FIRE: How Much Money Do You Need to Retire?

Spending that grows with inflation, a portfolio that earns a steady yield, and a horizon of $n$ years: under these assumptions the pot that is spent down to exactly zero has a closed formula. This note derives it, then lets you run your own numbers — including the year you can stop working.

The question

FIRE (Financial Independence, Retire Early) reduces to a single number: how much is enough? The popular 25× rule prices a near-perpetual portfolio. Here we answer a sharper version of the question: enough to fund exactly $n$ more years of inflation-growing spending — no more, no less.

The model

Write $S$ for this year's spending, $i$ for inflation, $r$ for the yield on investments, and $R_i = 1+i$, $R_y = 1+r$. Assume:

a full year of spending is withdrawn at the start of each year; whatever remains stays invested and earns $r$ over that year;
spending inflates at $i$: year $t$ costs $S\,R_i^{\,t-1}$ (year 1 is at today's prices);
the yield $r$ is the same every year;
the money must last exactly $n$ years — the pot reaches zero at the end of year $n$, leaving nothing behind;
while you are still working, spending is zero in the model: the yearly saving figure of §5 is what is left after all spending, so spending is already accounted for;
savings are deposited at the end of each year, and the deposit grows with inflation (income is assumed to keep pace with prices).

From recurrence to formula

Start with $P$; withdraw, let the rest grow, and repeat until nothing is left after $n$ withdrawals:

$$\Bigl(\bigl((P-S)\,R_y - S R_i\bigr)R_y - S R_i^{2}\Bigr)R_y - \cdots \;=\; 0. \tag{1}$$

Expanding the brackets:

$$P\,R_y^{\,n} \;-\; S\bigl(R_y^{\,n} + R_i R_y^{\,n-1} + R_i^{2} R_y^{\,n-2} + \cdots + R_i^{\,n-1} R_y\bigr) \;=\; 0. \tag{2}$$

Divide through by $R_y^{\,n}$: the sum turns geometric with ratio $x = R_i / R_y$, and

$$P \;=\; S\sum_{k=0}^{n-1} x^{\,k} \;=\; S\cdot\frac{1-x^{\,n}}{1-x}, \qquad x=\frac{1+i}{1+r}. \tag{3}$$

If $r = i$ then $x = 1$ and (3) collapses to $P = S\,n$: the yield only treads water against inflation, so you need the full $n$ years of spending up front.

How much do you need today?

Parameters

Monthly spending ($) Inflation i (%/yr) Yield r (%/yr) Years it must last n

Figure 1 — the pot is drawn down to exactly zero at the end of year $n$, while the yearly withdrawal keeps rising with inflation.

Table 1 — year-by-year drawdown for the parameters above.

Year	Spending	After withdrawal	Interest	Value at year end

When can you retire?

Now the saving side. Suppose you put aside $D$ per year — already net of all spending, per assumption (e) — growing with inflation, deposited at year end, compounding at $r$. Meanwhile the amount you would need in order to retire falls as the years still to be funded shrink: with $n-t$ years left, plug $n-t$ into (3):

$$\mathrm{Need}_t \;=\; S\cdot\frac{1-x^{\,n-t}}{1-x}. \tag{4}$$

The first year your savings overtake $\mathrm{Need}_t$ is your FIRE year — the highlighted row of Table 2. ($\mathrm{Need}_t$ is stated in today's spending power; see remark 3.)

Parameters — saving side

Saving per year ($) Already saved today ($)

Figure 2 — total saved vs. the amount still needed; the dot marks the earliest retirement year.

Table 2 — savings vs. the amount needed to retire (first sufficient year highlighted; rows stop a few years past the crossover).

Year	Saved this year	Total saved (year end)	Needed to retire	Surplus

Remarks

$r$ is assumed constant. Real portfolios deliver returns in a random order, and bad early years hurt a retiree disproportionately (sequence-of-returns risk) — prefer a conservative $r$.
$i$ and $r$ are nominal and before tax and fees; use your own net figures.
$\mathrm{Need}_t$ in Table 2 keeps spending at today's level. To state it in year-$t$ dollars instead, multiply by $R_i^{\,t}$ — the crossover then arrives a few years later.
$n$ is yours to choose: 70 suits a 30-year-old planning to age 100. Try 60 or 50 and watch $P$ fall.

Model numbers, not financial advice.