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TCOM501 Networking: Theory & Fundamentals Final Examination Professor Yannis A. Korilis April 26, 2002 Problem 1 [30 points]: Consider a ring network with nodes 1,2,, K . In this network, a customer that completes service at node i exits the network with probability p , or it is routed to node i +1 with probability 1- p , for i = 1,2,, K-1. Customers that complete service at node K , either exit the network, or are routed to node 1, with respective probabilities p and 1-p . At each node, external customers arrive according to a Poisson process with rate . The service times at each node are exponentially distributed with rate . The arrival processes and the service times at the various nodes are independent. 1. [7 points] Find the aggregate arrival rates , 1,2,..., i iK = . 2. [2 points] Under what conditions does the ring network have a stationary distribution? 3. [7 points] Assuming that the conditions of question 2 are satisfied, find the stationary distribution of the network. 4. [7 points] Find the average time that a customer spends in the network. 5. [7 points] Is this ring network reversible? Justify your answer. Solution 1: The network is shown in the following figure. Since there are external arrivals, it is an open Jackson network. p p p 1 p 1 p 1 p i 1 i + 1. The aggregate arrival rates satisfy: 1 of 15 1 21 32 1 (1 ) (1 ) (1 ) (1 ) K KK p p p p =+ =+ =+ =+ # Since all nodes are symmetric same arrival rates, service rates, and routing probabilities the aggregate arrival rates must be equal: 12 K = = = = . The above equations, then, give: (1 ) p p =+ = 2. The network has a stationary distribution if the aggregate arrival rate at each node is less than the service rate at the node. Therefore, we must have: p < 3. From Jacksons theorem for open networks, the stationary distribution is: 1 1 11 (, , ) () ( 1 ) ( 1 ) , i K KK n nn K Ki i ii pn n p n p ++ == = = = = " 4. The average number of customers at each queue is: ,1 , , 1 i N iK == and the average number of customers in the network: 1 NK = Since the arrival rate at the network is , Littles theorem implies that the average time a customer spends in the network is: K 11 1 N T Kp == = 5. The network is not time reversible. ... View Full Document

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