Project Management

1. Project Management

2. Projects • Projects are an interrelated set of activities with a definite starting and ending point, which results in a unique outcome from a specific allocation of resources • Projects are common in everyday life • The three main goals are to: • Complete the project on time • Not exceed the budget • Meet the specifications to the satisfactions of the customer

3. Projects • Project management is a systemized, phased approach to defining, organizing, planning, monitoring, and controlling projects • Projects often require resources from many different parts of the organization • Each project is unique • Projects are temporary • A collection of projects is called a program

4. Defining and Organizing Projects • Define the scope, time frame, and resources of the project • Select the project manager and team • Good project managers must be • Facilitators • Communicators • Decision makers • Project team members must have • Technical competence • Sensitivity • Dedication

5. Organizational Structure • Different structures have different implications for project management • Common structures are • Functional • Pure project • Matrix

6. Planning Projects • There are five steps to planning projects • Defining the work breakdown structure • Diagramming the network • Developing the schedule • Analyzing the cost-time trade-offs • Assessing risks

7. Work Breakdown Structure • A statement of all the tasks that must be completed as part of the project • An activity is the smallest unit of work effort consuming both time and resources that the project manager can schedule and control • Each activity must have an owner who is responsible for doing the work

8. Organizing and Site Preparation Physical Facilities and Infrastructure Select administration staff Purchase and deliver equipment Site selection and survey Construct hospital Select medical equipment Develop information system Prepare final construction plans Install medical equipment Bring utilities to site Train nurses and support staff Interview applicants fornursing and support staff Work Breakdown Structure Level 0 Relocation of St. John’s Hospital Figure 2.1 Level 1 Level 2

9. Diagramming the Network • Network diagrams use nodes and arcs to depict the relationships between activities • Benefits of using networks include • Networks force project teams to identify and organize data to identify interrelationships between activities • Networks enable the estimation of completion time • Crucial activities are highlighted • Cost and time trade-offs can be analyzed

10. Diagramming the Network • Precedent relationships determine the sequence for undertaking activities • Activity times must be estimated using historical information, statistical analysis, learning curves, or informed estimates • In the activity-on-node approach, nodes represent activities and arcs represent the relationships between activities

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

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

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

14. Developing the Schedule • Schedules can help managers achieve the objectives of the project • Managers can • Estimate the completion time by finding the critical path • Identify start and finish times for each activity • Calculate the amount of slack time for each activity

15. Critical Path • The sequence of activities between a project’s start and finish is a path • The critical path is the path that takes the longest time to complete

16. St. John’s Hospital Project Kramer Stewart Johnson Taylor Adams Taylor Burton Johnson Walker Sampson Casey Murphy Pike Ashton 0 12 9 10 10 24 10 35 40 15 4 6 0 START START A B B A C D A E, G, H F, I, J K Example 2.1

17. 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 I 15 A 12 F 10 K 6 C 10 G 35 Start Finish B 9 D 10 H 40 J 4 E 24 St. John’s Hospital Project Completion Time Figure 2.3 Example 2.1

18. I 15 A 12 F 10 K 6 C 10 G 35 Start Finish B 9 D 10 H 40 J 4 E 24 St. John’s Hospital Project Completion Time Figure 2.3 Example 2.1

19. I 15 A 12 F 10 K 6 C 10 G 35 Start Finish B 9 D 10 H 40 J 4 E 24 St. John’s Hospital Project Completion Time Figure 2.3 Example 2.1

20. Application 2.1 The following information is known about a project Draw the network diagram for this project

21. B 2 F 3 A 7 D 4 E 4 Start Finish C 4 G 5 Application 2.1

22. Project Schedule • The project schedule specifies start and finish times for each activity • Managers can use the earliest start and finish times, the latest start and finish times, or any time in between these extremes

23. Project Schedule • The earliest start time (ES) for an activity is the latest earliest finish time of any preceding activities • The earliest finish time (EF) is the earliest start time plus its estimated duration EF = ES + t • The latest finish time (LF) for an activity is the latest start time of any preceding activities • The latest start time (LS) is the latest finish time minus its estimated duration LS = LF –t

24. Early Start and Early Finish Times EXAMPLE 2.2 Calculate the ES, EF, LS, and LF times for each activity in the hospital project. Which activity should Kramer start immediately? Figure 2.3 contains the activity times. SOLUTION To compute the early start and early finish times, we begin at the start node at time zero. Because activities A and B have no predecessors, the earliest start times for these activities are also zero. The earliest finish times for these activities are EFA = 0 + 12 = 12 and EFB = 0 + 9 = 9

