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Math Help - finance question

  1. #1
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    Lightbulb finance question

    1. Paul plans to retire when he turns 55.
    a. How much money will he need in his retirement investment so he can withdraw $40,000 per year for 30 years when he retires? Assume the interest rate in his reitrement years will be fixed at 8%, compounded annually.

    b. Paul is 25 years old. How much does he need to save each year from now until he retires to fund his retirement plan? Assume the interest rate during his work years is fixed at 12%, compounded annually.

    $40000 / 12 = 3333,333(a month)
    A = P(1+i) exp n

    P = 40000?
    i = 0.08
    n =30
    A = 402, 506? Need feedback
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  2. #2
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    Re: finance question

    i dont think that is right.

    its an annuity problem, lookup your annuity formula
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  3. #3
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    Re: finance question

    Hello, terminator!

    If you've been taught Compound Interest only,
    . . you should not have been assigned this problem.


    1. Paul plans to retire when he turns 55.
    (a) How much money will he need in his retirement investment
    so he can withdraw $40,000 per year for 30 years when he retires?
    Assume the interest rate in his reitrement years will be 8%, compounded annually.

    This is a Sinking Fund problem.

    Formula: . A \;=\;P\,\frac{i(1+i)^n}{(1+i)^n - 1} \quad\Rightarrow\quad P \;=\;A\,\frac{(1+i)^n-1}{i(1+i)^n}

    . . where: . \begin{Bmatrix} A &=& \text{periodic withdrawl} \\ P &=& \text{principal invested} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}


    We have: . \begin{bmatrix} A &=& \$40,\!000 \\ i &=& 0.08 \\ n &=& 30 \end{bmatrix}

    Hence: . P \;=\;40,\!000\,\frac{(1.08)^{30}-1}{(0.08)(1.08)^{30}} \;=\;450,\!311.3337


    At age 55, Paul will need \$450,\!311.34.

    . . . Note that I rounded up. .(Why?)



    (b) Paul is 25 years old. .How much does he need to save each year
    from now until he retires to fund his retirement plan?
    Assume the interest rate during his work years is 12%, compounded annually.

    SpringFan25 is correct . . . This is an Annuity problem.

    Formula: . A \:=\:D\,\frac{(1+i)^n-1}{i} \quad\Rightarrow\quad D \:=\:A\,\frac{i}{(1+i)^n-1}

    . . where: . \begin{Bmatrix}A &=& \text{final balance} \\ D &=& \text{periodic deposit} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}


    We have: . \begin{bmatrix}A &=& 450,\!311.34 \\ i &=& 0.12 \\ n &=& 30 \end{bmatrix}

    Hence: . D \;=\;(450,\!311.34)\,\frac{0.12}{(1.12)^{30}-1} \;=\;1865.935985

    Until his retirement, Paul must save \$1,\!865.94 each year.

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