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PERT Program Evaluation and Review Technique

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PERT Program Evaluation and Review Technique. Tong Wang 511 D ERB. PERT Example. Consider a small project that involves the following activities. PERT Example (cont’d). (a) Determine the expected value and the variance of the completion time for each activity.

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PERT Program Evaluation and Review Technique

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  1. PERTProgram Evaluation and Review Technique Tong Wang 511DERB

  2. PERT Example 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 (a) END E 14.5 A6 D4 C 14 START B7

  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:   σ2cp = (4+256+4) / 36 = 264/36 so that the standard deviation is readily computed as σcp= 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  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

  7. 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.

  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(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.

  10. Questions?

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