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Project Management
Operations -- Prof. Juran
Operations -- Prof. Juran 2
Outline • Definition of Project Management
– Work Breakdown Structure– Project Control Charts– Structuring Projects
• Critical Path Scheduling– By Hand– Linear Programming
• PERT– By Hand
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
Operations -- Prof. Juran 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 andDevelopment
Engineering Manufacturing
ProjectA
ProjectB
ProjectC
ProjectD
ProjectE
ProjectF
ProjectG
ProjectH
ProjectI
Example: Project “B” is in the functional area of Research and
Development.
Example: Project “B” is in the functional area of Research and
Development.
Functional ProjectsAdvantages:• 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 andDevelopment
Engineering Manufacturing Marketing
ManagerProject A
ManagerProject B
ManagerProject C
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
Program
Project 1 Project 2
Task 1.1
Subtask 1.1.1
Work Package 1.1.1.1
Level
1
2
3
4
Task 1.2
Subtask 1.1.2
Work Package 1.1.1.2
A hierarchy of project tasks, subtasks, and work packages
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 Identification2. Activity Sequencing and Network
Construction3. Determine the critical path
– From the critical path all of the project and activity timing information can be obtained
Operations -- Prof. Juran 16
House-Building Example
Activity Description Predecessors Duration (days) Activity A Build foundation — 5 Activity B Build walls and ceilings A 8 Activity C Build roof B 10 Activity D Do electrical wiring B 5 Activity E Put in windows B 4 Activity F Put on siding E 6 Activity G Paint house C, F 3
Operations -- Prof. Juran 17
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.
Operations -- Prof. Juran 18
Activity-on-Arc Network1
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2
A5
B8
C10
E4
D5
G3
F6
5End
Operations -- Prof. Juran 19
Method 1: “By Hand”
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C10
E4
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G3
F6
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Operations -- Prof. Juran 20
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
Operations -- Prof. Juran 21
CPM Jargon
1
0Start
43
2
A5
B8
C10
E4
D5
G3
F6
5End
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.
Operations -- Prof. Juran 22
Conclusions
The project will take 26 days to complete.
The only activity that is not critical is the electrical wiring (Activity D).
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
Operations -- Prof. Juran 24
Crashing Parameters
Activity Cost per Day to
Reduce Duration Maximum Possible Reduction (Days)
Build foundation $30 2 Walls and ceilings $15 3 Build roof $20 1 Electrical wiring $40 2 Put in windows $20 2 Put on siding $30 3 Paint house $40 1
What is the least expensive way to finish this project in 20 days? (This is hard!)
Operations -- Prof. Juran 25
CPM with Three Activity Time Estimates
Activity Optimistic Most Likely Pessimistic A 3 5 7 B 5 8 11 C 9 10 11 D 3 5 7 E 2 4 6 F 3 6 9 G 2 3 4
Operations -- Prof. Juran 26
Expected Time Calculations
6Time Pess. + Time) Likely 4(Most + Time Opt.
= Time Expected
12345678
A B C D E F GActivity Optimistic Most Likely Pessimistic Expected
A 3 5 7 5B 5 8 11 8C 9 10 11 10D 3 5 7 5E 2 4 6 4F 3 6 9 6G 2 3 4 3
=(B2+C2*4+D2)/6
Operations -- Prof. Juran 27
Expected Time Calculations
12345678
A B C D E F GActivity Optimistic Most Likely Pessimistic Expected
A 3 5 7 5B 5 8 11 8C 9 10 11 10D 3 5 7 5E 2 4 6 4F 3 6 9 6G 2 3 4 3
=(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).
Operations -- Prof. Juran 28
1
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2
A5
B8
C10
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D5
G3
F6
5End
The sum of the critical path expected times is the expected duration of the whole project (in this case, 26 days).
Operations -- Prof. Juran 29
What is the probability of finishing this project in less than 25 days?
What is the probability of finishing this project in less than 25 days?
Z = D - TE
cp2
Operations -- Prof. Juran 30
22
6 Variance Activity Estimated
OptimisticcPessimisti
12345678
A B C D E F GActivity Optimistic Most Likely Pessimistic Expected Variance
A 3 5 7 5 0.4444B 5 8 11 8 1.0000C 9 10 11 10 0.1111D 3 5 7 5 0.4444E 2 4 6 4 0.4444F 3 6 9 6 1.0000G 2 3 4 3 0.1111
=((D2-B2)/6)^2
Operations -- Prof. Juran 31
Sum the variances along the critical path. (In this case, with two critical paths, we’ll take the one with the larger variance.)
