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Subhas Chandra Bose

The expected ...

Question

The expected time $(t_{e})$ of a PERT activity in terms of optimistic time $(t_{o}),$ pessimistic time $(t_{p})$ and most likely time $(t_{l})$ is given by

A

t_{e} = \frac{t_{o} + 4 t_{p} + t_{l}}{3}

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B

t_{e} = \frac{t_{o} + 4 t_{p} + t_{l}}{6}

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C

t_{e} = \frac{t_{o} + 4 t_{l} + t_{p}}{6}

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D

t_{e} = \frac{t_{o} + 4 t_{l} + t_{p}}{3}

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Solution

The correct option is C $t_{e} = \frac{t_{o} + 4 t_{l} + t_{p}}{6}$
The expected time $(t_{o})$ of PERT activity in terms of optimistic time $(t_{o})$ , pessimistic time $(t_{p})$ and most likely time $(t_{l})$ is given by,

$t_{o} = \frac{t_{o} + 4 t_{l} + t_{p}}{6}$

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Similar questions

Q. Consider the following PERT network:

The optimistic time, most likely time and pessimistic time of all the activities are given in the table below:

Activity	Optimistic time (days)	Most likely time (days)	Pessimistic time (days)
1 - 2	1	2	3
1 - 3	5	6	7
1 - 4	3	5	7
2 - 5	5	7	9
3 - 5	2	4	6
5 - 6	4	5	6
4 - 7	4	6	8
6 - 7	2	3	4

The standard deviation of the critical path is

Q. Consider the following PERT network:

The optimistic time, most likely time and pessimistic time of all the activities are given in the table below:

Activity	Optimistic time (days)	Most likely time (days)	Pessimistic time (days)
1 - 2	1	2	3
1 - 3	5	6	7
1 - 4	3	5	7
2 - 5	5	7	9
3 - 5	2	4	6
5 - 6	4	5	6
4 - 7	4	6	8
6 - 7	2	3	4

The critical path duration of the network (in days) is

Q. The project activities, precedence relationships and durations are described in the table. The critical path of the project is

Activity	Precedence	Duration (in days)
P	-	3
Q	-	4
R	P	5
S	Q	5
T	R, S	7
U	R, S	5
V	T	2
W	U	10

Q. A project consists of two activities A and B. The most optimistic and pessimistic time for completionn of project is shown in the table below.

Acitivity	Optimistic Time	Pessimistic Time
A	1 hr	3 hr
B	2 hr	5 hr

The standard deviation for the project is:

Q. In PERT, the distribution of activity time is assumed to be

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