Replacement and Retention Decisions

1. Replacement and Retention Decisions Chapter 11

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3. 3 Replacement or Retention? One of the most commonly performed engineering economy studies is that of replacement or retention of an asset or system that is currently installed. This differs from previous studies where all the alternatives are new. The fundamental question answered by a replacement study about a currently installed asset or system is: “Should it be replaced now or later?” If the decision is to replace, the study is complete. If the decision is to retain, the cost estimates and decision will be revisited each year. A replacement study is an application of the AW method of comparing unequal-life alternatives, first introduced in Chapter 6. In a replacement study with no specified time period, the AW values are determined by a technique of cost evaluation called the economic service life (ESL) analysis.

4. 4 Basics The need for a replacement study can develop from several sources: Reduced performance Physical deterioration, reduced reliability or productivity. This usually results in increased costs of operation, higher scrap and rework costs, lost sales, reduced quality, diminished safety, and larger maintenance expenses. Altered requirements New requirements of accuracy, speed, or other specifications cannot be met by the existing equipment or system (complete replacement, retrofitting or augmentation). Obsolescence International competition and rapidly changing technology make currently used systems and assets perform acceptably but less productively than equipment coming available. The decreased development cycle time to bring new products to market is often the reason for premature replacement studies, that is, studies performed before the estimated useful or economic life is reached.

5. 5 The Arkansas Division of ADM, a large agricultural products corporation, purchased a state-of-the-art ground-leveling system for rice field preparation 3 years ago for $120,000. When purchased, it had an expected service life of 10 years, an estimated salvage of $25,000 after 10 years, and AOC of $30,000. Current account book value is $80,000. The system is deteriorating rapidly; 3 more years of use and then salvaging it for $10,000 on the international used farm equipment network are now the expectations. The AOC is averaging $30,000. A substantially improved, laser-guided model is offered today for $100,000 with a trade-in of $70,000 for the current system. The price goes up next week to $110,000 with a trade-in of $70,000. The ADM division engineer estimates the laser-guided system to have a useful life of 10 years, a salvage of $20,000, and an AOC of 20,000. A $70,000 market value appraisal of the current system was made today. If no further analysis is made on the estimates, state the correct values to include if the replacement study is performed today.

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7. 7 Definitions Defender and challenger are the names for two mutually exclusive alternatives. The defender is the currently installed asset and the challenger is the potential replacement. A replacement study compares these two alternatives. AW values are used as the primary economic measure of comparison between the defender and challenger. The term EUAC (equivalent uniform annual cost) may be used instead of AW, often only costs are included in the evaluation; revenues generated by the defender or challenger are assumed to be equal. Since the equivalence calculations for EUAC are exactly the same as for AW, we use the term AW. Therefore, all values will be negative when only costs are involved. Salvage value is an exception; it is a cash inflow (positive).

8. First Costs Defender first cost is the initial investment amount (P) used for the defender. The current market value (MV) is the correct estimate to use for P for the defender in a replacement study. The fair market value may be obtained from professional appraisers, resellers, or liquidators who know the value of used assets. The estimated salvage value at the end of 1 year becomes the market value at the beginning of the next year, provided the estimates remain correct as the years pass. It is incorrect to use trade-in value that does not represent a FMV, or the depreciated book value taken from accounting records as MV for the defender first cost. If the defender must be upgraded or augmented to make it equivalent to the challenger (in speed, capacity, etc.), this cost is added to the MV to obtain the estimate of defender first cost. In the case of asset augmentation for the defender alternative, this separate asset and its estimates are included along with the installed asset estimates to form the complete defender alternative. This alternative is then compared with the challenger via a replacement study. Challenger first cost is the amount of capital that must be recovered (amortized) when replacing a defender with a challenger. This amount is almost always equal to P, the first cost of the challenger. On occasion, an unrealistically high trade-in value may be offered for the defender compared to its fair market value. In this event, the net cash flow required for the challenger is reduced, and this fact should be considered in the analysis. The correct amount to recover and use in the economic analysis for the challenger is its first cost minus the difference between the trade-in value (TIV) and market value (MV) of the defender. In equation form, this is P- (TIV - MV). This amount represents the actual cost to the company because it includes both the opportunity cost (market value of the defender) and the out-of-pocket cost (first cost minus trade-in) to acquire the challenger. Of course, when the trade-in and market values are the same, the challenger Pvalue is used in all computations.

