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Contemporary Engineering Economics, 4 th edition, © 2007 Replacement Decisions Lecture No. 58 Chapter 14 Contemporary Engineering Economics Copyright © 2006

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Contemporary Engineering Economics, 4 th edition, © 2007 Required Assumptions and Decision Frameworks Planning Horizon (Study Period) Infinite planning horizon Finite planning horizon Technology No technology improvement is anticipated over the planning horizon Technology improvement cannot be ruled out Revenue Cash Flow Information Decision Frameworks

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Contemporary Engineering Economics, 4 th edition, © Replace the defender now: The cash flows of the challenger estimated today will be used. An identical challenger will be used thereafter if replacement becomes necessary again in the future. This stream of cash flows is equivalent to a cash flow of AEC C* each year for an infinite number of years. 2. Replace the defender, say, x years later: The cash flows of the defender will be used in the first x years. Starting in year x+1,the cash flows of the challenger will be used indefinitely thereafter. Replacement Strategies under the Infinite Planning Horizon

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Contemporary Engineering Economics, 4 th edition, © 2007 Types of Typical Replacement Decision Frameworks

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Contemporary Engineering Economics, 4 th edition, © 2007 Example 14.4 Relevant Cash Flow Information (Defender) Cash flow diagram for defender when N = 4 years

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Contemporary Engineering Economics, 4 th edition, © 2007 Economic Service Life for the Defender (Example 14.4) Defender: Find the remaining useful (economic) service life.

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Contemporary Engineering Economics, 4 th edition, © 2007 AEC as a Function of the Life of the Defender (Example 14.4)

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Contemporary Engineering Economics, 4 th edition, © 2007 Economic Service Life for the Challenger (Example 14.3) N = 1 year: AE(15%) = $7,500 N = 2 years: AE(15%) = $6,151 N = 3 years: AE(15%) = $5,847 N = 4 years: AE(15%) = $5,826 N = 5 years: AE(15%) = $5,897 N C* =4 years AE C* =$5,826 Investment cost = $10,000 Salvage value N = 1: $6,000 N > 1: decreases at a 15% over previous year Operating cost N = 1: $2,000 N > 1: increases at a15% annual rate

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Contemporary Engineering Economics, 4 th edition, © 2007 Replacement Decisions Should replace the defender now? No, because AEC D < AEC C If not, when is the best time to replace the defender? Need to conduct the marginal analysis. N C* = 4 years AE C* =$5,826

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Contemporary Engineering Economics, 4 th edition, © 2007 Question: What is the additional (incremental) cost for keeping the defender one more year from the end of its economic service life, from Year 2 to Year 3? Financial Data: Opportunity cost at the end of year 2: Equal to the market value of $3,000 Operating cost for the 3 rd year: $5,000 Salvage value of the defender at the end of year 3: $2,000 Marginal Analysis to Determine when the Defender should be Replaced

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Contemporary Engineering Economics, 4 th edition, © $3000 $2000 $5000 Step 1: Calculate the equivalent cost of retaining the defender one more from the end of its economic service life, say 2 to 3. $3,000(F/P,15%,1) + $5,000 - $2,000 = $6,450 Step 2: Compare this cost with AEC C = $5,826 of the challenger. Conclusion: Since keeping the defender for the 3 rd year is more expensive than replacing it with the challenger, DO NOT keep the defender beyond its economic service life. $6,450 23

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Contemporary Engineering Economics, 4 th edition, © 2007 Example 14.5 Replacement Analysis under the Finite Planning Horizon N Annual Equivalent Cost ($) DefenderChallenger 15,1307,500 25,1166,151 35,5005,857 45,9615,826 56,4345,897 Some likely replacement patterns under a 8-year finite planning horizon

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Contemporary Engineering Economics, 4 th edition, © 2007 Replacement Analysis under the Finite Planning Horizon (Example 14.5) Option 1:(j 0, 0), (j, 4), (j, 4) PW(15%) 1 =$5,826(P/A, 15%, 8) =$26,143 Option 2:(j 0, 1), (j, 4), (j, 3) PW(15%) 2 =$5,130(P/F, 15%, 1) +$5,826(P/A, 15%, 4)(P/F, 15%, 1) +$5,857(P/A, 15%, 3)(P/F, 15%, 5) =$25,573

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Contemporary Engineering Economics, 4 th edition, © 2007 Option 3(j 0, 2), (j, 4), (j, 2) PW(15%) 3 =$5,116(P/A, 15%, 4)(P/F, 15%, 2) +$5,826(P/A, 15%, 4)(P/F, 15%, 2) +$6,151(P/A, 15%, 2)(P/F, 15%, 6) = $25,217minimum cost Option 4 (j 0, 3), (j, 5) PW(15%) 4 = $5,500(P/A, 15%, 3) +$5,897(P/A, 15%, 5)(P/F, 15%, 3) =$25,555 Example 14.5 continued

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Contemporary Engineering Economics, 4 th edition, © 2007 Example 14.5 continued Option 5:(j 0, 3), (j, 4), (j, 1) PW(15%) 5 = $5,500(P/A, 15%, 3) + $5,826(P/A, 15%, 4)(P/F, 15%, 3) + $7,500(P/F, 15%, 8) = $25,946 Option 6: (j 0, 4), (j, 4) PW(15%) 6 = $5,826(P/A, 15%, 4)(P/F, 15%, 4) + $5,826(P/A, 15%, 4)(P/F, 15%, 4) = $26,529

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Contemporary Engineering Economics, 4 th edition, © 2007 Graphical Representation of Replacement Strategies under a Finite Planning Horizon

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