Chapter 8 - Project Management
Chapter Topics
• The Elements of Project Management
• The Project Network
• Probabilistic Activity Times
• Project Crashing and Time-Cost Trade-Off
• Formulating the CPM/PERT Network as a Linear
Programming Model
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Project Management
Overview
• Uses networks for project analysis.
• Networks show how projects are organized and are used to
determine time duration for completion.
• Network techniques used are
- CPM (Critical Path Method)
- PERT (Project Evaluation and Review Technique)
• Developed during late 1950s.
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The Elements of Project Management
• Management is generally perceived as concerned with
planning, organizing, and control of an ongoing process or
activity.
• Project Management is concerned with control of an activity
for a relatively short period of time after which management
effort ends
• Primary elements of Project Management to be discussed:
- Project team
- Project planning
- Project control.
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The Elements of Project Management
The Project team
• Project team typically consists of a group of individuals from
various areas in an organization and often includes outside
consultants.
• Members of engineering staff often assigned to project work.
• Most important member of project team is the project manager.
• Project manager is often under great pressure because of
uncertainty inherent in project activities and possibility of failure.
• Project manager must be able to coordinate various skills of team
members into a single focused effort.
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Project Planning: PERT/CPM
• PERT
– Program Evaluation and Review Technique
– Developed by U.S. Navy for Polaris missile project
– Developed to handle uncertain activity times
• CPM
– Critical Path Method
– Developed by Du Pont & Remington Rand
– Developed for industrial projects for which activity
times generally were known
• Today’s project management software packages have
combined the best features of both approaches.
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PERT/CPM
• PERT and CPM have been used to plan, schedule, and
control a wide variety of projects:
– R&D of new products and processes
– Construction of buildings and highways
– Maintenance of large and complex equipment
– Design and installation of new systems
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PERT/CPM
• Project managers rely on PERT/CPM to help them answer
questions such as:
– What is the total time to complete the project?
– What are the scheduled start and finish dates for each
specific activity?
– Which activities are critical and must be completed exactly
as scheduled to keep the project on schedule?
– How long can noncritical activities be delayed before they
cause an increase in the project completion time?
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PERT Network: Activity-on-Node Approach
• A PERT network can be constructed to model the
precedence of the activities.
• The arcs of the network represent the precedence
relationships of the activities.
• The nodes (rectangles) of the network represent activities.
– You will need to add a “Start” and a “Finish” nodes.
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PERT Network: Activity-on-Node Approach
• Activity time estimates usually can not be made with
certainty.
• In the three-time estimate approach, the time to complete an
activity is assumed to follow a Beta distribution.
• An activity i’s mean completion time is:
ti = (ai + 4mi + bi)/6
• An activity’s completion time variance is:
s2i = ((bi-ai)/6)2
– ai = activitity i’s optimistic completion time estimate
– bi = activitity i’s pessimistic completion time estimate
– mi = activitity i’s most likely completion time estimate
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PERT Network: Activity-on-Node Approach
• In the three-time estimate approach, the critical path is
determined as if the mean times for the activities were
fixed times.
• The expected project time is the sum of the expected times
of the critical path activities.
• The project variance is the sum of the variances of the
critical path activities.
• The expected project time is assumed to be normally
distributed (based on central limit theorem).
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PERT Analysis Algorithm
• Step 1: Make a forward pass through the network as
follows: For each of these activities, i, compute:
– Earliest Start (ES) Time = the maximum of all
earliest finish times for all its immediate
predecessors. (For node “START”, this is 0.)
• ESi= Maximum (EFj) for all immediate proceeding activities j.
– Earliest Finish (EF) Time = (Earliest Start Time) +
(Time to complete activity i).
• EFi= ESi+ ti
The project completion time is the of the Earliest Finish
Times at the “FINISH” node.
– This will also be used as Latest Finish Time at
“FINISH” node in the next step.
