Chapter 15 Replacement Decisions Replacement Analysis Fundamentals Economic Service Life Replacement Analysis When Required Service is Long Replacement Analysis with Tax Consideration (c) 2001 Contemporary Engineering Economics

Replacement Analysis Fundamentals Replacement projects are decision problems involve the replacement of existing obsolete or worn-out assets. When existing equipment should be replaced with more efficient equipment. Examine replacement analysis fundamentals Approaches for comparing defender and challenger Determination of economic service life Replacement analysis when the required service period is long (c) 2001 Contemporary Engineering Economics

Replacement Terminology Defender: an old machine Challenger: new machine Current market value: selling price of the defender in the market place Sunk cost: any past cost that is unaffected by any future decisions Trade-in allowance: value offered by the vendor to reduce the price of a new equipment (c) 2001 Contemporary Engineering Economics

Sunk Cost associated with an Asset’s Disposal (example 15.1) Original investment $20,000 Lost investment (economic depreciation) Market value Repair cost $10,000 $10,000 $5000 Sunk costs = $15,000 $0 $5000 $10,000 $15,000 $20,000 $25,000 $30,000 (c) 2001 Contemporary Engineering Economics

(c) 2001 Contemporary Engineering Economics Replacement Decisions Two basic approaches to analyze replacement problems. Cash Flow Approach - Treat the proceeds from sale of the old machine as down payment toward purchasing the new machine. Opportunity Cost Approach Treat the proceeds from sale of the old machine as the investment required to keep the old machine. (c) 2001 Contemporary Engineering Economics

Cash Flow Approach Example 15.2 Replacement Analysis Sales proceeds from defender $10,000 $5500 $2500 0 1 2 3 0 1 2 3 $6000 $8000 (b) Replace the defender With Challenger (a) Keep the Defender $15,000 (c) 2001 Contemporary Engineering Economics

Annual Equivalent Cost Cash Flow Approach  Defender: PW(12%)D = $2,500 (P/F, 12%, 3) - $8,000 (P/A, 12%, 3) = - $17,434.90 AE(12%)D = PW(12%)D(A/P, 12%, 3) = -$7,259.10  Challenger: PW(12%)C = $5,500 (P/F, 12%, 3) - $5,000 - $6,000 (P/A, 12%, 3) = -$15,495.90 AE(12%)C = PW(12%)C(A/P, 12%, 3) = -$6,451.79 Annual difference is $807.31 Replace the defender now! (c) 2001 Contemporary Engineering Economics

Example 15.3 Opportunity Cost Approach Defender Challenger $2500 $5500 0 1 2 3 0 1 2 3 $8000 $6000 $10,000 Proceeds from sale viewed as an opportunity cost of keeping the asset $15,000 (c) 2001 Contemporary Engineering Economics

Opportunity Cost Approach  Defender: Charge $ 10,000 as an opportunity cost or incurred cost. PW(12%)D = -$10,000 - $8,000(P/A, 12%, 3) + $2,500(P/F, 12%, 3) = -$27,434.90 AE(12%)D = PW(12%)D(A/P, 12%, 3) = -$11,422.64  Challenger: PW(12%)C = -$15,000 - $6,000(P/A, 12%, 3) + $5,500(P/F, 12%, 3) = -$25,495.90 AE(12%)C = PW(12%)C(A/P, 12%, 3) = -$10,615.33 Because of the annual difference of $807.31 in favor of the challenger. Replace the defender now! (c) 2001 Contemporary Engineering Economics

(c) 2001 Contemporary Engineering Economics Economic Service Life Definition: The economic service of an asset is defined to be the period of useful life that minimizes the annual equivalent cost of owning and operating asset. We should use the economic service lives of the defender and the challenger when conducting a replacement analysis. Minimize Ownership (Capital) cost + Operating cost (c) 2001 Contemporary Engineering Economics

Economic Service Life Continue…. Capital cost have two components: Initial investment and the salvage value at the time of disposal. The initial investment for the challenger is its purchase price. For the defender, we should treat the opportunity cost 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. (c) 2001 Contemporary Engineering Economics

Mathematical Relationship AE of Capital Cost: AE of Operating Cost: Total AE Cost: Objective: Find n* that minimizes total AEC AEC OC(i) CR(i) n* (c) 2001 Contemporary Engineering Economics

