M/G/1 Cheng-Fu Chou

Residual Life Inter-arrival time of bus is exponential w/ rate l while hippie arrives at an arbitrary instant in time Question: How long must the hippie wait, on the average , till the bus comes along? Answer 1: Because the average inter-arrival time is 1/l, therefore 1/2l P. 2

Residual Life (cont.) Answer 2: because of memoryless, it has to wait 1/l General Result P. 3

Derivation P. 4

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M/G/1 M/G/1 a(t) = le –lt b(t) = general Describe the state [N(t), X0(t)] N(t): the no. of customers present at time t X0(t): service time already received by the customer in service at time t Rather than using this approach, we use “the method of the imbedded Markov Chain” P. 6

Imbedded Markov Chain [N(t), X0(t)] Select the “departure” points, we therefore eliminate X0(t) Now N(t) is the no. of customer left behind by a departure customer. (HW) For Poisson arrival pk(t) = rk(t) If in any system (even in non-Markovian) where N(t) makes discontinuous changes in size (plus or minus) one, then rk = dk = prob[departure leaves k customers behind] Therefore, for M/G/1 rk = dk = pk P. 7

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Mean Queue Length P. 10

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This is the famous Pollaczek – Khinchin Mean Value Formula

Examples P. 18

Mean Residual Service Time Wi: waiting time in queue of the i-th customer Ri: residual service time seen by the i-th customer Xi: service time of the i-th customer Ni: # of customers found waiting in the queue by the i-th customer upon arrival P. 19

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Residual Service Time rt x1 time t x1 x2 xM(t) P. 21

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Distribution of Number in the System P. 23

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Ex Q(z) for M/M/1 P. 25

Sol. P. 26

Waiting Time Distribution P. 27

Ex Response time distribution for M/M/1 Waiting time distribution for M/M/1 P. 28

Response Time Distribution

Waiting Time Distribution P. 30