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REPLACEMENT DECISIONS

REPLACEMENT DECISIONS. CHAPTER 12. Replacement Decisions. Replacement Analysis Fundamentals Economic Service Life Replacement Analysis When a Required Service is Long. Replacement Terminology. Sunk cost: any past cost unaffected by any future investment

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REPLACEMENT DECISIONS

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  1. REPLACEMENT DECISIONS CHAPTER 12

  2. Replacement Decisions • Replacement Analysis Fundamentals • Economic Service Life • Replacement Analysis When a Required Service is Long

  3. Replacement Terminology Sunk cost: any past cost unaffected by any future investment decisions. Trade-in allowance: value offered by the vendor to reduce the price of a new equipment Defender: an old machine Challenger: a new machine Current market value: selling price of the defender in the market place

  4. Replacement Analysis Fundamentals • Replacement projects are decision problems that involve the replacement of existing out of date or worn-out assets. • When existing equipment should be replaced with more efficient equipment? Examine three aspects of the replacement problem • Approaches for comparing defender and challenger • Determination of economic service life • Replacement analysis when the required service period is long

  5. Two basic approaches to analyze replacement problems Cash Flow Approach • Treat the proceeds (amount of money received) from sale of the old machine as down payment toward purchasing the new machine. (Use it to bye the new one) • This approach is meaningful when both the defender and challenger have the same service life. • Use PW or AE values in the analysis Opportunity Cost Approach • Treat the proceeds (amount of money received) from sale of the old machine as the investment required to keep the old machine. (Do not use the money to bye the new one). • This approach is more commonly practiced in replacement analysis.

  6. Economic Service Life • Economic service life is the remaining useful life of an asset that results in the minimum annual equivalent cost. • We should use the respective economic service lives of the defender and the challenger when conducting a replacement analysis. Minimize Capital (Ownership) cost Annual Equivalent Cost + Operating cost

  7. Economic Service Life Continue…. • Capital cost have two components: Initial investment ( I ) and the salvage value ( S ) at the time of disposal. • The initial investment for the challenger is its purchase price. For the defender, we should treat the opportunity cost (potential benefit that is given up as you seek an alternative action) as its initial investment. • Use N to represent the length of time in years the asset will be kept; ( I ) is the initial investment, and SN is the salvage value at the end of the ownership period of N years. • The operating costs of an asset include operating and maintenance (O&M) costs, labor costs, material costs and energy consumption costs.

  8. Mathematical RelationshipObjective: Find n* that minimizes total AEC AE of Capital Cost: AE of Operating Cost: Total AE Cost:

  9. Here are some special cases that the economic service life can be determined easily: • If the salvage value is constant and equal to the initial cost, and the annual operating cost increases with time, AEC is an increasing function of N and attains its minimum at N = 1. In this case, we should try to replace the asset as soon as possible. • If the annual operating cost is constant and the salvage value is less than the initial cost and decreases with time, AEC is a decreasing function of N. In this case, we would try to delay replacement of the asset as long as possible. • If the salvage value is constant and equal to the initial cost and the annual operating costs are constant, AEC will also be constant. In this case, the time at which the asset is replaced does not make any economic difference.

  10. Example 12.4 Economic Service Life for a Lift Truck A company considers buying a new electric forklift truck that would cost $20,000, have operating cost of $6,000 in the first year, and have a salvage value of $14,000 at the end of the first year. For the remaining years, operating costs increase each year by 25% over the previous year’s operating costs. Similarly, the salvage value declines each year by 30% from the previous year’s salvage value. The truck has a maximum life of eight years, without any major engine overhaul. The firm’s required rate of return is 12%. Find the economic service life of this new machine.

  11. N = 1 CR (12%)1 = I (A/P, 12%, 1) – S1 (A/F, 12%, 1) CR (12%)1= $20,000 (1.12) – $14,000 (1) = $8,400 OC (12%)1 =Σ [OC1 (P/F, 12%, 1)] (A/P, 12%, 1) OC (12%)1=Σ [$6,000(0.8929)] (1.12) = $6,000 AEC1 = CR 1 + OC1 = $8,400 + $6,000 AEC1 = $14,400

  12. N = 2 CR (12%)2 = I (A/P, 12%, 2) – S2 (A/F, 12%, 2) CR (12%)2= $20,000 (0.5917) – $9,800 (0.4717) = 11,834 – 4,623 = $7,211 OC (12%)2 =Σ [OC1 (P/F, 12%, 1) + OC2 (P/F, 12%, 2)] x (A/P, 12%, 2) OC (12%)2=Σ [$6,000(0.8929) + $7500 (0.7972)] x (0.5917) = $6,708 AEC2= CR (12%)2 + OC2 = $7,211 + $6,708 AEC2 = $13,919

