11 votes 11 votes In a token ring network the transmission speed is $10^7$ bps and the propagation speed is $200\;\text{meters}/\mu \text{s}.$ The $1$-bit delay in this network is equivalent to: $500$ meters of cable. $200$ meters of cable. $20$ meters of cable. $50$ meters of cable. Computer Networks gatecse-2007 computer-networks token-ring out-of-syllabus-now isro2016 + – Kathleen asked Sep 21, 2014 • edited Dec 4, 2022 by Lakshman Bhaiya Kathleen 17.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 25 votes 25 votes As, token ring network So, transmission delay = length / bandwidth transmission delay = 1/10^7 = 0.1micro sec. as, propagation speed =200m/micro sec. In , 1micro sec. it covers 200m than in 0.1micro sec. it is 20metres.(C) answer Vinay Yadav answered Aug 8, 2015 • selected Aug 21, 2016 by Prashant. Vinay Yadav comment Share Follow See all 3 Comments See all 3 3 Comments reply smartmeet commented Jan 16, 2017 reply Follow Share Can it be solve using, d<= L/2*V/B and by playing with units? 1 votes 1 votes Regina Phalange commented Nov 8, 2017 reply Follow Share i have the same doubt as @smartmeet. i tried using formula but couldn't approach. Anyone can help? 0 votes 0 votes Asim Siddiqui 4 commented Aug 2, 2020 reply Follow Share @smartmeet & @Regina Phalange It works on CSMA/CD and note that there we deal only with one sender and one receiver at distant location (so wait 2Tp for both collision S-R & R-S direction) but in ring each station(sender/receiver) is connected with another adjacent station. So we don't require that much time to wait for. L/B = d/v So it will work but remove 2 from your from your TRICK for another example working https://www.avatto.com/computer-science/test/mcqs/networking/questions/81/10.html navigate to last question THIS IS MY OBSERVATION ANY KIND OF CORRECTION IS WELCOMED 0 votes 0 votes Please log in or register to add a comment.