Unformatted text preview: onential servers.
With this modiﬁcation, the transition rates in (6.60) and (6.61) are modiﬁed by replacing
µi with µi,mi . The hypothesized backward transition rates are modiﬁed in the same way,
and the only eﬀect of these changes is to replace ρi and ρj for each i and j in (6.72)(6.74)
with ρi,mi = ∏i /µi,mi and ρj,mj = ∏j /µj,mj . With this change, (6.75) becomes
p(m ) = k
Y pi (mi ) = i=1 pi (0) = 1 + k
Y j =1
1
X m−1
Y m=1 j =0 pi (0)
−1 ρi,j mi
Y ρi,j (6.79) j =0 . (6.80) Thus, p(m ) is given by the product distribution of k individual birth death systems. 6.7.1 Closed Jackson networks The second generalization is to a network of queues with a ﬁxed number M of customers
in the system and with no exogenous inputs or outputs. Such networks are called closed
Jackson networks, whereas the networks analyzed above are often called open Jackson
networks. Suppose a k node closed network has routing probabilities Qij , 1 ≤ i, j ≤ k,
P
where j Qij = 1, and has exponential service times of rate µi (this can be generalized to
µi,mi as above). We make the same assumptions as before about independence of service
variables and routing variables, and assume that there is a path between each pair of nodes.
Since {Qij ; 1 ≤ i, j ≤ k} forms an irreducible stochastic matrix, there is a one dimensional
set of solutions to the steady state equations
X
∏j =
∏i Qij ;
1 ≤ j ≤ k.
(6.81)
i We interpret ∏i as the timeaverage rate of transitions that go into node i. Since this set
of equations can only be solved within an unknown multiplicative constant, and since this
constant can only be determined at the end of the argument, we deﬁne {πi ; 1 ≤ i ≤ k} as
the particular solution of (6.81) satisfying
πj = X
i πi Qij ; 1 ≤ j ≤ k; X πi = 1. (6.82) i Thus, for all i, ∏i = απi , where α is some unknown constant. The state of the Markov
P
process is again taken as m = (m1 , m2 , . . . , mk ) with the condition i mi = M . The
transition rates of the Markov process are the same as for open networks, except that there
are no exogenous arrivals or departures; thus (6.59)(6.61) are replaced by
qm ,m 0 = µi Qij for m 0 = m − e i + e j , mi > 0, 1 ≤ i, j ≤ k. (6.83) 6.7. JACKSON NETWORKS 265 We hypothesize that the backward time process is also a closed Jackson network, and as
before, we conclude that if the hypothesis is true, the backward transition rates should be
∗
qm ,m 0 = µi Q∗
ij where ∏i Qij = ∏j Q∗i
j for m 0 = m − e i + e j , mi > 0, 1 ≤ i, j ≤ k
for 1 ≤ i, j ≤ k. (6.84)
(6.85) In order to use Theorem 6.5 again, we must verify that a PMF p(m ) exists satisfying
p(m )P ,m 0 = p(m 0 P∗m 0 ,m for all possible states and transitions, and we must also verify
qm
)q
that m 0 qm ,m 0 = m 0 q ∗m ,m 0 for all possible m . This latter veriﬁcation is virtually the
same as before and is left as an exercise. The former veriﬁcation, with the use of (72), (73),
and (74), becomes
p(m )(µi /∏i ) = p(m 0 )(µj /∏j ) for m 0 = m − e i + e j , mi > 0. (6.86) Using the open network solution to guide our intuition, we see that the following choice of
P
p(m ) satisﬁes (6.86) for all possible m (i.e., all m such that i mi = M )
p(m ) = A k
Y (∏i /µi )mi ; for m such that i=1 X mi = M . (6.87) i The constant A is a normalizing constant, chosen to make p(m ) sum to unity. The problem
with (6.87) is that we do not know ∏i (except within a multiplicative constant independent
of i). Fortunately, however, if we substitute πi /α for ∏i , we see that α is raised to the power
−M , independent of the state m . Thus, letting A0 = Aα − M , our solution becomes
p(m ) = A0 K
Y (πi /µi )mi ; for m such that i=1 1
A0 =
m: X P k
Y i=1
i mi =M √ πi
µi !mi X mi = M . (6.88) i . (6.89) Note that the steady state distribution of the closed Jackson network has been found without
solving for the timeaverage transition rates. Note also that the steady state distribution
looks very similar to that for an open network; that is, it is a product distribution over the
nodes with a geometric type distribution within each node. This is somewhat misleading,
however, since the constant A0 can be quite diﬃcult to calculate. It is surprising at ﬁrst
that the parameter of the geometric distribution can be changed by a constant multiplier
in (6.88) and (6.89) (i.e., πi could be replaced with ∏i ) and the solution does not change;
the important quantity is the relative values of πi /µi from one value of i to another rather
than the absolute value.
In order to ﬁnd ∏i (and this is important, since it says how quickly the system is doing its
work), note that ∏i = µi Pr {mi > 0}). Solving for Pr {mi > 0} requires ﬁnding the constant
A0 in (6.83). In fact, the ma jor diﬀerence between open and closed networks is that the
relevant constants for closed networks are tedious to calculate (even by computer) for large
networks and large M . 2...
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This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.
 Spring '09
 R.Srikant

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