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Consider GBN with sender's window size of n . Suppose that at time t , the next in order packet the receiver is expecting has a sequence number of 'k' . Assume that the medium does not reorder messages , then what are the possible sets of sequence numbers at time t . Assume the sender already received ACK for k-1 packets .

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GBN after how many packet, a packet get lost matter, which u havenot mentioned in question

Like every 5 th packet a packet get get lost

Since, it is inorder packet receiving , only one order is possible

Some where I read In middle buffer may contain out of order packet, but at last it will be inorder only take starting sequence number either 0 0r 1 .......overall total n seq. number should be there at tym t.

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Now, at time t, there can be 2 clear demarcations of the frames available in the sender window side. One of them, consists of the frames numbered from k to n - 1 and the other set contains frames numbered from 0 to x, wherein, if we add up the total number of frames observed in both the sets, we get a total of n

So, number of frames available in the 1st set = (n - 1) - (k) + 1 = n - k = A (let)

and number of frames available in the 2nd set = (x) - (0) + 1 = x + 1 = B (let)

Now, A + B = n.

Hence, n - k + x + 1 = n.

Solving it, we get that the value of x is :-   x = k - 1.

Hence, according to me, the set of values found is :-  k, k + 1, ..., n - 1, 0, 1, ..., k - 1.

Now, since k can be any value from 0 to n - 1, so a total of n possible sequences exist.

Someone please verify my answer and do let me know if I have erred anywhere...

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I think you are right