1. Thyroid (Thyroid gland, right nooddles) - See the below list of Apollo doctors in Chennai
2. West tumer (cyst, in between chest bone and west) - Searching...
3. Leg pain (left leg, knee pain) - Searching...
4. Find a package (thorough check up) - Searching...
When going:
-02/04/2017 - 17/04/2017 Choto ma on holidays
-7 days on check up
-Going to chennai apollo hospital
-From khulna to banapol (plane)
-From Domdom airport to chennai(plane) - 2hours
when returning:
--Update information about this later.
List of Endocrinologist in Chennai Apollo :
1.
Dr. S.Ramkumar
Fees- 400
Experience- 11
Hospital- Apollo
Time - Call Apollo for this doctor
MBBS , DM - Endocrinology , DNB - Diplomate in Endocrinology, Diabetes, Metabolism , PDCC - Pediatric Endocrinology , Clinical Fellowship in Diabetes , Certificate Insulin Pump
Endocrinologist , Diabetologist ,
7:00 PM - 9:00 PM
Address: Apollo Children’s Hospital, No., 15, Shafee Mohammed Rd, Thousand Lights, Chennai, Tamil Nadu 600006, India
Phone: +91 99402 86196
More information on Web
2.
Dr. Narayanan N K
Fees- 500
Experience- 17
Hospital- Apollo
Time - Monday to Saturday 9am to 5 pm
More information on Web
3.
Dr. Shriram Mahadevan
Fees- 700
Experience- 17
Hospital- Apollo
Time - Monday to Saturday 9am to 5 pm
More information on Web
4.
Dr Sivagnana Sundaram
Fees - 1000
Experience - 32
Hospital- Apollo
Monday-Friday
2:00 PM - 8:00 PM
More information on Web
5.
Dr Sundaraman P G
Fees - 1000
Experience - 26
Hospital- Apollo
Monday-Friday
11:00 AM - 12:30 PM
More information on Web
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
List of Endocrinologist in Chennai Other Hospitals :
1.
Dr Boochandran T S - MBBS, MRCP, M.Sc
209 Recommendations
Can book appointment online
25 Years of experience - Endocrinology from 2001
Monday-Saturday
8:00 AM - 11:00 AM , 2:00 PM - 5:00 PM
2.
Dr. Anantha Baskar
20 years experience
No.106, Jawaharlal Nehru Salai,
NearKoyambedu Junction ,
Thirumangalam,AnnaNagar,
Chennai, Tamil Nadu,600040, India
More Information
3.
Dr. Krishna Seshadri (Fortis Malar Hospital Chennai)
Endocrinology consultant
Fees - 700 IN R
22 years experience
More information
4.
Dr. Usha Ayyagari (Lady)
MBBS, CCT (Endocrinology), MRCP (UK)
More Information
5.
Dr. Dhalapathy Sadacharan
Fees- 500
Experience- 8
Hospital- Vijaya goup of hospitals
Time - Monday to Saturday 9am to 5 pm
********************************************************************************************************************
Find more Endocrinologist here
Saturday, March 25, 2017
Thursday, October 29, 2015
Thursday, April 9, 2015
Media Disconnected issue resolved
netsh int ip reset reset.log
netsh winsock reset catalog
See the wifi password
netsh wlan show profile name="wifi id goes here" key=clear
above cmd will show you the password for the wifi connection you / other devices are hooked in to.
Renew Ip
ipconfig/release
ipconfig/renew
Thursday, September 25, 2014
Java - Try Catch
Very good example to understand basic Try Catch
The Call Stack Explained
This text refers to the concept the "call stack" in several places. By the call stack is meant the sequence of method calls from the current method and back to the Main method of the program. If a method A calls B, and B calls C then the call stack looks like this:
A
B
C
When method C returns the call stack only contains A and B. If B then calls the method D, then the call stack looks like this:
A
B
D
Understanding the call stack is important when learning the concept of exception propagation. Exception are propagated up the call stack, from the method that initially throws it, until a method in the call stack catches it. More on that later.
Throwing Exceptions
If a method needs to be able to throw an exception, it has to declare the exception(s) thrown in the method signature, and then include a throw-statement in the method. Here is an example:
public void divide(int numberToDivide, int numberToDivideBy)
throws BadNumberException{
if(numberToDivideBy == 0){
throw new BadNumberException("Cannot divide by 0");
}
return numberToDivide / numberToDivideBy;
}
When an exception is thrown the method stops execution right after the "throw" statement. Any statements following the "throw" statement are not executed. In the example above the "return numberToDivide / numberToDivideBy;" statement is not executed if a BadNumberException is thrown. The program resumes execution when the exception is caught somewhere by a "catch" block. Catching exceptions is explained later.
You can throw any type of exception from your code, as long as your method signature declares it. You can also make up your own exceptions. Exceptions are regular Java classes that extends java.lang.Exception, or any of the other built-in exception classes. If a method declares that it throws an exception A, then it is also legal to throw subclasses of A.
Catching Exceptions
If a method calls another method that throws checked exceptions, the calling method is forced to either pass the exception on, or catch it. Catching the exception is done using a try-catch block. Here is an example:
public void callDivide(){
try {
int result = divide(2,1);
System.out.println(result);
} catch (BadNumberException e) {
//do something clever with the exception
System.out.println(e.getMessage());
}
System.out.println("Division attempt done");
}
The BadNumberException parameter e inside the catch-clause points to the exception thrown from the divide method, if an exception is thrown.
If no exeception is thrown by any of the methods called or statements executed inside the try-block, the catch-block is simply ignored. It will not be executed.
If an exception is thrown inside the try-block, for instance from the divide method, the program flow of the calling method, callDivide, is interrupted just like the program flow inside divide. The program flow resumes at a catch-block in the call stack that can catch the thrown exception. In the example above the "System.out.println(result);" statement will not get executed if an exception is thrown fromt the divide method. Instead program execution will resume inside the "catch (BadNumberException e) { }" block.
If an exception is thrown inside the catch-block and that exception is not caught, the catch-block is interrupted just like the try-block would have been.
When the catch block is finished the program continues with any statements following the catch block. In the example above the "System.out.println("Division attempt done");" statement will always get executed.
