TM 661 Replacement Solutions 1

# TM 661 Replacement Solutions 1

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## TM 661 Replacement Solutions 1

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1. TM 661 Replacement Solutions 1 A firm is considering an optimal replacement policy for a truck. The initial cost of a new truck is \$100,000 and is expected to remain so for the forseeable future. Maintenance costs are expected to be \$10,000 per year and are expected to escalate by 10% per year. The following table estimates the salvage value of a new truck after n years of use. Using a minimum attractive rate of return of 20%, determine the optimal replacement interval for this truck. Years Salvage 1 \$50,000 2 \$30,000 3 \$20,000 4 \$10,000 5 \$ 5,000 > 5 \$ 0 Soln: Costs are represented as positive. For example, the cash flow 1 Yr replacement for t=1 is computed as 10,000 in maintenance less 50,000 for salvage. The cash flow for 2 Yr replacement, t=2, is computed as 10,000 x 1.1 less 30,000 salvage. Cash flows, NPV and EUAC for replacement intervals 1-5 are shown below.

2. 40,000 20,000 1 2 3 4 5 1 2 3 4 5 1,000 4,500 75,000 6,500 TM 661 Replacement Solutions 2 A company owns a 5-year old turret lathe that has a book value of \$28,000. The present market value for the lathe is \$20,000. The company has just finished closing out the books for the current year where maintenance costs totaled \$4,000. Costs for next year are estimated to be \$4,500 and are expected to increase by \$500 per year. The current lathe is expected to have a remaining life of 5 years. A new lathe can be purchased for \$75,000. Annual operating and maintenance costs are expected to average \$1,000 over the first 5 years with a salvage value of \$40,000 at that time. Determine if the company should retain the current machine or purchase a new lathe. MARR is 15%. Soln: I used the cash flow approach. Keep NPWK = -4,500(P/A,15,5) - 500(P/G,15,5) = -4,500(3.3522) - 500(5.7751) = -17,972 Replace NPWR = -55,000 - 1,000(P/A,15,5) + 40,000(P/F,15,5) = -55,000 - 1,000(3.3522) + 40,000(.4972) = -38,464 Keep the current Lathe