  13. N = 3 CR (12%)3 = I (A/P, 12%, 3) – S3 (A/F, 12%, 3) CR (12%)3 = $20,000 (0.4163) – $6,860 (0.2963) = 8,326 – 2,032 = $6,294 OC (12%)3 =Σ [ OC1 (P/F, 12%, 1) + OC2 (P/F, 12%, 2) + OC3 (P/F, 12%, 3)] (A/P, 12%, 3) OC (12%)3 =Σ [$6,000(0.8929) + $7,500 (0.7972) + $9,375 (0.7118)] (0.4163) = $7,498 AE3 = CR (12%)3 + OC3 AEC3 = $6,294 + $7,498 = $13,792

  14. N = 4 CR (12%)4 = I (A/P, 12%, 4) – S4 (A/F, 12%, 4) CR (12%)4 = $20,000 (0.3292) – $4,802 (0.2092) = $6,584 – $1,004 = $5,580 OC (12%)4 =Σ [OC1 (P/F, 12%, 1) + OC2 (P/F, 12%, 2) + OC3 (P/F, 12%, 3) + OC4 (P/F, 12%, 4)] (A/P, 12%, 4) OC (12%)4 =Σ [ $6,000(0.8929) + $7,500 (0.7972) + $9,375 (0.7118) + $11,719 (0.6355)] (0.3292) = $8,381 AE4 = CR (15%)4 + OC4 AEC4 = $5,580 + $8,381 = $13,961

  15. AEC if the Asset were Kept N Years N = 1, AEC1 = $14,400 N = 2, AEC2 = $13,919 N = 3, AEC3 = $13,792 N = 4, AEC4 = $13,961 N = 5, AEC5 = $14,387 N = 6, AEC6 = $15,044 N = 7, AEC7 = $15,920 N = 8, AEC8 = $17,008 Minimum cost If you purchase the asset, it is most economical to replace the asset every 3 years Economic Service Life

  16. Replacement Analysis When the Required Service Period is Long • We know how the economic service life of an asset is determined. • The next question is to decide whether now is the time to replace the defender. Consider the following factors: Planning horizon (study period) • By planning horizon, it is meant the service period required by the defender and future challengers. • The infinite planning horizon is used when we are unable to predict when the activity under consideration will be terminated. • In other situations, the project will have a definite and predictable duration. In these cases, replacement policy should be formulated based on a finite planning horizon.

  17. Decision Frameworks continue……. Relevant cash flow information • Many varieties of predictions can be used to estimate the pattern of revenue, cost and salvage value over the life of an asset. Decision Criterion • The AE method provides a more direct solution when the planning horizon is infinite. When the planning horizon is finite, the PW method is convenient to be used. • Although the economic service life of the defender is defined as the number of years of service that minimizes the annual equivalent cost (or maximizes the annual equivalent revenue), the end of the economic life is not necessarily the optimum time to replace the defender.

  18. Handling Unequal Service Life Problems in Replacement Analysis • Let us consider a situation where you are comparing a defender (D) with an economic service life of three years and a challenger (C) with an economic service life of six years. • This means that if we decide to keep the old machine (D) for three years, we will replace it at time 3 by an asset similar to the new machine. This asset will in turn be replaced six years later, at time 9, by another asset C. • There are two implied infinite sequences in this scenario: • Keep defender (D), buy a challenger (C) at time 3, buy another challenger (C) at time 9, buy another challenger (C) at time 15, etc. • Buy challenger (C) at time 0, buy a challenger (C) at time 6, buy a challenger (C) at time 12, and so on

  19. Handling Unequal Service Life Problems in Replacement Analysis • It is clear that the AE cost approach for either sequence of assets is the same after the remaining life of the defender. Therefore, we can directly compare the AEC for the remaining life of the defender with the AEC for the challenger over its economic service life.

  20. Replacement Strategies under the Infinite Planning Horizon • Compute the economic lives of both defender and challenger. Let’s use ND* and NC* to represent economic lives of the defender and the challenger. • The annual equivalent cost for the defender and the challenger at their respective economic lives are indicated by AED* and AEC* • Compare AED* and AEC*. If AED* is bigger than AEC*, it is more costly to keep the defender than to replace it with the challenger. Thus, the challenger should replace the defender now.

  21. Replacement Strategies under the Infinite Planning Horizon • If the defender is not to be replaced now, when should it be replaced? • First, we need to continue to use it until its economic life is over. Then, calculate the cost of running the defender for one more year. If this cost is greater than AEC*, the defender should be replaced at the end of its economic life. • This process should be continued until you find the optimal replacement time. This approach is called marginal analysis.

  22. Replacement Strategies under the Infinite Planning Horizon

  23. EXAMPLE 12.5 Replacement Analysis under the Infinite Planning Horizon General Engineering Company is considering replacing an old vertical cylinder honing (polishing) machine. They are considering two options: Option 1: Retain or keep the old machine. If it is kept, the old machine can be used for another six years with proper maintenance. The market value of the machine is expected to decline 25% annually over the previous years. The operating costs are estimated at $3,500 during the first year and are expected to increase by $1,000 per year thereafter.

