0 0 votes Computer Networks computer-networks token-ring + – Na462 2.4k views answer comment Share Follow Print See all 6 Comments 6 6 Comments reply Na462 commented Dec 3, 2018 reply Follow flag I am getting B as an answer 0 0 replyShare arya_stark commented Dec 3, 2018 reply Follow flag Propagation Speed :: $200m/μsec$ Bw :: $20Mbps = 20 * 10^6$ Distance :: $20km = 200 * 10 ^ 3m$ propagation delay : $(200 * 10 ^ 3m)/(200*10^6m/sec)$ L=2*Propagationsdelay*Bw $L = 20,000bits$ 0 0 replyShare Na462 commented Dec 4, 2018 reply Follow flag Are u taking The L = Bandwidth * RTT ? It should be simply L = Bandwidth * Propagation delay becasue here in token ring only one request can happen at a time. 0 0 replyShare Na462 commented Dec 4, 2018 reply Follow flag What i did is in 200km it will require 1000micro second So total capacity is = 20Mbps * 10,000 = 20,000. Assuming it to be half duplex 1 1 replyShare arya_stark commented Dec 7, 2018 reply Follow flag @Na462 I did efficiency = 1 0 0 replyShare nishant_magarde commented May 14, 2019 reply Follow flag when we take efficiency = 1 or 100% then, Length= $T_{p}*BW$ Am I right?? 0 0 replyShare Please log in or register to add a comment.
