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Unformatted text preview: ECE 407: Introduction to Computer Communications Homework #4 - Solutions Problem 24 Solution: There are 2 32 = 4 , 294 , 967 , 296 possible sequence numbers. a. The sequence number does not increment by one with each segment. Rather, is increments by the number of bytes of data sent. So the size of the MSS is irrelevant the maximum size file that can be sent from A to B is simply the number of bytes representable by 2 32 4 . 19 Gbytes. b. The number of segments is 2 32 536 = 8 , 012 , 999. 66 bytes of header get added to each segment giving a total of 528,857,913 bytes of header. The total number of bytes transmitted is 2 32 + 528 , 857 , 913 = 4 , 823 , 825 , 209. Thus it would take 248.97 seconds. Problem 25 Solution: a. In the second segment from Host A to B, the sequence number is 197, source port number is 302 and destination port number is 80. b. If the first segment arrives before the second, in the acknowledgement of the first arriving seg- ment, the acknowledgement number is 197, the source port number is 80 and the destination...
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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.
- Spring '08