Unformatted text preview: n +1) =
⇡ (n) = ⇡ (0)
If we take ⇡ (0) = e
/µ. /µ and q (n, n 1) = then this becomes the Poisson distriibution with mean 4.4. EXIT DISTRIBUTIONS AND HITTING TIMES 133 Example 4.17. M/M/s queue with balking. A bank has s tellers that
serve customers who need an exponential amount of service with rate µ nd
queue in a single line if all of the servers are busy. Customers arrive at times
of a Poisson process with rate but only join the queue with probability an if
there are n customers in line. Thus, the birth rate n = an for n 0, while
the death rate is
nµ 0 n s
sµ n s
for n 1. It is reasonable to assume that if the line is long the probability the
customer will join the queue is small. The next result shows that this is always
enough to prevent the queue length from growing out of control.
Theorem 4.6. If an ! 0 as n ! 1, then there is a stationary distribution.
Proof. It follows from (4.17) that if n
⇡ (n + 1) =
If N is large enough and n P n · ⇡ (n) = an · sµ · ⇡ (n) N , then an /(sµ) 1/2 and it follows that ⇡ (n + 1) This implies that n µn+1 s, then 1
⇡ (n) . . . 2 ✓ ◆n...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
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