Discrete-time stochastic processes

# B use part a to show that limn1 pr n n 0 and explain

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Unformatted text preview: ns, and L(t) is n Ln (t) − n Sn (t). 3.6. APPLICATIONS OF RENEWAL-REWARD THEORY 121 where Zi is the required service time of the ith customer in the queue at time t. Since the service times are independent of the arrival times and of the earlier service times, Lq (t) is independent of Z1 , Z2 , . . . , ZLq (t) , so, taking expected values, E [U (t)] = E [Lq (t)] E [Z ] + E [R(t)] . (3.64) Figure 3.15 illustrates how to ﬁnd the time-average of R(t). Viewing R(t) as a reward ❅ ❅ ❅ ❅ ❅ R(t) ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ t❅ ❅ ❅ ❅ ✛ Z1 ✲✛ Z2 ✲✛ Z3✲ 0 ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ S1 Figure 3.15: Sample value of the residual life function of customers in service. function, we can ﬁnd the accumulated reward up to time t as the sum of triangular areas. R First, consider R(τ )dτ from 0 to SN (t) , i.e., the accumulated reward up to the last renewal epoch before t. SN (t) is not only a renewal epoch for the renewal process, but also an arrival epoch for the arrival process; in particular, it is the A(SN (t) )th arrival epoch, and the A(SN (t) ) − 1 earlier arrivals are the customers that have received service up to time SN (t) . Thus, Z A(SN (t) )−1 SN (t) R(τ ) dτ = τ =0 X i=1 A(t) 2 XZ Zi2 i ≤ . 2 2 i=1 R SN We can similarly upper bound the term on the right above by τ =0(t)+1 R(τ ) dτ . We also Rt know (from going through virtually the same argument many times) that (1/t) τ =0 R(τ )dτ will approach a limit with probability 1 as t → 1, and that the limit will be unchanged if t is replaced with SN (t) or SN (t)+1 . Thus, taking ∏ as the arrival rate, lim t→1 Rt 0 R(τ ) dτ = lim t→1 t £§ ∏E Z 2 Zi2 A(t) = 2A(t) t 2 PA(t) i=1 W.P.1. From Corollary 3.2, we can replace the time average above with the limiting ensembleaverage, so that £§ ∏E Z 2 lim E [R(t)] = . (3.65) t→1 2 Finally, we can use Little’s theorem, in the limiting ensemble-average form of (3.61), to assert that limt→1 E [Lq (t)] = ∏W q . Substituting this plus (3.65) into (3.64), we get £§ ∏E Z 2 lim E [U (t)] = ∏E [Z ] W q + . (3.66) t→1 2 122 CHAPTER 3. RENEWAL PROCESSES This shows that limt→1 E [U (t)] exists, so that E [U (t)] is asymptotically independent of t. It is now important to distinguish between E [U (t)] and W q . The ﬁrst is the expected unﬁnished work at time t, which is the queue delay that a customer would incur by arriving at t; the second is the time-average expected queue delay. For Poisson arrivals, the probability of an arrival in (t, t + δ ] is independent of U (t)11 . Thus, in the limit t → 1, each arrival faces an expected delay limt→1 E [U (t)], so limt→1 E [U (t)] must be equal to W q . Substituting this into (3.66), we obtain the celebrated Pol laczek-Khinchin formula, £§ ∏E Z 2 Wq = . (3.67) 2(1 − ∏E [Z ]) This queueing delay has some of the peculiar features of residual life, and in particular, £§ if E Z 2 = 1, the limiting expected queueing delay is inﬁnite even though the expected service time is less than the expected interarrival interval. £§ In trying to visualize why the queueing delay is so large when E Z 2 is large, note that while a particularly long service is taking place, numerous arrivals are coming into the system, and all are being delayed by this single long service. In other words, the number of new customers held up by a long service is proportional to the length of the service, and the amount each of them are held up is also proportional to the length of the service. This visualization is rather crude, but does serve to explain the second moment of Z in (3.67). This phenomenon is sometimes called the “slow truck eﬀect” because of the pile up of cars behind a slow truck on a single lane road. For a G/G/1 queue, (3.66) is still valid, but arrival times are no longer independent of U (t), so that typically E [U (t)] 6= W q . As an example, suppose that the service time is uniformly distributed between 1 − ≤ and 1 + ≤ and that the interarrival interval is uniformly distributed between 2 − ≤ and 2 + ≤. Assuming that ≤ &lt; 1/2, the system has no queueing and W q = 0. On the other hand, for small ≤, limt→1 E [U (t)] ∼ 1/4 (i.e., the server is busy half the time with unﬁnished work ranging from 0 to 1). 3.7 Delayed renewal processes We have seen a certain awkwardness in our discussion of Little’s theorem and the M/G/1 delay result because an arrival was assumed, but not counted, at time 0; this was necessary for the ﬁrst interarrival interval to be statistically identical to the others. In this section, we correct that defect by allowing the epoch at which the ﬁrst renewal occurs to be arbitrarily distributed. The resulting type of process is a generalization of the class of renewal processes known as delayed renewal processes. The word delayed does not necessarily imply that the ﬁrst renewal epoch is in any sense larger than the other inter-renewal intervals. Rather, it means that the usual renewal...
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