The document discusses key aspects of project management including defining projects, organizing project teams, planning projects, developing schedules, and analyzing costs and time tradeoffs. Some key points are: - Projects have a definite start and end, involve resources, and aim to be completed on time, within budget and to specifications. - Project management involves systematically defining, organizing, planning, monitoring and controlling projects. - Planning projects involves work breakdown structures, network diagrams, schedules, and risk assessment. The critical path is the longest sequence of activities. - Tradeoffs between time and costs must be analyzed, as crashing activities may reduce time but increase costs.

1. Project Management Mehmet TANLAK For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education PowerPoint Slides by Jeff Heyl

11. Diagramming the Network Figure 2.2 S T U S precedes T, which precedes U. AON Activity Relationships S T U S and T must be completed before U can be started .

12. Diagramming the Network Figure 2.2 AON Activity Relationships T U S T and U cannot begin until S has been completed. S T U V U and V can’t begin until both S and T have been completed.

13. Diagramming the Network Figure 2.2 AON Activity Relationships S T U V U cannot begin until both S and T have been completed; V cannot begin until T has been completed. S T V U T and U cannot begin until S has been completed and V cannot begin until both T and U have been completed.

16. St. John’s Hospital Project Example 2.1 START START A B B A C D A E, G, H F, I, J K 0 12 9 10 10 24 10 35 40 15 4 6 0 Kramer Stewart Johnson Taylor Adams Taylor Burton Johnson Walker Sampson Casey Murphy Pike Ashton Activity Immediate Predecessors Activity Times (wks) Responsibility ST. JOHN’S HOSPITAL PROJECT START ORGANIZING and SITE PREPARATION A. Select administrative staff B. Select site and survey C. Select medical equipment D. Prepare final construction plans E. Bring utilities to site F. Interview applicants for nursing and support staff PHYSICAL FACILITIES and INFRASTRUCTURE G. Purchase and deliver equipment H. Construct hospital I. Develop information system J. Install medical equipment K. Train nurses and support staff FINISH

17. St. John’s Hospital Project Example 2.1 Completion Time Figure 2.3 Activity Immediate Predecessors Activity Times (wks) Responsibility ST. JOHN’S HOSPITAL PROJECT START ORGANIZING and SITE PREPARATION A. Select administrative staff B. Select site and survey C. Select medical equipment D. Prepare final construction plans E. Bring utilities to site F. Interview applicants for nursing and support staff PHYSICAL FACILITIES and INFRASTRUCTURE G. Purchase and deliver equipment H. Construct hospital I. Develop information system J. Install medical equipment K. Train nurses and support staff FINISH Finish K 6 I 15 F 10 C 10 H 40 J 4 A 12 B 9 Start G 35 D 10 E 24 Activity IP Time A START 12 B START 9 C A 10 D B 10 E B 24 F A 10 G C 35 H D 40 I A 15 J E, G, H 4 K F, I, J 6

18. St. John’s Hospital Project Example 2.1 Completion Time Figure 2.3 Activity Immediate Predecessors Activity Times (wks) Responsibility ST. JOHN’S HOSPITAL PROJECT START ORGANIZING and SITE PREPARATION A. Select administrative staff B. Select site and survey C. Select medical equipment D. Prepare final construction plans E. Bring utilities to site F. Interview applicants for nursing and support staff PHYSICAL FACILITIES and INFRASTRUCTURE G. Purchase and deliver equipment H. Construct hospital I. Develop information system J. Install medical equipment K. Train nurses and support staff FINISH Finish K 6 I 15 F 10 C 10 D 10 H 40 J 4 A 12 B 9 Start G 35 E 24 Path Estimated Time (weeks) A–I–K 33 A–F–K 28 A–C–G–J–K 67 B–D–H–J–K 69 B–E–J–K 43

19. St. John’s Hospital Project Example 2.1 Completion Time Figure 2.3 Activity Immediate Predecessors Activity Times (wks) Responsibility ST. JOHN’S HOSPITAL PROJECT START ORGANIZING and SITE PREPARATION A. Select administrative staff B. Select site and survey C. Select medical equipment D. Prepare final construction plans E. Bring utilities to site F. Interview applicants for nursing and support staff PHYSICAL FACILITIES and INFRASTRUCTURE G. Purchase and deliver equipment H. Construct hospital I. Develop information system J. Install medical equipment K. Train nurses and support staff FINISH Finish K 6 I 15 F 10 C 10 D 10 H 40 J 4 A 12 B 9 Start G 35 E 24 Path Estimated Time (weeks) A–I–K 33 A–F–K 28 A–C–G–J–K 67 B–D–H–J–K 69 B–E–J–K 43

