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    tabitha20 Posts: 7, Reputation: 1
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    May 14, 2011, 09:02 AM
    engineering economy: uniform gradient
    If $10,000 now is equivalent to 4Z at the end of year 2, 3Z at the end of year 3, 2Z at the end of year 4 and Z at the end of year 5, what is the value of Z when the interest rate is 8% per year?

    so I already solved this using the discrete compounding tables. My solution was:
    P=z(P/G,8%,5)
    10000=z(7.372)
    z=$1356.5
    so apparently that's the wrong answer. I assumed the 10000 was the present value... is it the cash flow at the end of year 1 instead of p? Since I can't get the correct answer which is $1256.05

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