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PERT Analysis Algorithm
• Step 2: Make a backwards pass through the network as
follows: Move sequentially backwards from the last
node, “FINISH” to its immediate predecessors, etc. At a
given node, j, consider all activities immediately
following it and compute:
– Latest Finish (LF) Time = the minimum of the latest
start times for all activities that immediately follow j.
(For node “FINISH”, this is the project completion
time.)
• LFj= Minimum (LSi) for all immediate following activities i.
– Latest Start (LS) Time = (Latest Finish Time) - (Time
to complete activity j).
• LSj= LFj - tj
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PERT Analysis Algorithm
• Step 3: Calculate the slack time for each activity by:
Slack = (Latest Start) - (Earliest Start) or
= (Latest Finish) - (Earliest Finish).
A critical path is a path of activities, from node “START”
to “FINISH”, with 0 slack times.
– Shared slack is slack available for a sequence of activities.
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Example: Riverwalk Associates
Riverwalk Associates is in the business of building
elaborate parade floats. Its crew has a new float to build
and want to use PERT/CPM to help them manage the
project .
The table on the next slide shows the activities that
comprise the project. Each activity’s estimated completion
time (in weeks) and immediate predecessors are listed as
well.
The project manager wants to know the total time to
complete the project, which activities are critical, and the
earliest and latest start and finish dates for each activity.
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Example: Riverwalk Associates
• Project activity initial information:
Immed. Optimistic
Most Likely Pessimistic
Activity (i) Predec. Time (weeks) Time (wk.) Time (wk.)
A
—
4
6
8
B
—
1
4.5
5
C
A
3
3
3
D
A
4
5
6
E
A
0.5
1
1.5
F
B,C
3
4
5
G
B,C
1
1.5
5
H
E,F
5
6
7
I
E,F
2
5
8
J
D,H
2.5
2.75
4.5
K
G,I
3
5
7
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Example: Riverwalk Associates
• Activity Expected Time and Variances
ti = (ai + 4mi + bi)/6 s2i = ((bi-ai)/6)2
Activity (i)
Expected Time Variance (week2)
A
6
4/9
B
4
4/9
C
3
0
D
5
1/9
E
1
1/36
F
4
1/9
G
2
4/9
H
6
1/9
I
5
1
J
3
1/9
K
5
4/9
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Example: Riverwalk Associates
• PERT Activity Node Representation
Earliest Start
Earliest Finish
A ES EF
6 LS LF
Expected
Duration of
the activity
Latest Start
Latest Finish
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Example: Riverwalk Associates
• PERT Network Representation
J
3
C
3
A
6
H
6
D
5
START
0
0
I
5
E
1
F
4
B
4
FINISH
K
5
G
2
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Example: Riverwalk Associates
• Earliest/Latest Times
Activity ES EF LS LF Slack
A
0 6
0 6
0 *critical
B
0
4
5
9
5
C
6
9
6
9
0*
D
6 11 15 20
9
E
6 7 12 13
6
F
9 13
9 13
0*
G
9 11 16 18
7
H
13 19 14 20
1
I
13 18 13 18
0*
J
19 22 20 23
1
K
18 23 18 23
0*
• The estimated project completion time is t0 = 23 (weeks) at
FINISH.
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Riverwalk Associates – Linear Programming Form
• Define variables for each activity in the following manner:
ES_i = Earliest Start time for activity i
EF_i = Earliest Finish time for activity i
LS_i = Latest Start time for activity i
LF_i = Latest Finish time for activity i
where i=A, B…K;
and FINISH = the earliest and also latest completion time of
the project.
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Riverwalk Associates – LP Model 1 for Earliest Times
Minimize ES_A + EF_A + ES_B + EF_B +
… + EF_K + FINISH
S.t.
