1 votes 1 votes In a token ring, if $a<1$, i.e $\frac{T_p}{T_t} < 1$ , then will not the data get corrupted because it will come in contact with the data that is still being transmitted ? Computer Networks token-ring computer-networks + – Mojo-Jojo asked Aug 27, 2015 Mojo-Jojo 743 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 1 votes 1 votes yes . it will corrupt the data thats why if we want to apply such type of topologies bit delayer were introduced in between so that the data may not collide . so we always calculate first . whether a is less than 1 then we calculate the bit delayer required and then apply it in somewhere in the middle . it delays the data for some time. Tendua answered Aug 30, 2015 Tendua comment Share Follow See all 2 Comments See all 2 2 Comments reply Purple commented Jan 24, 2016 reply Follow Share token ring is not in syllabus now right?? 0 votes 0 votes Tendua commented Jan 25, 2016 reply Follow Share yes . out of syllabus now . 1 votes 1 votes Please log in or register to add a comment.