0 votes 0 votes But I am getting 5.427 as Answer. And unfortunately that is given incorrect. Kindly Explain. Thnks. Computer Networks computer-networks round-trip-time virtual-gate-test-series + – Jason GATE asked Jan 8, 2017 • edited Apr 14, 2019 by Lakshman Bhaiya Jason GATE 587 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Kaushik.P.E commented Jan 8, 2017 reply Follow Share I think the questions asks for RTT delay time. You need not add transmission delays for that. 1 votes 1 votes Jason GATE commented Jan 8, 2017 reply Follow Share Why? 0 votes 0 votes Kaushik.P.E commented Jan 8, 2017 reply Follow Share By definition RT delay time is = 2 * Tp 0 votes 0 votes Jason GATE commented Jan 8, 2017 reply Follow Share No, the correct Definition is RTT=Tt+2*Tp. 1 votes 1 votes target2017 commented Jan 9, 2017 reply Follow Share 5.29? 0 votes 0 votes Jason GATE commented Jan 9, 2017 reply Follow Share Procedure ? 0 votes 0 votes target2017 commented Jan 9, 2017 reply Follow Share Not sure I'm right or wrong Transmission time = L/B; where B is bandwidth So Tt (at A) = (1024×8)/(3×109) sec = 0.0027 msec Tt (at B) = (1024×8)/(60×106) sec = 0.1365 msec Tp = (500km+15km)/(2×108) = 2.575 msec While returing ACK are neglegable Tt=0 Total time: A->B->C->A = 0.0027+0.1365+2.575×2 = 5.2892 = 5.29 2 votes 2 votes Jason GATE commented Jan 9, 2017 reply Follow Share Yes, I think you are Right. I did it the Same way but , calculated Ttrans Twice. 0 votes 0 votes Please log in or register to add a comment.
