5).Interpolationininteresttables153.Interestformulasforsingl(1).Compound-amountfactorGivenP,i,NFindFF=P(1+i)N(1+i)Nisknownasthecompound-amountfactorInteresttablesinappendixAHowtouseappendixAtofindFF=P(1+i)N=P(F/P,i,N)factornotation16(1).Compound-amountfactorGive(2).Present-worthfactorGivenF,i,NFindPP=F[1/(1+i)N]=F(P/F,i,N)

1/(1+i)Nisknownasthepresent-worthfactor17(2).Present-worthfactorGivenAnexampleGivenF=$1,000,i=6%,N=8FindPP=$1,000(P/F,6%,8)=$627.40P=?$1,00008yearsi=6%peryear18AnexampleGivenF=$1,000,i=6%(3).SolvingfortimeGivenF,i,PFindN19(3).SolvingfortimeGivenF,iAnexampleGivenF=$6,000,i=8%,P=$3,000FindNF=P(1+i)NUsingExcelN=NPER(8%,0,-3000,6000)=9$3,000$6,0000N=?yearsi=8%peryear20AnexampleGivenF=$6,000,i=8%(4).SolvingforinterestGivenF,N,PFindi21(4).SolvingforinterestGivenAnexampleGivenF=$20,N=5,P=$10FindiF=P(1+i)NUsingExceli=RATE(5,0,-10,20)=14.87%$10$2005yearsi=?22AnexampleGivenF=$20,N=5,P=Doublingtimeestimatesusingtheruleof72andtheactualtimeusingcompoundinterestcalculationsrateofreturn(%)rule-of-72estimatesactualyears1727023635.3514.414.3107.27.5203.63.9401.8223DoublingtimeestimatesusingCASEDeterminethevalueoftheA/Pfactorforaninterestrateof7.3%andnof10years.Thatis(A/P,7.3%,10)24CASEDeterminethevalueofthe(5).Interpolationininteresttables7%7.3%8%0.14238X0.14903線性內(nèi)插(7.3-7)/(8-7.3)=(x-0.14238)/(0.14903-X)x=0.14338DeterminethevalueoftheA/Pfactorforaninterestrateof7.3%andnof10years.Thatis(A/P,7.3%,10)25(5).Interpolationininterest4.Uneven-paymentseriesWhatisuneven-paymentseries264.Uneven-paymentseriesWhatiDecompositionofunevencashflowseries27DecompositionofunevencashfAnexampleGiven:unevencashflowasthepicture,i=10%peryearFindPP=25,000(P/F,10%,1)+3,000(P/F,10%,2)+5,000(P/F,10%,4)Result:$28,622P=?$25,0000$5,000yearsi=?$3,00028AnexampleGiven:unevencashfAfuture-valuecalculationwithchanginginterestrates29Afuture-valuecalculationwitF=300(1+0.05)(1+0.06)(1+0.06)(1+0.04)(1+0.04)+500(1+0.06)(1+0.04)(1+0.04)+400(1+0.04)Result:$1,372OrF={[300(F/P,5%,1)(F/P,6%,1)+500](F/P,6%,1)(F/P,4%,1)+400}(F/P,4%,1)30F=300(1+0.05)(1+0.06)(1+0.06)(5.Equal-paymentseries(1).Compound-amountfactor(復利系數(shù))(2).Sinking-fundfactor(償債基金系數(shù))(3).Capital-recoveryfactor(Annuityfactor)(資本回收系數(shù))(4).Present-worthfactor(現(xiàn)值系數(shù))(5).Presentvalueofperpetuities(永久年金)315.Equal-paymentseries(1).Com(1).Compound-amountfactorEqual-payment-seriescompound-amountfactorF=A{[(1+i)N-1]/i}=A(F/A,i,N)Howtogetit?(seetheprocess)Methodof