3=2
Activity Optimistic Most Likely Pessimistic Expected Variance Criticial Path 1 Criticial Path 2A 3 5 7 5 0.444 0.444 0.444B 5 8 11 8 1.000 1.000 1.000C 9 10 11 10 0.111 0.111D 3 5 7 5 0.444E 2 4 6 4 0.444 0.444F 3 6 9 6 1.000 1.000G 2 3 4 3 0.111 0.111 0.111
Expected 26.000 26.000Variance 3.000 1.667St Dev 1.732 1.291
Operations -- Prof. Juran 32
There is a 28.19% probability that this project will be completed in less than 25 days.There is a 28.19% probability that this project will be completed in less than 25 days.
p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)
0.5774- =326-25
=T - D
= Z2
cp
E
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.
Operations -- Prof. Juran 34
Method 2: Excel Solver(Linear Programming)
1
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A5
B8
C10
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D5
G3
F6
5End
Operations -- Prof. Juran 35
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 36
LP FormulationDecision VariablesWe are trying to decide when to begin and end each of the activities.ObjectiveMinimize the total time to complete the project.ConstraintsEach activity has a fixed duration.There are precedence relationships among the activities.We cannot go backwards in time.
Operations -- Prof. Juran 37
Formulation
Decision VariablesDefine 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.
ObjectiveMinimize t5.
Operations -- Prof. Juran 38
FormulationConstraintsThere 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,
xjxkx dtt
0it
Operations -- Prof. Juran 39
Solution Methodology
123
456
789
101112131415161718
A B C D E F G H I J K L M N O P
1
t0 t1 t2 t3 t4 t50 1 1 1 1 1
t0 t1 t2 t3 t4 t5A -1 1 0 0 0 0 1 >= 5B 0 -1 1 0 0 0 0 >= 8C 0 0 -1 0 1 0 0 >= 10D 0 0 -1 0 0 1 0 >= 5E 0 0 -1 1 0 0 0 >= 4F 0 0 0 -1 1 0 0 >= 6G 0 0 0 0 -1 1 0 >= 3
1
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2
A5
B8
C10
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D5
G3
F6
5End
=G6
=SUMPRODUCT($B$6:$G$6,B12:G12)
Operations -- Prof. Juran 40
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 41
Operations -- Prof. Juran 42
Optimal Solution123
456
789101112131415161718
A B C D E F G H I J K L M N O P
26
t0 t1 t2 t3 t4 t50 5 13 17 23 26
t0 t1 t2 t3 t4 t5A -1 1 0 0 0 0 5 >= 5B 0 -1 1 0 0 0 8 >= 8C 0 0 -1 0 1 0 10 >= 10D 0 0 -1 0 0 1 13 >= 5E 0 0 -1 1 0 0 4 >= 4F 0 0 0 -1 1 0 6 >= 6G 0 0 0 0 -1 1 3 >= 3
1
0Start
43
2
A5
B8
C10
E4
D5
G3
F6
5End
Operations -- Prof. Juran 43
CPM Jargon
Any activity for which
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).
xjxkx dtt
Operations -- Prof. Juran 44
CPM Jargon
Any activity x for which
is defined to be a “critical” activity, with zero slack time.
xjxkx dtt
Operations -- Prof. Juran 45
Excel Tricks: Conditional Formatting1
2
3
45
6
789101112131415161718
A B C D E F G H I J K L M N O P
26
t0 t1 t2 t3 t4 t5
0 5 13 17 23 26
t0 t1 t2 t3 t4 t5A -1 1 0 0 0 0 5 >= 5B 0 -1 1 0 0 0 8 >= 8C 0 0 -1 0 1 0 10 >= 10D 0 0 -1 0 0 1 13 >= 5E 0 0 -1 1 0 0 4 >= 4F 0 0 0 -1 1 0 6 >= 6G 0 0 0 0 -1 1 3 >= 3
1
0Start
43
2
A5
B8
C10
E4
D5
G3
F6
5End
Operations -- Prof. Juran 46
Critical Activities: Using the Solver Answer Report
ConstraintsCell Name Cell Value Formula Status Slack
$H$12 A 5 $H$12>=$J$12 Binding 0$H$13 B 8 $H$13>=$J$13 Binding 0$H$14 C 10 $H$14>=$J$14 Binding 0$H$15 D 13 $H$15>=$J$15 Not Binding 8$H$16 E 4 $H$16>=$J$16 Binding 0$H$17 F 6 $H$17>=$J$17 Binding 0$H$18 G 3 $H$18>=$J$18 Binding 0
Operations -- Prof. Juran 47
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 48
Crashing Parameters
Activity Cost per Day to
Reduce Duration Maximum Possible Reduction (Days)
Build foundation $30 2 Walls and ceilings $15 3 Build roof $20 1 Electrical wiring $40 2 Put in windows $20 2 Put on siding $30 3 Paint house $40 1
Operations -- Prof. Juran 49
Managerial Problem Definition
Find a way to build the house in 20 days.
