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Chapter 7: Project Management

Department of Business Administration. Chapter 7: Project Management. FALL 20 10 - 2011. Outline: What You Will Learn. Discuss the behavioral aspects of projects in terms of project personnel and the project manager.

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Chapter 7: Project Management

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  1. Department of Business Administration Chapter 7:Project Management FALL 2010-2011

  2. Outline: What You Will Learn . . . • Discuss the behavioral aspects of projects in terms of project personnel and the project manager. • Discuss the nature and importance of a work breakdown structure in project management. • Give a general description of PERT/CPM techniques. • Construct simple network diagrams. • List the kinds of information that a PERT or CPM analysis can provide. • Analyze networks with deterministic times. • Analyze networks with probabilistic times. • Describe activity “crashing” and solve typical problems.

  3. Build A A Done Build B B Done Build C C Done Build D Ship JAN FEB MAR APR MAY JUN On time! Projects Project:Unique, one-time operations designed to accomplish a specific set of objectives in a limited time frame.

  4. Project Management • How is it different? • Limited time frame • Narrow focus, specific objectives • Less bureaucratic • Why is it used? • Special needs • Pressures for new or improves products or services • What are the Key Metrics • Time • Cost • Performance objectives • What are the Key Success Factors? • Top-down commitment • Having a capable project manager • Having time to plan • Careful tracking and control • Good communication

  5. Project Management • What are the Major Administrative Issues? • Executive responsibilities • Project selection • Project manager selection • Organizational structure • Organizational alternatives • Manage within functional unit • Assign a coordinator • Use a matrix organization with a project leader • What are the tools? • Work breakdown structure • Network diagram • Gantt charts • Risk management

  6. Key Decisions • Deciding which projects to implement • Criteria-attractive-cost and benefit-available fund • Selecting a project manager • Central person • Selecting a project team • Person’s knowledge and skills-relationship with others • Planning and designing the project • Goals-timetable-budget-resources • Managing and controlling project resources • Personnel-equipment-budget • Deciding if and when a project should be terminated • Likelihood of success-costs-resources

  7. Project Manager Responsible for: Work Quality Human Resources Time Communications Costs

  8. Ethical Issues • Temptation to understate costs • Withhold information • Misleading status reports • Falsifying records • Comprising workers’ safety • Approving substandard work

  9. Feasibility Planning Management Concept Execution Termination Project Life Cycle • Concept: A proposal neededFeasibility: Cost, benefit and risk analysesPlanning: find out the necessary human resources, time and costExecution: control for time, available resource and costTermination: It should be reevaluated for the sake of project’s safety

  10. Project X Level 1 Level 2 Level 3 Level 4 Work Breakdown Structure

  11. Gantt Chart Locate new facilities Interview staff Hire and train staff Select and order furniture Remodel and install phones Move in/startup MAR APR MAY JUN JUL AUG SEP OCT NOV DEC Planning and Scheduling

  12. PERT and CPM • PERT: Program Evaluation and Review Technique • CPM: Critical Path Method • Both techniques are widely used for planning and coordinating large-scale projects. • Using the two techniques, manager are able to obtain: • Graphically displays project activities • Estimates how long the project will take • Indicates most critical activities • Show where delays will not affect project

  13. The Network Diagram • Network (precedence) diagram – diagram of project activities that shows sequential relationships by the use of arrows and nodes. • Activity-on-arrow (AOA) – a network diagram convention in which arrows designate activities. • Activity-on-node (AON) – a network diagram convention in which nodes designate activities. • Activities – steps in the project that consume resources and/or time. • Events – the starting and finishing of activities, designated by nodes in the AOA convention.

  14. The Network Diagram • Path • Sequence of activities that leads from the starting node to the finishing node • Critical path • The longest path; determines expected project duration • Critical activities • Activities on the critical path • Slack • Allowable slippage for path; the difference the length of path and the length of critical path

  15. Orderfurniture 4 Furnituresetup 2 Locatefacilities Remodel 1 5 6 Move in Interview Hire andtrain 3 Project Network – Activity on Arrow AOA

  16. Orderfurniture Furnituresetup Locatefacilities 2 6 1 Move in Remodel 5 S 7 Hire andtrain Interview 4 3 Project Network – Activity on Node AON

  17. Network Conventions a b c a c b a c a c Dummy activity b b d

  18. Time Estimates • The main determinant of the way PERT and CPM networks are analysed and interpreted is whether activity time estimates are probabilistic or deterministic. • Deterministic • Time estimates that are fairly certain • Probabilistic • Estimates of times that allow for variation

  19. Example-Bank Network convention • The following table contains information related to the major activities of a research project. Use the information to do the following: • Draw a precedence diagram using AOA and AON • Find the critical path based AOA. • Determine the expected length of the project.

