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Chapter 15 Replacement Decisions

Chapter 15 Replacement Decisions

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Chapter 15 Replacement Decisions

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

  2. 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 Replacement Terminology (c) 2001 Contemporary Engineering Economics

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

  4. 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. Replacement Decisions (c) 2001 Contemporary Engineering Economics

  5. Replacement Analysis – Cash Flow Approach Sales proceeds from defender $10,000 $5500 $2500 0 1 2 3 0 1 2 3 $6000 $8000 (a) Defender (b) Challenger $15,000 (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! (c) 2001 Contemporary Engineering Economics

  7. Opportunity Cost Approach Challenger Defender $5500 $2500 0 1 2 3 0 1 2 3 $6000 $8000 $10,000 Proceeds from sale viewed as an opportunity cost of keeping the asset $15,000 (c) 2001 Contemporary Engineering Economics

  8. Opportunity Cost Approach  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 Replace the defender now! (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. Minimize Ownership (Capital) Cost (init.+salvg.) + Operating cost (c) 2001 Contemporary Engineering Economics

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

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

  12. Economic Service Life Calculation (Example 15.4) • N = 1 AEC1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000 = $11,700 (c) 2001 Contemporary Engineering Economics

  13. N = 2 AEC2 = [$18,000 + $1,000(P/A,15%, 15%, 2)](A/P, 15%, 2) - $7,500 (A/F, 15%, 2) = $8,653 (c) 2001 Contemporary Engineering Economics

  14. N = 3, AEC3 = $7,406 N = 4, AEC4 = $6,678 N = 5, AEC5 = $6,642 N = 6, AEC6 = $6,258 N = 7, AEC7 = $6,394 Minimum cost Economic Service Life (c) 2001 Contemporary Engineering Economics

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

  16. Replacement Strategies under the Infinite Planning Horizon • 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 AEC* each year for an infinite number of years. • 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. (c) 2001 Contemporary Engineering Economics

  17. Example 15.5 • Defender: Find the remaining useful (economic) service life. (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 NC*=4 years AEC*=$5,826 (c) 2001 Contemporary Engineering Economics

  19. Replacement Decisions NC*=4 years AEC*=$5,826 • Should replace the defender now?No, because AED < AEC • If not, when is the best time to replace the defender? Need to conduct marginal analysis. (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 3rd year: $5,000 • Salvage value of the defender at the end of year 3: $2,000 (c) 2001 Contemporary Engineering Economics

  21. $2000 2 3 • 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 = $5,826 of the challenger. • Conclusion: Since keeping the defender for the 3rd year is more expensive than replacing it with the challenger, DO NOT keep the defender beyond its economic service life. $3000 $5000 2 3 $6,450 (c) 2001 Contemporary Engineering Economics

  22. Replacement Analysis under the Finite Planning Horizon Some likely replacement patterns under a finite planning horizon of 8 years (c) 2001 Contemporary Engineering Economics

  23. Example 15.6 Replacement Analysis under the Finite Planning Horizon (PW Approach) • Option 1: (j0, 0), (j, 4), (j, 4) PW(15%)1=$5,826(P/A, 15%, 8) =$26,143 • Option 2: (j0, 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 (c) 2001 Contemporary Engineering Economics

  24. Example 15.6 continued • Option 3 (j0, 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) = $25,217 minimum cost • Option 4 (j0, 3), (j, 5) PW(15%)4= $5,500(P/A, 15%, 3) +$5,897(P/A, 15%, 5)(P/F, 15%, 3) =$25,555 (c) 2001 Contemporary Engineering Economics

  25. Example 15.6 continued • Option 5: (j0, 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: (j0, 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 (c) 2001 Contemporary Engineering Economics

  26. Planning horizon = 8 years (j0, 0), (j, 4), (j, 4), (j0, 1), (j, 4), (j, 3), (j0, 2), (j, 4), (j, 2), (j0, 3), (j, 5), (j0, 3), (j, 4), (j, 1), (j0, 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 (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) (c) 2001 Contemporary Engineering Economics

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

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