Unformatted text preview: in equilibrium has a shifted geometric distribution so
L= 1
1 /µ 1= µ
µ µ
µ = µ By our fourth queueing equation, (3.6), the server’s busy periods have mean
✓
◆
✓
◆
1
1
1
µ
1
EB =
1=
1=
⇡ (0)
µ
µ
which agrees with (4.20).
Example 4.24. M/M/1 queue with a ﬁnite waiting room. In this system
customers arrive at the times of a Poisson process with rate . Customers enter
service if there are < N individuals in the system, but when there are N
customers in the system, the new arrival leaves never to return. Once in the
system, each customer requires an independent amount of service that has an
exponential distribution with rate µ.
Lemma 4.7. Let Xt be a Markov chain with a stationary distribution ⇡ that
satisﬁes the detailed balance condition. Let Yt be the chain constrained to stay
in a subset A of the state space. That is, jumps which take the chain out of
A are not allowed, but allowed jumps occur at the original P
rates. In symbols,
q (x, y ) = q (x, y ) if x, y 2 A and 0 otherwise. Let...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

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