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Unformatted text preview: then the reversed chain has transition probability pt (i, j ) = pt (i, j ). ˆ As we learned in Example 4.25, when < µs the M/M/s queue is a birth and death chain with a stationary distribution ⇡ that satisfies the detailed balance condition. Lemma 4.9 implies that if we take the movie of the Markov chain in equilibrium then we see something that has the same distribution as the M/M/s queue. Reversing time turns arrivals into departures, so the departures must be a Poisson process with rate . It should be clear from the proof just given that we also have: Theorem 4.10. Consider a queue in which arrivals occur according to a Poisson process with rate and customers are served at rate µn when there are n in the system. Then as along as there is a stationary distribution the output process will be a rate Poisson process. 143 4.6. QUEUEING NETWORKS* A second refinement that will be useful in the next section is Theorem 4.11. Let N (t) be the number of departures between time 0 and time t for the M/M/1 queue X (t) started from its equilibrium distribution. Then {N (s...
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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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