1Project Management:

PERT/CPM

Lecture 5

Lecturer : Dr. Dwayne Devonish

MGMT 2012: Introduction to Quantitative Methods

Learning Objectives

Students should be able to:

Identify, assess, and organise the various

elements/activities of a project,

Conduct a critical path analysis of a

project using PERT/CPM,

Compute a projects total completion time,

and determine the probability of

completion time.

2Project management scenarios

PERT/CPM tools can be used for:

Research and development of new

products or processes,

Construction of plants, buildings, and

highways

Designing and installation of new systems

Maintenance of large and complex

equipment

Planning complex events such as

weddings, conferences, festivals, etc

PERT/CPM Questions that can be addressed with

PERT/CPM include:

When will the entire project be completed?

What are the critical activities or tasks in the

project i.e. the ones that can delay the entire

project if they are late?

What are the noncritical activities in the project

i.e. the ones that can run late without delaying

the project

Is the project on schedule, behind schedule or

ahead of schedule?

3Steps of PERT/CPM

1. Define the project and all of its significant

activities and tasks,

2. Develop the relationships among the activities

decide which activities must precede others,

3. Draw the network connecting all of the

activities,

4. Assign time estimate to each activity

5. Compute the longest time path through

network critical path (first network diagram

input start and finish times).

6. Use network to help plan, schedule, and

monitor the project

Example of Project: Devonish

Building Project

Several activities must be completed for Devonish

Construction to expand an old shopping centre to

accommodate 8 more businesses. A list of

activities must be drawn up for this expansion

project which is Step 1 (define the tasks/activities

associated with overall project), and relationships

must be developed to highlight which activities

must precede other activities (Step 2).

An immediate predecessor (for a given activity) is

an activity that must be completed immediately

prior to the start of that given activity.

SEE TABLE NEXT SLIDE:

4Activity List for Devonish Project

Activity Description Immediate

Predecessor

A Prepare architectural drawings

-

B Identify potential tenants

-

C Select contractor A

D Prepare building permits A

E Obtain approval for permits D

F Perform construction C, E

G Finalise contracts with tenants B

H Tenants move in F,G

Step 3: Draw Network Diagram

There are two network diagrams that can be used:activity-on-node (AON) diagrams and activity-on-arrow(AOA) diagrams.

AON diagrams are more popular and easier tounderstand.

In AON diagrams, a separate activity is represented oneach node (rectangular box) in the diagram; the arcs(arrows) show the predecessors for the activities.

For example, arrows from A leading to C and Dindicate that A is an immediate predecessor (i.e.must be completed before) for both C and D.

5Draw Network Diagram

START

A

B

D

C

E F

G

H

FINISH

This network diagram shows activities on nodes (square

blocks). This diagram is not complete as time estimates

must be entered within various slots of each node which

are expressed in an upcoming table. The arrows show

which activities are related (i.e. must be completed after

certain activities).

Step 4 Assigning time estimates Each activity must be assigned a time estimate

which indicates the activity completion time.

In CPM, activity times are known with certainty.Hence, you assign a particular time estimate foreach activity.

However, in PERT (for uncertain times), acombined average of 3 time estimates for agiven activity is computed (i.e. expected time orT); these 3 estimates are:

1. Optimistic time (a) time an activity will take ifeverything goes as well as possible

2. Pessimistic time (b) time an activity will takeassuming very unfavourable conditions.

3. Most likely time (m) most realistic timeestimate for a given activity.

6Time Estimates Table (in weeks)Activity Optimistic

Time: a

Most Likely

Time: m

Pessimistic

Time: b

A 1 2 3

B 2 3 4

C 1 2 3

D 2 4 6

E 1 4 7

F 9 13 23

G 3 4 5

H 1 2 3

Note. Optimistic time for a given activity is always the shortest time estimate.

Pessimistic time is always the longest time estimate. Most likely time

estimate is always the one in the middle. Look at each activity above and see.

Expected Time and Variance

From this table, recall we need to obtain a combinedaverage which is referred to as the expected activitytime (denoted as t). We also have to compute thevariance of the activity time to determine the variationin completion time (given the times are uncertain).

A. For expected activity time, the formula is:

T = a + 4m + b , so for Activ. A :T = 1 + 4(2) + 3 = 2

6 6

B. The variance is computed by the formula:

V = (b a)2 so for Activ. A : V = (3 1)2 = 4

( 6 ) ( 6 ) 36

7Time Estimate Table (2)Activity Optimistic,

a

Most

Likely, m

Pessimistic

b

Exp.

