---------------------------------------------------------------------------
Mohamed G. Gouda						CS 356
Spring 2018							Homework 2
---------------------------------------------------------------------------
1. (5 points)
A client-server application is deployed on top of UDP in two hosts ch and 
sh. The client is deployed on top of UDP in ch and the server is deployed
on top of UDP in sh.

Execution of this application proceeds as follows. First the client sends
an application msg m to UDP in ch. Then, UDP in ch adds a UDP header
to msg m forming a UDP segment s and sends s. Eventually, segment s
arrives at UDP in sh which removes the UDP header from s and forwards the
the resuting application msg m to the application server in sh.

The application server in sh prepares a reply m' to msg m and forwards m'
to UDP in sh. UDP in sh adds a UDP header to m' forming a UDP segment s'. 
Eventually, segment s' arrives at UDP in ch which removes the UDP header
from s' and forwards the resulting reply m' to the application client in
ch.

The transfer of segments between UDP in ch and UDP in sh is "best effort".
Thus, segment s can be lost (or corrupted and later discarded) after it is 
sent by UDP in ch and before it is received by UDP in sh. Similarly,
segment s' can be lost (or corrupted and later discarded) after it is
sent by UDP in sh and before it is received by UDP in ch.

(a) Explain how does the loss of s or s' lead the communication between
the client and server to a "deadlock state".

(b) Show that the client in this application can be modified such that
the loss of s or s' does not lead the communication between the client
and server to a "deadlock state".
------------------------------------------------------------------------

2. (5 points)
Consider a TCP connection between two sides src and dst. Side src sends
data segments to dst and side dst sends ack segments to src.

Each data segment Di sent from src to dst has 64 data bytes. Consecutive 
data bytes in the data byte stream from src to dst have consecutive
sequence numbers that start with the sequence number 32.

Each data segment Di has a seq# which is the sequence number of the
first data byte in Di. Each ack segment Ai has an ack# which is the
sequence number of the next data byte expected to be received by
dst when Ai is sent (from dst to src).

Consider the following five-step scenario of this connection, where
two data segments D1 and D2 are sent from src to dst, and where two
ack segments A1 and A2 are sent from dst to src:

	1. src sends D1 to dst but D1 is lost
	2. src sends D2 to dst
	3. dst sends A1 to src
	4. src times out and resends D1 to dst
	5. dst sends A2 to src

(a) What are the values of seq#'s in D1 and in D2
 
(b) What are the values of ack#'s in A1 and in A2

-------------------------------------------------------------------------
Solutions
-------------------------------------------------------------------------

1.
(a) The loss of s or s' causes the client to wait indefinitely to receive
the reply msg m' which will never arrive. In other words, 
the communication between the client and server reaches a deadlock 
state.

(b) Consider the value (T1 + T2) where T1 is an upper bound on the time
needed to compute msg m and to transfer it from the client to the server,
and T2 is an upper bound on the time needed to compute msg m' and to 
transfer it from the server to the client.

Now, modify the client to wait at most (T1 + T2) time units after it sends
msg m and before it receives the reply m'. If the time period (T1 + T2)
expires before the client rcvs m', then the client is sure that either s
or s'is lost and so it resends msg m (one more time). THerefore, the
communication between the client and server never reaches a deadlock
state.
--------------------------------------------------------------------------

2.
(a) The value of seq# in D1 is 32
    The value of seq# in D2 is 96

(b) The value of ack# in A1 is 32
    The value of ack# in A2 is 160
--------------------------------------------------------------------------
 																																																														'.