Project Management
Outline
• Definition of Project Management
– Work Breakdown Structure
– Project Control Charts
– Structuring Projects
• Critical Path Scheduling
– By Hand
– Linear Programming
• PERT
– By Hand
Operations -- Prof. Juran
2
Project Management
• A Project is a series of related jobs usually
directed toward some major output and
requiring a significant period of time to
perform
• Project Management is the set of management
activities of planning, directing, and
controlling resources (people, equipment,
material) to meet the technical, cost, and time
constraints of a project
• An extreme case: Smallest possible
production volume, greatest possible
customization
Project Control Charts: Gantt Chart
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5
A pure project is where a self-contained team
works full-time on the project
Advantages:
• The project manager has full authority over the project
• Team members report to one boss
• Shortened communication lines
• Team pride, motivation, and commitment are high
Disadvantages:
• Duplication of resources
• Organizational goals and policies are ignored
• Lack of technology transfer
• Team members have no functional area "home"
Functional Projects
President
Research and
Development
Project Project Project
A
B
C
Engineering
Project Project Project
D
E
F
Manufacturing
Project Project Project
G
H
I
Example: Project “B” is in the functional
area of Research and Development.
Functional Projects
Advantages:
• A team member can work on several projects
• Technical expertise is maintained within the functional
area
• The functional area is a “home” after the project is
completed
• Critical mass of specialized knowledge
Disadvantages:
• Aspects of the project that are not directly related to the
functional area get short-changed
• Motivation of team members is often weak
• Needs of the client are secondary and are responded to
slowly
Matrix Project Organization Structure
President
Research and
Development
Manager
Project A
Manager
Project B
Manager
Project C
Engineering
Manufacturing
Marketing
Matrix Structure
Advantages:
• Enhanced communications between functional areas
• Pinpointed responsibility
• Duplication of resources is minimized
• Functional “home” for team members
• Policies of the parent organization are followed
Disadvantages:
• Too many bosses
• Depends on project manager’s negotiating skills
• Potential for sub-optimization
Work Breakdown Structure
A hierarchy of project tasks, subtasks, and work packages
Level
1
2
Program
Project 1
Project 2
Task 1.1
Task 1.2
3
Subtask 1.1.1
4
Work Package 1.1.1.1
Subtask 1.1.2
Work Package 1.1.1.2
Network Models
• A project is viewed as a sequence of activities that
form a “network” representing a project
• The path taking longest time through this network
of activities is called the “critical path”
• The critical path provides a wide range of
scheduling information useful in managing a
project
• Critical Path Method (CPM) helps to identify the
critical path(s) in the project network
Prerequisites for Critical Path Method
A project must have:
• well-defined jobs or tasks whose
completion marks the end of the project;
• independent jobs or tasks;
• and tasks that follow a given sequence.
Types of Critical Path Methods
• CPM with a Single Time Estimate
– Used when activity times are known with certainty
– Used to determine timing estimates for the project, each
activity in the project, and slack time for activities
• Time-Cost Models
– Used when cost trade-off information is a major
consideration in planning
– Used to determine the least cost in reducing total project
time
• CPM with Three Activity Time Estimates
– Used when activity times are uncertain
– Used to obtain the same information as the Single Time
Estimate model and probability information
CPM with Single Time Estimate
1. Activity Identification
2. Activity Sequencing and Network
Construction
3. Determine the critical path
–
From the critical path all of the project
and activity timing information can be
obtained
House-Building Example
Activity
Activity A
Activity B
Activity C
Activity D
Activity E
Activity F
Activity G
Description
Build foundation
Build walls and ceilings
Build roof
Do electrical wiring
Put in windows
Put on siding
Paint house
Operations -- Prof. Juran
Predecessors
—
A
B
B
B
E
C, F
Duration (days)
5
8
10
5
4
6
3
16
Managerial Problems
•Find the minimum number of days
needed to build the house.
