6 queueing networks canceling out the highest powers

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Unformatted text preview: ure process are independent. Since there is more than a little hand-waving going on in the proof of Theorem 4.11 and its application here, it is comforting to note that one can simply verify from the definitions that Lemma 4.12. If ⇡ (m, n) = c m+n /(µm µn ), where c = (1 /µ1 )(1 /µ2 ) 12 is a constant chosen to make the probabilities sum to 1, then ⇡ is a stationary distribution. Proof. The first step in checking ⇡ Q = 0 is to compute the rate matrix Q. To do this it is useful to draw a picture which assumes m, n > 0 (m 1, n + 1) (m, n + 1) I @ @ (a) @ @ µ1 @ µ2 @ @ (m 1, n) @ ? - (m, n) @ - (m + 1, n) I @ µ2 ? (m, n (c) @ 1) @ µ1 @ @ @ (b) (m + 1, n 1) The rate arrows plus the ordinary lines on the picture, make three triangles. We will now check that the flows out of and into (m, n) in each triangle balance. In symbols we note that (a) µ1 ⇡ (m, n) = c m+n = ⇡ (m µm 1 µn 1 2 (b) µ2 ⇡ (m, n) = c m+n = µ1 ⇡ (m + 1, n µm µn 1 12 (c) ⇡ (m, n) = c m+n+1 µm µn 12 1, n) 1) = µ2 ⇡ (m, n + 1) This shows that ⇡ Q = 0 when m, n > 0. There are three other cases to conside...
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