25. Early Start and Early Finish Times Because the earliest start time for activities I, F, and C is the earliest finish time of activity A, ESI = 12, ESF = 12, and ESC = 12 Similarly, ESD = 9 and ESE = 9 After placing these ES values on the network diagram, we determine the EF times for activities I, F, C, D, and E: EFI = 12 + 15 = 27, EFF = 12 + 10 = 22, EFC = 12 + 10 = 22, EFD = 9 + 10 = 19, and EFE = 9 + 24 = 33

26. Activity Earliest start time Earliest finish time A 0 12 Latest start time 2 14 12 Latest finish time Duration Early Start and Early Finish Times The earliest start time for activity G is the latest EF time of all immediately preceding activities. Thus, ESG = EFC = 22, ESH = EFD = 19 EFG = ESG + t = 22 + 35 = 57, EFH + t = 19 + 40 = 59

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

28. Early Start and Early Finish Times To compute the latest start and latest finish times, we begin by setting the latest finish activity time of activity K at week 69, which is the earliest finish time as determined in Figure 2.4. Thus, the latest start time for activity K is LSK = LFK – t = 69 – 6 = 63 If activity K is to start no later than week 63, all its predecessors must finish no later than that time. Consequently, LFI = 63, LFF = 63, and LFJ = 63

29. Early Start and Early Finish Times The latest start times for these activities are shown in Figure 2.4 as LSI = 63 – 15 = 48, LFF = 63 – 10 = 53, and LSJ = 63 – 4 = 59 After obtaining LSJ, we can calculate the latest start times for the immediate predecessors of activity J: LSG = 59 – 35 = 24, LSH = 59 – 40 = 19, and LSE = 59 – 24 = 35 Similarly, we can now calculate the latest start times for activities C and D: LSC = 24 – 10 = 14 and LSD = 19 – 10 = 9

30. Early Start and Early Finish Times Activity A has more than one immediately following activity: I, F, and C. The earliest of the latest start times is 14 for activity C. Thus, LSA = 14 – 12 = 2 Similarly, activity B has two immediate followers: D and E. Because the earliest of the latest start times of these activities is 9. LSB = 9 – 9 = 0

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

32. Gantt Chart Figure 2.5

33. Activity Slack • Activity slack is the maximum length of time an activity can be delayed without delaying the entire project • Activities on the critical path have zero slack • Activity slack can be calculated in two ways S = LS – ES or S = LF – EF

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

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

36. B 2 F 3 A 7 D 4 E 4 Start Finish C 4 G 5 Application 2.2 The critical path is A–C–D–E–G with a project duration of 24 days

37. B 2 F 3 A 7 D 4 E 4 Start Finish C 4 G 5 Application 2.2 The critical path is A–C–D–E–G with a project duration of 24 days

38. CC – NC NT – CT Cost to crash per period = Cost-Time Trade-Offs • Total project costs are the sum of direct costs and indirect costs • Projects may be crashed to shorten the completion time • Costs to crash 1. Normal time (NT) 3. Crash time (CT) 2. Normal cost (NC) 4. Crash cost (CC)

39. 8000 — 7000 — 6000 — 5000 — 4000 — 3000 — 0 — Crash cost (CC) Linear cost assumption Estimated costs for a 2-week reduction, from 10 weeks to 8 weeks Direct cost (dollars) 5200 Normal cost (NC) | | | | | | 5 6 7 8 9 10 11 (Crash time) (Normal time) Time (weeks) Cost-Time Relationships Figure 2.6

40. Cost-Time Relationships

41. A Minimum-Cost Schedule EXAMPLE 2.3 Determine the minimum-cost schedule for the St. John’s Hospital project. SOLUTION The projected completion time of the project is 69 weeks. The project costs for that schedule are $1,992,000 in direct costs, 69($8,000) = $552,000 in indirect costs, and (69 – 65)($20,000) = $80,000 in penalty costs, for total project costs of $2,624,000. The five paths in the network have the following normal times:

42. A Minimum-Cost Schedule STAGE 1 Step 1. The critical path is B–D–H–J–K. Step 2. The cheapest activity to crash per week is J at $1,000, which is much less than the savings in indirect and penalty costs of $28,000 per week. Step 3. Crash activity J by its limit of three weeks because the critical path remains unchanged. The new expected path times are A–C–G–J–K: 64 weeks and B–D–H–J–K: 66 weeks The net savings are 3($28,000) – 3($1,000) = $81,000. The total project costs are now $2,624,000 - $81,000 = $2,543,000.

