Lecture 12

# The mean number of packets in the system n b x j 0 jj

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Unformatted text preview: ved (from B-D model or from the special case M/M/1) that j ,1 Y  j = o i=0  jB  = 6= 1: 2 B 3,1 X  j5 0 = 41 + and, = For = 0  j j =1  1 , = 1 , =B+1 = 1, the following is true: j 1 = 1 , ,B+1 1 = B+1  = = 6= 1 j  B: Note that the packet loss probability is given by B . The mean number of packets in the system: N = B X j =0 jj = B +1 , B + 1B+1 1, 1,  6= 1 and N = B=2 if = 1: The mean number of packets at the server, E fNs g, is: E fNs g = P rfN = 0gE fNsjN = 0g + P rfN = 0  0 + 1 , 0  1: 0gE fNsjNs 0g 6= 1): 1+ B , 1 , B +1 : Nq = N , E fNs g = N , 1 , 0  = B 1, Thus, the mean number of packets in the queue (for while for = 1, we have Nq = B=2 , B=B + 1: Effective arrival rate is given by  0 = B1 X , j =0 j = 1 , B : The mean response time (average delay) is T The mean waiting time is W CS 522, v 0.94, d.medhi, W’99 = N= = N= 1 , B  : 0 = Nq= = Nq= 1 , B  : 0 17 13. M/M/m model: m parallel servers Another model of interest in communi...
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## This document was uploaded on 03/19/2014 for the course CS 6030 at Western Michigan.

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