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2 – 1Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Project ManagementProject Management2
For For Operations Management, 9eOperations Management, 9e by by Krajewski/Ritzman/Malhotra Krajewski/Ritzman/Malhotra © 2010 Pearson Education© 2010 Pearson Education
PowerPoint Slides PowerPoint Slides by Jeff Heylby Jeff Heyl
2 – 2Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
ProjectsProjects
Projects are an interrelated set of activities with a definite starting and ending point, which results in a unique outcome from a specific allocation of resources
Projects are common in everyday life
The three main goals are to: Complete the project on time Not exceed the budget Meet the specifications to the satisfactions of
the customer
2 – 3Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
ProjectsProjects
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
2 – 4Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Defining and Organizing ProjectsDefining and Organizing Projects
Define the scope, time frame, and resources of the project
Select the project manager and teamGood project managers must be
Facilitators Communicators Decision makers
Project team members must have Technical competence Sensitivity Dedication
2 – 5Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Organizational StructureOrganizational Structure
Different structures have different implications for project management
Common structures are Functional Pure project Matrix
2 – 6Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Planning ProjectsPlanning 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
2 – 7Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Work Breakdown StructureWork 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
2 – 8Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Purchase and deliver equipment
Construct hospital
Develop information system
Install medical equipment
Train nurses and support staff
Work Breakdown StructureWork Breakdown Structure
Figure 2.1
Select administration staff
Site selection and survey
Select medical equipment
Prepare final construction plans
Bring utilities to site
Interview applicants fornursing and support staff
Organizing and Site Preparation Physical Facilities and Infrastructure Level 1
Level 0
Level 2
Relocation of St. John’s Hospital
2 – 9Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Diagramming the NetworkDiagramming the Network
Network diagrams use nodes and arcs to depict the relationships between activities
Benefits of using networks include1. 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
2 – 10Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Diagramming the NetworkDiagramming 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
2 – 11Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
S T U S precedes T, which precedes U.
Diagramming the NetworkDiagramming the Network
AON Activity Relationships
S
T
US and T must be completed before U can be started.
Figure 2.2
2 – 12Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Diagramming the NetworkDiagramming the Network
AON Activity Relationships
T
U
ST 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
2 – 13Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Diagramming the NetworkDiagramming the Network
AON Activity Relationships
S
T
U
V
U cannot begin until both S and T have been completed; V cannot begin until T has been completed.
S T V
U
T and U cannot begin until S has been completed and V cannot begin until both T and U have been completed.
Figure 2.2
2 – 14Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Developing the ScheduleDeveloping the Schedule
Schedules can help managers achieve the objectives of the project
Managers can1. 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
2 – 15Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Critical PathCritical 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
2 – 16Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
St. John’s Hospital ProjectSt. John’s Hospital ProjectActivity Immediate
PredecessorsActivity 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
FINISHExample 2.1
START
START
A
B
B
A
C
D
A
E, G, H
F, I, J
K
0
12
9
10
10
24
10
35
40
15
4
6
0
Kramer
Stewart
Johnson
Taylor
Adams
Taylor
Burton
Johnson
Walker
Sampson
Casey
Murphy
Pike
Ashton
2 – 17Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Activity Immediate Predecessors
Activity Times (wks)
Responsibility
ST. JOHN’S HOSPITAL PROJECT
START
ORGANIZING and SITE PREPARATION
A. Select administrative staff
B. Select site and survey
C. Select medical equipment
D. Prepare final construction plans
E. Bring utilities to site
F. Interview applicants for nursing and support staff
PHYSICAL FACILITIES and INFRASTRUCTURE
G. Purchase and deliver equipment
H. Construct hospital
I. Develop information system
