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Activity Optimistic (a) Most likely (m) Pessimistic (b) Expected Time te = (a + 4m + b)/6 Variance 02 = (b - a)2/36 Standard Deviation =sqrt(variance) 2. 5 8 5 1 1 B 2 3 4 3 0.111 0.333 5 10 15 10 2.778 1.667 D 1 8 9 7 1.778 1333 E 5 10 5 8.333 0 0 F 2 5 8 5 1 1 G 3 10 11 9. 1.778 1.333 H 2 6 10 6 1.778 1.333 1 1 2 3 2 0.111 0.333 a) Assume the longest path thru this network (critical path) is C-G-H-I. Based on this path, what is the variance and standard deviation for this network. Variance = 6.445 = Standard Deviation = 2.5387 CALCULATIONS: The variance of the network (Sum of variance of all activities on the critical path Given that the critical path is C-G-H-I. Variance of Activity C = 2.778 Variance of Activity G = 1.778 Variance of Activity H = 1.778 Variance of Activity I = 0.111 Variance of the network (02) = 2.778 + 1.778 + 1.778 +0.111 = 6.445 So, the variance of the network = 6.445 , Standard deviation of the network (0) = V (Variance of the network) = V(6.445) = 2.5387 So, the standard deviation of the network (0) = 2.5387 b) Your boss is asking what the earliest and latest completion times are for getting this project completed. You rely on a 95% statistical confidence interval to estimate the latest time you are willing to say the project will be completed. What are you willing to tell the boss on the completion time interval based on this expected duration? Note: you should round the time to the nearest tenth of a day