Description
Question:
Dave Metal is examining two options – Machine A and Machine B – for a core-cutting machine. Machine A Initial Cost $ 250,000 Savings $28,000/year Maintenance Cost $10,000/year + $0.05/unit Labor $0.25/widget Life 10 years Salvage Value $22,500. Machine B Initial Cost $ 180,000 Savings $22,000/year Maintenance Cost $20,000 /year Labor $0.50/widget Life 5 years Salvage Value $ 17,000. MARR is 12 percent and the company’s sale projection is 30,000 widgets/year. Should either, both or neither machine be purchased? Use both present worth and annual worth comparisons.
Answer:
First we compute the npv normally. To compute equivalent annual worth, divide the npv with the present value annuity factor@12% for the respective years of life of the machines.
Machine A
Particulars | Year | PVF@12% | Amount | Present Value |
Savings | 1-10 | 5.65022 | 28000 | 158206.16 |
Less : maintenance cost | 1-10 | 5.65022 | 10000+0.05 x 30000 = 11500 | 64977.53 |
Less : labour | 1-10 | 5.65022 | 30000 x 0.25 = 7500 | 42376.65 |
Add: salvage vale | 10 | 0.32197 | 22500 | 7244.325 |
Present value of cash inflows | 58096.305 | |||
Less : initial cost | 0 | 1 | 250000 | 250000 |
Npv | (-)191903.695 |
Equivalent annual worth = (-)191903.695 / 5.65022 = (-)33963.93
Machine B
Particulars | Year | PVF@12% | Amount | Present Value |
Savings | 1-5 | 3.60478 | 22000 | 79305.16 |
Less : maintenance cost | 1-5 | 3.60478 | 20000 | 72095.60 |
Less : Labour | 1-5 | 3.60478 | 0.50 x 30000 = 15000 | 54071.70 |
Add : Salvage value | 5 | 0.56743 | 17000 | 9646.31 |
Present value of cash inflows | (-)37215.83 | |||
Less : initial cost | 180000 | |||
Npv | (-)217215.83 |
Equivalent annual worth = (217215.83) / 3.60478 = (-)60257.72
Machine A is better among the two as it’s equivalent annual worth is more. However, neither machine is recommended as both have a negative NPV and annual worth.
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