J. Install medical equipment
K. Train nurses and support staff
FINISH
St. John’s Hospital ProjectSt. John’s Hospital Project
Example 2.1
Completion Time
Finish
K6
I15
F10
C10
H40
J4
A12
B9
Figure 2.3
StartG
35
D10
E24
Activity IP Time
A START 12B START 9C A 10D B 10E B 24F A 10G C 35H D 40I A 15J E, G, H 4K F, I, J 6
2 – 18Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Activity Immediate Predecessors
Activity Times (wks)
Responsibility
ST. JOHN’S HOSPITAL PROJECT
START
ORGANIZING and SITE PREPARATION
A. Select administrative staff
B. Select site and survey
C. Select medical equipment
D. Prepare final construction plans
E. Bring utilities to site
F. Interview applicants for nursing and support staff
PHYSICAL FACILITIES and INFRASTRUCTURE
G. Purchase and deliver equipment
H. Construct hospital
I. Develop information system
J. Install medical equipment
K. Train nurses and support staff
FINISH
St. John’s Hospital ProjectSt. John’s Hospital Project
Example 2.1
Completion Time
Finish
K6
I15
F10
C10
D10
H40
J4
A12
B9
Figure 2.3
StartG
35
E24
Path Estimated Time (weeks)
A–I–K 33
A–F–K 28
A–C–G–J–K 67
B–D–H–J–K 69
B–E–J–K 43
2 – 19Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Activity Immediate Predecessors
Activity Times (wks)
Responsibility
ST. JOHN’S HOSPITAL PROJECT
START
ORGANIZING and SITE PREPARATION
A. Select administrative staff
B. Select site and survey
C. Select medical equipment
D. Prepare final construction plans
E. Bring utilities to site
F. Interview applicants for nursing and support staff
PHYSICAL FACILITIES and INFRASTRUCTURE
G. Purchase and deliver equipment
H. Construct hospital
I. Develop information system
J. Install medical equipment
K. Train nurses and support staff
FINISH
St. John’s Hospital ProjectSt. John’s Hospital Project
Example 2.1
Completion Time
Finish
K6
I15
F10
C10
D10
H40
J4
A12
B9
Figure 2.3
StartG
35
E24
Path Estimated Time (weeks)
A–I–K 33
A–F–K 28
A–C–G–J–K 67
B–D–H–J–K 69
B–E–J–K 43
2 – 20Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.1Application 2.1
The following information is known about a project
Draw the network diagram for this project
Activity Activity Time (days)Immediate
Predecessor(s)
A 7 —
B 2 A
C 4 A
D 4 B, C
E 4 D
F 3 E
G 5 E
2 – 21Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Finish
G5
F3
E4
D4
Application 2.1Application 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
B2
C4
Start A7
2 – 22Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Project ScheduleProject 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
2 – 23Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Project ScheduleProject 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
2 – 24Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish TimesEarly 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
2 – 25Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish TimesEarly 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
2 – 26Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish TimesEarly 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
Latest finish time
Latest start time
Activity
Duration
Earliest start timeEarliest finish time
0
2
12
14
A
12
2 – 27Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Network DiagramNetwork Diagram
K
6
C
10
G
35
J
4
H
40
B
9
D
10
E
24
I
15
FinishStart
A
12
F
10
0 9
9 33
9 19 19 59
22 5712 22
59 63
12 27
12 22 63 690 12
Figure 2.4
2 – 28Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish TimesEarly 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
2 – 29Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish TimesEarly 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
2 – 30Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish TimesEarly 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
2 – 31Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Network DiagramNetwork Diagram
Figure 2.4
K
6
C
10
G
35
J
4
H
40
B
9
D
10
E
24
I
15
FinishStart
A
12
F
10
0 9
9 33
9 19 19 59
22 5712 22
59 63
12 27
12 22 63 690 12
48 63
53 63
59 63
24 59
19 59
35 59
14 24
9 19
2 14
0 9
63 69
S = 36
S = 2 S = 41
S = 2
S = 26
S = 2
S = 0S = 0
S = 0S = 0
S = 0S = 0S = 0S = 0S = 0S = 0
2 – 32Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Gantt ChartGantt Chart
Figure 2.5
2 – 33Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Activity SlackActivity 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
2 – 34Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.2Application 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
The critical path is A–C–D–E–G with a project duration of 24 days
2 – 35Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.2Application 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
The critical path is A–C–D–E–G with a project duration of 24 days
7 9 9 11 9-7=2 No
7 7 11 11 7-7=0 Yes
19 21 22 24 21-19=2 No
19 19 24 24 19-19=0 Yes
11 11 15 15 11-11=0 Yes
15 15 19 19 15-15=0 Yes
