Case of Markov Chain

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Unit No.3.
DECISION SCIENCE
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
www.sanjivanimba.org.in

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302-DECISION SCIENCE
Unit No.3 Marko Chain & Simulation
3.1.3 Case 1: Markov Chain
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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MARKOV CHAIN & SIMULATION
 At the End of the Session Student will be able to
understand-
A. Case 1: Markov Chain

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MARKOV CHAIN
A market research organization studied the car purchasing trends in a
certain region, with a conclusion that a new car is purchased, on an
average, once every 4 years. The buying pattern of the customers is as
follows:
Of the current small car owners, 80% will replace the car again with a
small car and 20 percent with a large car. Similarly, 60% of the large car
owners will replace it with a small car while 40% will replace it with
another large car. Assuming the market and the preference remaining the
same-
a. Construct the Transition Matrix
b. If, there are currently 40,000 small cars and 20,000 large cars in the
region, what will be the distribution in 8 years from now.
c. Find the probability that a person presently using a small car will buy
a large car in the next to next purchase.

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MARKOV CHAIN
a. On the basis of given Information, Transition Matrix would
be-
S L
S 0.8 0.2
L 0.6 0.4
b. Present Share of Cars (i.e. For a Period n=0) will be-
Small Cars=
40000
40000+20000
= 2/3
Large Cars=
20000
40000+20000
= 1/3
Hence, R0 = 2/3 1/3

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MARKOV CHAIN
Distribution of Cars after 8 years i.e. after Period n=2, (as Car
is changed after every 4 years) will be-
R2 = R0 * P2
Hence, P2 will be-
0.8 0.2 0.8 0.2
0.6 0.4 * 0.6 0.4
0.64+0.12 0.16+0.08
P2 =
0.48+0.24 0.12+0.16
0.76 0.24
0.72 0.28

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MARKOV CHAIN
Now, R2 = R0 * P2
0.76 0.24
R2 = 2/3 1/3 *
0.72 0.28
P2 = 0.747 0.253
Hence,
No of Small Cars after 8 Years= 0.747*60000= 44820
No of Large Cars after 8 Years= 0.253*60000= 15,180

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MARKOV CHAIN
c. Probability that a person presently (i.e. at n=0) using a
small car will buy a large car in the next to next purchase (i.e.
at n=2) is the conditional probability reflected in Matrix P2
as-
S L
S 0.76 0.24
P2 =
L 0.72 0.28
Hence the required Probability is = 0.24

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EXERCISE
 Solve the Case 1 of Markov Chain in the Notebook as a
practice Exercise.

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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