Operations & Supply Chain
Three-Point Beta-PERT Expected Duration, Critical-Path Variance & P95 Buffer
Estimate asymmetric task durations with Beta-PERT weighted mean (O + 4M + P)/6 and variance ((P - O)/6)^2, then aggregate critical-path standard deviation via root-sum-of-squares.
Why PERT weights Most Likely 4x and adds task variances rather than standard deviations
Engineering cutovers and R&D milestones exhibit right-skewed risk: a 6-day task rarely finishes in 1 day, yet blocker discoveries can stretch it to 21 days. Approximating task completion on [O, P] with a four-parameter Beta-PERT distribution yields expected mean mu = (O + 4*M + P) / 6 (pulling the mean above M when P - M > M - O) and standard deviation sigma = (P - O) / 6.
Across independent sequential tasks on a critical path, expected durations sum linearly (SUM(mu_i)), whereas standard deviations combine in quadrature: sigma_path = SQRT(SUM(sigma_i^2)). Naive addition of task standard deviations over-buffers multi-task schedules dramatically.
| Critical-Path Task | (O, M, P) Days | PERT Mean (O+4M+P)/6 | Task Variance ((P-O)/6)^2 |
|---|---|---|---|
| Schema Migration | (3, 6, 21) | 8.00 days | 9.00 (SD = 3) |
| API Cutover | (2, 8, 26) | 10.00 days | 16.00 (SD = 4) |
| Smoke Freeze | (4, 4, 4) | 4.00 days | 0.00 (SD = 0) |
| Critical Path Rollup | Total = 22.00 days | Path Var = 25 -> Path SD = 5.00 | P95 = 22 + 1.645*5 = 30.23 days |
Task PERT Expected Mean & Variance Formulas
Compute weighted PERT mean (B2 + 4 * C2 + D2) / 6 in E2 and task variance ((D2 - B2) / 6) ^ 2 in F2.
=(B2 + 4 * C2 + D2) / 6Root-Sum-of-Squares Path SD & 95th Percentile Completion Date Buffer
Take SQRT(SUM(F2:F4)) for path standard deviation and add 1.645 * Path_SD to expected duration SUM(E2:E4) for a one-sided 95% confidence commitment.
=ROUND(SUM(E2:E4) + 1.645 * SQRT(SUM(F2:F4)), 2)Try the example
Use the original values or change them to test your understanding.