2 – 36Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.2Application 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
Start FinishA7
B2
C4
D4
E4
F3
G5
The critical path is A–C–D–E–G with a project duration of 24 days
2 – 37Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.2Application 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
Start FinishA7
B2
C4
D4
E4
F3
G5
The critical path is A–C–D–E–G with a project duration of 24 days
2 – 38Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Cost-Time Trade-OffsCost-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
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)
2 – 39Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Cost-Time RelationshipsCost-Time Relationships
Linear cost assumption
8000 —
7000 —
6000 —
5000 —
4000 —
3000 —
0 —
Dir
ect
cost
(d
oll
ars)
| | | | | |5 6 7 8 9 10 11
Time (weeks)
Crash cost (CC)
Normal cost (NC)
(Crash time) (Normal time)
Estimated costs for a 2-week reduction, from 10 weeks to 8 weeks
5200
Figure 2.6
2 – 40Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Cost-Time RelationshipsCost-Time Relationships
TABLE 2.1 | DIRECT COST AND TIME DATA FOR THE ST. JOHN’S HOSPITAL PROJECT
Activity Normal Time (NT) (weeks)
Normal Cost (NC)($)
Crash Time (CT)(weeks)
Crash Cost (CC)($)
Maximum Time Reduction (week)
Cost of Crashing per Week ($)
A 12 $12,000 11 $13,000 1 1,000
B 9 50,000 7 64,000 2 7,000
C 10 4,000 5 7,000 5 600
D 10 16,000 8 20,000 2 2,000
E 24 120,000 14 200,000 10 8,000
F 10 10,000 6 16,000 4 1,500
G 35 500,000 25 530,000 10 3,000
H 40 1,200,000 35 1,260,000 5 12,000
I 15 40,000 10 52,500 5 2,500
J 4 10,000 1 13,000 3 1,000
K 6 30,000 5 34,000 1 4,000
Totals $1,992,000 $2,209,500
2 – 41Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
EXAMPLE 2.3
Determine the minimum-cost schedule for the St. John’s Hospital project.
SOLUTION
The projected completion time of the project is 69 weeks. The project costs for that schedule are $1,992,000 in direct costs, 69($8,000) = $552,000 in indirect costs, and (69 – 65)($20,000) = $80,000 in penalty costs, for total project costs of $2,624,000. The five paths in the network have the following normal times:
A–I–K 33 weeks
A–F–K 28 weeks
A–C–G–J–K 67 weeks
B–D–H–J–K 69 weeks
B–E–J–K 43 weeks
2 – 42Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
STAGE 1
Step 1. The critical path is B–D–H–J–K.
Step 2. The cheapest activity to crash per week is J at $1,000, which is much less than the savings in indirect and penalty costs of $28,000 per week.
Step 3. Crash activity J by its limit of three weeks because the critical path remains unchanged. The new expected path times are
A–C–G–J–K: 64 weeks and B–D–H–J–K: 66 weeks
The net savings are 3($28,000) – 3($1,000) = $81,000. The total project costs are now $2,624,000 - $81,000 = $2,543,000.
2 – 43Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
STAGE 1
Step 1. The critical path is B–D–H–J–K.
Step 2. The cheapest activity to crash per week is J at $1,000, which is much less than the savings in indirect and penalty costs of $28,000 per week.
Step 3. Crash activity J by its limit of three weeks because the critical path remains unchanged. The new expected path times are
A–C–G–J–K: 64 weeks and B–D–H–J–K: 66 weeks
The net savings are 3($28,000) – 3($1,000) = $81,000. The total project costs are now $2,624,000 - $81,000 = $2,543,000.
Finish
K6
I15
F10
C10
D10
H40
J11
A12
B9
StartG
35
E24
2 – 44Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
STAGE 2
Step 1. The critical path is still B–D–H–J–K.
Step 2. The cheapest activity to crash per week is now D at $2,000.
Step 3. Crash D by two weeks. The first week of reduction in activity D saves $28,000 because it eliminates a week of penalty costs, as well as indirect costs. Crashing D by a second week saves only $8,000 in indirect costs because, after week 65, no more penalty costs are incurred. These savings still exceed the cost of crashing D by two weeks. Updated path times are
A–C–G–J–K: 64 weeks and B–D–H–J–K: 64 weeks
The net savings are $28,000 + $8,000 – 2($2,000) = $32,000. Total project costs are now $2,543,000 – $32,000 = $2,511,000.
2 – 45Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
STAGE 2
Step 1. The critical path is still B–D–H–J–K.
Step 2. The cheapest activity to crash per week is now D at $2,000.
Step 3. Crash D by two weeks. The first week of reduction in activity D saves $28,000 because it eliminates a week of penalty costs, as well as indirect costs. Crashing D by a second week saves only $8,000 in indirect costs because, after week 65, no more penalty costs are incurred. These savings still exceed the cost of crashing D by two weeks. Updated path times are
A–C–G–J–K: 64 weeks and B–D–H–J–K: 64 weeks
The net savings are $28,000 + $8,000 – 2($2,000) = $32,000. Total project costs are now $2,543,000 – $32,000 = $2,511,000.
