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Strategic Cost Management · Network Analysis - PERT, CPM

PERT Three Time Estimates and Probability of Completion

Updated 11 October 2026 · Fact-checked

PERT handles uncertain activity times using three estimates: optimistic (to), most likely (tm) and pessimistic (tp). Expected time is (to + 4tm + tp) ÷ 6 and variance is ((tp − to) ÷ 6)². Add expected times and variances along the critical path, then use Z = (T − TE) ÷ σ with the normal table for the probability.

Understand PERT: Three Time Estimates and Probability

In many projects, such as a new plant installation or an ERP roll-out, nobody knows exactly how long each activity will take. CPM uses one fixed time per activity. PERT (Programme Evaluation and Review Technique) admits the uncertainty and asks for three estimates per activity.

  • Optimistic time (to): the shortest time if everything goes well.
  • Most likely time (tm): the time you expect under normal conditions.
  • Pessimistic time (tp): the longest time if things go badly, excluding disasters.

PERT assumes activity times follow a beta distribution. It turns the three estimates into one expected time (te) that gives four times the weight to the most likely value. The spread between tp and to is treated as roughly six standard deviations, so the variance of an activity is ((tp − to) ÷ 6)². A wide gap between tp and to means a risky activity.

Once you have te for every activity, you draw the network and find the critical path exactly as in CPM, using te as the duration. The sum of te along the critical path is the expected project duration (TE).

For the probability, add the variances of the critical-path activities only. This assumes the activities are independent. The square root is the project standard deviation (σ). By the central limit theorem, the project duration is taken as approximately normal with mean TE and standard deviation σ. You convert any target date T into a Z value and read the probability from the normal table.

PERT vs CPM: PERT uses three time estimates, so it is probabilistic and suits new or uncertain projects. CPM uses one deterministic time and also deals with time-cost trade-off (crashing), so it suits repetitive projects with known durations. PERT is event-oriented; CPM is activity-oriented.

Key rules to remember

Expected time of an activity
te = (to + 4tm + tp) ÷ 6
Use te as the duration when finding the critical path.
Variance of an activity
σ² = ((tp − to) ÷ 6)²
Only to and tp enter the variance. tm does not.
Standard deviation of an activity
σ = (tp − to) ÷ 6
Do not add standard deviations across activities. Add variances.
Expected project duration
TE = Σ te of critical-path activities
If two paths tie as critical, treat the question as asking for one path unless it says otherwise.
Project variance and standard deviation
σ²(project) = Σ σ² of critical-path activities; σ(project) = √σ²(project)
Assumes activity times are independent.
Standard normal variable
Z = (T − TE) ÷ σ(project)
T is the target or scheduled completion time. Probability of finishing by T = Φ(Z) from the normal table.
Time for a given probability
T = TE + Z × σ(project)
Use for questions like: in how many days is completion 95% likely? Z is about 1.645 for 95% (one-sided).

How to solve PERT: Three Time Estimates and Probability questions

Follow the same sequence for every PERT question. Keep the workings in a small table so the examiner can award method marks.

  1. 1Tabulate each activity with to, tm and tp.
  2. 2Compute te = (to + 4tm + tp) ÷ 6 for every activity.
  3. 3Compute variance = ((tp − to) ÷ 6)² for every activity.
  4. 4Draw the network using te as the duration. Find all paths and pick the longest one. That is the critical path and its length is TE.
  5. 5Add the variances of the critical-path activities only. Take the square root to get σ of the project.
  6. 6Compute Z = (T − TE) ÷ σ, where T is the target time given in the question.
  7. 7Read Φ(Z) from the normal table. If Z is negative, the probability is 1 − Φ(|Z|). For a 'more than T' question, use 1 − Φ(Z).
  8. 8State the answer in words, for example 'There is about 86% probability that the project is completed within 18 days', and add a comment if a decision is asked.

Quickest way: Critical-path-only shortcut

When to use it: Use when the question gives three estimates and asks for probability. Skip variances of non-critical activities completely.

  1. Compute te for all activities first, since you need them to find the critical path.
  2. Find the critical path from te values.
  3. Compute variances only for activities on that path.
  4. Compute σ² total, then σ, then Z in one line.
  5. Read the table and write one sentence of conclusion.

Common mistakes in PERT: Three Time Estimates and Probability

  • Adding variances of all activities in the network instead of only critical-path activities.

    Students fill a variance column and add the whole column out of habit.

    Fix: Mark the critical path first. Add variances only for those activities.

  • Adding standard deviations instead of variances.

    Students treat σ like te, which is additive.

    Fix: Add variances, then take the square root of the total once.

  • Using ((tp − to) ÷ 6) without squaring, or dividing by 6 after squaring.

    Mixing up the standard deviation and the variance formulas.

    Fix: Write both lines: σ = (tp − to) ÷ 6, then σ² = σ squared. Check which one you are adding.

  • Choosing the critical path using tm or the most likely times.

    The most likely time looks like the 'normal' duration.

