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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 :

- $1$
- $2x$
- $2x - 1$
- $2(x - 1)$

@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^{k }- 1 )

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

The number N (=2^{k}-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

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@ Bikram

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..!!

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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!**