2 votes 2 votes Consider a network connecting two systems, ‘A’ and ‘B’ located 6000 km apart. The propagation speed of media is 2 × 106 mps. It is needed to design a Go-Back-7 sliding window protocol for this network. The average packet size is 107 bits. If network used as its full capacity, then the bandwidth of network is__________ Mbps Computer Networks computer-networks go-back-n sliding-window + – VS asked Dec 23, 2017 VS 965 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Saswat Senapati commented Dec 23, 2017 reply Follow Share hilarious answer given by made easy..... actually in the solution given by ME they have put N=127 instead of 7 . So there solution is wrong. 1 votes 1 votes Ajay Jadhav commented Dec 23, 2017 reply Follow Share 7 Tt/ Tt+2Tp=1 =10^7 bps 1 votes 1 votes Jai Kataria commented Jan 8, 2018 reply Follow Share @ VS Boss ma'am Do you get any resource for this, In GOBACK-N, they are not treating N as window size but number of bits, what I know N is the no of packet send without any acknowledgment 0 votes 0 votes VS commented Jan 8, 2018 reply Follow Share @Jai Kataria You are right ! Answer by Made easy is wrong. 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes Distance - 6000 * 10^3 m Speed - 2 * 10^6 mps Packet size = 10^7 bits Tp = 6000 * 10^3 / 2 * 10^6 sec = 3 sec 1 = W / ( 1 + 2a ) 1 + (2 * 3 / Tt) = 7 Tt = 1 sec Packet size / Bw = 1 sec Bw = 10^7 bits per sec Bw = 10 Mbps Yogesh Mandge answered Dec 23, 2017 Yogesh Mandge comment Share Follow See all 0 reply Please log in or register to add a comment.
