1. (c) 2001 Contemporary Engineering Economics 1 Chapter 15 Replacement Decisions Replacement Analysis Fundamentals Economic Service Life Replacement Analysis When Required Service is Long Replacement Analysis with Tax Consideration

2. (c) 2001 Contemporary Engineering Economics 2 Replacement Terminology Sunk cost: any past cost unaffected by any future decisions Trade-in allowance: value offered by the vendor to reduce the price of a new equipment Operating Cost Defender: an old machine Challenger: new machine Current market value: selling price of the defender in the market place

3. (c) 2001 Contemporary Engineering Economics 3 Sunk Cost associated with an Assets Disposal $0 $5000 $10,000 $15,000 $20,000 $25,000 $30,000 Original investment $10,000$5000 Market value $10,000 Lost investment (economic depreciation) Repair cost $20,000 Sunk costs = $15,000

4. (c) 2001 Contemporary Engineering Economics 4 Replacement Decisions Cash Flow Approach –Treat the proceeds from sale of the old machine as down payment toward purchasing the new machine. –Can be used in the analysis period is same for all alternatives. –Use NPW or AE analysis to decide Opportunity Cost Approach –Treat the proceeds from sale of the old machine as the investment required to keep the old machine.

5. (c) 2001 Contemporary Engineering Economics 5 Replacement Analysis – Cash Flow Approach 01230123 01230123 $8000 $2500 $15,000 $6000 $5500 (a) Defender(b) Challenger $10,000 Sales proceeds from defender

6. (c) 2001 Contemporary Engineering Economics 6 Annual Equivalent Cost - Cash Flow Approach Defender: PW(12%) D = $2,500 (P/F, 12%, 3) - $8,000 (P/A, 12%, 3) = - $17,434.90 AE(12%) D = PW(12%) D (A/P, 12%, 3) = -$7,259.10 Challenger: PW(12%) C = $5,500 (P/F, 12%, 3) - $5,000 - $6,000 (P/A, 12%, 3) = -$15,495.90 AE(12%) C = PW(12%) C (A/P, 12%, 3) = -$6,451.79 Replace the defender now!

7. (c) 2001 Contemporary Engineering Economics 7 Opportunity Cost Approach 01230123 01230123 $8000 $2500 $15,000 $6000 $5500Defender Challenger $10,000 Proceeds from sale viewed as an opportunity cost of keeping the asset

8. (c) 2001 Contemporary Engineering Economics 8 Defender: PW(12%) D = -$10,000 - $8,000(P/A, 12%, 3) + $2,500(P/F, 12%, 3) = -$27,434.90 AE(12%) D = PW(12%) D (A/P, 12%, 3) = -$11,422.64 Challenger: PW(12%) C = -$15,000 - $6,000(P/A, 12%, 3) + $5,500(P/F, 12%, 3) = -$25,495.90 AE(12%) C = PW(12%) C (A/P, 12%, 3) = -$10,615.33 Opportunity Cost Approach Replace the defender now!

9. (c) 2001 Contemporary Engineering Economics 9 Economic Service Life Def:Economic service life is the useful life of a defender, or a challenger, that results in the minimum equivalent annual cost Why do we need it?: We should use the respective economic service lives of the defender and the challenger when conducting a replacement analysis. Ownership (Capital) Cost (init.+salvg.) Operating cost + Minimize

10. (c) 2001 Contemporary Engineering Economics 10 Mathematical Relationship Capital Recov. Cost. Operating Cost: Total Cost: Objective: Find n* that minimizes AEC CR(i) OC(i) AEC n*n*

11. (c) 2001 Contemporary Engineering Economics 11 Economic Service Life for a Lift Truck

12. (c) 2001 Contemporary Engineering Economics 12 Economic Service Life Calculation (Example 15.4) N = 1 AEC 1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000 = $11,700

13. (c) 2001 Contemporary Engineering Economics 13 N = 2 AEC 2 = [$18,000 + $1,000(P/A,15%, 15%, 2)](A/P, 15%, 2) - $7,500 (A/F, 15%, 2) = $8,653

14. (c) 2001 Contemporary Engineering Economics 14 N = 3, AEC 3 = $7,406 N = 4, AEC 4 = $6,678 N = 5, AEC 5 = $6,642 N = 6, AEC 6 = $6,258 N = 7, AEC 7 = $6,394 Minimum cost Economic Service Life

15. (c) 2001 Contemporary Engineering Economics 15 Required Assumptions and Decision Frameworks Planning horizon (study period) Technology Relevant cash flow information Decision Frameworks

16. (c) 2001 Contemporary Engineering Economics 16 Replacement Strategies under the Infinite Planning Horizon 1.Replace the defender now: The cash flows of the challenger will be used from today and will be repeated because an identical challenger will be used if replacement becomes necessary again in the future. This stream of cash flows is equivalent to a cash flow of AE 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.

