1. © 2007 Pearson Education Project Management Chapter 3

2. © 2007 Pearson Education How Project Management fits the Operations Management Philosophy Operations As a Competitive Weapon Operations Strategy Project Management Process Strategy Process Analysis Process Performance and Quality Constraint Management Process Layout Lean Systems Supply Chain Strategy Location Inventory Management Forecasting Sales and Operations Planning Resource Planning Scheduling

3. © 2007 Pearson Education Bechtel Group, INC.  Bechtel is a $16.3 billion-a-year construction contractor that specializes in large projects.  It is successful because it is able to bring large projects in quickly and on time.  For each major project it organizes a project team and provides it with the supporting information systems and resources.  It utilizes a web-based communications system that provides access to project information electronically.  Team members have instant access to schedules, progress reports, drawings and messages.

4. © 2007 Pearson Education Projects  A project is an interrelated set of activities with a definite starting and ending point, which results in a unique outcome for a specific allocation of resources.  The three main goals of project management are… 1.Complete the project on time or earlier. 2.Do not exceed the budget. 3.Meet the specifications to the satisfaction of the customer.

5. © 2007 Pearson Education Project Management  Project management is a systemized, phased approach to defining, organizing, planning, monitoring, and controlling projects.  A collection of projects is called a program, which is an interdependent set of projects with a common strategic purpose.  A cross-functional effort: Even though a project may be under the overall purview of a single department, other departments likely should be involved.

6. © 2007 Pearson Education Planning Projects  Planning projects involves five steps: 1.Defining the work breakdown structure -- a statement of all work that has to be completed. 2.Diagramming the network -- a graphical network 3.Developing the schedule -- specifying start times for each activity 4.Analyzing cost-time trade-offs 5.Assessing risks

7. © 2007 Pearson Education A Work Breakdown Structure (three levels) for a new business © 2007 Pearson Education

8. Diagramming the Network  A Network Diagram visually displays the interrelated activities using nodes (circles) and arcs (arrows) that depict the relationships between activities.  Two network planning methods (PERT & CPM) were originally distinctive, but today the differences are minor and will be jointly referred to as PERT/CPM.  PERT (Program Evaluation and Review Technique) was utilized when activity times involved risk.  CPM (Critical Path Method) was used when activity times were certain.

9. © 2007 Pearson Education Precedence Relationships  Precedence relationships determine a sequence for undertaking activities, and specify that any given activity cannot start until a preceding activity has been completed. In the AON approach, the nodes (circles) represent activities, and the arcs represent the precedence relationships between them. AON STU Activity On Node approach “S” precedes “T” which precedes “U”

10. © 2007 Pearson Education Activity Relationships T U S T & U cannot begin until S has been completed. S T U S & T must be completed before U can be started.

11. © 2007 Pearson Education Activity Relationships S T U V U & V can’t begin until S & T have been completed. S T U V U cannot begin until S & T have been completed. V cannot begin until T has been completed.

12. © 2007 Pearson Education Activity Relationships STV U T & U cannot begin until S has been completed; V cannot begin until both T & U have been completed.

13. © 2007 Pearson Education St. Adolf’s Hospital Example 3.1 Immediate ActivityDescriptionPredecessor(s)Responsibility ASelect administrative and medical staff. BSelect site and do site survey. CSelect equipment. DPrepare final construction plans and layout. EBring utilities to the site. FInterview applicants and fill positions in nursing, support staff, maintenance, nursing, support staff, maintenance, and security. and security. GPurchase and take delivery of equipment. HConstruct the hospital. IDevelop an information system. JInstall the equipment. KTrain nurses and support staff.

