Homework 7 Solution

Problem 1 (5)

A process on host 1 has been assigned port p, and a process on host 2 has been assigned port q. Is it possible for there to be two or more TCP connections between these two ports at the same time?

No. A pair of ports uniquely sets up ONE connection

Problem 2 (40)

TCP provides reliable transfer through a mixture of sequence number, receiver buffer, cumulative acknowledgement, and fast retransmission. Answer the following True or False problems. If it's False, explain why.

a) Host A is sending host B a large file over a TCP connection. Assume host B has no data to send to A. Host B will not send acknowledgements to host A because host B cannot piggyback the acknowledgement on data.
False. Piggyback is only for efficiency. If there's no data packet to be piggybacked to, then B will just send the acknowledgement packet.
b) The size of the TCP advertised window (RcvWindow) never changes throughout the duration of the connection.
False. It is the size of the receiver's buffer that's never changed. RcvWindow is the part of the receiver's buffer that's changing all the time depending on the processing capability at the receiver's side and the network traffic.
c) Suppose host A is sending host B a large file over a TCP connection. The number of unacknowledged bytes that A sends cannot exceed the size of the receiver's buffer.
False and True. The number of unacknowleged bytes that A sends cannot exceed the size of the receiver's window. But if it can't exceed the receiver's window, then it surely has no way to exceed the receiver's buffer as the window size is always less than or equal to the buffer size. On the other hand, for urgent messages, the sender CAN send it in even though the receiver's buffer is full.
d) Suppose host A is sending host B a large file over a TCP connection. If the sequence number for a segment of this connection is m, then the sequence number for the subsequent segment will necessarily be m+1.
False. The sequence number of the subsequent segment depends on the number of 8-byte characters in the current segment.
e) Suppose host A sends host B one segment with sequence number 38 and 4 bytes of data. Then in the same segment the acknowledgement number is necessarily 42.
False. The acknowledgement number has nothing to do with the sequence number. The ack. number indicates the next sequence number A is expecting from B.
f) Suppose that the last sampleRTT in a TCP connection is equal to 1 second. Then timeout for the connection will necessarily be set to a value >= 1 second.
False. Next_RTT = alpha * last_estimated_RTT + (1-alpha)*newly_collected_RTT_sample. In this case even though the last sampleRTT which is the newly_collected_RTT_sample is 1sec, the next_RTT still depends on alpha and last_estimated_RTT. Therefore, the next_RTT is not necessarily greater than 1sec.
g) With the selective repeat protocol, it is possible for the sender to receive an ACK for a packet that falls outside of its current window.
False. SR uses selective acknowlegement. The ack. number has no way to fall outside the current window.
h) With the Go-Back-N, it is possible for the sender to receive an ACK for a packet that falls outside of its current window.
True. GBN uses cumulative acknowlegement. Imagine a scenario where ACK1 arrives AFTER ACK2. Once the sender receives ACK2, it would know that both packet1 and 2 were received correctly. So it can remove packet1 and 2 from its window. Now if ACK1 arrives, then ACK1 actually falls outside the current window.

Problem 3 (10)

TCP provides congestion control through slow start and AIMD ( additive increase and multiplicative decrease). Answer the following True or False problems. If it's False, explain why.

a) Consider congestion control in TCP. When a timer expires at the sender, the threshold is set to one half of its previous value.
False. The threshold is set to one half of the current congestion window size.
b) The slow start is really slow, which is one of the overhead introduced by congestion control.
False. The slow start isn't really slow. It grows exponentially.

Problem 4 (10)

Consider the effect of using slow start on a line with a 10-msec round-trip time and no congestion. The receive window is 24KB and the maximum segment size is 2KB. How long does it take before the first full window can be sent?
With slow start, the first RTT sends out 1 segment (or 2KB), the 2nd RTT sends out 2 segments (or 4KB), the 3rd 4 segments (or 8KB), the 4th 8 segments (or 16KB). The 5th RTT would have sent out 16 segments (or 32KB), however, it'll exceed the receiver's window. Therefore, the amont of time it takes BEFORE the 5th RTT (or full window, that is, 24KB) is 4*10=40msec.
Suppose the TCP congestion window is set to 18KB and a timeout occurs. How big will the window be if the next four transmission bursts are all successful? Assume that the maximum segment size is 1KB.
When a timeout occurs, three things happend. First, slow start will be initiated. Second, the congestion window would start at 1. Third, the threshold will be reset to 18KB/2=9KB. If the next four transmission are all successful, then
After these four successful transmissions, the window size is supposed to be 16. However, since the threshold is 9KB, the window size can only be 9KB.

Problem 5 (20)

Based on the sample codes given in class, write a client and server "echo" program which uses UDP and TCP. You need to turn in the following codes
- echoserv-tcp.c
- echocli-tcp.c
- echoserv-udp.c
- echocli-udp.c
It is important to understand what qualifies a REAL server. Following is a list of criteria,
Given the above criteria, the way you know that you have a perfect server is that open three windows. In one window, run the server program, in the other two windows, run two client program. Write messages in each of the client windows and see if the server can respond correctly. Then close one client connection, see if the server can server the other client correctly.
In the following, I've provided different versions of implementations:
- Non-iterative server and client using TCP: servertcp.c, clienttcp.c (The server and the client can only exchange ONE message, then both exit. This is the sample code I posted on slide)
- Non-iterative server and client using UDP: serverudp.c, clientudp.c (To modify TCP to UDP, three components are needed
  - change from SOCK_STREAM to SOCK_DGRAM (all of you did this right except one)
  - no need for listen(), accept() at the server's side, and no need for connect() at the client side
  - change from read()/write() to recvfrom()/sendto(). Many of you didn't have this part correct. Pay special attention to the arguments used, the socket ID used, etc. I used the original TCP code and commented in detail the changes.
- Iterative server and client using UDP: iserverudp.c, iclientudp.c (this is a lot easier than the TCP implementation)
- Iterative server and client using TCP. I provided three versions:
  - Merely adding an infinite loop can NOT make a correct server. Try the sample code and see what's wrong. iservertcp_v1.c, iclienttcp.c
  - Using fork(). iservertcp_fork.c, iclienttcp.c
  - Using select(). iservertcp_select.c, iclienttcp.c
tar file with all the codes: socket.tar
(+10)Design and develop an application scenario that tests the effect of having more requests than the allowed queuing length in "listen()". You need to turn in the code as well as the output
To be added.

Problem 6 (15)

Refresh concepts on ``big endian'' and ``little endian''

Write a program which can detect and report the hardware architecture it is running on. Give the test result for Intel PC and Sun SPARC (unix.cas.utk.edu). Turn in the code and the output.
To be added
Is there any relationship among ``big endian'', ``little endian'', ``network byte order'', ``host byte order''? Explain why or why not.
Network byte order is consistent with "big endian". "host byte order" can be either "big endian" or "little endian", depending on the computer architecture.

Problem 7 (10 - check for completion but not correctness): Task 1 of project 4

Write a server for node0 and call it node0.c. Write a client for node1 and call it node1.c. Initially, prepare the initial DV table for both node0 and node1. Let node1 send its DV to node0. Upon receiving node1's DV, node0 will modify its DV according to node1's input. Then it'll reply node1 with the updated DV.