Class notes: Project Management

The objective of these notes is to provide you with a detailed description of the case study

discussed in class and a summary of the project management procedures and models used in

class to approach the case. These notes do not include all the details, or discussions or

explanations given in class and they are not intended to replace the class lecture. You should

read chapter 9 of the textbook for full coverage of the topic.

Milwaukee Paper Manufacturing Case

Milwaukee Paper Manufacturing, Inc., located near downtown Milwaukee, has long been

delaying the expense of installing air pollution control equipment in its facility. The

Environmental Protection Agency (EPA) has recently given the manufacturer 16 weeks to install

a complex air filter system. Milwaukee Paper has been warned that it may be forced to close

the facility unless the device is installed in the allotted time. Joni Steinberg, the plant manager,

wants to make sure that installation of the filtering system progresses smoothly and on time.

Given the following information, develop a table showing activity precedence relationships.

Approach: Milwaukee Paper has identified the eight activities that need to be performed in

order for the project to be completed. When the project begins, two activities can be

simultaneously started: building the internal components for the device ( activity A) and the

modifications necessary for the floor and roof (activity B). The construction of the collection

stack ( activity C) can begin when the internal components are completed. Pouring the concrete

floor and installation of the frame (activity D) can be started as soon as the internal

components are completed and the roof and floor have been modified. After the collection stack

has been constructed, two activities can begin: building the high- temperature burner ( activity

E) and installing the pollution control system ( activity F). The air pollution device can be

installed (activity G) after the concrete floor has been poured, the frame has been installed, and

the high-temperature burner has been built. Finally, after the control system and pollution

device have been installed, the system can be inspected and tested (activity H).

Solution: Activities and precedence relationships may seem rather confusing when they are

presented in this descriptive form. It is therefore convenient to list all the activity information

in a table, as shown in Table 1. We see in the table that activity A is listed as an immediate

predecessor of activity C. Likewise, both activities D and E must be performed prior to starting

activity G.

Table 1

Insight: To complete a network, all predecessors must be clearly defined.

Learning Exercise: What is the impact on the sequence of activities if EPA approval is required

after Inspect and Test?

Answer: The immediate predecessor for the new activity would be H, Inspect and Test, with

EPA approval as the last activity.

Note that in the Example, it is enough to list just the immediate predecessors for each activity.

For instance, in Table 1, since activity A precedes activity C, and activity C precedes activity E,

the fact that activity A precedes activity E is implicit. This relationship need not be explicitly

shown in the activity precedence relationships. When there are many activities in a project with

fairly complicated precedence relationships, it is difficult for an individual to comprehend the

complexity of the project from just the tabular information. In such cases a visual

representation of the project using a project network is convenient and useful. A project

network is a diagram of all the activities and the precedence relationships that exist between

these activities in a project. The project network for Milwaukee Paper Manufacturing is as

follows.

Activity DescriptionImmediate

Predecessors

A Build internal components —

B Modify roof and floor —

C Construct collection stack A

D Pour concrete and install frame A, B

E Build high-temperature burner C

F Install pollution control system C

G Install air pollution device D, E

H Inspect and test F, G

Determining the Project Schedule

Perform a Critical Path Analysis

The critical path is the longest path through the network

The critical path is the shortest time in which the project can be completed

Any delay in critical path activities delays the project

Critical path activities have no slack time

Activity Description Time (weeks)

A Build internal components 2

B Modify roof and floor 3

C Construct collection stack 2

D Pour concrete and install frame 4

E Build high-temperature burner 4

F Install pollution control system 3

G Install air pollution device 5

H Inspect and test 2

Total Time (weeks) 25

Earliest start (ES) = earliest time at which an activity can start, assuming all predecessors

have been completed

Earliest finish (EF) = earliest time at which an activity can be finished

Latest start (LS) = latest time at which an activity can start so as to not delay the completion

time of the entire project

Latest finish (LF) = latest time by which an activity has to be finished so as to not delay the

completion time of the entire project

Possible representation of an activity node (Note: your textbook shows another possible

representation)

Following this representation and after performing the forward and backward passes of the

critical path method (CPM) the network model becomes.

