1. PERT Program Evaluation and Review Technique Tong Wang 511D ERB

2. PERT Example Precedin g Completion Times (days) Activit y OptimisticMost Likely Pessimistic a-567 b-4518 ca41520 db,c345 ea51618 Consider a small project that involves the following activities.

3. 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?

4. Exercise Solution Activit y MeanVariance a64/36 b7196/36 c14256/36 d44/36 e14.5169/36 (a) START A6A6 C 14 B7B7 E 14.5 D4D4 END 06 06 07 1320 24 2024 620 6 620.5 9.524

5. 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: σ 2 cp = (4+256+4) / 36 = 264/36 so that the standard deviation is readily computed as σ cp = 2.708.

6. 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 z 26 = (26-24) / 2.708=0.739 Then by looking up the normal table with z 26 =0.739, we have one result that P(t ≤ 26) = 0.770

7. Exercise Solution (cont’d) Secondly, at 18 we have z 18 = (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.

8. 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.

9. Exercise Solution (cont’d) (e)There are three paths in total. They are A-C-D, A- E, and B-D. The probabilities for the three paths to be completed 18 weeks are given below. P(X 1 ≤ 18) = P( z ≤ (18-24) / 2.708) = 0.013 P(X 2 ≤ 18) = P( z ≤ (18-20.5) / 2.192) = 0.127 P(X 3 ≤ 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(X 1 ≤ 18) P(X 2 ≤ 18) P(X 3 ≤ 18) = 0.0016. meanstandard deviation A-C-D242.708 A-E20.52.192 B-D112.356

10. Questions?