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Unit no.3
Programme Evaluation & Review
Technique
(P.E.R.T)
Mr. Kiran R. Patil
Assistant Professor,
Department of Civil Engineering,
D. Y. Patil College of Engineering & Technology, Kolhapur

 Introduction to PERT
• PERT was developed by U.S. Navy engineers while working on the Polaris Missile
Programme during 1957-58.
• PERT is used for planning and controlling the projects involving uncertainties. This
technique is usually used for non-repetitive project such as launching of satellites, research
and development (R & D) projects, etc. in which correct time estimation for various
activities cannot be made due to lack of past data.
• PERT is an event-oriented technique. This technique uses a network diagram consisting of
events which must be established to reach project objectives.
• More stress is given on time in this technique. The uncertainty in activity times is
measured by using the following three time estimates,
1. The optimistic time estimate
2. The pessimistic time estimate
3. The most likely time estimate
• For computation of critical path, the three-time are converted into a single-time in
network. This single time is called the expected time.
1. The Optimistic time estimate (tO):-
This is the shortest possible time in which an activity can be completed under ideal
conditions. It is possible when all the conditions are favor of project.
2. The Pessimistic time estimate (tP)
This is the maximum time that would be required to complete the activity. It represents

the time it might take to complete a particular activity if everything went wrong
about of project.
3. The Most likely time estimate (tL)
• It is the normal time required to complete an activity. It is in between the position
of optimistic & pessimistic time.
tO + 4tL + tP
• Expected Time, tE =
6

• Comparison between C.P.M. and P.E.R.T.
Sr.No. C.P.M P.E.R.T.
1 CPM is Critical Path Method PERT is Programme Evaluation and Review
Technique
2 CPM is used for repetitive type of projects
where accurate time estimates can be made
and costs can be calculated in advance.
e.g. construction projects
PERT is used for non-repetitive types of projects
where accurate time estimates cannot be calculated
in advance.
e.g. Research and Development projects,
Launching of missile or satellite.
3 In CPM, cost optimization is given prime
importance. The time duration for completion
depends on this cost optimization. The cost is
not directly proportional to time.
i.e. cost is the direct controlling factor.
In PERT, it is assumed that the cost varies directly
with time. Therefore, time minimization is given
more attention.
i.e. time is the controlling factor.
4 CPM uses activity-oriented network diagram.
i.e. importance is given on activities
PERT uses event-oriented network diagram.
i.e. importance is given on events
5 In CPM, critical path is the one which passes
through critical activities.
In PERT, critical path is the one which passes
through critical events.
6 In CPM, the critical path is determined by
activity-oriented float.
In PERT, the critical path is determined by event-
oriented slack.

• Frequency Distribution
• The most likely time (tL) can also be found out by using frequency distribution curve.
• The curve is symmetrical about its apex. Such a curve is known as ‘Normal Distribution
Curve’.
• The statistical data for varying time durations for the jobs of particular type consumed in
the past can be expressed in the form of a frequency distribution curve.
• The probability curve which is not symmetrical about its apex is known as ‘Beta
Distribution Curve’.

• The beta distribution is used in PERT because it satisfies the following requirements,
1. The distribution should have a small probability of reaching the optimistic time (shortest
time).
2. The distribution should have a small probability of reaching the pessimistic time (longest
time)
3. The distribution should have only one most likely time which would be free to move
between the two limits of to and tp.
• For beta distribution, the standard deviation is given by,
• 𝜎 =
𝑡𝑝−𝑡𝑜
6
• The variance 𝜎2 =
𝑡𝑝−𝑡𝑜
6
2
• Ex. 1
• On a construction project, the times required for digging 54 trenches of fixed dimensions
are recorded below. The trenches were excavated by different gangs, each consisting of the
same number of labours. Plot the frequency distribution curve.
Times of completion of trenches (days)
8 10 12 6 9 11 10 13 15 12
11 9 8 10 14 16 12 16 15 10
14 12 7 9 13 10 8 11 17 13
9 11 13 10 14 9 12 15 14 9
10 9 11 10 7 13 11 8 12 11
8 10 9 11

