Project Management Chapter 2 02-0 1
What is a Project? Project An interrelated set of activities with a definite starting and ending point, which results in a unique outcome for a specific allocation of resources. 02-0 2
What is Project Management? Project Management A systemized, phased approach to defining, organizing, planning, monitoring, and controlling projects. 02-03
Defining and Organizing Projects Defining the Scope and Objectives Selecting the Project Manager and Team Recognizing Organizational Structure 02-0 4
Planning Projects Defining the Work Breakdown Structure Diagramming the network Developing the schedule Analyzing cost-time trade-offs Assessing risks 02-0 5
Work Breakdown Structure Work Breakdown Structure A statement of all the work that has to be completed. Activity The smallest unit of work effort consuming both time and resources that can be planned and controlled. 02-06
Work Breakdown Structure Relocation of St. John s Hospital Level 0 Organizing and Site Preparation Physical Facilities and Infrastructure Level 1 Select administration staff Purchase and deliver equipment Site selection and survey Construct hospital Select medical equipment Develop information system Level 2 Prepare final construction plans Install medical equipment Bring utilities to site Train nurses and support staff Interview applicants for nursing and support staff Copyright 2013 Pearson Education, Inc. Publishing as. 02-07
Diagramming the Network Program Evaluation and Review Technique (PERT) Critical Path Method (CPM) Activity-on-Node approach 02-08
Establishing Precedence Relationships Precedence Relationships Determining the sequence for undertaking activities. 02-09
Example 2.1 02-10
Example 2.1 Immediate Predecessor I A B C A D B E B F A G C H D I A J E,G,H K F,I,J Start A B F C D G H K J Finish E 02-11
Application 2.1 The following information is known about a 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 Draw the network diagram for this project 02-12
Application 2.1 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 F 3 Start A 7 D 4 E 4 Finish C 4 G 5 02-13
Developing the Schedule Path The sequence of activities between a project s start and finish. Critical Path The sequence of activities between a start and finish that takes the longest time to complete. 02-14
Developing the Schedule Earliest start time (ES) - the latest earliest finish time of any immediately preceding activities Earliest finish time (EF) - the earliest start time plus its estimated duration EF = ES + t Latest finish time (LF) the earliest of the latest start times of any of the immediately following activities. Latest start time (LS) - the latest finish time minus its estimated duration LS = LF t Activity Slack - the maximum length of time an activity can be delayed without delaying the entire project LF-EF or LS-ES 02-15
Developing the Schedule Activity Earliest start time 0 A 12 Earliest finish time Latest start time 2 12 14 Latest finish time Duration 02-16
Example 2.2 Paths are the sequence of activities between a project s start and finish. I A F K Path Time (wks) Start C G Finish 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 B D E H J 02-17
Example 2.2 Earliest start time I 12 27 15 Earliest finish time 0 A 12 F K 12 22 63 69 12 10 6 Start C 12 22 10 G 22 57 35 Finish B 0 9 9 9 D 19 19 H 59 10 40 J 59 63 4 9 E 33 24 02-18
Example 2.2 I 12 27 15 The Critical Path takes 69 weeks A 0 12 12 12 F 22 63 K 69 10 6 Start C 12 22 10 G 22 57 35 Finish B 0 9 9 9 D 19 19 H 59 10 40 J 59 63 4 9 E 33 24 02-19
Example 2.2 Latest start time I 12 27 48 63 15 Latest finish time A 0 12 2 12 14 K 12 F 22 63 69 53 63 63 69 10 6 Start C 12 22 14 10 24 22 G 57 24 59 35 Finish 0 B 9 0 9 9 D H 9 19 19 59 9 19 10 19 40 59 J 59 63 59 4 63 9 E 33 35 24 59 02-20
Example 2.2 S = 36 12 I 27 48 15 63 S = 2 A 0 12 2 12 14 S = 0 S = 41 F K 12 22 63 69 53 10 63 63 69 6 S = 2 Start S = 2 12 C 22 14 10 24 22 G 57 24 59 35 Finish S = 0 0 B 9 0 9 9 S = 0 S = 0 9 D 19 19 H 59 9 19 10 19 40 59 J 59 63 59 63 4 S = 0 S = 26 9 E 33 35 24 59 02-21
Developing a Schedule Gantt chart 02-22
