OpenSpecEngineering

Estimates how long a project will take, and the chance of finishing by a target date, from three-point estimates of the activities on its critical path. Enter an optimistic, a most likely and a pessimistic time for each activity. Project managers, planners and engineers use it to set realistic dates and to show how much schedule risk a plan carries [2].

The PERT equations

t_e = (a + 4m + b) / 6    σ² = ((b − a) / 6)²

For each activity, a is the optimistic time, m the most likely and b the pessimistic time. t_e is its expected time and σ² its variance, from the beta distribution PERT assumes for activity times [1].

T_e = Σt_e    σ_T = √(Σσ²)    z = (T_s − T_e) / σ_T

T_e is the expected project duration, the sum along the critical path, and σ_T the project standard deviation. Treating the path total as normal, the probability of finishing by the target T_s is Φ(z), the standard normal probability [3]. The duration for 95 % confidence is T_e + 1.645·σ_T.

Limits and common mistakes

• Enter only the activities on the critical path, in one time unit throughout (days or weeks). The page does not find the critical path for you.
• The method ignores near-critical paths. When another path is almost as long and has a wider spread, the real chance of finishing on time is lower than the page shows [3].
• The activity times are assumed independent. A cause that delays several activities at once, such as a late supplier, widens the real spread.
• The normal approximation is weak for a path of only two or three activities.
• Ask for the pessimistic time as a real worst case, not just a cautious estimate. Estimates that are too narrow give false confidence.

Worked example

A tooling change-over has five activities on its critical path, estimated in weeks: design (2, 3, 6), tool build (4, 6, 10), tryout (3, 4, 8), qualification (5, 8, 14) and launch (1, 2, 3). The customer wants it in 27 weeks.

1. Choose Activities on the Critical Path 5 and enter Target Completion Time (same time unit) 27.
2. For each activity, enter the optimistic (a), most likely (m) and pessimistic (b) times above, in order: Activity 1 is 2, 3, 6, up to Activity 5, 1, 2, 3.
3. Press Solve. The Solution shows: Expected project duration: 24.6667
   Project variance: 4.5
   Project standard deviation: 2.12132
   z for target time: 1.09994
   Probability of finishing by target: 86.4322 %
   Duration for 95% confidence: 28.1562

Check: the expected times are 3.33 + 6.33 + 4.5 + 8.5 + 2 = 24.67 weeks. The variances are 0.444 + 1 + 0.694 + 2.25 + 0.111 = 4.5, so σ_T = 2.12. z = 2.33 / 2.12 = 1.10, and a normal table gives 0.864. To commit at 95 %, quote about 28 weeks.