- Consider the possible case described in following diagram in which the window size is same as the sequence number spaces, i.e S=N.
- Here, in above image, The sequence number spaces is denoted by: 2^m, where m is the size of the sequence number field of header in bits.
- Now, consider the first case when Sf is at 0 and Sn is at 1. The timer starts as soon as the first frame is sent. As the first frame receives, Receiver sends ACK. but assume that it is lost due to some reasons.
- But as, window size is 4 here, the sender will continue sending next frames, till it reaches the max window size. (4 in this case).
- As there is no loss in the frames, the frames will successfully reach the receiver, Rn will be incremented by 1 every time when frame arrives.
- Assume all the ACK. is lost. and after a while timer times out. So the sender will assume the frames are lost or the ACK are lost. So it will resent the data from beginning.
- But at the receiver end, the frames are successfully arrived, it will be waiting for next frame (in this case 0).
- And sender will also send the frame 0. Which is the previous frame. But It will be erroneously accepted by the receiver.
- So it is better to have window size(N) has to be smaller than the sequence number space(S).
- Consider the following diagram:
- In above diagram the S=N+1 or window size is 2^m-1. In this case also assume the same as we've assumed in the previous case. But the difference is, At last the receiver will be waiting for frame 3 and the sender will send frame 0. So it will be discarded. And hence the sender will send the whole set of frame again.
- Therefore in 2nd case there is more possibility of successful communication.
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