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
Thursday, January 28, 2010
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.
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:
(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.
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.
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.
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.
Saturday, January 2, 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.
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.
Exercise 3
a) Compare the destination port in the TCP packet in frame 3 with the destination port in the TCP packet in frame 12. What difference do you see? What does this tell you about the difference in the two requests?
The destination port in the TCP packet in frame 3 is HTTP (80) and in frame 12 is HTTPS (443). The request in frame 3 is a normal request but the request in frame 12 is a secure request.
The following table compares the two requests for web pages. For example, row i) shows that frames 1-2 and frames 8-9 represent the DNS lookups for each of the web requests.
Row www.yahoo.com
frames my.usf.com
frames Brief Explanation of Activity
i) 1-2 8-9 DNS Request to find IP address for common name & DNS Response
ii) 3-5 10-12 Three-way handshake
iii) -- 13-20
iv) 6 21 “Get” request for web page
v) 7 22 First packet from web server with web page content.
b) Explain what is happening in row “iii” above. Why are there no frames listed for yahoo in row “iii"?
Row “iii” in above table shows secure operations.
There are no frames listed for yahoo because it is not using SSL or secure page.
c) Look at the “Info” column on frame 6. It says: “GET / HTTP / 1.1. What is the corresponding Info field for the my.usf.com web request (frame 21)? Why doesn’t it read the same as in frame 6?
The corresponding Info field for the my.usf.com web request is “Application Data”.
It doesn’t read the same as in frame 6 because it is encrypted due to security reasons.
The destination port in the TCP packet in frame 3 is HTTP (80) and in frame 12 is HTTPS (443). The request in frame 3 is a normal request but the request in frame 12 is a secure request.
The following table compares the two requests for web pages. For example, row i) shows that frames 1-2 and frames 8-9 represent the DNS lookups for each of the web requests.
Row www.yahoo.com
frames my.usf.com
frames Brief Explanation of Activity
i) 1-2 8-9 DNS Request to find IP address for common name & DNS Response
ii) 3-5 10-12 Three-way handshake
iii) -- 13-20
iv) 6 21 “Get” request for web page
v) 7 22 First packet from web server with web page content.
b) Explain what is happening in row “iii” above. Why are there no frames listed for yahoo in row “iii"?
Row “iii” in above table shows secure operations.
There are no frames listed for yahoo because it is not using SSL or secure page.
c) Look at the “Info” column on frame 6. It says: “GET / HTTP / 1.1. What is the corresponding Info field for the my.usf.com web request (frame 21)? Why doesn’t it read the same as in frame 6?
The corresponding Info field for the my.usf.com web request is “Application Data”.
It doesn’t read the same as in frame 6 because it is encrypted due to security reasons.
Exercise 2
a) In the first few packets, the client machine is looking up the common name (cname) of a web site to find its IP address. What is the cname of this web site? Give two IP addresses for this web site.
CNAME: www.yahoo.com
IP addresses:
• 216.109.117.106
• 216.109.117.109
b) How many packets/frames does it take to receive the web page (the answer to the first http get request only)?
It takes 22 packets to receive the web page.
c) Does this web site use gzip to compress its data for sending? Does it write cookies? In order to answer these questions, look under the payload for the reassembled packet that represents the web page. This will be the last packet from question b above. Look to see if it has “Content-Encoding” set to gzip, and to see if it has a “Set-Cookie” to write a cookie.
This web site does not use qzip to compress its data for sending.
This web site does not write cookies.
d) What is happening in packets 26 and 27? Does every component of a web page have to come from the same server? See the Hint to the left.
In packet 26, the server is sending query to another server.
In packet 27, the next server is responding to the main server.
This concludes that every component of a webpage do not have to come from same server. It might need smaller components from other server as well.
e) In packet 37 we see another DNS query, this time for us.i1.yimg.com. Why does the client need to ask for this IP address? Didn’t we just get this address in packet 26? (This is a trick question; carefully compare the two common names in packet 26 and 37.)
