1. 2 – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Project Management Chapter 2

2. 2 – 2 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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 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. 2 – 3 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Projects Project management is a systemized, approach to defining, organizing, planning, monitoring, and controlling projects Projects often require resources from many different parts of the organization Each project is unique

4. 2 – 4 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Planning Projects There are five steps to planning projects 1.Defining the work breakdown structure 2.Diagramming the network 3.Developing the schedule 4.Analyzing the cost-time trade-offs 5.Assessing risks

5. 2 – 5 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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

6. 2 – 6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Diagramming the Network Network diagrams use nodes and arcs to depict the relationships between activities Benefits of using networks include 1.Networks force project teams to identify and organize data to identify interrelationships between activities 2.Networks enable the estimation of completion time 3.Crucial activities are highlighted 4.Cost and time trade-offs can be analyzed

7. 2 – 7 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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

8. 2 – 8 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. STU S precedes T, which precedes U. Diagramming the Network AONActivity Relationships S T U S and T must be completed before U can be started. Figure 2.2

9. 2 – 9 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Diagramming the Network AONActivity 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. Figure 2.2

10. 2 – 10 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Diagramming the Network AONActivity Relationships S T U V U cannot begin until both S and T have been completed; V cannot begin until T has been completed. STV U T and U cannot begin until S has been completed and V cannot begin until both T and U have been completed. Figure 2.2

11. 2 – 11 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Developing the Schedule Schedules can help managers achieve the objectives of the project Managers can 1.Estimate the completion time by finding the critical path 2.Identify start and finish times for each activity 3.Calculate the amount of slack time for each activity

12. 2 – 12 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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

13. 2 – 13 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. St. John’s Hospital Project (Modified) ActivityImmediate 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 Example 2.1 START A B A C D A E, G, H F, I J, K 0 12 9 10 24 10 35 40 15 4 6 0 Kramer Stewart Johnson Taylor Adams Taylor Burton Johnson Walker Sampson Casey Murphy Pike Ashton F,I,J in the text

14. 2 – 14 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. ActivityImmediate 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 St. John’s Hospital Project Example 2.1 Completion Time Finish K6K6 I 15 F 10 C 10 H 40 J4J4 A 12 B9B9 Figure 2.3 Start G 35 D 10 E 24 ActivityIP Time ASTART12 BSTART9 CA10 DB10 EB24 FA10 GC35 HD40 IA15 JE, G, H4 KF, I6

15. 2 – 15 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. ActivityImmediate 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 St. John’s Hospital Project Example 2.1 Completion Time Finish K6K6 I 15 F 10 C 10 D 10 H 40 J4J4 A 12 B9B9 Figure 2.3 Start G 35 E 24 PathEstimated Time (weeks) A–I–K33 A–F–K28 A–C–G–J61 B–D–H–J63 B–E–J37

16. 2 – 16 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. ActivityImmediate 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 St. John’s Hospital Project Example 2.1 Completion Time Finish K6K6 I 15 F 10 C 10 D 10 H 40 J4J4 A 12 B9B9 Figure 2.3 Start G 35 E 24 PathEstimated Time (weeks) A–I–K33 A–F–K28 A–C–G–J61 B–D–H–J63 B–E–J37

17. 2 – 17 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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

18. 2 – 18 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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

19. 2 – 19 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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 EF A = 0 + 12 = 12 and EF B = 0 + 9 = 9

20. 2 – 20 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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, ES I = 12, ES F = 12, and ES C = 12 Similarly, ES D = 9 and ES E = 9 After placing these ES values on the network diagram, we determine the EF times for activities I, F, C, D, and E: EF I = 12 + 15 = 27, EF F = 12 + 10 = 22, EF C = 12 + 10 = 22, EF D = 9 + 10 = 19, and EF E = 9 + 24 = 33

21. 2 – 21 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Early Start and Early Finish Times The earliest start time for activity G is the latest EF time of all immediately preceding activities. Thus, ES G = EF C = 22, ES H = EF D = 19 EF G = ES G + t = 22 + 35 = 57, EF H + t = 19 + 40 = 59 Latest finish time Latest start time Activity Duration Earliest start time Earliest finish time 0 2 12 14 A 12

