0 votes 0 votes Assume there are 100 nodes are connected to a 1000 meter length of the coaxial cable.using some protocol,each node can transmit 50 frames per second where the average frame is 2500bits.the transmission rate at each node is 10^8 .the efficiency of this protocol is Mayank Gupta 3 asked Dec 22, 2018 Mayank Gupta 3 353 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes For single node, Transmission rate = $50*2500=125*10^3\,bps$ For 100 nodes, = $100*125*10^2=125*10^5\,bps$ Efficiency $=\frac{125*10^5}{10^8}=0.125$ Shobhit Joshi answered Dec 22, 2018 Shobhit Joshi comment Share Follow See all 0 reply Please log in or register to add a comment.