This shows the critical path linking all critical activities and gives the total project completion

time of 15 weeks.

The following table summarizes these results and shows the computation of the slack times for

each of the project activities. Note that critical activities have slack zero.

Also, from the information above the following Gantt Charts can be obtained, which are

particularly helpful in the monitoring (or controlling) phase.

ES → EF Gantt Chart for Milwaukee Paper

Earliest Earliest Latest Latest OnStart Finish Start Finish Slack Critical

Activity ES EF LS LF LS – ES Path

A 0 2 0 2 0 Yes

B 0 3 1 4 1 No

C 2 4 2 4 0 Yes

D 3 7 4 8 1 No

E 4 8 4 8 0 Yes

F 4 7 10 13 6 No

G 8 13 8 13 0 Yes

H 13 15 13 15 0 Yes

A Build internal components

B Modify roof and floor

C Construct collection stack

D Pour concrete and install frame

E Build high-temperature burner

F Install pollution control system

G Install air pollution device

H Inspect and test

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

LS → LF Gantt Chart for Milwaukee Paper

Variability in Activity Times

CPM assumes we know a fixed time estimate for each activity and there is no variability in

activity times.

PERT uses a probability distribution for activity times to allow for variability.

Three time estimates are required

Optimistic time (a) – if everything goes according to plan

Most likely time (m) – most realistic estimate

Pessimistic time (b) – assuming very unfavorable conditions

Estimate follows Beta distribution

Expected time:

4

6

a m bt

Variance of times:

22

6

b a

A Build internal components

B Modify roof and floor

C Construct collection stack

D Pour concrete and install frame

E Build high-temperature burner

F Install pollution control system

G Install air pollution device

H Inspect and test

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Beta Distribution

Given estimated times (a, m, and b) each activity the following table shows the computation of

the corresponding expected times and variance.

Probability of Project Completion

Project variance is computed by summing the variances of critical activities:

σ2 = Project variance = (variances of activities on critical path)

( , )N

Project variance is computed by

Project variance

2 .11 .11 1.00 1.78 .11 3.11p

Project standard deviation

project variance 3.11 1.76 weeksp

PERT makes two more assumptions:

Total project completion times follow a normal probability distribution –

Activity times are statistically independent

In the example, if x denotes a random variable representing a due date, we have

x ~ N (15, 1.76)

Determining the Probability of Project Completion

Question 1: What is the probability this project can be completed on or before the 16 week

deadline?

Probabilities are given in Normal distribution tables for a random variable z that follows a

standard Normal distribution with mean 0 and standard deviation 1.

More formally, z ~ N (0, 1).

The transformation of x into z is given by

p

xz

In our example

16 150.57

1.76z

As a result, μ = 15 and x = 16 in the original Normal distribution correspond to μ = 0 and z =

0.57 in the standard Normal distribution.

( 16) ( 0.57)P x P z

From Appendix A (Normal curve areas), we get

( 0.57) 0.5 0.2157 0.7157,P z hence, approximately 72%.

Question 2: What’s the due date that gives the company’s project a 99% chance of on-time

completion?

The answer to this question consists of determining the value of x for the z-value that

corresponds to a 99% of probability.

x z

Again, from Appendix A we see that for a 99% of probability given by 0.50 (area up to the

mean) + 0.4901 (area above the mean as given in Table A.1) = 0.99, the corresponding z value

is 2.33.

Hence,

15 2.33(1.76) 19.1 weeks.x

In sum: What Project Management has given so far?

The project’s expected completion time is 15 weeks

There is a 71.57% chance the equipment will be in place by the 16 week deadline

There is a 99% chance the equipment will be in place within about 19 weeks

Five activities (A, C, E, G, and H) are on the critical path

Three activities (B, D, F) are not on the critical path and have slack time

A detailed schedule is available