9. 9 Economic Service Life The economic service life (ESL) is the number of years n that the equivalent uniform annual worth (AW) of costs is the minimum, considering the most current cost estimates over all possible years that the asset may provide a needed service. The ESL is also referred to as the economic life or minimum cost life.

10. 10 Calculating ESL The ESL is determined by calculating the total AW of costs if the asset is in service 1 year, 2 years, 3 years, and so on, up to the last year the asset is considered useful. Total AW of costs is the sum of capital recovery (CR), which is the AW of the initial investment and any salvage value, and the AW of the estimated annual operating cost (AOC), that is:

11. 11 Determining Economic Service Life (ESL)

12. 12 Decreasing cost of capital recovery The capital recovery is the AW of investment; it decreases with each year of ownership. Capital recovery is calculated using Equation [11.2], the salvage value S that usually decreases with time is the estimated market value (MV) in that year.

13. 13 A 3-year-old manufacturing process asset is being considered for early replacement. Its current market value is $13,000. Estimated future market values and annual operating costs for the next 5 years are given in Table 11–1, columns 2 and 3. What is the economic service life of this defender if the interest rate is 10% per year?

14. 14 Increasing cost of AW of AOC Since the AOC estimates usually increase over the years, the AW of AOC increases. To calculate the AW of the AOC series for the given term, determine the present worth of each AOC value with the P/F factor, then redistribute this P value over the years of ownership, using the A/P factor. The complete equation for total AW of costs over k years is:

15. 15 Solution Equation [11.3] is used to calculate total AWk for k = 1, 2, . . . , 5. Table 11–1, column 4, shows the capital recovery for the $13,000 current market value ( j = 0) plus 10% return. Column 5 gives the equivalent AW of AOC for k years. As an illustration, the computation of total AW for k = 3 from Equation [11.3] is: Total AW3 = -P(A/P, i, 3) + MV3(A/F, i, 3) - [PW of AOC1,AOC2, AOC3](A/P, i, 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) = - 3415 - 2717 = - $6132 A similar computation is performed for each year 1 through 5. The lowest equivalent cost (numerically largest AW value) occurs at k = 3. Therefore, the defender ESL is n = 3 years, and the AW value is - $6132. In the replacement study, this AW will be compared with the best challenger AW determined by a similar ESL analysis.

16. 16 It is reasonable to ask about the difference between the ESL analysis and the AW analyses performed in previous chapters. In all previous cases, there were no year-by-year MV estimates applicable over the years. When the expected life n is known for the challenger or defender, determine its AW over n years, using the first cost or current market value, estimated salvage value after n years, and AOC estimates. This AW value is the correct one to use in the replacement study.

17. 17 An engineer has determined that a 3-year-old manufacturing process asset has a market value of $13,000 now, and the estimated salvage/market values and AOC values shown in Table 11–1. Determine the AW of the marginal cost values by computer, and compare it with the total AW values in Figure 11–2. Use the marginal cost series to determine the correct values for n and AW if the asset is the defender in a replacement study.

18. 18 Marginal Costs Marginal costs (MC) are year-by-year estimates of the costs to own and operate an asset for that year. There are 3 components to each annual marginal cost estimate: Cost of ownership (loss in market value is the best estimate of this cost). Forgone interest on the market value at the beginning of the year. AOC for each year.

19. Comparison of Total AW and AW of Marginal Costs

20. 20 Performing A Replacement Study Replacement studies are performed in one of two ways: without a study period specified or with one defined. The following figure gives an overview of the approach taken for each situation. The procedure discussed in this section applies when no study period (planning horizon) is specified. A replacement study determines when a challenger replaces the existing defender. The complete study is finished if the challenger (C) is selected to replace the defender (D) now. If the defender is retained now, the study may extend over a number of years equal to the life of the defender nD.

