Mohamed G Gouda							CS 326E
Spring 2013							Homework 2
-------------------------------------------------------------------------------
1. A non-profit organization, named Gouda,Inc, has the following Authoratative
   DNS server and Web server: 
	
   	Alias name of the Authoratative DNS server = gouda.org
	Canonical name of the Authoratative DNS server = dns.texas.gouda.org
	IP address of the Authoratative DNS server = 128.83.115.27

	Canonical name of the Web server = www.gouda.org
	IP address of the Web server = 128.83.126.39

Describe how to register the web page of this organization in the DNS.

Sol:

   First, publish the following two Resource Records (RRs) in every .org  
   DNS server in the world.

	(gouda.org, dns.texas.gouda.org, NS, TTL)
	(dns.texas.gouda.org, 128.83.115.27, A, TTL)

   Second, publish the following RR in the Authoratative DNS server of
   the organization.


	(www.gouda.org, 128.83.126.39, A, TTL)
-------------------------------------------------------------------------------
2. Assume that the UDP socket table in the client host ch, whose IP address is 
   128.83.117.205, has the following entry:

	(dst port = 10345, socket = 4519)

1) Describe a scenario where the host ch is required to read this entry.

2) Describe a scenario where the host ch removes this entry from the 
   UDP socket table.

Sol:
    
   1) When host ch receives a UDP segment, whose dst port is 10345, then host 
      ch adds this segment at the tail of the in-buffer of socket 4519.
 
   2) That this entry is in the UDP socket table indicates that an application 
      client apc has allocated both UDP port 10345 and UDP socket 4519 and 
      has linked them. When apc de-allocates both port 10345 adn socket 4519,
      it also removes the corresponding entry from the UDP socket table.

-------------------------------------------------------------------------------
3. Assume that the TCP socket table in the server host sh, whose IP address is 
   128.83.110.34, has only two entries, namely :

 	(src     = (128.83.202.55, 2005),
	 dst     = (128.83.110.34, 80),
	 socket  = 1867)

	(src     = (any,any),
	 dst     = (128.83.110.34, 80),
	 socket  = 1572)

(Hint : Note that the second entry belongs to the Welcome Socket for port 80 in sh)

1) Describe a scenario where the host sh is required to read the first entry.

2) Describe a scenario where the host sh is required to read both entries.

Sol : 

1) When host sh receives a TCP segment whose src = (128.83.202.55, 2005) and whose 
   dst = (128.83.110.34, 80), then host sh adds this segment at the tail of the
   in-buffer of socket 1867.

2) When host sh receives a TCP segment whose src != (128.83.202.55, 2005) and whose 
   dst = (128.83.110.34, 80), then host sh adds this segment at the tail of the
   in-buffer of socket 1572.

-------------------------------------------------------------------------------
4. A client apc communicates with a server aps over a TCP connection as follows.

	1) Client apc sends a SYN segment whose seq# = 315 and whose 
	   ack# = "don't care"
	2) Server aps replies by sending a SYN-ACK segment whose seq# = 1009
	   and whose ack# = 316.
	3) Client apc replies by sending an ACK segment that has 512 data bytes
	4) Server aps replies by sending two segments. The first has no data bytes
	   and the second has 512 data bytes.
	5) Client apc replies by sending a FIN segment.
	6) Server aps replies by sending an ACK segment with 32 data bytes followed
	   by FIN segment.
	7) Client apc replies by sending an ACK segment with no data bytes.

Assuming that none of the segments is lost, corrpted or re-ordered with other 
segments, compute the seq# and ack# for each sent segment in this connection.

Sol:

SYN segment:
	seq# = 315
	ack# = "don't care"

SYN-ACK segment:
	seq# = 1009
	ack# = 316

ACK segment from apc:
	seq# = 316
	ack# = 1010

First data segment from aps:
	seq# = "don't care"
	ack# = 828

Second data segment from aps:
	seq# = 1010
	ack# = 828

FIN segment from apc:
	seq# = 828
	ack# = 1522

ACK segment from aps:
	seq# = 1522
	ack# = 829

FIN segment from aps:
	seq# = 1554
	ack# = 829

Final ACK segment from apc:
	seq# = "don't care"
	ack# = 1555

-------------------------------------------------------------------------------