Discrete-time stochastic processes

Hint use induction exercise 52 a for the markov chain

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Unformatted text preview: In the special case of an empty queue, s = φ, the state is unchanged if either no arrival occurs or an arrival requiring one increment of service arrives. Otherwise, the new state is s = (1, 1), i.e., the one customer in the system has received one increment of service. We next find the probability of each transition for s 6= φ. The probability of no arrival is 1 − ∏δ . Given no arrival, and given a non-empty system, s 6= φ, the probability of a departure is g (z1 ) = f (z1 + 1)/F (z1 ), i e., the probability that one more increment of service allows the customer at the front of the queue to depart. Thus the probability of a departure is (1 − ∏δ )g (z1 ) and the probability of a rotation is (1 − ∏δ )[1 − g (z1 )]. Finally, the probability of an arrival is ∏δ , and given an arrival, the new arrival will leave the system after one unit of service with probability g (0) = f (1). Thus the probability of an arrival and no departure is ∏δ [1 − f (1)] and the probability of an unchanged system is ∏δ f (1). To 220 CHAPTER 5. COUNTABLE-STATE MARKOV CHAINS summarize, for s 6= φ, Ps ,r(s ) = (1 − ∏δ )[1 − g (z1 )]; r(s ) = (m, z2 , . . . , zm , z1 + 1) Ps ,a(s ) = ∏δ [1 − f (1)]; a(s ) = (m + 1, z1 , z2 , . . . , zm , 1) Ps ,d(s ) = (1 − ∏δ )g (z1 ); Ps ,s d(s ) = (m − 1, z2 , . . . , zm ) = ∏δ f (1). (5.45) For the special case of the idle state, Pφ,φ = (1 − ∏δ ) + ∏δ f (1) and Pφ,(1,1) = ∏δ (1 − f (1)). We now find the steady-state distribution for this Markov chain by looking at the backward Markov chain. We will hypothesize backward transition probabilities, and then use Theorem 5.7 to verify that the hypothesis is correct. Consider the backward transitions corresponding to each of the forward transitions in (5.45). A rotation in forward time causes the elements z1 , . . . , zm in the state s = (m, z1 , . . . , zm ) to rotate left, and the left most element (corresponding to the front of the queue) is incremented while rotating to the right end. The backward transition from r(s ) to s corresponds to the elements z2 , . . . , zm , z1 + 1 rotating to the right, with the right most element being decremented while rotating to the left end. If we view the transitions in backward time as a kind of round-robin system, we see that the rotation is in the opposite direction from the forward time system. In the backward time system, we view the numbers z1 , . . . , zm in the state as the remaining service required before the corresponding customers can depart. Thus, these numbers decrease in the backward moving system. Also, since the customer rotation in the backward moving system is opposite to that in the forward moving system, zm is the remaining service of the customer at the front of the queue, and z1 is the remaining service of the customer at the back of the queue. We also view departures in forward time as arrivals in backward time. Thus the backward transition from d(s ) = (m − 1, z2 , . . . , zm ) to s = (m, z1 , . . . , zm ) corresponds to an arrival requiring z1 + 1 units of service; the arrival goes to the front of the queue, receives one increment of service, and then goes to the back of the queue with z1 increments of remaining service. The nicest thing we could now hope for is that the arrivals in backward time are Bernoulli. This is a reasonable hypothesis to make, partly because it is plausible, and partly because it is easy to check via Theorem 5.7. Fortunately, we shall find that it is valid. According to ∗ this hypothesis, the backward transition probability Pr(s ),s is given by 1 − ∏δ ; that is, given that Xn+1 is r(s ) = (m, z2 , . . . , zm , z1 + 1), and given that there is no arrival in the backward system at time (n + 1)δ , then the only possible state at time n is s = (m, z1 , . . . , zn ). Next consider a backward transition from d(s ) = (m − 1, z2 , . . . , zn ) to s = (m, z1 , z2 , . . . , zm ). This corresponds to an arrival in the backward moving system; the arrival requires z1 + 1 increments of service, one of which is provided immediately, leaving the arrival at the back of the queue with z1 required increments of service remaining. The probability of this ∗ transition is Pd(s ),s = ∏δ f (z1 + 1). Calculating the other backward transitions in the same way, the hypothesized backward transition probabilities are given by ∗ Pr(s ),s ∗ Pa(s ),s = 1 − ∏δ = 1 − ∏δ ∗ Pd(s ),s = ∏δ f (z1 + 1) ∗ Ps ,s = ∏δ f (1). (5.46) One should view (5.46) as an hypothesis for the backward transition probabilities. The arguments leading up to (5.46) are simply motivation for this hypothesis. If the hypothesis 5.6. ROUND-ROBIN AND PROCESSOR SHARING 221 is correct, we can combine (5.45) and (5.46) to express the steady-state equations of Theorem 5.7 (for s 6= f ) as ∗ πs Ps ,r(s ) = πr(s ) Pr(s ),s ; πs Ps ,d(s ) = πs Ps ,a(s ) = πs Ps ,s = (1 − ∏δ )[1 − g (z1 )]πs = (1 − ∏δ )πr(s ) (5.47) ∏δ [1...
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