Project A costs -$1500 amount of money to begin and -$95 amount to operate every year starting from end of year 5 and will last forever; also $1,000,000 of salvage value at year 100. Project B costs $-186 amount to begin and $-130 to operate every year starting from the end of year 1 to the end of year 5 and $70 of salvage value at the end of year 5. Use an annual rate of 6% to conduct a present worth analysis to determine the PW of both A and B.
We have the following information
Project A |
Project B |
|
First Cost ($) |
– 1500 |
– 186 |
Annual Operating Cost ($) |
– 95 starting from the end of Year 5 |
– 130 |
Salvage Value ($) |
1,000,000 |
70 |
Project A
Present Worth = – 1,500 – 95(A/F, 6%, 5)(P/A, 6%, n=infinite) + 1,000,000(P/F, 6%, 100)
Present Worth = – 1,500 – {95 × [0.06/((1 + 0.06)5 – 1)] × (1/0.06)} + 1,000,000/(1 + 0.06)100
Present Worth = – 1,500 – {95 × 0.1774 × 16.67} + (1,000,000 × 0.0029)
Present Worth of Project A = $1,166.29
Project B
Present Worth = – 186 – 130(P/A, 6%, 5) + 70(P/F, 6%, 5)
Present Worth = – 186 – 130[((1+0.06)5 – 1)/0.06(1+0.06)5] + 70/(1 + 0.06)5
Present Worth = – 186 – (130 × 4.212) + (70 × 0.7473)
Present Worth of Project B = – ($675.25)
Since, the present worth of Project A is higher, so Project A should be selected.
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