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1、ENGINEERING ECONOMY Fifth Edition 05 Present Worth AnalysisCHAPTER VPRESENT WORTH ANALYSISMcGrawHillENGINEERING ECONOMY Fifth EditionBlank and TarquinCHAPTER TOPICSFormulating AlternativesPW of Equal-Life AlternativesPW of Different-Life alternativesFuture Worth AnalysisPayback PeriodPW of BondsSpre
2、adsheet ApplicationsCHAPTER V5.1 FORMULATING ALTERNATIVESMcGrawHillENGINEERING ECONOMY Fifth EditionBlank and Tarquin5.1 FORMULATING MUTUALLY EXCLUSIVE ALTERNATIVESViable firms/organizations have the capability to generate potential beneficial projects for potential investmentTwo types of investment
3、 categoriesMutually Exclusive SetIndependent Project Set5.1 FORMULATING MUTUALLY EXCLUSIVE ALTERNATIVESMutually Exclusive set is where a candidate set of alternatives exist (more than one)Objective: Pick one and only one from the set.Once selected, the remaining alternatives are excluded.5.1 INDEPEN
4、DENT PROJECT SETGiven a set of alternatives (more than one)The objective is to:Select the best possible combination of projects from the set that will optimize a given criteria.Subjects to constraints5.1 Evaluating AlternativesPart of Engineering Economy is the selection and execution of the best al
5、ternative from among a set of feasible alternativesAlternatives must be generated from within the organizationOne of the roles of engineers!5.1 AlternativesProblemDoNothingAlt.1Alt.2Alt.mAnalysisSelectionExecution5.1 Type of AlternativesRevenue/Cost the alternatives consist of cash inflow and cash o
6、utflowsSelect the alternative with the maximum economic valueService the alternatives consist mainly of cost elementsSelect the alternative with the minimum cost valueCHAPTER V5.2 Present Worth ApproachMcGrawHillENGINEERING ECONOMY Fifth EditionBlank and Tarquin5.2 Present Worth ApproachA process of
7、 transforming all of the current and future estimated cash flow back to a point in time (time t = 0) called the Present Worth P(i%) = P( + cash flows) +P( - cash flows) 5.2 THE PRESENT WORTH METHODIf P(i%) 0 then the project is deemed acceptable.If P(i%) 0 the project is usually rejected.If P(i%) =
8、0 Present worth of costs = Present worth of revenues Indifferent!5.2 THE PRESENT WORTH METHOD Depends upon the Discount Rate UsedThe present worth is purely a function of the discount rate.If one changes the discount rate, a different present worth will result.5.2 Equal Lives Straightforward!Given t
9、wo or more alternatives with equal lives.Alt. 1Alt. 2Alt. NN = for all alternativesFind PW(i%) for each alternative then compareSmallest if cost and largest if profit.5.2 PRESENT WORTH: ExampleConsider: Machine A Machine BFirst Cost $2,500 $3,500Annual Operating Cost 900 700Salvage Value 200 350Life
10、 5 years 5 yearsi = 10% per year Which alternative should we select?Note: Where costs dominate a problem it is customary to assign a positive value to cost and negative to inflows5.2 PRESENT WORTH: Cash Flow Diagram0 1 2 3 4 5$2,500A = $900F5=$200MA0 1 2 3 4 5$3,500F5=$200A = $700MB5.2 PRESENT WORTH
11、: SolvingPA = 2,500 + 900 (P|A, .10, 5) 200 (P|F, .01, 5)= 2,500 + 900 (3.7908) - 200 (.6209) = 2,500 + 3,411.72 - 124.18 = $5,788PB = 3,500 + 700 (P|A, .10, 5) 350 (P|F, .10, 5)= 3,500 + 2,653.56 - 217.31 = $5,936SELECT MACHINE A: Lower PW cost!CHAPTER V5.3 PRESENT WORTH: Different LivesMcGrawHillE
12、NGINEERING ECONOMY Fifth EditionBlank and Tarquin5.3 PRESENT WORTH: Different LivesComparison must be made over equal time periodsCompare over the least common multiple, LCM, for their livesExample: 3,4, and 6 years. 5.3 PRESENT WORTH: Example Unequal LivesEXAMPLEMachine A Machine BFirst Cost $11,00
13、0 $18,000Annual Operating Cost 3,500 3,100Salvage Value 1,000 2,000Life 6 years 9 years i = 15% per yearWhich alternative should we select?5.3 PRESENT WORTH: Example Unequal LivesA common mistake is to compute the present worth of the 6-year project and compare it to the present worth of the 9-year
14、project.NO! NO! NO!5.3 PRESENT WORTH: Unequal Livesi = 15% per year0 1 2 3 4 5 6 $11,000F6=$1,000A 1-6 =$3,500Machine A0 1 2 3 4 5 6 7 8 9F6=$2,000A 1-9 =$3,100$18,000Machine BLCM(6,9) = 18 year study period will apply for present worth5.3 Unequal Lives: 2 Alternativesi = 15% per yearMachine ACycle
15、1 for ACycle 2 for ACycle 3 for A6 years6 years6 yearsCycle 1 for BCycle 2 for B18 years9 years9 yearsMachine B5.3 Example: Unequal Lives SolvingLCM = 18 yearsCalculate the present worth of a 6-year cycle for A PA = 11,000 + 3,500 (P|A, .15, 6) 1,000 (P|F, .15, 6)= 11,000 + 3,500 (3.7845) 1,000 (.4323)= $23,813, which occurs at time 0, 6 and 125.3 Example: Unequal LivesPA= 23,813+
