Question 5

An analog cellular system has a total of 33 MHz of bandwidth and uses
two 25-kHz simplex (one-way) channels to provide full duplex voice
and 
control channels. 

a) What is the number of channels available per cell for a 
frequency
reuse factor of 

(1) 4 cells 
 (2) 7 cells 
 and (3) 12
cells? 



Given,
Total bandwidth of the system = 33 MHz = 33000 kHz
Bandwidth of each channel = 25 kHz
Total bandwidth of the channel = 25*2 kHz = 50 kHz

We have,
Total available channels = bandwidth of system/bandwidth of channels
i.e. 33000 kHz / 50 kHz
= 660

Now,
When number of cells (N) = 4,
Channel per cell = 660/4 = 165

When N = 7,
Channel per cell = 660/7 = 95

When N = 12,
Channel per cell = 660/12 = 55


b) Assume that 1 MHz is dedicated to control channels but that 
only
one control channel is needed per cell. Determine a reasonable
distribution of control channels and voice channels in each cell for
the three frequency reuse factors of part (a).

Dedicated bandwidth to control channels = 1 MHz = 1000 kHz
Therefore,
No of control channels = dedicated bandwidth/bandwidth of channel
i.e. 1000 kHz/ 50 kHz
= 20
And,
No of voice channels = 660 – 20 = 640

When N = 4,
No of control channels per cell = 20/4 = 5
No of voice channels per cell = 165 – 5 = 160
When N = 7,
No of control channels per cell = 20/7 = 3 (approximately)
No of voice channels per cell = 95 – 3 = 92
When N = 12,
No of control channels per cell = 20/12 = 2 (approximately)
No of voice channels per cell = 55 – 2 = 53

Question 4

There are many free tools and applications available for helping decipher wireless networks. One of the most popular is Netstumbler. Obtain the software at www.netstumbler.com and follow the links for downloads. The site has a list of supported wireless cards. Using the Netstumbler software, determine the following:
(NOTE: DUE TO INCOMPATIBILITY OF Netstumbler WITH THE NETWORK ADAPTER OF MY COMPUTER, I HAVE USED SIMILAR SOFTWARE, CommView for WiFi

a) How many access points in your network have the same SSID? 


I found two access points in my metwork with the same SSID.

The highlighted part in the above figure shows access points with the same SSID.

b) What is your signal strength to your access point? 


The signal strength to my access point is 65%, as shown in the picture below:

c) How many other wireless networks and access points can you find?

I found altogether 24 wireless networks, in which, 21 were access points. While, remaining 3 were ad-hoc networks.

The list of wireless networks, strength and SSID are shown in the picture below:

Question 3

Broadband Integrated Services Digital Network (BISDN) is an optical broadband version of ISDN (which provides digital transmission over (POTS). BISDN services incorporate several technologies, including Asynchronous Transfer Mode (ATM) and Synchronous Optical Network (SONET). Provide a basic overview of SONET. Can SONET be used to extend the reach of Fibre Channel data center protocol for international data traffic?

Synchronous Optical Network (SONET) is an optical transmission interface which was developed to increasing the transmission capability of digital data through optical fiber. Similar to T-1, SONET is capable of multiplexing data using Synchronous Time Division Multiplexing to provide many data channels from different sources.
Since it uses fiber optic media for data transmission and effective framing techniques, SONET is much efficient and faster with the data rate 51.84 Mbps to 159.25248 Gbps, logically (STS-1/OC-1 to STS-3072).

SONET is considered as an evolutionary step in communication which ensures efficient, high volume transport of digitized voice, image or data communication on fiber optic networks. It can also be used by ISPs for high speed internet, Telecoms for providing 3G and 4G Networks, Television Broadcasters, and so on.

These facts show that SONET can be used to extend the reach of Fiber Channel data center protocol for international data traffic.

Though, it cannot be ignored that 10 Gigabit Ethernet is head-to-head with SONET, with some advantages like:
• Support of all kinds of topologies
• Use of shared bandwidth instead of dedicated bandwidth, as in SONET
• Use of Protection Bandwidth
• Cost effective due to simplicity and ease of management
• No bulk investment since the infrastructure for Ethernet has been introduced widely
• Easy migration and flexibility

References:
Woodcook, R (n.d). GAINING AN UNDERSTANDING OF SONET. Retrieved from http://misnt.indstate.edu/harper/Sonet.htm, January 2010.
Wikipedia. Synchronous optical networking. Retrieved from http://en.wikipedia.org/wiki/Synchronous_optical_networking#SONET.2FSDH_network_management_protocols, January 2010.
Jumboswitch. Advantage Ethernet Has Over SONET/SDH. Retrieved from http://www.jumboswitch.com/advantage_sonet-sdh.html, January 2010.

Question 2

Suppose you need to send one message to three different users:
us...@example.com, us...@example.com and us...@example.com. Is there any difference between sending one separate message per user and sending only one message with multiple (three) recipients? Explain.

There is not much difference between sending one message to multiple users and sending one separate message per user. Most of the effects come on the sender’s side.

