1. A company wants to develop a project using PERT to modify one of their products which has poor sales. They have determined the following activities, precedence relationships and time estimates in days for the project: Immediate Optimistic Most Likely Pessimistic Activity Predecessors to tm tp a none 4 6 7 b a 1 2 3 c a 5 5 5 d b 7 9 12 e b 2 4 8 f c, d 6 9 10 Draw the network. b. What is the critical path? c. What is the duration of the critical path in days? d. What is the probability that the project will take longer than 28 days to complete?

Variance of the duration of critical path
= Sum of variances of activities on critical path ( i.e. a,b,d,f)
= 0.25 + 0.11 + 0.69 + 0.44
= 1.49
Therefore, standard deviation of duration of critical path = Square root ( Variance of critical path) = Square root ( 1.49) =1.22
Let the Z value corresponding to the probability that project will take maximum 28 days = Z1
Therefore ,
Expected duration of critical path + Z1 x Standard deviation of duration of critical path = 28
Or, 25.67 + 1.22.Z1 = 28
Or , 1.22.Z1 = 2.33
Z1 = 1.91
Probability for Z = 1.91 as derived from standard table for normal distribution will be 0.97193
Probability that the project will take maximum 28 days = 0.97193
Therefore, probability that the project will take longer than 28 days
= 1 – Probability that the project will take maximum 28 days
= 1- 0.97193
= 0.02807
PROBABILITY THAT THE PROJECT WILL TAKE LONGER THAN 28 DAYS = 0.02807
 
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