AN UP- AND DOWNCROSSING TECHNIQUE

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AN UP- AND DOWNCROSSING TECHNIQUE

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In this section we discuss a generally applicable up- and downcrossing technique that, in conjunction with the PASTA property, can be used to establish relations between customer-average and time-average probabilities in queueing systems To illustrate this, we consider the so-called GI /M/1 queue In this single-server system, customers arrive according to a renewal process and the service times of the customers have a common exponential distribution The single server can handle only one customer at a time and there is ample waiting room for customers who nd the server busy upon arrival The service times of the customers are independent of each other and are also independent of the arrival process Denoting by the average arrival rate (1/ = the mean interarrival time) and by the service rate (1/ = the mean service time), it is assumed that < The continuous-time stochastic process {X(t), t 0} and the discrete-time stochastic process {Xn , n = 1, 2, } are de ned by X(t) = the number of customers present at time t, and Xn = the number of customers present just prior to the nth arrival epoch The stochastic processes {X(t)} and {Xn } are both regenerative The regeneration epochs are the epochs at which an arriving customer nds the system empty It is stated without proof that the assumption of / < 1 implies that the processes have a nite mean cycle length Thus we can de ne the time-average and the customer-average probabilities pj and j by pj = the long-run fraction of time that j customers are present and j = the long-run fraction of customers who nd j other customers present upon arrival for j = 0, 1, Time averages are averages over time, and customer averages t are averages over customers To be precise, pj = limt (1/t) 0 Ij (u) du and j = limn (1/n) n Ik (j ), where Ij (t) = 1 if j customers are present at k=1 time t and Ij (t) = 0 otherwise, and In (j ) = 1 if j other customers are present just before the nth arrival epoch and In (j ) = 0 otherwise The probabilities pj and j are related to each other by j 1 = pj , j = 1, 2, (271)

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The proof of this result is instructive and is based on three observations Before giving the three steps, let us say that the continuous-time process {X(t)} makes an upcrossing from state j 1 to state j if a customer arrives and nds j 1

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RENEWAL-REWARD PROCESSES

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other customers present The process {X(t)} makes a downcrossing from state j to state j 1 if the service of a customer is completed and j 1 other customers are left behind Observation 1 Since customers arrive singly and are served singly, the long-run average number of upcrossings from j 1 to j per time unit equals the long-run average number of downcrossings from j to j 1 per time unit This follows by noting that in any nite time interval the number of upcrossings from j 1 to j and the number of downcrossings from j to j 1 can differ at most by 1 Observation 2 The long-run fraction of customers seeing j 1 other customers upon arrival is equal to the long-run average number of upcrossings from j 1 to j per time unit the long-run average number of arrivals per time unit for j = 1, 2, In other words, the long-run average number of upcrossings from j 1 to j per time unit equals j 1 The latter relation for xed j is in fact a special case of the Little relation (241) by assuming that each customer nding j 1 other customers present upon arrival pays $1 (using this reward structure observation 2 can also be obtained directly from the renewal-reward theorem) Observations 1 and 2 do not use the assumption of exponential services and apply in fact to any regenerative queueing process in which customers arrive singly and are served singly Observation 3 For exponential services, the long-run average number of downcrossings from j to j 1 per time unit equals pj with probability 1 for each j 1 The proof of this result relies heavily on the PASTA property To make this clear, x j and note that service completions occur according to a Poisson process with rate as long as the server is busy Equivalently, we can assume that an exogenous Poisson process generates events at a rate of , where a Poisson event results in a service completion only when there are j customers present Thus, by part (a) of Theorem 241, E[Ij (t)] = E[Dj (t)] for t > 0 (272)

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for any j 1, where Ij (t) is de ned as the amount of time that j customers are present during (0, t] and Dj (t) is de ned as the number of downcrossings from j to j 1 in (0, t] Letting the constant dj denote the long-run average number of downcrossings from j to j 1 per time unit, we have by the renewal-reward theorem that limt Dj (t)/t = dj with probability 1 Similarly, limt Ij (t)/t = pj with probability 1 The renewal-reward theorem also holds in the expected-value version Thus, for any j 1,

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