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# CH8_102 - Machine X Machine Y Initial cost \$200 \$700...

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Unformatted text preview: Machine X Machine Y Initial cost \$200 \$700 Uniform annual beneﬁt 95 120 End—of—useful—life—salvage value 50 150 Useful life, in years 6 12 Noting that for both Machine X and Machine Y, the PW of cost reﬂects the convention of treating salvage value as a reduction in cost rather than a beneﬁt, and using a 12—year analysis period, we have Machine X PW ofcost = 200 + (200 — 50)(P/F, 10%, 6) — 50(P/F,10%,12) = 200 + 150(O.5645) — 50(0.3186) = 269 PW ofbeneﬁt = 95(P/A,10%,12) = 95(6.814) = 647 Machine Y PW of cost = 700 — 150(P/F,10%,12) = 700 — 150(0.3186) = 652 PW ofbeneﬁt = 120(P/A,10%,12) : 120(6.814) = 818 ”When the two alternatives plotted (Figure 8-4), we see that the increment Y—X has a slope much 6% than the 10% rate of return line. The rate of return on the increment of investment is less than 0%; hence, the increment is undesirable. This means that Machine X should be selected rather I Machine Y. URE 8-4 Beneﬁt—cost graph. \$38 888 Present Worth of Beneﬁts 99 DJ 8 I 5100 \$200 \$300 \$460 \$560 \$600 \$760 5300 Present Worth of Cost ...
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