Chapter 1

Chapter 1

2 Project Management PowerPoint Slides by Jeff Heyl Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. For Operations Management, 9e by Krajewski/Ritzman/Malhotra 2010 Pearson Education 21 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 Not the project on time exceed the budget Meet the specifications to the satisfactions of the customer Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 22 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 23

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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 24 Organizational Structure Different structures have different implications for project management Common structures are Functional Pure project Matrix Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 25 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 26 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 27 Work Breakdown Structure Relocation of St. Johns Hospital Organizing and Site Preparation Level 0 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 for nursing and support staff Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Level 1 Level 2 Figure 2.1 28 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 29 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 10 Diagramming the Network AON S Activity Relationships T U S U T S precedes T, which precedes U. S and T must be completed before U can be started. Figure 2.2 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 11 Diagramming the Network AON Activity Relationships T T and U cannot begin until S has been completed. S U S

U T V U and V cant begin until both S and T have been completed. Figure 2.2 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 12 Diagramming the Network AON Activity Relationships S U T V S T V U U cannot begin until both S and T have been completed; V cannot begin until T has been completed. 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.

2 13 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 14 Critical Path The sequence of activities between a projects start and finish is a path The critical path is the path that takes the longest time to complete Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 15 St. Johns Hospital Project Activity Immediate Predecessors Activity Times (wks) ST. JOHNS HOSPITAL PROJECT Responsibility Kramer START

0 ORGANIZING and SITE PREPARATION Stewart A. Select administrative staff START 12 B. Select site and survey START 9 Taylor C. Select medical equipment A 10 Adams D. Prepare final construction plans B 10 Taylor E. Bring utilities to site B

24 Burton F. Interview applicants for nursing and support staff A 10 Johnson G. Purchase and deliver equipment C 35 Walker H. Construct hospital D 40 Sampson I. Develop information system A 15 Casey J. Install medical equipment E, G, H 4

Murphy K. Train nurses and support staff F, I, J 6 Pike K 0 Ashton Johnson PHYSICAL FACILITIES and INFRASTRUCTURE FINISH Example 2.1 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 16 St. Johns Hospital Project Activity Immediate Predecessors ST. JOHNS HOSPITAL PROJECT START Completion Time ORGANIZING and SITE PREPARATION Activity IP Time staff A A. Select STARTadministrative 12

A B B. Select STARTsite and 9 survey 12 C A 10 C. Select medical equipment D B 10 final24construction plans E D. Prepare B F E. Bring A 10to site utilities Start G C 35 F. Interview applicants for nursing and H D support staff40 I A 15 PHYSICAL FACILITIES and INFRASTRUCTURE J E, G, H 4 B K G. Purchase F, I, J and6 deliver equipment 9 H. Construct hospital I. Develop information system J.

Install medical equipment Figureand 2.3 support staff Train nurses K. Activity Times (wks) Responsibility I 15 F 10 K 6 C 10 G 35 D 10 H 40 Finish J 4 E 24 FINISH Example 2.1 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 17 Estimated Time (weeks) St. Johns

Hospital Project AIK 33 Path AFK Activity 28 ACGJK 67 BDHJK START Completion 69 Time Immediate Predecessors ST. JOHNS HOSPITAL PROJECT Activity Times (wks) Responsibility I 15 BEJK 43 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. A 12 Start Interview applicants for nursing and support staff PHYSICAL FACILITIES and INFRASTRUCTURE B G. Purchase and deliver equipment H. Construct hospital I. Develop information system J. Install medical equipment Figureand 2.3 support staff Train nurses K. 9 F 10 K 6 C 10 G 35 D

10 H 40 Finish J 4 E 24 FINISH Example 2.1 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 18 Estimated Time (weeks) St. Johns Hospital Project AIK 33 Path AFK Activity 28 ACGJK 67 BDHJK START Completion 69 Time Immediate Predecessors ST. JOHNS HOSPITAL PROJECT Activity Times (wks) Responsibility

I 15 BEJK 43 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. A 12 Start Interview applicants for nursing and support staff PHYSICAL FACILITIES and INFRASTRUCTURE B G. Purchase and deliver equipment H. Construct hospital I. Develop information system J. Install medical equipment

Figureand 2.3 support staff Train nurses K. 9 F 10 K 6 C 10 G 35 D 10 H 40 Finish J 4 E 24 FINISH Example 2.1 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 19 Application 2.1 The following information is known about a project Activity A B C D E F

