1. Project Management Dr. Ron Tibben-Lembke Operations Management

2. What’s a Project? Bringing about change is hard Many related activities Hard to plan production A project focuses on the outcome Regular teamwork focuses on the work process

3. Examples of Projects Building construction New product introduction Software implementation Training seminar Research project

4. Why are projects hard? Resources- –People, materials Planning –What needs to be done? –How long will it take? –What sequence? –Keeping track of who is supposedly doing what, and getting them to do it

5. IT Projects Half finish late and over budget Nearly a third are abandoned before completion –The Standish Group, in Infoworld Get & keep users involved & informed Watch for scope creep/ feature creep

6. Pinion Pine Power Plant SPP Co. 1992-97 A year late, $25m over budget Experimental technology –Coal gasification –20% less water than other plants –Partnership with DOE Unfortunately, didn’t work “In the Reno demonstration project, researchers found an inherent problem with the design of IGCC technology available at that time such that it would not work above 300 feet from sea level elevations.” - Wikipedia “Chemistry helped kill Pinon Pine, a $400 million government- funded flop in Nevada.” – NJ Ledger

7. Project Scheduling Establishing objectives Determining available resources Sequencing activities Identifying precedence relationships Determining activity times & costs Estimating material & worker requirements Determining critical activities

8. Project Personnel Structure Pure project “Skunk Works” Functional Project Matrix Project

9. Work Breakdown Structure Hierarchy of what needs to be done, in what order For me, the hardest part –I’ve never done this before. How am I supposed to know what I’ll have to do and how long it’ll take?

10. Project Scheduling Techniques Gantt chart Critical Path Method (CPM) Program Evaluation & Review Technique (PERT)

11. Gantt Chart

12. PERT & CPM Network techniques Developed in 1950’s CPM by DuPont for chemical plants PERT by U.S. Navy for Polaris missile Consider precedence relationships & interdependencies Each uses a different estimate of activity times

13. Completion date? On schedule? Within budget? Probability of completing by...? Critical activities? Enough resources available? How can the project be finished early at the least cost? Questions Answered by PERT & CPM

14. PERT & CPM Steps Identify activities Determine sequence Create network Determine activity times Find critical path Earliest & latest start times Earliest & latest finish times Slack

15. Activity on Node (AoN) 2 4? Years Enroll Receive diploma Project: Obtain a college degree (B.S.) 1 month Attend class, study etc. 1 1 day 3

16. Activity on Arc (AoA) 4,5 ? Years Enroll Receive diploma Project: Obtain a college degree (B.S.) 1 month Attend class, study, etc. 1 1 day 234

17. AoA Nodes have meaning Graduating Senior Applicant Project: Obtain a college degree (B.S.) 1 Alum 234 Student

18. We’ll use Activity on Node 1-2 must be done before 2-3 or 3-4 can start 2 3 4 1

19. Activity Relationships 2-3 must be done before 3-4 or 3-5 can start 2 3 4 1 5

20. Activity Relationships 2-4 and 3-4 must be done before 4-5 can start 2 3 4 1 5

21. Activity Relationships When 5-6 is done, project is complete. 2 3 4 1 56

22. Network Example You’re a project manager for Bechtel. Construct the network. ActivityPredecessors A-- BA CA DB EB FC GD HE, F

23. Network Example - AON ACEFBDGHZ

24. Network Example - AOA 2 4 5136879 A C F E B D H G

25. AOA Diagrams 231 A C B D A precedes B and C, B and C precede D 241 A C B D 354 Add a phantom arc for clarity.

26. Critical Path Analysis Provides activity information Earliest (ES) & latest (LS) start Earliest (EF) & latest (LF) finish Slack (S): Allowable delay Identifies critical path Longest path in network Shortest time project can be completed Any delay on activities delays project Activities have 0 slack

27. Critical Path Analysis Example

28. Network Solution A A E E D D B B C C F F G G 1 6 2 3 1 43

29. But can you build an Apartment?

30. Earliest Start & Finish Steps Begin at starting event & work forward ES = 0 for starting activities ES is earliest start EF = ES + Activity time EF is earliest finish ES = Maximum EF of all predecessors for non-starting activities

31. Activity A Earliest Start Solution For starting activities, ES = 0. A A E E D D B B C C F F G G 1 6 2 3 1 43

32. Earliest Start Solution A A E E D D B B C C F F G G 1 6 2 3 1 43

33. Latest Start & Finish Steps Begin at ending event & work backward LF = Maximum EF for ending activities LF is latest finish; EF is earliest finish LS = LF - Activity time LS is latest start LF = Minimum LS of all successors for non-ending activities

34. Earliest Start Solution A A E E D D B B C C F F G G 1 6 2 3 1 4 3

35. Latest Finish Solution A A E E D D B B C C F F G G 1 6 2 3 1 43

36. Compute Slack

37. Critical Path A A E E D D B B C C F F G G 1 6 2 3 1 43

38. New notation Compute ES, EF for each activity, Left to Right Compute, LF, LS, Right to Left C 7 LSLF ESEF

39. Exhibit 2.6, p.35 A 21 E 5 D 2 B 5 C 7 F 8 G 2

40. Exhibit 2.6, p.35 A 21 E 5 D 2 B 5 C 7 F 8 G 2 21282836 3638 2833 26282126 021 F cannot start until C and D are done. G cannot start until both E and F are done.

