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Math Help - How do you do this?

  1. #1
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    How do you do this?

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    Last edited by margite; January 6th 2010 at 08:52 PM.
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  2. #2
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    You need to come up with a problem that's a bit like this:

    $x will be invested annually for 20 years at 8% compounded annually;
    this will permit 4 annual withdrawals (years 21 to 24) of $50,000;
    What is x?
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  3. #3
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    First you need to find the interest rates offered by each bank, and how often your money is compounded. I'll give you an example:

    If bank A offer 5% compounded yearly, and assuming the parents start investing the moment the child is born:

    Let Y = yearly payment.

    Year 0: Y
    The parents just started the first payment Y


    Year 1: Y(1.05) + Y
    The first instalment Y grows by 5% by this year, and the parents add another payment Y

    Year 2: (Y(1.05) + Y)(1.05) + Y
    The amount of money from year 2 (Y(1.05) + Y) grows 5%, and the parents add another payment Y. This can be factorised:
    (Y(1.05) + Y)(1.05) + Y = Y(1.05)+ Y(1.05) + Y
    = Y
    (1 + 1.05 + 1.05)

    Following this pattern, year 18, when the child goes to college will be
    Year 18: Y(1+ 1.05 + 1.05 + ... + 1.05^(18)) - 50 000
    We subtract 50 000 to note the withdrawal for the first year of college. Over 4 years, 50 000 each year equates to 200 000.

    In the following year:
    Year 19: {Y(1+ 1.05 + 1.05 +...+ 1.05^(18)) - 50 000}(1.05) - 50 000)
    = Y(1+ 1.05 + 1.05 +...+ 1.05^(19)) - (50 000(1.05) + 50 000)

    In the last year:
    Year 21: Y(1 + 1.05 + 1.05 + ... + 1.05^(21) - 50 000 (1 + 1.05 + 1.05 + 1.05) = 0
    By this time, we equate money in the bank to equal 0, as all the money has been withdrawn for the college tuition.

    Now 1+ 1.05 + 1.05 + ... + 1.05^(21) and 1 + 1.05 + 1.05 + 1.05 are geometric series, whose sum is given by the equation  \frac{a(r^{n+1} - 1)}{r-1} where a is the first term and r is the common ratio. After calculating these values, all you have to do is to substitute into your final year equation and solve for Y.

    Note that it gets a bit messier if the interest is compounded monthly, or biannually. But the idea is the same.
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  4. #4
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    Last edited by margite; January 6th 2010 at 08:53 PM.
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  5. #5
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    Quote Originally Posted by margite View Post
    I'm still confused, What would y be? and R would be 18, because the money has to be saved from year 1 to 18 (when the kid enters college) right?
    This is YOUR job; come up with:
    an interest rate
    number of years that money will be saved
    number of years in college
    amount required each year in college

    From that, we'll show you how to do the calculation.

    Do not ask us stuff like:
    "because the money has to be saved from year 1 to 18 (when the kid enters college) right?"
    That is for YOU to research and find out.
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  6. #6
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    Last edited by margite; January 6th 2010 at 08:54 PM.
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  7. #7
    Super Member bigwave's Avatar
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    your original statement says the withdrals would be at 21-24
    does this ment the student starts college at 21 (not 18)
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  8. #8
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    Quote Originally Posted by margite View Post
    Interest rate:
    number of years that money will be saved: 18 years..
    number of years in college: four.. the amount of money that is needed is $ 200,000
    amount required each year in college: $50,000
    OK. Let's say the interest rate YOU decide to use is 8% annual;
    "I have to find the average annual interest for three banks"
    we're sure not going to check out 3 banks for you!

    Formula for future value of a constant deposit:
    f = d[(1 + i)^n - 1] / i
    where:
    f = future value (?)
    d = deposit amount (?)
    n = number of years (18)
    i = interest rate (.08)
    So f = d(1.08^18 - 1) / .08

    Formula for present value of a constant withdrawal:
    p = w[1 - 1/(1 + i)^n] / i
    where:
    p = present value (?)
    w = withdrawal amount (50000)
    n = number of years (4)
    i = interest rate (.08)
    So p = 50000(1 - 1/1.08^4) / .08

    Since f = p, then:
    d(1.08^18 - 1) / .08 = 50000(1 - 1/1.08^4) / .08

    d = 50000(1 - 1/1.08^4) / (1.08^18 - 1)
    d = 4422.0364

    So $4,422 deposited annually for 18 years...
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