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Probabilistic Time Estimates

Probabilistic Time Estimates. Optimistic time Time required under optimal conditions Pessimistic time Time required under worst conditions Most likely time Most probable length of time that will be required. Variability in Activity Times.

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Probabilistic Time Estimates

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  1. Probabilistic Time Estimates • Optimistic time • Time required under optimal conditions • Pessimistic time • Time required under worst conditions • Most likely time • Most probable length of time that will be required

  2. Variability in Activity Times • CPM assumes we know a fixed time estimate for each activity and there is no variability in activity times • PERT uses a probability distribution for activity times to allow for variability

  3. Variability in Activity Times • Three time estimates are required • Optimistic time (a) – if everything goes according to plan • Most–likely time (m) – most realistic estimate • Pessimistic time (b) – assuming very unfavorable conditions

  4. a + 4m + b 6 Mean (expected time): t = 2 b - a 6 Variance: 2 = where a = optimistic estimate m = most likely time estimate b= pessimistic time estimate Probabilistic Time Estimates • Beta distribution • A probability distribution traditionally used in PERT

  5. Why Use the Beta Distribution? • Mean and variance can be estimated with 3 time estimates • Continuous, but no predetermined shape • Tradition

  6. P(time) P(time) a m t b a t m b Time Time P(time) a m = t b Time Examples of Beta Distributions

  7. Start Project Network with Probabilistic Time Estimates: Example Lay foundation Build house 4 2 Finish work 11,12,20 7,8,12 7 1 3,4,7 9,12,15 Design house and obtain financing 6 3 5 2,4,6 3,4,5 2,4,5 Select carpet Order and receive materials Select paint

  8. TIME ESTIMATES (WKS) MEAN TIME VARIANCE ACTIVITYa m b tб2 1 9 12 15 12 1.00 2 7 8 12 8.5 0.69 3 3 4 5 4 0.11 4 11 12 20 13.17 2.25 5 2 4 5 3.83 0.25 6 2 4 6 4 0.44 7 3 4 7 4.33 0.44 Activity Time Estimates

  9. ACTIVITYtбES EF LS LF S 1 12 1.00 0 12 0 12 0 2 8.5 0.69 12 20.5 12 20.5 0 3 4 0.11 12 16 16.5 20.5 4.5 4 13.17 2.25 20.5 33.7 20.5 33.7 0 5 3.83 0.25 20.5 24.3 25.8 29.7 5.4 6 4 0.44 24.3 28.3 29.7 33.7 5.4 7 4.33 0.44 33.7 38 33.7 38 0 Activity Early, Late Times, and Slack

  10. 2 12 20.5 4 20.5 33.7 2 12 20.5 3 20.5 33.7 1 0 12 7 33.7 38 3 0 12 Start 1 33.7 38 Earliest, Latest, and Slack Critical Path Lay foundation Build house Design house and obtain financing Finish work 6 24.3 28.3 3 12 16 1 29.7 33.7 16.5 20.5 1 5 20.5 24.3 Select carpet 29.7 1 25.8 Order and receive materials Select paint

  11. Total project variance • 2 = б12 + б22 +б42 +б72 • 2 = 1.00 + 0.694 + 2.25 + 0.444 • = 4.389 weeks

  12. Determine probability that project is completed within specified time where  = tp = project mean time  = project standard deviation x = proposed project time Z = number of standard deviations xis from mean x -   Z = Probabilistic Network Analysis

  13. Key Assumptions • Activity times are statistically independent • Allows summation of individual expected activity times and variances to get the overall projects expected time and variance • Network mean and variance normally distributed

  14. Probability Z  = tp x Time Normal Distribution Of Project Time

  15. Z = = = 0.9547 P(x 40 weeks)  2 = 4.389 weeks  = 4.389  = 2.095 weeks x -   40 - 38 2.095  = 38 x = 40 Time (weeks) House Example – 40 Weeks What is the probability that the project is completed within 40 weeks? From Table A.1, (appendix A) a Z score of 0.9547 corresponds to a probability of ~0.3289. Thus P(40) = 0.3289 + 0.5000 = 0.8289

  16. Probabilistic Estimates to tm te tp Activity start Optimistictime Most likely time (mode) Pessimistic time Beta Distribution

  17. te = to + 4tm +tp6 Expected Time te = expected time to = optimistic time tm = most likely time tp = pessimistic time

