CS3331/9331 Computer Networks and Applications
Transport Layer (2) – Answers
Flags = A Seq. No. = 4677 Ack. No. = 3335 Data: 100 Bytes
Flags = A Seq. No. = 3335 Ack. No. = 4777 Data: 40 Bytes
Copyright By https://powcoder.com 加微信 powcoder
a) The sequence number does not increment by one with each segment. Rather, it increments by the number of bytes of data sent. So for this part, 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 232 ¡Ö 4 Gbytes.
A2) Reordering only by 1 packet would cause an unnecessary retransmission. A3) There are 232 = 4,294,967,296 possible sequence numbers.
Flags = S Seq. No. = 4676 Ack. No. = Blank
Flags = S A Seq. No. = 3334 Ack. No. = 4677
Flags = F A Seq. No. = 4777 Ack. No. = 3375
Flags = F A Seq. No. = 3375 Ack. No. = 4778
Flags = A Seq. No. = 4778 Ack. No. = 3376
CS3331/9331 Computer Networks and Applications
b) Total number of segments generated is (2^32/536) = 8,012,999. 66 bytes get added as headers for each segment giving us 8012999 x 66 = 528,857,934 bytes of overhead. Total bytes that need to be transmitted = 4.824 x 109 bytes. Time required for transmission = (4.824 x 109 x 8)/155 x 106 = 249 sec.