Project Management

Operations -- Prof. Juran

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

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 and Development Engineering Manufacturing Project A Project B Project C Project D Project E Project F Project G Project H Project 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 Engineering Manufacturing Marketing Manager Project A Manager Project B Manager Project 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

A hierarchy of project tasks, subtasks, and work packages Level Program 1 Project 1 Project 2 2 Task 1.1

Task 1.2

3 Subtask 1.1.1

Subtask 1.1.2

4 Work Package 1.1.1.1

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 Description

Activity A Build foundation Activity B Build walls and ceilings Activity C Build roof Activity D Do electrical wiring Activity E Put in windows Activity F Put on siding Activity G Paint house

Predecessors

— A B B B E C, F

Duration (days)

5 8 10 5 4 6 3 Operations -- Prof. Juran 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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0 Start A 5

Activity-on-Arc Network

1 3 F 6 4 E 4 G 3 B 8 C 10 5 End D 5 2

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0 Start A 5

Method 1: “By Hand”

1 3 F 6 4 E 4 G 3 B 8 C 10 D 5 2 5 End

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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 Operations -- Prof. Juran 20

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 0 Star t A 5 B 8

Operations -- Prof. Juran

2 3 F 6 E 4 C 10 D 5 4 G 3 5 End

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!) Operations -- Prof. Juran 24

CPM with Three Activity Time Estimates Activity A B C D E F G Optimistic 3 5 9 3 2 3 2 Most Likely 5 8 10 5 4 6 3 Pessimistic 7 11 11 7 6 9 4 Operations -- Prof. Juran 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 C D E F G 5 9 3 2 3 2 8 10 5 4 6 3 11 11 7 6 9 4 8 10 5 4 6 3 F G

=(B2+C2*4+D2)/6

Expected Time = Opt.

Time + 4(Most Likely Time) + Pess.

Time 6 Operations -- Prof. Juran 26

Expected Time Calculations

5 6 7 8 1 2 3 4 A B C D E Activity Optimistic Most Likely Pessimistic Expected A B C 3 5 9 5 8 10 7 11 11 5 8 10 D E F G 3 2 3 2 5 4 6 3 7 6 9 4 5 4 6 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 3 F 6 4 A 5 G 3 E 4 0 Start B 8 C 10 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?

Operations -- Prof. Juran Z = D - T E   cp 2 29

Estimated Activity Variance   2 

Pessimisti c



Optimistic

6 2 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 C 5 9 8 10 11 11 8 10 1.0000

0.1111

D E 3 2 5 4 7 6 5 4 0.4444

0.4444

F G 3 2 6 3 9 4 6 3 1.0000

0.1111

G

=((D2-B2)/6)^2

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Activity Optimistic Most Likely Pessimistic Expected Variance Criticial Path 1 Criticial Path 2 A 3 5 7 5 0.444

0.444

0.444

B 5 8 11 8 1.000

1.000

1.000

C D E F G 9 3 2 3 2 10 5 4 6 3 11 7 6 9 4 10 5 4 6 3 0.111

0.444

0.444

1.000

0.111

Expected Variance St Dev 0.444

1.000

0.111

26.000

3.000

1.732

0.111

0.111

26.000

1.667

1.291

  2

= 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 31

Z = D T

E   cp 2

= 25 26 3 = - 0.5774

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.

0 Start A 5

Method 2: Excel Solver

(Linear Programming)

1 3 F 6 4 E 4 G 3 B 8 C 10 D 5 2

Operations -- Prof. Juran

5 End

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 t i to be the time at which node i occurs, and at which time all activities preceding node i have been completed.

Define t 0 to be zero.

Objective

Minimize t 5 .

Operations -- Prof. Juran 37

Formulation

Constraints

There is really one basic type of constraint. For each activity x, let the time of its starting node be represented by t jx and the time of its ending node be represented by

t

kx . Let the duration of activity x be represented as d x . For every activity x,

t kx



t jx



d x

For every node i,

t i

 0 Operations -- Prof. Juran 38

Solution Methodology

A B C D E F G H 1 2 3 4 5 6 13 14 15 16 17 18 7 8 9 10 11 12 t0 t1 t2 t3 t4 t5 0 1 1 1 1 1 1 t0 t1 t2 t3 t4 t5 A -1 1 0 0 0 0 B 0 -1 1 C 0 0 -1 0 0 0 1 0 0 D 0 E 0 F 0 G 0 0 -1 0 -1 0 0 0 0 0 1 -1 0 0 0 1 -1 1 0 0 1 I

=G6

J K

0 Start

L

A 5 1 B 8

M

E 4 =SUMPRODUCT($B$6:$G$6,B12:G12)

1 >= 5 0 >= 8 0 >= 10

2

0 >= 5 0 >= 4 0 >= 6 0 >= 3

3 C 10

N

F 6 D 5

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4 G 3 5 End

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 Solution

A B C D E F G H I J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 t0 t1 t2 t3 t4 t5 0 26 5 13 17 23 26 t0 t1 t2 t3 t4 t5 A -1 1 B 0 -1 1 C 0 D 0 0 0 -1 0 -1 0 0 0 0 0 0 1 0 0 0 0

