Answers
Machine A
Equivalent uniform annual worth = Capital recovery cost + Annual operating cost
Capital recovery cost = - First cost Capital recovery factor + Salvage value
Sinking fund factor
Capital recovery cost machine A = - 15,000 (A/P, 10%, 2 years ) + 3000
(A/F, 10% , 2 years)
Capital recovery cost machine A = - 15,000 0.5762 + $ 3000
0.4762 = - 7214.4
Equivalent uniform annual worth = - 7214.4 - 1500
Equivalent uniform annual worth machine A = $ - 8,714.4
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Machine B
Equivalent uniform annual worth = Capital recovery cost + Annual operating cost
Capital recovery cost machine B = - 25,000 (A/P, 10%, 10 years ) + 6000
(A/F, 10% , 10 years)
Capital recovery cost machine B = - 25000 0.1627 + 6000
0.0627
Capital recovery cost machine B = - 3691.3
Equivalent uniform annual worth = - 3691.3 - 400 - 50 ( P/A, 10% , 10 years)
( P/A, 10% , 10 years) = Uniform series present worth
Equivalent uniform annual worth machine B = - 3691.3 - 400 - 50 6.145
Equivalent uniform annual worth machine B = $- 4398.6
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Thus machine B must be selected because it has a lower equivalent uniform annual worth of cost when compared to machine A.