Question

Suppose that a typical customer at a local restaurant spends 20 minutes in the store. It is estimated that the average number of customers in the restaurant is 4. Assume that the service time in this restaurant is random with some known distribution. How many customers does the restaurant serve during a typical hour that it is open? Suppose the restaurant wants to decrease the average number of customers in the restaurant at any time. However, assume it cannot control the arrival patterns of its customers or its own service rate. How can it achieve its objective?

Answer 1

Answer-

(A) (a)How many customers does the restaurant serve during a typical hour that it is open?

We will expect a solitary server framework here. We have

W = 20 minutes or 1/3 hour

L = 4

Presently, according to M/M/1 framework we know

L = W*λ where λ is the appearance rate.

This implies λ = L/W = 4/(1/3) = 12 every hour

Presently, on the off chance that λ is 12, at that point we can get the estimation of µ by

W = 1/( µ - λ ) or 1/3 = 1/(µ - 12) or µ = 15

This implies the administration rate is 15 clients for every hour

a)

The restaurant serves 15 clients for each hour overall.

(B) Suppose the restaurant wants to decrease the average number of customers in the restaurant at any time. However, assume it cannot control the arrival patterns of its customers or its own service rate. How can it achieve its objective?

(b) If λ or µ both can not be changed, the best way to lessen the quantity of clients is to open one more channel of administration. This can be either a seperate line for every server or the two servers using he clients from a similar line.

Usage rho = λ/µ= 12/15 =0.8

Number of clients hanging tight for two channels with rho =0.8 will be 0.1524

L = Lq +rho = 0.1524+0.8 =0.9524

which is lower than the present estimation of 4.

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