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Chapter 5 present worth analysis
1.
© 2012 by
McGraw-Hill, New York, N.Y All Rights Reserved 5-1 Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin Chapter 5Chapter 5 Present WorthPresent Worth AnalysisAnalysis
2.
5-2 LEARNINGLEARNING OUTCOMESOUTCOMES 1. Formulate
Alternatives 2. PW of equal-life alternatives 3. PW of different-life alternatives 4. Future Worth analysis 5. Capitalized Cost analysis © 2012 by McGraw-Hill All Rights Reserved
3.
5-3 Formulating AlternativesFormulating Alternatives Two
types of economic proposals Mutually Exclusive (ME) Alternatives: Only one can be selected; Compete against each other Independent Projects: More than one can be selected; Compete only against DN © 2012 by McGraw-Hill All Rights Reserved Do Nothing (DN) – An ME alternative or independent project to maintain the current approach; no new costs, revenues or savings
4.
5-4 Revenue: Alternatives include
estimates of costs (cash outflows) and revenues (cash inflows) Two types of cash flow estimates Cost: Alternatives include only costs; revenues and savings assumed equal for all alternatives; also called service alternatives Formulating AlternativesFormulating Alternatives © 2012 by McGraw-Hill All Rights Reserved
5.
5-5 Convert all cash
flows to PW using MARR Precede costs by minus sign; receipts by plus sign For mutually exclusive alternatives, select one with numerically largest PW For independent projects, select all with PW > 0 PW Analysis of AlternativesPW Analysis of Alternatives © 2012 by McGraw-Hill All Rights Reserved For one project, if PW > 0, it is justified EVALUATION
6.
5-6 Project ID Present
Worth A $30,000 B $12,500 C $-4,000 D $ 2,000 Solution: (a) Select numerically largest PW; alternative A (b) Select all with PW > 0; projects A, B & D Selection of Alternatives by PWSelection of Alternatives by PW © 2012 by McGraw-Hill All Rights Reserved
7.
5-7 Example: PW Evaluation
of Equal-Life ME Alts.Example: PW Evaluation of Equal-Life ME Alts. Alternative X has a first cost of $20,000, an operating cost of $9,000 per year, and a $5,000 salvage value after 5 years. Alternative Y will cost $35,000 with an operating cost of $4,000 per year and a salvage value of $7,000 after 5 years. At an MARR of 12% per year, which should be selected? Solution: Find PW at MARR and select numerically larger PW value PWX = -20,000 - 9000(P/A,12%,5) + 5000(P/F,12%,5) = -$49,606 PWY = -35,000 - 4000(P/A,12%,5) + 7000(P/F,12%,5) = -$45,447 Select alternative Y © 2012 by McGraw-Hill All Rights Reserved
8.
5-8 PW of Different-Life
AlternativesPW of Different-Life Alternatives Must compare alternatives for equal service (i.e., alternatives must end at the same time) Two ways to compare equal service: (The LCM procedure is used unless otherwise specified) Least common multiple (LCM) of lives Specified study period © 2012 by McGraw-Hill All Rights Reserved
9.
Assumptions of LCM
approachAssumptions of LCM approach Service provided is needed over the LCM or more years Selected alternative can be repeated over each life cycle of LCM in exactly the same manner Cash flow estimates are the same for each life cycle (i.e., change in exact accord with the inflation or deflation rate) 1-9 © 2012 by McGraw-Hill All Rights Reserved
10.
5-10 Example: Different-Life AlternativesExample:
Different-Life Alternatives Compare the machines below using present worth analysis at i = 10% per year Machine A Machine B First cost, $ Annual cost, $/year Salvage value, $ Life, years 20,000 30,000 9000 7000 4000 6000 3 6 Solution: PWA = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) + 4000(P/F,10%,6) = $-68,961 PWB = -30,000 – 7000(P/A,10%,6) + 6000(P/F,10%,6) = $-57,100 LCM = 6 years; repurchase A after 3 years Select alternative B © 2012 by McGraw-Hill All Rights Reserved 20,000 – 4,000 in year 3
11.
