PERT Analysis: Methodology and Completion Estimation

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PERT Analysis: Methodology and Completion Estimation

Project Management · Chapter 5: Project Management Tools and Analytical TechniquesSection 5.5 · lesson 34 of 7312 min · free with a free account

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In this lesson

Using three-point estimates to put probabilities on completion dates.

  • Calculate PERT expected duration, standard deviation, and variance from three-point estimates
  • Combine activities along a path by adding expected durations and variances
  • Convert a target date into a z value and read the probability of finishing by it
  • Explain how independence and near-critical paths make single-path PERT probabilities optimistic

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Transcript

Everything said in the video, section by section (1,536 words). Read along, or search it.

Opening Question

Day 50, 55, or 60 for the Launch

Dana is launching a new online course at Larkspur Learning Systems. Marcus, the sponsor, wants one promise date, so sales can announce it. Her five-step path is expected to take 50 working days. Should she promise day 50, day 55, or day 60? To answer, she needs more than one number per activity. She needs to know how likely each date is.

Why Three Numbers

A Hard Floor and a Soft Ceiling

PERT, the Program Evaluation and Review Technique, was developed in the late 1950s, within a United States defense program. It describes each activity with three estimates. A single-point estimate is one number, like: the foundation will take 12 days. It's easy to add up, which is why most schedules use it. But two estimators can both say 12, while one expects 11 to 13, and the other anything from 6 to 30. The schedule looks identical, yet the second activity is a real risk to the finish. Work has a hard floor and a soft ceiling. There's a shortest time a job can take, but delays, rework and waiting can stretch it much further. So durations bunch near a typical value, with a long late tail. Three numbers capture that shape.

Three-Point Estimate

Optimistic, Most Likely, and Pessimistic

A three-point estimate describes one activity with three values, under the same assumed resources. The optimistic estimate is the shortest realistic duration, if the work goes unusually well. The most likely estimate is the duration that would occur most often, if the activity were done many times under normal conditions. The pessimistic estimate is the longest realistic duration, if the foreseeable difficulties happen. Rare disasters belong to risk management, not here. Ask the people doing the work, and ask for the two limits first, so the middle value doesn't anchor them. Then check the order: optimistic, most likely, pessimistic. And don't pad. Hidden safety time overstates the expected duration, and nobody can later say where the time went.

Expected Duration

The One-Four-One Weighted Average

The expected duration is a weighted average, with the most likely value counted four times as much as each limit. Expected equals optimistic, plus four times most likely, plus pessimistic, all divided by six. The weights 1, 4 and 1 add up to six. Take Harbourline Constructions, building a small community pavilion. Site preparation has estimates of 8, 11 and 32 days. 8, plus 4 times 11, which is 44, plus 32, gives 84. Divided by 6, that's 14 days. The foundation, at 6, 12 and 24, gives 78 over 6: 13 days. Notice that 14 is more than the most likely 11. That's not a mistake. The long late tail pulls the balance point later than the peak.

Measuring Spread

Standard Deviation and Variance

The expected duration says where the middle is, but not how wide the uncertainty is. The standard deviation measures spread, how far an outcome typically lies from the expected value. PERT estimates it as pessimistic minus optimistic, divided by six. Six, because nearly all outcomes of a bell-shaped curve fall within about three standard deviations, either side of the middle. For site preparation, 32 minus 8 is 24, over 6: 4 days. For the foundation, 24 minus 6 is 18, over 6: 3 days. The variance is the standard deviation squared: 16, and 9. Squared days can't be pictured, but variances of independent activities can be added, and standard deviations can't.

Worked Example

Pavilion Path of Fifty-Five Days

Here's the pavilion's critical path: five activities, each with its three estimates, expected duration, standard deviation and variance. Site preparation and the foundation you've seen. The frame gives 13, with a standard deviation of 3. Roofing and the interior each give 7 and 8, with a standard deviation of 1. For the path, add the expected durations. The path expected duration is 55 days. Add the variances: 36. The path standard deviation is the square root of 36, which is six. Adding the standard deviations, to get 12, is a common mistake. Independent delays partly cancel, and adding variances captures that. And the most likely values add to only 49 days. Remember that number.

Probability

Turning a Target Into a z Value

Now the chance of finishing by a target. First convert the target into z: the target minus the path expected duration, divided by the path standard deviation. Then read the probability from a standard normal table, the share of outcomes at or below that z on a bell curve. For a target of day 61, z is 61 minus 55, over 6. That's exactly 1, and the table gives 0.8413: about an 84 percent chance. Now the most-likely date, day 49. z is 49 minus 55, over 6: minus 1. By symmetry, the chance is 1 minus 0.8413, about 16 percent.

Choosing a Promise Date

Each Extra Three Days Buys Less

The real use is choosing a promise date. Day 55, the expected finish, has z of 0: a coin toss, 50 percent. Day 58 gives about 69 percent. Day 61, 84. Day 64, 93. And day 67, about 98. Each extra 3 days buys less than the one before: about 19 points, then 15, then 9, then 4. No date reaches certainty. The calculation supplies the odds. The sponsor, who bears the consequences, decides how much risk to accept. So show two or three dates with their chances, not one.

