Answer them all
In 2008, Amphenol Industrial purchased a new quality inspection system for $530,000. The esti- mated salvage value was $30,000 after 8 years. Currently the expected remaining life is 3 years with an AOC of $27,000 per year and an estimated salvage value of $30,000. The new president has recommended early replacement of the system with one that costs $400,000 and has a 5-year economic service life, a $45,000 salvage value, and an estimated AOC of $50,000 per year. If the MARR for the corporation is 12% per year, find the minimum trade-in value now necessary to make the president's re economically advantageous A CNC milling machine purchased by Proto Tool and Die 10 years ago for $75,000 can be used for } more years. Estimates are an annual operating cost of $63,000 and a salvage value of $25,000. A chal- longer will cost $130,000 with an economic life of 6 years and an operating cost of $32,000 per year Its salvage value will be $45,000. On the basis ofSolution The study period is fixed at 10 years, so the intent of the replacement study assumptions is not present. This means the defender follow-on estimates are very important to the analysis. Further, any analyses to determine the ESL values are unnecessary since alternative lives are already set and no projected annual market values are available. The first step of the replace- ment study procedure is to define the options. Since the defender will be replaced now or in 3 years, there are only two options: 1. Challenger for all 10 years. 1. Defender for 3 years, followed by the challenger for 7 years. Cash flows are diagrammed in Figure 11-7. For option I, the challenger is used for all 10 years. Equation [1 1.3] is applied to calculate AW using the following estimates: Challenger: P =$-750,000 AOC = $-50,000 # = 10 years S=0 AW, = -750,000(4/P,10%,10) - 507000 = $-172,063 10 ADC - $56000 End of study period fat Challenger |option IF 11 Year AOC -30 0U _Defender- current CR - 5146475 -Defender-fallow-an 4hy Defender alternative [option ?) Figure 11-7 Cash flow diagrams for a 10-year mudy period replacement study, Example II.6. The second option has more complex cost estimates. The AW for the in-place system is calcu- lated over the first 3 years. Added to this is the capital recovery for the defender follow-on for the next 7 years. However is this case, the CR amown is determined over its full 10-year Me. (It is not unusual for the recovery of invested capital to be moved between projects, especially for contract work.) Refer to the AW components as AWp (subscript DC for defender current) and AW (sub- script DF for defender follow-on). The final cash flows are shown in Figure 11-76. Defender current market value = $-70,000 AOC =$-30,000 a =3 years AWac = [-70,000 - 30,000(P/4, 10%,3) (4/P.10%,10) = $-23,534 Defender follow-on: P = $-900,000, a = 10 years for capital recovery calculation only, AOC = $-50,000 for years 4 through 10, 8 = 0. The CR and AW for all 10 years are CROF = -900,00014/P,10%%,101 = $-146,475 [11.6] AWDE =(-146,475 - 50,000)(F/4,10%,7)(4/F,10%,10) = $-116,966 Total AW, for the defender is the sum of the two annual worth values above. This is the AW for option 2 AW = AWac + AW = -23,534 - 116,956 =$-140,500 Option 2 has a lower cost ($-140,300 versus $-172,063). Retain the defender now and expect to purchase the follow-on system 3 years hence. Comment The capital recovery cost for the defender follow-on will be home by some yet-to-be-identified project for years 10 through 13. If this assumption were not made, its capital recovery cost would be calculated over 7 years, not 10, in Equation [1 1 6], increasing CR to $-184,869. This raises the annual worth to AW,, = $-163,357. The defender alternative (option 2) is still selected.\f\f
