compounding interest/ Roth IRA calculation help

This is a discussion on compounding interest/ Roth IRA calculation help within the C++ Programming forums, part of the General Programming Boards category; As a financial advisor, you are helping a young engineer, Rosie. In order to take advantage of the tax exempt ...

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    compounding interest/ Roth IRA calculation help

    As a financial advisor, you are helping a young engineer,
    Rosie. In order to take advantage of the tax exempt growth of Roth IRA.
    Rosie wishes to compare the growth of two different contribution
    options, A and B, after 40 years:

    Option A. Rosie contributes $800 every month.
    Option B. Rosie contributes $500 per month in the first year, then
    increases her contribution by 10% every year. For example,
    she will contribute $550 per month in the second year, and
    $605 per month in the third year, and so on.

    Each contribution is made at the beginning of month.
    Assuming that her account grows monthly at an annual rate of 7.5%, you
    are asked to compute and print Rosie's Roth IRA account balance,
    the total contributions, and the tax exempt growth
    (= balance - contributions) after 40 years for both options A and B.


    I already did option A, the code is as follows. However, I am stuck on how to do part B. I did part of it but it doesn't come out right.


    Code:
    int months, years = 40, k;
    double m_payment = 800.0, temp, bal, i_rate, t_payments, n_payment, bal_x;
    printf("Option A\n");
    i_rate = 0.075/12.0;
    months = 12 * years;
    t_payments = (double)months * m_payment;
    temp = pow(1.0 + i_rate, (double)months);
    bal = (temp - 1.0)/i_rate * m_payment;
    printf("Rosie's Roth IRA\n");
    printf("account balance = $%.2f\n", bal);
    printf("total contribution = $%.2f\n", t_payments);
    printf("tax exempt growth = $%.2f\n", bal-t_payments);
    
    printf("\nOption B\n");
    n_payment = 500.0;
    for (k=0; k<years; k++) {
        temp = pow(1.0 + i_rate, (double)months);
        bal_x = (temp - 1.0)/i_rate * n_payment;
        n_payment += 500.0 * .10;
        printf("%.2f\n", n_payment);
        printf("%.2f\n", bal_x);
    }
    Last edited by Salem; 12-01-2010 at 09:44 PM. Reason: Added code tags

  2. #2
    and the hat of wrongness Salem's Avatar
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    > n_payment += 500.0 * .10;
    1. You need 2 loops, for years and months. The payment per month increases once per year.

    And this isn't adding 10%
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    I tried to plan out my coding, but I just can't picture how I will be able to do option B

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