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2 1Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Project Management2
For Operations Management, 9e by Krajewski/Ritzman/Malhotra 2010 Pearson Education
PowerPoint Slides by Jeff Heyl
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2 2Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Projects
Projects are an interrelated set of activities with a definite starting and ending point, which results in a unique outcome from a specific allocation of resources
Projects are common in everyday life The three main goals are to:
Complete the project on timeNot exceed the budgetMeet the specifications to the satisfactions of
the customer
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2 3Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 temporaryA collection of projects is called a program
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2 4Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Defining and Organizing Projects
Define the scope, time frame, and resources of the project
Select the project manager and teamGood project managers must be
FacilitatorsCommunicatorsDecision makers
Project team members must have Technical competenceSensitivityDedication
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Organizational Structure
Different structures have different implications for project management
Common structures are FunctionalPure projectMatrix
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Planning Projects
There are five steps to planning projects1. Defining the work breakdown structure
2. Diagramming the network
3. Developing the schedule
4. Analyzing the cost-time trade-offs
5. Assessing risks
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2 7Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Work Breakdown Structure
A statement of all the tasks that must be completed as part of the project
An activity is the smallest unit of work effort consuming both time and resources that the project manager can schedule and control
Each activity must have an owner who is responsible for doing the work
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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 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. Johns Hospital
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2 9Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Diagramming the Network
Network diagrams use nodes and arcs to depict the relationships between activities
Benefits of using networks 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
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2 10Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Diagramming the Network
Precedent relationships determine the sequence for undertaking activities
Activity times must be estimated using historical information, statistical analysis, learning curves, or informed estimates
In the activity-on-node approach, nodes represent activities and arcs represent the relationships between activities
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2 11Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
S T U S precedes T, which precedes U.
Diagramming the Network
AON Activity Relationships
S
T
US and T must be completed before U can be started.
Figure 2.2
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Diagramming the Network
AON Activity Relationships
T
U
ST and U cannot begin until S has been completed.
S
T
U
V
U and V cant begin until both S and T have been completed.
Figure 2.2
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2 13Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Diagramming 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
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2 14Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Developing 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
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2 15Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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
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2 16Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
St. Johns Hospital ProjectActivity Immediate
PredecessorsActivity Times (wks)
Responsibility
ST. JOHNS HOSPITAL PROJECT
START
ORGANIZING and SITE PREPARATION
A. Select administrative staff
B. Select site and survey
C. Select medical equipment
D. Prepare final construction plans
E. Bring utilities to site
F. Interview applicants for nursing and support staff
PHYSICAL FACILITIES and INFRASTRUCTURE
G. Purchase and deliver equipment
H. Construct hospital
I. Develop information system
J. Install medical equipment
K. Train nurses and support staff
FINISH
Example 2.1
START
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
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2 17Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Activity Immediate Predecessors
Activity Times (wks)
Responsibility
ST. JOHNS 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. Johns Hospital Project
Example 2.1
Completion Time
Finish
K6
I15
F10
C10
H40
J4
A12
B9
Figure 2.3
Start G35
D10
E24
Activity IP TimeA START 12B START 9C A 10D B 10E B 24F A 10G C 35H D 40I A 15
J E, G, H 4K F, I, J 6
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2 18Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Activity Immediate Predecessors
Activity Times (wks)
Responsibility
ST. JOHNS 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. Johns Hospital Project
Example 2.1
Completion Time
Finish
K6
I15
F10
C10
D10
H40
J4
A12
B9
Figure 2.3
Start G35
E24
Path Estimated Time (weeks)
AIK 33
AFK 28
ACGJK 67
BDHJK 69
BEJK 43
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2 19Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Activity Immediate Predecessors
Activity Times (wks)
Responsibility
ST. JOHNS 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. Johns Hospital Project
Example 2.1
Completion Time
Finish
K6
I15
F10
C10
D10
H40
J4
A12
B9
Figure 2.3
Start G35
E24
Path Estimated Time (weeks)
AIK 33
AFK 28
ACGJK 67
BDHJK 69
BEJK 43
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2 20Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.1
The following information is known about a project
Draw the network diagram for this project
Activity Activity Time (days)Immediate
Predecessor(s)A 7 B 2 AC 4 AD 4 B, CE 4 DF 3 EG 5 E
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Finish
G5
F3
E4
D4
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
B2
C4
Start A7
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2 22Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Project Schedule
The project schedule specifies start and finish times for each activity
Managers can use the earliest start and finish times, the latest start and finish times, or any time in between these extremes
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2 23Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Project Schedule
The earliest start time (ES) for an activity is the latest earliest finish time of any preceding activities
The earliest finish time (EF) is the earliest start time plus its estimated duration
EF = ES + t
The latest finish time (LF) for an activity is the latest start time of any preceding activities
The latest start time (LS) is the latest finish time minus its estimated duration
LS = LF t
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2 24Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish Times
EXAMPLE 2.2
Calculate the ES, EF, LS, and LF times for each activity in the hospital project. Which activity should Kramer start immediately? Figure 2.3 contains the activity times.
