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2 May, 15:52

Hinson's Homegrown Farms needs a new irrigation system. System one will cost $145,000, have annual maintenance costs of $10,000, and need an overhaul at the end of year six costing $30,000. System two will have first-year maintenance costs of $5000 with increases of $500 each year thereafter. System two would not require an overhaul. Both systems will have no salvage value after 12 years. If Hinson's cost of capital is 4%, using annual worth analysis determine the maximum Hinson's should be willing to pay for system two. Contributed by Ed Wheeler, University of Tennessee at Martin

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  1. 2 May, 16:10
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    System I:

    Calculate the present worth as shown below:

    Present Worth = - Cost - Annual Cost (P/A, I, n) - overhaul (P/F, I, n)

    = - 145000 - 10000 (P / A, 4%, 12) - 30,000 (P / F, 4%, 12)

    = - 145000 - 10000 (9385) - 30000 (0.6246)

    = - 145000 - 93850 - 18738

    = - $257588

    Calculate the annual worth as shown below:

    Annual Worth PW (A/P, I, n)

    = - 257588 (A / P, 4%, 12)

    = - 257588 * 0.1066

    = - 27458.88

    System 2:

    Calculate the present worth as shown below:

    Plasma Worth = - Annual Cost (P/A, I, n) - Growth (P/G, I, n)

    = - 5000 (P/A, i, n) - 500 (P/G, 4%, I2)

    = - 5000 (9385) - 500 (47248)

    = - 46925 - 23624

    = - 70549

    Calculate the annual worth as shown below:

    Amoral Worth = PW (A/P, I, o)

    = - 70549 (A/P, 4%, 12)

    = - 70549 x 0.1066

    = - 7520.52

    Therefore, Hinson would be willing to pay 7520.52 far system two.
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