•Find the “critical path”.
•Find the least expensive way to finish
the house in a certain amount of time.
•Study the possible effects of randomness
in the activity times.
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Activity-on-Arc Network
1
3
A
5
0
Start
F
6
E
4
B
8
4
G
3
C
10
5
End
D
5
2
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Method 1: “By Hand”
A
5
0
Start
F
6
3
1
E
4
B
8
4
G
3
C
10
5
End
D
5
2
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Procedure
• Draw network
• Forward pass to find earliest times
• Backward pass to find latest times
• Analysis: Critical Path, Slack Times
• Extensions (really hard!):
– Crashing
– PERT
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CPM Jargon
Every network of this type has at least one critical path,
consisting of a set of critical activities.
In this example, there are two critical paths: A-B-C-G and
A-B-E-F-G.
1
A
5
0
Star
t
F
6
3
4
G
3
E
4
5
End
C
10
B
8
D
5
2
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21
Conclusions
The project will take 26 days to
complete.
The only activity that is not critical is
the electrical wiring (Activity D).
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Time-Cost Models
• Basic Assumption: Relationship between
activity completion time and project cost
• Time Cost Models: Determine the optimum
point in time-cost tradeoffs
–
–
–
Activity direct costs
Project indirect costs
Activity completion times
Crashing Parameters
Activity
Build foundation
Walls and ceilings
Build roof
Electrical wiring
Put in windows
Put on siding
Paint house
Cost per Day to
Reduce Duration
$30
$15
$20
$40
$20
$30
$40
Maximum Possible
Reduction (Days)
2
3
1
2
2
3
1
What is the least expensive way to finish this project
in 20 days? (This is hard!)
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CPM with Three Activity Time Estimates
Activity
A
B
C
D
E
F
G
Optimistic
3
5
9
3
2
3
2
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Most Likely
5
8
10
5
4
6
3
Pessimistic
7
11
11
7
6
9
4
25
Expected Time Calculations
1
2
3
4
5
6
7
8
A
B
C
D
E
Activity Optimistic Most Likely Pessimistic Expected
A
3
5
7
5
B
5
8
11
8
C
9
10
11
10
D
3
5
7
5
E
2
4
6
4
F
3
6
9
6
G
2
3
4
3
Expected Time =
F
G
=(B2+C2*4+D2)/6
Opt. Time + 4(Most Likely Time) + Pess. Time
6
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Expected Time Calculations
1
2
3
4
5
6
7
8
A
B
C
D
E
Activity Optimistic Most Likely Pessimistic Expected
A
3
5
7
5
B
5
8
11
8
C
9
10
11
10
D
3
5
7
5
E
2
4
6
4
F
3
6
9
6
G
2
3
4
3
F
G
=(B2+C2*4+D2)/6
Note: the expected time of an activity is not necessarily
equal to the most likely time (although that is true in this
problem).
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1
A
5
0
Start
F
6
3
4
G
3
E
4
C
10
B
8
5
End
D
5
2
The sum of the critical path expected times is the expected
duration of the whole project (in this case, 26 days).
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What is the probability of finishing this project in
less than 25 days?
Z =
Operations -- Prof. Juran
D - TE
2

 cp
29
Pessimisti c  Optimistic 
Estimated Activity Variance   2  

6


1
2
3
4
5
6
7
8
A
B
C
D
E
F
Activity Optimistic Most Likely Pessimistic Expected Variance
A
3
5
7
5
0.4444
B
5
8
11
8
1.0000
C
9
10
11
10
0.1111
D
3
5
7
5
0.4444
E
2
4
6
4
0.4444
F
3
6
9
6
1.0000
G
2
3
4
3
0.1111
Operations -- Prof. Juran
G
=((D2-B2)/6)^2
30
2
Activity Optimistic Most Likely Pessimistic Expected Variance Criticial Path 1
A
3
5
7
5
0.444
0.444
B
5
8
11
8
1.000
1.000
C
9
10
11
10
0.111
D
3
5
7
5
0.444
E
2
4
6
4
0.444
0.444
F
3
6
9
6
1.000
1.000
G
2
3
4
3
0.111
0.111
Expected
26.000
Variance
3.000
St Dev
1.732

2
Criticial Path 2
0.444
1.000
0.111
0.111
26.000
1.667
1.291
=3
Sum the variances along the critical path.
