Formula for compound spending and time to exhaust fixed account

Joe D.
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 02-14-2006, 04:51 PM
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.

Dave Dodson
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 02-14-2006, 06:16 PM
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account (\$10,000 in your example), A = the initial amount withdrawn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

Mark Freeland
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 02-14-2006, 07:35 PM
"Dave Dodson" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ups.com...
> 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 account (\$10,000 in your example), A = the initial amount withdrawn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
--
Mark Freeland
(E-Mail Removed)