22. 2 – 22 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Network Diagram 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 27 330 12 Figure 2.4

23. 2 – 23 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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 63, which is the earliest project finish time. Thus, the latest start time for activity K is LS K = LF K – t = 63 – 6 = 57 If activity K is to start no later than week 57, all its predecessors must finish no later than that time. Consequently, LF I = 57, LF F = 57, and LF J = 57

24. 2 – 24 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Early Start and Early Finish Times The latest start times for these activities are shown in Figure 2.4 as LS I = 57 – 15 = 42, LF F = 57 – 10 = 47, and LS j = 63 – 4= 59 After obtaining LS J, we can calculate the latest start times for the immediate predecessors of activity J: LS G = 59 – 35 = 24, LS H = 59 – 40 = 19, and LS E = 59 – 24 = 35 Similarly, we can now calculate the latest start times for activities C and D: LS C = 24 – 10 = 14 and LS D = 19 – 10 = 9

25. 2 – 25 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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, LS A = 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. LS B = 9 – 9 = 0

26. 2 – 26 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Network Diagram Figure 2.4 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 27 330 12 42 57 47 57 59 63 24 59 19 59 35 59 14 24 9 19 2 14 0 9 57 63 S = 30 S = 2S = 35 S = 2 S = 26 S = 2 S = 0 S = 30 S = 0

27. 2 – 27 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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

28. 2 – 28 Project Costs The total project costs are the sum of direct costs, indirect costs, penalty cost, and crashing cost. 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. Penalty cost is zero unless the project is completed late Crashing cost is the cost to decrease the activity time

29. 2 – 29 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Cost-Time Trade-Offs Projects may be crashed to shorten the completion time Costs to crash Cost to crash per period = CC – NC NT – CT 1.Normal time (NT)3.Crash time (CT) 2.Normal cost (NC)4.Crash cost (CC)

30. 2 – 30 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.

31. 2 – 31 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Cost-Time Relationships 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 Figure 2.6

32. 2 – 32 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Cost-Time Relationships TABLE 2.1 | DIRECT COST AND TIME DATA FOR THE ST. JOHN’S HOSPITAL PROJECT ActivityNormal Time (NT) (weeks) Normal Cost (NC)($) Crash Time (CT)(weeks) Crash Cost (CC)($) Maximum Time Reduction (week) Cost of Crashing per Week ($) A12$12,00011$13,00011,000 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

33. 2 – 33 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assessing Risk Risk is the measure of the probability and consequence of not reaching a defined project goal Risk-management plans are developed to identify key risks and prescribe ways to circumvent them

34. 2 – 34 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Simulation and Statistical Analysis When uncertainty is present, simulation can be used to estimate the project completion time Statistical analysis requires three reasonable estimates of activity times 1.Optimistic time (a) 2.Most likely time (m) 3.Pessimistic time (b)

35. 2 – 35 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Statistical Analysis ambMean Time Beta distribution amb Mean Time 3σ3σ3σ3σ Area under curve between a and b is 99.74% Normal distribution Figure 2.7

36. 2 – 36 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Statistical Analysis The mean of the beta distribution can be estimated by t e = a + 4m + b 6 The variance of the beta distribution for each activity is σ 2 = b – a 6 2

37. 2 – 37 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating Means and Variances EXAMPLE 2.4 Suppose that the project team has arrived at the following time estimates for activity B (site selection and survey) of the St. John’s Hospital project: a = 7 weeks, m = 8 weeks, and b = 15 weeks a.Calculate the expected time and variance for activity B. b.Calculate the expected time and variance for the other activities in the project.

