# J as t 1 detailed balance condition a su cient

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Unformatted text preview: ( 1+ 2 ) = r2 (1 2 p2 ) + r1 (1 p1 ) The ﬁrst two equations hold by (4.28). The third is the sum of the ﬁrst two, so it holds as well. This shows that ⇡ Q = 0 when m, n &gt; 0. As in the proof for the tandem queue, there are three other cases to consider: (i) m = 0, n &gt; 0, (ii) m &gt; 0, n = 0, and (iii) m = 0, n = 0. In these cases some of the rates are missing. However, since the rates in each group balance we have ⇡ Q = 0. Example 4.28. Network of M/M/1 queues. Assume now that there are stations 1 i K . Arrivals from outside the system occur to station i at rate i and service occurs there at rate µi . Departures go to station j with probability p(i, j ) and leave the system with probability X q (i) = 1 p(i, j ) (4.29) j To have a chance of stability we must suppose (A) For each i it is possible for a customer entering at i to leave the system. That is, for each i there is a sequence of states i = j0 , j1 , . . . jn with p(jm 1 , jm ) &gt; 0 for 1 m n and q (jn ) &gt; 0. Generalizing (...
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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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