Math

A firm is evaluating two machines. The first costs $250,000 and will require annual maintenance of $30,000 per year for 10 years. At the end of 10 years, the salvage value will be $75,000.
The second machine costs $400,000 and will require $225,000 at the end of the fifth year. The salvage value after 10 years will be $175,000. Which machine should the firm select if interest in 8.5% compounded annually?

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  1. cost of 1st machine
    = 250000 + 30000( 1 - 1.085^-10)/.085 - 75000(1.085)^-10
    = 413,669.04

    cost of 2nd machine
    = 400000 + 225000(1.085)^-5 - 175000(1.085)^-10
    = 472,235.27

    1st is cheaper, I used the present time as my "focal point in time"
    check my arithmetic

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