Dr. Henry Deng

1. IS 488 Information Technology Project Management Dr. Henry Deng Assistant Professor MIS Department UNLV

2. Today • Course schedule and dates • Questions from PERT lecture 1 • In class exercise for PERT lecture 1 • Review some of your ‘exam’ questions • Activity time example • Lecture 2 on PERT • Exercise 2 on PERT • Project team and topic

3. Let’s try this exercise before Lecture 2 -Calculate: ES,EF,LS,LF, Slacks, and CP E 5 4 5 D 3 2 F 1 C 1 B 2 8 3 6 A 2 I 2 1 G 2 H 1 7

4. Solution: ES,EF,LS,LF, Slacks, and CP E[6,11] 5[6,11] 4 5 D[3,6] 3[3,6] 2 F[11,12] 1[11,12] C[2,3] 1[2,3] B[0,2] 2[0,2] 8 3 6 A[0,2] 2[1,3] I[14,16] 2[14,16] 1 G[12,14] 2[12,14] H[0,1] 1[13,14] 7 CP: B,C,D,E,F,G, and I

5. PERT - Estimating activity time • Consider following question • What is the average waiting time in line at the registrar’s office? • How would you go about calculating an average score?

6. PERT - Estimating activity time • Consider following question • How long does it take to test codes for an accounts receivable program? • How would you get an average score? Different from previous question?

7. PERT - Estimating activity time • Consider following question • How long does it take to get sufficient responses to a RFP? • How would you estimate that?

8. PERT - Estimating activity time • Calculating the duration of the entire project and the scheduling of the specific activities depends on how we calculate time for each activity. • Obtaining estimates for projects that are repeat or projects that we have experience with is relatively easy. Estimating activity time for new and unique projects is significantly more difficult. • To factor uncertainly into the network analysis, often three estimates are used: Optimistic time (a), most probable time (m), and pessimistic time (b).

9. Estimating uncertain activity time • The three estimates (a, m, b) enable the systems analyst to develop the most likely activity time that ranges from the best possible (optimistic) time to the worst possible (pessimistic) time. • The expected time (t) can be calculated using the following formula: t = (a+4m+b)/6 • To measure the dispersion or variation in the activity time values, the common statistical measure of the variance can be used: 2 =[(b-a)/6]2 (This formula assumes that a standard deviation is approximately 1/6 of the difference between the extreme values of the distribution: (b-a)/6. The variance is simply the square of the standard deviation).

10. Example of estimating activity time Consider the optimistic, most probable, and pessimistic time estimates for a project that involves the following activities: Activity Optimistic Most probable Pessimistic (a) (m) (b) ------------------------------------------------------------------------- A 4 5 12 B 1 1.5 5 C 2 3 4 D 3 4 11 E 2 3 4 F 1.5 2 2.5 G 1.5 3 4.5 H 2.5 3.5 7.5 I 1.5 2 2.5 J 1 2 3

11. Estimating time for activity A Using the expected time (t) formula t = (a + 4m + b)/6 we have an estimated average or expected completion time of tA = [4 + 4(5) + 12]/6 = 36/6 = 6 weeks and using the variance formula 2 =[(b - a)/6]2 we can determine the measure of uncertainty or the variance for activity A: 2A =[(12 - 4)/6]2 = (8/6)2 = 1.78

12. Estimating time for all activities Activity Expected time Variance (in weeks) ------------------------------------------------------------------------- A [4 + 4(5) + 12]/6 6 [(12 - 4)/6]2 1.78 B 2 0.44 C [2 + 4(3) + 4]/6 3 [(4 - 2)/6]2 0.11 D 5 1.78 E 3 0.11 F 2 0.03 G 3 0.25 H [2.5 + 4(3.5) + 7.5]/6 4 [(7.5 – 2.5)/6]2 0.69 I 2 0.03 J 2 0.11 Total 32 Once expected activity times are calculated, we can proceed with the critical path calculations to determine the expected project completion time and a detailed activity schedule.

13. Network with expected activity times C 3 2 5 F 2 D 5 A 6 E 3 G 3 J 2 1 4 7 8 B 2 I 2 H 4 3 6

14. Network with ES & EF C [6,9] 3 2 5 F [9,11] 2 D [6,11] 5 A [0,6] 6 J [15,17] 2 G [11,14] 3 E [6,9] 3 1 4 7 8 B [0,2] 2 I [13,15] 2 H [9,13] 4 3 6

15. Network with ES, EF, LS & LF C [6,9] 3 [10,13] Earliest Start Time Earliest Finish Time 2 5 D [6,11] 5 [7,12] A [0,6] 6 [0,6] F [9,11] 2 [13,15] J [15,17] 2 [15,17] G [11,14] 3 [12,15] E [6,9] 3 [6,9] 1 4 7 8 B [0,2] 2 [7,9] I [13,15] 2 [13,15] Latest Start Time Latest Finish Time H [9,13] 4 [9,13] 3 6