Application 2.2 Calculate the four times for each activity in order to determine the critical path and project duration for the diagram in Application Problem 1. 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 The critical path is A C D E G with a project duration of 24 days. 02-23
Application 2.2 Calculate the four times for each activity in order to determine the critical path and project duration. 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 11 11 15 15 11-11=0 Yes 15 15 19 19 15-15=0 Yes 19 21 22 24 21-19=2 No 19 19 24 24 19-19=0 Yes 02-24
Application 2.2 The critical path is A C D E G with a project duration of 24 days. B 2 F 3 Start A 7 D 4 E 4 Finish C 4 G 5 02-25
Analyzing Cost-Time Trade-Offs Project Crashing Shortening (or expediting) some activities within a project to reduce overall project completion time. Project Costs Direct Costs Indirect Costs Penalty Costs 02-26
Analyzing Cost-Time Trade-Offs Project Costs Normal time (NT) is the time necessary to complete an 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. Cost to crash per period = CC NC NT CT 02-27
Direct cost (dollars) Cost-Time Relationships 8000 7000 6000 5200 5000 4000 Crash cost (CC) Linear cost assumption Estimated costs for a 2-week reduction, from 10 weeks to 8 weeks Figure 2.6 3000 0 5 6 7 8 9 10 11 (Crash time) (Normal time) Time (weeks) Copyright 2013 Pearson Education, Inc. Publishing as. Normal cost (NC) 02-28
Analyzing Cost-Time Trade-Offs Determining 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. 02-29
Example 2.3 Activity DIRECT COST AND TIME DATA FOR THE ST. JOHN S HOSPITAL PROJECT 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 02-30
Example 2.3 Determine the minimum-cost schedule for the St. John s Hospital project. Project completion time =69 weeks. Project cost = $2,624,000 Direct = $1,992,000 Indirect = 69($8,000) = $552,000 Penalty = (69 65)($20,000) = $80,000 A I K 33 weeks A F K A C G J K B D H J K B E J K 28 weeks 67 weeks 69 weeks 43 weeks 02-31
Example 2.3 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 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. 02-32
Example 2.3 STAGE 1 I 15 A 12 F 10 K 6 Start C 10 G 35 Finish B 9 D 10 H 40 J 1 E 24 02-33
Example 2.3 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. 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. 02-34
Example 2.3 STAGE 2 I 15 A 12 F 10 K 6 Start C 10 G 35 Finish B 9 D 8 H 40 J 1 E 24 02-35
Example 2.3 STAGE 3 Step 1. The critical paths are B D H J K and A-C-G-J-K Step 2. Activities eligible to be crashed: (A, B); (A, H); (C, B); (C, H); (G, B); (G, H) or to crash Activity K We consider only those alternatives for which the costs of crashing are less than the potential savings of $8,000 per week. We choose activity K to crash 1 week at $4,000 per week. Step 3. 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. 02-36
Example 2.3 STAGE 3 I 15 A 12 F 10 K 5 Start C 10 G 35 Finish B 9 D 8 H 40 J 1 E 24 02-37
Example 2.3 STAGE 4 Step 1. The critical paths are still B D H J K and A C G J K. Step 2. Activities eligible to be crashed: (B,C) @ $7,600 per week. Step 3. Crash activities B and C by two weeks. 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. 02-38
Example 2.3 STAGE 4 I 15 A 12 F 10 K 5 Start C 8 G 35 Finish B 7 D 8 H 40 J 1 E 24 02-39
Example 2.3 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 1 J 3 B-D-H-J- K 2 D 2 B-D-H-J- K A-C-G-J- K 3 K 1 B-D-H-J- K A-C-G-J- K 4 B, C 2 B-D-H-J- K A-C-G-J- K 69 1,992.0 552.0 80.0 2,624.0 66 1,992.0 3.0 528.0 20.0 2,543.0 64 1,995.0 4.0 512.0 0.0 2,511.0 63 1,999.0 4.0 504.0 0.0 2,507.0 61 2,003.0 15.2 488.0 0.0 2,506.2 02-40
Application 2.3 Indirect project costs = $250 per day Penalty cost = $100 per day past day 14. Activity Normal Time (days) Project Activity and Cost Data Normal Cost ($) 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 02-41
Application 2.3 5 6 C 2 7 8 ES LS ID DUR EF LF 0 A 5 13 F 15 1 5 6 19 2 21 Start 8 8 E 5 13 13 Finish 0 B 5 5 D 8 13 G 16 16 H 21 0 5 5 5 3 8 13 3 16 16 5 21 02-42
Application 2.3 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 Normal Total Costs = Total Indirect Costs = $7,500 $250 per day 21 days = $5,250 Penalty Cost = $100 per day 7 days = $700 Total Project Costs = $13,450 02-43
