Discrete-time stochastic processes

# Note the purpose of this exercise is to make you

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Unformatted text preview: − f (1)]πs = (1 − ∏δ )πa(s ) ∗ πd(s ) Pd(s ),s ; ∗ πa(s ) Pa(s ),s ; ∗ πs Ps ,s ; (5.49) (1 − ∏δ )g (z1 )πs = ∏δ f (z1 + 1)πd(s ) (5.48) ∏δ f (1)πs = ∏δ f (1)πs . (5.50) We next show that (5.48), applied repeatedly, will allow us to solve for πs (if ∏ is small enough for the states to be positive recurrent). Verifying that the solution also satisﬁes (5.47) and (5.49), will then verify the hypothesis. Since f (z1 + 1)/g (z1 ) is F (z1 ) from (5.43), we have πs = ∏δ F (z1 )πd(s ) . 1 − ∏δ (5.51) For m &gt; 1, d(s ) = (m − 1, z2 , . . . , zm ), so we can apply (5.51) to πd(s ) , and substitute the result back into (5.51), yielding πs = √ ∏δ 1 − ∏δ !2 F (z1 )F (z2 )πd(d(s )) , (5.52) where d(d(s )) = (m − 2, z3 , . . . , zm ). Applying (5.51) repeatedly to πd(d(s )) , πd(d(d(s ))) , and so forth, we eventually get √ !m m Y ∏δ πs = F (zj ) πφ . (5.53) 1 − ∏δ j =1 Before this can be accepted as a steady-state distribution, we must verify that it satisﬁes (5.47) and (5.49). The left hand side of (5.47) is (1 − ∏δ )[1 − g (z1 )]πs , and, from (5.43), 1 − g (z1 ) = [F (z1 ) − f (z1 + 1)]/F (z1 ) = F (z1 + 1)/(z1 ). Thus using (5.53), the left side of (5.47) is √ !m m √ !m m Y Y F (z1 +1) ∏δ ∏δ (1 − ∏δ ) F (zj ) πφ = (1−∏δ ) F (zj ) F (z1 +1)πφ . 1−∏δ 1−∏δ F (z1 ) j =1 j =2 This is equal to (1 − ∏δ )πr(s ) , verifying (5.47). Equation (5.49) is veriﬁed in the same way. We now have to ﬁnd P whether there is a solution for pf such that these probabilities sum to 1. First deﬁne Pm = z1 , . . . , zm π (m, z1 , . . . , zm ). This is the probability of m customers in the system. Whenever a new customer enters the system, it receives one increment of service immediately, so each zi ≥ 1. Using the hypothesized solution in (5.53), √ !m m 1 YX ∏δ Pm = F (i) πφ . (5.54) 1 − ∏δ j =1 i=1 222 CHAPTER 5. COUNTABLE-STATE MARKOV CHAINS Since F (i) = Pr {W &gt; iδ }, since W is arithmetic with span δ , and since the mean of a non-negative random variable is the integral of its complementary distribution function, we have δ 1 X i=1 F (i) = E [W ] − δ Pm = √ ∏ 1 − ∏δ !m (5.55) ≥ ¥m E [W ] − δ πφ . (5.56) Deﬁning ρ = [∏/(1 − ∏δ )]{E [W ] − δ }, we see Pm = ρm πφ . If ρ &lt; 1, then πφ = 1 − ρ, and Pm = (1 − ρ)ρm ; m ≥ 0. (5.57) The condition r &lt; 1 is required for the states to be positive-recurrent. The expected P number of customers in the system for a round-robin queue is m mPm = ρ/(1 − ρ), and using Little’s theorem, Theorem 3.8, the expected delay is ρ/[∏(1 − ρ)]. In using Little’s theorem here, however, we are viewing the time a customer spends in the system as starting when the number m in the state increases; that is, if a customer arrives at time nδ , it goes to the front of the queue and receives one increment of service, and then, assuming it needs more than one increment, the number m in the state increases at time (n + 1)δ . Thus the actual expected delay, including the original d when the customer is being served but not counted in the state, is δ + ρ/[∏(1 − ρ)]. The relation between ρ and ∏E [W ] is shown in Figure 5.7, and it is seen that ρ &lt; 1 for ∏E [W ] &lt; 1. The extreme case where ∏δ = ∏E [W ] is the case for which each customer requires exactly one unit of service. Since at most one customer can arrive per time increment, the state always remains at s = φ, and the delay is δ , i.e., the original increment of service received when a customer arrives. 1 ° ° ρ ° ° ∏δ ° ∏E [W ] 1 Figure 5.7: ρ as a function of ∏E [W ] for given ∏δ . Note that (5.57) is the same as the distribution of customers in the system for the M/M/1 Markov chain in (5.42), except for the anomaly in the deﬁnition of ρ here. We then have the surprising result that if round-robin queueing is used rather than FCFS, then the distribution of the number of customers in the system is approximately the same as that for an M/M/1 queue. In other words, the slow truck eﬀect associated with the M/G/1 queue has been eliminated. Another remarkable feature of round-robin systems is that one can also calculate the expected delay for a customer conditional on the required service of that customer. This 5.7. SEMI-MARKOV PROCESSES 223 is done in Exercise 5.15, and it is found that the expected delay is linear in the required service. Next we look at processor sharing by going to the limit as δ → 0. We ﬁrst eliminate the assumption that the service requirement distribution is arithmetic with span δ . Assume that the server always spends an increment of time δ on the customer at the front of the queue, and if service is ﬁnished before the interval of length δ ends, the server is idle until the next sample time. The analysis of the steady-state distribution above P still valid if is 1 we deﬁne F (j ) = Pr...
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## This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.

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