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Unformatted text preview: what is theavergage number of customers E(n) = sum(k*p_k,0,inf.) E(n) = U/1-U E(n) = Nbar what is the throughput? its mew when there is a job Sheet1 Page 2 its 0 when there is no job therefore X=U*mew + 0(1-U) X=lambda/mew * mew = lambda IE the throughput is the arival rate and it should be because customers are not created or lost all customers are serviced what is the response time? little's law E(n)= X * R (ie avergage # in the system = throughput * response time) therefore R=E(N)/X = mew/(1-mew)/lambda = lambda/mew/lambda(1-lambda/mew) = 1/(mew-lambda) R= 1/(mew - lambda) when U=0 R=time to process request when U=1 R=infinity wait time = R-mew things to think about what happens if you double the proccess speed...
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- Spring '08
- lim, #, Queueing theory, 0.02 seconds, steady state assume