#SlidingWindow

Question

In an ethernet using sliding window protocol ,sender window size is 5 and receiver window size is 3.How many bits are required to store maximum sequence number??

Answer 1

Min sequence bit required  = log(Sender Window Size + Reciever Window Size ) = log (5 + 3) = 3 bits
correct me if i am wrong :D

Comments

Answer is 4 bit

Because receivers window size>1

So this is selective repeat arq

And in this protocol sws<=2^(m-1)

As well as rws<=2^(m-1)

5<=2^(m-1)

So m=4