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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 > 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
q (i) = 1
p(i, j )
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 (...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
- Spring '10
- The Land