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tcom501-final02-sol - TCOM501 Networking: Theory &...

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tcom501-final02-sol

tcom501-final02-sol - TCOM501 Networking: Theory &...

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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 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 iK 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
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1 21 32 1 (1 ) ) ) ) K KK p p p p λ =γ+ λ λ λ λ # Since all nodes are symmetric – same arrival rates, service rates, and routing probabilities – the aggregate arrival rates must be equal: 12 K λ =λ = =λ =λ . The above equations, then, give: ) 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 Jackson’s theorem for open networks, the stationary distribution is: 1 1 11 (, , ) () ( 1 ) ( , i K n nn K Ki i ii pn 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 iK ρ and the average number of customers in the network: 1 NK ρ = Since the arrival rate at the network is , Little’s theorem implies that the average time a customer spends in the network is: K γ 1 N T Kp ρ = γγ− 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 ) 0 qnn e e p + += µ > while in the reversed network: * 1 qnne e + since customers cannot move from node i to node i +1 in the reversed network. 2 of 15
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Problem 2 [20 points]: Consider a closed Jackson network with K nodes and M customers. The normalization constant for the network is: 12 1 () K K nn n K M GM ++ = = ρρ ρ " (1) 1. [3 points] Show that the number of terms in the summation on the right-hand side of eq. (1) is 1 MK M +    2. [7 points] Show that the normalization constant can be calculated based on the following
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