The DNS query made in packet 26 and 37 is different.
f) In packet 42 we see a HTTP “Get” statement, and in packet 48 a new HTTP “Get” statement. Why didn’t the system need another DNS request before the second get statement? Click on packet 42 and look in the middle window. Expand the line titled “Hypertext Transfer Protocol” and read the “Host:” line. Compare that line to the “Host:” line for packet 48.
In both packets, 42 and 48, the host is same: us.i1.yimg.com\r\n
So it does not require for another DNS query in the same session.
g) Examine packet 139. It is one segment of a PDU that is reassembled with several other segments in packet 160. Look at packets 141, 142, and 143. Are these three packets also part of packet 160? What happens if a set of packets that are supposed to be reassembled do not arrive in a continuous stream or do not arrive in the proper order?
Packets 141 and 142 are not the part of packet 160.
Packet 143 is a part of packet 160.
If a set of packets that are supposed to be reassembled do not arrive in a continuous stream or do not arrive in the proper order, it does not effect the main packet.
h) Return to examine frames 141 and 142. Both of these are graphics (GIF files) from the same source IP address. How does the client know which graphic to match up to each get statement? Hint: Click on each and look in the middle window for the heading line that starts with “Transmission Control Protocol”. What difference do you see in the heading lines for the two files? Return to the original “Get” statements. Can you see the same difference in the “Get” statements?
Both files in frames 141 and 142 are similar and from the same source IP address. The client knows the graphic to match up to each get statement from their “Stream Index”. Each of them have different “Stream Index”.
CNAME: www.yahoo.com
IP addresses:
• 216.109.117.106
• 216.109.117.109
b) How many packets/frames does it take to receive the web page (the answer to the first http get request only)?
It takes 22 packets to receive the web page.
c) Does this web site use gzip to compress its data for sending? Does it write cookies? In order to answer these questions, look under the payload for the reassembled packet that represents the web page. This will be the last packet from question b above. Look to see if it has “Content-Encoding” set to gzip, and to see if it has a “Set-Cookie” to write a cookie.
This web site does not use qzip to compress its data for sending.
This web site does not write cookies.
d) What is happening in packets 26 and 27? Does every component of a web page have to come from the same server? See the Hint to the left.
In packet 26, the server is sending query to another server.
In packet 27, the next server is responding to the main server.
This concludes that every component of a webpage do not have to come from same server. It might need smaller components from other server as well.
e) In packet 37 we see another DNS query, this time for us.i1.yimg.com. Why does the client need to ask for this IP address? Didn’t we just get this address in packet 26? (This is a trick question; carefully compare the two common names in packet 26 and 37.)
The DNS query made in packet 26 and 37 is different.
f) In packet 42 we see a HTTP “Get” statement, and in packet 48 a new HTTP “Get” statement. Why didn’t the system need another DNS request before the second get statement? Click on packet 42 and look in the middle window. Expand the line titled “Hypertext Transfer Protocol” and read the “Host:” line. Compare that line to the “Host:” line for packet 48.
In both packets, 42 and 48, the host is same: us.i1.yimg.com\r\n
So it does not require for another DNS query in the same session.
g) Examine packet 139. It is one segment of a PDU that is reassembled with several other segments in packet 160. Look at packets 141, 142, and 143. Are these three packets also part of packet 160? What happens if a set of packets that are supposed to be reassembled do not arrive in a continuous stream or do not arrive in the proper order?
Packets 141 and 142 are not the part of packet 160.
Packet 143 is a part of packet 160.
If a set of packets that are supposed to be reassembled do not arrive in a continuous stream or do not arrive in the proper order, it does not effect the main packet.
h) Return to examine frames 141 and 142. Both of these are graphics (GIF files) from the same source IP address. How does the client know which graphic to match up to each get statement? Hint: Click on each and look in the middle window for the heading line that starts with “Transmission Control Protocol”. What difference do you see in the heading lines for the two files? Return to the original “Get” statements. Can you see the same difference in the “Get” statements?
Both files in frames 141 and 142 are similar and from the same source IP address. The client knows the graphic to match up to each get statement from their “Stream Index”. Each of them have different “Stream Index”.
