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CHAPTER 11 PowerPoint Presentation

CHAPTER 11

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CHAPTER 11

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  1. Mc Graw Hill ENGINEERING ECONOMYSixth Edition Blank and Tarquin CHAPTER 11 Replacement and Retention Decisions

  2. Chapter 11 Learning Objectives • Basics of Replacement Study; • Economic Service Life; • Performing a Replacement Study; • Additional Considerations in a Replacement Study; • Replacement Study over a Specified Study Period; • Chapter Summary

  3. CHAPTER 11 Section 11.1 Basics of Replacement Study

  4. 11.1 Why Replace Assets • Reduced Performance: • Wear and Tear; • Decreasing reliability and Productivity; • Increasing operating and maintenance costs. • Altered Requirements: • New production needs, accuracy, speed, etc. • Obsolescence: • Current assets may be less productive; • Not state of the art – meet competition.

  5. 11.1 Terminology • Defender Asset: • Current installed asset; • Challenger Asset: • The potential replacement or “challenging” asset; • Under consideration to replace the defender asset. • Together, the Defender and Challenger: • Constitute mutually exclusive alternatives; • Select one and reject the other.

  6. 11.1 Annual Worth Values • Analysis Approach for Replacement: • Annual Worth Approach; • EUAC – since costs tend to dominate the study (-) cash flows; • Salvage values – if any – are also part of the analysis (+) value.

  7. 11.1 Economic Service Life • Economic Service Life (ESL) • Number of years for an alternative for which the AW or EUAC is Minimum; • Implies that a period by period analysis is performed; • Computing the AW for 1 year; then 2 years; … until a minimum cost time period is found; • Performed manually or by spreadsheet.

  8. 11.1 Investment Concerns-Defender • For a replacement analysis two investment costs are critical: • The proper investment cost to apply to keeping the defender in service; • The proper investment cost to apply to any challenger asset that might replace the current defender asset.

  9. 11.1 Investment Concerns- Defender • While it may seem strange to charge an investment cost for keeping one’s own asset (the defender) this is what must occur. • Keeping the defender is not free! • Why? • Because the firm is giving up the opportunity to receive a possible cash flow from selling the current defender!

  10. 11.1 Investment Concerns • One must assign an investment cost to KEEPING the defender asset! • The appropriate investment cost to assign to the defender asset is: • The current fair market value of the defender at the time the replacement decision is being examined.

  11. 11.1 Defender First Cost • Defender First Cost: • Initial investment in the defender asset back in time; • This investment (cost) is considered “sunk” for analysis purposes; • A past cost that cannot be changed or altered; • The issue of the relevance of the investment cost in the analysis will be addressed soon.

  12. 11.1 Challenger First Cost • This is the total investment (Pchallenger) required in a new (challenger) asset that will possibly replace the current defender. • In a replacement study this investment is know with a fair amount of certainty. • What IF a trade in value is offered for the defender to apply to the challenger?

  13. 11.1 Trade In Concerns • Often a trade in value is offered by a vendor to take in the defender with a credit on the purchase towards the challenger. • Be careful how this is handled! • Points to focus upon….

  14. 11.1 Basic Principles • The past investment in the defender is “sunk” and not totally relevant to the analysis. • Only the Fair Market Value (FMV) of the defender is relevant. • FMV of the defender is the net economic worth of the current defender; • Sale or disposal price less any costs associated with removing the defender.

  15. 11.1 Basic Principles – Important Question • At times, a “high” trade-in value may be offered for the defender compared to its current fair market value. • If this is the case: • What should be the investment cost in the challenger for a replacement study analysis if a trade-in value is offered?

  16. 11.1 Trade-In Issues • If a trade-in is offered what should be the proper investment cost in the challenger? • Let PC = the cash sale price of the challenger with no trade-in; • Let TIV = the trade-in value for the challenger as offered by the vendor (take in the defender); • Let MVD = the Fair Market Value of the current defender;

  17. 11.1 Trade-In Issues • For a Trade-In, the correct investment cost to assign to the challenger is: Investment in the Challenger: PC – (TIV – MVD)

  18. 11.1 Trade-In Issues • Investment in the challenger PC – (TIV – MVD) Cash Price for the challenger less: (Trade-in Value – Market Value of the Defender) This represents the true investment in the challenger to the firm!