Propagating Exceptions
You don't have to catch exceptions thrown from other methods. If you cannot do anything about the exception where the method throwing it is called, you can just let the method propagate the exception up the call stack to the method that called this method. If you do so the method calling the method that throws the exception must also declare to throw the exception. Here is how the callDivide() method would look in that case.
public void callDivide() throws BadNumberException{
int result = divide(2,1);
System.out.println(result);
}
Notice how the try-catch block is gone, and the callDivide method now declares that it can throw a BadNumberException. The program execution is still interrupted if an exception is thrown from the divide method. Thus the "System.out.println(result);" method will not get executed if an exception is thrown from the divide method. But now the program execution is not resumed inside the callDivide method. The exception is propagated to the method that calls callDivide. Program execution doesn't resume until a catch-block somewhere in the call stack catches the exception. All methods in the call stack between the method throwing the exception and the method catching it have their execution stopped at the point in the code where the exception is thrown or propagated.
Example: Catching IOException's
If an exception is thrown during a sequence of statements inside a try-catch block, the sequence of statements is interrupted and the flow of control will skip directly to the catch-block. This code can be interrupted by exceptions in several places:
public void openFile(){
try {
// constructor may throw FileNotFoundException
FileReader reader = new FileReader("someFile");
int i=0;
while(i != -1){
//reader.read() may throw IOException
i = reader.read();
System.out.println((char) i );
}
reader.close();
System.out.println("--- File End ---");
} catch (FileNotFoundException e) {
//do something clever with the exception
} catch (IOException e) {
//do something clever with the exception
}
}
If the reader.read() method call throws an IOException, the following System.out.println((char) i ); is not executed. Neither is the last reader.close() or the System.out.println("--- File End ---"); statements. Instead the program skips directly to the catch(IOException e){ ... } catch clause. If the new FileReader("someFile"); constructor call throws an exception, none of the code inside the try-block is executed.
Example: Propagating IOException's
This code is a version of the previous method that throws the exceptions instead of catching them:
public void openFile() throws IOException {
FileReader reader = new FileReader("someFile");
int i=0;
while(i != -1){
i = reader.read();
System.out.println((char) i );
}
reader.close();
System.out.println("--- File End ---");
}
If an exception is thrown from the reader.read() method then program execution is halted, and the exception is passed up the call stack to the method that called openFile(). If the calling method has a try-catch block, the exception will be caught there. If the calling method also just throws the method on, the calling method is also interrupted at the openFile() method call, and the exception passed on up the call stack. The exception is propagated up the call stack like this until some method catches the exception, or the Java Virtual Machine does.
Finally
You can attach a finally-clause to a try-catch block. The code inside the finally clause will always be executed, even if an exception is thrown from within the try or catch block. If your code has a return statement inside the try or catch block, the code inside the finally-block will get executed before returning from the method. Here is how a finally clause looks:
public void openFile(){
FileReader reader = null;
try {
reader = new FileReader("someFile");
int i=0;
while(i != -1){
i = reader.read();
System.out.println((char) i );
}
} catch (IOException e) {
//do something clever with the exception
} finally {
if(reader != null){
try {
reader.close();
} catch (IOException e) {
//do something clever with the exception
}
}
System.out.println("--- File End ---");
}
}
No matter whether an exception is thrown or not inside the try or catch block the code inside the finally-block is executed. The example above shows how the file reader is always closed, regardless of the program flow inside the try or catch block.
Note: If an exception is thrown inside a finally block, and it is not caught, then that finally block is interrupted just like the try-block and catch-block is. That is why the previous example had the reader.close() method call in the finally block wrapped in a try-catch block:
} finally {
if(reader != null){
try {
reader.close();
} catch (IOException e) {
//do something clever with the exception
}
}
System.out.println("--- File End ---");
}
That way the System.out.println("--- File End ---"); method call will always be executed. If no unchecked exceptions are thrown that is. More about checked and unchecked in a later chapter.
You don't need both a catch and a finally block. You can have one of them or both of them with a try-block, but not none of them. This code doesn't catch the exception but lets it propagate up the call stack. Due to the finally block the code still closes the filer reader even if an exception is thrown.
public void openFile() throws IOException {
FileReader reader = null;
try {
reader = new FileReader("someFile");
int i=0;
while(i != -1){
i = reader.read();
System.out.println((char) i );
}
} finally {
if(reader != null){
try {
reader.close();
} catch (IOException e) {
//do something clever with the exception
}
}
System.out.println("--- File End ---");
}
}
Notice how the catch block is gone.
Catch or Propagate Exceptions?
You might be wondering whether you should catch or propate exceptions thrown in your program. It depends on the situation. In many applications you can't really do much about the exception but tell the user that the requested action failed. In these applications you can usually catch all or most exceptions centrally in one of the first methods in the call stack. You may still have to deal with the exception while propagating it though (using finally clauses). For instance, if an error occurs in the database connection in a web application, you may still have to close the database connection in a finally clause, even if you can't do anything else than tell the user that the action failed. How you end up handling exceptions also depends on whether you choose checked or unchecked exceptions for your application. There is more on that in other texts in the error handling trail.
Monday, May 12, 2014
Statistics
Corelations ratio:
Very intersing this to compare and understand individual results (e.g.graph pattern) to understand the internal effect.
Ref
Very intersing this to compare and understand individual results (e.g.graph pattern) to understand the internal effect.
Ref
BAU -IT issues
Cant copy and paste when using VPN
In order to use the copy/paste successfully in the remote connection to Windows 8, please check the following things:
1. On the Windows 8 machine, run the command "gpedit.msc" and navigate to Computer Configuration\Administrative Templates\Windows Components\Remote Desktop Services\Remote Desktop Session Host\Device and Resource Redirection
2. Disable the policy "Do not allow clipboard redirection" and run the command "gpupdate /force"
3. On the Windows 7 remote computer, open the Remote Desktop Connection and select Clipboard checkbox under the Local Resources.
This issue is also probably caused by the rdpclip.exe going out of sync in the RDP sharing chain. If this is the reason for the copy paste not to work, the quick solution is to restart the rdpclip.exe program in the remote computer.
Here are the steps:
1. Login to remote computer using Remote Desktop (RDP)
2. Open Task Manager in the remote machine
3. Click the "Process" tab
4. Locate a program called "rdpclip.exe"
5. Right click and select "End process" to kill this program
6. Click on "File" menu in the task manager and select "New Task (Run)"
7. Type rdpclip.exe and press the button to start the process.
Ref
How to open the task manager on a windows remote desktop session :
In some time we have to run task manager in remote session to stop some application which hang, producing continues error message or an abnormal behavior. Normally remote session does not support the Ctrl+Alt +Delete and we are unable to get the right click on Task bar. In this case we can use Ctrl + Shift + Esc or Ctrl + Alt + End to get Task Manager
Ref
In order to use the copy/paste successfully in the remote connection to Windows 8, please check the following things:
1. On the Windows 8 machine, run the command "gpedit.msc" and navigate to Computer Configuration\Administrative Templates\Windows Components\Remote Desktop Services\Remote Desktop Session Host\Device and Resource Redirection
2. Disable the policy "Do not allow clipboard redirection" and run the command "gpupdate /force"
3. On the Windows 7 remote computer, open the Remote Desktop Connection and select Clipboard checkbox under the Local Resources.
