1. Project Management Dr. Ron Lembke Operations Management

2. What’s a Project? Changing something from the way it is to the desired state Never done one exactly like this Many related activities Focus 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. Project Scheduling Establishing objectives Determining available resources Sequencing activities Identifying precedence relationships Determining activity times & costs Estimating material & worker requirements Determining critical activities

7. 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

8. 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

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

10. 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

11. 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

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

13. Precedence Relationships – Scenario 1a Hierarchy of what needs to be done, in what order –Time required for each thing For me, the hardest part –I’ve never done this before. How do I know what I’ll do when and how long it’ll take? –I think in phases –The farther ahead in time, the less detailed –Figure out the tricky issues, the rest is details –A lot will happen between now and then –It works not badly with no deadline

14. Precedence Relationships

15. Mudroom

16. Mudroom Remodel Big-picture sequence easy: –Demolition –Framing –Plumbing –Electrical –Drywall, tape & texture –Slate flooring –Cabinets, lights, paint Hard: can a sink fit? D W DW Before: After:

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

18. Gantt Chart

21. Day before Thanksgiving, 2015, late afternoon, 27ºF What do these 3 guys have in common?

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. A BEDCFHGZ

25. A BEDC F HGZ

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

27. 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.

28. Scenario 1a TaskDescriptionDaysPred ADrawings1-- BGet & Dye fabric3-- CBuild stage4A DSew fabric, frame back2B ERent, deliver, hang lights1A FPaint stage2C GTest assembly1D,F Hdisassemble, deliver1G IAssemble, touch-up1E,H

29. Scenario 1a Network Diagram 3 START A 1 B 3 E 1 C 4 D 2 F 2 G 1 H 1 I 1 Finish

30. Scenario 1a Longest Path 3 START A 1 B 3 E 1 C 4 D 2 F 2 G 1 H 1 I 1 Finish Paths:Days AEI3 ACFGHI10 BDGHI8

31. 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 critical activities delays the whole project Activities have 0 slack

32. Critical Path Analysis Example

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

34. 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

35. 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

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

37. 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

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

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

40. Compute Slack

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

42. Scenario 1b ES/EF times START A 1 0101B 3 0303E 1 1212C 4 1515D 2 3535F 2 5757G 1 7878 H 1 8989I 1 910 Finish

43. Scenario 1c LF/LS times START A 1 0101 0101 B 3 0303 2525 E 1 1212 8989 C 4 1515 1515 D 2 3535 5757 F 2 5757 5757 G 1 7878 7 8 H 1 8989 8989 I 1 910 Finish

44. Scenario 1c Finish START A 1 0101 0101 B 3 0303 2525 E 1 1212 8989 C 4 1515 1515 D 2 3535 5757 F 2 5757 5757 G 1 7878 7 8 H 1 8989 8989 I 1 910

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

46. Example #2 A 21 E 5 D 2 B 4 C 7 F 7 G 2 21282835 3537 2833 25272125 021

47. Example #2 A 21 E 5 D 2 B 4 C 7 F 7 G 2 21282835 3537 2833 25272125 021 F cannot start until C and D are done. G cannot start until both E and F are done.

48. Example #2 A 21 E 5 D 2 B 4 C 7 F 7 G 2 2226 021 26283035 3537 21282835 21282835 3537 2833 25272125 021 E just has to be done in time for G to start at 35, so it has slack. D has to be done in time for F to go at 28, so it has no slack.

49. Example #2 A 21 E 5 D 2 B 4 C 7 F 7 G 2 2226 021 26283035 3537 21282835 21282835 3537 2833 25272125 021 E just has to be done in time for G to start at 35, so it has slack. D has to be done in time for F to go at 28, so it has no slack.

50. Gantt Chart - ES 051015 20 25 30 35 40 A B C D E F G

51. Gantt Chart - LS 051015 20 25 30 35 40 A B C D E F G

52. Another Example A 1 B 4 C 3 D 7 E 6 F 2 H 9 I 4 G 7

53. Solved Problem 1 A 1 0101 0101 B 4 1515 1515 C 3 6969 1414 D 7 2929 1818 E 6 511 F 2 911 810 H 9 918 817 I 4 1822 G 7 1118

54. Can We Go Faster?

55. 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.