Operations -- Prof. Juran 50
Formulation
Decision VariablesNow the problem is not only when to schedule the activities, but also which activities to accelerate. (In CPM jargon, accelerating an activity at an additional cost is called “crashing”.)
ObjectiveMinimize the total cost of crashing.
Operations -- Prof. Juran 51
FormulationConstraintsThe project must be finished in 20 days.Each activity has a maximum amount of crash time.Each activity has a “basic” duration. (These durations were considered to have been fixed in Part a; now they can be reduced.)There are precedence relationships among the activities.We cannot go backwards in time.
Operations -- Prof. Juran 52
Formulation
Decision VariablesDefine the number of days that activity x is crashed to be Rx.
For each activity there is a maximum number of crash days Rmax, x
Define the crash cost per day for activity x to be Cx
ObjectiveMinimize Z =
7
1xxxCR
Operations -- Prof. Juran 53
FormulationConstraints
For every activity x,
For every activity x,
For every node i,
xxjxkx Rdtt
xx RR max,
0it
Operations -- Prof. Juran 54
Solution Methodology
123
456
7891011
12131415161718
A B C D E F G H I J K L M N O
30
t0 t1 t2 t3 t4 t50 5 13 17 23 26
20 <-- Max Completion Time
t0 t1 t2 t3 t4 t5 Improved Duration Basic Duration Crash Time Max Crash Cost/Time
A -1 1 0 0 0 0 5 >= 4 5 1 2 30$ B 0 -1 1 0 0 0 8 >= 8 8 0 3 15$ C 0 0 -1 0 1 0 10 >= 10 10 0 1 20$ D 0 0 -1 0 0 1 13 >= 5 5 0 2 40$ E 0 0 -1 1 0 0 4 >= 4 4 0 2 20$ F 0 0 0 -1 1 0 6 >= 6 6 0 3 30$ G 0 0 0 0 -1 1 3 >= 3 3 0 1 40$
1
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2
A5
B8
C10
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D5
G3
F6
5End
=SUMPRODUCT($B$6:$G$6,B12:G12) =L12-M12
=SUMPRODUCT(M12:M18,O12:O18)
Operations -- Prof. Juran 55
Solution Methodology
G3 now contains a formula to calculate the total crash cost. The new decision variables (how long to crash each activity x, represented by Rx) are in M12:M18.
G8 contains the required completion time, and we will constrain the value in G6 to be less than or equal to G8.
The range J12:J18 calculates the revised duration of each activity, taking into account how much time is saved by crashing.
Operations -- Prof. Juran 56
Operations -- Prof. Juran 57
Optimal Solution12
3
45
6
7891011
12131415161718
A B C D E F G H I J K L M N O
145
t0 t1 t2 t3 t4 t5
0 3 8 11 17 20
20 <-- Max Completion Time
t0 t1 t2 t3 t4 t5 Improved Duration Basic Duration Crash Time Max Crash Cost/Time
A -1 1 0 0 0 0 3 >= 3 5 2 2 30$ B 0 -1 1 0 0 0 5 >= 5 8 3 3 15$ C 0 0 -1 0 1 0 9 >= 9 10 1 1 20$ D 0 0 -1 0 0 1 12 >= 5 5 0 2 40$ E 0 0 -1 1 0 0 3 >= 3 4 1 2 20$ F 0 0 0 -1 1 0 6 >= 6 6 0 3 30$ G 0 0 0 0 -1 1 3 >= 3 3 0 1 40$
1
0Start
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2
A5
B8
C10
E4
D5
G3
F6
5End
Operations -- Prof. Juran 58
Crashing Solution
Basic Duration Crash Time Improved Duration Cost/Time Cost A 5 2 3 $30 $60 B 8 3 5 $15 $45 C 10 1 9 $20 $20 D 5 0 5 $40 $- E 4 1 3 $20 $20 F 6 0 6 $30 $- G 3 0 3 $40 $- $145
Operations -- Prof. Juran 59
Crashing SolutionIt is feasible to complete the project in 20 days, at a cost
of $145.
1
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A3
B5
C9
E3
D5
G3
F6
5End
Operations -- Prof. Juran 60
Summary• Definition of Project Management
– Work Breakdown Structure– Project Control Charts– Structuring Projects
• Critical Path Scheduling– By Hand– Linear Programming
• PERT– By Hand