Solution:
• From the above record, we find that the minimum time taken for completion of
trenches is 6 days, which is the optimistic time (tO). The maximum time is 17 days
which is the pessimistic time (tP).
• The following table gives the number of trenches completed in 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16 and 17 days respectively.
Days of
completion
No. of trenches
completed during
these days
Days of
completion
No. of trenches
completed during
these days
6 1 12 6
7 2 13 5
8 5 14 4
9 8 15 3
10 9 16 2
11 8 17 1

Program Evaluation & Review Technique for Civil Engineering

 P.E.R.T.: Network Analysis
• Slack:
• Slack means ‘time to spare’
• It is the difference between the latest allowable occurrence time and the earliest expected
time of an event
• S = TL – TE
• Slack can be +ve, zero or –ve
1. Positive Slack: when TL > TE
It is an indication of an ‘ahead of schedule’ situation (excess resources)
2. Zero Slack: when TL = TE
It is an indication of ‘on schedule’ situation (adequate resources)
3. Negative Slack: when TL < TE
It is an indication of a ‘behind schedule’ situation (lack of resources)
• Critical Path:
• It is the one which consumes maximum time. It is time wise longest path.
• It is time wise longest path who connects the events of zero slack.

• The network for a project is shown below. Determine the expected time for each path.
Which path is critical?
Solution: There are 4 paths
1. Path A: 1-2-7-8
2. Path B: 1-2-6-8
3. Path C: 1-3-6-8
4. Path D: 1-4-5-6-8
• Expected time tE = tO + 4tL + tP
6

Path Activity tO tL tP tE ∑= tE
A
1-2 6 8 11 8.17
26.34
2-7 8 10 12 10.00
7-8 5 8 12 8.17
B
1-2 6 8 11 8.17
26.83
2-6 4 8 14 8.33
6-8 7 10 15 10.33
C
1-3 3 7 9 6.67
27.00
3-6 8 10 12 10.00
6-8 7 10 15 10.33
D
1-4 5 7 10 7.17
28.33
4-5 4 6 8 6.00
5-6 3 5 6 4.83
6-8 7 10 15 10.33
Path D is critical since∑= tE for this path is maximum.= 28.33

• Finding the probability of meeting the scheduled time of completion of a
project
• Determine the standard deviation (𝜎) appropriate to the critical path.
• 𝝈 = 𝒔𝒖𝒎 𝒐𝒇 𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆𝒔 𝒂𝒍𝒐𝒏𝒈 𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒑𝒂𝒕𝒉
• 𝝈 = ∑𝝈𝒊𝒋𝟐
Where, 𝜎𝑖𝑗2 = variance for the activity i-j along the critical path
𝝈𝒊𝒋𝟐 =
𝒕𝒑𝒊𝒋−𝒕𝒐𝒊𝒋
𝟔
𝟐
• Find probability factor Z.
𝒁 =
𝑻𝒔−𝑻𝑬
𝝈
=
𝑻𝒔−𝑻𝑬
∑𝝈𝒊𝒋𝟐
Where, Ts = scheduled time of completion
TE = earliest expected time of completion
• Z can be +ve, 0 or –ve.
1. When Z is +ve, the chances of completing the project in scheduled time are
more than 50 %.
2. When Z is 0, the chances of completing the project in scheduled time are fifty-
fifty.
3. When Z is - ve, the chances of completing the project in scheduled time are
less than 50 %.
• Find % probability with respect to the normal deviation Z from the table.

Values of Standard Normal Distribution Function

Problem 1. A project is expected to take 15 months along the critical path, having a
standard deviation of 3 months. What is the probability of completing the project within
(i) 15 months, (ii) 18 months, (iii) 12 months?
• Solution:
• Given Data: TE= 15 months , 𝜎 = 3 months
• Probability Factor 𝑍 =
𝑇𝑠−𝑇𝐸
𝜎
1) Ts = 15 months;
𝑍 =
15 − 15
3
= 0
• For Z = 0, Probability = 50 % (from the Table)
2) Ts = 18 months;
𝑍 =
18 − 15
3
= 1
For Z = 1, Probability = 84.13 % (from the Table)
3) Ts = 12 months;
𝑍 =
12 − 15
3
= −1
• For Z = -1, Probability = 15.87 % (from the Table)

