question archive D Try to choose the correct answer, note that four wrong answers will cancel one correct answer (
D Try to choose the correct answer, note that four wrong answers will cancel one correct answer (
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D Try to choose the correct answer, note that four wrong answers will cancel one correct answer (.e. each wrong answer will be penalized by 1 pt.): 14 pts. ] An industrial engineer is considering to alterative robots to purchase by a liber-optie manufacturing company. If a 6 - years study period is considered, which one of the following value is closest to the future worth of incremental (Y-X) cash flows at an interest Investiment cost now. S 55,000 125,000 Annual operating cost. S 23.000 Salvage value at the end of study period, S. 15,000 26,000 OOO a -1,070 b. -588
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D)
Future value = Present value*(1+r)^n
Future value of annuity = P*[(1+r)^n - 1 / r ]
r = rate of interest
n = number of periods
P = annual cost
Future worth of X = 55000*(1+9%)^6 + 23000*[(1+9%)^6 - 1 / 9% ] - 15,000
= $250,277.20
Future worth of Y = 125000*(1+9%)^6 + 9000*[(1+9%)^6 - 1 / 9% ] - 26000
=$251,347.52
Y - X = 251,347.52 - 250,277.20
= $1070
Since it is a cash outflow it will be = -$1070
Option a is correct
