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Math Help - Accumulation question.

  1. #1
    DCU
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    Accumulation question.

    An investor is to receive a series of annual payments for a term of 15 years in which payments are
    increased by 2.5% compound each year to allow for inflation. The first payment is to be 14,400 on 1 January 2011. Find the accumulated value of the annuity payments as at 1 January 2031 if the investor achieves an effective rate of return 3.5% per annum effective.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by DCU View Post
    An investor is to receive a series of annual payments for a term of 15 years in which payments are
    increased by 2.5% compound each year to allow for inflation. The first payment is to be 14,400 on 1 January 2011. Find the accumulated value of the annuity payments as at 1 January 2031 if the investor achieves an effective rate of return 3.5% per annum effective.
    A bit hard to follow: 2011 to 2031 is not 15 years...
    anyway, I'll assume 15 years, 1st deposit end of 1st year:
    n = 15
    p = 14400
    x = 1 + .025
    y = 1 + .035

    F = p(y^n - x^n) / (y - x) = 14400(1.035^15 - 1.025^15) / (1.035 - 1.025) = 326952.9565...

    Account will look like:
    Code:
    YEAR    PAYMENT    INTEREST    BALANCE
     0                                 .00
     1     14400.00         .00   14400.00
     2     14760.00      504.00   29664.00
     3     15129.00     1038.24   45831.24
    ....
    15     20346.82    10368.32  326952.96
    Last edited by Wilmer; January 14th 2011 at 07:56 PM.
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  3. #3
    DCU
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    Sorry, I figured it out a few days ago. The annuity is paid over 15 years so you get the accumulation to 15 but you also have to find out what this is worth 5 years later, ie in 2031, without any payments, just interest. This was my equation of value:

    F = 14,400(y)(( 1 - (x/y)^15)/(1- (x/y)))
    This is the value after all of the annuity is paid after 15 years
    Then, we must find out what it accumulates to 5 years later which is simply:
    A = F(y)^5

    Thanks for the help though

    I'm using the notation you were using, x = 1 + .025
    y = 1+.035
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