Solutions_Homework4

Solutions_Homework4 - nections. Problem 3 The average...

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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 A schedule can be: λ 1 (1,3) (1,3) (1,3) (1,3) (4,3) (4,3) (1,2) (4,2) (1,4) (1,4) (1,4) (4,1) λ 2 (2,4) (2,4) (3,4) (3,4) (2,1) (2,1) (3,1) (3,1) (3,1) (2,3) (3,2) (3,2) time t t+1 t+2 t+3 t+4 t+5 t+6 t+7 t+8 t+9 t+10 t+11 Problem 2 1. The length of the CAS is 8. 2. The CAS can be λ 1 (1,2) (1,3) (1,4) λ 2 (2,3) (2,4) (2,4) λ 3 (3,1) (3,1) (3,2) (3,2) (3,2) (3,4) λ 4 (4,3) (4,2) (4,1) (4,1) (4,3) (4,2) (4,3) (4,3) time 1 2 3 4 5 6 7 8 3. The traffic balance is equal to 0.625 and so totally unbalanced. 4. The two transmitters generating the higher traffic are the third and fourth. We will add a transmitter to each so α 3 = α 4 = 2. The two receivers collecting the higher traffic are the second and third. We will add a receiver to each so β 2 = β 3 = 2. If we then re-calculate the balance traffic we obtain 1. 5. The connection table is 1
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connection number tr k r k t k 1 1 2 1 2 1 3 1 3 1 4 1 4 2 3 1 5 2 4 2 6 3 4 1 7 3 { 1,2 } 2 8 3 2 1 9 4 { 1,2,3 } 2 10 4 3 2 The average traffic is 14. The LC hypergraph has two multicast con-
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Unformatted text preview: nections. Problem 3 The average length of the CAS is 43. The wasted capacity is 0.04%. Problem 4 The protocol discussed here is a slotted Aloha. Each channel chooses randomly a wavelength among W available channels. The probabil-ity of transmission of k packets is given here by a law of Poisson P k = G k e-G k ! The throughput is given by S = GP ( success ) = Ge-G This value is maximum for G=1 so the throughput is 36%. Problem 5 In a slotted ALOHA, the max utilization of a channel is 1/e, so the max bandwidth of a channel is 2.5/e = 0.92 Gbps. There are 8 channels, so the max network bandwidth is 8 × 0.92 = 7.36 Gbps. The max average bandwidth of a single link is then the network bandwidth divided by the number of links, 7.36/(8 × 7)=131 Mbps. 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 Tech.

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Solutions_Homework4 - nections. Problem 3 The average...

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