This issue is also probably caused by the rdpclip.exe going out of sync in the RDP sharing chain. If this is the reason for the copy paste not to work, the quick solution is to restart the rdpclip.exe program in the remote computer.
Here are the steps:
1. Login to remote computer using Remote Desktop (RDP)
2. Open Task Manager in the remote machine
3. Click the "Process" tab
4. Locate a program called "rdpclip.exe"
5. Right click and select "End process" to kill this program
6. Click on "File" menu in the task manager and select "New Task (Run)"
7. Type rdpclip.exe and press the button to start the process.
Ref
How to open the task manager on a windows remote desktop session :
In some time we have to run task manager in remote session to stop some application which hang, producing continues error message or an abnormal behavior. Normally remote session does not support the Ctrl+Alt +Delete and we are unable to get the right click on Task bar. In this case we can use Ctrl + Shift + Esc or Ctrl + Alt + End to get Task Manager
Ref
Wednesday, April 30, 2014
Monday, April 7, 2014
Network and Transport Layer Part 2
Review Questions :
1. What is routing? And list some of the routing protocols.
Routing is the process of determining path through which a message travels from the sending computer to the receiving computer. In some network there could be a lot of paths available from the sender to receiver (e.g. Internet) in others there could be a very few options available however someone needs to make the decision of routing.
Routing protocol used inside an autonomous system are called interior routing protocol. Routing protocol used between autonomous systems are called exterior routing protocol.
There are five commonly used routing protocol :
#.RIP-Routing information protocol
#.BGP-Border Gateway Protocol
#.ICMP-Internet control message protocol
#.IS-IS- Intermediate system to Intermediate system protocol
#.OSPF-Open shortest path first.
2. How does decentralized routing differ from centralized routing?
There are two types of routing available :
# Centralized and De-centralized
Decentralized could be divided into two :
Static and Dynamic.
Centralized routing - In this process of routing, a centralized computer makes all the routing decisions. This mostly takes place in host base networks. All the computers are connected to this central computer. When a message needs to be routed, the message gets sent to the central computer which re transmits the message to the adequate circuit to the destination.
Static Routing:
Individual computers make their own routing decision based on the routing table. The routing table gets set by the network manager who shares the information. This is what happens in WAN and MAN. For LAN and BN the committee or network manager as an individual makes the routing decision. This routing system is pretty self adjusting. If it finds any computer as non responding one, it updates the respective routing table by keeping that path as a deactivated one.
Dynamic Routing:
This Routing system is mainly used where there are more than one routing option available in the notwork. Using dynamic routing methodology the best route gets chosen for sending and receiving messages by avoiding busy networks and computers. Initially the router table gets set by the network manager, however afterwords it gets changed or updated depending on the networks variable conditions.
There are two types of Dynamic routing which are as follows :
# Distance vector Dynamic routing
# Link state Dynamic routing
3. What is a session? In relation to data communications.
A session is a semi permanent interactive information interchange. Often to exchange data a Session is required between computers over the network. Sessions could be connection based and connectionless.
4. What is Quality of Service routing and why is it useful?
5. Compare and contrast unicast, broadcast, and multicast messages.
6. Explain how multicasting works and reasons for using multicasting?
7. Explain how the client computer in Figure 5.14 would obtain the data link layer address of its subnet gateway.
When a computer is installed on a TCP/IP network (or dials into a TCP/IP network), it knowsthe IP address of its subnet gateway. This information can be provided by a configuration file or via a bootp or DHCP server. However, the computer does not know the subnet gateway’sEthernet address (data link layer address). Therefore, TCP would broadcast an ARP request toall computers on its subnet, requesting that the computer whose IP address is 128.192.98.1 torespond with its Ethernet address.All computers on the subnet would process this request, but only the subnet gateway would respond with an ARP packet giving its Ethernet address. The network layer software on theclient would then store this address in its data link layer address table.
Ref
8. Explain why HTTP protocol uses the transport layer protocol TCP and why the DNS server uses the transport layer protocol UDP?
9. How does static routing differ from dynamic routing? When would you use static routing? When would you use dynamic routing?
10. What is the transmission efficiency of a 10-byte Web request sent using HTTP, TCP/IP, and Ethernet? Assume the HTTP packet has 100 bytes in addition to the 10-byte URL. Hint: Remember from Chapter 4 that efficiency = user data/total transmission size.
Data = 10 Byte
Total Transmission size = 192 Bytes
Efficiency = User data / Total Transmission size = 10/192 = 0.05
11. What is the transmission efficiency of a 1000 byte file sent in response to a web request HTTP, TCP/IP, and Ethernet? Assume the HTTP packet has 100 bytes in addition to the 1000-byte file. Hint: Remember from Chapter 4, that efficiency = user data / total transmission size.
12. What is the transmission efficiency of a 5000 byte file sent in response to a web request HTTP, TCP/IP, and Ethernet? Assume the HTTP packet has 100 bytes in addition to the 5000-byte file. Assume that the maximum packet size is 1200 bytes Hint:
############################################################################
****************************************************************************
############################################################################
Tutorial Questions :
FIT5135: Tutorial 5 in Week 6 - V1.2
Instructions
1. Form groups of 3 students to perform this week’s exercises.
2. Activities in this tutorial are based on material from, Topic Notes – Network and Transport Layers and chapter 5 of prescribed textbook
Part 1: Using Windows Network Commands
In this exercise we will be using a number of commands available in your windows operating system. We will be following the “Hands-on-activities” from the pages 188-194 of the prescribed textbook. If you do not have a textbook you can download the required pages from Moodle week 6.
You must open a Command Window to use the network commands.
1. IPCONFIG: Reading your computer’s network settings
a. Display the network settings, type ipconfig /all
done
b. What is the ip address of your PC?
IPV4 Address 118.138.160.207
c. What subnet is the PC in?
255.255.254.0
d. How many DNS Servers are there?
130.194.1.99
130.194.7.99
e. What does a DNS Server do?
