收益率计算(英文).ppt

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1、Chapter 7Rate of Return AnalysisqRate of ReturnqMethods for Finding RORqInternal Rate of Return(IRR)CriterionqIncremental AnalysisqMutually Exclusive AlternativesRate of ReturnDefinition A relative percentage method which measures the annual rate of return as a percentage of investment over the life

2、 of a project.In 1970,when Wal-Mart Stores,Inc.went public,an investment of 100 shares cost$1,650.That investment would have been worth$12,283,904 on January 31,2002.What is the rate of return on that investment?Example:Meaning of Rate of ReturnSolution:032$12,283,904$1,650Given:P=$1,650 F=$12,283,9

3、04 N=32Find i:$12,283,904=$1,650(1+i)32 i=32.13%Rate of ReturnSuppose that you invested that amount($1,650)in a savings account at 6%per year.Then,you could have only$10,648 on January,2002.What is the meaning of this 6%interest here?This is your opportunity cost if putting money in savings account

4、was the best you can do at that time!So,in 1970,as long as you earn more than 6%interest in another investment,you will take that investment.Therefore,that 6%is viewed as a minimum attractive rate of return(or required rate of return).So,you can apply the following decision rule,to see if the propos

5、ed investment is a good one.ROR(32.13%)MARR(6%)Return on InvestmentDefinition 1:Rate of return(ROR)is defined as the interest rate earned on the unpaid balance of an installment loan.Example:A bank lends$10,000 and receives annual payment of$4,021 over 3 years.The bank is said to earn a return of 10

6、%on its loan of$10,000.Loan Balance Calculation:A=$10,000(A/P,10%,3)=$4,021Unpaid Return onUnpaid balance unpaidbalanceat beg.balancePaymentat the endYearof year(10%)receivedof year0123-$10,000-$10,000-$6,979-$3,656-$1,000 -$698 -$366+$4,021+$4,021+$4,021-$10,000-$6,979-$3,6560A return of 10%on the

7、amount still outstanding at the beginning of each yearRate of Return:Definition 2:Rate of return(ROR)is the break-even interest rate,i*,which equates the present worth of a projects cash outflows to the present worth of its cash inflows.Mathematical Relation:Return on Invested CapitalDefinition 3:Re

8、turn on invested capital is defined as the interest rate earned on the unrecovered project balance of an investment project.It is commonly known as internal rate of return(IRR).Example:A company invests$10,000 in a computer and results in equivalent annual labor savings of$4,021 over 3 years.The com

9、pany is said to earn a return of 10%on its investment of$10,000.Project Balance Calculation:0123Beginningproject balanceReturn on invested capitalPayment receivedEnding projectbalance-$10,000-$6,979-$3,656 -$1,000 -$697 -$365-$10,000+$4,021+$4,021+$4,021-$10,000-$6,979-$3,656 0The firm earns a 10%ra

10、te of return on funds that remain internally invested in the project.Since the return is internal to the project,we call it internal rate of return.Methods for Finding Rate of ReturnnTypes of Investment(cash flow)ClassificationqSimple InvestmentqNon-simple InvestmentnOnce we identified the type of i

11、nvestment cash flow,there are several ways available to determine its rate of return.nComputational MethodsqDirect Solution MethodqTrial-and-Error MethodqComputer Solution Method13Investment ClassificationSimple InvestmentnDefinition:Initial cash flows are negative,and only one sign change occurs in

12、 the net cash flows series.nExample:-$100,250,$300 (-,+,+)nROR:A unique ROR nIf the initial flows are positive and one sign change occurs referred to simple-borrowing.Non-simple InvestmentnDefinition:Initial cash flows are negative,but more than one sign changes in the remaining cash flow series.nEx

13、ample:-$100,300,-$120 (-,+,-)nROR:A possibility of multiple RORsPeriod(N)Project AProject BProject C0-$1,000-$1,000+$1,0001-5003,900-4502800-5,030-45031,5002,145-45042,000Project A is a simple investment.Project B is a non-simple investment.Project C is a simple borrowing.Computational MethodsDirect

