Navigation Panel: Previous | Up | Forward | Graphical Version | PostScript version | U of T Math Network Home

# Regular Withdrawals on Compound Interest

Asked by An anonymous poster on July 18, 1997:
If a person has \$150,000 and it compounds at, let's say, eight per-cent per year, and that person (while all this is taking place) draws \$1,200 monthly from the fund, how long will the fund last until it is exhausted??? Is there a formula I can use to determine the time/amounts based on compounded interest etc.??? For example, if it was \$175,000 at 9 per-cent... how many years when drawing out \$1,300 per month?? I suspect you have a simple formula or way of doing it????? Appreciate it if you could e mail me your suggestion (short of taking a math class.) Thanks.
Your question is actually the same as the principles used in mortgage calculation. Here is how to derive the formulas:

Suppose the monthly interest rate is I, and that an amount W is withdrawn each month. What this means is that, during each month, the balance gets multiplied by 1+I (it becomes the original amount plus the interest, which is I times the original amount) and then has W subtracted from it.

Therefore, if B(0) is the starting balance, the balance after one month will be

B(1) = (1+I)B(0) - W.
For convenience, let J denote 1+I, so that B(1) = J B(0) - W. After two months the balance will be
B(2) = J B(1) - W = J (J B(0) - W) - W = J^2 B(0) - W - J W
and in general, after n months the balance will be
B(n) = J^n B(0) - W - J W - J^2 W - ... - J^(n-1) W.
There's a convenient formula for the sum 1 + J + ... + J^(n-1): it is (J^n-1)/(J-1) = (J^n - 1)/I. Therefore, abbreviating the starting balance B(0) to just "B",
B(n) = J^n B - (J^n - 1) W / I.
Your question is asking how large n has to be before the balance drops to zero; in other words, you want to solve for n in the equation
0 = J^n B - (J^n - 1) W / I.
Some basic algebra lets you rewrite this equation as J^n = W/(W-BI); taking logarithms of both sides gives
n log(J) = log(W) - log(W-BI)
so
n = log(W)/log(J) - log(W-BI)/log(J).
The only other thing you need to know before being able to solve your problems is what the monthly interest rate is. The monthly interest rate I and annual rate A are related by (1+I)^(12) = 1+A (because your balance is multiplied by 1+I each month and hence is multiplied by (1+I)^(12) each year, but we also know that each year it is multiplied by 1+A, so these two factors must be equal).

Therefore, J = 1+I = (1+A)^(1/12).

In your first example, A = 0.08 (8 percent), so J = 1.08^(1/12) = 1.0064... and I is roughly 0.0064. You have W = 1200 and B = 150000. Plugging these into the above formula gives

n = 254.30...
so the account would be exhausted during the 255th month (during the 22nd year).

In your second example, A = 0.09, W = 1300, B=175000, giving n = 489.28... so the account would be exhausted during the 41st year.

These formulas are usually used in mortgage calculations: the amount you owe is increased by the interest accruing on it, but is reduced by each of your monthly payments W (just as, in your example, your bank account increases by the interest earned but is recuced by your monthly withdrawals). In mortgages, however, n is known and W is what needs to be calculated. For example, in a 25-year mortgage at 8% on \$100,000, the bank needs to calculate a monthly payment amount W which will reduce the balance to zero after n=300 months. So the equation

0 = J^n B - (J^n - 1) W / I
needs to be solved for W. The solution is
W = J^n B I / (J^n - 1).
In this example, n=300, B = 100,000, J = 1.0064... and I=0.0064..., yielding W = 753.415... . Your monthly payments on such a mortgage would be \$753.42.

[ Submit Your Own Question ] [ Create a Discussion Topic ]

This part of the site maintained by (No Current Maintainers)
Last updated: April 19, 1999
Original Web Site Creator / Mathematical Content Developer: Philip Spencer
Current Network Coordinator and Contact Person: Joel Chan - mathnet@math.toronto.edu