3 age and residual life returns to 0 and this will

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

Ask a homework question - tutors are online