Find domain names
f. What does a DHCP Server do?
Dynamic Host Configuration Protocol assigns IP address to the computers in a network.
g. At what time does the Lease of the IP address expire?
Tuesday 8th april 2014 2:16:20 PM
h. What is the IP address of your PC’s default gateway?
118.138.161.254
i. What is the purpose of the default gateway?
In computer networking, a gateway is a node (a router) on a TCP/IP network that serves as an access point to another network. A default gateway is the node on the computer network that the network software uses when an IP address does not match any other routes in the routing table
2. ARP: Reading the physical and IP address mappings
a. Display the ARP table, type arp -a
done
b. What is the physical address of your PC’s default gateway?
00-00-0c-9f-f0-04
3. NSLOOKUP: Finding the IP address of a Web Server
a. Find IP address of www.google.com, type nslookup www.google.com
74.125.237.183
74.125.237.184
74.125.237.191
b. Find IP address of the web you are accessing in assignment 1
www.navy.gov.au 130.194.1.99
non authorative answer 103.11.78.154
4. DNS Cache: A record of the sites you have visited
a. Use IE Browser to visit the websites: www.cisco.com, www.racv.com.au, www.commbank.com.au
b. Save your DNS cache to a text file, type ipconfig /displaydns > dnscache.txt
done
c. The file will be saved in c:\users\
done
d. Open text file with Notepad and check the entries for the web sites you have visited
done
5. TRACERT: Finding routes through Internet
a. Trace the route to each of these web sites:
i. type tracert www.monash.edu.au
ii. type tracert www.cisco.com
b. For each site, if trace completed, type the last IP address into IE browser, what happens?
for monash it takes to the website however for the cisco one the last website doesnt work
2-2
Part 2: Protocols Analysis using Wireshark
This exercise aims to:
Examine captured packet trace files that demonstrate basic features of selected protocols: 3-way TCP handshake, ARP, DNS, DHCP and PING
(Extracted from "TCP/IP Analysis and Troubleshooting" by Laura A Chappell. Protocol Analysis Institute).
Download five files WSHxxx.cap from the Moodle Week 6 folder
i. 1. TCP Handshake
Run Wireshark on WSHtcpshake.cap
This trace shows the three-way handshake process used to establish a TCP connection.
a. In this trace, the source (130.57.20.10) sends a SYN (synchronize sequence number) request to the destination (130.57.20.1).
The second segment is the reply that contains a SYN request (since both sides maintain separate sequence numbers) with an ACK.
The final packet is the ACK that finishes up the process.
b. One interesting area to examine is the MSS (maximum segment size) negotiation that takes place in the first and second segments of the handshake process. Each side of this connection offers an MSS value of 1460 bytes. Where did that come from?
Well.... 1460 bytes + 20 bytes typical TCP header length + 20 bytes typical IP header length + 18 bytes Ethernet header and CRC = 1518 bytes. That is the maximum allowable packet size on an Ethernet network.
c. Can you tell what application has prompted this handshake process to occur? Do you recognize port value 524? (Basic port value list. A more complete list is available from www.iana.org).
NCP - Network core protocol
2-3
ii. 2. Basic ARP
Run Wireshark on WSHarp1.cap
These two frames show the basic ARP (Address Resolution Process) that enables one IP device to obtain the MAC (media access control) address of another local device. The MAC address is required to build the Ethernet frame and header for IP packets going from one local device to another local device.
a. In this example, you will notice that ARP packets do not have an IP header – ARP is actually protocol independent. The Ethernet type field value for ARP communications is 0x0806 (whereas IP packets use 0x0800).
b. Inside the ARP request, you'll notice that the target hardware address field is filled out with all 1s (0xFFFFFFFFFFFF). This is an indication that the source does not know the hardware address for 10.0.0.99 (the target's IP address).
c. In the reply, the source and target devices are now reversed with 10.0.0.99 being the source of the reply. The desired MAC address is seen in the reply.
iii. 3. DNS Configuration Fault
Run Wireshark on WSHdnsfault.cap
Assume the following:
i. The subnet mask is 255.0.0.0
ii. Three DNS servers are configured for the Client
This trace contains 9 packets that show what happens when DNS configurations are done improperly.
a. In this case, we can see in packets 1 and 2, the client (10.0.0.99) is ARPing for two devices that never answer. Those were the client's first choices as DNS servers, but they are not up currently.
b. In packet 3, we can see the client make a DNS query to 10.0.0.1.
c. The reply (ICMP port unreachable) indicates that the desired port number (53) is not in use at the destination -- 10.0.0.1 does not have the DNS server daemon loaded.
d. The client tries again to ARP for the first two DNS queries, but again receives no answer.
e. Then it tries again on 10.0.0.1 and gets rejected with the ICMP port unreachable message.
2-4
iv. 4. DHCP Bootup Sequence
Run Wireshark on WSHdhcpboot.cap
This trace depicts the typical four-packet startup sequence of a DHCP client.
a. The DHCP Discovery (sent to the broadcast address) indicates that this client last used
10.0.99.2 as its address (see 'Request Specific IP Address' in the trace file). You will also notice that the Discovery packet came from source IP address 0.0.0.0.
b. The offered address is 10.0.99.2 -- this packet is sent directly to the clients known hardware address and the assumed IP address (this won't conflict with any duplicate IP address yet, because it was directed to the correct host at the data link layer).
c. When the client sends the DHCP Request packet (packet #3), look at the source and destination addresses – the client does not assume anything yet.
d. Finally, the server ACKs the client and provides the desired information regarding default gateway server IP address, subnet mask, lease time, DNS server and DNS domain.
v. 5. Simple PING
A simple ping from one device (10.0.0.1) to another device (10.0.99.2) on the same network.
Run Wireshark on WSHping1.cap
a. Ping uses ICMP type 8 (echo) and type 0 (echo reply) packets.
b. This ping was executed on a server. Notice the matching identifiers in the ICMP header.
c. Do the captured packets comply with typical ICMP echo request/reply traffic? Look for the protocol specifications for ICMP. What do the specifications say about the size of echo request/reply messages?
2-5
vii. 6. For General Discussion, comment on the implications of using packet capture and protocol analyzer tools on the public Internet.
a. What are the implications of using a hub as against using a switch?
b. Should packet capture be allowed to anyone, or the operators of, a public Internet cafe that uses a hub?
c. How does the potential for unauthorized Internet eavesdropping affect the responsibilities of application developers?
d. What options are available to application developers and network designers in order to mitigate the risks of unauthorized eavesdropping?
e. What network standards would be useful in mitigating such risks?
