0 votes 0 votes Station A needs to send a message consisting of 10 packets to station B using a sliding window of size 4. All packets are ready and can be transferred immediately. Selective repeat and GBN are used at 2 different times and every 5th packet get lost for both protocols. (ACK’s from B never gets lost). Let x and y be the number of transmissions that A has to make in selective repeat and GBN respectively to ensure safe delivery to B. Then x+y = ? Ans given=31...........but i got 38(x=12 and y=26) dileswar sahu asked Dec 25, 2016 dileswar sahu 647 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Aghori commented Dec 25, 2016 reply Follow Share x = 12. y = 18. 0 votes 0 votes Prabhanjan_1 commented Dec 25, 2016 reply Follow Share 1 2 3 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 7 8 9 10 8 9 10 y = 25. 0 votes 0 votes Aghori commented Dec 25, 2016 reply Follow Share I think next 4 packets won't be resent rather all outstanding packets will be resent. 0 votes 0 votes Digvijaysingh Gautam commented Dec 26, 2016 reply Follow Share @dileshwar sahu. You are correct 1 2 3 4 5 6 7 8 5 9 10 9 ------------------------- so x = 12 1 2 3 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 7 8 9 10 8 9 10 10 -------------------------- so y=26 2 votes 2 votes target2017 commented Dec 28, 2016 reply Follow Share yes i'm also getting x=12, y=26. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Answer 31 is correct as 1,2,3,4,5,6,7,8,5,6,7,8,6,7,8,9,8,9,10 resulting in 19 packets so 12 plus 19 equals 31 Kaluti answered Dec 26, 2016 Kaluti comment Share Follow See all 2 Comments See all 2 2 Comments reply Neeraj7375 commented Jul 7, 2017 reply Follow Share please explain in detail 0 votes 0 votes ashudevjava commented Jul 1, 2018 reply Follow Share Window size is given as 4 and you put only 3 packets where 2 nd lost occurred wrong answer plz correct it 0 votes 0 votes Please log in or register to add a comment.