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

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

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

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

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

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

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

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

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

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

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

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

18. Replacement (Outsider View) 0 1 2 3 4 5 7,000 11,000 2,000 Keep 9,000 0 1 2 3 4 5 4,000 12,000 Replace 36,000 1,000 NPW K = ($37,246) NPW R = ($35,809) Note: NPW R - NPW K = $ 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. Replacement (Cash Flow) 0 1 2 3 4 5 10 7,000 11,000 2,000 Keep 31,000 1,000 4,000 15,000 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. Replacement (Cash Flow) 0 1 2 3 4 5... 10 4,000 Replace 36,000 9,000 1,000 9,000 3,000.. NPW= -27,000 - 1,000(P/G,15,10) + 3,000(P/F,15,10) = ($43,237)

23. 10-Year Horizon 0 1 2 3 4 5... 10 4,000 Replace 36,000 9,000 1,000 9,000 3,000.. 0 1 2 3 4 5 10 7,000 11,000 2,000 Keep 31,000 1,000 4,000 15,000 NPW K = (42,821)NPW R = (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

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 value15,000... Estimated useful life, years 5 15 Estimated value, 5 years$7,000$10,000 Salvage after 15 years...5,000 Annual operating cost5,0003,000

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