2-6
True location fig for networking and routing
1. What is routing? And list some of the routing protocols.
Routing is the process of determining path through which a message travels from the sending computer to the receiving computer. In some network there could be a lot of paths available from the sender to receiver (e.g. Internet) in others there could be a very few options available however someone needs to make the decision of routing.
Routing protocol used inside an autonomous system are called interior routing protocol. Routing protocol used between autonomous systems are called exterior routing protocol.
There are five commonly used routing protocol :
#.RIP-Routing information protocol
#.BGP-Border Gateway Protocol
#.ICMP-Internet control message protocol
#.IS-IS- Intermediate system to Intermediate system protocol
#.OSPF-Open shortest path first.
2. How does decentralized routing differ from centralized routing?
There are two types of routing available :
# Centralized and De-centralized
Decentralized could be divided into two :
Static and Dynamic.
Centralized routing - In this process of routing, a centralized computer makes all the routing decisions. This mostly takes place in host base networks. All the computers are connected to this central computer. When a message needs to be routed, the message gets sent to the central computer which re transmits the message to the adequate circuit to the destination.
Static Routing:
Individual computers make their own routing decision based on the routing table. The routing table gets set by the network manager who shares the information. This is what happens in WAN and MAN. For LAN and BN the committee or network manager as an individual makes the routing decision. This routing system is pretty self adjusting. If it finds any computer as non responding one, it updates the respective routing table by keeping that path as a deactivated one.
Dynamic Routing:
This Routing system is mainly used where there are more than one routing option available in the notwork. Using dynamic routing methodology the best route gets chosen for sending and receiving messages by avoiding busy networks and computers. Initially the router table gets set by the network manager, however afterwords it gets changed or updated depending on the networks variable conditions.
There are two types of Dynamic routing which are as follows :
# Distance vector Dynamic routing
# Link state Dynamic routing
3. What is a session? In relation to data communications.
A session is a semi permanent interactive information interchange. Often to exchange data a Session is required between computers over the network. Sessions could be connection based and connectionless.
4. What is Quality of Service routing and why is it useful?
5. Compare and contrast unicast, broadcast, and multicast messages.
6. Explain how multicasting works and reasons for using multicasting?
7. Explain how the client computer in Figure 5.14 would obtain the data link layer address of its subnet gateway.
When a computer is installed on a TCP/IP network (or dials into a TCP/IP network), it knowsthe IP address of its subnet gateway. This information can be provided by a configuration file or via a bootp or DHCP server. However, the computer does not know the subnet gateway’sEthernet address (data link layer address). Therefore, TCP would broadcast an ARP request toall computers on its subnet, requesting that the computer whose IP address is 128.192.98.1 torespond with its Ethernet address.All computers on the subnet would process this request, but only the subnet gateway would respond with an ARP packet giving its Ethernet address. The network layer software on theclient would then store this address in its data link layer address table.
Ref
8. Explain why HTTP protocol uses the transport layer protocol TCP and why the DNS server uses the transport layer protocol UDP?
9. How does static routing differ from dynamic routing? When would you use static routing? When would you use dynamic routing?
10. What is the transmission efficiency of a 10-byte Web request sent using HTTP, TCP/IP, and Ethernet? Assume the HTTP packet has 100 bytes in addition to the 10-byte URL. Hint: Remember from Chapter 4 that efficiency = user data/total transmission size.
Data = 10 Byte
Total Transmission size = 192 Bytes
Efficiency = User data / Total Transmission size = 10/192 = 0.05
11. What is the transmission efficiency of a 1000 byte file sent in response to a web request HTTP, TCP/IP, and Ethernet? Assume the HTTP packet has 100 bytes in addition to the 1000-byte file. Hint: Remember from Chapter 4, that efficiency = user data / total transmission size.
12. What is the transmission efficiency of a 5000 byte file sent in response to a web request HTTP, TCP/IP, and Ethernet? Assume the HTTP packet has 100 bytes in addition to the 5000-byte file. Assume that the maximum packet size is 1200 bytes Hint:
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Tutorial Questions :
FIT5135: Tutorial 5 in Week 6 - V1.2
Instructions
1. Form groups of 3 students to perform this week’s exercises.
2. Activities in this tutorial are based on material from, Topic Notes – Network and Transport Layers and chapter 5 of prescribed textbook
Part 1: Using Windows Network Commands
In this exercise we will be using a number of commands available in your windows operating system. We will be following the “Hands-on-activities” from the pages 188-194 of the prescribed textbook. If you do not have a textbook you can download the required pages from Moodle week 6.
You must open a Command Window to use the network commands.
1. IPCONFIG: Reading your computer’s network settings
a. Display the network settings, type ipconfig /all
done
b. What is the ip address of your PC?
IPV4 Address 118.138.160.207
c. What subnet is the PC in?
255.255.254.0
d. How many DNS Servers are there?
130.194.1.99
130.194.7.99
e. What does a DNS Server do?
Find domain names
f. What does a DHCP Server do?
Dynamic Host Configuration Protocol assigns IP address to the computers in a network.
g. At what time does the Lease of the IP address expire?
Tuesday 8th april 2014 2:16:20 PM
h. What is the IP address of your PC’s default gateway?
118.138.161.254
i. What is the purpose of the default gateway?
In computer networking, a gateway is a node (a router) on a TCP/IP network that serves as an access point to another network. A default gateway is the node on the computer network that the network software uses when an IP address does not match any other routes in the routing table
2. ARP: Reading the physical and IP address mappings
a. Display the ARP table, type arp -a
done
b. What is the physical address of your PC’s default gateway?
00-00-0c-9f-f0-04
3. NSLOOKUP: Finding the IP address of a Web Server
a. Find IP address of www.google.com, type nslookup www.google.com
74.125.237.183
74.125.237.184
74.125.237.191
b. Find IP address of the web you are accessing in assignment 1
www.navy.gov.au 130.194.1.99
non authorative answer 103.11.78.154
4. DNS Cache: A record of the sites you have visited
a. Use IE Browser to visit the websites: www.cisco.com, www.racv.com.au, www.commbank.com.au
b. Save your DNS cache to a text file, type ipconfig /displaydns > dnscache.txt
done
c. The file will be saved in c:\users\
done
d. Open text file with Notepad and check the entries for the web sites you have visited
done
5. TRACERT: Finding routes through Internet
a. Trace the route to each of these web sites:
i. type tracert www.monash.edu.au
ii. type tracert www.cisco.com
b. For each site, if trace completed, type the last IP address into IE browser, what happens?
for monash it takes to the website however for the cisco one the last website doesnt work
2-2
Part 2: Protocols Analysis using Wireshark
This exercise aims to:
Examine captured packet trace files that demonstrate basic features of selected protocols: 3-way TCP handshake, ARP, DNS, DHCP and PING
(Extracted from "TCP/IP Analysis and Troubleshooting" by Laura A Chappell. Protocol Analysis Institute).
