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M c. Graw. Hill. ENGINEERING ECONOMY Sixth Edition. Blank and Tarquin. CHAPTER 11. Replacement and Retention Decisions. Chapter 11 Learning Objectives. Basics of Replacement Study; Economic Service Life; Performing a Replacement Study; Additional Considerations in a Replacement Study;
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Mc Graw Hill ENGINEERING ECONOMYSixth Edition Blank and Tarquin CHAPTER 11 Replacement and Retention Decisions
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
CHAPTER 11 Section 11.1 Basics of Replacement Study
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.
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.
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.
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.
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.
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!
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.
11.1 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.
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?
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….
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.
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?
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;
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)
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!
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!
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.
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.
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.
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!
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!
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!
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!
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!
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!
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.
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.
11.1 Study Period for Replacement • The study Period for a replacement problem may be: • Finite or, • Considered infinite.
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!
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.
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”
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, … }
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.
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.
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
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
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, …. }
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.
11.2 Typical ESL Plot Min. Total AW of costs life
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.
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.
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.
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.
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!)
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.
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.
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