# 3 age and residual life returns to 0 and this will

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Unformatted text preview: ous customers, so it is independent of her service time, i.e., E (si qi ) = Esi · WQ and we have Y = (Esi )WQ + E (s2 /2) i PASTA implies that arriving customers see the long run average behavior so the workload they see Z = WQ , so we have WQ = (Esi )WQ + E (s2 /2) i Solving for WQ now gives E (s2 /2) i (3.7) 1 E si the so-called Pollaczek-Khintchine formula. Using formula (3.3), and Theorem 3.6, we can now compute WQ = W = WQ + Esi L= W Example 3.8. We see a number of applications of the equations from this section to Markovian queues in Chapter 4. Customers arrive at the CIT help desk at rate 1/6 per minute, i.e., the mean time between arrivals is 6 minutes. Suppose that each service takes a time with mean 5 and standard deviation p 59. (a) In the long run what is the fraction of time, ⇡ (0), that the server is idle? = 1/6, Esi = 5 = 1/µ, so by (3.5) ⇡ (0) = 1 (1/6)/(1/5) = 1/6. (b) What is the average waiting W time for a customer (including their service time)? Es2 = 52 + 59 = 84, so (3.7) implies i WQ = E s2 /2 (1/6) · 84/2 i = = 42 1...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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