PERT CPM assumes duration of activity is known with certainty. PERT assumes duration of activity as a random variable. For an activity, PERT estimates three quantities: a : optimistic duration under most favorable condition b : pessimistic duration under least favorable condition m : most likely value for activity’s duration

Probabilistic Estimates Activity start Optimistic time Most likely time (mode) Pessimistic time to tp tm te

Let Tij be the duration of activity (i,j) Let Tij be the duration of activity (i,j). PERT assumes that Tij follows a beta distribution and, mean and variance of Tij may be approximated by : expected duration of activities on any path : variance of duration of activities on any path

Example variance of PROJECT = variance of CRITICAL PATH if more than one critical path, PROJECT VARIANCE=largest of CRITICAL Example Activity a b m (1,2) 5 13 9 (1,3) 2 10 6 (3,5) 3 8 (3,4) 1 7 (4,5) 12 (5,6) 15

Project Diagram

Specified time – Path mean Path standard deviation critical path = B-D-E-F. Thus, E(CP) = 9 + 0 + 7 + 10 + 12 = 38 var(CP) = 1.78 + 0 + 4 + 0.44 + 1 = 7.22 standard deviation for CP is (7.22)1/2=2.69 What is the chance of completing project within 35 days ? there is a 13% chance of completing project within 35 days. Z = Specified time – Path mean Path standard deviation Z indicates how many s.d the specified time is beyond the expected path duration.

PERT Example activity duration predecessor te A 2 / 3 / 6 weeks - 3.33 B 3 / 6 / 10 weeks A 6.17 C 1 / 1 / 2 week A 1.17 D 3 / 3 / 3 weeks B, C 3.00 CRITICAL PATH: A-B-D EXPECTED DURATION: 3.33 + 6.17 + 3 = 12.5 VARIANCE: { (6-2) / 6}^2 + {(10-3) / 6}^2 + {(3-3) / 6}^2 = 1.805 STD = 1.344 e.g. if critical path expected = 9.5, and STD = 1.354 target=10 , Z = (10-9.5) / 1.354 = .369 probability = .644

EXAMPLE

EXAMPLE (Cont.) Identify all paths and their estimated completion times and variances.

Probability of Critical Path Being Completed in 39 Weeks or Less

PERT Example 1 2 4 6 7 3 5 9 8 Manual Testing Dummy System Training System Testing Orientation Position recruiting System development Equipment testing and modification Final debugging System changeover Job training Equipment installation 25

Example (Cont.) Time estimates (wks) Mean Time Variance Activity a b c t s2 1 - 2 6 8 10 8 .44 1 - 3 3 6 9 6 1.00 1 - 4 1 3 5 3 .44 2 - 5 0 0 0 0 .00 2 - 6 2 4 12 5 2.78 3 - 5 2 3 4 3 .11 4 - 5 3 4 5 4 .11 4 - 8 2 2 2 2 .00 5 - 7 3 7 11 7 1.78 5 - 8 2 4 6 4 .44 7 - 8 0 0 0 0 .00 6 - 9 1 4 7 4 1.00 7 - 9 1 10 13 9 4.00 26

Example (Cont.) Activity t s2 ES EF LS LF S 1 - 2 8 0.44 0 8 1 9 1 1 - 2 8 0.44 0 8 1 9 1 1 - 3 6 1.00 0 6 0 6 0 1 - 4 3 0.44 0 3 2 5 2 2 - 5 0 0.00 8 8 9 9 1 2 - 6 5 2.78 8 13 16 21 8 3 - 5 3 0.11 6 9 6 9 0 4 - 5 4 0.11 3 7 5 9 2 4 - 8 2 0.00 3 5 14 16 11 5 - 7 7 1.78 9 16 9 16 0 5 - 8 4 0.44 9 13 12 16 3 7 - 8 0 0.00 13 13 16 16 3 6 - 9 4 1.00 13 17 21 25 8 7 - 9 9 4.00 16 25 16 25 0 27

Example (Cont.) ( ) ( ) ( ) ( ) 1 2 4 6 7 3 5 9 8 ( ) ES=8, EF=8 LS=9, LF=9 ES=6, EF=9 LS=6, LF=9 ( ) ES=3, EF=5 LS=14, LF=16 ES=0, EF=3 LS=2, LF=5 ES=0, EF=6 LS=0, LF=6 ES=3, EF=7 LS=5, LF=9 ES=9, EF=13 LS=12, LF=16 LS=9, LF=16 ES=13, EF=13 LS=16 LF=16 ES=13, EF=25 LS=16 LF=25 ES=8, EF=13 LS=16 LF=21 ( ) ES=0, EF=8 LS=1, LF=9 ( ) ES=13, EF=25 LS=16 LF=25 s2 = s2 13 + s2 35 + s2 57 + s2 79 = 1.00 +0.11 + 1.78 + 4.00 = 6.89 weeks 28

. Determining Probability From Z Value s Z 0.00 0.01 … 0.09 What is the probability that project is completed within 30 weeks? s2 = 6.89 weeks, s = 6.89 = 2.62 weeks Z = x - m = 30 - 25 = 1.91 P(Z < 1.91) = 0.9719 2.62 s P( x < 30 weeks) = 0.50+ 0.4767 = 0.9767 . Z 0.00 0.01 … 0.09 1.9 0.4713 0.4719 … 0.4767 m = 25 Time (weeks) x = 30 33

What is the probability that project will be completed within 22 weeks ? Time (weeks) x = 22 P( x< 22 weeks) = 0.1271 22 - 25 -3 2.62 2.62 P(Z< -1.14) = 0.1271 = Z = = -1.14 34

PERT Probability Example A project has an expected completion time of 40 weeks, with a standard deviation of 5 weeks. What is the probability of finishing the project in 50 weeks or less? T = 40 s = 5 50 X Normal Distribution Z = - 40 5 2 . m z = 0 = 1 2.0 Standardized Normal Distribution 48

Obtaining the Probability m z = 0 s Z = 1 2.0 .00 .01 .02 0.0 .50000 .50399 .50798 : .97725 .97784 .97831 2.1 .98214 .98257 .98300 Standardized Normal Probability Table (Portion) Probabilities in body 49