41 41 votes A channel has a bit rate of $4$ $kbps$ and one-way propagation delay of $20$ $ms$. The channel uses stop and wait protocol. The transmission time of the acknowledgment frame is negligible. To get a channel efficiency of at least $50$$\text{%}$, the minimum frame size should be $80$ $\text{bytes}$ $80$ $\text{bits}$ $160$ $\text{bytes}$ $160$ $\text{bits}$ Computer Networks gateit-2005 computer-networks stop-and-wait normal + – Ishrat Jahan 18.7k views answer comment Share Follow Print See all 3 Comments 3 3 Comments reply iit2012193 commented Dec 24, 2019 reply Follow flag instead of Stop and Wait , if GBN or selective repeat given,then how to solve? 0 0 replyShare SaiKo commented Oct 23, 2024 reply Follow flag Depends on window size. just multiply L with size of Window as that is the number of bits sent. 0 0 replyShare SaiKo commented Oct 23, 2024 reply Follow flag BW is given as bit-rate therefore answer units will be in bits. 0 0 replyShare Please log in or register to add a comment.
0 0 votes Answer is option D Srken answered 3 days ago Srken comment Share Follow 0 reply Please log in or register to add a comment.