Finish
K6
I15
F10
C10
D88
H40
J11
A12
B9
StartG
35
E24
2 – 46Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA 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.
2 – 47Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
STAGE 3
Step 3. We crash activity K to the greatest extent possible—a reduction of one week—because it is on both critical paths. Updated path times are
A–C–G–J–K: 63 weeks and B–D–H–J–K: 63 weeks
Net savings are $8,000 - $4,000 = $4,000. Total project costs are $2,511,000 – $4,000 = $2,507,000.
2 – 48Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
STAGE 3
Step 3. We crash activity K to the greatest extent possible—a reduction of one week—because it is on both critical paths. Updated path times are
A–C–G–J–K: 63 weeks and B–D–H–J–K: 63 weeks
Net savings are $8,000 - $4,000 = $4,000. Total project costs are $2,511,000 – $4,000 = $2,507,000.
Finish
K55
I15
F10
C10
D88
H40
J11
A12
B9
StartG
35
E24
2 – 49Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
STAGE 4
Step 1. The critical paths are still B–D–H–J–K and A–C–G–J–K.
Step 2. The only viable alternative at this stage is to crash activities B and C simultaneously at a cost of $7,600 per week. This amount is still less than the savings of $8,000 per week.
Step 3. Crash activities B and C by two weeks, the limit for activity B. Updated path times are
A–C–G–J–K: 61 weeks and B–D–H–J–K: 61 weeks
The net savings are 2($8,000) – 2($7,600) = $800. Total project costs are now $2,507,000 – $800 = $2,506,200.
2 – 50Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
STAGE 4
Step 1. The critical paths are still B–D–H–J–K and A–C–G–J–K.
Step 2. The only viable alternative at this stage is to crash activities B and C simultaneously at a cost of $7,600 per week. This amount is still less than the savings of $8,000 per week.
Step 3. Crash activities B and C by two weeks, the limit for activity B. Updated path times are
A–C–G–J–K: 61 weeks and B–D–H–J–K: 61 weeks
The net savings are 2($8,000) – 2($7,600) = $800. Total project costs are now $2,507,000 – $800 = $2,506,200.
Finish
K55
I15
F10
C88
D88
H40
J11
A12
B77
StartG
35
E24
2 – 51Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A Minimum-Cost ScheduleA Minimum-Cost Schedule
Stage Crash Activity
Time Reduction (weeks)
Resulting Critical Path(s)
Project Duration (weeks)
Project Direct Costs, Last Trial ($000)
Crash Cost Added ($000)
Total Indirect Costs ($000)
Total Penalty Costs ($000)
Total Project Costs ($000)
0 — — B-D-H-J-K 69 1,992.0 — 552.0 80.0 2,624.0
1 J 3 B-D-H-J-K 66 1,992.0 3.0 528.0 20.0 2,543.0
2 D 2 B-D-H-J-K
A-C-G-J-K
64 1,995.0 4.0 512.0 0.0 2,511.0
3 K 1 B-D-H-J-K
A-C-G-J-K
63 1,999.0 4.0 504.0 0.0 2,507.0
4 B, C 2 B-D-H-J-K
A-C-G-J-K
61 2,003.0 15.2 488.0 0.0 2,506.2
2 – 52Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3Application 2.3
Indirect project costs = $250 per day and penalty cost = $100 per day for each day the project lasts beyond day 14.
Project Activity and Cost Data
ActivityNormal Time (days)
Normal Cost ($)
Crash Time (days)
Crash Cost ($)
Immediate Predecessor(s)
A 5 1,000 4 1,200 —
B 5 800 3 2,000 —
C 2 600 1 900 A, B
D 3 1,500 2 2,000 B
E 5 900 3 1,200 C, D
F 2 1,300 1 1,400 E
G 3 900 3 900 E
H 5 500 3 900 G
2 – 53Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3Application 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 = 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
2 – 54Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3Application 2.3
Step 1: The critical path is , and the project duration is
B–D–E–G–H21 days.
Step 2: Activity E on the critical path has the lowest cost of crashing ($150 per day). Note that activity G cannot be crashed.