14. © 2007 Pearson EducationImmediate ActivityDescriptionPredecessor(s)Responsibility ASelect administrative and medical staff.—Johnson BSelect site and do site survey.—Taylor CSelect equipment.AAdams DPrepare final construction plans and layout.BTaylor EBring utilities to the site.BBurton FInterview applicants and fill positions inAJohnson nursing, support staff, maintenance, nursing, support staff, maintenance, and security. and security. GPurchase and take delivery of equipment.CAdams HConstruct the hospital.DTaylor IDevelop an information system.ASimmons JInstall the equipment.E,G,HAdams KTrain nurses and support staff.F,I,JJohnson St. Adolf’s Hospital Example 3.1

15. © 2007 Pearson Education St. Adolf’s Hospital Diagramming the Network Finish Start A B C D E FGHI J K A— B— CA DB EB FA GC HD IA JE,G,H KF,I,J Immediate Predecessor

16. © 2007 Pearson Education St. Adolf’s Hospital FinishStart A B C D E F G H I J K Path Time (wks) A-I-K33 A-F-K28 A-C-G-J-K67 B-D-H-J-K69 B-E-J-K43 Paths are the sequence of activities between a project’s start and finish.

17. © 2007 Pearson Education St. Adolf’s Hospital FinishStart A B C D E F G H I J K Path Time (wks) A-I-K33 A-F-K28 A-C-G-J-K67 B-D-H-J-K69 B-E-J-K43 Project Expected Time is 69 wks. The critical path is the longest path!

18. © 2007 Pearson Education Application 3.1

19. © 2007 Pearson Education Application 3.1 Solution

20. © 2007 Pearson Education  The project team must make time estimates for each activity.  Activity times may be risky, in which case a probability distribution can be used (CPM).  For this project the times will be certain.  Activity slack is the maximum length of time that an activity can be delayed without delaying the entire project.  For St. Adolf’s we can’t go beyond 69 weeks. St. Adolf’s Hospital Developing the Schedule

21. © 2007 Pearson Education  Earliest Start Time (ES) is the latest earliest finish time of the immediately preceding activities.  Earliest Finish Time (EF) is an activity’s earliest start time plus its estimated duration.  Latest Start Time (LS) is the latest finish time minus the activity’s estimated duration.  Latest Finish Time (LF) is the earliest latest start time of the activities that immediately follow.  For simplicity, all projects start at time zero. St. Adolf’s Hospital Developing the Schedule

22. © 2007 Pearson Education What AON Nodes look like Latest Finish Latest Start Activity Activity Duration Slack The earliest you can complete an activity -- determined by adding the activity time to the earliest start time. The latest you can finish an activity without delaying the project completion date. It is the same as the Latest Start time of the next activity. If there are two or more subsequent activities, this time is the same as the earliest of those “Latest Start” times. Determined by the earliest finish time of the precedent activity. If there are two or more precedent activities, this time is the same as precedent activity with the latest “Earliest Finish” time. This is the Latest Finish time minus the activity time. Slack is the difference, if any, between the earliest start and latest start times (or the earliest finish and latest finish times). S = LS – ES or S = LF– EF © 2007 Pearson Education Earliest Start Earliest Finish

23. © 2007 Pearson Education Earliest Start and Earliest Finish Times K6K6 C 10 G 35 J4J4 H 40 B9B9 D 10 E 24 I 15 Finish Start A 12 F 10 © 2007 Pearson Education 0 Earliest start time 12 Earliest finish time 0 9 9 33 9 1919 59 22 57 12 22 59 63 12 27 12 22 63 69 Example 3.2

24. © 2007 Pearson Education Earliest Start and Earliest Finish Times © 2007 Pearson Education Critical Path The Critical Path takes 69 weeks K6K6 C 10 G 35 J4J4 H 40 B9B9 D 10 E 24 I 15 Finish Start A 12 F 10 0 9 9 33 9 1919 59 22 57 12 22 59 63 12 27 12 22 63 690 12 Example 3.2

25. © 2007 Pearson Education K6K6 C 10 G 35 J4J4 H 40 B9B9 D 10 E 24 I 15 Finish Start A 12 F 10 0 9 9 33 9 1919 59 22 57 12 22 59 63 12 27 12 22 63 690 12 Latest Start and Latest Finish Times © 2007 Pearson Education 48 63 53 63 59 63 24 59 19 59 35 59 14 24 9 19 2 14 0 9 Latest finish time 63 69 Latest start time Example 3.2

26. © 2007 Pearson Education Example 3.2

27. © 2007 Pearson Education Project Schedule  A Gantt Chart is a project schedule, usually created by the project manager using computer software, that superimposes project activities, with their precedence relationships and estimated duration times, on a time line.  Activity slack is useful because it highlights activities that need close attention.  Free slack is the amount of time an activity’s earliest finish time can be delayed without delaying the earliest start time of any activity that immediately follows.  Activities on the critical path have zero slack and cannot be delayed without delaying the project completion.

