0 votes 0 votes Parul Agarwal asked Dec 29, 2014 Parul Agarwal 9.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 9 votes 9 votes 2 * propagation delay + transmission time for jamming signal <= transmission time for frame 64 μs + 48/100 μs <= transmission time for frame 64.48 μs <= l/100 μs l >= 6448 bits >= 806 bytes. Arjun answered Dec 29, 2014 selected Dec 26, 2017 by rahul sharma 5 Arjun comment Share Follow See all 2 Comments See all 2 2 Comments reply Shreyans Dhankhar commented Jan 2, 2015 reply Follow Share it shud be 800 only coz jamming signal is not a part of frame size in ethernet so y we are counting transmission time of that 2 votes 2 votes Bhagirathi commented Nov 20, 2015 reply Follow Share Yes I do feel the same 0 votes 0 votes Please log in or register to add a comment.
6 votes 6 votes L>= 2* Tp *B L-> Length of frame Tp -> Propagation time = RTT/2 = 64 * 10- 6/2 = 32 * 10-6 B -> Bandwidth = 108 bps So, L>= 2 * 32 * 10-6 * 108 L>= 6400 b Converting in Bytes 6400/8= 800 B So option 2 Sohini answered Jun 19, 2016 Sohini comment Share Follow See all 0 reply Please log in or register to add a comment.