To prove that there is one we return to the general

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Unformatted text preview: wait for an exponential amount of service with rate µ2 . - µ1 - µ2 - Our main problem is to find conditions that guarantee that the queue stabilizes, i.e., has a stationary distribution. This is simple in the tandem queue. The first queue is not a↵ected by the second, so if < µ1 , then (4.23) tells us that the equilibrium probability of the number of customers in the first queue, 1 Xt , is given by the shifted geometric distribution ✓ ◆m ✓ ◆ 1 P (Xt = m) = 1 µ1 µ1 In the previous section we learned that the output process of an M/M/1 queue in equilibrium is a rate Poisson process. This means that if the first 2 queue is in equilibrium, then the number of customers in the queue, Xt , is itself an M/M/1 queue with arrivals at rate (the output rate for 1) and service rate µ2 . Using the results in (4.23) again, the number of individuals in the second queue has stationary distribution ✓ ◆n ✓ ◆ 2 P (Xt = n) = 1 µ2 µ2 To specify the stationary distribution of the system, we need to know the 1 2 joint distribution of Xt and Xt . The answer is somewhat remarkable: in 144 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS equilibrium the two queue lengths are independent. ✓ ◆m ✓ ◆ ✓ ◆n ✓ 1 2 P (Xt = m, Xt = n) = 1 · 1 µ1 µ1 µ2 µ2 ◆ (4.27) Why is this true? Theorem 4.11 implies that the queue length and the depart...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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