---
title: "PERT with example"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:pert-define-example"
source: https://www.bctengineer.com/q/semester_7/project_engineering/pert-define-example
site: BCT Engineering
---

# PERT with example

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2071 Chaitra, 4 marks
- Video: [What is a PERT Chart? Project Management in Under 5](https://www.youtube.com/watch?v=i160aaBX7mE) — Online PM Courses - Mike Clayton
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/project_engineering/pert-define-example
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

What is PERT? Discuss with an example.

## Short answer

PERT is a probabilistic network technique using three time estimates (optimistic, most likely, pessimistic) to compute expected duration and variance. For the example with activities A, B, C in series, expected project time = 12 days, standard deviation = 1.83 days, and probability of finishing within 14 days is about 86%.

## Answer

**PERT (Programme Evaluation and Review Technique)** is a probabilistic network technique for planning and controlling projects in which activity times are uncertain. Each activity has three estimates: optimistic time $t_o$, most likely time $t_m$ and pessimistic time $t_p$. Times are assumed to follow a beta distribution.

$$
t_e = \frac{t_o + 4t_m + t_p}{6}, \qquad \sigma = \frac{t_p - t_o}{6}, \qquad \text{Variance} = \sigma^2
$$

For the project, the critical path is found with $t_e$ values, the expected duration is the sum of $t_e$ on the path, and the project variance is the sum of the variances of the critical activities.

#### Example

A project has three activities in series.

| Activity | $t_o$ | $t_m$ | $t_p$ | $t_e$ | $\sigma$ | Variance |
|---|---|---|---|---|---|---|
| A | 2 | 4 | 6 | 4.00 | 0.67 | 0.44 |
| B | 3 | 5 | 13 | 6.00 | 1.67 | 2.78 |
| C | 1 | 2 | 3 | 2.00 | 0.33 | 0.11 |

Expected project time $= 4 + 6 + 2 = 12$ days.
Project variance $= 0.444 + 2.778 + 0.111 = 3.333$, so $\sigma_P = \sqrt{3.333} = 1.83$ days.

Probability of finishing in 14 days: $Z = \dfrac{14 - 12}{1.826} = 1.095 \approx 1.10$, so $P \approx 0.864$ (86.4 %) from the normal table.

**Answer: expected time 12 days, SD 1.83 days, probability of completion in 14 days $\approx$ 86 %.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which set of estimates is made for every activity in PERT?
   A. One fixed duration and its cost
   B. Optimistic, most likely and pessimistic times
   C. Early start, late start and float
   D. Normal time, crash time and crash cost
   Answer: B. PERT handles uncertain durations using t_o, t_m and t_p for each activity, assumed to follow a beta distribution. (Key point: Three time estimates)

2. Activity B has t_o = 3, t_m = 5 and t_p = 13 days. What is its expected time t_e?
   A. 5.00 days
   B. 7.00 days
   C. 6.00 days
   D. 4.33 days
   Answer: C. t_e = (3 + 4x5 + 13)/6 = 36/6 = 6.00 days. (Key point: Expected time formula)

3. How is the project standard deviation found from the activities on the critical path?
   A. Add the standard deviations of the activities
   B. Take the square root of the sum of the variances
   C. Average the variances of the activities
   D. Take the largest activity standard deviation
   Answer: B. Variances add along the critical path; the project SD is the square root of this sum (sqrt(3.333) = 1.83 days in the example). (Key point: Project standard deviation)

4. For expected time 12 days and project SD 1.83 days, what is Z for completion in 14 days?
   A. 0.55
   B. 1.64
   C. 2.00
   D. 1.10
   Answer: D. Z = (14 - 12)/1.83 = 1.10, which gives a probability of about 86.3% from the normal table. (Key point: Probability of completion)