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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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plz explain?
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@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

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3 Answers

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Best answer

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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dhruvkc123

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

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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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dhruvkc123  

Yes, your interpretation was correct..

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

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As per previous question your answer was correct..

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

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4 votes
4 votes

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.

0 votes
0 votes

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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