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From Algorithms to Z-Scores - matloff - University of California, Davis

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21.5. NONEXPONENTIAL SERVICE TIMES 437<br />

21.4.1.2 Going Beyond Finding the π<br />

One can calculate a number <strong>of</strong> interesting quantities from the πi:<br />

• The probability <strong>of</strong> a hand<strong>of</strong>f call being rejected is πn.<br />

• The probability <strong>of</strong> a new call being dropped is<br />

n<br />

k=n−g<br />

πk<br />

(21.47)<br />

• Since the per-channel utilization in state i is i/n, the overall long-run per-channel utilization<br />

is<br />

n<br />

i=0<br />

πi<br />

i<br />

n<br />

(21.48)<br />

• The long-run proportion <strong>of</strong> accepted calls which are hand<strong>of</strong>f calls is the rate at which hand<strong>of</strong>f<br />

calls are accepted, divided by the rate at which calls are accepted:<br />

λ1<br />

n−1 λ2<br />

i=0 πi<br />

n−g−1 i=0<br />

πi + λ2<br />

21.5 Nonexponential Service Times<br />

n−1<br />

i=0 πi<br />

(21.49)<br />

The Markov property is <strong>of</strong> course crucial <strong>to</strong> the analyses we made above. Thus dropping the<br />

exponential assumption presents a major analytical challenge.<br />

One queuing model which has been found tractable is M/G/1: Exponential interarrival times,<br />

general service times, one server. In fact, the mean queue length and related quantities can be<br />

obtained fairly easily, as follows.<br />

Consider the residence time R for a tagged job. R is the time that our tagged job must first wait<br />

for completion <strong>of</strong> service <strong>of</strong> all jobs, if any, which are ahead <strong>of</strong> it—queued or now in service—plus<br />

the tagged job’s own service time. Let T1, T2, ... be i.i.d. with the distribution <strong>of</strong> a generic service<br />

time random variable S. T1 represents the service time <strong>of</strong> the tagged job itself. T2, ..., TN represent<br />

the service times <strong>of</strong> the queued jobs, if any.<br />

Let N be the number <strong>of</strong> jobs in the system, either being served or queued; B be either 1 or 0,<br />

depending on whether the system is busy (i.e. N > 0) or not; and S1,resid be the remaining service

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