Homework Answers
b) Annual worth analysis
MARR = 12%
AW of machine A = -52000*(A/P,12%,4) + 38000 - 15000 + 13000*(A/F,12%,4)
= -52000*0.329234 + 23000 + 13000*0.209344
= 8599.86
AW of machine B = -67000*(A/P,12%,12) + 37000 - 12000 + 22000*(A/F,12%,12)
= -67000*0.161436 + 25000 + 22000*0.0414368
= 15095.34
AW of machine C = -63000*(A/P,12%,6) + 31000 - 9000 - 250*(A/G,12%,6) + 19000*(A/F,12%,6)
= -63000*0.243225 + 22000 - 250*2.172047 + 19000*0.123225
= 8475.06
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please answer all the parts and show the formulas where possible. - + Fit to page...
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