Discrete-time stochastic processes

The function r has a strictly positive second

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Unformatted text preview: B ) is irreducible, show that the new process is reversible and find its steady state probabilities. Exercise 6.24. Consider a queueing system with two classes of customers. Type A customer arrivals are Poisson with rate ∏A and Type B customer arrivals are Poisson with rate ∏B . The service time for type A customers is exponential with rate µA and that for type B is exponential with rate µB . Each service time is independent of all other service times and of all arrival epochs. a) First assume there are infinitely many identical servers, and each new arrival immediately enters an idle server and begins service. Let the state of the system be (i, j ) where i and j are the numbers of type A and B customers respectively in service. Draw a graph of the state transitions for i ≤ 2, j ≤ 2. Find the steady state PMF, {p(i, j ); i, j ≥ 0}, for the Markov process. Hint: Note that the type A and type B customers do not interact. b) Assume for the rest of the exercise that there is some finite number m of servers. Customers who arrive when all servers are occupied are turned away. Find the steady state PMF, {p(i, j ); i, j ≥ 0, i + j ≤ m}, in terms of p(0, 0) for this Markov process. Hint: Combine part (a) with the result of Exercise 6.23. c) Let Qn be the probability that there are n customers in service at some given time in steady state. Show that Qn = p(0, 0)ρn /n! for 0 ≤ n ≤ m where ρ = ρA + ρB , ρA = ∏A /µA , and ρB = ∏B /µB . Solve for p(0, 0). Exercise 6.25. a) Generalize Exercise 6.24 to the case in which there are K types of customers, each with independent Poisson arrivals and each with independent exponential service times. Let ∏k and µk be the arrival rate and service rate respectively for the kh user type, 1 ≤ k ≤ K . Let ρk = ∏k /µk and ρ = ρ1 + ρ2 + · · · + ρK . In particular, show, as before, that the probability of n customers in the system is Qn = p(0, . . . , 0)ρn /n! for 0 ≤ n ≤ m. 6.9. EXERCISES 277 b) View the customers in part (a) as a single type of customer with Poisson arrivals of rate P P ∏ = k ∏k and with a service density k (∏k /∏)µk exp(−µk x). Show that the expected service time is ρ/∏. Note that what you have shown is that, if a service distribution can be represented as a weighted sum of exponentials, then the distribution of customers in the system is the same as for the M/M/m,m queue with equal mean service time. Exercise 6.26. Consider a sampled-time M/D/m/m queueing system. The arrival process is Bernoulli with probability ∏ << 1 of arrival in each time unit. There are m servers; each arrival enters a server if a server is not busy and otherwise the arrival is discarded. If an arrival enters a server, it keeps the server busy for d units of time and then departs; d is some integer constant and is the same for each server. Let n, 0 ≤ n ≤ m be the number of customers in service at a given time and let xi be the number of time units that the ith of those n customers (in order of arrival) has been in service. Thus the state of the system can be taken as (n, x ) = (n, x1 , x2 , . . . , xn ) where 0 ≤ n ≤ m and 1 ≤ x1 < x2 < . . . < xn ≤ d. Let A(n, x ) denote the next state if the present state is (n, x ) and a new arrival enters service. That is, A(n, x ) = (n + 1, 1, x1 + 1, x2 + 1, . . . , xn + 1) A(n, x ) = (n, 1, x1 + 1, x2 + 1, . . . , xn−1 + 1) for n < m and xn < d (6.90) for n ≤ m and xn = d. (6.91) That is, the new customer receives one unit of service by the next state time, and all the old customers receive one additional unit of service. If the oldest customer has received d units of service, then it leaves the system by the next state time. Note that it is possible for a customer with d units of service at the present time to leave the system and be replaced by an arrival at the present time (i.e., (6.91) with n = m, xn = d). Let B (n, x ) denote the next state if either no arrival occurs or if a new arrival is discarded. B (n, x ) = (n, x1 + 1, x2 + 1, . . . , xn + 1) for xn < d B (n, x ) = (n − 1, x1 + 1, x2 + 1, . . . , xn−1 + 1) for xn = d. (6.92) (6.93) a) Hypothesize that the backward chain for this system is also a sampled-time M/D/m/m queueing system, but that the state (n, x1 , . . . , xn )(0 ≤ n ≤ m, 1 ≤ x1 < x2 < . . . < xn ≤ d) has a different interpretation: n is the number of customers as before, but xi is now the remaining service required by customer i. Explain how this hypothesis leads to the following steady state equations: ∏πn,x ∏πn,x (1 − ∏)πn,x (1 − ∏)πn,x = (1 − ∏)πA(n,x ) = ∏πA(n,x ) = ∏πB (n,x ) = (1 − ∏)πB (n,x ) ; n < m, xn < d ; n ≤ m, xn = d ; n ≤ m, xn = d ; n ≤ m, xn < d. (6.94) (6.95) (6.96) (6.97) b) Using this hypothesis, find πn,x in terms of π0 , where π0 is the probability of an empty system. Hint: Use (6.96) and (6.97); your answer should depend on n, but not x . 278 CHAPTER 6. MARKOV PROCESSES WITH C...
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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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