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Solve as Program Using Microsoft Excel To calculate the below question Subject name: Performance Evaluation of Info communication Networks Book name: Basic Queueing Theory by

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Solve as Program Using Microsoft Excel To calculate the below question

Subject name: Performance Evaluation of Info communication Networks

Book name: Basic Queueing Theory by Dr. Jnos Sztrik

image text in transcribed
M/M/2 with heterogeneous servers and fastest free server service discipline Consider a variant of the M/ M/2 queue where the service rates of the two servers are not identical. Denote the service rate of the first server by / and the service rate of the second server by /2, where u1 2 /2 . In the case of heterogeneous servers, the rule is that when both servers are idle, the faster server is scheduled for service before the slower one, that is called FFS - Fastest Free Server But if there is only one server free when an arrival occurs, it enters service with the free server regardless of the service rate. If both servers are busy, the arriving customer waits in common line for service in the order of arrival. Define the utilization, a, for this system to be a = >/(M1 + 12). Let us determine the mean number of jobs in the system E(N), mean response and waiting time E(T), E(W), respectively. It was proved, for example in Harchol-Balter [46] and Trivedi [120] that E(N) = A(1 -a)2 where A = ! HIM2(1 + 2a) * (1+ 162 ) 1 - a Using Little law E(T) = E(N) /X. It can be shown that P(Q = 2) = atti/A, i = 1, 2, ...co and thus a 2 E(W) = - E(Q) E(S) = E(N) - E(@), P(W > 0) = ( 1 -a)2A a E(Q) = A(1 - a)?' It is easy to see if u1 = 12 = u then a = p/2 and there is no difference between the servers thus the corresponding measures are the same as in the homogeneous M/M/2 system, that is 4 p p3 E(N) = E(Q) E(S) M/M/2 with heterogeneous servers and random free server service discipline Let us see the provious system with the exception that and arriving customer select between the free servers with the same probability, that is the selection probability is 0.5. Let us call this discipline as RFS -Random Free Server However, this small modification makes the calculations rather complicated, but it can be treated numerically. To do so, let us introduce the following notations: bet (or, 62, 4) be the state of the system, where & is the number of customers in the queue, and a is I if server . is busy and 0 otherwise. Let us denote by Hogs, Ho,to, Hoo, It,1,t, 4 - 0,1, 2,. O the steady state distribution of the custom which exists if a

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