Tcom501-final02-sol

tcom501-final02-sol
Download Document
Showing pages : 1 - 3 of 15
This preview has blurred sections. Sign up to view the full version! View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 i K = . 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 2 1 3 2 1 (1 ) (1 ) (1 ) (1 ) K K K p p p p = + = + = + = + # Since all nodes are symmetric same arrival rates, service rates, and routing probabilities the aggregate arrival rates must be equal: 1 2 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 1 1 ( , , ) ( ) (1 ) (1 ) , i K K K n n n K K i i i i p n n p n p + + = = = = = = " 4. The average number of customers at each queue is: , 1, , 1 i N i K = = and the average number of customers in the network: 1 N K = Since the arrival rate at the network is , Littles theorem implies that the average time a customer spends in the network is: K 1 1 1 N T K p = = = 5. The network is not time reversible. To see this, note that a transition of a customer from node i to node i +1 in the original network corresponds to a transition from node i +1 to node i in the reversed network. Considering state n with n i >0, in the original network, we have: 1 ( , ) (1 ) i i q n n e e p + + = > while in the reversed network: * 1 ( , ) i i q n n e e + + = since customers cannot move from node i to node i +1 in the reversed network. ...
View Full Document