3 votes 3 votes bandwidth of a link is 1000 Mbps and round trip time is 250 micro seconds.If the frame size is 500 bits the link utilization ( IN %) of channel using stop and wait protocol is ? shreshtha5 asked Oct 19, 2015 shreshtha5 2.7k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes Bandwidth-delay product $= 1000 \times 10^6 \times 250 \times 10^{-6} = 250000$ But the system is sending only 500 bits so, Link utilisation is $500/250000 \times 100 = 0.2\%$ Prabhanjan_1 answered Dec 15, 2015 • selected Dec 15, 2015 by Arjun Prabhanjan_1 comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Efficiency=transmission time / (transmission time + 2propogation time) Efficiency=tt/(tt+2tp) or Efficiency=1/ 1+ 2pt/tt =1/1+250/0.5 =1/501 = 1.9960 * 10-4 =0.000199 In % = 0.000199 *100 =0.01996 is anythng wrong? Prashant. answered Nov 15, 2015 Prashant. comment Share Follow See all 2 Comments See all 2 2 Comments reply Himanshu1 commented Dec 15, 2015 reply Follow Share @ Anirudh , when R.T.T. given in question , then consider it as R.T.T = T.T + 2 P.T 2 votes 2 votes Sanjay Sharma commented Feb 27, 2017 reply Follow Share 1/501 =1.9960 x10-3 in % 1.9960x 10-1 =0.199% 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Efficiency=transmission time/(RTT) Efficiency=tt/(RTT) =0.5us/250us =0.2 %. Pooja Palod answered Oct 19, 2015 • edited Dec 15, 2015 by Arjun Pooja Palod comment Share Follow See 1 comment See all 1 1 comment reply Desert_Warrior commented Dec 13, 2015 reply Follow Share Calculation is wrong. correct ans should be 0.199% 2 votes 2 votes Please log in or register to add a comment.