Application 2.3 Step 1: The critical path is B D E G H, and the project duration is 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: Normal 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 02-44
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 = $7,500 + $300 (the added crash costs) = $7,800 Crash Cost Added = $200 2 days = $400 Total Indirect Costs = $250 per day 17 days = $4,250 Penalty Cost = $100 per day 3 days = $300 Total Project Cost = $12,750 02-45
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 = $7,800 + $400 (the added crash costs) = $8,200 Crash Cost Added = $500 1 day = $500 Total Indirect Costs = $250 per day 16 days = $4,000 Penalty Cost = $100 per day 2 days = $200 Total Project Cost = $12,900 which is greater than the last trial. Hence we stop the crashing process. 02-46
Application 2.3 The recommended completion date is day 17. 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 Further reductions will cost more than the savings in indirect costs and penalties. The critical path is B D E G H and the duration is 17 days. 02-47
Assessing Risks Risk-management Plans Strategic Fit Service/Product Attributes Project Team Capability Operations 02-48
Assessing Risks Statistical Analysis Optimistic time (a) Most likely time (m) Pessimistic time (b) 02-49
Statistical Analysis Area under curve between a and b is 99.74% a m Mean b Time Beta distribution a 3σ m 3σ b Mean Time Normal distribution 02-50
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 02-51
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. 02-52
Example 2.4 a. The expected time for activity B is 7 + 4(8) + 15 54 t e = = = 9 weeks 6 6 The variance for activity B is 2 15 7 8 σ 2 = = = 1.78 6 6 2 02-53
Example 2.4 b. The following table shows the expected activity times and variances for this project. Activity Optimistic (a) Time Estimates (week) Most Likely (m) Pessimistic (b) Activity Statistics Expected Time (t e ) Variance (σ 2 ) A 11 12 13 12 0.11 B 7 8 15 9 1.78 C 5 10 15 10 2.78 D 8 9 16 10 1.78 E 14 25 30 24 7.11 F 6 9 18 10 4.00 G 25 36 41 35 7.11 H 35 40 45 40 2.78 I 10 13 28 15 9.00 J 1 2 15 4 5.44 K 5 6 7 6 0.11 02-54
Application 2.4 The director of continuing education at Bluebird University just approved the planning for a sales training seminar. Her administrative assistant identified the various activities that must be done and their relationships to each other: 02-55
Application 2.4 The Network Diagram is: A D G J Start B E I Finish C F H 02-56
Application 2.4 For the Bluebird University sales training seminar activities, calculate the means and variances for each activity. Activity Immediate Predecessor(s) Optimistic (a) Most Likely (m) Pessimistic (b) 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 Expected Time (t) Variance (σ) 02-57
Application 2.4 For the Bluebird University sales training seminar activities, calculate the means and variances for each activity. Activity Immediate Predecessor(s) Optimistic (a) Most Likely (m) Pessimistic (b) 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 Expected Time (t) Variance (σ) 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 02-58
Analyzing Probabilities Because the central limit theorem can be applied, the mean of the distribution is the earliest expected finish time for the project Expected activity times T E = on the critical path = Because the activity times are independent Mean of normal distribution σ p2 = (Variances of activities on the critical path) Using the z-transformation z = T T E σ p where T = due date for the project 02-59
Example 2.5 Calculate the probability that St. John s Hospital will become operational in 72 weeks, using (a) the critical path and (b) path A C G J K. a. The critical path B D H J K has a length of 69 weeks. From the table in Example 2.4, we obtain the variance of path B D H J K: σ 2 = 1.78 + 1.78 + 2.78 + 5.44 + 0.11 = 11.89 Next, we calculate the z-value: z 72 69 11.89 3 3.45 0.87 02-60