T Var.

A 1 2 3 2 4/36

B 2 3 4 3 4/36

C 1 2 3 2 4/36

D 2 4 6 4 16/36

E 1 4 7 4 36/36

F 9 13 23 14 196/36

G 3 4 5 4 4/36

H 1 2 3 2 4/36

Step 5: Find Critical Path Now, we have to find the critical path i.e. longest time

path through the network. This gives us the total

completion time of the project.

Firstly, for each node (i.e. activity), we compute four

important quantities:

Earliest start time: (ES): the earliest time an activity

can begin (without violating immediate predecessor

requirements)

Earliest finish time: (EF): the earliest time at which an

activity can end.

Latest start time: (LS): the latest time an activity can

begin without delaying the entire project

Latest finish time: (LF): the latest time an activity can

end without delaying the entire project

8AN ACTIVITY NODES

FORMAT

ACTIVITY EXPECTED TIME

ESTIMATE

EARLIEST START TIME

(ES)

EARLIEST FINISH TIME

(EF)

LASTEST START TIME

(LS)

LATEST FINISH TIME

(LF)

Six pieces of information must be entered with six

slots of an activity node: Activity Type, Expected

Time, ES, EF, LS, and LF.

Activity List for Devonish ProjectActivity Description Immediate

Predecessor

T

A Prepare architectural drawings

-

2

B Identify potential tenants

-

3

C Select contractor A 2

D Prepare building permits A 4

E Obtain approval for permits D 4

F Perform construction C, E 14

G Finalise contracts with tenants B 4

H Tenants move in F,G 2

9Computing Quantities in Network

We first begin by calculating (earliest start) ES and (earliest finish) EF times. Activity A and Activity B can start as soon as the project starts, so we set the ES for these activities to zero: 0 or Week 0.

Now Activity A has a completion time of 2 weeks, and Activity B has a completion time of 3 weeks.

Since we have the ES for both of these activities, we need calculate the EF by adding the time (t) to the ES. Hence, EF = ES + t. So for Activity A, the EF is 0 + 2 = 2 (end of week 2); and for Activity B, the EF is 0 + 3 = 3 (end of week 3).

Network Diagram

A 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

REMEMBER THE EF times of A

and B = ES + T

10

Network Diagram

A 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

C 2

2 4

D 4

2 6

Given that A is an immediate

predecessor of

C and D; their ES times

correspond to the EF time of A (=2).

ES time of C is added to its T (=2) = EF= 4

ES time of D is added to its T (=4)=EF= 6

A 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

C 2

2 4

D 4

2 6

E 4

6 10

F 14

10 24

Now E : its ES time is the EF time of D (its

predecessor); and Es EF time = its ES + t

C has an EF time of 4 and E has an EF time of 10.

F must start after both of these activities are

completed; hence, the ES time for F will have to

be 10 (bigger or later of the two immediate

predecessors).

11

A 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

C 2

2 4

D 4

2 6

E 4

6 10

F 14

10 24

B is an immediate predecessor of G; so the

ES of G is the EF of B = 3. Hence, EF of G = 7

G 4

3 7

A 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

C 2

2 4

D 4

2 6

E 4

6 10

F 14

10 24

H must be completed after G and F is done;

hence, its ES must 24 (i.e. after the later of

these two predecessor activities)

G 4

3 7

H 2

24 26

12

A 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

C 2

2 4

D 4

2 6

E 4

6 10

F 14

10 24

10 24

Now we have to work backwards to find LS

and LF times (below). Last activity (H) has

the same LS and LF as EF and ES. The LF

times of its immediate predecessors (F and

G) can be worked out: their LFs = LS of H

(=24)

G 4

3 7

20 24

H 2

24 26

24 26

Now the LS times of F and G are worked out

after their LFs are found as: LS = LF T. LS

of F = 24-14 = 10; LS of G = 24 -4 =20.

A 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

C 2

2 4

D 4

2 6

2 6

E 4

6 10

6 10

F 14

10 24

10 24

E and D are worked out nicely.

G 4

3 7

20 24

H 2

24 26

24 26

13

A 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

17 20

C 2

2 4

8 10

D 4

2 6

2 6

E 4

6 10

6 10

F 14

10 24

10 24

G 4

3 7

20 24

H 2

24 26

24 26

B and C are worked out nicely. WAIT: A

has two activities following it, D and C.