20. Application 2.1 The following information is known about a project Draw the network diagram for this project Activity Activity Time (days) Immediate Predecessor(s) A 7 — B 2 A C 4 A D 4 B, C E 4 D F 3 E G 5 E

21. Application 2.1 Finish G 5 F 3 E 4 D 4 Activity Activity Time (days) Immediate Predecessor(s) A 7 — B 2 A C 4 A D 4 B, C E 4 D F 3 E G 5 E B 2 C 4 Start A 7

27. Network Diagram Figure 2.4 K 6 C 10 G 35 J 4 H 40 B 9 D 10 E 24 I 15 Finish Start A 12 F 10 0 9 9 33 9 19 19 59 22 57 12 22 59 63 12 27 12 22 63 69 0 12

31. Network Diagram Figure 2.4 K 6 C 10 G 35 J 4 H 40 B 9 D 10 E 24 I 15 Finish Start A 12 F 10 0 9 9 33 9 19 19 59 22 57 12 22 59 63 12 27 12 22 63 69 0 12 48 63 53 63 59 63 24 59 19 59 35 59 14 24 9 19 2 14 0 9 63 69 S = 36 S = 2 S = 41 S = 2 S = 26 S = 2 S = 0 S = 0 S = 0 S = 0 S = 0

32. Gantt Chart Figure 2.5

34. Application 2.2 Calculate the four times for each activity in order to determine the critical path and project duration. The critical path is A–C–D–E–G with a project duration of 24 days Activity Duration Earliest Start (ES) Latest Start (LS) Earliest Finish (EF) Latest Finish (LF) Slack (LS-ES) On the Critical Path? A 7 0 0 7 7 0-0=0 Yes B 2 C 4 D 4 E 4 F 3 G 5

35. Application 2.2 Calculate the four times for each activity in order to determine the critical path and project duration. The critical path is A–C–D–E–G with a project duration of 24 days Activity Duration Earliest Start (ES) Latest Start (LS) Earliest Finish (EF) Latest Finish (LF) Slack (LS-ES) On the Critical Path? A 7 0 0 7 7 0-0=0 Yes B 2 C 4 D 4 E 4 F 3 G 5 7 9 9 11 9-7=2 No 7 7 11 11 7-7=0 Yes 19 21 22 24 21-19=2 No 19 19 24 24 19-19=0 Yes 11 11 15 15 11-11=0 Yes 15 15 19 19 15-15=0 Yes

36. Application 2.2 The critical path is A–C–D–E–G with a project duration of 24 days Activity Duration Earliest Start (ES) Latest Start (LS) Earliest Finish (EF) Latest Finish (LF) Slack (LS-ES) On the Critical Path? A 7 0 0 7 7 0-0=0 Yes B 2 7 9 9 11 9-7=2 No C 4 7 7 11 11 7-7=0 Yes D 4 11 11 15 15 11-11=0 Yes E 4 15 15 19 19 15-15=0 Yes F 3 21 21 22 24 21-19=2 No G 5 19 19 24 24 19-19=0 Yes Start Finish A 7 B 2 C 4 D 4 E 4 F 3 G 5

37. Application 2.2 The critical path is A–C–D–E–G with a project duration of 24 days Activity Duration Earliest Start (ES) Latest Start (LS) Earliest Finish (EF) Latest Finish (LF) Slack (LS-ES) On the Critical Path? A 7 0 0 7 7 0-0=0 Yes B 2 7 9 9 11 9-7=2 No C 4 7 7 11 11 7-7=0 Yes D 4 11 11 15 15 11-11=0 Yes E 4 15 15 19 19 15-15=0 Yes F 3 21 21 22 24 21-19=2 No G 5 19 19 24 24 19-19=0 Yes Start Finish A 7 B 2 C 4 D 4 E 4 F 3 G 5

39. Cost-Time Relationships Figure 2.6 Linear cost assumption 8000 — 7000 — 6000 — 5000 — 4000 — 3000 — 0 — Direct cost (dollars) | | | | | | 5 6 7 8 9 10 11 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

40. Cost-Time Relationships TABLE 2.1 | DIRECT COST AND TIME DATA FOR THE ST. JOHN’S HOSPITAL PROJECT Activity Normal Time (NT) (weeks) Normal Cost (NC)($) Crash Time (CT)(weeks) Crash Cost (CC)($) Maximum Time Reduction (week) Cost of Crashing per Week ($) A 12 $12,000 11 $13,000 1 1,000 B 9 50,000 7 64,000 2 7,000 C 10 4,000 5 7,000 5 600 D 10 16,000 8 20,000 2 2,000 E 24 120,000 14 200,000 10 8,000 F 10 10,000 6 16,000 4 1,500 G 35 500,000 25 530,000 10 3,000 H 40 1,200,000 35 1,260,000 5 12,000 I 15 40,000 10 52,500 5 2,500 J 4 10,000 1 13,000 3 1,000 K 6 30,000 5 34,000 1 4,000 Totals $1,992,000 $2,209,500

