The Vendor API Estimate and the 'Worst Case' Date
Activity 'Integrate vendor API' has three-point inputs: optimistic 2 days (SDK works on first attempt), most likely 5 days (one round of credential debugging), pessimistic 14 days (the vendor's sandbox lacks a required endpoint and a workaround must be hand-built). It is the long-lead item on a 9-activity path. The sponsor wants a single committed date for a board presentation and proposes using 14 days, arguing that 'we agreed that is the worst case, so we cannot be late.' Separately, a scheduler proposes simply averaging the three numbers to 7 days and calling it the duration. The estimate is currently the only item on the path with a wide range.
1. Compute the PERT expected duration and standard deviation, compare with the scheduler's 7-day average, and explain why the two differ in the direction they do.
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PERT (beta-weighted):
- Mean = (O + 4M + P) / 6 = (2 + 4×5 + 14) / 6 = (2 + 20 + 14) / 6 = 36 / 6 = 6.0 days
- Standard deviation = (P − O) / 6 = (14 − 2) / 6 = 2.0 days
- Variance = 4.0
Triangular (simple average): (2 + 5 + 14) / 3 = 21 / 3 = 7.0 days
Why they diverge: the two formulas weight the mode differently. PERT gives the most likely value a weight of 4 out of 6 - two-thirds of the total - so a mode that sits well below the midpoint of the range pulls the mean down toward it. The triangular formula treats all three points equally, so the long pessimistic tail has more pull. Here M = 5 sits below the midpoint of 8, so PERT lands 1 day under the triangular result.
That gap is a modelling signal, not trivia: a mode below the range midpoint means the distribution is right-skewed - the common shape for integration work, where 'normal' is quick and failure is slow. In that case PERT is usually the better central estimate, and the triangular figure double-counts the tail into the plan.
Either way, do not report 6.0 as a promise: it is the central tendency of a distribution with σ = 2, not a date.
2. The sponsor wants to commit to 14 days on the grounds that it is the worst case. What commitment date would you offer, and how do you justify it numerically?
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Refuse 14 as the plan, and explain that it is not a worst case either - it is an assumption that the tail is bounded.
What 14 actually means: under the PERT beta approximation the range is treated as roughly ±3σ, so P = μ + 3σ = 6 + 6 = 12... which is already short of 14. Practically: a date at μ + 3σ corresponds to roughly a 99.9% confidence for a normal-shaped tail, and PERT explicitly assumes the extremes are achievable-but-unlikely, not impossible. Fourteen days is therefore an expensive over-commitment, and it still carries non-zero tail risk (vendors miss deadlines that were 'agreed').
A defensible ladder from μ = 6.0, σ = 2.0:
- 8 days (μ + 1σ): ≈ 84% confidence. Reasonable internal target with a contingency reserve held elsewhere.
- 10 days (μ + 2σ): ≈ 98% confidence. Reasonable external commitment for a board-level date on a long-lead item.
- 12 days (μ + 3σ): ≈ 99.9%. Reserve territory, not plan territory.
What to say: 'Eight days is our working plan with 84% confidence; ten days is the date we will publish externally. Fourteen is not a worst case - it is a number that will cost us a week of float on every successor and still will not be a guarantee.' Keep the difference between the plan date and the commitment date visible as explicit contingency reserve, not as padding baked into the activity's duration.
3. Mid-project the vendor confirms the sandbox endpoint will not be provided. P is revised from 14 to 26 days, O and M unchanged. What happens to the mean and the standard deviation, and why is the σ change the more important one?
Read the worked answer
Recompute both:
- Mean = (2 + 20 + 26) / 6 = 48 / 6 = 8.0 days (up 2.0)
- σ = (26 − 2) / 6 = 24 / 6 = 4.0 days (up from 2.0 - doubled)
- Variance = 16.0 (up from 4.0 - quadrupled)
Why σ matters more than the 2-day mean shift:
- Variance adds, means just add. On a path, expected durations sum linearly, but variances sum and σ is the square root of the total. A single activity with σ jumping from 2 to 4 injects +12 days of variance into the path - it can dominate every other activity's contribution combined.
- Probability is driven by σ. A confidence date at μ + 2σ moves from 6 + 4 = 10 days to 8 + 8 = 16 days. The committed date moves 6 days for a 2-day change in the mean, because the confidence band widened at the same time.
- It changes the risk response. A mean shift is handled by re-baselining and re-sequencing; a variance shift is handled by removing the uncertainty (build the workaround in a timeboxed spike, get the vendor's commitment in writing with a date, or split the activity so the uncertain part is isolated behind a decision gate).
Report to the sponsor in exactly those terms: 'the expected date moved two days; the range moved six. Re-baseline the date, and here is the spike that shrinks the range.'