G Activity Time (days) 7 2 4 4 4 3 5 Immediate Predecessor(s) A A B, C D E E Draw the network diagram for this project Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 20 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 Start A 7 F 3 D 4 C 4 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. E 4 Finish G 5

2 21 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 22 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 23 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 24 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 25 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 Activity Earliest start time 0 Latest start time 2 A 12 Earliest finish time 12 14 Latest finish time Duration Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 26 Network Diagram 12 I 27 15 0

A 12 12 F 22 63 K 69 10 6 12 12 Start C 22 22 10 0 B 9 9 9 D 19 10 G 57 Finish 35 19 H 59 40 59 J 63 4

9 E 33 Figure 2.4 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 24 2 27 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 28 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 29 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 30

Network Diagram 12 48 I 27 15 63 S = 36 A 0 12 2 12 14 12 F 22 53 63 S=2 S = 41 10 C 12 22 14 10 24 Start S=2 0 0 B 9 9 9 S=0 63 K 69 63 6 69 9 9 D 19 10 19 S=0

S=0 22 24 G 57 Finish 35 59 S=2 19 19 H 59 40 59 S=0 J 59 63 59 4 63 S=0 9 E 33 35 59 Figure 2.4 24 S = 26 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 31 Gantt Chart Figure 2.5 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 32 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. or S = LF EF 2 33 Application 2.2 Calculate the four times for each activity in order to determine the critical path and project duration. Activity A Duration 7 B 2 C 4 D 4 E 4 F 3 G 5 Earliest Start

(ES) 0 Latest Start (LS) 0 Earliest Finish (EF) 7 Latest Finish (LF) 7 Slack (LS-ES) 0-0=0 On the Critical Path? Yes The critical path is ACDEG with a project duration of 24 days Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 34 Application 2.2 Calculate the four times for each activity in order to determine the critical path and project duration. Activity A Duration 7 B 2 C 4

D 4 E 4 F 3 G 5 Earliest Start (ES) 0 Latest Start (LS) 0 Earliest Finish (EF) 7 Latest Finish (LF) 7 Slack (LS-ES) 0-0=0 On the Critical Path? Yes 7 7 9

7 9 11 11 11 9-7=2 7-7=0 No Yes 11 15 11 15 15 19 15 19 11-11=0 15-15=0 Yes Yes 19 19 21 19 22 24 24 24 21-19=2 19-19=0 No Yes

The critical path is ACDEG with a project duration of 24 days Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 35 Application 2.2 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 The critical path is ACDEG with a project duration of 24 days B 2 Start

A 7 F 3 D 4 C 4 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. E 4 Finish G 5 2 36 Application 2.2 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 The critical path is ACDEG with a project duration of 24 days B 2 Start A 7 F 3 D 4 C 4 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. E 4 Finish G 5 2 37 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) 2. Normal cost (NC) 3. Crash time (CT) 4. Crash cost (CC) CC NC Cost to crash per period =

NT CT Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 38 Cost-Time Relationships Direct cost (dollars) 8000 Crash cost (CC) 7000 Linear cost assumption 6000 Estimated costs for a 2-week reduction, from 10 weeks to 8 weeks 5000 5200 4000 3000 0 Normal cost (NC) | 5 | 6 (Crash time) Figure 2.6 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. | 7 8 | 9 | 10

| 11 (Normal time) Time (weeks) 2 39 Cost-Time Relationships TABLE 2.1 | DIRECT COST AND TIME DATA FOR THE ST. JOHNS HOSPITAL PROJECT Activity Normal Time (NT) (weeks) A 12 B Crash Time (CT)(weeks) Crash Cost (CC)($) Maximum Time Reduction (week) $12,000 11 $13,000 1 1,000 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 Normal Cost (NC)($) $1,992,000 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Cost of Crashing per Week ($) $2,209,500 2 40

A Minimum-Cost Schedule EXAMPLE 2.3 Determine the minimum-cost schedule for the St. Johns 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: AIK AFK ACGJK BDHJK BEJK Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 33 weeks 28 weeks 67 weeks 69 weeks 43 weeks 2 41 A Minimum-Cost Schedule STAGE 1 Step 1. The critical path is BDHJK. 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 ACGJK: 64 weeks and BDHJK: 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. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 42 A Minimum-Cost Schedule STAGE 1 I Step 1. The critical path is BDHJK. 15 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

A F K costs of $28,000 per week. 12 10 6 Step 3. Crash activity J by its limit of three weeks because the critical path remains unchanged. The new expected path times C G are Start Finish 10 35 ACGJK: 64 weeks and BDHJK: 66 weeks B 3($1,000) D H J The net savings are 3($28,000) = $81,000. The total 9 10 40 1 project costs are now $2,624,000 - $81,000 = $2,543,000. E 24 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 43 A Minimum-Cost Schedule STAGE 2 Step 1. The critical path is still BDHJK. 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 ACGJK: 64 weeks and BDHJK: 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.

Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 44 A Minimum-Cost Schedule STAGE 2 I Step 1. The critical path is still BDHJK. 15 Step 2. The cheapest activity to crash per week is now D at $2,000. A 12 weeks. F 10 first K 6 reduction Step 3. Crash D by two The week of 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 C G Start Finish saves only $8,000 in indirect costs because, after week 65, no 10 35 more penalty costs are incurred. These savings still exceed the cost of crashing D by two weeks. Updated path times are ACGJK: 64 B weeks 9

and D H BDHJK: 8 40 64 J weeks 1 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. E 24 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 45 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. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 46 A Minimum-Cost Schedule STAGE 3 Step 3. We crash activity K to the greatest extent possiblea reduction of one weekbecause it is on both critical paths. Updated path times are ACGJK: 63 weeks and BDHJK: 63 weeks Net savings are $8,000 - $4,000 = $4,000. Total project costs are $2,511,000 $4,000 = $2,507,000.

Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 47 A Minimum-Cost Schedule STAGE 3 I Step 3. We crash activity K to the greatest extent possiblea reduction of one weekbecause it is15on both critical paths. Updated path times are A 12 F 10 K 5 ACGJK: 63 weeks and BDHJK: 63 weeks Start C 10 G 35 D 8 H 40 Finish Net savings are $8,000 - $4,000 = $4,000. Total project costs are $2,511,000 $4,000 = $2,507,000. B 9 J 1 E 24 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.

2 48 A Minimum-Cost Schedule STAGE 4 Step 1. The critical paths are still BDHJK and ACGJK. 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 ACGJK: 61 weeks and BDHJK: 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. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 49 A Minimum-Cost Schedule STAGE 4 I Step 1. The critical paths are still BDHJK and ACGJK. 15 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. A F K This amount is still less than the savings of $8,000 per 12 10 5 week. Step 3. Crash activities B and C by two weeks, the limit for activity B. Updated path times are Start ACGJK: 61 weeks and C G 8 35

BDHJK: 61 weeksFinish The net savings are 2($8,000) 2($7,600) = $800. Total project B D H J costs are now $2,507,000 $800 = $2,506,200. 7 8 40 1 E 24 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 50 A Minimum-Cost Schedule Stage Crash Activity Time Reduction (weeks) Resulting Critical Path(s) Project Duration (weeks) 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Project

Direct Costs, Last Trial ($000) Crash Cost Added ($000) Total Indirect Costs ($000) Total Penalty Costs ($000) Total Project Costs ($000) 2 51 Application 2.3 Indirect project costs = $250 per day and penalty cost = $100 per day for each day the project lasts beyond day 14. Project Activity and Cost Data Activity A Normal Time Normal (days) Cost ($) 5 1,000 Crash Time Crash (days) Cost ($) 4 1,200 Immediate Predecessor(s)

B 5 800 3 2,000 C 2 600 1 900 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. A, B 2 52 Application 2.3 Direct cost and time data for the activities: 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 Solution: Original costs: Normal Total Costs = $7,500 Total Indirect Costs = $250 per day 21 days = $5,250 Penalty Cost = $100 per day 7 days = $700 Total Project Costs = $13,450

Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 53 Application 2.3 Step 1: The critical path is BDEGH , 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: Costs Last Trial = $7,500 Crash Cost Added = $150 2 days = $300 Total Indirect Costs = $250 per day 19 days = $4,750 Penalty Cost = $100 per day 5 days = $500 Total Project Cost = $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). Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 54 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 Note that the cost to crash ($200 per day) is less than the combined indirect cost and the penalty cost per day savings ($350). Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 55 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. Note that the cost to crash ($500 per day) is greater than the combined indirect cost and the penalty cost per day savings ($350). Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 56 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. Trial Crash Activity Resulting Critical Paths Reduction (days) Project Duration (days) 0 B-D-E-G-H 21 1 E B-D-E-G-H 2 2