43. I 15 A 12 F 10 K 6 C 10 G 35 Start Finish B 9 D 10 H 40 J 1 E 24 A Minimum-Cost Schedule STAGE 1 Step 1. The critical path is B–D–H–J–K. Step 2. The cheapest activity to crash per week is J at $1,000, which is much less than the savings in indirect and penalty costs of $28,000 per week. Step 3. Crash activity J by its limit of three weeks because the critical path remains unchanged. The new expected path times are A–C–G–J–K: 64 weeks and B–D–H–J–K: 66 weeks The net savings are 3($28,000) – 3($1,000) = $81,000. The total project costs are now $2,624,000 - $81,000 = $2,543,000.

44. A Minimum-Cost Schedule STAGE 2 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. Step 3. Crash D by two weeks. The first week of reduction in activity D saves $28,000 because it eliminates a week of penalty costs, as well as indirect costs. Crashing D by a second week saves only $8,000 in indirect costs because, after week 65, no more penalty costs are incurred. These savings still exceed the cost of crashing D by two weeks. Updated path times are A–C–G–J–K: 64 weeks and B–D–H–J–K: 64 weeks The net savings are $28,000 + $8,000 – 2($2,000) = $32,000. Total project costs are now $2,543,000 – $32,000 = $2,511,000.

45. I 15 A 12 F 10 K 6 C 10 G 35 Start Finish B 9 D 8 H 40 J 1 E 24 A Minimum-Cost Schedule STAGE 2 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. Step 3. Crash D by two weeks. The first week of reduction in activity D saves $28,000 because it eliminates a week of penalty costs, as well as indirect costs. Crashing D by a second week saves only $8,000 in indirect costs because, after week 65, no more penalty costs are incurred. These savings still exceed the cost of crashing D by two weeks. Updated path times are A–C–G–J–K: 64 weeks and B–D–H–J–K: 64 weeks The net savings are $28,000 + $8,000 – 2($2,000) = $32,000. Total project costs are now $2,543,000 – $32,000 = $2,511,000.

46. A Minimum-Cost Schedule STAGE 3 Step 1. After crashing D, we now have two critical paths. Both critical paths must now be shortened to realize any savings in indirect project costs. Step 2. Our alternatives are to crash one of the following combinations of activities—(A, B); (A, H); (C, B); (C, H); (G, B); (G, H)—or to crash activity K, which is on both critical paths (J has already been crashed). We consider only those alternatives for which the costs of crashing are less than the potential savings of $8,000 per week. The only viable alternatives are (C, B) at a cost of $7,600 per week and K at $4,000 per week. We choose activity K to crash.

47. A Minimum-Cost Schedule STAGE 3 Step 3. We crash activity K to the greatest extent possible—a reduction of one week—because it is on both critical paths. Updated path times are A–C–G–J–K: 63 weeks and B–D–H–J–K: 63 weeks Net savings are $8,000 - $4,000 = $4,000. Total project costs are $2,511,000 – $4,000 = $2,507,000.

48. I 15 A 12 F 10 K 5 C 10 G 35 Start Finish B 9 D 8 H 40 J 1 E 24 A Minimum-Cost Schedule STAGE 3 Step 3. We crash activity K to the greatest extent possible—a reduction of one week—because it is on both critical paths. Updated path times are A–C–G–J–K: 63 weeks and B–D–H–J–K: 63 weeks Net savings are $8,000 - $4,000 = $4,000. Total project costs are $2,511,000 – $4,000 = $2,507,000.

49. A Minimum-Cost Schedule STAGE 4 Step 1. The critical paths are still B–D–H–J–K and A–C–G–J–K. Step 2. The only viable alternative at this stage is to crash activities B and C simultaneously at a cost of $7,600 per week. This amount is still less than the savings of $8,000 per week. Step 3. Crash activities B and C by two weeks, the limit for activity B. Updated path times are A–C–G–J–K: 61 weeks and B–D–H–J–K: 61 weeks The net savings are 2($8,000) – 2($7,600) = $800. Total project costs are now $2,507,000 – $800 = $2,506,200.

50. I 15 A 12 F 10 K 5 C 8 G 35 Start Finish B 7 D 8 H 40 J 1 E 24 A Minimum-Cost Schedule STAGE 4 Step 1. The critical paths are still B–D–H–J–K and A–C–G–J–K. Step 2. The only viable alternative at this stage is to crash activities B and C simultaneously at a cost of $7,600 per week. This amount is still less than the savings of $8,000 per week. Step 3. Crash activities B and C by two weeks, the limit for activity B. Updated path times are A–C–G–J–K: 61 weeks and B–D–H–J–K: 61 weeks The net savings are 2($8,000) – 2($7,600) = $800. Total project costs are now $2,507,000 – $800 = $2,506,200.