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TM 661 Engineering Economics

TM 661 Engineering Economics. Replacement Analysis. Replacement / Challenge. Example Car grows older and needs repairs at engine overhaul time should we fix or replace?. Replacement / Challenge. Example Car grows older and needs repairs

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TM 661 Engineering Economics

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  1. TM 661 Engineering Economics Replacement Analysis

  2. Replacement / Challenge • Example Car grows older and needs repairs • at engine overhaul time should we fix or replace?

  3. Replacement / Challenge • Example Car grows older and needs repairs • at engine overhaul time should we fix or replace? • Note: sunk costs are unrecoverable • Example Just put $800 in car, engine needs overhaul, should we repair or replace? • The $800 just invested has no bearing number is not part of analysis.

  4. Example: Replacement Chemical Plant owns filter press purchased 3 years ago. Operating expense started at $4,000 per year 2 years ago and has increased by $1,000 per year. The press could last 5 more years with an estimated salvage of $2,000 at that time. Current market value of the press is $9,000. A new press can be purchased for $36,000 with an estimated life of 10 years. Annual operating costs are 0 in year 1 growing by $1,000 per year.

  5. Cash Flow Approach

  6. Keep 2,000 0 1 2 3 4 5 7,000 11,000 Replacement (Cash Flow)

  7. Keep Replace 12,000 9,000 2,000 0 1 2 3 4 5 0 1 2 3 4 5 7,000 1,000 36,000 11,000 4,000 Replacement (Cash Flow)

  8. Keep 2,000 0 1 2 3 4 5 7,000 11,000 Replacement (Cash Flow) • NPW = -7,000 (P/A, 15,5) • - 1,000 (P/G, 15, 5) • + 2,000 (P/F, 15, 5) • = ($28,246)

  9. Replace 12,000 9,000 0 1 2 3 4 5 1,000 36,000 4,000 Replacement (Cash Flow) NPW = 9,000 - 36,000 -1,000 (P/G, 15,5) + 12,000 (P/F, 15, 5) = ($26,809)

  10. Keep Replace 12,000 9,000 2,000 0 1 2 3 4 5 0 1 2 3 4 5 7,000 1,000 36,000 11,000 4,000 Replacement (Cash Flow) NPWK = ($28,246) NPWR = ($26,809)

  11. Keep Replace 12,000 9,000 2,000 0 1 2 3 4 5 0 1 2 3 4 5 7,000 1,000 36,000 11,000 4,000 Replacement (Cash Flow) NPWK = ($28,246) NPWR = ($26,809) Choose Replace

  12. Keep Replace 12,000 9,000 2,000 0 1 2 3 4 5 0 1 2 3 4 5 7,000 1,000 36,000 11,000 4,000 Replacement (Cash Flow) NPWK = ($28,246) NPWR = ($26,809) Note: NPWR - NPWK = $ 1,437

  13. Keep Replace 12,000 2,000 0 1 2 3 4 5 0 1 2 3 4 5 7,000 1,000 9,000 36,000 11,000 4,000 Replacement (Outsider View)

  14. Keep 2,000 0 1 2 3 4 5 7,000 9,000 11,000 Replacement (Outsider View) • NPW = - 9,000 • -7,000 (P/A, 15,5) • - 1,000 (P/G, 15, 5) • + 2,000 (P/F, 15, 5) • = ($37,246)

  15. Replace 12,000 0 1 2 3 4 5 1,000 36,000 4,000 Replacement (Outsider View) NPW = - 36,000 -1,000 (P/G, 15,5) + 12,000 (P/F, 15, 5) = ($35,809)

  16. Keep Replace 12,000 2,000 0 1 2 3 4 5 0 1 2 3 4 5 7,000 1,000 9,000 36,000 11,000 4,000 Replacement (Outsider View) NPWK = ($37,246) NPWR = ($35,809)

  17. Keep Replace 12,000 2,000 0 1 2 3 4 5 0 1 2 3 4 5 7,000 1,000 9,000 36,000 11,000 4,000 Replacement (Outsider View) NPWK = ($37,246) NPWR = ($35,809) Choose Replace

  18. Keep Replace 12,000 2,000 0 1 2 3 4 5 0 1 2 3 4 5 7,000 1,000 9,000 36,000 11,000 4,000 Replacement (Outsider View) NPWK = ($37,246) NPWR = ($35,809) Note: NPWR - NPWK = $ 1,437

  19. With 10 year Horizon Suppose we now consider a 10 year planning horizon. We estimate that the old press will still have a salvage value of $2,000 5 years from now but that the new press will only cost $31,000 5 years from now. Further, estimated salvage 5 years hence is $15,000. Then:

  20. With 10 Year Planning Horizon

  21. Keep 15,000 2,000 0 1 2 3 4 5 10 1,000 7,000 11,000 4,000 31,000 Replacement (Cash Flow) NPW = -7,000(P/A, 15,5) - 1,000(P/G,15,5) -29,000(P/F,15,5) -1,000(P/A,15,5)(P/F,15,5) + 12,000(P/F,15,10) = ($42,821)

  22. Replace 3,000 9,000 0 1 2 3 4 5 . . . 10 . . . . 1,000 36,000 4,000 9,000 Replacement (Cash Flow) NPW = -27,000 - 1,000(P/G,15,10) + 3,000(P/F,15,10) = ($43,237)

  23. Replace Keep 15,000 3,000 9,000 2,000 0 1 2 3 4 5 10 0 1 2 3 4 5 . . . 10 1,000 . . . . 7,000 1,000 11,000 4,000 36,000 4,000 9,000 31,000 10-Year Horizon NPWK = (42,821) NPWR = (43,237) Choose Keep, trade in 5 years

  24. Multiple Alternatives Suppose Dealer offers a $10,000 trade-in. In addition, we identify 2 new alternatives: 3. New press for $40,000 with salvage after 5 years of $13,000. Trade-in on this machine is $12,000. 4. Lease a press for $7,500 per year during the 5 year horizon. Existing press will be sold on the open market.

  25. Trade - In / Lease Options

  26. Outsider Viewpoint Approach

  27. Optimal Replacement Suppose we have a compressor which costs $2,000 and has annual maintenance costs of $500 increasing by $100 per year. MARR=20%. Then:

  28. Optimal Replacement

  29. Optimal Replacement

  30. Class Problem The new president of Angstrom Technologies feels the company must use the newest and finest equipment in its labs. He has recommended that a 2-year-old piece of precision measurement equipment be replaced immediately. Besides, he feels it can be shown that his proposed equipment is economically advantageous at a 15%-per-year return and a planning horizon of 5 years. Perform the replacement analysis for a 5-year period.

  31. Check Out Replacement Excel File

  32. Class Problem • Current Proposed • Original purchase price $30,000 $40,000 • Current market value 15,000 ... • Estimated useful life, years 5 15 • Estimated value, 5 years $7,000 $10,000 • Salvage after 15 years ... 5,000 • Annual operating cost 5,000 3,000

  33. 7,000 15,000 10,000 0 1 2 3 4 5 0 1 2 3 4 5 3,000 5,000 40,000 Solution Keep Replace EUAW = -25,000(A/P,15,5) -3,000 + 10,000(A/F,15,5) = ($8,975) EUAW = -5,000 + 7,000(A/F,15,5) = ($3,962)

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