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 16.9k 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.