  19. 11.1 Trade-In Issues • Investment in the challenger PC – (TIV – MVD) The Cash price for The challenger with No trade-in The Opportunity Cost Given up by not Selling the Defender outright!

  20. 11.1 Trade-In Issues - Example • Bought a system 3 years ago for $120,000. (Defender); • A fair market value of the current defender is $70,000 right now; • A challenger can be purchased for cash for $100,000 now! • The vendor selling the challenger offers a trade-in of $80,000 on the current defender.

  21. 11.1 Trade-In Issues - Example • What should be the proper investment cost for the challenger to the firm if the defender is traded? • PC = $100,000; • TIV = $80,000 • FMVD = $70,000 • InvestmentChallenger now = $100,000 – ($80,000 - $70,000) = $90,000. • This represents the “true” investment in the challenger with the trade.

  22. 11.1 Other Issues… • Investment in the challenger asset must include: • Actual cash price for purchase; • Transportation costs; • Installing/make-ready for use costs; • Other one-time costs at time t = 0 associated with placing the challenger in-service.

  23. 11.1 Warning! What Not to Do! • At times a decision maker might do the following: • Take the investment cost in the challenger; • Then add the remaining book value of the defender to that investment; • This is wrong! • Overly penalizes the challenger with a sunk cost associated with the defender asset! • Do not do this!

  24. 11.1 Sunk Costs • A “sunk” cost is any cost that has occurred in the past and cannot be changed or altered by a current decision. • The past investment in any defender or its remaining book value is not relevant! • Unless, an after-tax replacement analysis is being conducted!

  25. 11.1 Allocating Costs • Allocate only those costs associated with: • The Defender (opportunity cost of not disposing – giving up a salvage value if kept.) • Those costs associated with obtaining the challenger – do not penalize the challenger with costs from the defender!

  26. 11.1 The Suggested Viewpoint • It is customary to take the following philosophical approach: • This is termed: • The Outsider’s view or, • The Consultant’s view. One assumes that the analyst is an outsider to the firm and owns neither the defender or the challenger!

  27. 11.1 Outsider’s or Consultant’s View • One assumes that you own neither of the assets in question; • The service provided by the defender can be “purchased” with an investment of the firm’s money equal to the current fair market value of the defender. • Buy your own asset! • Hard to comprehend? Perhaps…but this is an objective approach to costing the defender!

  28. 11.1 Costing the Defender • The consultant’s view assumes: • If the defender is retained in service, the firm is giving up (forgone opportunity) a potential cash inflow – IF NOT REPLACED now! • This view attempts to minimize any bias towards either the defender or the challenger!

  29. 11.1 Replacement Approach • The traditional approach to conducting a replacement analysis is: • The Annual Cost or Annual Worth approach! • With an assumed interest rate; • Assumed lives for each alternative.

  30. 11.1 Assumptions • The traditional approach to conducting a replacement analysis is: • The Annual Cost or Annual Worth approach! • With an assumed interest rate; • Assumed lives for each alternative.

  31. 11.1 Study Period for Replacement • The study Period for a replacement problem may be: • Finite or, • Considered infinite.

  32. 11.1 Study Period for Replacement • If Infinite then: • The required services needed are needed indefinitely; • The challenger is the best available and if selected will have the same repeated cycles of costs forever! • Cost estimates for every future cycle will be the same!

  33. 11.1 Study Period for Replacement • Study Period Finite: • The previous assumptions do not hold! • See Section 11.5 for a fixed study period analysis.

  34. Section 11.2 Economic Service Life (ESL) • The best value for “n” is not known in this type of problem. • The ESL for a given asset is: • The number of years where the AW of the future costs is minimum; • Using the cost estimates of all possible years that the asset may provide a needed service! • Termed, “The minimum cost life”

  35. 11.2 ESL Analysis • One estimates the ESL for the challenger and, • The ESL for the current defender. • Requires estimates of future operating and maintenance costs and any salvage value. • Apply an Annual Worth Analysis; • For assumed values of n = { 1, 2, … }

  36. 11.2 ESL – General Format • Compute: • AW(i%)t =: • -Capital Recovery • - AW of operating costs • Salvage values may be incorporated into the capital recovery term. • Do this for n = 1 then n = 2, then n = … and observe the min cost “n” value.

