Question

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Use the data in the table to calculate the expected activity times, di, and then compute the critical path for each event using di as the completion time for activity (i, i). Assume that the solution that you obtain is the planned” time to complete each event, and then find the probabilities that the events will occur without delay. TABLE 9.16 Project (a) Activity (a, b, m) Activity (a, b, m) 1,2 3,6 4,6 (3,5,4) 4,7 1,4 1,5 2,3 2,5 2,6 3,4 (5,8,6) (1,4,3) (2,5,4) (4,6,5) (7,10,8) (8,13,9) (5, 10,9) 5,7 (5,8,6) (9, 15, 10) (4,8,6) (3,5,4) 6,7 Project (b) Activity (a, b, m) Activity (a, b, m) 3,7 1,2 1,3 4,5 1,4 (1,4,3) (5,8,7) (6,9,7) (1,3,2) (3,5,4) (7,9,8) (10,20, 15) (12, 14, 13) (10, 15, 12) (8, 12, 10) (7,11,8) (2,8, 4) (5,7,6) 1,6 5,6 2,3 5.7 2,5 6,7 3.4 8 10 Project (a) 15 22 210 12 / Project (b)

Answer 1

Project (a)

For each activity, calculate the expected time u =(a + 4m + b)/6 For each activity, Variance o* =(b-a)/36. Standard deviation of each activity is square root of Variance Find the Early Start (ES), Early Finish (EF), Late Start (LS), Late Finish (LF) times and Slack for each activity. Early Start is calculated by forward pass method as latest of the predecessors Early Finish. Late Finish is calculated using reverse pass method as earliest of the successors Late Start. Early Finish = Early Start + Activity Duration Late Start=Late Finish - Activity Duration Slack = Late Finish - Early Finish, or Late Start - Early Start Activity ES LS EF LF Slack Critical Activity Yes 1,2 1,4 i Expected variance Time (u) (?) 6.17 0.25 2.83 0.25 3.83 0.25 5.00 0.11 8.17 0.25 9.50 0.69 8.50 0.69 4.00 0.11 7.67 1.00 6.17 0.25 10.67 1.00 6.00 0.44 4.00 0.11 0.00 0.00 0.00 6.17 6.17 6.17 11.17 11.17 19.67 19.67 14.33 14.33 27.33 0.00 16.83 12.83 6.17 8.50 17.83 11.17 23.33 19.67 25.17 16.67 25.33 27.33 6.17 2.83 3.83 11.17 14.33 15.67 19.67 15.17 27.33 25.83 25.00 20.33 31.33 6.17 0.00 19.67 16.83 16.67 12.83 11.17 0.00 16.67 2.33 27.33 11.67 19.67 0.00 27.33 12.17 27.33 0.00 31.33 5.50 27.33 2.33 31.33 11.00 31.330.00 Yes 6,7 4

Critical path is the longest path in the network.

It can be identified by 0 slack time.

Critical path is :  1-2-3-4-6-7

Expected time and planned time of project completion = 31.33

Variance = Sum of variances of critical activities = (0.25+0.11+0.69+1+0.11) = 2.17

Standard deviation = sqrt(2.17) = 1.47

z value = (31.33-31.33) / 1.47 = 0

Probability, P(z) = NORMSDIST(0) = 0.5

Probabilities that the events will occur without delay = 0.5