Formula for compound spending and time to exhaust fixed account

Discussion in 'Financial Planning' started by Joe D., Feb 14, 2006.

  1. Joe D.

    Joe D. Guest

    Does anybody know the formula for time to exhaust a fixed
    account value based on compound spending?

    E.g, you have a fixed $10,000 account, and spend $500
    the first year, increasing at 5% per year. What's the formula
    to calculate how long the account will last?

    I looked around and can't find it; only formulas for
    compound interest.

    Looking for the actual formula, not an automatic
    calculator.

    -- Joe D.
     
    Joe D., Feb 14, 2006
    #1
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  2. Joe D.

    Dave Dodson Guest

    Let's define some symbols: P0 = the initial principal in the fixed
    account ($10,000 in your example), A = the initial amount withdrawn
    ($500), r = the rate of increase of the withdrawal (5%), and n = the
    number of years. Then we need to solve the equation

    A * ((1+r)^n - 1) / r = P0.

    Thus, n = int(log(P0*r/A + 1) / log(1+r))

    Here, int() is the greatest integer function, and log is the base-10
    logarithm.

    In your example, n = 14 years. There would be a small balance in the
    account that would be insufficient to make the next year's
    distribution.

    Dave
     
    Dave Dodson, Feb 14, 2006
    #2
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  3. "Dave Dodson" <> wrote in message
    news:...
    > Let's define some symbols: P0 = the initial principal in the fixed
    > account ($10,000 in your example), A = the initial amount withdrawn
    > ($500), r = the rate of increase of the withdrawal (5%), and n = the
    > number of years. Then we need to solve the equation
    >
    > A * ((1+r)^n - 1) / r = P0.


    Yes, but where did that come from? - solving it is the easy part :).

    The withdrawal at the end of year 1 is $500
    The withdrawal at the end of year 2 is $500 * (1 + 5%)
    The withdrawal at the end of year 3 is $500 * (1 + 5%) ^ 2, ...

    The total withdrawals are $500 * [1 + (1 + 5%)^1 + (1+5%)^2 + ... +
    (1 + 5%)^(n-1) ]

    or algebraically, A * [ 1 + (1+r)^1 + (1+r)^2 + ... + (1+r)^(n-1)]

    Using the formula for the value of a finite geometric series
    http://mathworld.wolfram.com/GeometricSeries.html

    we get the expression on the left hand side.

    > Thus, n = int(log(P0*r/A + 1) / log(1+r))
    >
    > Here, int() is the greatest integer function, and log is the base-10
    > logarithm.


    Log can be any base - base 10, base e, base 2, it doesn't matter.

    As this is misc.invest.financial-plan, where would such a situation arise,
    where a drawing account were effectively kept under a pillow (i.e. in a
    non-interest bearing account)? I can't see that happening in a trust
    (fiduciary duty to invest prudently), but I suppose it might happen in some
    business accounts?
    --
    Mark Freeland
     
    Mark Freeland, Feb 14, 2006
    #3
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