21. 21 Replacement and Retention Decisions

22. 22 Two years ago, Toshiba Electronics made a $15 million investment in new assembly line machinery. It purchased approximately 200 units at $70,000 each and placed them in plants in 10 different countries. The equipment sorts and tests electronic components for printed circuit boards. This year, new international industry standards will require a $16,000 retro-fit on each unit, in addition to the expected operating cost. Due to the new standards, coupled with rapidly changing technology, a new system is challenging the retention of these 2-year-old machines. The chief engineer at Toshiba USA realizes that the economics must be considered, so he has asked that a replacement study be performed this year and each year in the future. Given the estimates on the following slide: determine the AW values and economic service lives necessary to perform the replacement study.

23. The results of the ESL analysis (Table 11–2) include the MV and AOC estimates for the challenger in part (a) of the table. Note that P = $50,000 is also the MV in year 0. The total AW of costs is for each year, should the challenger be placed into service for that number of years. As an example, the year k = 4 amount of -$19,123 is determined using Equation [11.3]. The A/G factor is applied instead of the P/F and A/P factors to find the AW of the arithmetic gradient series in the AOC. 23

24. Other Considerations in a Replacement Study There are several additional aspects of a replacement study that may be introduced. Future-year replacement decisions at the time of the initial replacement study. Opportunity-cost versus cash-flow approaches to alternative comparison. Anticipation of improved future challengers. In most cases when management initiates a replacement study, the question is best framed as, “Replace now, 1 year from now, 2 years from now, etc.?” The procedure answers this question provided the estimates for C and D do not change with time. If estimates change over time, the decision to retain the defender may be prematurely reversed in favor of a better challenger. The first costs (P values) for the challenger and defender have been correctly taken as the initial investment for the challenger C and current market value for the defender D. This is called the opportunity-cost approach because it recognizes that a cash inflow of funds equal to the market value is forgone if the defender is selected. 24

25. Replacement study information for Amoco Canada’s oil field equipment placed into service 5 years ago: The current equipment will have to serve for either 2, 3 or 4 more years before replacement. The equipment has a current market value of $100,000; expected to decrease by $25,000 per year. The AOC is constant now and is expected to remain so, at $25,000 per year. The replacement challenger is a fixed-price contract to provide the same services at $60,000 per year for a minimum of 2 years and a maximum of 5 years. Use MARR of 12% per year to perform a replacement study over a 6-year period to determine when to sell the current equipment and purchase the contract services. 25

26. Solution Since the defender will be retained for 2, 3 or 4 years, there are three viable options (X,Y and Z). The defender annual worth values are identified with subscripts D2, D3, and D4 for the number of years retained. AWD2 = - 100,000(A/P,12%,2) + 50,000(A/F,12%,2) - 25,000 = - $60,585 AWD3 = -100,000(A/P,12%,3) + 25,000(A/F,12%,3) - 25,000 = - $59,226 AWD4 = -100,000(A/P,12%,4) - 25,000 = - $57,923 For all options, the challenger has an annual worth of AWC = - $60,000 26

27. Solution - continued Table 11–3 presents the cash flows and PW values for each option over the 6-year study period. A sample PW computation for option Y is: PWY = - 59,226(P/A,12%,3) - 60,000(F/A,12%,3)(P/F,12%,6) = - $244,817 Option Z has the lowest cost PW value ($240,369). Keep the defender all 4 years, then replace it. 27

28. 28 Summary It is important in a replacement study to compare the best challenger with the defender. Best (economic) challenger is described as the one with the lowest annual worth (AW) of costs for some period of years. If the expected remaining life of the defender and the estimated life of the challenger are specified, the AW values over these years are determined and the replacement study proceeds. However, if reasonable estimates of the expected market value (MV) and AOC for each year of ownership can be made, these year-by-year (marginal) costs help determine the best challenger. The economic service life (ESL) analysis is designed to determine the best challenger’s years of service and the resulting lowest total AW of costs. The resulting nC and AWC values are used in the replacement study procedure. The same analysis can be performed for the ESL of the defender. Replacement studies where no study period (planning horizon) is specified utilize the annual worth method of comparing two unequal-life alternatives. The better AW value determines how long the defender is retained before replacement. When a study period is specified for the replacement study, it is vital that market value and cost estimates for the defender be as accurate as possible. When the defender’s remaining life is shorter than the study period, it is critical that the cost for continuing service be estimated carefully. All the viable options for using the defender and challenger are enumerated, and their AW equivalent cash flows are determined. For each option, the PW or AW value is used to select the best option. This option determines how long the defender is retained before replacement.