Now remember, the LF of A = LS of

following activity. LS of D = 2; LS of C = 8.

Choose the smaller (earlier) of these

times = 2.

A 2

0 2

0 2

Act T

ES EF

LS LF

START

B 3

0 3

17 20

C 2

2 4

8 10

D 4

2 6

2 6

E 4

6 10

6 10

F 14

10 24

10 24

If you had chosen 8 for the LF for A; this

would have delayed D as its LS (the latest

it can start) must be 2 (End of Week 2).

The delay = As latest finish time would

have been at the end of Week 8.

G 4

3 7

20 24

H 2

24 26

24 26

14

Now we can work out critical path

Once we work out our ES, EF, LS, and LF times, we

would have come to our total completion time: i.e. 26

weeks.

We need to determine which activities are our critical

activities i.e. the critical path. We do this by

calculating slack for each activity.

Slack is the length of time an activity can be delayed

without affecting the entire project.

Hence, activities without slack are the critical

activities (i.e. these cant be delayed)

Slack is computed as: LS minus ES or LF minus EF

Critical Path??

Activity

ES EF LS LF

Slack

LS-ES

Critical

?

A 0 2 0 2 0 Yes

B 0 3 17 20 17

C 2 4 8 10 6

D 2 6 2 6 0 Yes

E 6 10 6 10 0 Yes

F 10 24 10 24 0 Yes

G 3 7 20 24 17

H 24 26 24 26 0 Yes

15

Critical Path Analysis Results Hence, the path A-D-E-F-H is the critical path.

The sum of these expected times = 26 weeks.

None of these activities can be delayed without

delaying the entire project.

We are able to answer the following questions:

1. How long will the project take to complete?

26 weeks

2. What are the scheduled start and completion times for

each activity?

Look at the ES, LS, and EF, LF times

3. Which activities are critical and must be completed

exactly as scheduled to keep the project on schedule?

A,D,E, F and H

Variation in Project Time We recognise that the total project will run for 26

weeks.

Recall we calculated the variance estimates for each

activity.

In particular, the variances of the critical activities

are important as these denote the variation in the

critical activities which can affect the entire project.

To determine overall variation in the entire project,

sum the variances of the critical activities:

4/36 + 16/36 + 36/36 + 196/36 + 4/36 OR

.11 + .44 + 1 + 5.44 + .11 = 7.11

The variance for entire project is 7.11 weeks

squared we need to convert these to weeks by

determining the standard deviation.

16

Variances of critical activities

Activity Optimistic,

a

Most

Likely, m

Pessimistic

b

Exp.

T Var.

A 1 2 3 2 4/36

B 2 3 4 3 4/36

C 1 2 3 2 4/36

D 2 4 6 4 16/36

E 1 4 7 4 36/36

F 9 13 23 14 196/36

G 3 4 5 4 4/36

H 1 2 3 2 4/36

Probability of Project Completion We have to find the square root of the variance

(7.11) to obtain the standard deviation (SD) which is

a better estimate of project completion time

variation.

The SD is 2.67 weeks. This means we are

expecting the project to be completed in 26 weeks

(plus or minus 2.67 weeks).

Now we can do more with this information: Lets say

we have a deadline for our project of 29 weeks.

We can calculate the probability of meeting this

deadline by the following formula:

Z = Due date Expected completion date

SD

17

Probability of Project Completion

So: Z = 29 26 = 1.12

2.67

The z-value of 1.12 is based on the normaldistribution, and represents the number of SDs thatthe due date lies from the expected date.

Look at your normal distribution table (this is in yourtext as well), look for 1.12 under z column, read theprobability - which is left of this value on the normalcurve. This value is .868.

We have about an 86.8% chance of completing theproject on or before the 29-week deadline.

SoM

Lets use QM for Windows to work out

this example.

You will see how QM software can

make our jobs easier as quantitative

analysts.

We will do some practice in tutorials.

18

Projects with known times

We were using project scheduling analysis

for a project with uncertain times (e.g. we

had to average 3 time estimates).

However, there are situations in which

task times are known (i.e. certain).

Hence, you dont have to worry about

pessimistic, optimistic or most likely time

estimates. You can generate your

network diagram with certainty. In these

cases, variation is not considered.

A PROJECT EXAMPLE WITH KNOWN TIMES:

LETS DO THIS ONE

Activity Immediate Predecessor COMPLETION

TIME

A

- 10 days

B

- 4 days

C A 3 days

D A, B 2 days

E D 1 day