41. Exhibit 2.6, p.35 A 21 E 5 D 2 B 5 C 7 F 8 G 2 2126 021 26283136 3638 21282836 21282836 3638 2833 26282126 021 E just has to be done in time for G to start at 36, so it has slack. D has to be done in time for F to go at 28, so it has no slack.

42. Exhibit 2.6, p.35 A 21 E 5 D 2 B 5 C 7 F 8 G 2 2126 021 26283136 3638 21282836 21282836 3638 2833 26282126 021

43. Time-Cost Models 1. Identify the critical path 2. Find cost per day to expedite each node on critical path. 3. For cheapest node to expedite, reduce it as much as possible, or until critical path changes. 4. Repeat 1-3 until no feasible savings exist.

44. Time-Cost Example ABC is critical path=30 Crash costCrash per weekwks avail A5002 B8003 C5,0002 D1,1002 C 10 B 10 A 10 D 8 Cheapest way to gain 1 Week is to cut A

45. Time-Cost Example ABC is critical path=29 Crash costCrash per weekwks avail A5001 B8003 C5,0002 D1,1002 C 10 B 10 A 9 D 8 Cheapest way to gain 1 wk Still is to cut A Wks IncrementalTotal GainedCrash $Crash $ 1500500

46. Time-Cost Example ABC is critical path=28 Crash costCrash per weekwks avail A5000 B8003 C5,0002 D1,1002 C 10 B 10 A 8 D 8 Cheapest way to gain 1 wk is to cut B Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000

47. Time-Cost Example ABC is critical path=27 Crash costCrash per weekwks avail A5000 B8002 C5,0002 D1,1002 C 10 B 9 A 8 D 8 Cheapest way to gain 1 wk Still is to cut B Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800

48. Time-Cost Example Critical paths=26 ADC & ABC Crash costCrash per weekwks avail A5000 B8001 C5,0002 D1,1002 C 10 B 8 A 8 D 8 To gain 1 wk, cut B and D, Or cut C Cut B&D = $1,900 Cut C = $5,000 So cut B&D Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600

49. Time-Cost Example Critical paths=25 ADC & ABC Crash costCrash per weekwks avail A5000 B8000 C5,0002 D1,1001 C 10 B 7 A 8 D 7 Can’t cut B any more. Only way is to cut C Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600 51,9004,500

50. Time-Cost Example Critical paths=24 ADC & ABC Crash costCrash per weekwks avail A5000 B8000 C5,0001 D1,1001 C 9 B 7 A 8 D 7 Only way is to cut C Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600 51,9004,500 65,0009,500

51. Time-Cost Example Critical paths=23 ADC & ABC Crash costCrash per weekwks avail A5000 B8000 C5,0000 D1,1001 C 8 B 7 A 8 D 7 No remaining possibilities to reduce project length Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600 51,9004,500 65,0009,500 75,00014,500

52. Time-Cost Example C 8 B 7 A 8 D 7 No remaining possibilities to reduce project length Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600 51,9004,500 65,0009,500 75,00014,500 Now we know how much it costs us to save any number of days Customer says he will pay $2,000 per day saved. Only reduce 5 days. We get $10,000 from customer, but pay $4,500 in expediting costs Increased profits = $5,500

53. Benefits of PERT/CPM Useful at many stages of project management Mathematically simple Use graphical displays Give critical path & slack time Provide project documentation Useful in monitoring costs

54. Limitations of PERT/CPM Clearly defined, independent, & stable activities Specified precedence relationships Activity times (PERT) follow beta distribution Subjective time estimates Over emphasis on critical path

55. Conclusion Explained what a project is Summarized the 3 main project management activities Drew project networks Compared PERT & CPM Determined slack & critical path Computed project probabilities

56. PERT Activity Times 3 time estimates Optimistic times (a) Most-likely time (m) Pessimistic time (b) Follow beta distribution Expected time: t = (a + 4m + b)/6 Variance of times: v = (b - a) 2 /36  

57. Project Times Expected project time (T) Sum of critical path activity times, t Project variance (V) Sum of critical path activity variances, v

58. Example ActivityambE[T]variance A2484.331 B36.111.56.482 C48107.671 Project18.54 C C B B A A 4.33 6.48 7.67

59. Sum of 3 Normal Random Numbers 102030405060 Average value of the sum is equal to the sum of the averages Variance of the sum is equal to the sum of the variances Notice curve of sum is more spread out because it has large variance

60. Back to the Example: Probability of <= 21 wks 18.5 21 Average time = 18.5, st. dev = 2 21 is how many standard deviations above the mean? 21-18.5 = 2.5. St. Dev = 2, so 21 is 2.5/2 = 1.25 standard deviations above the mean Book Table says area between mean and 1.25 st dv is 0.3944 Probability <= 17 = 0.5+0.3944 = 0.8944 = 89.44%