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Managing Project Time Estimates with PERT and CPM: Understanding Variance and Probability

Chapter 10, Part 2 of the project management series delves into the techniques of Critical Path Method (CPM) and Program Evaluation Review Technique (PERT) to handle uncertainties in project durations. This section explains the use of three time estimates—optimistic, most likely, and pessimistic—to calculate expected project completion times and variances. Additionally, it demonstrates how these statistics help assess probabilities of finishing projects within specific time frames. Practical examples illustrate how to apply formulas and interpret results accurately.

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Managing Project Time Estimates with PERT and CPM: Understanding Variance and Probability

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  1. Chapter 10 part 2 Project Management

  2. CPM With 3 Time Estimates (PERT) • In past, time estimate is firm • Now, uncertain as to task duration • Natural variance • Use 3 time estimates to deal with uncertainty • a = optimistic estimate • m = most likely • b = pessimistic

  3. Critical Path • Path length is sum of expected times of activities on path (not sum of most likely times) • Longest expected path length is critical • ET = expected time of project • Path length is uncertain, and so is project duration • Path has variance equal to sum of variances of individual activities on path

  4. Formulas • ET (expected time) = (a + 4m + b)/6 • σ2 (variance) = [(b - a)/6]2 • σp (st dev of project) = sqrt(S (variances on CP)) • Z = (D - ET)/σ

  5. PERT Probability Example You’re a project planner for Apple. A new Ipod project has an expected completion time of 40 weeks, with a standard deviation of 5 weeks. What is the probability of finishing the sub in 50 weeks or less?

  6. Converting to Standardized Variable - - X ET 50 40 = = = Z 2 . 0 s 5 Normal Distribution Standardized Normal Distribution s = 1 s = 5 Z m = 40 50 X T Z = 0 2.0 z

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

  8. Example 1. CPM with Three Activity Time Estimates

  9. Example 1. Expected Time Calculations ET(A)= 3+4(6)+15 6 ET(A)=42/6=7

  10. Ex. 1. Expected Time Calculations ET(B)= 2+4(4)+14 6 ET(B)=32/6=5.333

  11. Ex 1. Expected Time Calculations ET(C)= 6+4(12)+30 6 ET(C)=84/6=14

  12. Duration = 54 Days C(14) E(11) H(4) A(7) D(5) F(7) I(18) B (5.333) G(11) Example 1. Network

  13. D=53 Example 1. Probability Exercise What is the probability of finishing this project in less than 53 days? p(t < D) t TE = 54

  14. (Sum the variance along the critical path.)

  15. p(t < D) t TE = 54 D=53 p(Z < -.156) = .438, or 43.8 % There is a 43.8% probability that this project will be completed in less than 53 weeks. (from the z table)

  16. Ex 1. Additional Probability Exercise • What is the probability that the project duration will exceed 56 weeks?

  17. p(t < D) t TE = 54 D=56 Example 1. Additional Exercise Solution p(Z > .312) = .378, or 37.8 %

  18. Activity List for Example Problem Required Activity Immediate Activity Description Predecessors Time (weeks) A 3 Select office site - B - 5 Create organization and financial plan C B 3 Determine personnel requirements D A,C 4 Design facility E D 8 Construct the interior F 2 Select personnel to move C G F 4 Hire new employees H F 2 Move records, key personnel, etc. I B 5 Make financial arrangements J 3 H,E,G Train new personnel

  19. Example 2 • ET = (a + 4m + b)/6 • σ2 = (b - a)2/36 = [(b - a)/6]2

  20. Example 2 • T = (a + 4m + b)/6 • σ2 = (b - a)2/36 = [(b - a)/6]2

  21. Example 2 • T = (a + 4m + b)/6 • σ2 = (b - a)2/36 = [(b - a)/6]2

  22. Calculations • ET = 23 weeks • Σσ2CP = σ 2B + σ 2C + σ 2D + σ 2E + σ 2J = 5.8056 • Z = (D - ET)/σ CP = (22-23)/2.4095 = -0.42 • Take .42 to the Z table => 1 – Z[.42] = .337 • When negative – subtract from 1 • Beware - never add standard deviations!

  23. ET = 23 22 34% chance < 22 50% chance > 23 16% chance between 22 and 23 (from table) 22 is 0.42 SD below Mean 0.4 0.3 0.2 0.1 0

  24. 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.

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