38. 2 – 38 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating Means and Variances SOLUTION a.The expected time for activity B is Note that the expected time does not equal the most likely time. These will only be the same only when the most likely time is equidistant from the optimistic and pessimistic times. The variance for activity B is t e = = = 9 weeks 7 + 4(8) + 15 6 54 6 σ 2 = = = 1.78 15 – 7 6 2 8686 2

39. 2 – 39 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating Means and Variances b.The following table shows the expected activity times and variances for this project. Time Estimates (week)Activity Statistics ActivityOptimistic (a)Most Likely (m)Pessimistic (b)Expected Time (t e )Variance (σ 2 ) A111213120.11 B 7 815 91.78 C 51015102.78 D 8 916101.78 E142530247.11 F 6 918104.00 G253641357.11 H354045402.78 I101328159.00 J 1 215 45.44 K 5 6 7 60.11

40. 2 – 40 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Network Diagram Figure 2.4 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 27 330 12 42 57 47 57 59 63 24 59 19 59 35 59 14 24 9 19 2 14 0 9 57 63 S = 30 S = 2S = 35 S = 2 S = 26 S = 2 S = 0 S = 30 S = 0

41. 2 – 41 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Analyzing Probabilities Because the central limit theorem can be applied, the mean of the distribution is the earliest expected finish time for the project T E = = Expected activity times on the critical path  Mean of normal distribution Because the activity times are independent σ 2 =  (Variances of activities on the critical path) z = T – T E σ2σ2  Using the z-transformation where T = due date for the project

42. 2 – 42 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating the Probability EXAMPLE Calculate the probability that St. John’s Hospital will become operational in 65 weeks, using (a) the critical path and (b) path A–C–G–J. SOLUTION a.The critical path B–D–H–J has a length of 63 weeks. From slide 50, we obtain the variance of path B–D–H–J: σ 2 = 1.78 + 1.78 + 2.78 + 5.44 = 11.78. Next, we calculate the z-value:

43. 2 – 43 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating the Probability Using the Normal Distribution appendix, we go down the left- hand column to 0.5 and then across to 0.08. This gives a value of 0.7190. Thus the probability is about 0.719 that the length of path B–D–H–J will be no greater than 65 weeks. Length of critical path Probability of meeting the schedule is 0.7190 Normal distribution: Mean = 63 weeks; σ = 3.432 weeks Probability of exceeding 65 weeks is 0.2810 Project duration (weeks) 63 65 Because this is the critical path, there is a 28.1 percent probability that the project will take longer than 65 weeks.

44. 2 – 44 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. From the normal table for 98%, z = 2.055 T E = 63 weeks,  3.432 weeks Critical Path = B - D - H - J T = T E + z  Expected project completion time EXAMPLE Determine the completion time of the critical path with 98% probability. T = 63 + 2.055 (3.432) = 70.05 weeks

45. 2 – 45 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating the Probability SOLUTION b.From the table in Example 2.4, we determine that the sum of the expected activity times on path A–C–G–J is 12 + 10 + 35 + 4 = 61 weeks and that σ 2 = 0.11 + 2.78 + 7.11 + 5.44 = 15.44. The z-value is The probability is about 0.8461 that the length of path A–C–G–J will be no greater than 65 weeks. EXAMPLE Calculate the probability that St. John’s Hospital will become operational in 65 weeks, using (a) the critical path and (b) path A–C–G–J.

46. 2 – 46 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Near-Critical Paths Project duration is a function of the critical path Since activity times vary, paths with nearly the same length can become critical during the project Project managers can use probability estimates to analyze the chances of near- critical paths delaying the project

47. 2 – 47 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 2.1 The following information is known about a project Draw the network diagram for this project ActivityActivity Time (days) Immediate Predecessor(s) A7— B2A C4A D4B, C E4D F3E G5E

48. 2 – 48 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Finish G5G5 F3F3 E4E4 D4D4 Application 2.1 ActivityActivity Time (days) Immediate Predecessor(s) A7— B2A C4A D4B, C E4D F3E G5E B2B2 C4C4 Start A7A7

49. 2 – 49 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 2.2 Calculate the four times for each activity in order to determine the critical path and project duration. ActivityDuration Earliest Start (ES) Latest Start (LS) Earliest Finish (EF) Latest Finish (LF) Slack (LS-ES) On the Critical Path? A700770-0=0Yes B2 C4 D4 E4 F3 G5 The critical path is A–C–D–E–G with a project duration of 24 days