Download five files WSHxxx.cap from the Moodle Week 6 folder
i. 1. TCP Handshake
Run Wireshark on WSHtcpshake.cap
This trace shows the three-way handshake process used to establish a TCP connection.
a. In this trace, the source (130.57.20.10) sends a SYN (synchronize sequence number) request to the destination (130.57.20.1).
The second segment is the reply that contains a SYN request (since both sides maintain separate sequence numbers) with an ACK.
The final packet is the ACK that finishes up the process.
b. One interesting area to examine is the MSS (maximum segment size) negotiation that takes place in the first and second segments of the handshake process. Each side of this connection offers an MSS value of 1460 bytes. Where did that come from?
Well.... 1460 bytes + 20 bytes typical TCP header length + 20 bytes typical IP header length + 18 bytes Ethernet header and CRC = 1518 bytes. That is the maximum allowable packet size on an Ethernet network.
c. Can you tell what application has prompted this handshake process to occur? Do you recognize port value 524? (Basic port value list. A more complete list is available from www.iana.org).
NCP - Network core protocol
2-3
ii. 2. Basic ARP
Run Wireshark on WSHarp1.cap
These two frames show the basic ARP (Address Resolution Process) that enables one IP device to obtain the MAC (media access control) address of another local device. The MAC address is required to build the Ethernet frame and header for IP packets going from one local device to another local device.
a. In this example, you will notice that ARP packets do not have an IP header – ARP is actually protocol independent. The Ethernet type field value for ARP communications is 0x0806 (whereas IP packets use 0x0800).
b. Inside the ARP request, you'll notice that the target hardware address field is filled out with all 1s (0xFFFFFFFFFFFF). This is an indication that the source does not know the hardware address for 10.0.0.99 (the target's IP address).
c. In the reply, the source and target devices are now reversed with 10.0.0.99 being the source of the reply. The desired MAC address is seen in the reply.
iii. 3. DNS Configuration Fault
Run Wireshark on WSHdnsfault.cap
Assume the following:
i. The subnet mask is 255.0.0.0
ii. Three DNS servers are configured for the Client
This trace contains 9 packets that show what happens when DNS configurations are done improperly.
a. In this case, we can see in packets 1 and 2, the client (10.0.0.99) is ARPing for two devices that never answer. Those were the client's first choices as DNS servers, but they are not up currently.
b. In packet 3, we can see the client make a DNS query to 10.0.0.1.
c. The reply (ICMP port unreachable) indicates that the desired port number (53) is not in use at the destination -- 10.0.0.1 does not have the DNS server daemon loaded.
d. The client tries again to ARP for the first two DNS queries, but again receives no answer.
e. Then it tries again on 10.0.0.1 and gets rejected with the ICMP port unreachable message.
2-4
iv. 4. DHCP Bootup Sequence
Run Wireshark on WSHdhcpboot.cap
This trace depicts the typical four-packet startup sequence of a DHCP client.
a. The DHCP Discovery (sent to the broadcast address) indicates that this client last used
10.0.99.2 as its address (see 'Request Specific IP Address' in the trace file). You will also notice that the Discovery packet came from source IP address 0.0.0.0.
b. The offered address is 10.0.99.2 -- this packet is sent directly to the clients known hardware address and the assumed IP address (this won't conflict with any duplicate IP address yet, because it was directed to the correct host at the data link layer).
c. When the client sends the DHCP Request packet (packet #3), look at the source and destination addresses – the client does not assume anything yet.
d. Finally, the server ACKs the client and provides the desired information regarding default gateway server IP address, subnet mask, lease time, DNS server and DNS domain.
v. 5. Simple PING
A simple ping from one device (10.0.0.1) to another device (10.0.99.2) on the same network.
Run Wireshark on WSHping1.cap
a. Ping uses ICMP type 8 (echo) and type 0 (echo reply) packets.
b. This ping was executed on a server. Notice the matching identifiers in the ICMP header.
c. Do the captured packets comply with typical ICMP echo request/reply traffic? Look for the protocol specifications for ICMP. What do the specifications say about the size of echo request/reply messages?
2-5
vii. 6. For General Discussion, comment on the implications of using packet capture and protocol analyzer tools on the public Internet.
a. What are the implications of using a hub as against using a switch?
b. Should packet capture be allowed to anyone, or the operators of, a public Internet cafe that uses a hub?
c. How does the potential for unauthorized Internet eavesdropping affect the responsibilities of application developers?
d. What options are available to application developers and network designers in order to mitigate the risks of unauthorized eavesdropping?
e. What network standards would be useful in mitigating such risks?
2-6
True location fig for networking and routing
Tuesday, April 1, 2014
Assignment 1
1. HTTP Scenario
Use your web browser to connect to website of an Australian government department (www.****.gov.au). You must obtain a website from your tutor to ensure no students are using the same website. Using the Wireshark protocol analyser capture all HTTP packets related to, the request to and response from the Home page of the above website.
a. Include the Wireshark screen containing the first request HTTP packet. The packet number and the HTTP packet request line (including the host) should be clearly visible in the packet details pane.
(2 marks)
b. Find out the HTTP version number used by both client & server?
Which languages does the browser support?
When was the received webpage last updated?
(3 marks)
c. Both HTTP request & response header include a “Connection” field. It has a value of either “keep–alive” or “close”. Explain the purpose of this field, and meanings of keep–alive or close values. Also discuss if these values are relevant in the latest HTTP specifications.
(5marks)
d. Under an HTTP response header, check the field “Content-Encoding:gzip”. Explain what are meanings & advantages of using field.
[Note: you might not find Content-Encoding in an HTTP response message which is delivering an image. Look for it in a frame which contains textual data.]
(4 marks)
e. Capture and include the TCP packets that establish the connection to the web site using the three way handshake.
(2 marks)
f. Select an HTTP request‐response pair which transfers an image object. Calculate how many TCP transmissions it takes to download the object. Give the number of frames and size of the segments downloaded. What is the total size (in bytes) of the object downloaded?
