Periodic Withdrawals Calculator

Accumulation and decumulation are not mirror images. While you are saving, a bad year is an inconvenience you can wait out. While you are withdrawing, a bad year permanently removes capital that would otherwise have compounded, because the withdrawal happens whatever the balance has done.

This calculator models the drawdown period by period and reports the result three ways. Ask how long the money lasts and it returns a duration and the exact withdrawal number at which the balance reaches zero. Ask how much you can withdraw over a chosen number of years and it solves for the initial amount that lands on zero at the end. Ask what will be left after a fixed duration and it returns the ending balance, in nominal and in today's money.

Two things it makes explicit that most withdrawal calculators leave implicit. The first is timing: withdrawing at the start of a period rather than the end costs the balance a period of growth every time, and over three decades that is worth months of longevity. The second is escalation: a withdrawal that rises each year to keep pace with prices depletes a balance far faster than a level one, and the difference is not small.

It is decumulation arithmetic, not a retirement plan. It does not ask your age, model Social Security or a pension, order withdrawals across account types, or apply required minimum distributions. If the question is whether you have enough to retire, that is the accumulation problem and belongs to the retirement savings projection.

Every headline is a total of the schedule below it, not a separate formula. Total withdrawn, total growth, the ending balance and the depletion point all come from the same period loop, so a figure in the summary and the same figure in the table can never disagree by more than the rounding applied at display.

The year-one withdrawal rate is the annualised first-year withdrawal divided by the starting balance — nothing more. It is labelled a modeled rate deliberately: it is not a safe withdrawal rate, because safety is a claim about outcomes under uncertainty and this model contains no uncertainty at all.

Watch for the partial final withdrawal. When the balance cannot meet the scheduled amount, the model pays out what is actually there, marks the payment partial, and stops. On the default figures the last payment is $2,203.06 against a scheduled $6,139.22 — a detail that matters, because a plan whose last year is a third funded is not a plan that lasted 30 years.

"Not depleted" is a statement about the arithmetic, not about durability. When the return covers the withdrawal, a deterministic model never exhausts the balance, and the calculator says so against a stated 100-year calculation horizon rather than inventing a date centuries away.

The 3×3 matrix is where the honest answer lives. Half a point of return or half a point of annual increase moves the answer by years, and reading the corners tells you far more about whether a plan is robust than the central figure does.

The schedule is the model. Each period the balance is either drawn on and then grown, or grown and then drawn on, depending on the timing you choose, and the loop continues until the balance is exhausted or the run reaches its end. Every published total is accumulated inside that loop at full precision and rounded once, at the display boundary.

The expected return is treated as an effective annual rate. The period rate is its exact n-th root, so twelve monthly periods compound back to precisely the annual figure you entered — the nominal rate is not simply divided by twelve, which would quietly overstate the return.

Depletion is detected, not estimated. If a scheduled withdrawal exceeds the available balance, the model pays out only what is there, records the payment as partial, sets the ending balance to exactly zero and stops. If the requested amount is met precisely as the balance reaches zero, that is recorded as an exact depletion instead.

On the calculation horizon: the loop is bounded at 100 years and never runs unbounded. When the return covers the withdrawal the balance is never exhausted, and reporting a depletion date centuries out would be false precision — so the model reports that it does not deplete within the horizon and leaves it there.

On the inflation field: it restates figures in today's money and nothing else. It never changes a withdrawal. The annual increase is the field that does that, and keeping the two apart is deliberate — a withdrawal rising 1% a year against 3% inflation looks stable in dollars while losing a third of its purchasing power over a decade, and conflating the two inputs would hide exactly that.

Take $750,000, $3,000 withdrawn at the start of each month, a 5% expected return and a 2.5% annual increase in the withdrawal. Year one takes out $36,000, which is a modeled year-one withdrawal rate of 4.80%.

The balance lasts 29 years 5 months. It is exhausted at withdrawal 353, during year 30, having paid out $1,533,586.60 in total — of which $783,586.60 was growth earned during the drawdown rather than the original capital. The final payment is partial: $2,203.06 against a scheduled $6,139.22, which is the detail that separates a plan that lasted 30 years from one that ran out inside it.

The first year is instructive on its own. Withdrawals of $36,000 are outrun by $36,532.27 of growth, so the balance actually rises to $750,532.27 by the end of year one. It is the annual increase that eventually overtakes the return: by the final year the scheduled withdrawal is $6,139.22 a month, more than double where it started.

Switching the timing to end-of-period, with everything else unchanged, extends the life to 29 years 7 months. Two extra withdrawals is not a rounding difference — it is what a period of growth taken 353 times is worth.

Asked the second question instead — what can be withdrawn over exactly 30 years — the solver returns $2,956.45 a month, $35,477.40 in year one, a 4.73% modeled year-one rate. By year 30 that has escalated to $6,050.10 a month, worth $2,956.45 in year-one dollars at 2.5% inflation: the increase exactly preserves purchasing power, which is what it was set to do. Re-running the schedule at the solved figure leaves $1.35 after 30 years.

Set the annual increase to zero and the same solve returns $3,959.28 a month. That case has a closed form, and the standard annuity calculation returns $3,959.28 as well — the solver and the formula agree to the cent, which is the point of running both.

The table below isolates the assumption doing the most work.