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105Chapter7InternalRateofReturn7-1Tonyinvested$15,000inahighyieldaccount.Attheendof30yearsheclosedtheaccountandreceived$539,250.Computetheeffectiveinterestratehereceivedontheaccount.SolutionRecallthatF=P(1+i)n539,250=15,000(1+i)30⇒539,250/15,000=(1+i)3035.95=(1+i)303595111268112681268%30....=+=+==iiii7-2Theheatlossthroughtheexteriorwallsofaprocessingplantisestimatedtocosttheowner$3,000nextyear.AsalesmanfromSuperfiber,Inc.claimshecanreducetheheatlossby80%withtheinstallationof$15,000ofSuperfibernow.Ifthecostofheatlossrisesby$200peryear,afternextyear(gradient),andtheownerplanstokeepthebuildingtenmoreyears,whatishisrateofreturn,neglectingdepreciationandtaxes?SolutionNPW=0attherateofreturnTry12%NPW=-15,000+.8(3,000)(P/A,12%,10)+.8(200)(P/G,12%,10)=$1,800.64Try15%NPW=-$237.76Byinterpolationi=14.7%106Chapter7InternalRateofReturn7-3Doesthefollowingprojecthaveapositiveornegativerateofreturn?Showhowthisisknowntobetrue.Investmentcost$2,500Netbenefits$300inYear1,increasingby$200/yearSalvage$50Usefullife4yearsSolutionYearBenefits1300TotalBenefitsobtainedareless2500thantheinvestment,sothereturn3700ontheinvestmentisnegative4900550Total=2,450Cost7-4Atwhatinterestratewould$1000attheendof1995beequivalentto$2000attheendof2002?Solution(1+i)7=2;i=(2)1/7-1=0.1041or10.41%7-5Apainting,purchasedonemonthagofor$1000,hasjustbeensoldfor$1700.Whatnominalannualrateofreturndidtheownerreceiveonhisinvestment?Solutioni=7,000/1,000=70%IRR=70x12=840%7-8Findtherateofreturnfora$10,000investmentthatwillpay$1000/yearfor20years.Solution10,000=1,000(P/A,i%,20)(P/A,i%,20)=10Fromtables:7%i8%∴interpolatei=7.77%Chapter7InternalRateofReturn1077-9Ayoungengineerhasamortgageloanata15%interestrate,whichhegotsometimeago,foratotalof$52,000.Hehastopay120moremonthlypaymentsof$620.72each.Asinterestratesaregoingdown,heinquiresabouttheconditionsunderwhichhecouldrefinancetheloan.Ifthebankchargesanewloanfeeof4%oftheamounttobefinanced,andifthebankandtheengineeragreeonpayingthisfeebyborrowingtheadditional4%underthesametermsasthenewloan,whatpercentageratewouldmakethenewloanattractive,iftheconditionsrequirehimtorepayitinthesame120payments?SolutionTheamounttoberefinanced:a)PWof120monthlypaymentsleftP=A(P/A,i,n)=620.72(P/A,15%/12,120)=$38,474.09b)Newloanfee(4%)38,474.09x0.04=$1,538.96⇒Totalamounttorefinance=38,474.09+1,538.96=40,013.05Thenewmonthlypaymentsare:ANEW=40,013.05(A/P,i,120)Thecurrentpaymentsare:AOLD=620.72WewantANEWAOLDSubstituting⇒40,013.05(A/P,i,120)620.72(A/P,i,120)620.72/40,013.05=0.0155fori=1%(A/P,1%,120)=0.0143fori=1¼%(A/P,1¼%,120)=0.01611%i1¼%∴interpolatei=1.1667%Thiscorrespondstoanominalannualpercentagerateof12x0.011667=14%Therefore,shehastowaituntilinterestratesarelessthan14%.108Chapter7InternalRateofReturn7-10Yourcompanyhasbeenpresentedwithanopportunitytoinvestinaproject.Thefactsontheprojectarepresentedbelow:Investmentrequired$60,000,000AnnualGrossincome20,000,000AnnualOperatingcosts:Labor2,500,000Materials,licenses,insurance,etc.1,000,000Fuelandothercosts1,500,000Maintenancecosts500,000Salvagevalueafter10years0Theprojectisexpectedtooperateasshownfortenyears.Ifyourmanagementexpectstomake25%onitsinvestmentsbeforetaxes,wouldyourecommendthisproject?SolutionNetincome=20M–(2.5M+1M+1.5M+.5M)=$14,500,000NPW=0attherateofreturn0=-60,000,000+14,500,000(P/A,i,10)(P/A,i,10)=60/14.5=4.138@20%P/A=4.192@25%P/A=3.57120%i25%∴interpolatei=20.43%IRR25%∴RejectProject7-11Considerthefollowinginvestmentinapieceofland.Purchaseprice$10,000Annualmaintenance:100Expectedsalepriceafter5years:$20,000Determine:(a)Atrialvaluefori(b)Therateofreturn(to1/100percent)(c)Whatisthelowestsalepricetheinvestorshouldacceptifshewishestoearnareturnof10%afterkeepingthelandfor10years?Chapter7InternalRateofReturn109Solution(a)(F/P,i%,5)=20,000/10,000=2i=15%(b)NPW=-10,000-100(P/A,i%,5)+20,000(P/F,i%,5)=0Tryi=15%:=-391.2Tryi=12%:=+987.515%i12%∴interpolatei=14.15%(c)NFW=0=-10,000(F/P,10%,10)-100(F/A,10%,10)+SalePriceSalePrice=$27,5347-12Calculatetherateofreturnofthefollowingcashflowwithaccuracytothenearest1/10percent.SolutionNPW=0attherateofreturn0=-3,100+1,000(P/A,i,5)(P/A,i,5)=3.1(P/A,20%,5)=2.991(P/A,18%,5)=3.12718%i20%∴interpolatei=18.4%012345$3,100A=$1,000110Chapter7InternalRateofReturn7-13Aninvestmentthatcost$1000issoldfiveyearslaterfor$1261.Whatisthenominalrateofreturnontheinvestment(interestrate)?SolutionF=P(F/P,i%,5)1,261=1,000(F/P,i%,5)(F/P,i%,5)=1,261/1,000=1.2610Fromtables:(F/P,4½%,5)=1.246(F/P,5%,5)=1.2764½%i5%∴interpolatei=4.75%7-14Elizabethmadeaninitialinvestmentof$5000inatradingaccountwithastockbrokeragehouse.Afteraperiodof17monthsthevalueoftheaccounthadincreasedto$8400.Whatisthenominalannualinterestraterealizedontheinitialinvestmentifitisassumedtherewerenoadditionsorwithdrawalsfromtheaccount?SolutionF=P(F/P,i,17)F/P=8,400/5,000=1.68(1+i)17=1.681+i=(1.68)1/17=1.031Annualinterestrate=3.1x12=37.2%7-15Youborrowed$25fromafriendandafterfivemonthsrepaid$27.Whatinterestratedidyoupay(annualbasis)?Chapter7InternalRateofReturn111SolutionF=P(1+i)n27=25(1+i)5(27/25)1/5=1+i=1.0155i=