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25 votes
25 votes

The maximum window size for data transmission using the selective reject protocol with $n\text{-bit}$ frame sequence numbers is:

  1. $2^n$
  2. $2^{n-1}$
  3. $2^n-1$
  4. $2^{n-2}$
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Best answer
54 votes
54 votes

Answer is b)

In $\text{selective reject protocol},$ the maximum window size must be half
the Sequence number space $=\dfrac{2^n}{2}=2^{n-1}$.

For Go-back n, the maximum window size can be $2^n-1$.

http://webmuseum.mi.fh-offenburg.de/index.php?view=exh&src=73 or archive

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20 votes
20 votes

NUMBER OF SEQUENCE NUMBERS AVAILABLE >= SENDER WINDOW SIZE+RECEIVER WINDOW SIZE.

Thus with n bits 2^n sequence numbers are available.

In SR protocol SENDER WINDOW SIZE = RECEIVER WINDOW SIZE..

thus window size= 2^n/2= 2^n-1

5 votes
5 votes
Here n bit has been used for sequence no.. so total possible sequence no is = 2^n

As we know sender window size= receiver window size (Selective repeat / Reject protocol)..

So window size= (2^n)/2   => 2^(n-1)
3 votes
3 votes

n bits in sequence no means 2^n sequence numbers are there
we know that in SR ,sender and reciever's window size are same...so divide these 2^n sequence no's equally..
so sender can have maximum 2^(n-1) sequence nos...so maximum possible window size is 2^(n-1)...

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