17. (c) 2001 Contemporary Engineering Economics 17 Example 15.5 Defender: Find the remaining useful (economic) service life.

18. (c) 2001 Contemporary Engineering Economics 18 Challenger: find the economic service life. 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

19. (c) 2001 Contemporary Engineering Economics 19 Replacement Decisions Should replace the defender now? No, because AE D < AE C If not, when is the best time to replace the defender? Need to conduct marginal analysis. N C* =4 years AE C* =$5,826

20. (c) 2001 Contemporary Engineering Economics 20 Marginal Analysis 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

21. (c) 2001 Contemporary Engineering Economics 21 2 3 $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 AE 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

22. (c) 2001 Contemporary Engineering Economics 22 Replacement Analysis under the Finite Planning Horizon Annual Equivalent Cost ($) NDefenderChallenger 15,1307,500 25,1166,151 35,5005,857 45,9615,826 56,4345,897 Some likely replacement patterns under a finite planning horizon of 8 years

23. (c) 2001 Contemporary Engineering Economics 23 Example 15.6 Replacement Analysis under the Finite Planning Horizon (PW Approach) 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

24. (c) 2001 Contemporary Engineering Economics 24 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 15.6 continued

25. (c) 2001 Contemporary Engineering Economics 25 Example 15.6 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

26. (c) 2001 Contemporary Engineering Economics 26 (j 0, 0), (j, 4), (j, 4), (j 0, 1), (j, 4), (j, 3), (j 0, 2), (j, 4), (j, 2), (j 0, 3), (j, 5), (j 0, 3), (j, 4), (j, 1), (j 0, 4), (j, 4), Option 1 Option2 Option 3 Option 4 Option 5 Option 6 0 1 2 3 4 5 6 7 8 Years in service Planning horizon = 8 years

27. (c) 2001 Contemporary Engineering Economics 27 Replacement Analysis with Tax Consideration Whenever possible, replacement decisions should be based on the cash flows after taxes. (Example 15.8) When computing the net proceeds from sale of the old asset, any gains or losses must be identified to determine the correct amount of the opportunity cost. (Example 15.7) All basic replacement decision rules including the way of computing economic service life remain unchanged. (Example 15.10)

28. (c) 2001 Contemporary Engineering Economics 28 $0 $4000 $8000 $12,000 $16,000 $20,000 $10,000 $14,693 $20,000 $4693 Depreciation basis Book value Total depreciation $5307 Market value Loss tax credit Book loss Net proceeds from disposal ($11,877)

29. (c) 2001 Contemporary Engineering Economics 29 Summary In replacement analysis, the defender is an existing asset; the challenger is the best available replacement candidate. The current market value is the value to use in preparing a defenders economic analysis. Sunk costspast costs that cannot be changed by any future investment decisionshould not be considered in a defenders economic analysis.

30. (c) 2001 Contemporary Engineering Economics 30 Two basic approaches to analyzing replacement problems are the cash flow approach and the opportunity cost approach. –The cash flow approach explicitly considers the actual cash flow consequences for each replacement alternative as they occur. –The opportunity cost approach views the net proceeds from sale of the defender as an opportunity cost of keeping the defender.

31. (c) 2001 Contemporary Engineering Economics 31 Economic service life is the remaining useful life of a defender, or a challenger, that results in the minimum equivalent annual cost or maximum annual equivalent revenue. We should use the respective economic service lives of the defender and the challenger when conducting a replacement analysis. Ultimately, in replacement analysis, the question is not whether to replace the defender, but when to do so. The AE method provides a marginal basis on which to make a year-by-year decision about the best time to replace the defender. As a general decision criterion, the PW method provides a more direct solution to a variety of replacement problems, with either an infinite or a finite planning horizon, or a technological change in a future challenger.

32. (c) 2001 Contemporary Engineering Economics 32 The role of technological change in asset improvement should be weighed in making long-term replacement plans Whenever possible, all replacement decisions should be based on the cash flows after taxes.