Question

Pert project method Activity Predecessors A group 5 optimistic Most likely Pessimistic Expected Variance a m b Time (t) o 2 3
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Answer #1

1.

The project network diagram is given below :

с 2 4 () F А н. o E 6 7 1 B D G 3 5

2.

The table for the time estimate and the variance are :

2 Activity to p to +4-1m tlo 6 --1939 А 1 2 3 2 0.11 B 2 3 4 3 0.11 с 1 2 3 2 0.11 D 2 4 6 4 0.44 E 1 4 7 4 1 F 1 2 9 3 1.78

where :

to = optimistic time

tp = pessimistic time

tm = most likely time

te = expected time

σ2 = Variance

3.

The critical path of the project is the one which has the longest project duration and all the activities which lie on the critical path are the critical activities.

In the above problem the critical path is represented by the red line.

The critical path is :  1-2-4-5-6-7

Critical Activites are :  A,C,E,G,H

THE TOTAL CRITICAL PATH TIME = 17

HENCE THE PROJECT COMPLETION TIME = 17 WEEKS

C (2) 2 4 F (3) A(2) H(3) E(4) 6 7 1 B(3) kle Za D(4) G(6) 3 5

4.

Critical Activity Optimistic Time : to

Most Likely Time : tm

Pessimistic Time : tp Range : tp - to Standard Deviation t-t o 9

Variance :

o? P 6

A 1 2 3 2 1/3 1/9
C 1 2 3 2 1/3 1/9
E 1 4 7 6 1 1
G 4 5 12 8 4/3 16/9
H 2 3 4 2 1/3 1/9

STANDARD DEVIATION OF ( CRITICAL PATH ) =( 19/9 + 1 ) ^ 0.5 = 1.76

Z = [GIVEN TIME - EXPECTED TIME ( CRITICAL PATH )] / SD OF CRITICAL PATH = (20-17)/1.76 = 3/1.76 = 1.70

ACCORDING TO THE Z TABLE THE VALUE FOR Z VALUE = 1.70 IS 0.9554

HENCE THE PROBABLITY IS 0.9554*100 = 95.54 % THAT THE PROJECT WILL GET COMPLETED ON OR BEFORE 20 WEEKS

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