PROJECT MANAGEMENT: PERT AMAT 167
PROBABILISTIC TIME ESTIMATES We need three time estimates for each activity: Optimistic time (t o ): length of time required under optimum conditions; Most likely time (t m ): most probable amount of time required; and Pessimistic time (t p ): length of time required under the worst conditions;
BETA DISTRIBUTION This is used to describe the inherent variability in time estimates. 0 t o t m t e t p Activity start Optimistic time Most likely time (mode) Pessimistic time
BETA DISTRIBUTION There is no established theoretical justification for using the beta distribution. It has certain features that make it attractive in practice: ØThe distribution can be symmetrical or skewed to either the right or the left according to the nature of a particular activity; Øthe mean and variance of the distribution can be readily obtained from the three time estimates (t o, t m, t p ); and Øthe distribution is unimodal with a high concentration of probability surrounding the most likely time estimate.
AVERAGE TIME Expected time of an activity: t e t 4 t t 6 o m p Expected duration of a path: Path mean of expected times of activities on the path
Variance of an activity s time: The larger the variance, the greater the uncertainty. VARIANCE 2 2 2 ( tp to) ( tp to) or 6 36 Note: the standard deviation of each activity s time is estimated as one-sixth of the difference between the pessimistic and optimistic time estimates Variance of a path s time: (Variances of activities on path)
EXAMPLE AOA diagram 1-3-4 a Optimistic time Most likely time 2-4-6 b Pessimistic time 2-3-5 c 3-4-5 d 2-3-6 g 3-5-7 e 4-6-8 h 5-7-9 f 3-4-6 i
EXAMPLE AON diagram Start 1-3-4 2-4-6 2-3-5 a b c 3-4-5 3-5-7 5-7-9 d e f 2-3-6 4-6-8 3-4-6 g h i Finish
REMARK (TRIANGULAR DISTRIBUTION) http://www.epixanalytics.com/modelassist/atrisk/images/15/image219.gif
PATH PROBABILITY Although activity times are represented by a beta distribution, the path distribution is represented by a normal distribution. The central limit theorem tells us that the summing of activity times (random variables) results in a normal distribution. The normal tendency improves as the number of random variables increases. However, even when the number of items being summed is fairly small, the normal approximation provides a reasonable approximation to the actual distribution.
FIGURE 17.9 PATH PROBABILITY Path Activity Activity Activity Beta Beta Beta Normal Probability that a given path will be completed in a specified length of time, can be determined using: z Specified time Path mean Path standard deviation
PATH PROBABILITY Probability of completing the path by the specified time 0 Expected path duration z Specified time
PATH PROBABILITY The resulting value of z indicates how many standard deviations of the path distribution the specified time is beyond the expected path duration. The more positive the value, the better. Rule of thumb: treat the path probability as being equal to 100 percent if the value of z is +3.00 or more. Note: probability when z = +3.00 is.9987.
PATH PROBABILITY A project is not completed until all of its activities have been completed, not only those on the critical path. It sometimes happens that another path ends up taking more time to complete than the critical path. This requires determining the probability that all paths will finish by a specified time. To do that, find the probability that each path will finish by the specified time, and then multiply those probabilities. The result is the probability that the project will be completed by the specified time.
INDEPENDENCE This assumes independence of each path duration times: find the probability that each path will finish by the specified time, and then multiply those probabilities. To use this, activity times should be independent of each other, and each activity is only on one path. However, this assumption is usually considered to be met if only a few activities in a large project are on multiple paths.
REMARK (NOT INDEPENDENT) In practice, when dependent cases occur, project planners often use Monte Carlo Simulation.
EXERCISE What is the probability that the project can be completed within 17 weeks of its start? AON diagram 1-3-4 2-4-6 2-3-5 a b c 3-4-5 3-5-7 5-7-9 What is the probability that the project will not be completed within 17 weeks of its start? Start d e f 2-3-6 4-6-8 3-4-6 g h i Finish
SOLUTION Path 17 weeks 1.00 a-b-c 10.0 Weeks d-e-f 16.0 Weeks 1.00 g-h-i 13.5 Weeks
SOLUTION Path z 17 Expectedpath duration Path standarddeviation Probability of Completion in 17 Weeks a-b-c 17 10 722. 097. d-e-f 17 16 100. 100. 1.00.8413 g-h-i 17 13. 5 107. 327. 1.00 P( Finish by week 17) P( Path a-b-c finish ) P( Path d-e-f finish ) P( Path g-h-i finish) 100.. 8413 100.. 8413