SOLUTION :
r = MARR = 10% = 0.1
=> (1 + r) = 1.1
PV of costs of Machine A :
= 24311 + 13353/ 1.1 + 13353/1.1^2 + 13353/1.1^3 - 6323/1.1^3
= 52767.37 ($)
PV of costs of Machine B :
= 100000 + 7/0.1
= 100070 ($)
Therefore, as per CC, Machine A has lower cost. So Machine A should be selected.
(ANSWER)
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> Please correct : 7th line as : = 100000+ 7000/0.1 ; 8th line as : = 170000 ($)
Tulsiram Garg Mon, Oct 11, 2021 2:32 AM