Estimate with buffer.

Worked example

A task has a best case of 4 hours, a most likely time of 8 and a worst case of 18. The expected time is (4 + 4 x 8 + 18) / 6 = 9.0 hours and the standard deviation is (18 - 4) / 6 = 2.33 hours. The 90% range is 9.0 +/- 1.645 x 2.33, so 5.16 to 12.84 hours; the 95% range is 9.0 +/- 2 x 2.33, so 4.33 to 13.67 hours. A flat 50% buffer on the most likely 8 hours gives 12 hours.

How it works

For each task the tool uses the three-point formulas: expected = (optimistic + 4 x most likely + pessimistic) / 6 and standard deviation = (pessimistic - optimistic) / 6. For a list of tasks the expected times add up, and the total standard deviation is the square root of the sum of the squared task deviations, which assumes the tasks are independent.

The 68%, 90% and 95% ranges are expected +/- 1, 1.645 or 2 standard deviations, and the upper end of each is shown. The source notes these ranges assume the combined estimate is roughly normal, which typically takes 20 to 30 tasks, so with a handful of tasks treat them as rough guides rather than guarantees.

The flat buffer is the simpler habit of padding the most likely time by a percentage. It is a rule of thumb, not a published standard, and is shown only so you can compare it with the three-point result.

Sources and basis (checked 2026-10-08): Three-point estimation, Wikipedia (opened 2026-10-08), for the expected value and standard deviation formulas, the project-level combination and the 1, 1.645 and 2 standard-deviation ranges. The flat buffer is arithmetic on the number you enter.

FAQ

What is a three-point estimate?

It is an estimate built from an optimistic, a most likely and a pessimistic value. The usual weighting is (optimistic + 4 x most likely + pessimistic) / 6.

How much buffer should I add to an estimate?

The three-point method derives it from your own range: the gap between the expected time and the 90% or 95% figure. If you prefer a flat percentage, the tool shows that too, but no percentage is a standard.

What does the 90% figure mean?

About 90% of outcomes would fall inside expected +/- 1.645 standard deviations, so the upper end is a cautious planning figure. The figure shown is the expected time plus 1.645 standard deviations. With few tasks that is an approximation.

Why does the total spread grow more slowly than the task spreads added together?

Independent task errors partly cancel, so the total standard deviation is the square root of the sum of squares rather than the plain sum. This assumes the tasks do not all overrun for the same reason.

What unit should I use?

Any, as long as it is the same for every number. Hours and days both work.