Discrete-time stochastic processes

6 9 section 111 theorem 31 shows that it has a unique

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Unformatted text preview: lity 1 by the strong law for renewal processes, applied to {A(t) − 1; t ≥ 0}. This limit is called the arrival rate ∏, and is equal to the reciprocal of the mean interarrival interval for {A(t)}. Since these two limits exist with probability 1, the first limit on the right, which is the sample-path-average waiting time per customer, denoted W , also exists with probability 1. We have thus proved Little’s theorem, Theorem 3.8 (Little). For a FCFS G/G/1 queue in which the expected inter-renewal interval is finite, the time-average number of customers in the system is equal, with probability 1, to the sample-path-average waiting time per customer multiplied by the customer arrival rate, i.e., L = ∏W . The mathematics we have brought to bear here is quite formidable considering the simplicity of the idea. At any time t within an idle period, the sum of customer waiting periods up to time t is precisely equal to t times the time-average number of customers in the system up to t (see Figure 3.13). Renewal theory informs us that the limits exist and that the edge effects (i.e., the customers in the system at an arbitrary time t) do not have any effect in the limit. Recall that we assumed earlier that customers departed from the queue in the same order in which they arrived. From Figure 3.14, however, it is clear that FCFS order is not required for the argument. Thus the theorem generalizes to systems with multiple servers and arbitrary service disciplines in which customers do not follow FCFS order. In fact, all that the argument requires is that the system has renewals (which are IID by definition of a renewal) and that the inter-renewal interval is finite with probability 1. A(τ ) ✛ ✛ ✛ 0 W1 W2 ✲ W3 ♣♣ ♣♣ ♣♣ ✲ ♣♣ ♣♣ ♣♣ ♣♣ ♣♣ ♣ ♣ ♣ ♣ ♣ ♣♣ ♣♣ ♣♣ ♣♣ ♣♣ ♣♣ ✲ S1 t Figure 3.14: Arrivals and departures in non-FCFS systems. The aggregate reward (integral of number of customers in system) up to time t is the enclosed area to the left of t; the sum of waits of customers arriving by t includes the additional shaded area to the right of t. Finally, suppose the inter-renewal distribution is non-arithmetic; this occurs if the interarrival distribution is non-arithmetic. Then L, the time-average number of customers in the 120 CHAPTER 3. RENEWAL PROCESSES system, is also equal to10 limt→1 E [L(t)]. It is also possible (see Exercise 3.30) to replace the time-average waiting time W with limn→1 E [Wn ]. This gives us the following variant of Little’s theorem: lim E [L(t)] = ∏W = lim ∏E [Wn ] . n→1 t→1 (3.59) The same argument as in Little’s theorem can be used to relate the average number of customers in the queue (not counting service) to the average wait in the queue (not counting service). Renewals still occur on arrivals to an empty system, and the integral of customers in queue over a busy period is still equal to the sum of the queue waiting times. Let Lq (t) be Rt the number in the queue at time t and let Lq = limt→1 (1/t) 0 Lq (τ )dτ be the time-average queue wait. Letting W q be the time-average waiting time in queue, Lq = ∏W q . (3.60) If the inter-renewal distribution is non-arithmetic, then lim E [Lq (t)] = ∏W q . t→1 (3.61) The same argument can also be applied to the service facility. The time-average of the number of customers in the server is just the fraction of time that the server is busy. Denoting this fraction by ρ and the expected service time by Z , we get ρ = ∏Z . 3.6.2 (3.62) Expected queueing time for an M/G/1 queue For our last example of the use of renewal-reward processes, we consider the expected queueing time in an M/G/1 queue. We again assume that an arrival to an empty system occurs at time 0 and renewals occur on subsequent arrivals to an empty system. At any given time t, let Lq (t) be the number of customers in the queue (not counting the customer in service, if any) and let R(t) be the residual life of the customer in service. If no customer is in service, R(t) = 0, and otherwise R(t) is the remaining time until the current service will be completed. Let U (t) be the waiting time in queue that would be experienced by a customer arriving at time t. This is often called the unfinished work in the queueing literature and represents the delay until all the customers currently in the system complete service. Thus the rv U (t) is equal to R(t), the residual life of the customer in service, plus the service times of each of the Lq (t) customers currently waiting in the queue. Lq (t) U (t) = X Zi + R(t). (3.63) i=1 10 To show this mathematically requires a little care. One approach is to split the reward function into many individual terms. Let Ln (t) = 1 if the nth arrival since the beginning of the busy period has arrived by time t, and let Ln (t) = 0 otherwise. Let Sn (t) = 1 if the nth departure since the beginning of the busy period occurs by time t. It is easy to show that P direct Riemann integrability condition holds for each of the P these reward functio...
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