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I am trying to come up with a formula to determine how much a person will need in retirement. So every year, they will take their inflation-adjusted withdrawals out of their retirement savings, and the remainder of their savings will accrue a return based on the anticipated return rate of their investments. I am content to run this calculation only one time per year. I am trying to solve for that initial amount, the total that they need to have saved in year 0 of their retirement, in order to have 0 dollars left in the year they expect their retirement to end?

For some extra context, most retirement calculators will let a person input how much they are contributing, their retirement expenses, etc. and then will tell them if they will meet their retirement goals. To me, that's totally backwards. Why would you make a user fiddle with sliders until they find something that works, instead of just getting the correct inputs and telling them what works and how to get there?

rook218
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Although formulas for this exist already, you can kind of reason it out. Suppose you retire with $P$ dollars in your account, which generate interest at a rate of $i$ (if, say, $i=0.05$, then your interest rate is 5%). Suppose you anticipate your retirement to end in $t$ years, and at the end of each year (right before your interest is calculated), you spend $S$ dollars on rent and insurance and whatever.

At the end of year 1, you will have $(P-S)(1+i)$ dollars.

At the end of year 2, you will have $((P-S)(1+i)-S)(1+i) = (P-S)(1+i)^2-S(1+i)$ dollars.

At the end of year 3, you will have $((P-S)(1+i)^2 - S(1+i) - S)(1+i) = (P-S)(1+i)^3 - S(1+i)^2 - S(1+i)$ dollars.

Hopefully you can see the pattern now. You subtract your expenses ($S$), then multiply by the interest rate $(1+i)$ each time. With some simplification, you can see that the amount of money you'll have after $t$ years is $(P-S)(1+i)^t -S(1+i)^{t-1} - S(1+i)^{t-2} - ... - S(1+i)\\ = P(1+i)^t - S((1+i)^t + (1+i)^{t-1} + ... + (1+i))\\ = P(1+i)^t - S\left( \sum_{n=1}^t(1+i)^n \right)\\ = P(1+i)^t - S\left(\sum_{n=0}^t(1+i)^n-1 \right)\\ = P(1+i)^t - S\left( \frac{1-(1+i)^{t+1}}{1-(1+i)}-1 \right)\\ = P(1+i)^t - S\left( \frac{(1+i)^{t+1}-1}{i}-1 \right)\\ = P(1+i)^t - S\left( \frac{(1+i)^{t+1}-1-i}{i} \right)\\ = P(1+i)^t - S\left( \frac{(1+i)^{t+1}-(1+i)}{i} \right)\\ = P(1+i)^t - S\left( \frac{(1+i)((1+i)^{t}-1)}{i} \right)$

Maybe I made a mistake somewhere, who knows. But usually formulas like this can be derived with a few iterations and some summation formulas (usually the geometric series formula works fine).

IAAW
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  • Thank you very much, that's exactlyt he kind of answer I was looking for! You said, "formulas for this exist already" - do you know if this formula already has a name? – rook218 Feb 05 '23 at 16:16
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    Not sure if this answer will be useful, since this sort of math is honestly too applied for me, but here might be a good resource. In general, just look up "compound interest with investment" or something like that. If you are taking money out instead of putting it in, make your "investment" a negative number I guess. – IAAW Feb 05 '23 at 18:53