  20. c d a f g h Answer-Bank Network convention • Activities with no predecessors are at the beginning (life side) of the network. • Activities with multiple predecessors are located at path intersections. • Use first AOA 8 2 b 7 i S 10 5 e 6 m 3 8 1 k End 17 2

  21. Answer-Bank Network convention • Activities with no predecessors are at the beginning (life side) of the network. • Activities with multiple predecessors are located at path intersections. • Use Second AON c d a b i b f m S e f End k h g

  22. Example-Bank Network convention • (b)Find the critical path based AOA. • a-c-d-i-m*=5+8+2+10+8=33# • a-b-i-m=5+7+10+8=30 • e-f-m= 3+6+8=17 • g-h-k=1+2+17=20 • a-c-d-i-m*-Critical path • (c) Determine the expected length of the project. • 33 # -Expected project duration

  23. 6 weeks 4 Orderfurniture 3 weeks 2 Furnituresetup 8 weeks Locatefacilities Remodel Move in 11 weeks 1 5 6 1 week Interview Hire and train 4 weeks 9 weeks 3 Example-Bank Network Figure Bank Network question

  24. Example-Bank Network Figure • Given the information on the bank network: • Determine • The length of each path • The critical path • The expected length of the project • The amount of slack time for each path • Knowledge of slack times provides managers with information for planning allocation of scarce resources and for directing control efforts toward those activities that may be most susceptible to delaying the project.

  25. Critical Path Answer-Bank Network Figure

  26. Computing Algorithm • Network activities • ES: early start • EF: early finish-EF=ES+t • LS: late start-LS=LF-t • LF: late finish • Used to determine • Expected project duration • Slack time-LS-ES or LF-EF • Critical path ES t EF LS LF ES t EF

  27. 4 2 1 5 6 3 Example-ES-EF-LS-LF-slack Required: Compute slack time, ES, EF, LS and LF Forward pass 8 6 14 ES t EF 14 3 17 8 11 19 0 8 8 19 1 20 0 4 4 4 9 13 LS LF ES t EF • EF: early finish-EF=ES+t • LS: late start-LS=LF-t • Slack time-LS-ES or LF-EF Backward pass

  28. 4 2 1 5 6 3 Answer-ES-EF-LS-LF-slack Forward pass 10 6 16 8 14 2 ES t EF 8 6 14 16 3 19 14 17 2 0 8 8 0 8 0 14 3 17 8 11 19 0 8 8 • 11 19 • 8 19 • 0 19 1 20 19 1 20 19 20 0 0 4 4 4 9 13 6 4 10 0 4 6 10 9 19 4 13 6 ES t EF LS LF EF: early finish-EF=ES+t LS: late start-LS=LF-t Slack time-LS-ES or LF-EF Backward pass

  29. Probabilistic Time Estimates • Optimistic time • Time required under optimal conditions • Pessimistic time • Time required under worst conditions • Most likely time • Most probable length of time that will be required

  30. to tm te tp Activity start Optimistictime Most likely time (mode) Pessimistic time Probabilistic Estimates Beta Distribution is generally used to describe the inherent variability in time Estimates. Although there is no real theoretical justification for using the Beta Distribution, it has certain features that make it attractive in practice.

  31. te = to + 4tm +tp6 Expected Time te = expected time to = optimistic time tm = most likely time tp = pessimistic time The knowledge of the expected path times and their std. Deviation enables a manager to compute probabilistic estimates of the project completion time as such specific time and scheduled time

  32. (tp – to)2 36 2 = Variance 2= variance to = optimistic time tp = pessimistic time • The size of Variance reflects the degree of uncertainty associated with an activity’s time: The large the variance, the greater the uncertainty.

  33. Optimistic time Most likely time Pessimistic time 2-4-6 b 2-3-5 c 1-3-4 a 3-4-5 d 3-5-7 e 5-7-9 f 2-3-6 g 3-4-6 i 4-6-8 h Example-Probabilistic Time Estimates • Given the following diagram: • Compute • The expected time • The expected duration • Identify the critical path • The variance • The std. deviation

  34. Answer-Probabilistic Time Estimates

  35. 4.00 b 3.17 c 2.83 a 4.00 d 5.0 e 7.0 f 3.33 g 4.17 i 6.0 h Answer-Probabilistic Time Estimates Tabc = 10.0Tdef = 16.0Tghi = 13.50

  36. Specified time – Path mean Path standard deviation Z = Path Probabilities • Z indicates how many standard deviationsof the path distribution the specified timeis beyond the expected path duration. The more positive the value, the better. A negative value of z indicates that the specified time is earlier than the expected path duration. • Z=+3.00-probability 100%- • From the relevant table +3.00 is almost equal to 0.9987.

  37. Example-The Path probability • Given the information on the example of probabilistic time estimates (the previous example): • Determine • The probability that the project can be completed within 17 weeks of its start. • The probability that the project will be completed within 15 weeks of its start. • The probability that the project will not be completed within 15 weeks of its start.

