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THERE IS NO EXHIBIT. PLEASE FIX OR REMOVE. Refer to the exhibit. Thre figure represents an example of simplified operation of Windowing in TCP. The PC on the left (Sending) has a window size of 4, and the PC on the right (Receiver) has a window size of 3. The Sender sends a total of 4 bytes to the Receiver as shown in step 1. The Receiver PC then received some of the data. In step 3, what should be the byte values for segments 5 and 6 (indicated by a "?") when the Sender receives the acknowledgment (ACK 4) sent by the PC on the right (Receiver)? ADMIN PLEASE FIX OR REMOVE
none, the packet is dropped
There is no picture.
bytes 5 and 6
both byte values on the segments will have a value of 5
bytes 4 and 5
For the segments in step 3, byte values should be 4 and 5. As shown in the graphic, Receiver's window size is 3 while the Sender has a window size of 4. This means that the sender is able to send 4 bytes but the receiver can only accept 3 so it will drop some of the bytes (4). Notice that the ACK sent by the receiver to the sender is 4, which means the receiver is now ready to accept byte 4 from the sender (retransmitted). In step 4 (Receiver side), the ACK is 6, which means it receives bytes 4 and 5 so it is now ready to accept byte 6 (..or this is what the receiver expects to receive). (http://www.firewall.cx/networking-topics/protocols/tcp/137-tcp-window-size-checksum.html)
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