# 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.

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