Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50 % is ?

For this question many of my friends and me are getting 320B as answer but the official answer which the exam has published given 160B as answer. I am not able to understand how the answer is 160B. My approach was that if we have to get efficiency at least 50% in Stop and Wait then we have the relation Transmission time>=2*Propagation time. But from this I am getting 320.

  • $\begingroup$ i get 320 too. do you by any chance have a reference formula for stop-and-wait link utilization (i mean the one that is used in the course for the exam). Maybe for some unknown reason they think that stop and wait can send ACKs before the whole packet is receiveed ( and then you will get transmission time > propagation time), which should be wrong though. $\endgroup$
    – Effie
    Dec 27, 2021 at 13:03
  • $\begingroup$ No I have no reference to that. That exam just tell the option which is correct and not a single explanation. $\endgroup$
    – Sameer Raj
    Dec 27, 2021 at 16:32
  • $\begingroup$ Maybe the discussion here will be useful for you. gateoverflow.in/8363/Gate-cse-2015-set-1-question-53 $\endgroup$
    – Sameer Raj
    Dec 27, 2021 at 16:34


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