SOLUTION
To compute the early start and early finish times, we begin at the start node at time zero. Because activities A and B have no predecessors, the earliest start times for these activities are also zero. The earliest finish times for these activities are
EFA = 0 + 12 = 12 and EFB = 0 + 9 = 9
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2 25Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish Times
Because the earliest start time for activities I, F, and C is the earliest finish time of activity A,
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
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2 26Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish Times
The earliest start time for activity G is the latest EF time of all immediately preceding activities. Thus,
ESG = EFC = 22, ESH = EFD = 19
EFG = ESG + t = 22 + 35 = 57, EFH + t = 19 + 40 = 59
Latest finish timeLatest start time
Activity
Duration
Earliest start time Earliest finish time
0
2
12
14
A
12
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2 27Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Network 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
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2 28Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish Times
To compute the latest start and latest finish times, we begin by setting the latest finish activity time of activity K at week 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
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2 29Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish Times
The latest start times for these activities are shown in Figure 2.4 as
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
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2 30Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Early Start and Early Finish Times
Activity A has more than one immediately following activity: I, F, and C. The earliest of the latest start times is 14 for activity C. Thus,
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
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2 31Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Network 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 = 0
S = 0
S = 0S = 0S = 0
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2 32Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Gantt Chart
Figure 2.5
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2 33Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Activity Slack
Activity slack is the maximum length of time an activity can be delayed without delaying the entire project
Activities on the critical path have zero slack
Activity slack can be calculated in two ways
S = LS ES or S = LF EF
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2 34Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.2
Calculate the four times for each activity in order to determine the critical path and project duration.
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 YesB 2C 4D 4E 4F 3G 5
The critical path is ACDEG with a project duration of 24 days
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2 35Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.2
Calculate the four times for each activity in order to determine the critical path and project duration.
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 YesB 2C 4D 4E 4F 3G 5
The critical path is ACDEG with a project duration of 24 days
7 9 9 11 9-7=2 No7 7 11 11 7-7=0 Yes
19 21 22 24 21-19=2 No19 19 24 24 19-19=0 Yes
11 11 15 15 11-11=0 Yes15 15 19 19 15-15=0 Yes
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2 36Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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
Start FinishA7
B2
C4
D4
E4
F3
G5
The critical path is ACDEG with a project duration of 24 days
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2 37Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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
Start FinishA7
B2
C4
D4
E4
F3
G5
The critical path is ACDEG with a project duration of 24 days
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2 38Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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
Cost to crash per period =CC NCNT CT
1. Normal time (NT) 3. Crash time (CT)2. Normal cost (NC) 4. Crash cost (CC)
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2 39Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Cost-Time Relationships
Linear cost assumption
8000
7000
6000
5000
4000
3000
0
D
i
r
e
c
t
c
o
s
t
(
d
o
l
l
a
r
s
)
| | | | | |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
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2 40Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Cost-Time Relationships
TABLE 2.1 | DIRECT COST AND TIME DATA FOR THE ST. JOHNS 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
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2 41Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 33 weeksAFK 28 weeksACGJK 67 weeksBDHJK 69 weeksBEJK 43 weeks
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2 42Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
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2 43Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
Finish
K6
I15
F10
C10
D10
H40
J1
A12
B9
Start G35
E24
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2 44Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
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2 45Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
Finish
K6
I15
F10
C10
D8
H40
J1
A12
B9
Start G35
E24
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2 46Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
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2 47Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
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2 48Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
Finish
K5
I15
F10
C10
D8
H40
J1
A12
B9
Start G35
E24
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2 49Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
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2 50Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
Finish
K5
I15
F10
C8
D8
H40
J1
A12
B7
Start G35
E24
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2 51Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
A 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
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2 52Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3
Indirect project costs = $250 per day and penalty cost = $100 per day for each day the project lasts beyond day 14.