50. 2 – 50 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 2.2 Calculate the four times for each activity in order to determine the critical path and project duration. ActivityDuration Earliest Start (ES) Latest Start (LS) Earliest Finish (EF) Latest Finish (LF) Slack (LS-ES) On the Critical Path? A700770-0=0Yes B2 C4 D4 E4 F3 G5 The critical path is A–C–D–E–G with a project duration of 24 days 799119-7=2No 7711 7-7=0Yes 1921222421-19=2No 19 24 19-19=0Yes 11 15 11-11=0Yes 15 19 15-15=0Yes

51. 2 – 51 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 2.2 ActivityDuration Earliest Start (ES) Latest Start (LS) Earliest Finish (EF) Latest Finish (LF) Slack (LS-ES) On the Critical Path? A700770-0=0Yes B2799119-7=2No C47711 7-7=0Yes D411 15 11-11=0Yes E415 19 15-15=0Yes F321 222421-19=2No G519 24 19-19=0Yes StartFinish A7A7 B2B2 C4C4 D4D4 E4E4 F3F3 G5G5 The critical path is A–C–D–E–G with a project duration of 24 days

52. 2 – 52 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 2.2 ActivityDuration Earliest Start (ES) Latest Start (LS) Earliest Finish (EF) Latest Finish (LF) Slack (LS-ES) On the Critical Path? A700770-0=0Yes B2799119-7=2No C47711 7-7=0Yes D411 15 11-11=0Yes E415 19 15-15=0Yes F321 222421-19=2No G519 24 19-19=0Yes StartFinish A7A7 B2B2 C4C4 D4D4 E4E4 F3F3 G5G5 The critical path is A–C–D–E–G with a project duration of 24 days

53. 2 – 53 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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) A51,00041,200— B580032,000— C26001900A, B D31,50022,000B E590031,200C, D F21,30011,400E G39003 E H55003900G

54. 2 – 54 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 2.3 Direct cost and time data for the activities: Project Activity and Cost Data ActivityCrash Cost/DayMaximum Crash Time (days) A2001 B6002 C3001 D5001 E1502 F1001 G00 H2002 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

55. 2 – 55 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Activity Immediate Predecessor(s) Optimistic (a) Most Likely (m) Pessimistic (b) Expected Time (t) Variance (σ) A—578 B—6812 C—345 DA111725 EB81012 FC, E345 GD489 HF579 IG, H81117 JG444 Application 2.4 Bluebird University: activity for sales training seminar 6.830.25 8.331.00 4.000.11 17.335.44 10.000.44 4.000.11 7.500.69 7.000.44 11.502.25 4.000.00

56. 2 – 56 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 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:

57. 2 – 57 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. = = = 1.30 3.83 2.94 47 – 43.17 8.63  Application 2.5 T = 47 days T E = 43.17 days And the sum of the variances for the critical activities is: 8.63 z = T – T E σ 2  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.

58. 2 – 58 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 Your company has just received an order from a good customer for a specially designed electric motor. The contract states that, starting on the thirteenth day from now, your firm will experience a penalty of $100 per day until the job is completed. Indirect project costs amount to $200 per day. The data on direct costs and activity precedent relationships are given in Table 2.2. a.Draw the project network diagram. b.What completion date would you recommend?

59. 2 – 59 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 TABLE 2.2 | ELECTRIC MOTOR PROJECT DATA ActivityNormal Time (days) Normal Cost ($) Crash Time (days) Crash Cost ($) Immediate Predecessor(s) A41,00031,300None B71,40042,000None C52,00042,700None D61,20051,400A E390021,100B F112,50063,750C G480031,450D, E H33001500F, G

60. 2 – 60 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 SOLUTION a.The network diagram is shown in Figure 2.10. Keep the following points in mind while constructing a network diagram. 1.Always have start and finish nodes. 2.Try to avoid crossing paths to keep the diagram simple. 3.Use only one arrow to directly connect any two nodes. 4.Put the activities with no predecessors at the left and point the arrows from left to right. 5.Be prepared to revise the diagram several times before you come up with a correct and uncluttered diagram. Start Finish A4A4 B7B7 C5C5 D6D6 E3E3 F 11 G4G4 H3H3 Figure 2.10

