Discrete-time stochastic processes

It also provided an excellent example of how renewal

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: on an alternating basis between messages. A round-robin service system is a system in which, if there are m customers in the system, say c1 , c2 , . . . , cm , then c1 is served for an incremental interval δ , followed by c2 being served for an interval δ , and so forth up to cm . After cm is served for an interval δ , the server returns and starts serving c1 for an interval δ again. Thus the customers are served in a cyclic, or “round-robin” order, each getting a small increment of service on each visit from the server. When a customer’s service is completed, the customer leaves the system, m is reduced, and the server continues rotating through the now reduced cycle as before. When a new customer arrives, m is increased and the new customer must be inserted into the cycle of existing customers in a way to be discussed later. Processor sharing is the limit of round-robin service as the increment δ goes to zero. Thus, with processor sharing, if m customers are in the system, all are being served simultaneously, but each is being served at 1/m times the basic server rate. For the example of two customers with service requirement 1 and 10, each customer is initially served at rate 1/2, so one customer departs at time 2. At that time, the remaining customer is served at rate 1 and departs at time 11. For round-robin service with an increment of 1, the customer with unit service requirement departs at either time 1 or 2, depending on the initial order of service. With other increments of service, the results are slightly different. We first analyze round-robin service and then go to the processor-sharing limit as δ → 0. As the above example suggests, the results are somewhat cleaner in the limiting case, but more realistic in the round-robin case. Round robin provides a good example of the use of backward transition probabilities to find the steady-state distribution of a Markov chain. The techniques used here are quite similar to those used in the next chapter to analyze queueing networks. Assume a Bernoulli arrival process in which the probability of an arrival in an interval δ is ∏δ . Assume that the ith arriving customer has a service requirement Wi . The random variables Wi , i ≥ 1, are IID and independent of the arrival epochs. Thus, in terms of the arrival process and the service requirements, this is the same as an M/G/1 queue (see Section 3.6), but with M/G/1 queues, the server serves each customer completely before going on to the next customer. We shall find that the round-robin service here avoids the “slow truck effect” identified with the M/G/1 queue. For simplicity, assume that Wi is arithmetic with span δ , taking on only values that are positive integer multiples of δ . Let f (j ) = Pr {Wi = j δ } , j ≥ 1 and let F (j ) = Pr {Wi > j δ }. Note that if a customer has already received j increments of service, then the probability that that customer will depart after 1 more increment is f (j + 1)/F (j ). This probability of departure on the next service increment after the j th is denoted by g (j ) = f (j + 1)/F (j ); j ≥ 1. (5.43) 5.6. ROUND-ROBIN AND PROCESSOR SHARING 219 The state s of a round-robin system can be expressed as the number, m, of customers in the system, along with an ordered listing of how many service increments each of those m customers have received, i.e., s = (m, z1 , z2 , . . . , zm ), (5.44) where z1 δ is the amount of service already received by the customer at the front of the queue, z2 δ is the service already received by the next customer in order, etc. In the special case of an idle queue, s = (0), which we denote as φ. Given that the state Xn at time nδ is s 6= φ, the state Xn+1 at time nδ + δ evolves as follows: • A new arrival enters with probability ∏δ and is placed at the front of the queue; • The customer at the front of the queue receives an increment δ of service; • The customer departs if service is complete. • Otherwise, the customer goes to the back of the queue It can be seen that the state transition depends, first, on whether a new arrival occurs (an event of probability ∏δ ), and, second, on whether a departure occurs. If no arrival and no departure occurs, then the queue simply rotates. The new state is s 0 = r(s ), where the rotation operator r(s ) is defined by r(s ) = (m, z2 , . . . , zm , z1 + 1). If a departure but no arrival occurs, then the customer at the front of the queue receives its last unit of service and departs. The new state is s 0 = δ (s ), where the departure operator δ (s ) is defined by δ (s ) = (m − 1, z2 , . . . , zm ). If an arrival occurs, the new customer receives one unit of service and goes to the back of the queue if more than one unit of service is required. In this case, the new state is s 0 = a(s ) where the arrival operator a(s ) is defined by a(s ) = (m + 1, z1 , z2 , . . . , zm , 1). If only one unit of service is required by a new arrival, the arrival departs and s 0 = s ....
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