In: Economics
A new machine will last for ten years. It will save the company $9984 per year and the savings will increase by $2660 each year. What is the present worth of this machine assuming an interest rate of 7.4%/year compounded annually.
year | Cash flow | Present Value factor | Present Value |
1 | $ 9,984.00 | 0.931098696 | $ 9,296.09 |
2 | $ 12,644.00 | 0.866944783 | $ 10,961.65 |
3 | $ 15,304.00 | 0.807211157 | $ 12,353.56 |
4 | $ 17,964.00 | 0.751593256 | $ 13,501.62 |
5 | $ 20,624.00 | 0.699807501 | $ 14,432.83 |
6 | $ 23,284.00 | 0.651589852 | $ 15,171.62 |
7 | $ 25,944.00 | 0.606694462 | $ 15,740.08 |
8 | $ 28,604.00 | 0.564892422 | $ 16,158.18 |
9 | $ 31,264.00 | 0.525970598 | $ 16,443.94 |
10 | $ 33,924.00 | 0.489730538 | $ 16,613.62 |
Present Worth= | $ 140,673.20 |
Present worth is sum of the present value of all cash flows defined at a given year.
Present Value factor = 1/(1+interest rate)^t. where t = years
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