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Let the number of bits used to represent the frame sequence number is $x$ then the total number of frames in the sliding window can be send from sender side using Go Back $N$ is :

  1. $1$
  2. $2x$
  3. $2x - 1$
  4. $2(x - 1)$
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Best answer
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Since sender window size < (2^X)      ; {where x is the no. of bits used to represent frame}

total no. of frames from sender side or window size of sender = (2^X) -1

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available sequence no. ≥ (sender window + receiver window)
in gbn sender window is n and receiver is 1.
now let suppose x bits are needed to represent sequence no. then
2x ≥ n+1
2- 1 ≥ n.

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In GBN, sequence numbers used are $n$ for the sender, and $1$ for the receiver.

Hence, total sequence numbers are $n + 1$.


In the question, number of bits for sequence numbers = $x$.

Total sequence numbers possible out of it = $2^x$

Reserve 1 number(not 1 bit) for the receiver, so we get $2^x-1$ numbers for the Sender.

 

None of the Options is correct!

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