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Your Savings Will Last About
122
months
Lasts indefinitely? No
Duration in years 10.15 years
Duration in months 121.84 months

Estimate assumes a constant interest rate, constant monthly withdrawals taken at period end, and no taxes or inflation adjustments.

What this calculator does

The Savings Withdrawal Duration Calculator tells you how long a lump sum of savings will last if you withdraw a fixed amount every month while the remaining balance keeps earning interest. It answers the everyday retirement and budgeting question: "If I have this much saved and take out this much each month, when does it run out?"

Declining savings balance over time with monthly withdrawals
Regular withdrawals draw down the balance while interest slows the decline, until the money runs out.

How to use it

Enter your current savings balance, the amount you plan to withdraw each month, and the annual interest rate your account earns. The calculator converts the annual rate to a monthly rate, applies the annuity-depletion formula, and reports the result in both months and years. If the interest earned each month is equal to or greater than your withdrawal, the balance never falls — the tool reports that your savings last indefinitely.

The formula explained

With a positive monthly rate \(r\), the number of withdrawals is $$n = \frac{-\ln\!\left(1 - \dfrac{r \cdot P}{\text{PMT}}\right)}{\ln(1 + r)},$$ where \(P\) is the starting balance and \(\text{PMT}\) is the monthly withdrawal. The term \(rP\) is the first month's interest; if \(\text{PMT}\) exceeds it, the balance shrinks and the logarithm is defined. When the rate is zero, the formula simplifies to $$n = \frac{P}{\text{PMT}}.$$

Diagram of variables in the withdrawal duration formula
The formula combines starting balance \(P\), monthly withdrawal \(\text{PMT}\), and monthly rate \(r\) to find the number of months \(n\).

Worked example

Suppose you have $100,000, withdraw $1,000 per month, and earn 4% annually. The monthly rate is \(0.04 / 12 = 0.0033333\). First-month interest is \(100{,}000 \times 0.0033333 = \$333.33\), which is less than $1,000, so the balance depletes. Plugging in: $$n = \frac{-\ln(1 - 0.0033333 \times 100{,}000 / 1{,}000)}{\ln(1.0033333)} = \frac{-\ln(0.66667)}{0.0033278} \approx \frac{0.405465}{0.0033278} \approx 121.8 \text{ months},$$ or about 10.2 years.

FAQ

What if my withdrawal is small? If your monthly withdrawal is less than or equal to the monthly interest, the principal grows or stays flat, so the calculator shows your savings last indefinitely.

Does it account for inflation or taxes? No. It assumes a constant rate, constant withdrawals, and ignores taxes and inflation. Real results may be shorter once those are considered.

When are withdrawals assumed to occur? At the end of each period (ordinary annuity), with interest credited monthly.

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