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retransmission, is still required.
24.9 The receiving TCP queues the data, but it cannot be passed to the application until
the three-way handshake is complete: when the receiving TCP moves into the
24.10 Five segments are exchanged:
1. Client to server: SYN, data (request), and FIN. The server must queue the data as
described in the previous exercise.
2. Server to client: SYN and ACK of client's SYN. file:///D|/Documents%20and%20Settings/bigini/Doc...omenet2run/tcpip/tcp-ip-illustrated/append_d.htm (16 of 20) [12/09/2001 14.48.04] Appendix D: Solutions to Selected Exercises 3. Client to server: ACK of server's SYN and client FIN (again). This causes the
server to move to the ESTABLISHED state, and the queued data from segment 1
is passed to the server application.
4. Server to client: ACK of client FIN (which also acknowledges client data), data
(server's reply), and server's FIN. This assumes that the SPT is short enough to
allow this delayed ACK. When the client TCP receives this segment, the reply is
passed to the client application, but the total time has been twice the RTT plus the
5. Client to server: ACK of server's FIN.
24.11 16,128 transactions per second (64,512 divided by 4).
24.12 The transaction time using T/TCP cannot be faster than the time required to
exchange a UDP datagram between the two hosts. T/TCP should always take longer,
since it still involves state processing that UDP doesn't do. Chapter 25
25.1 If a system is running both a manager and agent, they are probably different
processes. The manager listens on UDP port 162 for traps, and the agent listens on UDP
port 161 for requests. If the same port were used for both traps and requests, separating
the manager from the agent would be hard.
25.2 Refer to the section "Table Access" in Section 25.7. Chapter 26
26.1 We expect segments 2, 4, and 9 from the server to be delayed. The time difference
between segments 2 and 4 is 190.7 ms and the time difference between segments 2 and 9
is 400.7 ms.
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- Spring '12