5 votes 5 votes In Go-Back-N protocol, if the maximum window size is 127, what is the range of the sequence number? A. 0 to 127 B. 0 to 128 C. 1 to 127 D. 1 to 128 Answer given is A Please explain :) Thanks. Computer Networks computer-networks go-back-n sliding-window + – sunil sarode asked Oct 6, 2017 sunil sarode 13.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 11 votes 11 votes Here in Go Back - N : N refers to the senders window size So window size here = 127 as it is Go back - 127.. But we know number of distinct sequence numbers for Go back N = max senders window size + 1 = 128 And starting sequence number = 0.. Hence range of sequence numbers = 0 to 127 Habibkhan answered Oct 6, 2017 selected Oct 6, 2017 by sourav. Habibkhan comment Share Follow See all 7 Comments See all 7 7 Comments reply sunil sarode commented Oct 6, 2017 reply Follow Share so you mean Sender window +receiver window =minimum sequence number, so N+1=minimum sequence number. but then bit require for sequence number will be log(N+1)/log2 ,but we in general we consider only sender window size for sequence number. 0 votes 0 votes Ashwani Kumar 2 commented Oct 6, 2017 reply Follow Share In other words For any sliding window protocol Sender's WS + Receiver's WS <= Available sequence no.'s(ASN) In GBN sender's window size is N= 127(here) Receiver's window size = 1 ASN>= 127+1 >= 128 Range of seq no is 0 to 127 2 votes 2 votes vamp_vaibhav commented Oct 6, 2017 reply Follow Share Sunil.. for sequence number we must consider both sws and rws in all the three protocols .. 2 votes 2 votes sunil sarode commented Oct 6, 2017 reply Follow Share ok got it number of bit for sequence number SR ------log(Ws+Wr) or log(2Ws) as Ws=Wr GBN------log(Ws+1) as Wr=1 Stop and Wait ------log(1+2a) base two for all :) 0 votes 0 votes Nandkishor Nangre commented Sep 19, 2018 reply Follow Share is this approach correct there window size is 127 so numbering is from 0 to 126 = 0 to (2^n)-1 so, (2^n)-1=126 2^n=127 so, n=7 now, the actual range will be 0 to (2^n)-1 == 0 to (2^7)-1 i.e= 0 to 127 0 votes 0 votes vupadhayayx86 commented Jan 6, 2019 reply Follow Share @Nandkishor Nangre 2^7 is 128 not 127 0 votes 0 votes mehul vaidya commented Dec 7, 2019 reply Follow Share why sequence number must start from 0, here D can also be correct 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Sequence number >= sws+rws for any protocol.. If you go through gbn rws=1 Therefore seq n >= 127+1 Therefore the range would be from 0 to 127 vamp_vaibhav answered Oct 6, 2017 vamp_vaibhav comment Share Follow See 1 comment See all 1 1 comment reply Pratik Mishra commented Dec 5, 2017 reply Follow Share Since 128 sequence numbers are there, then option d can also be the answer or is it that the sequence numbers start from 0 always? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes https://stackoverflow.com/questions/28444055/go-back-n-window-size Please refer this it should clear all doubts vupadhayayx86 answered Jan 6, 2019 vupadhayayx86 comment Share Follow See all 0 reply Please log in or register to add a comment.