28. © 2007 Pearson Education

31. Activity Slack Analysis Node DurationESLS Slack A12022 B9000 C1012142 D10990 E2493526 F10125341 G3522242 H4019190 I15124836 J459590 K663630 Example 3.3

32. © 2007 Pearson Education Analyzing Cost-Time Trade-Offs  There are always cost-time trade-offs in project management.  You can completing a project early by hiring more workers or running extra shifts.  There are often penalties if projects extend beyond some specific date, and a bonus may be provided for early completion.  Crashing a project means expediting some activities to reduce overall project completion time and total project costs.

33. © 2007 Pearson Education Project Costs  The total project costs are the sum of direct costs, indirect costs, and penalty costs.  Direct costs include labor, materials, and any other costs directly related to project activities.  Indirect costs include administration, depreciation, financial, and other variable overhead costs that can be avoided by reducing total project time.  The shorter the duration of the project, the lower the indirect costs will be.

34. © 2007 Pearson Education Cost to Crash  To assess the benefit of crashing certain activities, either from a cost or a schedule perspective, the project manager needs to know the following times and costs.  Normal time (NT) is the time necessary to complete and activity under normal conditions.  Normal cost (NC) is the activity cost associated with the normal time.  Crash time (CT) is the shortest possible time to complete an activity.  Crash cost (CC) is the activity cost associated with the crash time.

35. © 2007 Pearson Education Cost to Crash per Period The Cost to Crash per Period = CC − NC NT − CT Crash Cost − Normal Cost Normal Time − Crash Time

36. © 2007 Pearson Education Linear cost assumption 8000 — 7000 — 6000 — 5000 — 4000 — 3000 — 0 — Direct cost (dollars) |||||| 567891011 Time (weeks) Crash cost (CC) Normal cost (NC) (Crash time)(Normal time) Estimated costs for a 2-week reduction, from 10 weeks to 8 weeks 5200 St. Adolf’s Hospital St. Adolf’s Hospital Cost-Time Relationships in Cost Analysis

37. © 2007 Pearson Education  The objective of cost analysis is to determine the project schedule that minimizes total project costs.  A minimum-cost schedule is determined by starting with the normal time schedule and crashing activities along the critical path in such a way that the costs of crashing do not exceed the savings in indirect and penalty costs. St. Adolf’s Hospital Minimizing Costs

38. © 2007 Pearson Education  Use these steps to determine the minimum cost schedule: 1.Determine the project’s critical path(s). 2.Find the activity or activities on the critical path(s) with the lowest cost of crashing per week. 3.Reduce the time for this activity until… a.It cannot be further reduced or b.Until another path becomes critical, or c.The increase in direct costs exceeds the savings that result from shortening the project (which lowers indirect costs). 4.Repeat this procedure until the increase in direct costs is larger than the savings generated by shortening the project. St. Adolf’s Hospital Minimum Cost Schedule

39. © 2007 Pearson Education Direct Cost and Time Data for the St. Adolf’s Hospital Project A12$ 12,00011$ 13,0001$ 1,000 B950,000764,00027,000 C104,00057,0005600 D1016,000820,00022,000 E24120,00014200,000108,000 F1010,000616,00041,500 G35500,00025530,000103,000 H401,200,000351,260,000512,000 I1540,0001052,50052,500 J410,000113,00031,000 K630,000534,00014,000 Totals$1,992,000$2,209,500 MaximumCost of NormalNormalCrashCrashTimeCrashing per TimeCostTimeCostReductionWeek Activity(NT)(NC)(CT)(CC)(wk)(CC-NC) Shorten first

40. © 2007 Pearson Education St. Adolf’s Hospital Finding the minimum cost schedule: Stage 1 Step 1: The critical path is: B-D-H-J-K. Step 2: The cheapest activity to crash is “J” at $1000. Step 3: Crash activity J by its limit of three weeks because the critical path remains unchanged. The new project length becomes 66 weeks. The project completion time is 69 weeks. The direct costs for that schedule are $1,992,000. The indirect costs are $8000 per week. Penalty costs after week 65 are $20,000 per week. Total cost is $2,624,000 for 69 weeks ($1,992,000 + 69($8000) + (69 –65)($20,000) Example 3.4