Project Activity and Cost Data
ActivityNormal Time (days)
Normal Cost ($)
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, BD 3 1,500 2 2,000 BE 5 900 3 1,200 C, DF 2 1,300 1 1,400 EG 3 900 3 900 EH 5 500 3 900 G
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2 53Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3Direct 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
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2 54Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.3
Step 1: The critical path is , and the project duration is
BDEGH21 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).
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2 55Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 = 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).
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2 56Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 = 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).
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2 57Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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.
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.
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2 58Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Assessing Risk
Risk is the measure of the probability and consequence of not reaching a defined project goal
Risk-management plans are developed to identify key risks and prescribe ways to circumvent them
Project risk can be assessed byStrategic fitService/product attributesProject team capabilitiesOperations
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2 59Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Simulation and Statistical Analysis
When uncertainty is present, simulation can be used to estimate the project completion time
Statistical analysis requires three reasonable estimates of activity times
1. Optimistic time (a)2. Most likely time (m)3. Pessimistic time (b)
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2 60Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Statistical Analysis
a m bMeanTime
Beta distribution
a m bMeanTime
3 3
Area under curve between a and bis 99.74%
Normal distributionFigure 2.7
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2 61Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 =b a
6
2
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2 62Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating Means and Variances
EXAMPLE 2.4
Suppose that the project team has arrived at the following time estimates for activity B (site selection and survey) of the St. 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.
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2 63Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating Means and Variances
SOLUTION
a. The expected time for activity B is
Note that the expected time does not equal the most likely time. These will only be the same only when the most likely time is equidistant from the optimistic and pessimistic times.
The variance for activity B is
te = = = 9 weeks7 + 4(8) + 15
6546
2 = = = 1.7815 7
6
286
2
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2 64Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating Means and Variances
b. The following table shows the expected activity times and variances for this project.
Time Estimates (week) Activity Statistics
Activity Optimistic (a) Most Likely (m) Pessimistic (b) Expected Time (te) Variance (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
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2 65Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
ActivityImmediate
Predecessor(s)Optimistic
(a)
Most Likely
(m)Pessimistic
(b)Expected Time (t)
Variance ()
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 2.4
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
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2 66Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Analyzing Probabilities
Because the central limit theorem can be applied, the mean of the distribution is the earliest expected finish time for the project
TE = =Expected activity times
on the critical path Mean of normal distribution Because the activity times are independent
2 = (Variances of activities on the critical path)
z =T TE
2
Using the z-transformationwhere
T = due date for the project
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2 67Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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:
0.873.45
311.89
6972 z
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2 68Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating the Probability
Using the Normal Distribution appendix, we go down the left-hand column to 0.8 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.
Length of critical path
Probability of meeting the schedule is 0.8078
Normal distribution:Mean = 69 weeks; = 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
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2 69Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Calculating the Probability
SOLUTION
b. From the table in Example 2.4, we determine that the sum of the expected activity times on path ACGJK is 67 weeks and that 2 = 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 ACGJK will be no greater than 72 weeks.
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.
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2 70Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Application 2.5
The director of the continuing education at Bluebird University wants to conduct the seminar in 47 working days from now. What is the probability that everything will be ready in time?
The critical path is
and the expected completion time is
T =
TE is:
ADGI, 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:
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2 71Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
= = = 1.303.832.94
47 43.178.63
Application 2.5
T = 47 daysTE = 43.17 daysAnd the sum of the variances for the critical activities is: 8.63
z = T TE
2
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.