Step 3: Reduce the time (crashing 2 days will reduce the project duration to 19 days) and re-calculate costs:
Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost =
$7,500$150 2 days = $300
$250 per day 19 days = $4,750$100 per day 5 days = $500
$13,050
Note that the cost to crash ($250 per day) is less than the combined indirect cost and the penalty cost per day savings ($350).
2 – 55Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3Application 2.3
Step 4: Repeat until direct costs greater than savings
(step 2) Activity H on the critical path has the next lowest cost of crashing ($200 per day).
(step 3) Reduce the time (crashing 2 days will reduce the project duration to 17 days) and re-calculate costs:
Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost =
$7,500 + $300 (the added crash costs) = $7,800$200 2 days = $400
$250 per day 17 days = $4,250$100 per day 3 days = $300
$12,750
Note that the cost to crash ($200 per day) is less than the combined indirect cost and the penalty cost per day savings ($350).
2 – 56Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3Application 2.3
(step 4) Repeat
(step 2) Activity D on the critical path has the next lowest crashing cost ($500 per day).
(step 3) Reduce the time (crashing 1 day will reduce the project duration to 16 days) and re-calculate costs:
Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost =
$7,800 + $400 (the added crash costs) = $8,200$500 1 day = $500
$250 per day 16 days = $4,000$100 per day 2 days = $200
$12,900 which is greater than the last trial. Hence we stop the crashing process.
Note that the cost to crash ($500 per day) is greater than the combined indirect cost and the penalty cost per day savings ($350).
2 – 57Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3Application 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.
TrialCrash 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.
2 – 58Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Assessing RiskAssessing 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
2 – 59Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Simulation and Statistical AnalysisSimulation 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)
2 – 60Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Statistical AnalysisStatistical Analysis
a m bMean
Time
Beta distribution
a m bMeanTime
3σ 3σ
Area under curve between a and b is 99.74%
Normal distributionFigure 2.7
2 – 61Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Statistical AnalysisStatistical 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
σ2 =b – a
6
2
2 – 62Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating Means and VariancesCalculating 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.
2 – 63Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating Means and VariancesCalculating 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
te = = = 9 weeks7 + 4(8) + 15
6
54
6
σ2 = = = 1.7815 – 7
6
28
6
2
2 – 64Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating Means and VariancesCalculating Means and Variances
b. The following table shows the expected activity times and variances for this project.
Time Estimates (week) Activity Statistics
Activity Optimistic (a) Most Likely (m) Pessimistic (b) Expected Time (te) Variance (σ2)
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
2 – 65Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
ActivityImmediate
Predecessor(s)Optimistic
(a)
Most Likely
(m)Pessimistic
(b)Expected Time (t)
Variance (σ)
A — 5 7 8
B — 6 8 12
C — 3 4 5
D A 11 17 25
E B 8 10 12
F C, E 3 4 5
G D 4 8 9
H F 5 7 9
I G, H 8 11 17
J G 4 4 4
Application 2.4Application 2.4
Bluebird University: activity for sales training seminar
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
2 – 66Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Analyzing ProbabilitiesAnalyzing 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
on the critical path Mean of normal distribution
Because the activity times are independent
σ2 = (Variances of activities on the critical path)
z =T – TE
σ2
Using the z-transformation
where
T = due date for the project
2 – 67Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating the ProbabilityCalculating the Probability
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.
SOLUTION
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: σ2 = 1.78 + 1.78 + 2.78 + 5.44 + 0.11. Next, we calculate the z-value:
0.873.45
311.89
6972 z
2 – 68Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating the ProbabilityCalculating the Probability
Using the Normal Distribution appendix, we go down the left-hand 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 B–D–H–J–K will be no greater 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
Project duration (weeks)
69 72
Because this is the critical path, there is a 19 percent probability that the project will take longer than 72 weeks.
Figure 2.8
2 – 69Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating the ProbabilityCalculating 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–K is 67 weeks and that σ2 = 0.11 + 2.78 + 7.11 + 5.44 + 0.11 = 15.55. The z-value is
1.273.94
515.55
6772 z
The probability is about 0.90 that the length of path A–C–G–J–K will be no greater than 72 weeks.
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.
2 – 70Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.5Application 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 =
TE is:
A–D–G–I,
43.17 days.
47 days
43.17 days
(0.25 + 5.44 + 0.69 + 2.25) = 8.63And the sum of the variances for the critical activities is:
2 – 71Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
= = = 1.303.83
2.94
47 – 43.17
8.63
Application 2.5Application 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
σ2
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.