(In this case, with two critical paths, we’ll take the one
with the larger variance.)
Operations -- Prof. Juran
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Z=
D - TE

2
cp
25 - 26
=
= - 0.5774
3
p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)
There is a 28.19% probability that this project will
be completed
in
less
than
25
days.
Operations -- Prof. Juran
32
CPM Assumptions/Limitations
•Project activities can be identified as entities (there
is a clear beginning and ending point for each
activity).
•Project activity sequence relationships can be
specified and networked.
•Project control should focus on the critical path.
•The activity times follow the beta distribution, with
the variance of the project assumed to equal the
sum of the variances along the critical path.
Method 2: Excel Solver
(Linear Programming)
1
3
A
5
0
Start
F
6
E
4
B
8
4
G
3
C
10
5
End
D
5
2
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34
Procedure
•
•
•
•
•
•
•
Define Decision Variables
Define Objective Function
Define Constraints
Solver Dialog and Options
Run Solver
“Sensitivity” Analysis: Critical Path, Slack
Extensions:
– Crashing (not too bad)
– PERT (need simulation, not Solver)
Operations -- Prof. Juran
35
LP Formulation
Decision Variables
We are trying to decide when to begin and end each of
the activities.
Objective
Minimize the total time to complete the project.
Constraints
Each activity has a fixed duration.
There are precedence relationships among the activities.
We cannot go backwards in time.
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Formulation
Decision Variables
Define the nodes to be discrete events. In other words, they
occur at one exact point in time. Our decision variables will
be these points in time.
Define ti to be the time at which node i occurs, and at which
time all activities preceding node i have been completed.
Define t0 to be zero.
Objective
Minimize t5.
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Formulation
Constraints
There is really one basic type of constraint. For each
activity x, let the time of its starting node be represented
by tjx and the time of its ending node be represented by
tkx.
Let the duration of activity x be represented as dx.
For every activity x,
For every node i,
Operations -- Prof. Juran
t kx  t jx  d x
ti  0
38
Solution Methodology
A B C D E
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
F G H
I
J
K
L
t0 t1 t2 t3 t4 t5
0 1 1 1 1 1
1
A
5
0
Start
A
B
C
D
E
F
G
t1
1
-1
0
0
0
0
0
t2
0
1
-1
-1
-1
0
0
t3
0
0
0
0
1
-1
0
t4
0
0
1
0
0
1
-1
t5
0
0
0
1
0
0
1
N
O
P
=G6
1
t0
-1
0
0
0
0
0
0
M
1
0
0
0
0
0
0
E
4
B
8
=SUMPRODUCT($B$6:$G$6,B12:G12)
>= 5
>= 8
2
>= 10
>= 5
>= 4
>= 6
>= 3
Operations -- Prof. Juran
F
6
3
4
G
3
C
10
5
End
D
5
39
Solution Methodology
The matrix of zeros, ones, and negative ones
(B12:G18) is a means for setting up the constraints.
The sumproduct functions in H12:H18 calculate
the elapsed time between relevant pairs of nodes,
corresponding to the various activities.
The duration times of the activities are in J12:J18.
