hw7-sol - ECE 407 Introduction to Computer Communications...

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ECE 407: Introduction to Computer Communications Homework #7 - Solutions Problem 5 Solution: If we divide 10011 into 10101010100000 we get 1011011100, with a remainder of R = 0100. Problem 8 Solution: Problem 10 Solution: a. Let R A is the throughput of node A when there is no other transmission. A’s average throughput = R A · p A (1 - p B ). The total efficiency = p A (1 - p B ) + p B (1 - p A ) b. When p A = 2 p B , Node A’s average throughput = R A · 2 p B (1 - p B ). Node B’s average throughput = R B · p B (1 - 2 p B ). Although R A = R B , node A’s average throughput is not twice as large as that of node B. To make this happen, p A (1 - p B ) = 2 p B (1 - p A ) p A - p A p B = 2 p B - 2 p A p B p A = 2 p B 1+ p B 1
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c. Assume that average throughput of each node when the others are not transmitting is R. A’s average throughput = R · 2 p (1 - p ) N - 1 Any other node’s throughput = R · p (1 - 2 p )(1 - p ) N - 2 Problem 14 Solution: a.b. IP address and MAC address assignment Interface IP address MAC address A 192.168.1.1 11-11-11-11-11-11 B 192.168.1.2 22-22-22-22-22-22 R1-1(left) 192.168.1.3
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This note was uploaded on 09/07/2010 for the course ECE 407 taught by Professor Eun during the Spring '08 term at N.C. State.

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hw7-sol - ECE 407 Introduction to Computer Communications...

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