  37. 11.2 Minimum Cost Life • The minimum cost life is: • That value of “n” that yields the lowest annual cost over the range of “n” values applied. • Capital Recover topic: See Chapter 6, section 2 to review.

  38. Sn 0 1 2 . . . n-1 n / / / / Investment at t = 0 (P) 11.2 Components of ESL • Capital Recovery Costs (CRC) • CRS’s generally decrease with each year of operation; • The longer one uses an asset the costs associated with owning the asset are spread out over more and more time periods. Diagram for Capital Recovery

  39. Sn 0 1 2 . . . n-1 n / / / / $P 11.2 Capital Recovery Formula • CRC Setup CRC(i%) = -P(A/P,i%,n) + S(A/F,i%,n) CRC(i%) is the annual cost of “owing” an asset over “n” time periods

  40. 11.2 Annual Operating Cost Component • Annual Operating Costs (AOC); • End-of-year estimated costs of operating the asset in question. • AOC’s tend to increase over time; • One wants to distribute the AOC over a range of assumed number of years; • “n” = {1, then 2, then 3, …. }

  41. 11.2 Plotting ESL • The ESL can be visualized by plotting three curve forms: • 1. Plot the CRC’s over assumed values of “n”; • 2. Plot the AOC’s over the same assumed values of “n”; • Plot the sum of the CRC and AOC over the same assumed values of “n” (Total AW of AOC’s) • Examine the AW plot to observe the minimum cost life of the respective asset.

  42. 11.2 Typical ESL Plot Min. Total AW of costs life

  43. 11.2 AW over “k” Years • Notation: • P = initial investment in the asset; • Sk = estimated salvage value after “k” years; • AOCj = annual operating costs for year j (j = 1 to k) • “k” the number of years for the analysis.

  44. 11.2 Closed Form of AWk • Analytical Form for Total AWk: Procedure: Year-by-year analysis for “k” years – where “k” is given or assumed.

  45. 11.2 Example 11.2 - Overview • Defender Asset; • 3 years old now; • Market value now: $13,000; • 5-year study period assumed; • Require Estimates of the future salvage values and annual operating costs for the 5-year period.

  46. 11.2 Example: Future Market Values • Estimated Future Market Values and AOC’s: MktVtAOCt • t = 1: $9000 $-2500 • t = 2: $8000 -2700 • t = 3: $6000 -3000 • t = 4: $2000 -3500 • t = 5: $0 -4500 Mkt. Values are decreasing: AOC’s are increasing: Assume the interest rate is 10% per year.

  47. S1 = $9000 0 1 AOC1 = -2500 P=$13,000 11.2 Example: Find the ESL • Period – by – period analysis • For “k” = 1 year: AW(10%)1 = (-$13,000)(A/P,10%,1) + $9000(A/F,10%,1) -2500 = -$7800 ( for one year!)

  48. 11.2 Example: Find the ESL • Period – by – period analysis • For “k” = 2 years: S2 = $8000 0 1 2 AOC1 = -2500 AOC2 = -$2700 P=$13,000 AW(10%)2 = (-13,000)(A/P,10%,2) + 8000(A/F,10%,2) -[2500(P/F,10%,1) + 2700(P/F,10%,2)](A/P,10%,2) = -$6276/yr for 2 years.

  49. 11.2 Example: Find the ESL • Period – by – period analysis • For “k” = 3 years: S3 = $6000 0 1 2 3 AOC1 = -2500 AOC2 = -$2700 P=$13,000 AOC3 = -$3000 AW(10%)3 = (-13,000)(A/P,10%,3) +6000(A/F,10%,3) -[2500(P/F,10%,1) + 2700(P/F,10%,2) + 3000(P/F,10%,3](A/P,10%,3) = -$6132/yr for 3 years.

  50. 11.2 Example - continued • A similar analysis for k = 4 and 5 is conducted; • The AW(10)k, K = {1,2,3,4,5} are tabulated as: Total AWk k=1: -7800 k=2: -6276 k=3: -6132 k=4: -6556 k=5: -6579 Min. Cost Year = 3 years