1 / 31

ENGM 661

ENGM 661. Replacement Analysis. Learning Objectives.

morrison
Download Presentation

ENGM 661

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ENGM 661 Replacement Analysis

  2. Learning Objectives Chapter 11 - Replacement Given a cash flow profile for current and replacement equipment, be able to use either the cash flow or the outsider viewpoint to determine the better alternative. Given a cash flow profile, be able to determine the optimal replacement interval for a piece of equipment.

  3. Replacement decisions occur for a variety of reasons, including: • Current asset, which we call the defender, has developed several deficiencies, • Potential replacement assets, which we call the challengers, are available which have a number of advantages over the defender • Changing external environment, including • customer preferences and expectations • requirements • new, alternative ways of obtaining the functionality provided by the defender • increased demand that cannot be met with the current equipment

  4. Two approaches for replacement analyses: • the cash flow approach or insider approach • opportunity cost approach or outsider viewpoint approach. If performed correctly, the two approaches will yield the same recommendation. (The essential difference in the two relates to how the salvage value of the defender is treated.)

  5. As equipment ages, O&M costs increase and CR cost decreases.

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

  7. 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.

  8. 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.

  9. Cash Flow Approach

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

  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)

  12. 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)

  13. 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)

  14. 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)

  15. 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

  16. 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

  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)

  18. 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)

  19. 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)

  20. 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)

  21. 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

  22. 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

  23. 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:

  24. Optimal Replacement

  25. Optimal Replacement

  26. ESL Example An existing process has a market value of $15 000, but to provide the required physical service level for the next 5 years will require an immediate overhaul for $4 000. The salvage value at the end of the current year is estimated to be $1 100, and that will decrease by $100/year thereafter. Operating and Maintenance costs are currently $4 000 yearly, but these costs will increase by $2 100/year after this year. Determine the economic service life and resultant total EAC of the asset if the MARR is 8%, compounded annually.

  27. DIAGRAM: $1 100 – $100(n-1) 1 2 3 n = 1, 2, … 5 yrs 0 $4 000 $4 000 $2 100 $15 000 (19 000) (1.0800) (0) (1.0000) =$15 184 (.5608) (.4808) (.4808) =$13 087 (.3880) (.9487) (.3080) =$12 507 (.3019) (1.4040) (.2219) =$12 518 (.2505) (1.8465) (.1705) ESL = 4 yrs, this is the EACDefender if used through the end ofthe 4th year! Finding EAC for lifetimes of 1 through 5 years: EAC1=(4000+15000)(A/P,8%,1)+4000+2100(A/G,8%,1) -1100(A/F,8%,1) =$23 420 EAC2=(19 000)(A/P,8%,2)+4000+2100(A/G,8%,2) -1000(A/F,8%,2) EAC3=(19 000)(A/P,8%,3)+4000+2100(A/G,8%,3) -900(A/F,8%,3) EAC4=(19 000)(A/P,8%,4)+4000+2100(A/G,8%,4) -800(A/F,8%,4) EAC5=(19 000)(A/P,8%,5)+4000+2100(A/G,8%,5) -700(A/F,8%,5)

  28. ESL Example w/ Challenger The Best Challenger process to replace the Defender has a 1st cost of $65 000, and an expected physical lifetime of 15 years. The same MARR of 8% (compounded annually) has been used to run an ESL study on the Challenger using the best data available today, and it looks like the ESL is 8 years. For the 8 year Challenger ESL, the operating and maintenance costs are expected to be $3 500 yearly; and the salvage value at the end of the 8 years is estimated to be $5 000. Determine how to replace the existing process.

  29. Defender: $800 1 2 3 n = 4 yrs 0 $4 000 $4 000 $2 100 $15 000 Challenger: $5 000 1 2 3 n = 8 yrs 0 $3 500 $65 000 Comparing Best Challenger Opportunity Cost View for “Outsider”:

  30. Challenger: $5 000 1 2 3 n = 8 yrs 0 Minimum EACDefender $3 500 $65 000 Ownership Life – b/c less than EACChallenger Finding EAC for Challenger w/ lifetime of 8 years: EACChallenger 8=(65 000)(A/P,8%,8) + 3 500 - 5 000(A/F,8%,8) =$14 340 (.1740) (.0940) Choose to keep Defender today, plan to replace defender at start of year 6 (but recheck next year to see if anything has changed!)

  31. References: • Principles of Engineering Economics 6th Edition (White, Case, Pratt) • D. H Jensen & K.D. Douglas • Dr. Stuart Kellogg

More Related