Quick Check

Vessel Refit by Week 44

Quick check, on a new project. A vessel refit's critical path has an expected duration of 39 weeks, and a path variance of 16. The sponsor wants it by week 44. Using the same method, is the chance about 69, 89, or 98 percent? Pause and predict. About 89 percent. The standard deviation is the square root of 16, which is 4. So z is 44 minus 39, over 4: 1.25. That reads as 0.8944. Dividing by 16 instead of 4 is the trap, and it would give a much lower figure. Always take the square root of the variance first.

Raising the Odds

Narrowing Site Preparation's Range

There are three ways to raise a probability. Move the target later. Or shorten the path through a real change, like another crew, not new numbers on paper. Or reduce the variance. Site preparation contributes 16 of the 36 units of path variance, 44 percent. Investigate it first. Suppose a site investigation narrows its pessimistic estimate from 32 to 20 days. Its expected duration becomes 12, and its variance 4. The path becomes 53 days, with variance 24, and a standard deviation of about 4.9. Day 61 is now about 1.63 standard deviations out, roughly a 95 percent chance, up from 84, without moving the date.

Assumptions

Convention, Independence, Normality, and Other Paths

Every number here rests on assumptions. First, the 1, 4, 1 weighting and the divide-by-6 are a convention, a widely used approximation, not a law of nature. Second, independence. Adding variances assumes one activity's duration doesn't affect another's. A wet season can delay site work, foundation and frame together. When activities run late together, the true variance is bigger, and the reported probability is too high. The arithmetic gives no warning. Third, the normal approximation. A sum of many activities tends toward a bell shape, but five skewed activities are a modest basis, especially in the tails. Fourth, near-critical paths. The method looks at one path, but the project finishes only when every path has finished. Two independent paths, each with an 84 percent chance by day 61, give 0.8413 times 0.8413: about 71 percent that both make it.

Quick Check

Two Independent Paths at Ninety Percent

Quick check. A project has two paths that share no activities. Each has a 90 percent chance of finishing by day 80. Assuming they're independent, what's the chance the whole project finishes by day 80: about 81, exactly 90, or about 95 percent? Pause and predict. About 81 percent. Both paths must finish, so multiply: 0.9 times 0.9 is 0.81. A project can never be more likely to finish on time than its critical path alone, which is why a single-path figure is optimistic.

Debate

Are Three Estimates Worth the Effort

Are three estimates worth the effort? One side says they make uncertainty visible, and give a sponsor a probability to discuss, for two extra numbers per activity. The other says the extra numbers are guesses too, and the formulas lend them a precision they don't deserve. Whether three-point estimates forecast better in practice is an empirical question, and depends on how well estimators judge their own uncertainty. No result is claimed here. The multi-path weakness leads to simulation, which runs the schedule many times and tracks every path at once.

Back to Dana

Promise Day 55, Aim for Day 50

Back to Dana. Her path has an expected duration of 50 days, and a variance of 25, so a standard deviation of 5. Day 50 is a coin toss. Day 55 is z of 1, about 84 percent. Day 60 is z of 2, about 98 percent. She warns Marcus the figures assume independent activities, so the real odds are probably a little lower. Day 50 risks a public miss half the time. Day 60 delays revenue by ten working days. She recommends day 55 as the promise, with day 50 as an internal aim. Marcus accepts: five days of waiting buy a rise from 50 to about 84 percent. Dana records the decision and its assumptions.

Key Takeaways

Formulas, Path Totals, and Cautions

To recap. Expected duration is optimistic, plus four times most likely, plus pessimistic, over six. The standard deviation is the range over six. Along a path, add expected durations and variances, then take the square root. z turns a target into a probability. Treat it as optimistic, because of independence and near-critical paths.

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FAQ

Questions students ask us

About PERT Analysis: Methodology and Completion Estimation

What is the PERT formula for expected activity duration?
The PERT expected duration is the optimistic estimate plus four times the most likely estimate plus the pessimistic estimate, divided by six. For estimates of 8, 11, and 32 days, that is 8 plus 44 plus 32, or 84, divided by 6, which gives 14 days. The result sits above the most likely value because the long late tail pulls the average later.
How do you calculate the standard deviation of a PERT path?
Find each activity's standard deviation as pessimistic minus optimistic over six, square it to get the variance, add the variances along the path, and take the square root. Five activities with variances of 16, 9, 9, 1, and 1 give a path variance of 36 and a standard deviation of 6 days. Adding the standard deviations directly, which gives 12, is a common mistake.
How do you find the probability of finishing a project by a target date using PERT?
Compute z as the target minus the path expected duration, divided by the path standard deviation, then read the probability from a standard normal table. With an expected duration of 55 days and a standard deviation of 6, a target of day 61 gives z of 1 and about an 84 percent chance. A target of day 55 gives only 50 percent.
Why are PERT completion probabilities often too optimistic?
PERT probabilities tend to be optimistic because the method assumes independent activities and looks at a single path. When weather or a shared crew delays several activities together, the true variance is larger than calculated. The project also finishes only when every path finishes, so two independent paths each with an 84 percent chance give only about 71 percent that both are done.

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