calculationInteresttablesinappendixA32(1).Compound-amountfactorEquaTheprocessF=A+A(1+i)+A(1+i)2+…+A(1+i)N-1[Eq.1]Multiplyingitby(1+i)resultinF(1+i)=A(1+i)+A(1+i)2+…+A(1+i)N[Eq.2]Subtracting[Eq.1]from[Eq.2]getsF(1+i)-F=-A+A(1+i)NSolvingforFyieldsF=A{[(1+i)N-1]/i}=A(F/A,i,N)FA0NN-133TheprocessF=A+A(1+i)+A(1+i)2+AcaseGiven:A=$5,000,N=5,i=6%FindFF=$5,000(F/A,6%,5)=$28,185.46CanalsoutilizeExcel,thefollowingfinancialcommandFV(6%,5,5000,0)yearsi=6%0$5,000F=?$5,00034AcaseGiven:A=$5,000,N=5,i=(2).Sinking-fundfactorSinking-fundfactorFromEq.F=A{[(1+i)N-1]/i},wecangetA=F{i/[(1+i)N-1]/}=F(A/F,i,N)35(2).Sinking-fundfactorSinkingAcaseGiven:cashflowasthepicture,N=5,i=7%,F=10,000FindAA=$10,000(A/F,7%,5)=$1,739yearsi=7%0A=?F=10,000536AcaseGiven:cashflowasthe(3).Capital-recoveryfactor(Annuityfactor)Capital-recoveryfactor(Annuityfactor)FromA=F{i/[(1+i)N-1]/}andF=P(1+i)NwegetA=P{i(1+i)N/[(1+i)N-1]}=P(A/P,i,N)37(3).Capital-recoveryfactor(AAcaseGiven:P=$21,061.82,N=5,i=6%FindAA=$21,061.82(A/P,6%,5)=$5,000yearsi=6%0A=?P=$21,061.82538AcaseGiven:P=$21,061.82,N=5(4).Present-worthfactorPresentworthfactorBecauseA=P{i(1+i)N/[(1+i)N-1]},soP=A{[(1+i)N-1]/i(1+i)N}=A(P/A,i,N)39(4).Present-worthfactorPresenAcaseGiven:A=$7.92million,N=25,i=8%FindPP=$7.92(P/A,8%,25)=$84.54millionyearsi=8%0A=$7.92millionP=?2540AcaseGiven:A=$7.92million,(5).PresentvalueofperpetuitiesP=A{[(1+i)N-1]/i(1+i)N}TakealimitontheEq.bylettingN→∞,wegettheclosed-formsulutionP=A/iyears0AP=?N→∞41(5).Presentvalueofperpetuit6.Dealingwithgradientseries(1).Lineargradientseries(線性梯度系列)(2).geometricgradientseries(幾何梯度系列)426.Dealingwithgradientserie(1).LineargradientseriesStrictgradientseries:Cashflowincreaseordecreasebyasetofamount,G,thegradientamountThepresent-worthfactorofgradientlinearLineargradientseriesascompositeseries43(1).LineargradientseriesStriStrictgradientseries44Strictgradientseries44Twotypesoflinear-gradientseriesascompositesofauniformseriesofNpaymentsofA1andthegradientseriesofincrementsofconstantamountG.45Twotypesoflinear-gradientsThepresent-worthfactoroflineargradientP=0+G/(1+i)2+2G/(1+i)3+…+(N-1)G/(1+i)NP=G{[(1+i)N-iN-1]/i2(1+i)N}P=G(P/G,i,N)Thegradient-seriespresent-worthfactor:P=G{[(1+i)N-iN-1]/i2(1+i)N}46Thepresent-worthfactorofliLineargradientseriesascompositeseriesA1+GA1+(N-1)GG01234N-1N(N-1)G+A1=01234N-1N01234N-1N47Lineargradientseriesascomp(2).geometricgradientseriesGeometricgradientseriesThegeometric-gradient-seriespresent-worthfactorAnalternativeforthegeometric-gradient-seriespresent-worthfactor48(2).geometricgradientseriesGAgeometricallyincreasingordecreasinggradientseriesataconstantrateg49AgeometricallyincreasingorThegeometric-gradient