2 – 72Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Near-Critical PathsNear-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
2 – 73Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Monitoring and Controlling ProjectsMonitoring 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
2 – 74Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Project Life CycleProject Life Cycle
Start Finish
Res
ou
rce
req
uir
emen
ts
Time
Definition and
organization
Planning Execution Close out
Figure 2.9
2 – 75Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Monitoring and Controlling ProjectsMonitoring 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 team’s recommendations for improving the project process
2 – 76Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1Solved 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 $200 per day. The data on direct costs and activity precedent relationships are given in Table 2.2.
a. Draw the project network diagram.
b. What completion date would you recommend?
2 – 77Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1Solved Problem 1
TABLE 2.2 | ELECTRIC MOTOR PROJECT DATA
Activity Normal Time (days)
Normal Cost ($)
Crash Time (days)
Crash Cost ($)
Immediate Predecessor(s)
A 4 1,000 3 1,300 None
B 7 1,400 4 2,000 None
C 5 2,000 4 2,700 None
D 6 1,200 5 1,400 A
E 3 900 2 1,100 B
F 11 2,500 6 3,750 C
G 4 800 3 1,450 D, E
H 3 300 1 500 F, G
2 – 78Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1Solved 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 comeup with a correct and uncluttered diagram.
Start
Finish
A4
B7
C5
D6
E3
F11
G4
H3
Figure 2.10
2 – 79Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1Solved 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 costNormal time – Crash time
CC – NCNT – CT
= = $300$1,300 – $1,0004 days – 3 days
2 – 80Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1Solved Problem 1
Activity Crash Cost per Day ($) 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
2 – 81Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1Solved Problem 1
TABLE 2.3 | PROJECT COST ANALYSIS
Stage Crash Activity
Time Reduction (days)
Resulting Critical Path(s)
Project Duration (days)
Project Direct Costs, Last Trial ($)
Crash Cost Added ($)
Total Indirect Costs ($)
Total Penalty Costs ($)
Total Project Costs ($)
0 — — C-F-H 19 10,100 — 3,800 700 14,600
1 H 2 C-F-H 17 10,100 200 3,400 500 14,200
2 F 2 A-D-G-H 15 10,300 500 3,000 300 14,100
B-E-G-H
C-F-H
2 – 82Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1Solved 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.
2 – 83Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2Solved 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
2 – 84Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2Solved Problem 2
Time Estimate (weeks)
Activity Optimistic Most Likely Pessimistic Immediate Predecessor(s)
A 1 4 7 —
B 2 6 7 —
C 3 3 6 B
D 6 13 14 A
E 3 6 12 A, C
F 6 8 16 B
G 1 5 6 E, F
2 – 85Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2Solved 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 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
2 – 86Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2Solved 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) 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
2 – 87Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2Solved 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) 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
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Solved Problem 2Solved Problem 2
A
4.0
0.0
4.0
4.0
8.0
D
12.0
4.0
8.0
16.0
20.0
E
6.5
9.0
9.0
15.5
15.5
G
4.5
15.5
15.5
20.0
20.0
C
3.55.5
5.5
9.0
9.0
F
9.05.5
6.5
14.5
15.5
B
5.5
0.0
0.0
5.5
5.5
Finish
Start
Figure 2.12
2 – 89Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2Solved Problem 2
Start (weeks) Finish (weeks)
Activity Earliest Latest Earliest Latest Slack Critical Path
A 0 4.0 4.0 8.0 4.0 No
B 0 0.0 5.5 5.5 0.0 Yes
C 5.5 5.5 9.0 9.0 0.0 Yes
D 4.0 8.0 16.0 20.0 4.0 No
E 9.0 9.0 15.5 15.5 0.0 Yes
F 5.5 6.5 14.5 15.5 1.0 No
G 15.5 15.5 20.0 20.0 0.0 Yes
Path Total Expected Time (weeks) Total Variance
A–D 4 + 12 = 16 1.00 + 1.78 = 2.78
A–E–G 4 + 6.5 + 4.5 = 15 1.00 + 2.25 + 0.69 = 3.94
B–C–E–G 5.5 + 3.5 + 6.5 + 4.5 = 20 0.69 + 0.25 + 2.25 + 0.69 = 3.88
B–F–G 5.5 + 9 + 4.5 = 19 0.69 + 2.78 + 0.69 = 4.16
2 – 90Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2Solved 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 = = = 1.52T – TE
σ223 – 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
2 – 91Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.