16. Activity schedule (in weeks) Earliest Latest Earliest Latest Start Start Finish Finish Slack Critical Activity (ES) (LS) (EF) (LF) (LS - ES) Path? ---------------------------------------------------------------------------------- A 0 0 6 6 0 Yes B 0 7 2 9 7 C 6 10 9 13 4 D 6 7 11 12 1 E 6 6 9 9 0 Yes F 9 13 11 15 4 G 11 12 14 15 1 H 9 9 13 13 0 Yes I 13 13 15 15 0 Yes J 15 15 17 17 0 Yes Critical path - A, E, H, I, and J Project duration - 17 weeks

17. Variance in critical path activities • Variation in critical path activities can cause variation in the project completion date. • If a non-critical activity is delayed beyond its slack time, then that activity would become part of the new critical path, and further delays would affect the project completion date. • Variation in critical path activities resulting in shorter critical path will result in an earlier than expected completion date. • The variance in the project duration is the same as the sum of the variance of the critical path activities.

18. Probability of meeting deadline • The expected (E) project time (T) for the previous example is E(T) = tA+ tE + tH+ tI+ tJ = 6 + 3 + 4 + 2 + 2 = 17 weeks • The variance (2) for that example is Var (T) = 2 = 2A + 2E + 2H + 2I + 2J • Since standard deviation is the square root of the variance, then  =  2=  2.72 = 1.65

19. Estimating time for all activities Activity Expected time Variance Variance (in weeks) 2 (for critical path) ------------------------------------------------------------------------- A (CP) 6 1.78 1.78 B 2 0.44 C 3 0.11 D 5 1.78 E (CP) 3 0.11 0.11 F 2 0.03 G 3 0.25 H (CP) 4 0.69 0.69 I (CP) 2 0.03 0.03 J (CP) 2 0.11 0.11 Total 32 2.72 Var (T) Standard deviation for critical path activities:  =  2=  2.72 = 1.65

20. Probability of meeting deadline • Assuming a normal (bell-shaped) distribution of the project completion time allows us to compute the probability of meeting a specified project completion date. • Suppose the management has allowed 20 weeks for the previous project. What is the probability that we will meet the 20-week deadline? • We are looking for the probability of T <=20. • The z value for the normal distribution of T = 20 is z = (20 - 17)/1.65 = 1.82 • We need to use the normal distribution table.

21. Normal distribution of project time -------------------------------------------------------- 17 20 Time (weeks)

23. Normal distribution of project time 0.4656 + 0.5000 z = = 0.9656 (20 -17)/1.65 = 1.82 p(T<= 20) -------------------------------------------------------- 17 20 Time (weeks)

24. Summary • PERT procedure can be used to schedule projects with uncertain activity times. • The three time estimates (optimistic, most likely, pessimistic) help calculate an expected time and variance for each activity. • The time for critical path activities provides the expected project completion time. • The sum of the variances of activities on the critical path provides the variance in the project completion time. • Normal probability distribution assumption and procedures are used to compute the probability of the project being completed by a specific time.

26. Time-Cost Trade-off: Crashing Video 6 Time Crashing

27. Resource Limitations critical path crashing (cost/time tradeoff) other methods

28. Crashing • can shorten project completion time by adding extra resources (costs) • start off with NORMAL TIME CPM schedule • get expected duration Tn, cost Cn • Tn should be longest duration • Cn should be most expensive in penalties, cheapest in crash costs

29. Time Reduction • to reduce activity time, pay for more resources • develop table of activities with times and costs • for each activity, usually assume linear relationship for relationship between cost & time

30. Crash Example Activity: programming Tn: 7 weeks Cn: $14,000(7 weeks, 2 programmers) if you add a third programmer, done in 6 weeks Tc: 6 weeks Cn: $15,000 cost slope = (15000-14000)/(6-7)=-$1000/week

31. Example Problem activity Pred Tn Cn Tc Cc slope max A requirements none 3 can’t crash B programming A 7 14000 6 15000 -1000 1 week C get hardware A 1 50000 .5 51000 -2000 .5 week D train users B,C 3 can’t crash Crashing Algorithm: 1 crash only critical activitiesB only choice 2 crash cheapest currently criticalB is cheapest 3 after crashing one time period, recheck critical

32. Crash Example Import critical software from Australia: late penalty $500/d > 12 d A get import license 5 days no predecessor B ship 7 days A is predecessor C train users 11 days no predecessor D train on system 2 days B,C predecessors can crash C: $2000/day more than current for up to 3 days B: faster boat 6 days $300 more than current bush plane 5 days $400 more than current commercial 3 days $500 more than current

33. Crash Example Original schedule: 14 days, $1,000 in penalties = $1000 crash B to 6 days:13 days, $500 penalties, $300 cost = $800* crash B to 5 C to 10: 12 days, no penalties, $400+2000 cost = $2400 to 11 days is worse NOW A SELECTION DECISION risk versus cost

34. Crashing Limitations • assumes linear relationship between time and cost • not usually true (indirect costs don’t change at same rate as direct costs) • requires a lot of extra cost estimation • time consuming • ends with tradeoff decision

35. Resource Constraining • CPM & PERT both assume unlimited resources NOT TRUE • may have only a finite number of systems analysts, programmers • RESOURCE LEVELING - balance the resource load • RESOURCE CONSTRAINING - don’t exceed available resources