0 votes 0 votes Consider the effect of using slow start on a line with 5 millisec round trip time and no congestion. The receiver window is 36 KB and Maximum segement size = 2KB. The time (in milliseconds) before the first full window can be sent is_____ Vasu_gate2017 asked Jan 25, 2017 Vasu_gate2017 471 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply santhoshdevulapally commented Jan 25, 2017 reply Follow Share RTT=5ms. MSS:2--4---8---16---32-----36.[No congestion] time before full window=4*RTT=4*5=20ms. Time to send full 36KB is equall to=5*5=25ms 0 votes 0 votes Vasu_gate2017 commented Jan 25, 2017 reply Follow Share @santoshdevulapally This explanation is correct or wrong? they they had sum up RTT and calculate it equal to 64KB. Is it of any use here? 0 votes 0 votes santhoshdevulapally commented Jan 26, 2017 reply Follow Share No need of 64KB. Upto 32 KB only calculate RTT .He is asking before full window sent but not entire window. 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes The Answer Should be 25. It is time to sendfull window. You can minus 5 if you want time before to send full window. vishwa ratna answered Jan 25, 2017 vishwa ratna comment Share Follow See all 0 reply Please log in or register to add a comment.
