EE555-USC-Spring08-Silvester-HW3 Remaining Solutions

EE555-USC-Spring08-Silvester-HW3 Remaining Solutions -...

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EE555 – Broadband Network Architectures (Spring 2009) Homework 3 Solutions (Partial) : Optical Networking Q1. a) From node 000 there are 3 paths of length 1 (to 001, 010, 100); 3 paths of length 2 (to 011, 101, 110); and 1 path of length 3 (to 111). So h = 12/7. b) There are 12 edges in a 3-D hypercube but we need to have both directions so there are 24 (directional) links. Thus 24 waves are needed. c) Each wavelength can carry an OC-48 channel. What is the maximum end-to end traffic that the network can handle (uniform traffic pattern). Let the total traffic be γ . Then the total flow (over all links) is Since we have a uniform traffic pattern and a regular network we assume that the flows on all links are equal. So the link flow is given by: Q2. Virtual Topology # of circuit s (A) Average circuit length in segments (B) # of circuits per segment (A*B/5) Max capacity per circuit Average path length in router hops (exclude exit router) Flow per circui t Bi-directional Ring 10 5/2 5 OC 38 3/2 3 γ /2 0 FC Mesh 20 5/2 10 OC 19 1 γ /20 Since all partitioning of OC192 is possible, an OC 38 can be implemented using 38 OC 1, and an OC 19 can be implemented using 19 OC 1. Delay Calculations: Bi-directional ring:
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The routing matrix is given by: To calculate the flow matrix consider the logical links and find out the flows that go through those links. You can list the flows for the links: in a similar way. So the average delay is: We multiply by 10 because there are total 10 logical links in a bidirectional ring. We divide by 20
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EE555-USC-Spring08-Silvester-HW3 Remaining Solutions -...

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