EXPLANATORY CASE ON PERT

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Unit No.4.
NETWORK ANALYSIS
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean, EDP & Associate Professor MBA
1
Sanjivani College of Engineering, Kopargaon
Department of MBA

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302-DECISION SCIENCE
Unit No.4. Network Analysis
4.13 Explanatory Case on Program
Evaluation and Review Technique
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean EDP & Associate Professor MBA
2
Sanjivani College of Engineering, Kopargaon
Department of MBA
www.sanjivanimba.org.in

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 At the End of the Session Student will be able to
understand-
A. Explanatory Case on Program Evaluation and Review
Technique under Network Calculations.
Program Evaluation and Review Technique

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A project is represented by the following network diagram and has
following time estimates in weeks-
1. Find the Critical Path and Expected Project Length.
2. Find the probability that the project is completed 2 weeks before
the expected time.
3. If a 30 weeks deadline is imposed, what is the probability that the
project will be finished within the limit.
4. If failing to meet the above deadline costs Rs. 1000 per week as a
penalty to the company, what is the probability that a penalty,
but not exceeding Rs. 3,000 will be paid.
Activity A B C D E F G H
Most Optimistic Time (a) 2 10 8 10 7 9 3 5
Most Likely Time (m) 4 12 9 15 7.5 9 3.5 5
Most Pessimistic Time (b) 12 26 10 20 11 9 7 5
Program Evaluation and Review Technique

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5. What should be the due date to have 0.90 probability
of project completion.
6. If the project manager wants to be 99% sure that the
project is completed on the scheduled date, how many
weeks before that date should he start the project?
1
A E
G
2
4
5 6
D
3
B
H
FC
Program Evaluation and Review Technique

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Activity Time
Estimates
Expected Time
tei =
a + 4m + b
6
Variance
σ2
i =
b − a
6
2
a m b
A 2 4 12 =
𝟐+𝟒∗𝟒+𝟏𝟐
𝟔
= 𝟓 25/9
B 10 12 26 =
𝟏𝟎+𝟒∗𝟏𝟐+𝟐𝟔
𝟔
= 𝟏𝟒 64/9
C 8 9 10 =
𝟖+𝟒∗𝟗+𝟏𝟎
𝟔
= 𝟗 1/9
D 10 15 20 =
𝟏𝟎+𝟒∗𝟏𝟓+𝟐𝟎
𝟔
= 𝟏𝟓 25/9
E 7 7.5 11 =
𝟕+𝟒∗𝟕.𝟓+𝟏𝟏
𝟔
= 𝟖 4/9
F 9 9 9 =
𝟗+𝟒∗𝟗+𝟗
𝟔
= 𝟗 0
G 3 3.5 7 =
𝟑+𝟒∗𝟑.𝟓+𝟕
𝟔
= 𝟒 4/9
H 5 5 5 =
𝟓+𝟒∗𝟓+𝟓
𝟔
= 𝟓 0
Program Evaluation and Review Technique

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Program Evaluation and Review Technique
1
A(5) E(8)
G(4)
2
4
5 6
D(15)
3
B(14)
H(5)
F(9)C(9)
Path Duration
1-2-4-5-6 5 + 15 + 4 + 5 = 29
1-2-5-6 5 + 8 + 5 = 18
1-2-3-5-6 5 + 9 + 9 + 5 = 28
1-3-5-6 14 + 9 + 5 = 28

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1. The expected time for completion of the project is-
Te= = 𝑡𝑒 𝐴 + 𝑡 𝑒𝐷 + 𝑡 𝑒𝐺 + 𝑡 𝑒𝐻
Te=5+15+4+5=29 Weeks
Also Project Duration Variance is-
𝜎T
2 = 𝜎A
2 + 𝜎D
2 + 𝜎G
2 + 𝜎H
2
𝜎T
2 = 25/9 + 25/9 + 4/9 + 0 = 54/9 = 6
Project Standard Duration is-
𝜎T =
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒
=
6
= 2.45 Weeks
Program Evaluation and Review Technique

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Now, the project follows normal distribution with mean, Te=
29 weeks and standard deviation = 2.45
The standard normal variate is-
z =
𝑇−𝑇𝑒
𝜎t
=
𝑇−29
2.45
2. Probability that the project is completed 2 weeks before the expected
time of 29 weeks is-
P(T≤ 27) = P(z ≤ zs) where, z =
𝑇−29
2.45
and zs =
27−29
2.45
= −0.82
= P(z ≤ -0.82)
= P(z ≤ 0.82) – (By Symmetry of Standard Normal Curve)
= 0.5 - P(0 ≤ z ≤ 0.82)
= 0.5 – 0.2930 – (Reading value from std. normal curve)
= 0.207
Program Evaluation and Review Technique

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3. Similarly, here Ts = 30 weeks.
Probability that the project is completed in 30 weeks is-
P(T≤ 30) = P(z ≤ zs) where, z =
𝑇−29
2.45
and zs =
30−29
2.45
= 0.41
= P(z ≤ 0.41)
= P(-∞<z ≤ 0.41)
= 0.5 + P(0 ≤ z ≤ 0.41)
= 0.5 + 0.1591 – (Reading value from std. normal curve)
= 0.6591
Program Evaluation and Review Technique

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4. We have to find the probability that the company pays a fine but it is not
more than Rs. 3,000. It will pay a fine of Rs. 1,000 per week delay in excess of
30 weeks. Thus, at most it should exceed by 3000/1000 = 3 days. Also, it must
take minimum 30 weeks (as only then it has to pay the fine)
Hence, Required Probability is-
P(30 ≤T≤ 33) = P(z1 ≤ z ≤ z2) where, z1 =
30−29
2.45
= 0.41 and z2 =
33−29
2.45
= 1.63
= P(0.41 ≤z ≤ 1.63)
= P(0< z ≤ 1.63) - P(0 < z < 0.41)
= 0.4484 - 0.1591 – (Reading value from std. normal curve)
= 0.2893
Program Evaluation and Review Technique

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5. Here, we have to first work out scheduled duration, so that the probability of
the project completion within duration is 0.90
P(T< Ts) = 0.90
P(z< zs) = 0.90
0.5 + P(0<z<zs) = 0.90
P(0<z <zs) = 0.4
Now reading the value of z from area under curve in reverse order, we see that
for probability value of 0.90, z=1.28
zs=1.28
𝑇 𝑠
−29
2.45
= 1.28
Ts = 32.13 Weeks, Thus the due date should be 32.13 weeks from start.
Program Evaluation and Review Technique

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6. The project manager wants to be 99% sure that the project is completed on
the scheduled date.
We have to first find Ts such that,
P(T< Ts) = 0.99
P(z< zs) = 0.99
0.5 + P(0<z<zs) = 0.99
P(0<z <zs) = 0.49
Now reading the value of z from area under curve in reverse order, we see that
for probability value of 0.99, z=2.33
zs=2.33
𝑇 𝑠
−29
2.45
= 2.33
Ts = 34.7 Weeks, Thus he should start 34.7 weeks before the due date.
Program Evaluation and Review Technique

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EXERCISE
 Solve the Explanatory Case on Program Evaluation and
Review Technique.

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For More Details Contact
Dr. V M Tidake
tidkevishal@gmail.com
tidkevishalmba@sanjivani.org.in

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Thank You