(4 marks)
[Total 20 marks]
2. Email Scenario
Assume that:
You have an email account at Monash domain (username@monash.edu.au)
You use Mozilla Thunderbird client at Monash to send an email to Mike’s email address Mike.Smith@gmail.com.
Mike is using web email to check his email and respond to your email.
You use POP protocol to check Mike’s reply.
Given the above scenario, answer the following:
a. Illustrate in a single diagram:
i. The flow of packets between the computers involved. Include the following packet types (if required): SMTP, HTTP, IMAP, POP.
ii. Annotate all servers and client computers involved (e.g. MyPC, MikePC, etc.).
iii. Number the steps and related packets in the diagram.
(10 marks)
b. Give a brief step‐by‐step explanation of the flow of packets.
(5 marks)
[Total 15 marks]
3. Voice-over-IP (VoIP)
Use the Internet to search for information on the VoIP products Skype and Ekiga.
a. What application layer protocols does each product use?
(2 marks)
b. Compare and contrast them in terms of their capabilities and performance.
(8 marks)
[Total 10 marks]
4. Modulation Techniques
Draw a diagram to show how the bit pattern 1 0 1 1 0 0 1 0 would be sent using:
a. Two‐bit FM (i.e., four frequencies) (2 marks)
b. Two‐bit PM (i.e., four different phases) (2 marks)
c. Single‐bit AM combined with single‐bit PM (2 marks)
d. Two‐bit AM combined with two‐bit PM. (4 marks)
e. Digital 16-QAM (read Wikipedia), compare with (d) above. (5 marks)
[Total 15 marks]
5. The Address Resolution Protocol (ARP)
Describe ARP and answer the following questions:
a. Why is this protocol needed? (1 mark)
b. At what layer of OSI Model does ARP operate? (1 mark)
c. Compare ARP request and reply messages. (4 marks)
d. Describe the ARP poisoning attack. (2 marks)
e. Describe the ARP proxy feature. (2 marks)
[Total 10 marks]
6. Forward Error Correction (FEC)
Very useful link for converting HEXADECIMAL to BINARY
Convert numbers
Answer the following questions:
a. Define FEC, and list three attributes of it.
(3 marks)
b. Suppose data of 4 bytes is to be sent over a network. Calculate and compare the overhead in bits using the Hamming Single-Bit Code versus using the 1 bit Parity Check.
(3 marks)
c. Suppose the hexadecimal representation of the 4 bytes of data is
0x1F351A02:
i. Determine the codeword for this data
ii. The codeword is transmitted and received at the destination with its data corrupted to 0x1F351B02. Show how this can be detected and fixed at the destination using the Hamming Single-Bit Code.
(4 marks)
[Total 10 marks]
7. Technical Report
The company you work for wants to purchase broadband capability for its 20 staff members, so that they can access the office systems from home. You have been assigned the task of contacting telecommunications service suppliers to compare their offerings, in terms of technical capability and cost. You are required to write a technical report that compares ADSL2+ services versus Cable modem services. Select two service suppliers and compare their offerings, determine which service supplier offers the best solution.
a. Technical Capability
In tabular form compare the following:
Service Supplier
Device
Modulation type
Modulation bit rates achievable
What error control strategy is used if any?
Is the modulation sensitive to crosstalk?
What cable and connector types are required for the line interface?
What host side interfaces are provided? (Hint: Ethernet, Firewire, USB3)
How many ports are provided?
What protocol support is provided (i.e. PPP or other)?
The protocol transmission efficiency of uploading a 20Mbyte file.
(10 marks)
b. Service Costs Comparison
Compare the service costs — e.g. installation, ongoing maintenance, technical support, warranty, leasing, purchasing etc.
(6 marks)
c. Recommendation
Which service do you recommend? Justify your answer.
(4 marks)
Note: Report word count should be in range of 600 to 1000 words.
[Total 20 marks] 
PSK Modulation Diagram link
Use your web browser to connect to website of an Australian government department (www.****.gov.au). You must obtain a website from your tutor to ensure no students are using the same website. Using the Wireshark protocol analyser capture all HTTP packets related to, the request to and response from the Home page of the above website.
a. Include the Wireshark screen containing the first request HTTP packet. The packet number and the HTTP packet request line (including the host) should be clearly visible in the packet details pane.
(2 marks)
b. Find out the HTTP version number used by both client & server?
Which languages does the browser support?
When was the received webpage last updated?
(3 marks)
c. Both HTTP request & response header include a “Connection” field. It has a value of either “keep–alive” or “close”. Explain the purpose of this field, and meanings of keep–alive or close values. Also discuss if these values are relevant in the latest HTTP specifications.
(5marks)
d. Under an HTTP response header, check the field “Content-Encoding:gzip”. Explain what are meanings & advantages of using field.
[Note: you might not find Content-Encoding in an HTTP response message which is delivering an image. Look for it in a frame which contains textual data.]
(4 marks)
e. Capture and include the TCP packets that establish the connection to the web site using the three way handshake.
(2 marks)
f. Select an HTTP request‐response pair which transfers an image object. Calculate how many TCP transmissions it takes to download the object. Give the number of frames and size of the segments downloaded. What is the total size (in bytes) of the object downloaded?
(4 marks)
[Total 20 marks]
2. Email Scenario
Assume that:
You have an email account at Monash domain (username@monash.edu.au)
You use Mozilla Thunderbird client at Monash to send an email to Mike’s email address Mike.Smith@gmail.com.
Mike is using web email to check his email and respond to your email.
You use POP protocol to check Mike’s reply.
Given the above scenario, answer the following:
a. Illustrate in a single diagram:
i. The flow of packets between the computers involved. Include the following packet types (if required): SMTP, HTTP, IMAP, POP.
ii. Annotate all servers and client computers involved (e.g. MyPC, MikePC, etc.).
iii. Number the steps and related packets in the diagram.
(10 marks)
b. Give a brief step‐by‐step explanation of the flow of packets.
(5 marks)
[Total 15 marks]
3. Voice-over-IP (VoIP)
Use the Internet to search for information on the VoIP products Skype and Ekiga.
a. What application layer protocols does each product use?
(2 marks)
b. Compare and contrast them in terms of their capabilities and performance.