  18. (tp – to)2 36 2 = Variance 2= variance to = optimistic time tp = pessimistic time

  19. Optimistic time Most likely time Pessimistic time 2-4-6 b 2-3-5 c 1-3-4 a 3-4-5 d 3-5-7 e 5-7-9 f 2-3-6 g 3-4-6 i 4-6-8 h Example 5

  20. 4.00 b 3.17 c 2.83 a 4.00 d 5.0 e 7.0 f 3.33 g 4.17 i 6.0 h Example 5 Time Estimates Tabc = 10.0Tdef = 16.0Tghi = 13.50

  21. Specified time – Path mean Path standard deviation Z = Path Probabilities Z indicates how many standard deviations of the path distribution the specified tine is beyond the expected path duration.

  22. 17 Weeks 1.00 a-b-c Weeks 10.0 d-e-f Weeks 16.0 1.00 g-h-i Weeks 13.5 Example 6

  23. Earliest start (ES) = earliest time at which an activity can start, assuming all predecessors have been completed Earliest finish (EF) = earliest time at which an activity can be finished Latest start (LS) = latest time at which an activity can start so as to not delay the completion time of the entire project Latest finish (LF) = latest time by which an activity has to be finished so as to not delay the completion time of the entire project Activity Description Time (weeks) A Build internal components 2 B Modify roof and floor 3 C Construct collection stack 2 D Pour concrete and install frame 4 E Build high-temperature burner 4 F Install pollution control system 3 G Install air pollution device 5 H Inspect and test 2 Total Time (weeks) 25 Determining the Project Schedule Perform a Critical Path Analysis Table 3.2

  24. Forward Pass Begin at starting event and work forward Earliest Start Time Rule: • If an activity has only a single immediate predecessor, its ES equals the EF of the predecessor • If an activity has multiple immediate predecessors, its ES is the maximum of all the EF values of its predecessors ES = Max {EF of all immediate predecessors}

  25. Forward Pass Begin at starting event and work forward Earliest Finish Time Rule: • The earliest finish time (EF) of an activity is the sum of its earliest start time (ES) and its activity time EF = ES + Activity time

  26. Backward Pass Begin with the last event and work backwards Latest Finish Time Rule: • If an activity is an immediate predecessor for just a single activity, its LF equals the LS of the activity that immediately follows it • If an activity is an immediate predecessor to more than one activity, its LF is the minimum of all LS values of all activities that immediately follow it LF = Min {LS of all immediate following activities}

  27. Backward Pass Begin with the last event and work backwards Latest Start Time Rule: • The latest start time (LS) of an activity is the difference of its latest finish time (LF) and its activity time LS = LF – Activity time

  28. Computing Slack Time After computing the ES, EF, LS, and LF times for all activities, compute the slack or free time for each activity • Slack is the length of time an activity can be delayed without delaying the entire project Slack = LS – ES or Slack = LF – EF

  29. Earliest Earliest Latest Latest On Start Finish Start Finish Slack Critical Activity ES EF LS LF LS – ES Path A 0 2 0 2 0 Yes B 0 3 1 4 1 No C 2 4 2 4 0 Yes D 3 7 4 8 1 No E 4 8 4 8 0 Yes F 4 7 10 13 6 No G 8 13 8 13 0 Yes H 13 15 13 15 0 Yes Computing Slack Time

  30. Variability in Activity Times • CPM assumes we know a fixed time estimate for each activity and there is no variability in activity times • PERT uses a probability distribution for activity times to allow for variability

  31. Variability in Activity Times • Three time estimates are required • Optimistic time (a) – if everything goes according to plan • Pessimistic time (b) – assuming very unfavorable conditions • Mostlikely time (m) – most realistic estimate

  32. Expected time: Variance of times: t = (a + 4m + b)/6 v = [(b – a)/6]2 Variability in Activity Times Estimate follows beta distribution

  33. Probability of 1 in 100 of < a occurring Probability of 1 in 100 of > b occurring Probability Activity Time Optimistic Time (a) Most Likely Time (m) Pessimistic Time (b) Variability in Activity Times Estimate follows beta distribution Expected time: Variance of times: t = (a + 4m + b)/6 v = [(b − a)/6]2