EF_A - ES_A >= 6 (“=“ OK)
EF_B - ES_B >= 4
EF_C - ES_C >= 3
EF_D - ES_D >= 5
EF_E - ES_E >= 1
EF_F - ES_F >= 4
EF_G - ES_G >= 2
EF_H - ES_H >= 6
EF_I - ES_I >= 5
EF_J - ES_J >= 3
EF_K - ES_K >= 5 (“=“ OK)
ES_C - EF_A >= 0 (Not “=“ )
ES_D - EF_A >= 0
ES_E - EF_A >= 0
ES_F - EF_B >= 0
ES_G - EF_B >= 0
ES_F - EF_C >= 0
ES_G - EF_C >= 0
ES_J - EF_D >= 0
ES_J - EF_H >= 0
ES_H - EF_E >= 0
ES_H - EF_F >= 0
ES_I - EF_E >= 0
ES_I - EF_F >= 0
ES_K - EF_G >= 0
ES_K - EF_I >= 0
FINISH - EF_J >= 0
FINISH - EF_K >= 0
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Riverwalk Associates – LP Model 2 for Latest Times
Maximize LS_A + LF_A + LS_B + LF_B +
… + LF_K
S.t.
LF_A - LS_A >= 6 (“=“ OK) LF_B LS_B >= 4
LF_C - LS_C >= 3
LF_D - LS_D >= 5
LF_E - LS_E >= 1
LF_F - LS_F >= 4
LF_G - LS_G >= 2
LF_H - LS_H >= 6
LF_I - LS_I >= 5
LF_J - LS_J >= 3
LF_K - LS_K >= 5 (“=“ OK)
LS_C - LF_A >= 0 (Not just “=“)
LS_D - LF_A >= 0
LS_E - LF_A >= 0
LS_F - LF_B >= 0
LS_G - LF_B >= 0
LS_F - LF_C >= 0
LS_G - LF_C >= 0
LS_J - LF_D >= 0
LS_J - LF_H >= 0
LS_H - LF_E >= 0
LS_H - LF_F >= 0
LS_I - LF_E >= 0
LS_I - LF_F >= 0
LS_K - LF_G >= 0
LS_K - LF_I >= 0
FINISH - LF_J >= 0
FINISH - LF_K >= 0
FINISH = 23 (from the optimal results
of Model 1) – this is the main diff.)
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Example: Riverwalk Associates
• Probability the project will be completed within t1=24
weeks
project time variance s2 = s2A + s2C + s2F + s2I + s2K
= 4/9 + 0 + 1/9 + 1 + 4/9
= 2 (weeks-squared)
project time standard deviation s = 1.414 (weeks).
z1 = (24 - 23)/ s = (24-23)weeks/1.414weeks = .71
From the Standard Normal Distribution table:
P(z < z1=.71) = .5 + .2611 = .7611
More precisely,
P(t < t1) = P(t-t0 < t1-t0) = P[(t-t0)/s < (t1-t0)/s]
= P(z < z1=.71) = .7611
if we define z= (t-t0)/s and z1 = (t1-t0)/s.
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Probability Analysis of the Project: Example 2
Question: If the mean project completion time is X0 = 25, what is the probability that the
project will be completed within X1=30 weeks?
s2 = 6.9, s = 2.63.
Z1 = (X1 - X0)/ s = (30 -25)/2.63 = 1.90
•Z1 value of 1.90 corresponds to probability of .4713 in Table A.1, appendix A.
Probability of completing project in 30 weeks or less : (.5000 + .4713) = .9713.
•More precisely, P(x < X1) = P(x- X0 < X1 - X0) = P[(x- X0)/ s < (X1 - X0)/ s ]
= P(z < Z1=1.90) = .9713
Figure 8.14
Probability the network will be
completed in 30 weeks or less
if we define
z= (x - X0)/ s (new variable)
and
Z1 = (X1 - X0)/ s (constant).
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Probability Analysis of the Project: Example 3
Question: If the mean project completion time is X0 = 25, what is the probability that the
project will be completed within X1=22 weeks?
Z1 = (22 - 25)/2.63 = -1.14
Where Z1 value of 1.14 (ignore negative) corresponds to probability of 0.3729 in Table A.1,
appendix A. Probability that customer will be retained is .1271
Figure 8.15
Probability the network
will be completed in 22
weeks or less
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Probability Analysis of the Project Network
CPM/PERT Analysis with QM for Windows
Exhibit 8.1
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Project Crashing and Time-Cost Trade-Off:
Definition
• Project duration can be reduced by assigning more resources to
project activities.