Problem 2. With the information given in the table draw the network for a construction
project. Determine,
1. Critical path and its standard deviation
2. Probability of completion of project in 40 days.
3. Time duration that will provide 95 % probability of its completion in time.
• Solution:
1. Calculate expected time for each activity,
Activity Optimistic time Most likely time Pessimistic time
1-2 2 5 8
2-3 8 11 20
3-4 0 0 0
2-4 4 7 16
2-5 4 9 20
4-6 7 10 13
5-6 3 7 17
3-7 3 5 13
6-7 2 3 10
7-8 2 4 6

Activity Optimistic
time
tO
Most likely
time
tL
Pessimistic time
tP
Expected Time
tE = tO + 4tL + tP
6
𝜎2
=
𝑡𝑝 − 𝑡𝑜
6
2
1-2 2 5 8 5 1
2-3 8 11 20 12 4
3-4 0 0 0 0 0
2-4 4 7 16 8 4
2-5 4 9 20 10 7.11
4-6 7 10 13 10 1
5-6 3 7 17 8 5.44
3-7 3 5 13 6 2.78
6-7 2 3 10 4 1.77
7-8 2 4 6 4 0.44
2. Draw the Network with expected time
Critical path: 1-2-3-4-6-7-8
Project duration = 35 days

1. Standard deviation along critical path = 𝜎 = ∑𝜎𝑖𝑗2
∑𝜎𝑖𝑗2
= 1 + 4 + 0 + 1 + 1.77 + 0.44 = 8.2
𝝈 = 8.21 = 2.87
• Probability Factor Z= (Ts-TE)/σ
2. Here, Ts = 40 days and σ = 2.87 months
• Z= (40-35)/2.87=1.74
• From the table,
• For Z = 1.7, Pr = 95.54 %, For Z = 1.8, Pr = 96.41 %
• For Z = 1.74, Pr = ?
1.7 = 95.54
1.8 = 96.41
0.1 = 0.87
• 0.04 =
0.04 𝑥 0.87
0.1
= 0.35
• For Z = 1.74, Pr = 95.54 + 0.35 = 95.89 %
3. For Pr = 95 %, Ts =?
• For 95.54 = 1.7
• For 94.52 = 1.6
• For 95, Z=?
Pro. 95.54-94.52= 1.02
Z 1.7-1.6=0.1
(0.54x1.02)/0.1= 0.05
Z = 1.6 + 0.05 = 1.65

• 𝑍 =
𝑇𝑠−𝑇𝐸
𝜎
• Ts = (𝜎 𝑋 Z) + TE
• = (2.87 x 1.65) + 35
• = 39.7
Ts = 40 days
Problem 3. For the information given in the table determine the critical path and standard
deviation for the network. Determine the probability of completing the project in 35 days.
• ( For Z = 0.9, Pr = 81.59 %. For Z = 1.0, Pr = 84.13 % and For Z = 1.1, Pr = 86.43 %)
Activity Optimistic time
tO
Most likely time
tL
Pessimistic time
tP
1-2 6 9 18
1-3 5 8 17
2-4 4 7 22
3-4 4 7 16
4-5 4 10 22
2-5 4 7 10
3-5 2 5 8

• Solution:
1. Draw the Network
2. Find expected time & Variance
Activity Optimistic
time
tO
Most likely
time
tL
Pessimistic
time
tP
Expected Time
tE = tO + 4tL + tP
6
𝜎2
=
6
2
1-2 6 9 18 10 4
1-3 5 8 17 9 4
2-4 4 7 22 9 9
3-4 4 7 16 8 4
4-5 4 10 22 11 9
2-5 4 7 10 7 1
3-5 2 5 8 5 1

• Standard deviation along critical path = 𝜎 = ∑𝜎𝑖𝑗2
• ∑𝜎𝑖𝑗2
= 4 + 9 + 9 = 22
• 𝜎 = 22 = 4.69
 Here, Ts = 35 days and 𝜎 = 4.69
• Probability Factor 𝑍 =
𝑇𝑠−𝑇𝐸
𝜎
• 𝑍 =
35−30
4.69
= 1.07
• From the table,
• For Z = 1.0, Pr = 84.13 %
• For Z = 1.1, Pr = 86.43%
• For Z= 1.07, Pr= ?
• 0.07= 2.3
• (0.07x 2.3)/0.1= 1.61
• For Z = 1.07, Pr = 84.13 + 1.61 = 85.74 %