  34. Most Expected Optimistic Likely Pessimistic Time Variance Activity a m b t = (a + 4m + b)/6 [(b – a)/6]2 A 1 2 3 2 .11 B 2 3 4 3 .11 C 1 2 3 2 .11 D 2 4 6 4 .44 E 1 4 7 4 1.00 F 1 2 9 3 1.78 G 3 4 11 5 1.78 H 1 2 3 2 .11 Computing Variance

  35. s2 = Project variance = (variances of activities on critical path) p Probability of Project Completion Project variance is computed by summing the variances of critical activities

  36. Project variance s2 = .11 + .11 + 1.00 + 1.78 + .11 = 3.11 Project standard deviation sp = Project variance = 3.11 = 1.76 weeks p Probability of Project Completion Project variance is computed by summing the variances of critical activities

  37. Probability of Project Completion PERT makes two more assumptions: • Total project completion times follow a normal probability distribution • Activity times are statistically independent

  38. 15 Weeks (Expected Completion Time) Probability of Project Completion Standard deviation = 1.76 weeks

  39. Z = – /sp = (16 wks – 15 wks)/1.76 = 0.57 due expected date date of completion Probability of Project Completion What is the probability this project can be completed on or before the 16 week deadline? Where Z is the number of standard deviations the due date or target date lies from the mean or expected date

  40. From Appendix .00 .01 .07 .08 .1 .50000 .50399 .52790 .53188 .2 .53983 .54380 .56749 .57142 .5 .69146 .69497 .71566 .71904 .6 .72575 .72907 .74857 .75175 Z = − /sp = (16 wks − 15 wks)/1.76 = 0.57 due expected date date of completion Probability of Project Completion What is the probability this project can be completed on or before the 16 week deadline? Where Z is the number of standard deviations the due date or target date lies from the mean or expected date

  41. 0.57 Standard deviations Probability(T ≤ 16 weeks)is 71.57% 15 16 Weeks Weeks Time Probability of Project Completion

  42. Probability of 0.99 Probability of 0.01 Z 2.33 Standard deviations From Appendix I 0 2.33 Determining Project Completion Time

  43. Variability of Completion Time for Noncritical Paths • Variability of times for activities on noncritical paths must be considered when finding the probability of finishing in a specified time • Variation in noncritical activity may cause change in critical path

  44. What Project Management Has Provided So Far • The project’s expected completion time is 15 weeks • There is a 71.57% chance the equipment will be in place by the 16 week deadline • Five activities (A, C, E, G, and H) are on the critical path • Three activities (B, D, F) are not on the critical path and have slack time • A detailed schedule is available

  45. Project Planning, Controlling and Scheduling Project Planning: 1. Setting goals. 2. Defining the project. 3. Tying needs into timed project activities. 4. Organizing the team. Project Scheduling: 1. Tying resources to specific activities. 2. Relating activities to each other. 3. Updating and revising on regular basis. Before Project Project Controlling: 1. Monitoring resources, costs, quality and budgets. 2. Revising and changing plans. 3. Shifting resources to meet demands. During Project

  46. Six Steps of PERT/CPM • Define the project and all of its significant activities or tasks • Develop the relationships among the activities and decide which activities must precede others • Draw the network connecting all of the activities • Assign time and/or cost estimates to each activity • Compute the longest time path through the network; this is called the critical path • Use the network to help plan, schedule, monitor, and control the project • The critical path is important since any delay in these activities can delay the completion of the project

  47. PERT/CPM • Given the large number of tasks in a project, it is easy to see why the following questions are important • When will the entire project be completed? • What are the critical activities or tasks in the project, that is, the ones that will delay the entire project if they are late? • Which are the non-critical activities, that is, the ones that can run late without delaying the entire project’s completion? • If there are three time estimates, what is the probability that the project will be completed by a specific date?

  48. PERT/CPM • At any particular date, is the project on schedule, behind schedule, or ahead of schedule? • On any given date, is the money spent equal to, less than, or greater than the budgeted amount? • Are there enough resources available to finish the project on time? • If the project is to be finished in a shorter amount of time, what is the best way to accomplish this at the least cost?

  49. General Foundry Example of PERT/CPM • General Foundry, Inc. has long been trying to avoid the expense of installing air pollution control equipment • The local environmental protection group has recently given the foundry 16 weeks to install a complex air filter system on its main smokestack • General Foundry was warned that it will be forced to close unless the device is installed in the allotted period • They want to make sure that installation of the filtering system progresses smoothly and on time

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