## 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.**