program evaluation and review technique

1

PROGRAM EVALUATION

AND REVIEW TECHNIQUE

(PERT)

CRITICAL PATH METHOD

(CPM)

CHAPTER – 14

2

• PERT was developed by the Navy

Special Project Office in

cooperation with Booz, Allen, and

Hamilton.

HISTORICAL BACKGROUND

3

• It was specifically directed at

planning and controlling the Polaris

missile program, a massive project

which had 250 prime contractors

and 9,000 subcontractors.

HISTORICAL BACKGROUND

4

Dummy activity - an activity which

requires no time and which is used to

establish a precedence relation of a

network.

Immediate predecessors – the activities that

must immediately precede a given activity

in a project.

DEFINITION OF TERMS

5

ACTIVITY TIME IN A PERT

NETWORK

Time is often expressed in calendar weeks because most of the activity in a PERT network will take considerable time to accomplish.

6

CHARACTERISTIC OF THE DISTRIBUTION

USED TO EXPRESS THE VARIATION OF TIME

1. A small probability (1 in 100) of reaching the most optimistic time (shortest time), symbolized a

2. A small probability (1 in 100) of reaching the most pessimistic time (longest time), symbolized b

3. One and only one most likely time, symbolized m, which would be free to move between the two extremes mentioned in 1 and 2 above

4. The ability to measure uncertainty in the estimating

7

6

bm4at

6DeviationStandard

ab

8

The earliest-start-time rule

The earliest time for an activity leaving any

node is equal to the largest earliest finish time of

all activities entering that same node.

The earliest start time for activities D and E is 2, the

earliest finish time for activity A.

Forward pass

9 Critical path - the longest path through the network

A path is defined as a sequence of connected activities that

leads from the beginning of the project to the end of the project.

Earliest finish time = earliest start time + expected time

EF = ES + t

10

The latest-finish-time rule

The second step in finding the critical path is to compute a latest start time and latest finish time for each activity.

Latest finish time is simply the latest time at which an activity can be completed without extending the completion time of the network.

The latest time start is the latest time at

which an activity can begin without extending the

completion time on the project

Latest start time = latest finish time - expected time

LS = LF - t

11

Activity Description Develop the advertising plan (a detailed plan of project radio,

television, and newspaper advertising) Develop the promotion and training materials plan (a

detailed study of the materials that will be required for training the store manager for the final in-store introduction)

Develop the training plan (the design of the training program that the store manager will undertake prior to the final in-store introduction)

Schedule the radio, television, and newspaper advertisements that will appear prior to the final introduction

Develop the advertising copy that will be required A “dummy” activity (one that takes no time) which lets us

draw networks using only straight line or which establishes precedence relationships (What activities precede each other)

Prepare promotion materials which will be used during the in-store introduction

Prepare materials which will be used in the training program for the store managers

Conduct the pre-introduction advertising campaign in the media

Screen and select the store managers who will undergo training

Conduct the training program The final in-store introduction of the Response 1000

Activity Symbol

A B

C

D

E F

G

H

I J

K L

Immediate

Predecessors --

--

--

A

A

E

B

B

D, F

C

H, J

G, I, K

12

Activity Description Develop the advertising plan (a detailed plan of project radio,

television, and newspaper advertising) Develop the promotion and training materials plan (a

detailed study of the materials that will be required for training the store manager for the final in-store introduction)

Develop the training plan (the design of the training program that the store manager will undertake prior to the final in-store introduction)

Schedule the radio, television, and newspaper advertisements that will appear prior to the final introduction

Develop the advertising copy that will be required A “dummy” activity (one that takes no time) which lets us

draw networks using only straight line or which establishes precedence relationships (What activities precede each other)

Prepare promotion materials which will be used during the in-store introduction

Prepare materials which will be used in the training program for the store managers

Conduct the pre-introduction advertising campaign in the media

Screen and select the store managers who will undergo training

Conduct the training program The final in-store introduction of the Response 1000

Activity Symbol

A B

C

D

E F

G

H

I J

K L

Immediate

Predecessors --

--

--

A

A

E

B

B

D, F

C

H, J

G, I, K

13

PERT NETWORK FOR THE RESPONSE 1000

INTRODUCTION

14

Response 1000 network with all earliest start (ES)

and all earliest finish (EF) time

15

Response 1000 network with all latest start (LS) and latest finish (LF) times.

16

Slack is the length of time we can delay an activity

without interfering with the project completion.

