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Unformatted text preview: e that customers pay $1 per
minute in the queue and repeat the derivation of Little’s formula, then
LQ = (3.4) a WQ The length of the queue is 1 less than the number in the system, except when
the number in the system is 0, so if ⇡ (0) is the probability of no customers, then
LQ = L 1 + ⇡ (0) Combining the last three equations with our ﬁrst cost equation:
⇡ (0) = LQ (L 1) = 1 + a (WQ W) = 1 a Esi (3.5) Recalling that Esi = 1/µ, we have a simple proof that the inequality in Theorem
3.5 is sharp. 108 3.2.3 CHAPTER 3. RENEWAL PROCESSES M/G/1 queue Here the M stands for Markovian input and indicates we are considering the
special case of the GI/G/1 queue in which the inputs are a rate Poisson
process. The rest of the set-up is as before: there is a one server and the ith
customer requires an amount of service si , where the si are independent and
have a distribution G with mean 1/µ.
When the input process is Poisson, the system has special properties that
allow us to go further. We learned in Theorem 3.5 that if < µ then a GI/G/1
queue will repeatedly return to...
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