PROGRAM EVALUATION AND REVIEW TECHNIQUE
Transcript of PROGRAM EVALUATION AND REVIEW TECHNIQUE
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PROGRAM EVALUATION
AND REVIEW TECHNIQUE
(PERT)
CRITICAL PATH METHOD
(CPM)
CHAPTER – 14
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• PERT was developed by the Navy
Special Project Office in
cooperation with Booz, Allen, and
Hamilton.
HISTORICAL BACKGROUND
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• 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
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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
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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.
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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
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bm4at
6DeviationStandard
ab
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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
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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
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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
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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
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PERT NETWORK FOR THE RESPONSE 1000
INTRODUCTION
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Response 1000 network with all earliest start (ES)
and all earliest finish (EF) time
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Response 1000 network with all latest start (LS) and latest finish (LF) times.
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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.
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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
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2
3
4
5
6
7
8
9
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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
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1
2
3
4
5 5
2
4
3
4
3
5
2 2
5
5
6
8
9
8
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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
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1
2
3
4
5 5
2
4
3
4
3
5
4 4
5
7
8
8
8
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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
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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
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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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CRASHING AND COST PLAN
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.
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CRASHING AND COST PLAN
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.
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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
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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?
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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
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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
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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.
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
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
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Step 3. Next path is with 30 weeks time duration
Along this path the least expensive activity is with P600.00
cost per week.
Crash this activity to its minimum time of 3 weeks.
Subtract: 8 – 3 = 5 weeks
Crash Cost = 5 weeks x P600.00 per week = P3,000.00
1
2 4 6
3 5 7
A
B
C
D
E
F
G
H
I
6
3
7
8
7
3
5
11
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
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Step 4. Next path is with 29 weeks time duration
Along this path the least expensive activity is with P1,000.00
cost per week.
Crash this activity to its minimum time of 6 weeks.
Subtract: 10 – 6 = 4 weeks
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
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Step 5. Next path is with 26 weeks time duration
Along this path the least expensive activity is with P400.00
cost per week.
Crash this activity to its minimum time of 4 weeks.
Subtract: 7 – 4 = 3 weeks
Crash Cost = 3 weeks x P400.00 per week = P1,200.00
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
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Step 6. Next path is with 26 weeks time duration
Along this path the least expensive activity is with P800.00
cost per week.
Crash this activity to its minimum time of 3 weeks.
Subtract: 7 – 3 = 4 weeks
Crash Cost = 4 weeks x P800.00 per week = P3,200.00
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
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Step 7. Next path is with 25 weeks time duration
Along this path the least expensive activity is with P700.00
cost per week.
Crash this activity to its minimum time of 2 weeks.
Subtract: 5 – 2 = 3 weeks
Crash Cost = 3 weeks x P700.00 per week = P2,100.00
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
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Step 8. Next path is with 20 weeks time duration
Along this path the least expensive activity is with P2,000.00
cost per week.
Crash this activity to its minimum time of 1 weeks.
Subtract: 3 – 1 = 2 weeks
Crash Cost = 2 weeks x P2,000.00 per week = P4,000.00
1
2 4 6
3 5 7
A
B
C
D
E
F
G
H
I
6
3
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
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Step 9. Next path is with 19 weeks time duration
Along this path the least expensive activity is with P2,000.00
cost per week.
Crash this activity to its minimum time of 2 weeks.
Subtract: 6 – 2 = 4 weeks
Crash Cost = 4 weeks x P2,000.00 per week = P8,000.00
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
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Step 10. Next path is with 17 weeks time duration
Along this path the least expensive activity is with P900.00
cost per week.
Crash this activity to its minimum time of 7 weeks.
Subtract: 11 – 7 = 4 weeks
Crash Cost = 4 weeks x P900.00 per week = P3,600.00
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
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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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