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4. PERT (PROGRAM EVALUATION AND REVIEW TECHNIQUE) Objective: To understand how to apply the PERT method to planning construction projects that are subject to uncertainty: Summary: 4.1 Introduction 4.2 PERT Procedure 1 4.1 INTRODUCTION PERT is usually used to assess risk in very risky projects, for example, where: – some activities have a high degree of uncertainty about their duration (such as in design-build); – the cost of not meeting a deadline is very high (such as very high penalty payments); 2 • Basic assumptions of PERT: – uncertainty is only important along the deterministically derived critical path; – there is no correlation between the durations of different activities (basic Monte Carlo CPM assumes this as well); 5 activity activity ‘X’ 6 ‘Y’ 5 9 3 11 • the durations of ‘X’ and ‘Y’ are assumed to be independent; • yet, if they were performed by the same efficient crew, then if ‘X’ completed quickly, maybe ‘Y’ would be as well; – the duration of an activity is a random variable that can be described by the Beta distribution; 3 probability density a 4m b the mean t 6 shaded area = 1.0 ba the variance v 6 2 activity duration a = optimistic duration m = most likely duration b = pessimistic duration Fig. 4-1: Simplified Beta Distribution • the optimistic duration is your estimate of the shortest the activity could take to complete (say 1 in 100 observations); • the pessimistic duration is your estimate of the longest the activity could take to complete (say 1 in 100); 4 • the most likely duration is your estimate of how long it is most likely to take. 4.1 PERT PROCEDURE The basis of the PERT approach is: – expected project duration = deterministic project duration; – variance on project duration (the square of the standard deviation) is equal to the sum of each activity’s variance along the critical path; – by the Central Limit Theorem, it is assumed that the distribution for the project duration is Normal; – with the above parameters, it is now possible to answer questions such as: what is the probability of completing the project on time? 5 Step 1: Determine the layout of the activity network and assign durations (a, b, and m) to each activity: KEY: ES TF FF activity LS ‘X’ a m 4 EF IF LF b activity activity activity ‘B’ 6 ‘D’ 5 ‘F’ 5 14 2 14 3 7 activity 3 ‘A’ 4 end 11 5 activity activity activity ‘C’ 10 ‘E’ 11 ‘G’ 6 15 4 12 4 14 Fig. 4-2: Example PERT Network 6 Step 2: Find the deterministic critical path: 1: compute the mean durations: t a 4m b 6 2: perform a deterministic forward pass to find the deterministic ES and EF 3: perform a deterministic backward pass to find the deterministic LS and LF 4: find the deterministic critical path(s) 12 5 12 activity 0 5 8 4 activity 0 3 ‘A’ 4 t=5 5 11 ‘B’ 6 t=7 15 14 activity 5 5 ‘C’ 10 t = 10 25 activity 15 5 18 21 2 ‘D’ 5 t=6 15 activity 27 14 25 27 3 ‘E’ 15 4 11 t = 10 ‘F’ 5 t=5 25 activity 15 15 30 32 7 32 end 32 32 activity 25 12 25 4 ‘G’ 6 t=7 32 14 Fig. 4-3: Deterministic Critical Path 7 Step 3: Determine the expected (mean) duration and variance for the project: 1: the PERT expected project duration is the deterministic critical path: T = 32 days 2: compute the variance for each duration (2 decimal places): ba v 6 2 3: the PERT variance for the project duration is the sum of the variances along the critical path (if multiple paths, use the largest for safety): V = 9.12 days 12 5 12 activity 0 5 8 4 activity 0 3 ‘A’ 5 4 11 t = 5 v = 1.78 21 2 15 15 activity 5 5 25 activity ‘B’ 15 6 14 t = 7 v = 2.78 5 18 ‘C’ 15 10 15 t = 10 v = 2.78 30 activity ‘D’ 27 5 14 t = 6 v = 4.00 25 27 3 25 activity ‘E’ 25 15 4 11 12 t = 10 v = 1.78 Fig. 4-4: Project Duration: Expected and Variance ‘F’ 32 5 7 t = 5 v = 0.44 32 end 32 32 activity 25 4 ‘G’ 32 6 14 t = 7 v = 2.78 8 Step 4: Evaluate the Project: Question 1: For the above project, what is the probability of completing within 35 days? 1: Assume the Central Limit Theorem applies: the combined distributions along the critical path are assumed to approximate a Normal distribution. t T 35 32 2: calculate the ‘z’ value: z V 9.12 0.99 3: use the z-value in the standard normal variable table to find the probability: probability of completing within t = 35 days: p = 83.9% probability density expected (mean) project duration T = 32 days shaded area equals probability of completing within t = 35 days required project completion t = 35 days increasing the variance increases the width of the distribution V = 9.12 days Fig. 4-5: Normal Distribution Project duration 9 Question 2: For the above project, what is the probability of completing within 25 days? t T 25 32 1: calculate the ‘z’ value: z V 9.12 2.32 2: use the z-value in the standard normal variable table to find the probability: probability of completing within 25 days = 1.02% probability density expected (mean) project duration T = 32 days shaded area equals probability of completing within t = 25 days required project completion t = 25 days V = 9.12 days Fig. 4-6: Normal Distribution Project duration 10 Question 3: For the above project, what project duration has a 90% probability of being satisficed? (reverse of previous questions) 1: calculate the ‘z’ value from the standard normal variable tables: z = 1.28 t T t 32 2: solve for t in the z-value formula: z V 1.28 9.12 therefore t = 35.87 days. probability density expected (mean) project duration T = 32 days shaded area equals 90% probability of completing within t = ? days required project completion t = ? days V = 9.12 days Fig. 4-7: Normal Distribution Project duration 11 Question 4: For the above project, what project duration has a 99% probability of being satisficed? 1: calculate the ‘z’ value from the standard normal variable tables: z = 2.33 t T t 32 2: solve for t in the z-value formula: z V 2.33 9.12 therefore t = 39.0 days. probability density expected (mean) project duration T = 32 days shaded area equals 99% probability of completing within t = ? days required project completion t = ? days V =9.12 days Fig. 4-7: Normal Distribution Project duration 12