-seriespresent-worthfactorPn=An(1+i)-n=A1(1+g)n-1(1+i)-nThegeometric-gradient-seriespresent-worthfactorIfi=g,P=A1N/(1+i)Ifi≠g,P=A1[1-(1+g)N(1+i)-N]/(i-g)50Thegeometric-gradient-seriesAnothermethodforthegeometric-gradient-seriesP=[A1/(1+g)]∑[(1+g)/(1+i)]^nAssume[1/(1+g’)]^n=[(1+g)/(1+i)]^nYieldsg’=(i-g)/(1+g)RewritepasfollowsP=[A1/(1+g)]∑[(1+g)/(1+i)]^n=[A1/(1+g)]∑{1/[1+(i-g)/(1+g)]^n}=[A1/(1+g)]∑(1+g’)^(-n)=[A1/(1+g)](P/A,g’,N)51Anothermethodforthegeometr5252SummaryofDiscreteCompoundingFormulaswithDiscretePayments53SummaryofDiscreteCompoundin7.CompositecashflowsInfact,manyinvestmentprojectscontainmixedtypesofcashflows.547.CompositecashflowsInfactStartsavingmoneyassoonaspossibleTwosavingsplansstartingattheageof21Option1:Save2000ayearfor10years,attheendof10years,makenofutureinvestments,butinvesttheamountaccumulatedattheendof10yearsuntilyoureachtheageof65.assumethatthefirstdepositwillbemadewhenyouare22.Option2:Donothingforthefirst10years,startsaving2000ayeareveryyearthereafteruntilyoureachtheageof65.assumethefirstdepositwillbemadewhenyouturn32.Theinterestrateis8%,whichoptionwouldresultinmoremoney?55Startsavingmoneyassoonas222331…65Option1213233…2,0002,0002,000F?222331…65Option2213233…2,0002,0002,000F?Answer56222331…65Option1213233…2,0002Foroption1F=2000(F/A,8%,10)(F/P,8%,34)=2000*14.4866*13.6901=396,646Foroption2F=2000(F/A,8%,34)=2000*158.6267=317,253Sooption1resultsinmoremoney.Anyothercomparisonway?57Foroption1AnyothercomparisFindpresentworthforalineargradientseriesChoosinghowtoreceiveyourwinningsOption1:receive3.44millioninonelumpsumOption2:receiveingradientannualpayments,theinitialis175,000,thefirstis189,000,therestswouldincreaseeachyearby7,000toafinalpaymentof357,000.Iftheinterestrateis4.5%.Whichonewouldyouprefer?58Findpresentworthforalinea3.44million123…25tOption1123…25tOption2175,000189,000189,000+7000189,000+7000*2189,000+7000*(t-1)357,000Answer2593.44million123…25tOption1123ForthesecondoptionP=175,000+G(P/G,4.5%,25)+A1(P/A,4.5%,25)(P/G,4%,25)=156.1040;(P/G,5%,25)=134.2275--

(P/G,4.5%,25)=145.1658(P/A,4%,25)=15.6221;(P/A,5%,25)=14.0939--

(P/A,4.5%,25)=14.858P=175,000+7000*145.1658+189,000*14.858=3,999,322.6>3.44millionSochoosethesecondoption60Forthesecondoption60Requiredcost-of-livingadjustmentcalculationSupposethatyourretirementbenefitsduringyourfirstyearofretirementare50,000.assumethatthisamountisjustenoughtomeetyourcostoflivingduringthefirstyear.However,yourcostoflivingisexpectedtoincreaseatanannualrateof5%,duetoinflation.Supposeyoudonotexpecttoreceiveanycostoflivingadjustmentinyourr