J as t 1 detailed balance condition a su cient

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 > 0. As in the proof for the tandem queue, there are three other cases to consider: (i) m = 0, n > 0, (ii) m > 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 ) > 0 for 1 m n and q (jn ) > 0. Generalizing (...
View Full Document

This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

Ask a homework question - tutors are online