Mohamed G Gouda							CS 326E
Spring 2013							Midterm 2
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1. Mention two objectives that both transport protocols UDP and TCP share.

Sol:

   First, both UDP and TCP use their port numbers to name their applications.
   For example, the UDP port number 53 names the server of the DNS application
   (runnning on top of UDP). Also, the TCP port number 25 names the server
   of the SMTP application (running on top of TCP).

   Second, both UDP and TCP have checksums that can be used for end-to-end
   corruption detection.

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2.Mention two reasons why the DNS application is designed to run on top of UDP.

Sol:

   First, it is cheaper (in terms of time and storage) to run DNS on top of 
   UDP rather than on top of TCP.

   Second, the communication style in DNS is simple, consisting of sending 
   one query then waiting to receive a reply. This communication style can 
   be made reliable in a straightforward fashion without having to rely on
   the "heavy" reliable data transfer capability of TCP.

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3. 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". This segment is received correctly.
	2) Server aps replies by sending a SYN-ACK segment whose seq# = 1009
	   and whose ack# = 316. This segment is received correctly.
	3) Client apc replies by sending an ACK segment that has 512 data bytes. 
           This segment is received correctly.
	4) Server aps replies by sending one segment that has 512 data bytes.
	   This segment is lost.	   
	5) Server aps times out and resends its last segment which had 512 data 
	   bytes. This segment is received correctly.

The seq# and ack# for the first 3 segments are as follows:

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

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

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

Compute the seq# and ack# for the last 2 segments.

Sol:

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

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

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4. Consider a TCP data byte stream from a client apc to a server aps. Assume 
   that this data stream is in the following state:

	The seq# of the next data byte to be acknowledged (na) by aps = 4500
	The seq# of the next data byte to be sent (ns) by apc = 5000
	The seq# of the next data byte to be received (nr) by aps = 4700
	The seq# of the next data byte to be read (nd) by aps = 4400
	The size of the circular buffer to store the received data bytes from 
	apc = 2000 bytes

   Compute the maximum number of data bytes that apc can send without 
   receiving any acknowledgements from aps.

Sol:
	
	The size wr of the receiving window in the circular buffer = 
	buffer size - (nr - nd) which is equal to 1700.

	The size w of the window = wr which is equal to 1700.

	Without receiving any acknowledgements from aps, 
	the maximum value of ns = na + wr which is equal to 6200.

	The maximum number of data bytes that apc can send without reeciving any
	aknowledgements from aps = na + wr - ns which is equal to 1200.

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