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(1) Least common multiple(LCM) of lives. (2) Specified planning period. Different Life Alternatives. Must compare alts for equal service ( i.e. alts must end at the same time ). Two ways to compare for equal service:. (The LCM procedure is used unless otherwise specified).
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(1) Least common multiple(LCM) of lives (2) Specified planning period Different Life Alternatives Must compare alts for equal service(i.e. alts must end at the same time) Two ways to compare for equal service: (The LCM procedure is used unless otherwise specified)
Compare the machines shown below on the basis of their (a) present worth, and (b) future worth. Use i =10% Machine B Machine A 30,000 First cost,$ 20,000 Annual cost,$/yr 9000 7000 Salvage value,$ 4000 6000 Life, yrs 3 6 Different Life Example Solution: (a) PWA = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) + 4000(P/F,10%,6) = -$68,961 PWB = -$30,000 – 7000(P/A,10%,6) + 6000(P/F,10%,6) = -$57,100 (b) FWA = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3) + 4000 = -$122,168 FWB = -30,000(F/P,10%,6) –7000(F/A,10%,6) +6000 = -$101,157 (both methods will always result in the same selection; in this case, machine B)