PERT and probability of completion

Three-time estimates, beta-based expected time and variance, critical-path variance and the normal approximation for the probability of meeting a due date.

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Why it matters

In research, development, tooling and first-time installations, nobody knows activity durations exactly. PERT (Program Evaluation and Review Technique) replaces single estimates with three, gives each activity a mean and a variance, and turns the project duration into a probability statement: "we have an 88 % chance of finishing in 23 weeks". That is what a customer or management actually needs when a due date is negotiated.

Key ideas

Three time estimates per activity.

  • Optimistic time a (or tₒ): shortest time if everything goes right.
  • Most likely time m: the mode — the duration that would occur most often.
  • Pessimistic time b (or tₚ): longest time if things go wrong (excluding catastrophes).

Beta distribution assumption. Each activity time is assumed to follow a beta distribution between a and b with mode m. This gives the approximations te = (a + 4m + b)/6 for the mean and σ = (b − a)/6 for the standard deviation (the range a to b is taken as about six standard deviations).

Project duration.

  1. Use te as the activity duration and run the CPM forward and backward passes to find the expected project duration Tₑ and the critical path.
  2. Variance of the project duration = sum of the variances of the activities on the critical path (activities are assumed independent).
  3. By the central limit theorem, the sum of several independent activity times is approximately normal, even though each is beta.

Probability of completion. For a scheduled (due) date T_s, compute Z = (T_s − Tₑ)/σ_T and read P(T ≤ T_s) = Φ(Z) from the standard normal table. Z = 0 gives 50 %; negative Z means less than 50 %. Conversely, the due date for a required probability p is T_s = Tₑ + Z_p·σ_T (Z = 1.645 for 95 %).

Limitations to state in an answer.

  • Only the critical path's variance is used; a near-critical path with a large variance can actually be more likely to delay the project, so PERT tends to be optimistic. Check near-critical paths separately.
  • Independence of activity times and the beta approximation are assumptions.
  • The CLT approximation needs several activities on the path.

PERT vs CPM. PERT: probabilistic times, event-oriented, used for new, uncertain projects. CPM: deterministic times, activity-oriented, includes time-cost trade-off (crashing), used for repetitive construction or maintenance work.

Formulas

  • Expected time: te = (a + 4m + b) / 6
  • Standard deviation of an activity: σ = (b − a) / 6
  • Variance of an activity: σ² = ((b − a) / 6)²
  • Project mean: Tₑ = Σ te (critical path)
  • Project variance: σ_T² = Σ σ² (critical path)
  • Standard normal variate: Z = (T_s − Tₑ) / σ_T
  • Probability: P(T ≤ T_s) = Φ(Z)
  • Due date for probability p: T_s = Tₑ + Z_p · σ_T

Symbols: a, m, b = optimistic, most likely and pessimistic times (days or weeks); te = expected activity time; σ, σ² = activity standard deviation and variance (days, days²); Tₑ = expected project duration; σ_T = project standard deviation; T_s = scheduled date; Φ = standard normal cumulative probability (from the normal table).

Worked examples

Example 1 (standard). Activities (a, m, b in weeks; predecessors): A (2, 4, 6; –), B (3, 5, 13; –), C (4, 6, 8; A), D (1, 4, 7; A), E (2, 5, 14; B, D), F (3, 6, 9; C, E). Find the probability of finishing within 23 weeks.

  1. te: A = (2 + 16 + 6)/6 = 4; B = (3 + 20 + 13)/6 = 6; C = (4 + 24 + 8)/6 = 6; D = (1 + 16 + 7)/6 = 4; E = (2 + 20 + 14)/6 = 6; F = (3 + 24 + 9)/6 = 6 weeks.
  2. Variances: A (4/6)² = 0.444; B (10/6)² = 2.778; C 0.444; D (6/6)² = 1.000; E (12/6)² = 4.000; F 1.000 weeks².
  3. Paths: A–C–F = 16; A–D–E–F = 4 + 4 + 6 + 6 = 20; B–E–F = 18. Critical path A–D–E–F, Tₑ = 20 weeks.
  4. σ_T² = 0.444 + 1.000 + 4.000 + 1.000 = 6.444 weeks²; σ_T = 2.539 weeks.
  5. Z = (23 − 20)/2.539 = 1.18; Φ(1.18) = 0.881.
  6. P(finish within 23 weeks) ≈ 88 %. Check the near-critical path B–E–F: mean 18, σ = √7.778 = 2.789, Z = 5/2.789 = 1.79, P = 0.963 — it is unlikely to govern.

Example 2 (GATE-type). The critical path of a project has three activities with expected times 12, 18 and 10 days and variances 4, 9 and 3 days². Find (a) P(completion within 46 days), (b) P(completion within 36 days), (c) the due date for 95 % confidence.