5 8 10

>= 5 >= 8 >= 10 1 13 >= 5 E 0 F 0 G 0 0 -1 0 0 0 0 1 -1 0 0 1 -1 0 0 1

4 6 3

>= 4 >= 6 >= 3 K L

1 0 S ta r t A 5 B 8 2

M

E 4 3 C 1 0

N

F 6 D 5

O

4 G 3 5 E n d

P Operations -- Prof. Juran 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

K A B C D E F G H I J 6 14 15 16 17 18 7 8 9 10 11 12 13 1 2 3 4 5 t0 t1 t2 t3 t4 t5 0 26 5 13 17 23 26 t0 t1 t2 t3 t4 t5 A -1 1 0 B 0 -1 1 0 0 0 0 0 0 C 0 D 0 0 -1 0 -1 0 0 1 0

5 8

>= 5 >= 8 0

10

>= 10 1 13 >= 5 E 0 F 0 G 0 0 -1 0 0 0 0 1 -1 0 0 1 -1 0 0 1

4 6 3

>= 4 >= 6 >= 3

0 S ta r t A 5

L

1 B 8 2

M

E 4 3 C 1 0

N

F 6 D 5

O

4 G 3

P

5 E n d

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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 Formula Status

5 $H$12>=$J$12 Binding 8 $H$13>=$J$13 Binding 10 $H$14>=$J$14 Binding 13 $H$15>=$J$15 Not Binding 4 $H$16>=$J$16 Binding 6 $H$17>=$J$17 Binding 3 $H$18>=$J$18 Binding

Slack

0 0 0 0 0 0 8 Operations -- Prof. Juran 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

Cost per Day to Reduce Duration

$30 $15 $20 $40 $20 $30 $40

Maximum Possible Reduction (Days)

2 3 1 2 2 3 1 Operations -- Prof. Juran 48

Managerial Problem Definition

Find a way to build the house in 20 days.

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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.

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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.

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Formulation

Decision Variables

Define the number of days that activity x is crashed to be R x . For each activity there is a maximum number of crash days

R

max, x Define the crash cost per day for activity x to be C x

Objective

Minimize Z =

x

7   1

R x C x

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Formulation

Constraints

For every activity x, For every activity x, For every node i,

t kx



t jx



d x



R x R x



R

max,

x t i

 0 Operations -- Prof. Juran 53

Solution Methodology

A B C D E F G H I J K L 7 8 9 10 11 12 13 14 15 16 17 18 1 2 3 4 5 6 t0 t1 t2 t3 t4 0 5 13 17 23 26

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

t0 t1 t2 t3 t4 A -1 1 F 0 G 0 0 0 0 B 0 -1 1 C 0 0 -1 D 0 E 0 0 -1 0 -1 0 0 0 0 0 0 1 -1 0 0 0 1 0 0 1 -1 30 t5 20 <- t5 0 0 0 1 0 0 1

=SUMPRODUCT(M12:M18,O12:O18)

5 >= 8 >= 10 >= 13 >= 4 >= 6 >= 3 >= Max Completion Time Improved Duration 4 8 10 5 4 6 3

=L12-M12 0 Start A 5 1 B 8 2 E 4

Basic Duration 5 8 10 5 4 6 3

3 C 10

M

F 6 D 5 4

N

G 3 5 End

O Crash Time Max Crash Cost/Time 1 0 0 0 0 0 0 2 3 1 2 2 3 1 $ 30 $ 15 $ 20 $ 40 $ 20 $ $ 30 40 Operations -- Prof. Juran 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 R x ) 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

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Optimal Solution

A B C D E F G H I J K 7 8 9 10 11 12 13 14 15 16 17 18 1 2 3 4 5 6 t0 t1 t2 t3 t4 0 3 t0 t1 t2 t3 t4 A -1 1 0 B 0 -1 1 C 0 0 -1 D 0 E 0 F 0 G 0 0 -1 0 -1 0 0 0 0 0 0 0 0 1 -1 0 0 0 1 0 0 1 -1 145 t5 8 11 17 20 20 <--

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

t5 0 0 0 1 0 0 1 3 5 9

=SUMPRODUCT(M12:M18,O12:O18)

>= >= >= 12 >= 3 6 3 >= >= >= Max Completion Time Improved Duration 5 3 6 3 3 5 9

0 S ta r t =L12-M12 A 5 1 B 8 2

L

E 4

Basic Duration 5 8 10 5 4 6 3

3 C 1 0

M

F 6 D 5 4

N

G 3 5 E n d

O Crash Time Max Crash Cost/Time 2 3 1 0 1 0 0 2 3 1 2 2 3 1 $ $ $ $ $ $ $ 30 15 20 40 20 30 40 Operations -- Prof. Juran 57

Crashing Solution

Basic Duration Crash Time Improved Duration Cost/Time A 5 2 3 $30 B C D 8 10 5 3 1 0 5 9 5 $15 $20 $40 E F G 4 6 3 1 0 0 3 6 3 $20 $30 $40 Cost $60 $45 $20 $- $20 $- $- $145 Operations -- Prof. Juran 58

Crashing Solution

It is feasible to complete the project in 20 days, at a cost of $145.

1 3 F 6 4 0 Star t A 3 B 5 E 3 C 9 G 3 5 End 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