 Business
6 September, 11:00

# An injection molding system has a first cost of \$175,000 and an annual operating cost of \$87,000 in years 1 and 2, increasing by \$4,000 per year thereafter. The salvage value of the system is 25% of the first cost regardless of when the system is retired within its maximum useful life of 5 years. Using a MARR of 14% per year, determine the ESL and the respective AW value of the system.

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Answers (1)
1. 6 September, 11:11
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The ESL is 5 years and annual worth is \$143,711

Explanation:

If negative values are not allowed, you can enter 143,711 as the annual worth

DF = Discounting factors are calculated by using the formula 1/1.14.

CF = cash flows. 3500 is added on annual basis from 3rd year, since the increase is per year.

Fifth year CF = 45000 is obtained as - 97500 + salvage value (210000 * 25%) 52500 = 45000.

AWF = Annual worth factor is obtained by dividing each year DF with the Total of DF.

In the last step we multiply CF and AWF to get equivalent annual worth.

Use the following formula:

AW = - 210000 / PVIFA - 87000 [ PV (1) / PVIFA] - 87000 [ PV (2) / PVIFA] - 90500 [ PV (3) / PVIFA] - 94000 [ PV (4) / PVIFA] - 45000 [ PV (5) / PVIFA].
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