class10 - what is theavergage number of customers E(n) =...

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Sheet1 Page 1 consider a birth-death process draw fig 6.9 assume all lambdas are equal assume all mews are equal to find steady state assume lim(as t -> inf) of P_n'(t) = 0 lim(as t -> inf) of P_n(t) = P_n for all n P_n'(t)= 0 = p(coming in)+P(coming back) - P(Leaving) see previous lecture for derivation examine just E_0 and E_1 P_0'(t)=0=mew_1*P_1-lambda_0*p_0 rearange P_1=lambda_0*P_0/mew_1 real example (infitine queue) -customers arive at 30/sec -each customer takes 0.02 seconds to be serviced what is the utilization? lambda=30 mew=1/(.02)=50 utilization = 1-p_0 p_0=1-lambda/mew = 1-30/50=1-.6=.4 U=lambda/mew utilzation=1-.4=.6= 30/50= .6 Utilization=60% *assumes lambda<mew infinite queue single server uniform rates what is p_k p_1=p_0*lambda/mew.4*.6 =.24 p_2=p_0*(lambda/mew)^2.4*.6^2 =.14 p_k=p_0*(lambda/mew)^k.4*.6^k
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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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class10 - what is theavergage number of customers E(n) =...

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