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2 72Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Near-Critical Paths
Project duration is a function of the critical path
Since activity times vary, paths with nearly the same length can become critical during the project
Project managers can use probability estimates to analyze the chances of near-critical paths delaying the project
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2 73Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Monitoring and Controlling Projects
Tracking systems collect information on three topicsOpen issues that require resolutionRisks that might delay the project completionSchedule status periodically monitors slack
time to identify activities that are behind schedule
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2 74Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Project Life Cycle
Start Finish
R
e
s
o
u
r
c
e
r
e
q
u
i
r
e
m
e
n
t
s
Time
Definition and
organization
Planning Execution Close out
Figure 2.9
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2 75Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Monitoring and Controlling Projects
Problems can be alleviated throughResource levelingResource allocationResource acquisition
Project close out includes writing final reports, completing remaining deliverables, and compiling the teams recommendations for improving the project process
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2 76Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1
Your company has just received an order from a good customer for a specially designed electric motor. The contract states that, starting on the thirteenth day from now, your firm will experience a penalty of $100 per day until the job is completed. Indirect project costs amount to $200 per day. The data on direct costs and activity precedent relationships are given in Table 2.2.
a. Draw the project network diagram.
b. What completion date would you recommend?
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2 77Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1
TABLE 2.2 | ELECTRIC MOTOR PROJECT DATA
Activity Normal Time (days)
Normal Cost ($)
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
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2 78Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1SOLUTION
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
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2 79Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1
b. With these activity times, the project will be completed in 19 days and incur a $700 penalty. Using the data in Table 2.2, you can determine the maximum crash-time reduction and crash cost per day for each activity. For activity A
Maximum crash time = Normal time Crash time =4 days 3 days = 1 day
Crash cost per day = =
Crash cost Normal costNormal time Crash time
CC NCNT CT
= = $300$1,300 $1,0004 days 3 days
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2 80Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1
Activity Crash Cost per Day ($) 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
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2 81Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved 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
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2 82Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 1
Table 2.3 summarizes the analysis and the resultant project duration and total cost. The critical path is 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.
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2 83Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2
An advertising project manager developed the network diagram in Figure 2.11 for a new advertising campaign. In addition, the manager gathered the time information for each activity, as shown in the accompanying table.
a. Calculate the expected time and variance for each activity.
b. Calculate the activity slacks and determine the critical path, using the expected activity times.
c. What is the probability of completing the project within 23 weeks?
Figure 2.11
Start
Finish
A
B
C
D
E
F
G
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2 84Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved 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
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2 85Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2SOLUTION
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 1.00B 5.5 0.69C 3.5 0.25D 12.0 1.78E 6.5 2.25F 9.0 2.78G 4.5 0.69
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2 86Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2SOLUTION
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 = 4.0B 0 0 + 5.5 = 5.5C 5.5 5.5 + 3.5 = 9.0D 4.0 4.0 + 12.0 = 16.0E 9.0 9.0 + 6.5 = 15.5F 5.5 5.5 + 9.0 = 14.5G 15.5 15.5 + 4.5 = 20.0
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2 87Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 2
Based on expected times, the earliest finish date for the project is week 20, when activity G has been completed. Using that as a target date, we can work backward through the network, calculating the latest start and finish times
Activity Latest Start (weeks) Latest Finish (weeks)
G 15.5 20.0F 6.5 15.5E 9.0 15.5D 8.0 20.0C 5.5 9.0B 0.0 5.5A 4.0 8.0
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2 88Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved 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
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2 89Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Solved 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
AD 4 + 12 = 16 1.00 + 1.78 = 2.78
AEG 4 + 6.5 + 4.5 = 15 1.00 + 2.25 + 0.69 = 3.94
BCEG 5.5 + 3.5 + 6.5 + 4.5 = 20 0.69 + 0.25 + 2.25 + 0.69 = 3.88
BFG 5.5 + 9 + 4.5 = 19 0.69 + 2.78 + 0.69 = 4.16
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2 90Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 = = = 1.52T TE
223 20
3.88Using 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