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PERT Program Evaluation and Review Technique Tong Wang 511D ERB PERT Example Consider a small project that involves the following activities. Precedin g Activit Activity y Completion Times (days) Optimistic Most Likely Pessimistic a - 5 6 7 b - 4 5 18 c a 4 15 20 d b,c 3 4 5 e a 5 16 18 PERT Example (cont’d) • (a) Determine the expected value and the variance of the completion time for each activity. • (b) Use the expected times from (a) to find the critical path. • (c) Assuming that the normal distribution applies, determine the probability that the critical path will take between 18 and 26 days to complete. • (d) How much time must be allowed to achieve a 90% probability of timely completion? • (e)By using modified probability of completion method, what is the probability that all paths will take before 18 weeks? Exercise Solution (a) 0 6 0 6 6 20.5 9.5 24 END E 14.5 START B 7 0 7 13 20 D 4 C 14 A 6 6 20 6 20 20 24 20 24 Activit Mean Variance y a 6 4/36 b 7 196/36 c 14 256/36 d 4 4/36 e 14.5 169/36 Exercise Solution (cont’d) (b) By using the expected time (mean) of each activity, we find that the critical path is A-C-D. Remark: For this simple project, we can find the longest path (A-C-D) has the largest expected time, which is the critical path. The mean critical path duration is μ= 6 + 14 + 4 = 24. The variance of the critical path duration is the sum of the variances along the path: σ2cp = (4+256+4) / 36 = 264/36 so that the standard deviation is readily computed as σcp = 2.708. Exercise Solution (cont’d) (c)The interval probability may be computed as the difference between two cumulative probabilities as follows: P(18 ≤ t ≤ 26) = P(t ≤ 26) - P(t ≤ 18). Two separate z computations are required. First at 26 we have z26= (26-24) / 2.708=0.739 Then by looking up the normal table with z26=0.739, we have one result that P(t ≤ 26) = 0.770 Exercise Solution (cont’d) Secondly, at 18 we have z18= (18-24) / 2.708= -2.216 and P(t ≤ 18) = 0.013 Combining these two results yields the desired probability as P(18 ≤ t ≤ 26) = 0.770 - 0.013= 0.757. Exercise Solution (cont’d) (d) For 90% probability, we must pick a z value corresponding to 90% of the area under the normal curve, 50% left of mean and 40% right of mean, so z = 1.282. Then solving for t we have t = 24 +1.282*2.708 = 27.47 days. Exercise Solution (cont’d) (e)There are three paths in total. They are A-C-D, AE, and B-D. mean mean standard deviation standard deviation A-C-D A-C-D 24 24 2.708 2.708 A-E A-E 18.5 20.5 2.192 2.192 11 2.356 B-D B-D 11 2.356 The probabilities for the three paths to be completed 18 weeks are given below. P(X1 ≤ 18) = P( z ≤ (18-24) / 2.708) = 0.013 P(X2 ≤ 18) = P( z ≤ (18-20.5) / 2.192) = 0.127 P(X3 ≤ 18) = P( z ≤ (18-11) / 2.356) = 0.999 Then the probability of completing all the paths in 18 weeks is P(X ≤18) = P(X1 ≤ 18) P(X2 ≤ 18) P(X3 ≤ 18) = 0.0016. Questions?