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Unformatted text preview: Host B Host A Seq = 127, 70 byt es Seq = 197, 50 bytes Ack = 197 Timeout inter val Ack = 247 Seq = 127, 70 byt es Ack = 247 Timeout inter val Problem 3 Note, wrap around if overflow. One's complement = 1 1 1 0 0 0 1 1. To detect errors, the receiver adds the four words (the three original words and the checksum). If the sum contains a zero, the receiver knows there has been an error. All one-bit errors will be detected, but two-bit errors can be undetected (e.g., if the last digit of the first word is converted to a 0 and the last digit of the second word is converted to a 1). Problem 4 (a) Adding the two bytes gives 10110010. Taking the one’s complement gives 01001101. (b) Adding the two bytes gives 00010001; the one’s complement gives 11101110. (c) First byte = 01011110 ; second byte = 01010100. Problem 24 There are possible sequence numbers. a) The sequence number does not increment by one with each segment. Rather, it 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 . b) The number of segments is . 66 bytes of header get added to each segment giving a total of 528,857,934 bytes of header. The total number of bytes transmitted is bytes....
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- Fall '09
- Transmission Control Protocol, congestion window size, Congestion Window