41. © 2007 Pearson Education St. Adolf’s Hospital Finding the minimum cost schedule: Stage 1 The project completion time is 69 weeks. The direct costs for that schedule are $1,992,000. The indirect costs are $8000 per week. Penalty costs after week 65 are $20,000 per week. Total cost is $2,624,000 for 69 weeks Crashing by 3 weeks saves $81,000 for a new total cost of $2,543,000. Savings is 3 weeks of indirect costs (3 * $8000 = $24,000) plus 3 weeks of penalties (3 * $20,000 = $60,000) less the cost of crashing (3 * $1,000 = $3,000) Example 3.4

42. © 2007 Pearson Education Step 1: The critical path is still B-D-H-J-K. Step 2: The cheapest activity to crash per week is now D at $2,000 a week. Step 3: Crash D by 2 weeks. The first week of reduction saves $28,000 by eliminating both the penalty and indirect costs (but $2,000 goes toward crashing costs.) The second week of reduction had no penalty, so it saves only the indirect costs of $8,000. Total cost is now $2,511,00 ($2,543,00 - $28,000 - $8,000 + $4,000) St. Adolf’s Hospital Finding the minimum cost schedule: Stage 2 The indirect costs are $8000 per week. Penalty costs after week 65 are $20,000 per week. Example 3.4

43. © 2007 Pearson Education St. Adolf’s Hospital Finding the minimum cost schedule: Stage 3 Start A 12 B9B9 D8D8 E 24 F 10 Finish C 10 G 35 H 40 I 15 J11J111 K6K6 Shortening D and J have created a second critical path, A-C-G-J-K. Both critical paths are 64 weeks. Both must now be shortened to realize any savings in indirect costs.  If not, the length of the project remians unchanged! Example 3.4

44. © 2007 Pearson Education St. Adolf’s Hospital Finding the minimum cost schedule: Stage 3 The alternatives are to crash one of the following combination of activities: A-B, A-H, C-B, C-H, G-B, G-H, or Crash activity K which is on both critical paths. (J and D have already been crashed.) The cheapest alternative is to crash activity K. It can only be crashed by one week at a cost of $4,000 The net savings are $8,000 − $4,000 = $4,000 Total project cost now becomes $2,507,000 A$ 1,000 B7,000 C600 D2,000 E8,000 F1,500 G3,000 H12,000 I2,500 J1,000 K4,000 The indirect costs are $8000 per week. Example 3.4

45. © 2007 Pearson Education St. Adolf’s Hospital Finding the minimum cost schedule: Stage 4 A$ 1,000 B7,000 C600 D2,000 E8,000 F1,500 G3,000 H12,000 I2,500 J1,000 K4,000 Start A 12 B9B9 D8D8 E 24 F 10 Finish C 10 G 35 H 40 I 15 J11J111 K5K5 63 wks The critical paths remain the same but are now both 63 weeks. Example 3.4

46. © 2007 Pearson Education B and C are the only remaining activities that can be crashed simultaneously without exceeding the potential savings of $8000 per week in indirect costs. Crash activities B and C by two weeks (the limit for activity B) Net savings are 2($8,000) − 2($7,600) = $800 Total project costs are now $2,506,200 i.e. Now, further crash costs exceed weekly indirectly costs, any other combination of activities will result in a net increase in total project costs  Stop crashing A$ 1,000 B7,000 C600 D2,000 E8,000 F1,500 G3,000 H12,000 I2,500 J1,000 K4,000 St. Adolf’s Hospital Finding the minimum cost schedule: Stage 4 The indirect costs are $8000 per week. Example 3.4

47. © 2007 Pearson Education St. Adolf’s Hospital Summary The minimum cost schedule is 61 weeks. Activities J, D, K, B, and C were crashed for a total savings of $117,800 (2624.0- 2506.2) Example 3.4