Slack = LF – EF or LS – ES

LF – EF for activity A = 7 – 2 = 5

LS – ES for activity A = 5 – 0 = 5

Latest-finish-time rule would be that the latest finish time for an activity entering any node is equal to the smallest latest start time for all activities entering that same node.

17

Example 1. From the following table of information, (a) Draw a precedence diagram, (b) Find the critical path, (c) determine the expected duration of the project

Activity Precedes Expected Time (days)

A C,B 4

C D 12

D I 2

B I 5

E F 3

F J 8

I J 12

J end 9

G H 1

H K 3

K end 15

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SOLUTION

1. Use pencil in constructing precedence diagram;

2. Activities without predecessors are placed at the start of

the network (left side);

3. Activities with multiple predecessors are located at path

intersections;

4. Start with a single node and end with a single node;

5. Avoid having paths that cross with each other;

6. Number nodes from left to right;

7. Activities should go from left to right;

8. Use only one arrow between any pair of nodes.

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Activities without predecessors:

A precedes C, B

E precedes F

G precedes H

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A = 4

E = 3

G = 1

C = 12

B = 5

F = 8

H = 3

D = 2

I = 12

K = 15

J = 9

Path:

A, C, D, I, J = 4 + 12 + 2 + 12 + 9 = 39*

A, B, I, J = 4 + 5 + 12 + 9 = 30

E, F, J = 3 + 8 + 9 = 20

G, H, K = 1 + 3 + 15 = 19

The Critical Path:

A, C, D, I, J

Duration of Project:

39 days

1

2

3

4

5

6

7

8

9

21

Example 2. From the following network diagram,

determine the following:

a. The early start (ES) in days

b. The early finish (EF) in days

c. The latest start (LS) in days

d. The latest finish (LF) in days

1

2

3

4

5 5

0

2

4

3

4

3

22

1

2

3

4

5 5

2

4

3

4

3

5

2 2

5

5

6

8

9

8

11

Numbers inside the blue box is the Early Finish Time of the activity

Numbers inside the red box is the Early Start Time of the activity

The project duration is 11 days

To determine the Early Start (ES) and Early

Finish (EF) Time

0

23

1

2

3

4

5 5

2

4

3

4

3

5

4 4

5

7

8

8

8

11

Numbers inside the blue box is the Latest Finish Time of the activity

Numbers inside the red box is the Latest Start Time of the activity

The project duration is 11 days

To determine the Latest Start (LS) and Latest

Finish (LF) Time

0

2

24

To compute the slack time using LS – ES or LF - EF

Activity LS ES SLACKS LF EF SLACKS

1-2 0 0 0 5 5 0

1-3 2 0 2 4 2 2

2-4 5 5 0 8 8 0

2-5 7 5 2 11 9 2

3-4 4 2 2 8 6 2

4-5 8 8 0 11 11 0

The activities with zero slack time indicate the critical path.

The critical paths are activities 1-2-4-5 having 11 days duration

25

CRASHING AND COST PLAN

1. Reducing Time and Cost

Purpose:

•To avoid penalties for not completing the project on time.

•To take advantage of monetary incentives for completing the project

on or before the target date.

•To free the resources such as, money, equipment and men for use on

other projects.

•Reduce indirect costs associated with the project such as:

1.Facilities and equipment

2.Supervision cost

3.Labor Cost

4.Personnel Cost

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1. Reducing Time and Cost

The following information should be

needed:

Regular time and crash time estimates.

Regular cost and crash estimate for

each activity.

A list of activities are on the critical

path.

27


2. Procedure in Crashing Project Time

Obtain an estimate of regular and crash time plus the costs of each activity.

Determine the length of all paths and their float/slack time.

Determine which activities are on the critical path.

Crash the critical activities in the order of increasing costs as long as crashing costs do not exceed benefits.

28

Example 3. From the following data, perform

crashing the project.

Activity Time in Weeks Cost Cost

Per Week Normal Crash Normal Crash

A 6 2 2,000.00 10,000.00 2,000.00

B 8 3 4,000.00 7,000.00 600.00

C 7 4 2,000.00 3,200.00 400.00

D 12 8 8,000.00 10,000.00 500.00

E 7 3 2,000.00 5,200.00 800.00

F 3 1 10,000.00 14,000.00 2,000.00

G 5 2 6,000.00 8,100.00 700.00

H 11 7 6,000.00 9,600.00 900.00

I 10 6 4,000.00 8,000.00 1,000.00

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1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I 6

8

7

12

7

3

5

11

10

Question: Can we finish the project in 18 weeks instead of 33

weeks at a cost of less than P75,100.00 in direct cost?

