Discrete-time stochastic processes

Explain why this does not mean that an innite number

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: em. Assuming that ∏ > µ1 and ∏ > µ2 , the departures from each queue are Poisson of rate ∏. Let X(t) be the state of queueing system 1 and Y (t) be the state of queueing system 2. Since X (t) at time t is independent of the departures from system 1 prior to t, X (t) is independent of the arrivals to system 2 prior to time t. Since Y (t) depends only on the arrivals to system 2 prior to t and on the service times that have been completed prior to t, we see that X (t) is independent of Y (t). This leaves a slight nit-picking question about what happens at the instant of a departure from system 1. We have considered the state X (t) at the instant of a departure to be the number of customers remaining in system 1 not counting the departing customer. Also the state Y (t) is the state in system 2 including the new arrival at instant t. The state X (t) then is independent of the departures up to and including t, so that X (t) and Y (t) are still independent. Next assume that both systems use FCFS service. Consider a customer that leaves system 1 at time t. The time at which that customer arrived at system 1, and thus the waiting time in system 1 for that customer, is independent of the departures prior to t. This means that the state of system 2 immediately before the given customer arrives at time t is independent of the time the customer spent in system 1. It therefore follows that the time that the customer spends in system 2 is independent of the time spent in system 1. Thus the total system time that a customer spends in both system 1 and system 2 is the sum of two independent random variables. This same argument can be applied to more than 2 queueing systems in tandem. It can also be applied to more general networks of queues, each with single servers with exponentially distributed service times. The restriction here is that there can not be any cycle of queueing systems where departures from each queue in the cycle can enter the next queue in the cycle. The problem posed by such cycles can be seen easily in the following example of a single queueing system with feedback (see Figure 6.11). 258 CHAPTER 6. MARKOV PROCESSES WITH COUNTABLE STATE SPACES M/M/1 ∏ ✻ Q ✲ µ ∏ ✲ 1−Q Figure 6.11: A queue with feedback. Assuming that µ > ∏/Q, the exogenous output is Poisson of rate ∏. We assume that the queueing system in Figure 6.11 has a single server with IID exponentially distributed service times that are independent of arrival times. The exogenous arrivals from outside the system are Poisson with rate ∏. With probability Q, the departures from the queue leave the entire system, and, alternatively, with probability 1 − Q, they return instantaneously to the input of the queue. Successive choices between leaving the system and returning to the input are IID and independent of exogenous arrivals and of service times. Figure 6.12 shows a sample function of the arrivals and departures in the case in which the service rate µ is very much greater than the exogenous arrival rate ∏. Each exogenous arrival spawns a geometrically distributed set of departures and simultaneous re-entries. Thus the overall arrival process to the queue, counting both exogenous arrivals and feedback from the output, is not Poisson. Note, however, that if we look at the Markov process description, the departures that are fed back to the input correspond to self loops from one state to itself. Thus the Markov process is the same as one without the self loops with a service rate equal to µQ. Thus, from Burke’s theorem, the exogenous departures are Poisson with rate ∏. Also the steady state distribution of X (t) is P {X (t) = i} = (1 − ρ)ρi where ρ = ∏/(µQ) (assuming, of course, that ρ < 1). exogenous arrival re-en ✄ ° tries ✄° ❄❄ ❄ ✂✁ ❄ ❄ ✂✁ endogenous departures ❄ ❄ endogenous departures ✄° ❄ ❄ ✂✁ ❄ Figure 6.12: Sample path of arrivals and departures for queue with feedback. The tandem queueing system of Figure 6.10 can also be regarded as a combined Markov process in which the state at time t is the pair (X (t), Y (t)). The transitions in this process correspond to, first, exogenous arrivals in which X (t) increases, second, exogenous departures in which Y (t) decreases, and third, transfers from system 1 to system 2 in which X (t) decreases and Y (t) simultaneously increases. The combined process is not reversible since there is no transition in which X (t) increases and Y (t) simultaneously decreases. In the next section, we show how to analyze these combined Markov processes for more general 6.7. JACKSON NETWORKS 259 networks of queues. 6.7 Jackson networks In many queueing situations, a customer has to wait in a number of different queues before completing the desired transaction and leaving the system. For example, when we go to the registry of motor vehicles to get a driver’s license, we must wait in one queue to have the application processed, in another queue to pay for the license, and in yet a third queue to obtain a photograph for the license. In...
View Full Document

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.

Ask a homework question - tutors are online