Exercise 1
a) What is the IP address of the client that initiates the conversation?
131.247.95.216
b) Use the first two packets to identify the server that is going to be contacted. List the common name, and three IP addresses that can be used for the server.
Server going to be contacted: www.google.com
The IP addresses that can be used for the server are:
• 64.233.161.99
• 64.233.161.104
• 64.233.161.147
c) What is happening in frames 3, 4, and 5?
In frame 3, connection establish request to the server 64.233.161.99
In frame 4, connection establish acknowledge to client 131.247.95.216. Acknowledge number=1
In frame 5, acknowledge returned to the server. Sequence number=1
d) What is happening in frames 6 and 7?
In frame 6, client is requesting server for URI (Uniform Resource Identifier)
In frame 7, server sends acknowledge for the request in frame 6
e) Ignore frame eight. However, for your information, frame eight is used to manage flow control.
f) What is happening in frames nine and ten? How are these two frames related?
In frame 9, acknowledge is set and forwarded to client
In frame 10, server is sending requested URI to client
g) What happens in packet 11?
Packet 11 is the acknowledge to the packets received in frame 10
h) After the initial set of packets is received, the client sends out a new request in packet 12. This occurs automatically without any action by the user. Why does this occur? See the first “hint” to the left.
The requested URI contains an image file which was not sent by the server in frame 10, which was in text format. So the client automatically asks for the image in another packet.
i) What is occurring in packets 13 through 22?
Packet 13 is acknowledge to packet 12. Packets 14 to 21 are requests and acknowledge related to the requested image file.
Packet 22 contains the image file that is finally sent to the client.
j) Explain what happens in packets 23 through 26. See the second “hint” to the left.
Frame 23 is an automatic request sent by the client
Frame 24 is the acknowledge to frame 23.
Frame 25 contains the image file requested by client.
Frame 26 is the acknowledge for received packet in frame 25.
k) In one sentence describe what the user was doing (Reading email? Accessing a web page? FTP? Other?).
The user was accessing the web page www.google.com.
131.247.95.216
b) Use the first two packets to identify the server that is going to be contacted. List the common name, and three IP addresses that can be used for the server.
Server going to be contacted: www.google.com
The IP addresses that can be used for the server are:
• 64.233.161.99
• 64.233.161.104
• 64.233.161.147
c) What is happening in frames 3, 4, and 5?
In frame 3, connection establish request to the server 64.233.161.99
In frame 4, connection establish acknowledge to client 131.247.95.216. Acknowledge number=1
In frame 5, acknowledge returned to the server. Sequence number=1
d) What is happening in frames 6 and 7?
In frame 6, client is requesting server for URI (Uniform Resource Identifier)
In frame 7, server sends acknowledge for the request in frame 6
e) Ignore frame eight. However, for your information, frame eight is used to manage flow control.
f) What is happening in frames nine and ten? How are these two frames related?
In frame 9, acknowledge is set and forwarded to client
In frame 10, server is sending requested URI to client
g) What happens in packet 11?
Packet 11 is the acknowledge to the packets received in frame 10
h) After the initial set of packets is received, the client sends out a new request in packet 12. This occurs automatically without any action by the user. Why does this occur? See the first “hint” to the left.
The requested URI contains an image file which was not sent by the server in frame 10, which was in text format. So the client automatically asks for the image in another packet.
i) What is occurring in packets 13 through 22?
Packet 13 is acknowledge to packet 12. Packets 14 to 21 are requests and acknowledge related to the requested image file.
Packet 22 contains the image file that is finally sent to the client.
j) Explain what happens in packets 23 through 26. See the second “hint” to the left.
Frame 23 is an automatic request sent by the client
Frame 24 is the acknowledge to frame 23.
Frame 25 contains the image file requested by client.
Frame 26 is the acknowledge for received packet in frame 25.
k) In one sentence describe what the user was doing (Reading email? Accessing a web page? FTP? Other?).
The user was accessing the web page www.google.com.
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