Answer:
Half of the window size is 2, so it count from 0 to 2;
- PC1 sends frames 0, 1, 2 to PC2.
- PC2 receives all three frames and tries to acknowledge with the third packet received.
3. Because of a noise burst, the acknowledgement for the third packet is lost.
4. PC1 times out and re-transmits the third packet.
5. PC2 has already increased its window size to accept frames 3, 0,
1, 2. Thus it assumes that frame 3 has been lost and that this is a
new frame 0, which it accepts.
Step-by-step explanation:
The problem in this situation is that there is an increase from 2 to 3 bits between the sending and receiving windows. To resolve this issue, the range of sequence numbers should be more than half of the maximum window size .