14、 SolutionDirect SolutionTrial&Error MethodComputer Solution MethodLogQuadraticnProject A Project BProject CProject D0-$1,000-$2,000-$75,000-$10,000101,30024,40020,000201,50027,34020,0003055,76025,00041,500Example 7.2 Direct Solution Methods Project AProject BTrial and Error Method Project CnStep 1:G

15、uess an interest rate,say,i=15%nStep 2:Compute PW(i)at the guessed i value.PW(15%)=$3,553nStep 3:If PW(i)0,then increase i.If PW(i)MARR,AcceptThis rule does not work for a situation where an investment has multiple rates of returnComparing Mutually Exclusive Alternatives Based on IRRIssue:Can we ran

16、k the mutually exclusive projects by the magnitude of its IRR?nA1A201IRR-$1,000-$5,000$2,000$7,000100%40%$818 MARRB-ANOTE:Make sure that both IRRA and IRRB are greater than MARR.Example 7.7-Incremental Rate of ReturnnB1B2B2-B10123-$3,0001,3501,8001,500-$12,0004,2006,2256,330-$9,0002,8504,4254,830IRR

17、25%17.43%15%Given MARR=10%,which project is a better choice?Since IRRB2-B1=15%10%,and also IRRB2 10%,select B2.IRR on Increment Investment:Three AlternativesnD1D2D30-$2,000-$1,000-$3,00011,5008001,50021,0005002,00038005001,000IRR34.37%40.76%24.81%Step 1:Examine the IRR for each project to eliminate

18、any project that fails to meet the MARR.Step 2:Compare D1 and D2 in pairs.IRRD1-D2=27.61%15%,so select D1.D1 becomes the current best.Step 3:Compare D1 and D3.IRRD3-D1=8.8%15%,so select D1 again.Here,we conclude that D1 is the bestAlternative.ItemsCMS OptionFMS Option Investment$4,500,000$12,500,000

19、Total annual operating costs$7,412,920$5,504,100 Net salvage value$500,000$1,000,000Example 7.9 Incremental Analysis for Cost-Only Projects The firms MARR is 15%.Which alternative would be a better choice,based on the IRR criterion?Discussion:Since we can assume that both manufacturing systems would

20、 provide the same level of revenues over the analysis period,we can compare these alternatives based on cost only.(these systems are service projects).Although we can not compute the IRR for each option without knowing the revenue figures,we can still calculate the IRR on incremental cash flows.Sinc

21、e the FMS option requires a higher initial investment than that of the CMS,the incremental cash flow is the difference(FMS CMS)Example 7.9 Incremental Analysis for Cost-Only Projects(cost are itemized)ItemsCMS OptionFMS OptionAnnual O&M costs:Annual labor cost$1,169,600$707,200 Annual material cost8

22、32,320598,400 Annual overhead cost3,150,0001,950,000 Annual tooling cost470,000300,000 Annual inventory cost141,00031,500 Annual income taxes1,650,0001,917,000Total annual operating costs$7,412,920$5,504,100Investment$4,500,000$12,500,000Net salvage value$500,000$1,000,000Example 7.9 Incremental Cas

23、h Flow(FMS CMS)nCMS OptionFMS OptionIncremental(FMS-CMS)0-$4,500,000-$12,500,000-$8,000,0001-7,412,920-5,504,1001,908,8202-7,412,920-5,504,1001,908,8203-7,412,920-5,504,1001,908,8204-7,412,920-5,504,1001,908,8205-7,412,920-5,504,1001,908,8206-7,412,920-5,504,100$2,408,820Salvage+$500,000+$1,000,000S

24、olution:nAlthough the FMS would provide an incremental annual savings of$1,908,820 in operating costs,the savings do not justify the incremental investment of$8,000,000.COMMENTS:nNote that the CMS option was marginally preferred to the FMS option.nHowever,there are dangers in relying solely on the e

25、asily quantified savings in input factors such as labor,energy,and materials from FMS and in not considering gains from improved manufacturing performance that are more difficult and subjective to quantify.nFactors such as improved product quality,increased manufacturing flexibility(rapid response t

26、o customer demand),reduced inventory levels,and increased capacity for product innovation are frequently ignored in financial analysis because we have inadequate means for quantifying benefits.nIf these intangible benefits were considered,as they ought to be,however,the FMS option could come out better than the CMS option.