1 votes 1 votes A selective repeat ARQ uses a window size of $26$ to send the data. The number of bits needed to define the sequence number is _____. Computer Networks tbb-cn-1 numerical-answers + – Bikram asked Nov 26, 2016 • edited Aug 25, 2019 by go_editor Bikram 421 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes IN selective repeat ARQ , sender window <= (2^m)/2 where m is no. of bits so ,in this case it is given 26 <= 2^m / 2 which gives .... m-1 = 5 , so m= 6 bits priyanka gautam-piya answered Dec 18, 2016 • selected Oct 28, 2018 by Manoja Rajalakshmi A priyanka gautam-piya comment Share Follow See all 3 Comments See all 3 3 Comments reply amitqy commented Oct 30, 2018 reply Follow Share in selective arq ---> sender window size = receiver window size. min seq no >= window size + sender size min seq no >=52. For 52 seq no, 6 bits are required. Is this approach correct ? . 0 votes 0 votes logan1x commented Nov 25, 2019 reply Follow Share @Bikram Sir, can you confirm why answer is 6 ? or how 26 <= 2^m / 2 which gives .... m-1 gives you m-1 ? 0 votes 0 votes JashanArora commented Dec 23, 2019 reply Follow Share logan1x In Selective Repeat whatever the size of Sender's window is, same is the size of the Receiver's Window. It is given that the Sender's Window size is 26 (26 numbers not bits) => Total numbers = 26 + 26 = 52. To uniquely identify 52 numbers, we need 6 bits. 2 votes 2 votes Please log in or register to add a comment.