61. 2 – 61 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 b.With these activity times, the project will be completed in 19 days and incur a $700 penalty. Using the data in Table 2.2, you can determine the maximum crash-time reduction and crash cost per day for each activity. For activity A Maximum crash time = Normal time – Crash time = 4 days – 3 days = 1 day Crash cost per day = Crash cost – Normal cost Normal time – Crash time CC – NC NT – CT = = $300 $1,300 – $1,000 4 days – 3 days

62. 2 – 62 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 ActivityCrash Cost per Day ($)Maximum Time Reduction (days) A3001 B2003 C7001 D2001 E 1 F2505 G6501 H1002

63. 2 – 63 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 Table 2.3 summarizes the analysis and the resultant project duration and total cost. The critical path is C–F–H at 19 days, which is the longest path in the network. The cheapest activity to crash is H which, when combined with reduced penalty costs, saves $300 per day. Crashing this activity for two days gives A–D–G–H: 15 days, B–E–G–H: 15 days, and C–F–H: 17 days Crash activity F next. This makes all activities critical and no more crashing should be done as the cost of crashing exceeds the savings.

64. 2 – 64 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 An advertising project manager developed the network diagram in Figure 2.11 for a new advertising campaign. In addition, the manager gathered the time information for each activity, as shown in the accompanying table. a.Calculate the expected time and variance for each activity. b.Calculate the activity slacks and determine the critical path, using the expected activity times. c.What is the probability of completing the project within 23 weeks? Figure 2.11 Start Finish A B C D E F G

65. 2 – 65 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 Time Estimate (weeks) ActivityOptimisticMost LikelyPessimisticImmediate Predecessor(s) A147— B267— C336B D61314A E3612A, C F6816B G156E, F

66. 2 – 66 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 SOLUTION a.The expected time and variance for each activity are calculated as follows t e = a + 4m + b 6 ActivityExpected Time (weeks)Variance A4.01.00 B5.50.69 C3.50.25 D12.01.78 E6.52.25 F9.02.78 G4.50.69

67. 2 – 67 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 SOLUTION b.We need to calculate the earliest start, latest start, earliest finish, and latest finish times for each activity. Starting with activities A and B, we proceed from the beginning of the network and move to the end, calculating the earliest start and finish times. ActivityEarliest Start (weeks)Earliest Finish (weeks) A00 + 4.0 =4.0 B00 + 5.5 =5.5 C5.55.5 + 3.5 =9.0 D4.04.0 + 12.0 =16.0 E9.09.0 + 6.5 =15.5 F5.55.5 + 9.0 =14.5 G15.515.5 + 4.5 =20.0

68. 2 – 68 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 Based on expected times, the earliest finish date for the project is week 20, when activity G has been completed. Using that as a target date, we can work backward through the network, calculating the latest start and finish times ActivityLatest Start (weeks)Latest Finish (weeks) G15.520.0 F6.515.5 E9.015.5 D8.020.0 C5.59.0 B0.05.5 A4.08.0

69. 2 – 69 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 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 Finish Start Figure 2.12

70. 2 – 70 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 Start (weeks)Finish (weeks) ActivityEarliestLatestEarliestLatestSlackCritical Path A04.0 8.04.0No B00.05.5 0.0Yes C5.5 9.0 0.0Yes D4.08.016.020.04.0No E9.0 15.5 0.0Yes F5.56.514.515.51.0No G15.5 20.0 0.0Yes PathTotal Expected Time (weeks)Total Variance A–D4 + 12 = 161.00 + 1.78 = 2.78 A–E–G4 + 6.5 + 4.5 = 151.00 + 2.25 + 0.69 = 3.94 B–C–E–G5.5 + 3.5 + 6.5 + 4.5 = 200.69 + 0.25 + 2.25 + 0.69 = 3.88 B–F–G5.5 + 9 + 4.5 = 190.69 + 2.78 + 0.69 = 4.16

71. 2 – 71 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 So the critical path is B–C–E–G with a total expected time of 20 weeks. However, path B–F–G is 19 weeks and has a large variance. c.We first calculate the z-value: z = = = 1.52 T – T E σ2σ2  23 – 20 3.88  Using the Normal Distribution Appendix, we find the probability of completing the project in 23 weeks or less is 0.9357. Because the length of path B–F–G is close to that of the critical path and has a large variance, it might well become the critical path during the project