Problem 4. For the information given in the table determine the critical path and standard
deviation for the network. Determine the probability of completing the project in 50 days.
( For Z = 0.7, Pr = 75.80 %. For Z = 0.8, Pr = 78.81% and For Z = 0.6, Pr = 72.57 %)
1. Find Expected time and variance
tO
Most likely time
tL
Pessimistic time
tP
1-2 4 8 18
1-3 4 5 18
2-4 3 7 23
2-5 1 11 33
3-4 6 6 18
3-5 3 9 21
4-5 7 12 35
5-6 9 8 31

tO
Most likely time
tL
Pessimistic time
tP
Expected Time
TE = tO + 4tL + tP
6
𝜎2
=
6
2
1-2 4 8 18 9 5.44
1-3 4 5 18 7 5.44
2-4 3 7 23 9 11.11
2-5 1 11 33 13 28.44
3-4 6 6 18 8 4
3-5 3 9 21 10 9
4-5 7 12 35 15 21.77
5-6 9 8 31 12 13.44
2. Draw network & find expected time
1
2
3
4
5 6
9
7
9
8
13
10
15
12
0
9
7
18
33 45
45
33
18
10
9
0
• Critical Path 1-2-4-5-6
• Project Duration = TE= 45 days

• Standard deviation along critical path = 𝜎 = ∑𝜎𝑖𝑗2
∑𝜎𝑖𝑗2
= 5.44+11.11+21.77+13.44 = 51.76
𝜎 = 51.76 = 7.19
• Probability of project completed in 50 days = Ts
𝑍 =
𝑇𝑠 − 𝑇𝐸
𝜎
𝑍 =
50−45
7.19
=0.69
• From table
Z= 0.6, Pr.= 72.57 Z= 0.7, Pr.= 75.80
0.6 72.57
0.7 75.80
0.1 3.23
0.09 x
0.1x= 0.09 X 3.23
x= 2.097
• Probability of completing project within 50days= 72.57+2.907= 75.48%

 Precedence Networks :
• In CPM or PERT networks, activities are connected according to the finish-to-start logic,
i.e. an activity starts only after its preceding activity is completed. But in reality, there will
be a certain overlapping of time in between the adjoining activities.
• e.g. In a multi-housing construction project, the CPM network may show the activity of
wall plastering as starting after the completion of masonry work of a particular building.
In practice, however, plastering can start as soon as the first room is complete. It need not
wait for completion of masonry work of all the rooms.
• The precedence network technique keeps the number of activities same as CPM networks
(A-O-A networks) but eliminates the dummies. In the precedence networks nodes
represent the activities and lines represent their interdependencies or precedence
relationships.
 Representation of nodes in Precedence Networks:
Description Code
EST Dur. EFT
LST No. LFT
Steel Fabrication
C
0 5 5
0 3 5
Rebar Transport
D
5 1 6
5 4 6
Rebar Fixing
E
6 3 9
6 5 9

 Logic of Precedence Diagram
1. Finish to Start
Activity B cannot start until activity A has been completed.
2. Start to Start
Activity B can start at the same time A but not before.
3. Finish to Finish
Activity B cannot be finished until activity A has been finished.
4 Start to start & Finish to Finish:
A B
Time lag
A
B
A
B
A
B
Time lag
Time lag
Time lag
Time lag

• Activity B cannot be started and cannot be finished earlier than the stated time lag after the
start and finish of activity A. This sequence is used to define overlapping independent
activities.
 Advantages of Precedence Networks:
• Precedence network can show activities which should be allowed to overlap each other or
must be separated by a time delay.
• Precedence network is self sufficient as it contains the necessary information regarding the
project. Time analysis results are incorporated in the drawing itself. This facilitates efficient
project scheduling and control.
• Revisions and modifications can be carried out easily without affecting most of the
activities
• No arrow is required to show the logical sequence of activities. From the drawings, it is
clear that all activities on the left precede those on the right.
• Precedence network adopts simple notations similar to flow charts and can be easily
understood by any person.

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Program Evaluation & Review Technique for Civil Engineering

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