  24. EXAMPLE 12.5 continue.......... Option 2: Alternatively, the firm can sell the machine to another firm in the industry now for $4,000 and buy new honing machine. The new machine costs $12,000 and will have operating costs of $2,300 in the first year, increasing by 20% per year thereafter. The expected salvage value is $8,000 after one year and will decline 30% each year thereafter. The company requires rate of return of 12%. Find the economic life for each option, and determine when the defender should be replaced.

  25. EXAMPLE 12.5 Replacement Analysis under the Infinite Planning Horizon

  26. SOLUTION: Economic Service Life for Defender N = 1 CR (12%)N = I (A/P, 12%, N) – SN (A/F, 12%, N) and AEOC (12%)N =Σ [OCn (P/F, 12%, N)] (A/P, 12%, N) CR (12%)1 = $4,000 (1.12) – 3,000 (1.0) = 4,480 – 3,000 = $1,480 AEOC1 = 3,500 (0.8929) (1.12) = $3,500 Σ AEC1= CR (15%)1 + AEOC1 = 1,480 + 3,500 = $4,980

  27. SOLUTION - Economic Service Life for Defender

  28. Economic Service Life Calculation For Defender

  29. SOLUTION - Economic Service Life for Challenger N = 1 CR (12%)N = I (A/P, 12%, N) – SN (A/F, 12%, N) AEOC =Σ [OCn (P/F, 12%, N)] (A/P, 12%, N) CR (12%)1 = $12,000 (1.12) – 8,000 (1.0) = 13,440 – 8,000 = $5,440 AEOC1 = 2,300 (0.8929) (1.12) = $2,300 Σ AEC1= CR (15%)1 + AEOC1 = 5,440 + 2,300 = $7,740

  30. Economic Service Life Calculation for Challenger

  31. Replacement Decisions NC* = 5 years AECC* = $6,312 Should replace the defender now? No, because AECD* < AECC* If not, when is the best time to replace the defender? Need to conduct the marginal analysis. ND* = 1 year AECD* = $4,980

  32. Marginal Analysis – When to Replace the Defender Question: What is the additional (incremental) cost for keeping the defender one more year from the end of its economic service life, from Year 1 to Year 2? Financial Data: Opportunity costat the end of year 1: $3,000 (market value of the defender at the end year 1) Operating costfor the 2nd year: $4,500 Salvage valueof the defender at the end of year 2: $2,250

  33. $2250 1 Step 1: Calculate the equivalent cost of retaining the defender one more year from the end of its economic service life, say 1 to 2. $3,000 (F/P,12%,1) + $4,500 - $2,250 = $5,610 Step 2: Compare this cost with AECC* = $6,312 of the challenger. Conclusion: Since keeping the defender for the 2nd year is less expensive than replacing it with the challenger, keep the defender beyond its economic service life. 2 $3000 $4,500 1 2 $5,610

  34. $1688 2 Step 1: Calculate the equivalent cost of retaining the defender one more year from year 2 to 3. $2,250 (F/P,12%,1) + $5,500 - $1,688 $2,250 x 1.12 + $5,500 - $1,688 =$6,332 Step 2: Compare this cost with AECC* = $6,312 of the challenger. Conclusion: For year three, keeping the defender is more expensive. This means that we should replace the defender at the end of year two. 3 $2250 $5,500 2 3 $6,332

  35. SOLVED PRACTICE PROBLEMS

  36. 12.1 NEWNAN Furniture owns and operates an industrial lift truck in their warehousing operation. The record indicates that the lift truck was purchased four years ago at $15,000. The estimated salvage value is $4,000 after four years of operation. First-year O&M expenses were $2,000, but the O&M expenses have increased by $400each year for the first four years of operation. Usingi= 10% compute the annual equivalent costs of the lift truck for four years. 12.1 SOLUTION

  37. 12.5 A firm is considering the replacement of a 2,000 kg capacity pressing machine. The machine was purchased five years ago at a cost of $22,000. The machine was originally expected to have a useful life of 10 years and a $2,000 estimated salvage value at the end of that period. However, the machine has not been dependable and is frequently out of service while awaiting repairs. The maintenance expenses of the pressing machine have been rising steadily and currently amount about $5,000 per year. The machine could be sold now for $6,000. If it is retained or kept the machine will require an immediate $2,500 overhaul to keep it in operable condition. This overhaul will neither extend the originally estimated service life nor increase the value of the machine. The updated annual operating costs, engine overhaul cost, and market values over the next five years are estimated as follows:

  38. 12.5

  39. 12.5 A drastic increase in operating costs during the fourth year is expected as a result of another overhaul, that is about $3,000 which will be required in order to keep the machine in operating condition. The firm’s MARR is 15%. If the machine is to be sold now, what will be its sunk cost? What is the opportunity cost of not replacing the machine now? What is the equivalent annual cost of owning and operating the machine for two more years? What is the equivalent annual cost of owning and operating the machine for five more years?

  40. 12.5) SOLUTION (a) Purchase cost = $22,000, market value = $6,000, sunk cost = $22,000 - $6,000 = $16,000 (b) Opportunity cost = $6,000 (c) (d)

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