Calculating Fixed Monthly Withdrawals from an Interest-Bearing Balance

Summary

The document presents a savings drawdown problem: determine a level monthly withdrawal that exhausts an initial bank balance after a fixed number of months, with a stated monthly effective interest rate and no further deposits. The answer treats the withdrawals as an ordinary annuity and equates the starting balance to the present value of the payment stream. Solving the annuity formula for the payment gives the monthly amount.

The example substitutes the stated balance, rate, and number of payments and reports a monthly withdrawal of $432.5186. The result depends on the timing convention: this formula assumes each withdrawal occurs at the end of a month, after interest accrues. Withdrawals at the beginning of each month would require an annuity-due adjustment. This is a basic time-value-of-money calculation rather than a trading strategy, but it illustrates how periodic cash flows and compounding determine a sustainable drawdown.

Key ideas

• A fixed withdrawal stream can be valued as an ordinary annuity when payments occur at period end.
• The present value of the payments must equal the initial account balance to exhaust it exactly.
• Solving the annuity present-value formula for the payment gives the constant monthly withdrawal.
• The calculation depends on the stated periodic interest rate and number of withdrawals.
• Beginning-of-period withdrawals require a different timing adjustment.

Tags

# Withdrawing monthly from a bank for 40 years


# Withdrawing monthly from a bank for 40 years












Consider you have $\$104107.4099$ in the bank with a $.33\%$ monthly effective interest rate. You plan to withdraw a fixed amount X every month for 40 years, such that you make 480 withdrawals in total, without making any deposits.

I need to find X such that there will be $\$0$ in the bank after the last withdrawal.

My (tentative) work:

So after the first month we have $(104107.4099-X)(1+.0033)=Y_1$. After the second month we have $(Y_1-X)(1+.0033)=Y_2$. After the third month we have $(Y_2-X)(1+.0033) = Y_3$, and so on until we get to the last withdrawal $(Y_{479}-X)$.

I was thinking of using the future cash flows formula in some way:

$P(480)=104107.4099-X\sum_{k=1}^{479}(1+.0033)^k=0$. But I know this does not work because we would only have interest on X.

Or

$P(480)=(104107.4099-X)\sum_{k=1}^{479}(1+.0033)^k$ but them solving for this gives you that $X=104107.4099$.

I am finding this problem very difficult, any help would be appreciated. Thank you.

## Answer by AlRacoon (score 3, accepted)

https://quant.stackexchange.com/a/37963

This is a annuity calculation.

Present Value of Annuity $= \text{Payment} \cdot \frac{1-(1+r)^{-n}}{r}$

Therefore:

Payment = Present Value of Annuity $\cdot \frac{r}{1-(1+r)^{-n}}$

Present Value of Annuity $= \$\,104,107.4099\,;\,\,\, r = 0.33\,\%;\,\,\, n = 480$

Monthly Payment $ = \$\, 432.5186$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.