While sending a message to multiple recipients, the message is first sent to the outgoing email server. The email server then makes a copy for each recipient and forwards them to corresponding domain. Since there are multiple recipients of same message, the server has to send them at the same time. In this case, the usage of processor of the server is higher than usual, but there is benefit of saving disc space because the server keeps only one copy of the message.

On the other hand, while sending separate message per user, there is no pressure on email server. The server forwards the message as it receives from the sender. But the disadvantage is that the messages with different email addresses have to be saved on the server even if the message is same.

Question 1

An 800 x 600 image with 24-bit colour depth needs to be stored on disc. Even though the image might contain 2^24 (2 in the power of 24) different colours, only 256 colours are actually present. This image could be encoded by means of a table (palette) of 256 24-bit elements and, for each pixel, an index of its RGB value in the table. This type of encoding is usually called Colour Look Up Table (CLUT) encoding.

a) How many bytes are needed to store the image raw information?

Given,
Image Resolution (R) = 800*600 pixel
Colour Depth (D) = 24 bits/pixel (i.e. 3 bytes/pixel)
We have,
Size of an image = R*D
i.e. 800*600*3 bytes
= 1440000 bytes

Therefore, 1440000 bytes are required to store the image raw information.

b) How many bytes are needed to store the image using CLUT encoding?

Given,
Image Resolution (R) = 800*600 pixel
Colour Depth (D) = 24 bits/pixel (i.e. 3 bytes/pixel)
No of available colours (N) = 256 (i.e. )
Therefore, Size of each pixel (D1) = 8 bits (i.e. 1 byte)

We have,
Size of Image = R*D1
i.e. 800*600*1 bytes
= 480000 bytes
And,
Size of Pallet Map (table) = N*D
i.e. 256*3 bytes
= 768 bytes

Therefore,
Total size required to store the image using CLUT encoding =
Size of Image + Size of Pallet Map
i.e. 480000 bytes + 768 bytes
= 480768 bytes.



c) What's the compression ratio achieved by using this simple encoding method?
The compression ratio achieved by using this simple encoding method is
Total size of image using CLUT : Size of image raw information
i.e. 480768 : 1440000
= 1 : 3 (approximately)

References:
HSC Online. Calculating image file size. Retrieved from http://www.hsc.csu.edu.au/ipt/mm_systems/3289/image_file_size.htm, January 2010.
Wikipedia. Indexed Color. Retrieved from http://en.wikipedia.org/wiki/Indexed_color#Colors_and_palettes, January 2010.

Exercise 5

In this exercise, you are going to capture live traffic from your email service. Begin by opening up your email service and preparing an email from you to yourself – you can use a subject line of “Homework” and a body of “Homework” if you chose. Don’t hit the send button yet.
Open up Wireshark, go to the “Capture” menu and tell it to “Start” capturing data. Now return to your email service and hit the send button. After the email is sent, tell Wireshark to “Stop” capturing data.
Look in the protocol column and see what protocol your email server uses.


What protocol did your system use?

The system I had used was a local client. It does not support secure communication and hence, I could only see TCP. But it used SSL while logging in to my email account. So as a conclusion, the user login is secure but not the data of email.

As we can see above figure, after hitting “Send” button, the email is sent to the server in smaller packets. In frame 8, the email is finally received by the server. If we see further inside the HTTP part, we can find the encapsulated data. We can use the “binary view” to view this encapsulated data.

Above diagram shows the email has finally been sent to the destination address. This occurred in frame 17 which sends “successful” message to client.

Exercise 4

In this exercise, you are going to capture live traffic from your computer. Open up Wireshark and use the “Capture” menu to save live traffic. The Wireshark “QuickStart” guide distributed with these exercises contains more instructions on using Wireshark.

Start capturing data, visit a live web site using your standard Internet browser, and stop capturing data.

The above image shows captured data of www.cnn.com.
The device used in above capture was a part of a private network with IP address 10.1.1.4. The router is working as an initial Server in this network, with IP address 10.1.1.1.
The first packet, 26, is a query sent by Client to the initial DNS for address www.cnn.com.
Packet 28 is an acknowledge sent to client in response to packet 26.
Packet 29 is the connection-establish request sent to server, 157.166.255.18.
Packet 32 is the connection-establish acknowledge from www.cnn.com to the client.
Packet 33 responses to the acknowledge of packet 32.
Packets 34 and 35 are the actions of query and responses, as in frames 26 and 28.
Frame 36 is the request for the URI to server 157.166.255.18.
Server sends the acknowledge to client in frame 37.
Frames 38 and 39 contain TCP segment to client that is to be reassembled in up-coming frames.
Frame 40 is sent by client as response to frame 39. But there is no acknowledgement to frame 38. This happened because frame 38 might not have well received by client, or lost.

The frames 41 to 79 are similar processes of sending TCP packets and replying to the server.
In frame 80, the final packet is sent by server to be reassembled in frame 81.
Frame 81 is the main HTTP packet requested by client in frame 36.

This process continues from frame 82 to frame 429 till the client get all files required by the requested URI.
Frame 448 sends final acknowledge to the server and the connection is closed.
Frame 451 acknowledges to the final acknowledge in frame 448.