  38. Answer-The Path probability • Determine • The probability that the project can be completed within 17 weeks of its start. Path: a-b-c 17 – 10 0.97 Z = =7.22 Prob.comp in 17 week=1.00 Appendix B, Table B, p.p 884/5 • Determine • The probability that the project will be completed within 17 weeks of its start. Path: d-e-f 17 – 16 1 =1 Prob.comp in 17 week=0.8413 Appendix B, Table B, p.p 885 Z =

  39. Answer-The Path probability • Determine • The probability that the project will be completed within 17 weeks of its start. Path: g-h-i 17 – 13.5 1.07 Prob.comp in 17 week=1.00 Appendix B, Table B, p.p 884/5 Z = =3.27 Prob finish in 17 week=1.00 X 0.8413 X 1.00= 0.8413

  40. Answer-The Path probability • Determine • The probability that the project can be completed within 15 weeks of its start. Path: a-b-c 15 – 10 0.97 Z = =5.15 Prob.comp in 15 week=1.00 Appendix B, Table B, p.p 884/5 • Determine • The probability that the project will be completed within 15 weeks of its start. Path: d-e-f 15 – 16 1 =-1.00 Prob.comp in 15 week=0.1587 Appendix B, Table B, p.p 885 Z =

  41. Answer-The Path probability • Determine • The probability that the project will be completed within 15 weeks of its start. Path: g-h-i 15 – 13.5 1.07 Prob.comp in 15 week=0.9192 Appendix B, Table B, p.p 884/5 Z = =1.40 Prob finish in 15 week=1.00 X 0.1587 X 0.9192= 0.1459 • The probability that the project will not be completed within 15 weeks of its start: 1- 0.1459=0.8541

  42. Answer-The Path probability-Graphically 17 Weeks 1.00 a-b-c Weeks 10.0 0.8413 d-e-f Weeks 16.0 1.00 g-h-i Weeks 13.5

  43. Answer-The Path probability-Graphically 15 Weeks 1.00 a-b-c Weeks 10.0 0.1587 d-e-f Weeks 16.0 0.9192 g-h-i Weeks 13.5

  44. Time-cost Trade-offs: Crashing • In many projects, it is possible to reduce the length of a project by injecting additional resources. The impetus to shorten projects may reflect efforts to avoid late penalties, or/ to take advantage of monetary incentives for timely completion of a project, or/ to free resources for use on other projects. This is called crashing. • Crash – briefly, shortening activity duration • Procedure for crashing • Crash the project one period at a time • Only an activity on the critical path • Crash the least expensive activity • Multiple critical paths: find the sum of crashing the least expensive activity on each critical path

  45. Total cost Expected indirect costs Shorten CRASH Cumulative (direct) cost of crashing Shorten Optimum Time-Cost Trade-Offs: Crashing

  46. Example-Crashing • Using the following information, develop the optimal time cost solution. Indirect costs are $ 1000 per day. • Determine which activities are on the critical path, its length, and the length of the other path • Rank the critical activities in order of lowest crashing cost, and datermine the number of days each can be crashed. • Determine the critical path after each reduction by shortening the project.

  47. 10 b 6 a 2 f 5 c 9 e 4 d • (a) Determine the critical path.

  48. Answer-Crashing • Determine which activities are on the critical path, its length, and the length of the other path • Path length • a-b-f 18 • c-d-e-f 20 (critical path) • (b) Rank the critical activities in order of lowest crashing cost, and datermine the number of days each can be crashed. • Activity Cost per day to crash Available days • c $ 300 1 • e 600 2 • d 700 3 • f 800 1

  49. Answer-Crashing • (c) Determine the critical path after each reduction by shortening the project. • (1) Shorten activity c one day at a cost of $ 300. The length of the critical path becomes 19 days. • (2) Activity c cannot be shorten any more. Shorten activity e one day at cost of $ 600. The length of the critical path c-d-e-f becomes 18 days which is the same as length of path a-b-f. • (3) The path are now both critical, further improvement will necesitate shortening both paths. • Path Activity Cost per day to crash • a-b-f a no reduction possible b $ 500 • f 800 • c-d-e-f c no further reduction possible d $ 700 • e 600 • f 800

  50. Answer-Crashing • At the first glance, it would seem that crashing f would not be advantageous, because it has the highest crashing cost. However, f is on both paths, so shortening f by one day would shorten both paths by one day for a cost of $ 800. The option of shortening the least expensive activity on each path would cost $ 500 for b and $ 600 for e or $ 1100. Thus shorten f by one day. The project duration is now 17 days. • (4) At this point, no additional improvement is feasible. The cost to crash b is $ 500 and the cost to crash e is $ 600, for a total of $ 1100 and that would exceed the indirect costs of $ 100 per day. • (5) The crashing sequence is summarized below: • Length after crashing n days • Path n=0 1 2 3 • a-b-f 18 18 18 17 • c-d-e-f 20 19 18 17 • activity crashed c e f • cost $300 600 800

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