56. 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

57. 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

58. 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

59. 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

60. 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

61. 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

62. 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

63. 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

64. 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 weeks Customer says he will pay $2,000 per week saved. Only reduce 5 weeks. We get $10,000 from customer, but pay $4,500 in expediting costs Increased profits = $5,500

65. Ex. AOA

66. AON Paths: ABF: 18 CDEF: 20 C5 B 10 A 6 D 4 E 9 F2 Activity AvailCost A0---- B2500 C1300 D3700 E2600 F 1800

67. Gantt charts to keep track Original A 6 B 10 C 5 D 4 E 9 2 F 2 Activity AvailCost A0---- B2500 C1300 D3700 E2600 F 1800 Consider CDEF C is cheapest, crash 1 day A 6B 10 C 4 D 4 E 9 1 F 2 C5 B10 A 6 D 4 E 9 F2 C4 B10 A 6 D 4 E 9 F2 Revised Pictures: Activity AvailCost A0---- B2500 C0 D3700 E2600 F 1800 Consider DEF E cheapest Crash E 1 day

68. Keep Track with Gantt Charts A 6B 10 C 4 D 4 E 8 F 2 C4 B10 A 6 D 4 E8 F2 New Pictures: Activity AvailCost A0---- B2500 C0 D3700 E1600 F1800 All Activities Critical ABF: 18 CDEF: 18 Options: Crash: F $800 – last one B, AND D or E E is cheaper than D $500+$600 = $1,100 > 1,000 Cost > Benefit A 6B 10 C 4 D 4 E 8 F 1 C4 B10 A 6 D 4 E8 F1

69. Scenario 2 Now, need to do it in 7 days! TaskNormalCrash$/day B32$1,000 C412,000 D21500 F21800 START A 1 0101 0101 B 3 0303 2525 E 1 1212 8989 C 4 1515 1515 D 2 3535 5757 F 2 5757 5757 G 1 7878 7 8 H 1 8989 8989 I 1 910 Not critical Cheapest $800 gain 1 day Finish

70. Scenario 2 Now, need to do it in 7 days! TaskNormalCrash$/day B32$1,000 C412,000 D21500 F21800 START A 1 0101 0101 B 3 0303 1414 E 1 1212 8 C 4 1515 1515 D 2 3535 4646 F 1 5656 5656 G 1 6767 67 H 1 7878 7878 I 1 8989 8 9 Not critical Used up Only option Day 1 $800 Day 2 $2,000 Finish

71. Scenario 2 Now, need to do it in 7 days! TaskNormalCrash$/day B32$1,000 C312,000 D21500 F21800 START A 1 0101 0101 B 3 0303 0303 E 1 1212 8 C 3 1414 1414 D 2 3535 3535 F 1 4545 4545 G 1 5656 56 H 1 6767 6767 I 1 7878 7 8 Used up Day 1 $800 Day 2 $2,000 Something on both paths C, and B or D, D cheaper Day 3 $2,500 Total$5,300 Finish

72. Scenario 2 Now, Done in 7 days! TaskNormalCrash$/day B32$1,000 C212,000 D11500 F21800 START A 1 0101 0101 B 3 0303 0303 E 1 1212 5656 C 2 1313 1313 D 1 3434 3434 F 1 3434 3434 G 1 4545 45 H 1 5656 5656 I 1 6767 6 7 Used up Day 1 $800 Day 2 $2,000 Day 3 $2,500 Total$5,300 Used up Finish

73. What about Uncertainty?

74. 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  

75. Example Activity a = 2, m = 4, b = 6 E[T] = (2 + 4*4 + 6)/6 = 24/6 = 4.0 σ 2 = (6 – 2) 2 / 36 = 16/36 = 0.444

76. Example ActivityambE[T]variance A2484.331 B36.111.56.482 C48107.671 Project18.54 Complete in 18.5 days, with a variance of 4. C C B B A A 4.33 6.48 7.67

77. 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

78. 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 Z=1.25 and –infinity is 0.8944 Probability <= 21 wks = 0.8944 = 89.44%

79. 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

80. 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

81. Conclusion Explained what a project is Drew project networks Compared PERT & CPM Determined slack & critical path Found profit-maximizing crash decision Computed project probabilities