H B-D-E-G-H 2 Costs Last Trial Crash Cost Added Total Indirect Costs Total Penalty Costs Total Project Costs $7,500 $5,250 $700 $13,450 19 $7,500 $300 $4,750 $500 $13,050 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. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 57 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 Project risk can be assessed by Strategic fit Service/product attributes Project team capabilities Operations Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 58 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) Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 59 Statistical Analysis Area under curve between a and b is 99.74% a m Mean b Time Beta distribution a Figure 2.7 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 3 m Mean Time 3 b Normal distribution 2 60 Statistical Analysis The mean of the beta distribution can be estimated by te = a + 4m + b

6 The variance of the beta distribution for each activity is = 2 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. ba 2 6 2 61 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. Johns 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. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 62 Calculating Means and Variances SOLUTION a. The expected time for activity B is te = 7 + 4(8) + 15 6 = 54 6 = 9 weeks 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 2 = 15 7 6 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 = 8 6 2 = 1.78 2 63 Calculating Means and Variances b. The following table shows the expected activity times and variances for this project. Time Estimates (week) Activity Statistics Optimistic (a) Most Likely (m) Pessimistic (b) Expected Time (te) 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 Activity Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Variance (2) 2 64 Application 2.4 Bluebird University: activity for sales training seminar Activity Immediate Predecessor(s) Optimistic (a) Most Likely (m) A 5 7 8

B 6 8 12 8.33 1.00 C 3 4 5 4.00 0.11 D A 11 17 25 17.33 5.44 E B 8 10

12 10.00 0.44 F C, E 3 4 5 4.00 0.11 G D 4 8 9 7.50 0.69 H F 5 7 9 7.00 0.44 I

G, H 8 11 17 11.50 2.25 J G 4 4 4 4.00 0.00 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Pessimistic (b) Expected Time (t) 6.83 Variance () 0.25 2 65 Analyzing Probabilities Because the central limit theorem can be applied, the mean of the distribution is the earliest expected finish time for the project TE =

Expected activity times Mean of normal = on the critical path distribution Because the activity times are independent 2 = (Variances of activities on the critical path) Using the z-transformation z= T TE 2 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. where T = due date for the project 2 66 Calculating the Probability EXAMPLE 2.5 Calculate the probability that St. Johns Hospital will become operational in 72 weeks, using (a) the critical path and (b) path ACGJK. SOLUTION a. The critical path BDHJK has a length of 69 weeks. From the table in Example 2.4, we obtain the variance of path BDHJK: 2 = 1.78 + 1.78 + 2.78 + 5.44 + 0.11. Next, we calculate the z-value: z 72 69 3 0.87 11.89 3.45 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 67 Calculating the Probability Using the Normal Distribution appendix, we go down the lefthand column to 0.8 and then across to 0.07. This gives a value

of 0.8078. Thus the probability is about 0.81 that the length of path BDHJK 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. Normal distribution: Mean = 69 weeks; = 3.45 weeks Length of critical path Probability of meeting the schedule is 0.8078 Figure 2.8 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Probability of exceeding 72 weeks is 0.1922 69 72 Project duration (weeks) 2 68 Calculating the Probability EXAMPLE 2.5 Calculate the probability that St. Johns Hospital will become operational in 72 weeks, using (a) the critical path and (b) path ACGJK. SOLUTION b. From the table in Example 2.4, we determine that the sum of the expected activity times on path ACGJK is 67 weeks and that 2 = 0.11 + 2.78 + 7.11 + 5.44 + 0.11 = 15.55. The z-value is z 72 67 5 1.27 15.55 3.94

The probability is about 0.90 that the length of path ACGJK will be no greater than 72 weeks. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 69 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? T = 47 days TE is: 43.17 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 ADGI, and the expected completion time is 43.17 days. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 70 Application 2.5 T = 47 days TE = 43.17 days And the sum of the variances for the critical activities is: 8.63 z= T TE 2 = 47 43.17 = 8.63 3.83 = 1.30 2.94 Assuming the normal distribution applies, we use the table for the normal probability distribution. Given z = 1.30, the probability that activities ADGI can be completed in 47 days or less is 0.9032. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.

