# Formula for retirement withdrawal amount same as principal decreases

I created a retirement calculator

The idea is if you save enough you can so when you withdraw say 4% you don't eat too much into your savings.

But based on my calculation 4% of 500000 is \$20k but next year it will be 4% of 480000 and so on.

What formula will determine what % of withdrawal will keep the withdrawal amount the same over a set time period or until the principal is exhausted?

• If you are assuming that there will be no gains during this same period of time, then it is impossible to withdraw both the same percentage and the same dollar amount every year. – GendoIkari Apr 4 at 18:40
• Look here: money.stackexchange.com/a/94307/11768 – Chris Degnen Apr 4 at 18:40
• The 4% rule is based on capital invested in such a way as to gain (on average) at least 4%+inflation. So after withdrawing the \$20k, you'd actually be at more like \$515k thanks to capital/investment gains. – Kevin Apr 4 at 18:52
• Also, the 4% rule is to take 4% of the initial capital every year, indexed to inflation, regardless of what happens to the investments. – Tavian Barnes Apr 4 at 19:17

The formula that "will determine what % of withdrawal [of the initial principal] will keep the withdrawal amount the same over a set time period or until the principal is exhausted" is

``````percentage = 100/number of years

i.e. with savings     s = 500000
no. years   n = 25

percentage = 100/n = 4

Check: annual withdrawal  w = 0.04 s = 20000
n w = s
``````

However, taking into account interest earned on the savings

``````with interest   r = 5% annually

percentage = 100 r (1 + 1/((1 + r)^n - 1)) = 7.095246

i.e. annual withdrawal  w = 0.07095246 s = 35476.23
`````` This depends on the withdrawals being at year-end, so the last withdrawal exhausts the principal at the end of the last year. It also means the first withdrawal happens at the end of the first year, so the savings interest accrual of the first year is included in the calculation. If you want immediate withdrawal you can adjust the start point.

``````For example, with 25 withdrawals starting immediately

no. years   n = 24
r = 0.05
s = 5000000

percentage = 100 r (1 + 1/((1 + r)^n - 1)) = 7.24709

annual withdrawal  w = 0.0724709 (s - w)

∴  w = 0.0724709 s/(1 + 0.0724709) = 33786.88
`````` • You seem to forget the income tax incurred by the withdraw. – scaaahu Apr 5 at 9:30
• It wasn't specified in the question and would be country specific anyway. – Chris Degnen Apr 5 at 9:45
• @scaaahu Income tax would also be account-type dependent. Some tax-advantaged accounts have no taxes on qualified withdrawals. Plus, income tax, if any, would be dependent on total taxable income including from other sources (social security benefits, company pensions, etc.) and should be calculated once those values are known. – Chris W. Rea Apr 5 at 11:58

The way the question is written, the answer is that your withdrawal rate must be whatever the constant growth rate of the account is. In other words, if your account is worth \$100 and makes 5% each year, you can safely withdraw \$5, or about 4.762%, of the EOY balance (or 5% of the BOY balance) each year perpetually.

However, you what you probably mean to ask is what the withdrawal amount should be to amortize the account over a given period. The balance B at month m = 0 is B(0) = p0. Assume a monthly interest rate of r and a total number of months M. The balance at month B(m + 1) = r * p0 - X, where X is the monthly withdrawal rate. If we want to exhaust the money after M months then we get B(M) = 0. Let's write out some terms:

``````m    B(m)
---------
0    p0
1    r * p0 - X
2    r * (r * p0 - X) - X = r^2 * p0 - r * X - X
3    r * (r^2 * p0 - r * X - X) - X = r^3 * p0 - r^2 * X - r * X - X
…
k    r^k * p0 - X * (r^(k-1) + r^(k-2) + … + 1)
``````

We can use the partial sum formula to simplify the sum of powers of r to get:

``````B(k) = r^k * p0 - X * (r^k - 1) / (r - 1)
``````

When k = M we require that B(k) = 0:

``````B(M) = r^M * p0 - X * (r^M - 1) / (r - 1) = 0
r^M * p0 = X * (r^M - 1) / (r - 1)
r^M * p0 * (r - 1) / (r^M - 1) = X
``````

So, if p0 = \$100k, r = 5% per year = 1.004074 and M = 240 (20 years), then the monthly withdrawal rate is:

``````X = 2.653219 * \$100k * (0.004074) / (1.653219)
~ \$653.82
``````

This is basically the same way mortgage amortization is done. Note that this is taking out the same payment but different percentages, which is probably a lot closer to what you're really looking for.

• Yes, same method as my answer. With `s = 100k`, `n = 240`, `r = (1 + 0.05)^(1/12) - 1` then `w = r (1 + 1/((1 + r)^n - 1)) s = 653.84` (with very slight rounding difference) – Chris Degnen Apr 5 at 15:33