  1. Tₑ = 12 + 18 + 10 = 40 days; σ_T² = 4 + 9 + 3 = 16 days²; σ_T = 4 days.
  2. (a) Z = (46 − 40)/4 = 1.5 → P = Φ(1.5) = 0.933.
  3. (b) Z = (36 − 40)/4 = −1.0 → P = Φ(−1) = 1 − 0.841 = 0.159.
  4. (c) T_s = 40 + 1.645 × 4 = 46.58 days (about 47 days).
  5. Note: standard deviations are not added — 2 + 3 + 1.73 = 6.73 would be wrong; variances are added, then the root taken.

Common mistakes

  • Adding standard deviations of activities instead of variances.
  • Using all activities' variances instead of only those on the critical path.
  • Using the formula σ = (b − a)/6 but forgetting to square it for the variance.
  • Mixing up a and b, which makes σ negative inside the bracket but hides the error after squaring — always check a ≤ m ≤ b.
  • Reading Φ(Z) for a negative Z directly from a table that lists only positive Z without using Φ(−Z) = 1 − Φ(Z).
  • Treating te as the most likely time m; they differ unless the distribution is symmetric.

For GATE PI

  • te and σ² of an activity from three estimates (quick NAT).
  • Project mean, standard deviation and probability of completion by a given date using a supplied Z-table value.
  • Due date for a given probability; reasoning about which path governs.
  • Practise the arithmetic of sixths and squares; express probabilities to three decimals.

Quick check

  1. a = 4, m = 7, b = 16 days. Find te and σ².
  2. Tₑ = 30 days, σ_T = 3 days. What is P(T ≤ 30)?
  3. Critical-path variances are 1, 4 and 4. What is σ_T?
  4. Is the PERT estimate of completion probability usually optimistic or pessimistic? Why?

Answers: 1. te = (4 + 28 + 16)/6 = 8 days; σ² = (12/6)² = 4 days². 2. 0.5. 3. √9 = 3. 4. Optimistic — it ignores near-critical paths that may run longer.

Interview questions

All Operations Research interview questions

Try answering each one aloud before you open it.

  1. 1.What is PERT in operations research?Concept

    PERT stands for Program Evaluation and Review Technique. It is a project management tool used to schedule, organize, and coordinate tasks within a project. PERT is particularly useful for projects where the time required to complete different tasks is uncertain. It helps in identifying the critical path and estimating the minimum time needed to complete the entire project.

  2. 2.Explain the difference between PERT and CPM.Concept

    Both use a precedence network and the critical-path logic. PERT treats activity times as random, with three estimates per activity giving a mean and a variance, and answers probability questions about meeting a due date; it suits new, uncertain R&D-type projects and is event-oriented. CPM uses single deterministic durations, is activity-oriented, and adds the time-cost trade-off (crashing) for repetitive work such as construction and maintenance where durations and costs are well known.

  3. 3.What are the three time estimates used in PERT?Concept

    PERT uses three time estimates for each activity: optimistic time (O), which is the shortest time in which the activity can be completed; pessimistic time (P), which is the longest time the activity might take; and most likely time (M), which is the best estimate of the time required to complete the activity, assuming everything proceeds as normal. These estimates are used to calculate the expected time for each activity.

  4. 4.How is the expected time for an activity calculated in PERT?Numerical

    Assuming the activity time follows a beta distribution between the optimistic time a and pessimistic time b with mode m, the expected time is te = (a + 4m + b)/6 — a weighted average with weight 4 on the most likely time. The standard deviation is approximated as (b − a)/6, so the variance is ((b − a)/6)².

  5. 5.Explain how PERT can be used to assess the probability of completing a project by a certain deadline.Application

    Find the critical path using expected times; the project mean Tₑ is the sum of their te values and the project variance is the sum of their variances (activities assumed independent). By the central limit theorem the project time is taken as normal, so Z = (deadline − Tₑ)/σ_T and the probability is Φ(Z) from the normal table. The answer is somewhat optimistic because near-critical paths with large variance are ignored.

  6. 6.Calculate the expected time for an activity with optimistic time of 3 days, most likely time of 5 days, and pessimistic time of 9 days.Numerical

    Using the PERT formula for expected time: TE = (O + 4M + P) / 6, where O = 3 days, M = 5 days, and P = 9 days. TE = (3 + 4*5 + 9) / 6 = (3 + 20 + 9) / 6 = 32 / 6 = 5.33 days.

  7. 7.How does PERT handle uncertainty in project scheduling?Application

    PERT handles uncertainty in project scheduling by using probabilistic time estimates for each activity. By considering optimistic, pessimistic, and most likely times, PERT provides a more realistic estimate of project duration. This approach allows project managers to assess risks and plan for potential delays, making it easier to manage projects with uncertain timelines.

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