51. A Minimum-Cost Schedule Stage Crash Activity Time Reduction (weeks) Resulting Critical Path(s) Project Duration (weeks) Project Direct Costs, Last Trial ($000) Crash Cost Added ($000) Total Indirect Costs ($000) Total Penalty Costs ($000) Total Project Costs ($000) 0 — — B - D - H - J - K 69 1,992.0 — 552.0 80.0 2,624.0 1 J 3 B - D - H - J - K 66 1,992.0 3.0 528.0 20.0 2,543.0 2 D 2 B - D - H - J - K A - C - G - J - K 64 1,995.0 4.0 512.0 0.0 2,511.0 3 K 1 B - D - H - J - K A - C - G - J - K 63 1,999.0 4.0 504.0 0.0 2,507.0 4 B, C 2 B - D - H - J - K A - C - G - J - K 61 2,003.0 15.2 488.0 0.0 2,506.2

52. Application 2.3 Indirect project costs = $250 per day and penalty cost = $100 per day for each day the project lasts beyond day 14. Project Activity and Cost Data Activity Normal Time (days) Normal Cost ($) Crash Time (days) Crash Cost ($) Immediate Predecessor(s) A 5 1,000 4 1,200 — B 5 800 3 2,000 — C 2 600 1 900 A, B D 3 1,500 2 2,000 B E 5 900 3 1,200 C, D F 2 1,300 1 1,400 E G 3 900 3 900 E H 5 500 3 900 G

53. Application 2.3 Direct cost and time data for the activities: Solution: Original costs: Normal Total Costs = Total Indirect Costs = Penalty Cost = Total Project Costs = $7,500 $250 per day  21 days = $5,250 $100 per day  7 days = $700 $13,450 Project Activity and Cost Data Activity Crash Cost/Day Maximum Crash Time (days) A 200 1 B 600 2 C 300 1 D 500 1 E 150 2 F 100 1 G 0 0 H 200 2

54. Application 2.3 Step 1: The critical path is , and the project duration is B–D–E–G–H 21 days. Step 2: Activity E on the critical path has the lowest cost of crashing ($150 per day). Note that activity G cannot be crashed. Step 3: Reduce the time (crashing 2 days will reduce the project duration to 19 days) and re-calculate costs: Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost = $7,500 $150  2 days = $300 $250 per day  19 days = $4,750 $100 per day  5 days = $500 $13,050 Note that the cost to crash ($250 per day) is less than the combined indirect cost and the penalty cost per day savings ($350).

55. Application 2.3 Step 4: Repeat until direct costs greater than savings (step 2) Activity H on the critical path has the next lowest cost of crashing ($200 per day). (step 3) Reduce the time (crashing 2 days will reduce the project duration to 17 days) and re-calculate costs: Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost = $7,500 + $300 (the added crash costs) = $7,800 $200  2 days = $400 $250 per day  17 days = $4,250 $100 per day  3 days = $300 $12,750 Note that the cost to crash ($200 per day) is less than the combined indirect cost and the penalty cost per day savings ($350).

56. Application 2.3 (step 4) Repeat (step 2) Activity D on the critical path has the next lowest crashing cost ($500 per day). (step 3) Reduce the time (crashing 1 day will reduce the project duration to 16 days) and re-calculate costs: Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost = $7,800 + $400 (the added crash costs) = $8,200 $500  1 day = $500 $250 per day  16 days = $4,000 $100 per day  2 days = $200 $12,900 which is greater than the last trial. Hence we stop the crashing process. Note that the cost to crash ($500 per day) is greater than the combined indirect cost and the penalty cost per day savings ($350).

57. Application 2.3 The summary of the cost analysis follows. The recommended completion date is day 17 by crashing activity E by 2 days and activity H by 2 days. Further reductions will cost more than the savings in indirect costs and penalties. The critical path is B – D – E – G – H. Trial Crash Activity Resulting Critical Paths Reduction (days) Project Duration (days) Costs Last Trial Crash Cost Added Total Indirect Costs Total Penalty Costs Total Project Costs 0 — B-D-E-G-H — 21 $7,500 — $5,250 $700 $13,450 1 E B-D-E-G-H 2 19 $7,500 $300 $4,750 $500 $13,050 2 H B-D-E-G-H 2 17 $7,800 $400 $4,250 $300 $12,750

60. Statistical Analysis Figure 2.7 a m b Mean Time Beta distribution a m b Mean Time 3 σ 3 σ Area under curve between a and b is 99.74% Normal distribution