Operations -- Prof. Juran
40
Operations -- Prof. Juran
41
Optimal Solution
A B C D E
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
F G H
I
J
K
L
M
N
O
F
6
4
P
26
1
t0 t1 t2 t3 t4 t5
0 5 13 17 23 26
A
5
0
S ta rt
A
B
C
D
E
F
G
t0
-1
0
0
0
0
0
0
t1
1
-1
0
0
0
0
0
t2
0
1
-1
-1
-1
0
0
t3
0
0
0
0
1
-1
0
t4
0
0
1
0
0
1
-1
t5
0
0
0
1
0
0
1
3
5
8
10
13
4
6
3
Operations -- Prof. Juran
>=
>=
>=
>=
>=
>=
>=
5
8
10
5
4
6
3
E
4
B
8
G
3
C
10
5
End
D
5
2
42
CPM Jargon
Any activity for which
t kx  t jx  d x
is said to have slack time, the amount of time by
which that activity could be delayed without
affecting the overall completion time of the
whole project. In this example, only activity D
has any slack time (13 – 5 = 8 units of slack
time).
Operations -- Prof. Juran
43
CPM Jargon
Any activity x for which
t kx  t jx  d x
is defined to be a “critical” activity, with zero
slack time.
Operations -- Prof. Juran
44
Excel Tricks: Conditional Formatting
A B C D E
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
F G H
I
J
K
L
M
N
O
F
6
4
P
26
1
t0 t1 t2 t3 t4 t5
0 5 13 17 23 26
A
5
0
S ta rt
A
B
C
D
E
F
G
t0
-1
0
0
0
0
0
0
t1
1
-1
0
0
0
0
0
t2
0
1
-1
-1
-1
0
0
t3
0
0
0
0
1
-1
0
t4
0
0
1
0
0
1
-1
t5
0
0
0
1
0
0
1
3
5
8
10
13
4
6
3
>=
>=
>=
>=
>=
>=
>=
5
8
10
5
4
6
3
Operations -- Prof. Juran
E
4
B
8
G
3
C
10
5
End
D
5
2
45
Critical Activities: Using the
Solver Answer Report
Constraints
Cell Name
$H$12 A
$H$13 B
$H$14 C
$H$15 D
$H$16 E
$H$17 F
$H$18 G
Cell Value
Operations -- Prof. Juran
5
8
10
13
4
6
3
Formula
$H$12>=$J$12
$H$13>=$J$13
$H$14>=$J$14
$H$15>=$J$15
$H$16>=$J$16
$H$17>=$J$17
$H$18>=$J$18
Status
Slack
Binding
0
Binding
0
Binding
0
Not Binding
8
Binding
0
Binding
0
Binding
0
46
House-Building Example, Continued
Suppose that by hiring additional workers,
the duration of each activity can be reduced.
Use LP to find the strategy that minimizes
the cost of completing the project within 20
days.
Operations -- Prof. Juran
47
Crashing Parameters
Activity
Build foundation
Walls and ceilings
Build roof
Electrical wiring
Put in windows
Put on siding
Paint house
Operations -- Prof. Juran
Cost per Day to
Reduce Duration
$30
$15
$20
$40
$20
$30
$40
Maximum Possible
Reduction (Days)
2
3
1
2
2
3
1
48
Managerial Problem Definition
Find a way to build the house in 20 days.
Operations -- Prof. Juran
49
Formulation
Decision Variables
Now the problem is not only when to schedule the
activities, but also which activities to accelerate. (In
CPM jargon, accelerating an activity at an additional
cost is called “crashing”.)
Objective
Minimize the total cost of crashing.
Operations -- Prof. Juran
50
Formulation
Constraints
The project must be finished in 20 days.
Each activity has a maximum amount of crash time.
Each activity has a “basic” duration. (These durations
were considered to have been fixed in Part a; now they
can be reduced.)
There are precedence relationships among the
activities.
We cannot go backwards in time.
Operations -- Prof. Juran
51
Formulation
Decision Variables
Define the number of days that activity x is crashed to be Rx.
