Which set of estimates is made for every activity in PERT?

Optimistic, most likely and pessimistic times

One fixed duration and its cost

Early start, late start and float

Normal time, crash time and crash cost

How is the project standard deviation found from the activities on the critical path?

Take the square root of the sum of the variances

Add the standard deviations of the activities

Average the variances of the activities

Take the largest activity standard deviation

2071 Chaitra · 4 marks

PERT with 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%.

What is PERT? Discuss with an example.

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 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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