    Fix: The critical path is found from te values. Always compute te before drawing paths.

  • Using Z = (TE − T) ÷ σ, which flips the sign.

    Memory slip about which value is the mean.

    Fix: Z = (target − expected) ÷ σ. If the target is later than TE, Z is positive and the probability is above 50%.

  • Reading the table for the wrong tail when Z is negative.

    The table gives probabilities only for positive Z.

    Fix: For negative Z, probability = 1 − Φ(|Z|). Sketch a quick bell curve to check the answer is below 50%.

Worked examples

Example 1

A project has the following activities with time estimates in days (to, tm, tp): A (1-2): 2, 4, 12; B (2-3): 3, 5, 7; C (2-4): 4, 6, 14; D (3-5): 2, 3, 4; E (4-5): 1, 2, 9. Find the expected project duration and the probability of completing the project within 18 days. Use Φ(1.11) = 0.8665.

Show the solution
  1. Expected times: A = (2 + 16 + 12) ÷ 6 = 5; B = (3 + 20 + 7) ÷ 6 = 5; C = (4 + 24 + 14) ÷ 6 = 7; D = (2 + 12 + 4) ÷ 6 = 3; E = (1 + 8 + 9) ÷ 6 = 3.
  2. Variances: A = (10 ÷ 6)² = 100/36; B = (4 ÷ 6)² = 16/36; C = (10 ÷ 6)² = 100/36; D = (2 ÷ 6)² = 4/36; E = (8 ÷ 6)² = 64/36.
  3. Paths: A-B-D = 5 + 5 + 3 = 13 days. A-C-E = 5 + 7 + 3 = 15 days. The critical path is A-C-E, so TE = 15 days.
  4. Project variance = (100 + 100 + 64) ÷ 36 = 264 ÷ 36 = 7.33. σ = √7.33 = 2.71 days.
  5. Z = (18 − 15) ÷ 2.71 = 1.11.
  6. Probability = Φ(1.11) = 0.8665.

Answer: Expected duration is 15 days along A-C-E. The probability of completion within 18 days is about 86.7%.

Example 2

The critical path of a project has four activities with (to, tm, tp) in weeks: P: 6, 9, 18; Q: 10, 13, 22; R: 5, 8, 17; S: 6, 9, 18. Assume all other paths are shorter and activity times are independent. (a) Find the probability of completing in 47 weeks. Use Φ(1.00) = 0.8413. (b) Find the time by which completion is 95% likely. Use Z = 1.645.

Show the solution
  1. Expected times: P = (6 + 36 + 18) ÷ 6 = 10; Q = (10 + 52 + 22) ÷ 6 = 14; R = (5 + 32 + 17) ÷ 6 = 9; S = (6 + 36 + 18) ÷ 6 = 10.
  2. TE = 10 + 14 + 9 + 10 = 43 weeks.
  3. Each activity has tp − to = 12, so each variance = (12 ÷ 6)² = 4.
  4. Project variance = 4 × 4 = 16. σ = √16 = 4 weeks.
  5. (a) Z = (47 − 43) ÷ 4 = 1.00. Probability = Φ(1.00) = 0.8413.
  6. (b) T = 43 + 1.645 × 4 = 43 + 6.58 = 49.58 weeks.

Answer: (a) Probability of completion within 47 weeks is about 84.1%. (b) The project should be scheduled for about 49.6 weeks (say 50 weeks) to be 95% sure of completion.

Exam tips

  • Show a table with to, tm, tp, te and variance. Method marks are given even if a calculation slip follows.
  • Check which activities are critical before computing project variance. Most lost marks come from adding the wrong variances.
  • Keep the Z-table values handy. Questions usually supply them, so use exactly the figure given and round Z to two decimals as the table needs.
  • End with a one-line conclusion in plain words, and add a recommendation if the question asks whether the deadline is safe.
  • For MCQs, test quick checks: if the target equals TE, the probability is 50%. If the target is earlier than TE, it is below 50%.

Practice questions from Network Analysis - PERT, CPM

PERT: Three Time Estimates and Probability in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

PERT: Three Time Estimates and Probability: frequently asked questions

What is the formula for PERT expected time and variance?

Expected time is te = (to + 4tm + tp) ÷ 6. Variance is ((tp − to) ÷ 6)². Standard deviation is the square root of the variance, which equals (tp − to) ÷ 6.

How do I calculate the probability of project completion in PERT?

Find the critical path using expected times. Add the variances of its activities and take the square root to get σ. Compute Z = (T − TE) ÷ σ and read the probability from the normal table.

What is the difference between PERT and CPM?

PERT uses three time estimates and is probabilistic, so it suits projects with uncertain durations. CPM uses a single known time per activity and includes time-cost trade-off analysis. PERT is event-oriented while CPM is activity-oriented.

Why do we use only critical-path variances for the project?

The project finishes when the longest path finishes, so its duration is set by the critical path. Variances of other activities do not change that path's length. This is a simplification, because a near-critical path could also delay the project.