Economic Service Life for a Lift Truck Example 15.4 (c) 2001 Contemporary Engineering Economics

Economic Service Life Calculation (Example 15.4) N = 1 (if you replace the asset every year) AEC1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000 = $11,700 (c) 2001 Contemporary Engineering Economics

(c) 2001 Contemporary Engineering Economics (Truck will be used for two years and disposed of at the end of year 2) AEC2 = [$18,000 + $1,000(P/A1, 15%, 15%, 2)](A/P, 15%, 2) - $7,500 (A/F, 15%, 2) = $8,653 (c) 2001 Contemporary Engineering Economics

(c) 2001 Contemporary Engineering Economics N = 3, AEC3 = $7,406 N = 4, AEC4 = $6,678 N = 5, AEC5 = $6,642 N = 6, AEC6 = $6,258 N = 7, AEC7 = $6,394 Minimum cost Economic Service Life (c) 2001 Contemporary Engineering Economics

Required Assumptions and Decision Frameworks Now we understand 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) The project will have a exact a predictable duration. Therefore, the replacement policy should be formulated based on a finite planning horizon. (c) 2001 Contemporary Engineering Economics

Decision Frameworks continue……. Technology Predictions of technological patterns over the planning horizon refer to the development of types of challengers that may replace those under study. If we assume that all future machines will be same as those now in service, there is no technological progress in the area will occur. In other cases, we may recognize the possibility of machines become available, that will be more efficient, reliable, productive than those in the current market. 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 (endless). When the planning horizon is finite (fixed), the PW method is convenient to be used. (c) 2001 Contemporary Engineering Economics

Replacement Strategies under the Infinite Planning Horizon Compute the economic lives of both defender and challenger. Let’s use ND* and NC* to indicate the economic lives of the defender and the challenger, respectively. 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*, we know that it is more costly to keep the defender than to replace it with the challenger. Thus, the challenger should replace the defender now. If the defender should not be replaced now, when should it be replaced? First, we need to continue to use until its economic life is over. Then, we should calculate the cost of running the defender for one more year after its economic life. If this cost is greater than AEC*, the defender should be replaced at the end of is economic life. This process should be continued until you find the optimal replacement time. This approach is called marginal analysis, that is, to calculate the incremental cost of operating the defender for just one more year. (c) 2001 Contemporary Engineering Economics

Replacement Analysis under the Infinite Planning Horizon Example 15.5 Step 1: Find the remaining useful (economic) service life of the defender. (c) 2001 Contemporary Engineering Economics

(c) 2001 Contemporary Engineering Economics Step 2: find the economic service life of the challenger. N = 1 year: AE(15%) = $7,500 N = 2 years: AE(15%) = $6,151 N = 3 years: AE(15%) = $5,847 N = 4 years: AE(15%) = $5,826 N = 5 years: AE(15%) = $5,897 NC*=4 years AEC*=$5,826 (c) 2001 Contemporary Engineering Economics

Replacement Decisions Should replace the defender now? No, because AED < AEC If not, when is the best time to replace the defender? Need to conduct marginal analysis. NC*=4 years AEC*=$5,826 (c) 2001 Contemporary Engineering Economics

(c) 2001 Contemporary Engineering Economics Marginal Analysis Question: What is the additional (incremental) cost for keeping the defender one more year from the end of its economic service life, from Year 2 to Year 3? Financial Data: Opportunity cost at the end of year 2: Equal to the market value of $3,000 Operating cost for the 3rd year: $5,000 Salvage value of the defender at the end of year 3: $2,000 (c) 2001 Contemporary Engineering Economics

(c) 2001 Contemporary Engineering Economics Step 1: Calculate the equivalent cost of retaining the defender one more year from the end of its economic service life, say 2 to 3. AED = $3,000(F/P,15%,1) + $5,000 - $2,000 = $6,450 Step 2: Compare this cost with AEC = $5,826 of the challenger. Conclusion: Since keeping the defender for the 3rd year is more expensive than replacing it with the challenger, DO NOT keep the defender beyond its economic service life. $2000 2 3 $3000 $5000 2 3 $6,450 (c) 2001 Contemporary Engineering Economics