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homework2_solution - EC441: Introduction to Computer...

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EC441: Introduction to Computer Networks Fall 2009 Homework 2 Solutions Problem 4 Sol: (a) Adding the two bytes gives 10011101. Taking the one’s complement gives 01100010 (b) Adding the two bytes gives 00011110. Plus a carrier wrapped around, the one’s complement gives 11100000. (c) For example, change the first byte to 00110101 and the second byte to 01101000. The one’s complement remains the same although a 1-bit error has occurred. Problem 11 Sol: The protocol would still work, since a retransmission would be what would happen if the packet received with errors has actually been lost (and from the receiver standpoint, it never knows which of these events, if either, will occur). To get at the more subtle issue behind this question, one has to allow for premature timeouts to occur. In this case, if each extra copy of the packet is ACKed and each received extra ACK causes another extra copy of the current packet to be sent, the number of times packet n is sent will increase without bound as n approaches infinity. P0 Prematur e timeout P0 P1 Prematur e timeout P1 P1 P0 .
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Problem 12 Sol: P0 x . S1 A0 A 1 A0 S 0 S 1 S 0 S 0 Tou P0 P0 P1 P1 P2 P0 P3 P3 The figure above shows one situation that may cause the protocol work incorrectly: Packet 0 is sent with sequence number 0 and is then repeated due to timeout. But this repeated packet is somehow badly delayed that it arrives at the receiver after Packet 2 is sent out and gets lost. Because P0 and P2 share the same sequence number, the receiver simply accepts it as P2 without knowing the fact. This means error in transmission which cannot be detected by the protocol itself.
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homework2_solution - EC441: Introduction to Computer...

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