30

Path Duration in Weeks

A-C-F-I

B-E-H

B-D-F-I

B-E-G-I

6 + 7 + 3 + 10 = 26

8 + 7 + 11 = 26

8 + 12 + 3 + 10 = 33*

8 + 7 + 5 + 10 = 30

Step 1. Determine the different lengths and identify the

critical path.

Path B-D-F-I with 33 weeks is the critical path

31

Activity Time in Weeks Cost Cost

Per Week Normal Crash Normal Crash

A 6 2 2,000.00 10,000.00 2,000.00

B 8 3 4,000.00 7,000.00 600.00

C 7 4 2,000.00 3,200.00 400.00

D 12 8 8,000.00 10,000.00 500.00

E 7 3 2,000.00 5,200.00 800.00

F 3 1 10,000.00 14,000.00 2,000.00

G 5 2 6,000.00 8,100.00 700.00

H 11 7 6,000.00 9,600.00 900.00

I 10 6 4,000.00 8,000.00 1,000.00

Total 33 18 44,000.00 75,100.00

Rank the critical path activities in the order of crashing the lowest

cost and find out how many number of days can be crashed.

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Step 2. Starting from path with 33 weeks time duration

Along this path the least expensive activity is with P500.00

cost per week.

Crash this activity to its minimum time of 8 weeks.

Subtract: 12 – 8 = 4 weeks

Crash Cost = 4 weeks x P500.00 per week = P2,000.00

1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I

6

8

7

12

7

3

5

11

10

T = 33

A-C-F-I = 26 weeks

B-E-H = 26 weeks

B-D-F-I = 33 weeks

B-E-G-I = 30 weeks

33

Step 2. Starting from path B-D-F-I with 33 weeks time duration

Along this path the least expensive activity is D with P500.00

cost per week.




1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I

6

8

7

8

7

3

5

11

10

T = 29

A-C-F-I = 26 weeks

B-E-H = 26 weeks

B-D-F-I = 29 weeks

B-E-G-I = 30 weeks

34

Step 3. Next path is with 30 weeks time duration


cost per week.




1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I

6

3

7

8

7

3

5

11

10

T = 25

A-C-F-I = 26 weeks

B-E-H = 26 weeks

B-D-F-I = 29 weeks

B-E-G-I = 25 weeks

35


Along this path the least expensive activity is with P1,000.00

cost per week.



Crash Cost = 4 weeks x P1,000.00 per week = P4,000.00

1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I

6

3

7

8

7

3

5

11

6

T = 20

A-C-F-I = 26 weeks

B-E-H = 26 weeks

B-D-F-I = 20 weeks

B-E-G-I = 25 weeks

36



cost per week.




1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I 6

3

4

8

7

3

5

11

6

T = 19

A-C-F-I = 19 weeks

B-E-H = 26 weeks

B-D-F-I = 20 weeks

B-E-G-I = 25 weeks

37



cost per week.




1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I

6

3

4

8

3

3

5

11

6

T = 17

A-C-F-I = 19 weeks

B-E-H = 17 weeks

B-D-F-I = 20 weeks

B-E-G-I = 25 weeks

38



cost per week.




1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I

6

3

4

8

3

3

2

11

6

T = 14

A-C-F-I = 19 weeks

B-E-H = 17 weeks

B-D-F-I = 20 weeks

B-E-G-I = 14 weeks

39



cost per week.




1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I

6

3

4

8

3

1

2

11

6

T = 18

A-C-F-I = 19 weeks

B-E-H = 17 weeks

B-D-F-I = 18 weeks

B-E-G-I = 14 weeks

40



cost per week.




1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I 2

3

4

8

3

1

2

11

6

T = 13

A-C-F-I = 13 weeks

B-E-H = 17 weeks

B-D-F-I = 18 weeks

B-E-G-I = 14 weeks

41



cost per week.




1

2 4 6

3 5 7

A

B

C

D

E

F

G

H

I

2

3

4

8

3

1

2

7

6

T = 13

A-C-F-I = 13 weeks

B-E-H = 13 weeks

B-D-F-I = 18 weeks

B-E-G-I = 14 weeks

42

Therefore;

Total Crash Cost =

Total Crash Cost = 31,100.00

Total Cost = Normal Cost + Total Crash Cost

= 44,000.00 + 31,100.00

Total Cost = 75,100.00

Length of Time per Path;

A-C-F-I =

B-E-H =

B-D-F-I =

B-E-G-I =

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