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

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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

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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

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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 Network1

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B8

C10

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D5

G3

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Method 1: “By Hand”

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C10

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G3

F6

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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

1

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43

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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.

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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

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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!)

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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

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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

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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).

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1

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A5

B8

C10

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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?

What is the probability of finishing this project in less than 25 days?

Z = D - TE

cp2

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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

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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

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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.

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Method 2: Excel Solver(Linear Programming)

1

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C10

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F6

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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)

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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.

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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

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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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43

2

A5

B8

C10

E4

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G3

F6

5End

=G6

=SUMPRODUCT($B$6:$G$6,B12:G12)

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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.

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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

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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

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CPM Jargon

Any activity x for which

is defined to be a “critical” activity, with zero slack time.

xjxkx dtt

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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

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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

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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.

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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

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Managerial Problem Definition

Find a way to build the house in 20 days.

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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.

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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.

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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

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FormulationConstraints

For every activity x,

For every activity x,

For every node i,

xxjxkx Rdtt

xx RR max,

0it

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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

0Start

43

2

A5

B8

C10

E4

D5

G3

F6

5End

=SUMPRODUCT($B$6:$G$6,B12:G12) =L12-M12

=SUMPRODUCT(M12:M18,O12:O18)

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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.

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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

43

2

A5

B8

C10

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D5

G3

F6

5End

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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

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Crashing SolutionIt is feasible to complete the project in 20 days, at a cost

of $145.

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A3

B5

C9

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F6

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Summary• Definition of Project Management

– Work Breakdown Structure– Project Control Charts– Structuring Projects

• Critical Path Scheduling– By Hand– Linear Programming

• PERT– By Hand