Example 2.5 Using the Normal Distribution appendix, we find a value of 0.8078. Thus the probability is about 0.81 the length of path B D H J K will be no greater than 72 weeks. Because this is the critical path, there is a 19 percent probability that the project will take longer than 72 weeks. Length of critical path Probability of meeting the schedule is 0.8078 Normal distribution: Mean = 69 weeks; σ = 3.45 weeks Probability of exceeding 72 weeks is 0.1922 69 72 Project duration (weeks) 02-61
Example 2.5 b. The sum of the expected activity times on path A C G J K is 67 weeks and that σ 2 = 0.11 + 2.78 + 7.11 + 5.44 + 0.11 = 15.55. The z-value is z 72 67 15.55 5 3.94 1.27 The probability is about 0.90 that the length of path A C G J K will be no greater than 72 weeks. 02-62
Application 2.5 The director of the continuing education at Bluebird University wants to conduct the seminar in 47 working days from now. Using the activity data from Application 2.4, what is the probability that everything will be ready in time? T = 47 days T E is: 43.16 days And the sum of the variances for the critical activities is: (0.25 + 5.44 + 0.69 + 2.25) = 8.63 The critical path is A D G I, and the expected completion time is 43.16 days. 02-63
Application 2.5 The Network Diagram is: A D G J Start B E I Finish C F H The critical path is A-D-G-I 02-64
Application 2.5 T = 47 days T E = 43.16 days And the sum of the variances for the critical activities is: 8.63 z = T T E σ 2 47 43.16 3.84 = = = 1.31 8.63 2.94 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 0.9049. 02-65
Monitoring and Controlling Projects Monitoring Project Status Open Issues and Risks Schedule Status Monitoring Project Resources 02-66
Solved Problem 2.1 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 $200 per day. The data on direct costs and activity precedent relationships are given in Table 2.2. a. Draw the project network diagram. b. What completion date would you recommend? 02-67
Solved Problem 2.1 Acti vity Normal Time (days) ELECTRIC MOTOR PROJECT DATA Normal Cost ($) 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 02-68
Solved Problem 2.1 a. The network diagram is shown in Figure 2.10. Keep the following points in mind while constructing a network diagram. A 4 D 6 G 4 Finish Start B 7 E 3 H 3 C 5 F 11 02-69
Solved Problem 2.1 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 = = $300 $1,300 $1,000 4 days 3 days CC NC NT CT 02-70
Solved Problem 2.1 Activity Crash Cost per Day ($) 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 02-71
Solved Problem 2.1 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 02-72
Solved Problem 2.1 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. 02-73
Solved Problem 2.2 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. A D E Finish b. Calculate the activity slacks and determine the critical path, using the expected activity times. Start B C F G c. What is the probability of completing the project within 23 weeks? 02-74
Solved Problem 2.2 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 02-75
Solved Problem 2.2 a. The expected time and variance for each activity are calculated as follows t e = a + 4m + b 6 Activity Expected Time (weeks) Variance A 4.0 1.00 B 5.5 0.69 C 3.5 0.25 D 12.0 1.78 E 6.5 2.25 F 9.0 2.78 G 4.5 0.69 02-76
Solved Problem 2.2 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. Activity Earliest Start (weeks) Earliest Finish (weeks) A 0 0 + 4.0 = 4.0 B 0 0 + 5.5 = 5.5 C 5.5 5.5 + 3.5 = 9.0 D 4.0 4.0 + 12.0 = 16.0 E 9.0 9.0 + 6.5 = 15.5 F 5.5 5.5 + 9.0 = 14.5 G 15.5 15.5 + 4.5 = 20.0 02-77
Solved Problem 2.2 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 Activity Latest Start (weeks) Latest Finish (weeks) G 15.5 20.0 F 6.5 15.5 E 9.0 15.5 D 8.0 20.0 C 5.5 9.0 B 0.0 5.5 A 4.0 8.0 02-78
Solved Problem 2.2 4.0 8.0 D 12.0 16.0 20.0 Finish 0.0 A 4.0 9.0 E 15.5 4.0 4.0 8.0 9.0 6.5 15.5 Start 5.5 C 9.0 5.5 3.5 9.0 0.0 0.0 B 5.5 5.5 5.5 5.5 F 14.5 15.5 15.5 G 4.5 20.0 20.0 6.5 9.0 15.5 02-79
Solved Problem 2.2 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 02-80
Solved Problem 2.2 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: T T E 23 20 z = = = 1.52 σ 2 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 02-81
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