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Ngara tailoring shop has one tailor specialized in men’s shirts. The number of c

ID: 3253648 • Letter: N

Question

Ngara tailoring shop has one tailor specialized in men’s shirts. The number of customers requiring stitching of shirts appears to follow a Poisson distribution with a mean arrival rate of 12 per hour. Customers are attended to by the tailor on a first come first serve basis and they are willing to wait for service if there be a queue. The time the tailor takes to attend a customer is exponentially distributed with a mean of four minutes. You are required to determine:

The probability that the system is busy                                                   [1 Marks]

The average time a customer spends in the system                                 [1 Marks]

The average length of the queue                                                             [2 Marks]

The probability that there will be five customers in the shop at a point in time

[2 Marks]

QUESTION 2

Define the following terms used in game theory:

Dominance                                                                                                      [1 mark]

Optimal strategy                                                                                             [1 mark]

Give four characteristics of zero-sum game.                                                       [4 marks]

Consider the two person zero sum game between players A and B given the following pay-off table:

              Player B Strategies

B1

B2

B3

B4

A1

3

2

4

0

Player A Strategies

A2

3

4

2

4

A3

4

2

4

0

A4

0

4

0

8

Required:

Use the concept of dominance to determine the optimal strategies and find the value of the game.                                                            

              Player B Strategies

B1

B2

B3

B4

A1

3

2

4

0

Player A Strategies

A2

3

4

2

4

A3

4

2

4

0

A4

0

4

0

8

Explanation / Answer

1) a) The probability that the System is Busy = Probability that an arriving customer will have to wait for service

Pw = /µ = 0.2 customers per min/0.25 customer per min = 0.8

b) W = Average time a customer spends in the system.
W = 1 /(µ – ) = 1/(0.25 - 0.2) = 20 minutes

c) Lq = Average number of customers in the queue.
Lq = 2/[µ(µ-)] = 0.22/[0.25(0.25-0.2)] = 3.2

d) Pn = [1 – (/µ)](/µ)n = [1-(0.2/0.25)](0.2/0.25)5 =0.0655

Note : Please post the 2nd question separately if it requires solving

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