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**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 γ . The service times at each node are exponentially distributed with rate μ. 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 λ = . 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. γ γ γ 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 λ = γ + λ λ = γ + λ λ = γ + λ λ = γ + λ # Since all nodes are symmetric same arrival rates, service rates, and routing probabilities the aggregate arrival rates must be equal: 1 2 K λ = λ = = λ = λ . The above equations, then, give: (1 ) p p γ λ = γ + λ λ = 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 γ < µ 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 + + = = γ = = ρ ρ = ρ ρ ρ = µ " 4. The average number of customers at each queue is: , 1, , 1 i N i K ρ = = ρ and the average number of customers in the network: 1 N K ρ = ρ Since the arrival rate at the network is , Littles theorem implies that the average time a customer spends in the network is: K γ 1 1 1 N T K p ρ = = = γ γ ρ µ γ 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 + + = µ > 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. ...

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