SOLUTION :
r = MARR = 10% = 0.1
=> (1 + r) = 1.1
PV of costs of Machine A :
= 21439 + 13160/ 1.1 + 13160/1.1^2 + 13160/1.1^3 - 5419/1.1^3
= 50094.60 ($)
PV of costs of Machine B :
= 124845 + 7237 / 0.1
= 197215 ($)
Therefore, as per CC, Machine A has lower cost. So Machine A should be selected.
(ANSWER)
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