# To do this we let ty minn 1 xn y and let xy px ty

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Unformatted text preview: ue of people waiting to use an automated teller machine. Let Xn denote the number of people in line at the moment of the departure of the nth customer. To model this as a Markov chain we let ak be the probability that k customers arrive during one service time and write down the transition probability p(0, k ) = ak and p(i, i 1 + k ) = ak for k 0 with p(i, j ) = 0 otherwise. To explain this, note that if there is a queue, it is reduced by 1 by the departure of a customer, but k new customers will come with probability k . On the other hand if there is no queue, we must ﬁrst wait for a customer to come and the queue that remains at her departure is the number of customers that arrived during her service time. The pattern becomes clear if we write out a few rows and columns of the matrix: 0 1 2 3 4 0 a0 a0 0 0 0 1 a1 a1 a0 0 0 2 a2 a2 a1 a0 0 3 a3 a3 a2 a1 a0 4 a4 a4 a3 a2 a1 5 a5 a5 a4 a3 a2 ... If we regard the customers that arrive during a person’s service time to be her children, then this queueing process gives rise to a branching process. From t...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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