30 votes 30 votes Consider a CSMA/CD network that transmits data at a rate of $100\;\textsf{Mbps}\; (10^8\;\text{bits}$ per second) over a $1\;\textsf{km}$ (kilometre) cable with no repeaters. If the minimum frame size required for this network is $1250\;\text{bytes},$ What is the signal speed $\textsf{(km/sec)}$ in the cable? $8000$ $10000$ $16000$ $20000$ Computer Networks gatecse-2015-set3 computer-networks congestion-control csma-cd normal + – go_editor asked Feb 14, 2015 • edited Jun 21, 2021 by Lakshman Bhaiya go_editor 14.2k views answer comment Share Follow See 1 comment See all 1 1 comment reply harypotter0 commented Nov 8, 2020 reply Follow Share Similar Questions: https://gateoverflow.in/3834/gate2005-it-71 https://gateoverflow.in/39589/gate2016-2-53 https://gateoverflow.in/3376/gate2008-it-65 1 votes 1 votes Please log in or register to add a comment.
4 votes 4 votes Answer D. Since its a CSMA/CD protocol, the formula TT=2*PT applies. TT is transmission time , PT is propagation time TT= data/bandwidth PT=distance/ velocity. putting values we get 20000 Amiti7 answered Jul 23, 2015 Amiti7 comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes It is correct avadh answered Nov 12, 2017 avadh comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes CSMA/CD bandwidth ,b=100 Mbps distance , d= 1km no repeaters l=1250 * 8 bits tf>= 2tp l/b >= 2 * d/v v=2 * 1000 * 100,000000/(1250*8) v = 20000 km/sec d option akankshadewangan24 answered Apr 29, 2017 akankshadewangan24 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes http://pet.ece.iisc.ernet.in/course/E2223/Problems.pdf Regina Phalange answered Apr 29, 2017 Regina Phalange comment Share Follow See all 0 reply Please log in or register to add a comment.