• Doing this however increases project cost.
• Decision is based on analysis of trade-off between time and
cost.
• Project crashing is a method for shortening project duration by
reducing one or more critical activities to a time less than
normal activity time.
• Crashing achieved by devoting more resources to crashed
activities.
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Crashing Activity Times
• In the Critical Path Method (CPM) approach to project
scheduling, it is assumed that the normal time to
complete an activity, tj , which can be met at a normal
cost, cj , can be crashed to a reduced time, tj’, under
maximum crashing for an increased cost, cj’.
• It is assumed that its cost per unit reduction, Kj , is
linear and can be calculated by:
Kj = (cj' - cj)/(tj - tj').
E.g.: in the example on the right,
Kj = total crash cost/total crash time
= $2000/5 = $400/wk
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Crashing Example for Riverwalk Associates

Normal Costs and Crash Costs
Normal
Crash
Activity
Time Cost Time Cost
A) Study Feasibility
6 $ 80,000
5 $100,000
B) Purchase Building
4 100,000
4 100,000
C) Hire Project Leader
3
50,000
2 100,000
D) Select Advertising Staff
5 150,000
2 300,000
E) Purchase Materials
1 180,000
1 180,000
F) Hire Manufacturing Staff
4 300,000
1 480,000
G) Manufacture Prototype
2 100,000
2 100,000
H) Produce First 50 Units
6 450,000
5 800,000
I) Advertising Product
5 350,000
1 650,000
J) Assessing User Feedback
3 300,000
3 300,000
K) Distributing Product
5 550,000
5 550,000
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Crashing Example for Riverwalk Associates
Crashing: The completion time for this project using
normal times is 23 weeks. Which activities should be
crashed, and by how many weeks, in order for the
project to be completed in a Target of 20 weeks?
Let: Yi = the amount of time activity i is crashed.
Then borrow from the LP formulation for the Earliest
Times, we have the following…
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Riverwalk Associates – LP Model for Crashing
Min 20YA + 50YC + 50YD + 60YF
+ 350YH + 75YI
S.t.
EF_A - ES_A >= 6 - YA
EF_B - ES_B >= 4
EF_C - ES_C >= 3 - YC
EF_D - ES_D >= 5 - YD
EF_E - ES_E >= 1
EF_F - ES_F >= 4 - YF
EF_G - ES_G >= 2
EF_H - ES_H >= 6 - YH
EF_I - ES_I >= 5 - YI
EF_J - ES_J >= 3
EF_K - ES_K >= 5
ES_C - EF_A >= 0 (Not “=“ )
ES_D - EF_A >= 0
ES_E - EF_A >= 0
ES_F - EF_B >= 0
ES_G - EF_B >= 0
ES_F - EF_C >= 0
ES_G - EF_C >= 0
ES_J - EF_D >= 0
ES_J - EF_H >= 0
ES_H - EF_E >= 0
ES_H - EF_F >= 0
ES_I - EF_E >= 0
ES_I - EF_F >= 0
ES_K - EF_G >= 0
ES_K - EF_I >= 0
FINISH - EF_J >= 0
FINISH - EF_K >= 0
FINISH <= 20 (Target)
Chapter 8 - Project Management
YA <=1
YC <=1
YD <=3
YF <=3
YH <=1
YI <=4
31
Project Crashing and Time-Cost Trade-Off
Project Crashing with QM for Windows
Exhibit 8.2
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Project Crashing and Time-Cost Trade-Off
General Relationship of Time and Cost
• Project crashing costs and indirect costs have an inverse relationship.
• Crashing costs are highest when the project is shortened.
• Indirect costs increase as the project duration increases.
• Optimal project time is at minimum point on the total cost curve.
Figure 8.20
The time–cost trade-off
Chapter 8 - Project Management
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