48. © 2007 Pearson Education Statistical Analysis  The Statistical Analysis approach requires that activity times be stated in terms of three reasonable time estimates for each activity. 1.Optimistic Time (a) is the shortest time in which a activity can be completed if all goes exceptionally well. 2.Most Likely Time (m) is the probable time for an activity. 3.Pessimistic Time (b) is the longest time required.  The expected time for an activity thus becomes… te =te =te =te = a + 4m + b 6

49. © 2007 Pearson Education Probabilistic Time Estimates Mean mab Time Probability Beta Distribution Pessimistic Optimistic

50. © 2007 Pearson Education Time Probability Normal Distribution Mean abm 3333 3333 Area under curve between a and b is 99.74% Probabilistic Time Estimates

51. © 2007 Pearson Education AF I CG Finish D E HBJ K Start t e = a + 4m + b 6 Mean 2 =2 =2 =2 = ( ) b – ab – a66b – ab – a666 2Variance St. Adolf’s Hospital Probabilistic Time Estimates Example 3.5 Calculating Means and Variances

52. © 2007 Pearson Education Activity B Most OptimisticLikelyPessimistic (a)(m)(b) 7815 AF I CG Finish D E HBJ K Start St. Adolf’s Hospital Probabilistic Time Estimates t e = = 9 weeks 7 + 4(8) + 15 6  2 = = 1.78 ( ) 15 - 7 6 2 Example 3.5 Calculating Means and Variances

53. © 2007 Pearson Education OptimisticLikelyPessimisticExpectedVariance Activity(a)(m)(b)Time (t e )(  2 ) Time Estimates (wk)Activity Statistics A111213120.11 B781591.78 C51015102.78 D8916101.78 E142530247.11 F6918104.00 G253641357.11 H354045402.78 I101328159.00 J121545.44 K56760.11 St. Adolf’s Hospital Probabilistic Time Estimates Example 3.5

54. © 2007 Pearson Education Application 3.4

55. © 2007 Pearson Education  2 =  (variances of activities) z = T – T E  2  2 = 1.78 + 1.78 + 2.78 + 5.44 + 0.11 = 11.89 z = 72 – 69 11.89 Probabilities Critical Path = B - D - H - J - K T = 72 weeks T E = 69 weeks St. Adolf’s Hospital Analyzing Probabilities From Normal Distribution appendix P z =.8078 .81 Example 3.6

56. © 2007 Pearson Education Project duration (weeks) 6972 Normal distribution: Mean = 69 weeks;  = 3.45 weeks Probability of exceeding 72 weeks is 0.1922 St. Adolf’s Hospital Probability of Completing Project On Time Example 3.6 Probability of meeting the schedule is 0.8078 Length of critical path

57. © 2007 Pearson Education  2 =  (variances of activities) z = T – T E  2  2 = 0.11 + 2.78 + 7.11 + 5.44 + 0.11 = 15.55 z = = 1.27 72 – 67 15.55 Probabilities Critical Path = A - C - G - J - K T = 72 weeks T E = 67 weeks From Normal Distribution appendix P z =.8980 .90 St. Adolf’s Hospital Probability of Completing Project On Time Example 3.6

58. © 2007 Pearson Education Application 3.5

59. © 2007 Pearson Education Application 3.5

60. © 2007 Pearson Education Project Life Cycle StartFinish Resource requirements Time Definition and organization Planning Execution Close out

61. © 2007 Pearson Education Self-Exercise

62. © 2007 Pearson Education Solved Problem 1

63. © 2007 Pearson Education Solved Problem 1

64. © 2007 Pearson Education Solved Problem 2 What is the probability of completing the project in 23 weeks?

65. © 2007 Pearson Education Solved Problem 2

66. © 2007 Pearson Education Solved Problem 2 Finish Start A 4.0 0.0 4.0 8.0 D 12.0 4.0 8.0 16.0 20.0 E 6.5 9.0 15.5 G 4.5 15.5 20.0 C 3.5 5.5 9.0 F 9.0 5.5 6.5 14.5 15.5 B 5.5 0.0 5.5

67. © 2007 Pearson Education Solved Problem 2 Using the Normal Distribution appendix, we find that the probability of completing the project in 23 weeks or less is 0.9357.