In the computer networking book (James F. Kurose, Keith W. Ross), it is written that we need 83,333 in-flight segments(window size) in order to reach 10 Gbps throughput.

(We have 1500 byte segments & 100 ms RTT)

Looks funny, but I can not calculate this number! Can somebody help me to determine how the average congestion window size is calculated?

  • A TCP window is not normally the size of a single segment unless you have a very bad connection.
    – Ron Maupin
    Feb 15 '18 at 17:26

James F. Kurose, Keith W. Ross are writing about theoretical throughput, which means if everything is fault free and at it's optimal, then you could theoretical get 83,333 segments per second at 100 ms RTT.

In order to calculate the PPS capabilities of a device, the best way is using the throughput ratings on the spec sheets. Remember the result is just theoretical:

  • 10 Gbps = bits into bytes = = bytes
  • Default MTU size is 1500, then divide by 1500 bytes = 833.333,33 packets per second.

Because it's 100 ms RTT you need to multiply 833.333,33 by 0.1 = 83.333,333

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.