0 0 votes Computer Networks ethernet computer-networks lan-technologies + – Na462 980 views answer comment Share Follow Print See all 3 Comments 3 3 Comments reply Shubhgupta commented Sep 15, 2018 reply Follow flag 50? 0 0 replyShare Na462 commented Sep 16, 2018 reply Follow flag Yes can u explain the solution 0 0 replyShare Prateek Raghuvanshi commented Nov 27, 2018 reply Follow flag https://gateoverflow.in/171527/jam-signal 0 0 replyShare Please log in or register to add a comment.
0 0 votes Propagation delay (Tp) = 225-bit times = 225 bit / 10 Mbps = 22.5 x 10-6 sec = 22.5 μsec At t = 0, Nodes A and B start transmitting their frame. Since both, the stations start simultaneously, so collision occurs in the midway. Time after which collision occurs = Half of the propagation delay. So, time after which collision occurs = 22.5 μsec / 2 = 11.25 μsec. At t = 11.25 μsec, After the collision occurs at t = 11.25 μsec, collided signals start traveling back. Collided signals reach the respective nodes after time = Half of propagation delay Collided signals reach the respective nodes after time = 22.5 μsec / 2 = 11.25 μsec. Thus, at t = 22.5 μsec, collided signals reach the respective nodes. At t = 22.5 μsec, As soon as nodes discover the collision, they immediately release the jamming signal. Time taken to finish transmitting the jam signal = 50 bit time = 50 bits/ 10 Mbps = 5 μsec. Thus, Time at which the jamming signal is completely transmitted = 22.5 μsec + 5 μsec = 27.5 μsec or 275-bit times Mohitdas answered May 31, 2020 Mohitdas comment Share Follow 0 reply Please log in or register to add a comment.
