2 votes 2 votes Bandwidth of a link is 1000 Mbps and round trip time is given as 250 μ sec. If frame size is 500 bits, the utilization (in percentage) of channel when STOP and WAIT ARQ is used is _______. Computer Networks made-easy-test-series computer-networks flow-control-methods stop-and-wait + – Akanksha Kesarwani asked Feb 1, 2016 • edited Mar 5, 2019 by akash.dinkar12 Akanksha Kesarwani 1.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 4 votes 4 votes Efficiency = Useful Time / Total Time = TT / RTT TT = 500b/ 1000Mb = 0.5 μs RTT = 250μs Efficiency = 0.5/250 = 0.5/250 = 0.002 Efficiency (in %) = 00.2 % Digvijay Pandey answered Feb 1, 2016 • selected Feb 2, 2016 by Akanksha Kesarwani Digvijay Pandey comment Share Follow See all 4 Comments See all 4 4 Comments reply Prashant Sharma 1 commented Jun 24, 2016 reply Follow Share Is it correct to say that: For Channel Efficiency:Consider only the data bits not the overhead bits. For Channel Utilization/Link Utilization/Sender Utilization: Consider both the data bits as well as overhead bits. Source:https://en.wikipedia.org/wiki/Throughput#Channel_utilization_and_efficiency 0 votes 0 votes BILLY commented Jul 19, 2016 reply Follow Share But total cycle time is TT+RTT.....plz explain ... 1 votes 1 votes bharti commented Jun 23, 2017 reply Follow Share @Digvijay can u please explain why have u not taken total time =TT +2PT. 0 votes 0 votes Sumit Singh Chauhan commented Aug 3, 2018 reply Follow Share He might be ignored it because it's value is bit small. But what I think that it has to be 250.5, but anyway when we divide TT by (TT+RTT)*100 the output will be 0.199 which is nothing but 0.2 1 votes 1 votes Please log in or register to add a comment.