65. Application 2.4 Bluebird University: activity for sales training seminar Activity Immediate Predecessor(s) Optimistic (a) Most Likely (m) Pessimistic (b) Expected Time ( t ) Variance ( σ ) A — 5 7 8 B — 6 8 12 C — 3 4 5 D A 11 17 25 E B 8 10 12 F C, E 3 4 5 G D 4 8 9 H F 5 7 9 I G, H 8 11 17 J G 4 4 4 6.83 0.25 8.33 1.00 4.00 0.11 17.33 5.44 10.00 0.44 4.00 0.11 7.50 0.69 7.00 0.44 11.50 2.25 4.00 0.00

70. Application 2.5 The director of the continuing education at Bluebird University wants to conduct the seminar in 47 working days from now. What is the probability that everything will be ready in time? The critical path is and the expected completion time is T = T E is: A–D–G–I, 43.17 days. 47 days 43.17 days (0.25 + 5.44 + 0.69 + 2.25) = 8.63 And the sum of the variances for the critical activities is:

71. Application 2.5 T = 47 days T E = 43.17 days And the sum of the variances for the critical activities is: 8.63 Assuming the normal distribution applies, we use the table for the normal probability distribution. Given z = 1.30, the probability that activities A–D–G–I can be completed in 47 days or less is 0.9032. = = = 1.30 3.83 2.94 47 – 43.17 8.63  z = T – T E σ 2 

74. Project Life Cycle Figure 2.9 Start Finish Resource requirements Time Definition and organization Planning Execution Close out

77. Solved Problem 1 TABLE 2.2 | ELECTRIC MOTOR PROJECT DATA Activity Normal Time (days) Normal Cost ($) Crash Time (days) Crash Cost ($) Immediate Predecessor(s) A 4 1,000 3 1,300 None B 7 1,400 4 2,000 None C 5 2,000 4 2,700 None D 6 1,200 5 1,400 A E 3 900 2 1,100 B F 11 2,500 6 3,750 C G 4 800 3 1,450 D, E H 3 300 1 500 F, G

80. Solved Problem 1 Activity Crash Cost per Day ($) Maximum Time Reduction (days) A 300 1 B 200 3 C 700 1 D 200 1 E 200 1 F 250 5 G 650 1 H 100 2

81. Solved Problem 1 TABLE 2.3 | PROJECT COST ANALYSIS Stage Crash Activity Time Reduction (days) Resulting Critical Path(s) Project Duration (days) Project Direct Costs, Last Trial ($) Crash Cost Added ($) Total Indirect Costs ($) Total Penalty Costs ($) Total Project Costs ($) 0 — — C-F-H 19 10,100 — 3,800 700 14,600 1 H 2 C-F-H 17 10,100 200 3,400 500 14,200 2 F 2 A-D-G-H 15 10,300 500 3,000 300 14,100 B-E-G-H C-F-H

84. Solved Problem 2 Time Estimate (weeks) Activity Optimistic Most Likely Pessimistic Immediate Predecessor(s) A 1 4 7 — B 2 6 7 — C 3 3 6 B D 6 13 14 A E 3 6 12 A, C F 6 8 16 B G 1 5 6 E, F

88. Solved Problem 2 Figure 2.12 A 4.0 0.0 4.0 4.0 8.0 D 12.0 4.0 8.0 16.0 20.0 E 6.5 9.0 9.0 15.5 15.5 G 4.5 15.5 15.5 20.0 20.0 C 3.5 5.5 5.5 9.0 9.0 F 9.0 5.5 6.5 14.5 15.5 B 5.5 0.0 0.0 5.5 5.5 Finish Start

89. Solved Problem 2 Start (weeks) Finish (weeks) Activity Earliest Latest Earliest Latest Slack Critical Path A 0 4.0 4.0 8.0 4.0 No B 0 0.0 5.5 5.5 0.0 Yes C 5.5 5.5 9.0 9.0 0.0 Yes D 4.0 8.0 16.0 20.0 4.0 No E 9.0 9.0 15.5 15.5 0.0 Yes F 5.5 6.5 14.5 15.5 1.0 No G 15.5 15.5 20.0 20.0 0.0 Yes Path Total Expected Time (weeks) Total Variance A–D 4 + 12 = 16 1.00 + 1.78 = 2.78 A–E–G 4 + 6.5 + 4.5 = 15 1.00 + 2.25 + 0.69 = 3.94 B–C–E–G 5.5 + 3.5 + 6.5 + 4.5 = 20 0.69 + 0.25 + 2.25 + 0.69 = 3.88 B–F–G 5.5 + 9 + 4.5 = 19 0.69 + 2.78 + 0.69 = 4.16