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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. $2(x - 1)$

plz explain?

@kavita

Go to page number 10 in the reference link ,

3.3.3.2    Go-back-N ARQ

and read the total number of frames in the sliding window using Go Back N is ( 2- 1 )

the maximum window size is limited  for a k-bit sequence number field it is limited to 2k-1.

The number N (=2k-1) specifies how many frames can be sent without receiving acknowledgement.

Reference

http://nptel.ac.in/courses/Webcourse-contents/IIT%20Kharagpur/Computer%20networks/pdf/M3L3.pdf

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

How you get 1 for receiver side ? any source or reference ?

as question want to say maximum window size only ..which is sender window size = 2x - 1

https://en.wikipedia.org/wiki/Go-Back-N_ARQ

2-3  no. line...just check and let me know if i am wrong....and if right kindly consider my anwer..!!

Now i change the question little bit  to remove the ambiguity .

But it was ambiguous previously so i changed it little bit , now you see the question again.

Total number of frames from sender side is asked , so it is now 2x - 1

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.

by

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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