Chapter 10 part 2

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.