The maximum achievable throughput

TCP has a 32-bit sequence number field and 16-bit advertised window field. Assume that RTT is 64 (26) ms, transmission speed is 10 Mbps (10×220bps) and each segment transmitted is 1B byte. Note that since not two identical sequence numbers can be unacknowledged in the pipe, half of the sequence numbers can be used (231).

Each 5 points

  1. (a) How long does it take for the sequence numbers to warp around?
  2. (b) Now, instead of sending 1B segment, let’s send a 8 B segment. How long does it take for the sequence numbers to warp around?
  3. (c) What is the drawback in using large segments?
  4. (d) What is the maximum achievable throughput?

Solution:

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