Program Evaluation and Review Technique

Also known as PERT. Estimates project duration when activity times are uncertain. It uses 3 time estimates for each activity and models the resulting project duration with a normal distribution.

Assumptions

PERT uses the following approximations:

Activity times follow beta distributions.
Activity durations are statistically independent.
The critical path remains unchanged near the target completion time.
The project duration is approximately normally distributed.

The normal approximation becomes less reliable when several paths have similar expected durations or when activity durations are strongly correlated.

Time Estimates

PERT uses 3 estimates for every activity:

Optimistic Time

The shortest realistic duration when execution conditions are favourable. Denoted by $a$ .

Most Likely Time

The duration expected under normal execution conditions. Denoted by $m$ .

Pessimistic Time

The longest realistic duration under unfavourable conditions, excluding catastrophes. Denoted by $b$ .

Expected Activity Time

The expected time of an activity is a weighted average.

t_e = \frac{a + 4m + b}{6}

The most likely time receives 4 times the weight of either extreme estimate.

Activity Standard Deviation

Denoted by $\sigma$ .

\sigma = \left(\frac{b-a}{6}\right)

$\sigma^2$ is the activity variation. Greater uncertainty between the optimistic and pessimistic times produces a larger variance.

Project Duration

Construct the project network using the expected activity times. Apply the critical path method to identify the critical path.

The expected project duration is the sum of expected activity times on the critical path:

T_e = \sum_{i \in C} t_{e,i}

The project variance is approximated by the sum of the variances on the critical path:

\sigma_T^2 = \sum_{i \in C} \sigma_i^2

The project standard deviation is:

\sigma_T = \sqrt{\sigma_T^2}

Completion Probability

For a target completion time $D$ , the standard normal score is:

Z = \frac{D-T_e}{\sigma_T}

The probability of completing by $D$ is approximated by the standard normal cumulative distribution function:

P(T \le D) = \Phi(Z)

A positive $Z$ means the target is later than the expected project duration. A negative $Z$ means the target is earlier.

Steps

List every activity, its predecessors, and its optimistic, most likely, and pessimistic times.
Calculate the expected time and variance of every activity.
Construct the network using the expected activity times.
Apply the critical path method.
Sum the expected times and variances on the critical path.
Calculate the standard normal score for any target completion time.
Read the completion probability from the standard normal distribution.

Worked Example

A project has 5 activities with the following precedence relationships and time estimates.

Activity	Predecessor	$a$	$m$	$b$
$A$	none	2	4	6
$B$	$A$	1	2	3
$C$	$A$	2	5	8
$D$	$B$ , $C$	1	3	5
$E$	$C$	1	2	3

Calculate the expected project duration and the probability of completing the project within $14$ time units.

Solution Reveal Hide

For each activity:

t_e = \frac{a + 4m + b}{6}, \qquad \sigma^2 = \left(\frac{b-a}{6}\right)^2

Activity	$t_e$	$\sigma^2$
$A$	$\dfrac{2+4(4)+6}{6}=4$	$\left(\dfrac{6-2}{6}\right)^2=\dfrac{4}{9}$
$B$	$\dfrac{1+4(2)+3}{6}=2$	$\left(\dfrac{3-1}{6}\right)^2=\dfrac{1}{9}$
$C$	$\dfrac{2+4(5)+8}{6}=5$	$\left(\dfrac{8-2}{6}\right)^2=1$
$D$	$\dfrac{1+4(3)+5}{6}=3$	$\left(\dfrac{5-1}{6}\right)^2=\dfrac{4}{9}$
$E$	$\dfrac{1+4(2)+3}{6}=2$	$\left(\dfrac{3-1}{6}\right)^2=\dfrac{1}{9}$

The possible paths and their expected durations are:

Path	Expected duration
$A \to B \to D$	$4 + 2 + 3 = 9$
$A \to C \to D$	$4 + 5 + 3 = 12$
$A \to C \to E$	$4 + 5 + 2 = 11$

The critical path is $A \to C \to D$ . Therefore:

T_e = 4 + 5 + 3 = 12

\sigma_T^2 = \frac{4}{9} + 1 + \frac{4}{9} = \frac{17}{9}

\sigma_T = \sqrt{\frac{17}{9}} = \frac{\sqrt{17}}{3} \approx 1.37

For the target duration $D=14$ :

Z = \frac{14-12}{1.37} \approx 1.46

The standard normal cumulative distribution gives:

P(T \le 14) = \Phi(1.46) \approx 0.928

The probability of completing the project within $14$ time units is approximately $92.8\%$ .