PW Evaluation Using
a Study PeriodPW Evaluation Using a Study Period Once a study period is specified, all cash flows after this time are ignored Salvage value is the estimated market value at the end of study period Short study periods are often defined by management when business goals are short-term Study periods are commonly used in equipment replacement analysis 1-11 © 2012 by McGraw-Hill All Rights Reserved
12.
5-12 Example: Study Period
PW EvaluationExample: Study Period PW Evaluation Compare the alternatives below using present worth analysis at i = 10% per year and a 3-year study period Machine A Machine B First cost, $ Annual cost, $/year Salvage/market value, $ Life, years -20,000 -30,000 -9,000 -7,000 4,000 6,000 (after 6 years) 10,000 (after 3 years) 3 6 Solution: PWA = -20,000 – 9000(P/A,10%,3) + 4000(P/F,10%,3) = $-39,376 PWB = -30,000 – 7000(P/A,10%,3) + 10,000(P/F,10%,3)= $-39,895 Study period = 3 years; disregard all estimates after 3 years Marginally, select A; different selection than for LCM = 6 years © 2012 by McGraw-Hill All Rights Reserved
13.
5-13 Future Worth AnalysisFuture
Worth Analysis Must compare alternatives for equal service (i.e. alternatives must end at the same time) Two ways to compare equal service: (The LCM procedure is used unless otherwise specified) Least common multiple (LCM) of lives Specified study period FW exactly like PW analysis, except calculate FW © 2012 by McGraw-Hill All Rights Reserved
14.
5-14 FW of Different-Life
AlternativesFW of Different-Life Alternatives Compare the machines below using future worth analysis at i = 10% per year Machine A Machine B First cost, $ Annual cost, $/year Salvage value, $ Life, years -20,000 -30,000 -9000 -7000 4000 6000 3 6 Solution: FWA = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3) + 4000 = $-122,168 FWB = -30,000(F/P,10%.6) – 7000(F/A,10%,6) + 6000 = $-101,157 LCM = 6 years; repurchase A after 3 years Select B (Note: PW and FW methods will always result in same selection) © 2012 by McGraw-Hill All Rights Reserved
15.
Capitalized Cost (CC)
AnalysisCapitalized Cost (CC) Analysis 5-15 CC refers to the present worth of a project with a very long life, that is, PW as n becomes infinite Basic equation is: CC = P = A i “A” essentially represents the interest on a perpetual investment For example, in order to be able to withdraw $50,000 per year forever at i = 10% per year, the amount of capital required is 50,000/0.10 = $500,000 For finite life alternatives, convert all cash flows into an A value over one life cycle and then divide by i © 2012 by McGraw-Hill All Rights Reserved
16.
5-16 Example: Capitalized CostExample:
Capitalized Cost Solution: Compare the machines shown below on the basis of their capitalized cost. Use i = 10% per year Machine 1 Machine 2 First cost,$ Annual cost,$/year Salvage value, $ Life, years -20,000 -100,000 -9000 -7000 4000 ----- 3 ∞ Convert machine 1 cash flows into A and then divide by i A1 = -20,000(A/P,10%,3) – 9000 + 4000(A/F,10%,3) = $-15,834 CC1 = -15,834 / 0.10 = $-158,340 CC2 = -100,000 – 7000/ 0.10 = $-170,000 Select machine 1 © 2012 by McGraw-Hill All Rights Reserved
17.
5-17 Summary of Important
PointsSummary of Important Points PW method converts all cash flows to present value at MARR PW comparison must always be made for equal service Alternatives can be mutually exclusive or independent Cash flow estimates can be for revenue or cost alternatives Equal service is achieved by using LCM or study period Capitalized cost is PW of project with infinite life; CC = P = A/i © 2012 by McGraw-Hill All Rights Reserved
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