2 71 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 nearcritical paths delaying the project Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 72 Monitoring and Controlling Projects Tracking systems collect information on three topics Open issues that require resolution Risks that might delay the project completion Schedule status periodically monitors slack time to identify activities that are behind schedule Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 73 Resource requirements Project Life Cycle Definition and organization Planning Start Figure 2.9 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Execution

Time Close out Finish 2 74 Monitoring and Controlling Projects Problems can be alleviated through Resource leveling Resource allocation Resource acquisition Project close out includes writing final reports, completing remaining deliverables, and compiling the teams recommendations for improving the project process Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 75 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? Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 76 Solved Problem 1 TABLE 2.2

| ELECTRIC MOTOR PROJECT DATA Activity Normal Time (days) Normal Cost ($) 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Crash Time (days) Crash Cost ($) Immediate Predecessor(s) 2 77 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. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. A 4 D 6 Finish G 4

Start B 7 E 3 C 5 F 11 H 3 Figure 2.10 2 78 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 Normal cost CC NC Crash cost = = per day Normal time Crash time NT CT = $1,300 $1,000 4 days 3 days Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. = $300 2 79 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 80 Solved Problem 1 TABLE 2.3 Stage | PROJECT COST ANALYSIS Crash Activity Time Reduction (days) 0 1 H 2 F Resulting Critical Path(s) Project Duration (days) Project Direct Costs, Last Trial ($)

Crash Cost Added ($) Total Indirect Costs ($) Total Penalty Costs ($) Total Project Costs ($) C-F-H 19 10,100 3,800 700 14,600 2 C-F-H 17 10,100 200 3,400 500 14,200

2 A-D-G-H 15 10,300 500 3,000 300 14,100 B-E-G-H C-F-H Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 81 Solved Problem 1 Table 2.3 summarizes the analysis and the resultant project duration and total cost. The critical path is CFH 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 ADGH: 15 days, BEGH: 15 days, and CFH: 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. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 82 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? Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Finish D A Start E C G B F Figure 2.11 2 83 Solved Problem 2 Time Estimate (weeks) Activity Optimistic Most Likely 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 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Pessimistic Immediate Predecessor(s) 2 84 Solved Problem 2 SOLUTION a. The expected time and variance for each activity are calculated as follows te = a + 4m + b 6 Activity Expected Time (weeks) Variance A B C D E F G 4.0 5.5 3.5 12.0 6.5 9.0 4.5 1.00 0.69 0.25 1.78 2.25 2.78 0.69

Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 85 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. Activity Earliest Start (weeks) A B C D E F G 0 0 5.5 4.0 9.0 5.5 15.5 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Earliest Finish (weeks) 0 + 4.0 = 0 + 5.5 = 5.5 + 3.5 = 4.0 + 12.0 = 9.0 + 6.5 = 5.5 + 9.0 = 15.5 + 4.5 = 4.0 5.5 9.0 16.0 15.5

14.5 20.0 2 86 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 Activity Latest Start (weeks) G F E D C B A 15.5 6.5 9.0 8.0 5.5 0.0 4.0 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Latest Finish (weeks) 20.0 15.5 15.5 20.0 9.0 5.5 8.0 2 87 Solved Problem 2 4.0 D 16.0

Finish 8.0 12.0 20.0 0.0 A 4.0 9.0 4.0 4.0 8.0 9.0 E 15.5 6.5 15.5 C Start 5.5 5.5 9.0 3.5 9.0 G 0.0 B 5.5 15.5 0.0 5.5

5.5 15.5 4.5 20.0 20.0 F Figure 2.12 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 5.5 14.5 6.5 9.0 15.5 2 88 Solved Problem 2 Start (weeks) Activity Earliest Finish (weeks) 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 AD AEG BCEG BFG Total Expected Time (weeks) Total Variance 4 + 12 = 16

1.00 + 1.78 = 2.78 4 + 6.5 + 4.5 = 15 1.00 + 2.25 + 0.69 = 3.94 5.5 + 3.5 + 6.5 + 4.5 = 20 0.69 + 0.25 + 2.25 + 0.69 = 3.88 5.5 + 9 + 4.5 = 19 0.69 + 2.78 + 0.69 = 4.16 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 89 Solved Problem 2 So the critical path is BCEG with a total expected time of 20 weeks. However, path BFG is 19 weeks and has a large variance. c. We first calculate the z-value: z= T TE 2 = 23 20 3.88 = 1.52 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 BFG is close to that of the critical path and has a large variance, it might well become the critical path during the project Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 90

Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 91

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