1. (20 points) In each one-second interval at a convenience store, a new custome

Question

1. (20 points) In each one-second interval at a convenience store, a new customer arrives with probability p, independent of the number of customers in the store and other arrivals at other times. The clerk gives each arriving customer a friendly "Hello". Suppose that In each unit of time in which there is no arrival, the clerk can provide a unit of service to a waiting customer. Once received a unit of service, a customer departs the store with probability q When the store is empty, the clerk sits idle. (a) (10 pts) Sketch a Markov chain for the number of customers in the store. (b) (10 pts) Under what conditions on p and q do limiting probabilities exist? Find the limiting probabilities under those conditions

Explanation / Answer

ANSWER:

The number of customers in the ”friendly” store is given by the Markov chain (1-p)(1-q) (1-p)(1-q) 1 (1-p)q (1-p)(1-q) p p p 0 (1-p)(1-q) (1-p)q i+1 i !!! (1-p)q !!! (1-p)q In the above chain, we note that (1 p)q is the probability that no new customer arrives, an existing customer gets one unit of service and then departs the store. By applying Theorem 13 with the state space partition S = {0, 1, . . . , i} , we see that for any state i 0, i p = i+1(1 S 0 = {i + 1, i + 2, . . .} , (1) p)q. (2) i (3) This implies i+1 = p (1 p)q Since Equation (3) holds for i = 0, 1, . . ., we have that i = 0 i where p = . (1 p)q (4) Requiring the state probabilities to sum to 1, we have that for < 1, 1 X i = 0 i=0 1 X i = i=0 0 1 = 1. (5) Thus for < 1, the limiting state probabilities are i = (1 In addition, for not exist. )i , 1 or, equivalently, p i = 0, 1, 2, . . . q/(1 (6) q), the limiting state probabilities

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