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Unformatted text preview: queue, this
allowed us to show that if
the average interarrival time Eti = 1/ ,
the average service time Esi = 1/µ,
the average waiting time in the queue is L,
the long run rate at which customers enter the system is
the average waiting time in the system is W ,
and the fraction of time the queue is empty is ⇡ (0) a, then we have
L= aW ⇡ (0) = 1 a µ
In the M/G/1 case, the expected duration of busy periods and the average
waiting time in the queue satisfy:
⇡ (0) = 1/
1/ + EB WQ = E (s2 /2)
/µ The ﬁrst formula is a simple consequence of our result for alternating renewal.
The more sophisticated second formula uses “Poisson Arrivals See Time Averages” along with cost equation reasoning. 3.5 Exercises 3.1. The weather in a certain locale consists of alternating wet and dry spells.
Suppose that the number of days in each rainy spell is a Poisson distribution
with mean 2, and that a dry spell follows a geometric distribution with mean 7.
Assume that the successive durations of rainy and dry spells are i...
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- Spring '10
- The Land