Consider the figure below, where a TCP sender sends 8 TCP segments at t = 1, 2, 3, 4, 5, 6, 7, 8. Suppose the initial value of the sequence number is 0 and every segment sent to the receiver each contains 100 bytes. The delay between the sender and receiver is 5 time units, and so the first segment arrives at the receiver at t = 6. The ACKs sent by the receiver at t = 6, 7, 8, 10, 11, 12 are shown. The TCP segments (if any) sent by the sender at t = 11, 13, 15, 16, 17, 18 are not shown.The segment sent at t=4 is lost, as is the ACK segment sent at t=7.What is the sequence number of the segment sent at t=2? What is the ACK value carried in the receiver-to-sender ACK sent at t = 6? What is the ACK value carried in the receiver-to-sender ACK sent at t = 8? What is the ACK value carried in the receiver-to-sender ACK sent at t = 10?

TCP sequence numbers and ACKs (2). Suppose that as shown in the figure below, a TCP sender is sending segments with 100 bytes of payload.  The TCP sender sends five segments with sequence numbers 100, 200, 300, 400, and 500.  Suppose that the segment with sequence number 300 is lost.  The TCP receiver will buffer correctly-received but not-yet-in-order segments for later delivery to the application layer (once missing segments are later received). Complete the sentences below ....After receiving segment 100, the receiver responds with an ACK with value: After receiving segment 200, the receiver responds with an ACK with value: After receiving segment 500, the receiver responds with an ACK with value: After receiving the retransmitted segment 300, the receiver responds with an ACK with value:

TCP sequence numbers and ACKs (1). Consider the TCP Telnet scenario below (from Fig. 3.31 in text). Why is it that the receiver sends an ACK that is one larger than the sequence number in the received datagram?Group of answer choicesBecause TCP sequence numbers always increase by 1, with every new segment, and the TCP receiver always send the sequence number of the next expected segmentBecause the send-to receiver segment carries only one byte of data, and after that segment is received, the next expected byte of data is just the next byte (i.e., has an index that is one larger) in the data stream.

A TCP receiver must send an acknowledgement for every segment it receives.Group of answer choicesTrue: for the sender to know that the bytes in each segment have been received the receiver must ACK each segment individually.False: TCP uses cumulative acknowledgements, so the receiver need only send an acknowledgment for the bytes in the segment with the largest sequence number that has been successfully received.

c. TCP congestion control example (c). Consider again the figure above (in question 3.7-1a), where a TCP sender sends 8 TCP segments at t = 1, 2, 3, 4, 5, 6, 7, 8 and the segment sent at t=4 is lost, as is the ACK segment sent at t=7.What does the sender do at t=16? You can assume for this question that no timeouts have occurred.Group of answer choicesInform the upper layer that the connection is terminated, and close the socket.Do nothing except increment the number of duplicate ACKs received by 1.Cut its value of cwnd in half, and retransmit the segment with sequence number 300Sets its cwnd window value to 1, and retransmit the segment with sequence number 300

c) In a TCP connection, the initial sequence number at the client site is 2717. The client opens the connection, sends three segments, the second of which carries 1000 bytes of data, and closes the connection. What is the value of the sequence number in each of the following segments sent by the client? [2 Marks] i) The SYN segment ii) The data segment iii) The FIN segment