For each activity there is a maximum number of crash days
Rmax, x
Define the crash cost per day for activity x to be Cx
Objective
Minimize Z =
7
R C
x 1
Operations -- Prof. Juran
x
x
52
Formulation
Constraints
For every activity x,
t kx  t jx  d x  R x
For every activity x,
Rx  Rmax,x
For every node i,
ti  0
Operations -- Prof. Juran
53
Solution Methodology
A B C D
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
E
F
G
H
I
J
K
L
1
=SUMPRODUCT(M12:M18,O12:O18)
30
0
Start
20 <--
A
B
C
D
E
F
G
t1
1
-1
0
0
0
0
0
t2 t3 t4
0 0 0
1 0 0
-1 0 1
-1 0 0
-1 1 0
0 -1 1
0 0 -1
t5
0
0
0
1
0
0
1
5
8
10
13
4
6
3
B
8
>=
>=
>=
>=
>=
>=
>=
Operations -- Prof. Juran
=L12-M12
Improved Duration
4
8
10
5
4
6
3
O
4
G
3
C
10
5
End
D
5
Max Completion Time
=SUMPRODUCT($B$6:$G$6,B12:G12)
t0
-1
0
0
0
0
0
0
E
4
N
F
6
3
A
5
t0 t1 t2 t3 t4 t5
0 5 13 17 23 26
M
2
Basic Duration Crash Time Max Crash
5
1
2
8
0
3
10
0
1
5
0
2
4
0
2
6
0
3
3
0
1
Cost/Time
$
30
$
15
$
20
$
40
$
20
$
30
$
40
54
Solution Methodology
G3 now contains a formula to calculate the total crash
cost.
The new decision variables (how long to crash each
activity x, represented by Rx) are in M12:M18.
G8 contains the required completion time, and we will
constrain the value in G6 to be less than or equal to G8.
The range J12:J18 calculates the revised duration of each
activity, taking into account how much time is saved by
crashing.
Operations -- Prof. Juran
55
Operations -- Prof. Juran
56
Optimal Solution
A B C D
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
E
F
G
H
I
J
K
L
M
1
=SUMPRODUCT(M12:M18,O12:O18)
A
5
145
0
S ta rt
t0 t1 t2 t3 t4 t5
0 3 8 11 17 20
20 <--
Max Completion Time
=SUMPRODUCT($B$6:$G$6,B12:G12)
A
B
C
D
E
F
G
t0
-1
0
0
0
0
0
0
t1
1
-1
0
0
0
0
0
t2 t3 t4
0 0 0
1 0 0
-1 0 1
-1 0 0
-1 1 0
0 -1 1
0 0 -1
t5
0
0
0
1
0
0
1
3
5
9
12
3
6
3
>=
>=
>=
>=
>=
>=
>=
Operations -- Prof. Juran
F
6
3
E
4
B
8
N
O
4
G
3
C
10
5
End
D
5
2
=L12-M12
Improved Duration
3
5
9
5
3
6
3
Basic Duration Crash Time Max Crash
5
2
2
8
3
3
10
1
1
5
0
2
4
1
2
6
0
3
3
0
1
Cost/Time
$
30
$
15
$
20
$
40
$
20
$
30
$
40
57
Crashing Solution
A
B
C
D
E
F
G
Basic Duration Crash Time Improved Duration Cost/Time
5
2
3
$30
8
3
5
$15
10
1
9
$20
5
0
5
$40
4
1
3
$20
6
0
6
$30
3
0
3
$40
Operations -- Prof. Juran
Cost
$60
$45
$20
$$20
$$$145
58
Crashing Solution
It is feasible to complete the project in 20 days, at a cost of $145.
1
A
3
0
Star
t
F
6
3
E
3
B
5
4
G
3
5
End
C
9
D
5
2
Operations -- Prof. Juran
59
Summary
• Definition of Project Management
– Work Breakdown Structure
– Project Control Charts
– Structuring Projects
• Critical Path Scheduling
– By Hand
– Linear Programming
• PERT
– By Hand
Operations -- Prof. Juran
60