title: Link Utilization Stop and Wait Protocol
body: <p>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 acknowledgement 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_________________.</p>
  
  <blockquote>
    <p>Here is my solution:</p>
  </blockquote>
  
  <p>Link Utilization = Amount of data sent/Max. amount of data that could be sent.</p>
  
  <p>Let x be the frame size in bits.</p>
  
  <p>In stop-and-wait protocol, once a frame is sent, next frame won't be sent until ACK is received. Time for this, 
  RTT = Propagation delay for frame + Transmission time for frame + Propagation delay for ACK + Transmission time for ACK
  = 20 ms + x /64 ms + 20 ms + 0(as given in question)
  = (40 + x/64) ms. </p>
  
  <p>Amount of data sent during RTT = x</p>
  
  <p>Max. amount of data that could be sent = (40 + x/64) * 64 = 2560 + x bits. </p>
  
  <p>So, link utilization, 0.5 = x/(2560 + x)</p>
  
  <p>x = 2560 bits = 320 bytes.</p>
  
  <p>I have checked with many people and they all agree with this. But unfortunately the answer to this was given as 160 in an exam taken by over 100,000 students in India. I would like to know if I'm missing something in the question. Please help. </p>