Solutions_Homework5 - (1-p 1 / ( N-1)) N-2 , therefore the...

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GEORGIA TECH LORRAINE GEORGIA INSTITUTE OF TECHNOLOGY School of Electrical and Computer Engineering ECE 6543 Fiber Optic Network Spring Semester 2009 Problem 1 The new traffic matrix is T = 0 2 2 0 0 0 1 3 2 0 1 1 0 0 4 0 1. The channel allocation scheduling with four wavelengths is, the conflicts are specified in italic λ 1 (1,2) (1,3) (1,2) (1,3) λ 2 (2,4) (2,3) (2,4) (2,4) λ 1 (3,1) (3,4) (3,1) (3,3) λ 1 (4,3) (4,3) (4,3) (4,3) time 1 2 3 4 2. The lossles scheduling will then be λ 1 (1,2) (1,3) (1,2) (1,3) λ 2 (2,3) (2,4) (2,4) λ 1 (3,1) (3,4) (3,1) (3,3) λ 1 (4,3) (4,3) (4,3) time 1 2 3 4 5 6 7 The length of the new CAS is 8. 3. The perfect scheduling will allocate 4 queues per NAS per wavelength. At node 4, all the packets will be on the queue of λ 4 . Problem 2 Pr(a given station has an arrival for same destination) = p*1/(N-1). Since the original source node and the destination node will not transmit a colliding packet, a collision will only occur if one of the remaining N-2 stations transmits a packet to the same destination. Pr (no collision) =
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Unformatted text preview: (1-p 1 / ( N-1)) N-2 , therefore the Pr (collision) = 1-(1-p 1 / ( N-1)) N-2 Problem 3 Each packet traverses h hops. In a (2,2) ShueNet, h = 2. At each hop, the packet experiences a queueing delay and a transmission de-lay.The average transmission delay at each node is simply 1 / . The queueing delay can be found using a standard M/M/1 queueing model. The aggregate 1 arrival rate to an arbitrary node is given by: tot = 7 h . Thus, the queueing delay is given by: D q = tot / ( - tot ) . The total delay at each node, includ-ing the transmission delay is: D = 1 /- tot . Since each packet experiences an average of two hops, the average delay is given by: D avg = 2 D = 2 /-14 Problem 4 1. In the dedicated wavelength operation, we need 54 wavelengths and 6 in the shared channel operation. 2. The average number of hops is 2.17 3. The total network capacity is 24.88 2...
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This note was uploaded on 09/01/2010 for the course ECE 6543 taught by Professor Boussert during the Spring '09 term at Georgia Institute of Technology.

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Solutions_Homework5 - (1-p 1 / ( N-1)) N-2 , therefore the...

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