(8 marks)
[Total 10 marks]
4. Modulation Techniques
Draw a diagram to show how the bit pattern 1 0 1 1 0 0 1 0 would be sent using:
a. Two‐bit FM (i.e., four frequencies) (2 marks)
b. Two‐bit PM (i.e., four different phases) (2 marks)
c. Single‐bit AM combined with single‐bit PM (2 marks)
d. Two‐bit AM combined with two‐bit PM. (4 marks)
e. Digital 16-QAM (read Wikipedia), compare with (d) above. (5 marks)
[Total 15 marks]
5. The Address Resolution Protocol (ARP)
Describe ARP and answer the following questions:
a. Why is this protocol needed? (1 mark)
b. At what layer of OSI Model does ARP operate? (1 mark)
c. Compare ARP request and reply messages. (4 marks)
d. Describe the ARP poisoning attack. (2 marks)
e. Describe the ARP proxy feature. (2 marks)
[Total 10 marks]
6. Forward Error Correction (FEC)
Very useful link for converting HEXADECIMAL to BINARY
Convert numbers
Answer the following questions:
a. Define FEC, and list three attributes of it.
(3 marks)
b. Suppose data of 4 bytes is to be sent over a network. Calculate and compare the overhead in bits using the Hamming Single-Bit Code versus using the 1 bit Parity Check.
(3 marks)
c. Suppose the hexadecimal representation of the 4 bytes of data is
0x1F351A02:
i. Determine the codeword for this data
ii. The codeword is transmitted and received at the destination with its data corrupted to 0x1F351B02. Show how this can be detected and fixed at the destination using the Hamming Single-Bit Code.
(4 marks)
[Total 10 marks]
7. Technical Report
The company you work for wants to purchase broadband capability for its 20 staff members, so that they can access the office systems from home. You have been assigned the task of contacting telecommunications service suppliers to compare their offerings, in terms of technical capability and cost. You are required to write a technical report that compares ADSL2+ services versus Cable modem services. Select two service suppliers and compare their offerings, determine which service supplier offers the best solution.
a. Technical Capability
In tabular form compare the following:
Service Supplier
Device
Modulation type
Modulation bit rates achievable
What error control strategy is used if any?
Is the modulation sensitive to crosstalk?
What cable and connector types are required for the line interface?
What host side interfaces are provided? (Hint: Ethernet, Firewire, USB3)
How many ports are provided?
What protocol support is provided (i.e. PPP or other)?
The protocol transmission efficiency of uploading a 20Mbyte file.
(10 marks)
b. Service Costs Comparison
Compare the service costs — e.g. installation, ongoing maintenance, technical support, warranty, leasing, purchasing etc.
(6 marks)
c. Recommendation
Which service do you recommend? Justify your answer.
(4 marks)
Note: Report word count should be in range of 600 to 1000 words.
[Total 20 marks]
PSK Modulation Diagram link
Monday, March 31, 2014
Network and Transport Layer Part 1
Tutorial:
Part 1: Wireshark. TCP - Transmission Control Protocol
In this practical part we would like to demonstrate how the basic functions of the TCP, namely,
• Linking to the Application Layer
• Segmenting
• Session management
are implemented in practical situations. We also need to know the structure of the TCP segment that you can conveniently access from http://en.wikipedia.org/wiki/Transmission_Control_Protocol#TCP_segment_structure
Download again from Moodle the Wireshark HTTP demo file WiresharkHTTPdemo090402.pcap and place it in a known folder, say, desktop. Invoke wireshark from the folder Start\Programs\Developments and open the demo file.
1. Linking to the Application Layer:
We start at the familiar frame 35. After selecting this frame in the top pane, in the middle pane select and expand the “Transmission Control Protocol” line.
a. Note the values of the source and destination ports.
b. Go to the frame 38 and again record the values of the source and destination port.
c. Describe the meaning of these values.
2. Segmenting
Start with the frame 42 (get Andrew’s photo)
a. How many TCP segments it takes to transmit the photo?
#Line no 112 to 43 = 71 and then ACK are 23. 71-23 = 48 -1 = 47
b. What is the total size of the photo?
66694
c. Read first about the “continuous ARQ” and then answer how many times the transmission of the photo segments have been acknowledged.
23
d. Give the numbers of the acknowledgment frames.
46,49,52,55,58,61,64,67,70,73,76,79,82,85,88,91,94,97,100,103,106,109,112
2-2
3. Session management
a. Identify from which packet or frame the start of TCP session between the
client (192.6168.1.2) and Server (130.194.64.145) is established. List the steps
required for session layer to identify the start and establishment of a session
between the communicating peers.
ip.src == 130.194.64.145 and ip.dst == 192.168.1.2 starts from line number 33
ip.src == 192.168.1.2 and ip.dst == 130.194.64.145 starts from line number 30
Wireshark Queried
3 WAYS HANDSHAKE
b. Identify from which packet or frame the ending of TCP session between the
client (192.6168.1.2) and Server (130.194.64.145) is finished. List the steps
required for session layer to identify the termination and finishing of a session
between the communicating peers.
3 WAYS HANDSHAKE -ACK AND SYN 30,33,34 STEPS PROCESS
4 WAYS ENDING WITH THE KEYOWRD - FIN
4. The length of the Ethernet frame
Verify that the Ethernet II frame does not have any information about its length.
Instead, it is dependent on the information supplied by the IP header:
8000 - IPV4
86DD - IPV6
TOTAL LENGHT - UNDER IP + 14 = 54 IPV4
TOTAL LENGHT - UNDER IP + 14 + 40 = 54 IPV6
1. What does the transport layer do?
Links Application to the network
2. What does the network layer do?
Routing and addressing
3. Compare and contrast the types of addresses used in a network.
Application layer address (www.google.com)Protocol Http
Network layer address (546.45.23.14 four bytes - protocol IP)
Data link layer address (6 bytes hexa decimal addressing 00-55-d4-gf-gg-55) Ethernet addressing
4. What is the function of Transport layer protocols TCP and UDP, Distinguish TCP from UDP protocol?
TCP - A typical TCP segment has 24 beyts or 192 bit header of control information which has information about the sender / receiver 's application layer. The message in receivers end tells in which application program it should be sent to. It also says which application was used at senders end to generate the message. TCP header also has a seq which helps to assemble tcp segments and complete the message. This makes sure no packet gets missed or lost. Requires connection ,Example web surfing.
UDP - TCP / IP has a second type of transport layer protocol which is called User datagram Protocol. It gets used for sending small messages which dont require to be segmented. It holds it has only 4 fields (8 bytes) plus application layer packet. It doesnt check for sequence so occasionally data gets lost. Connection less protocol. Example - IPTV, multilayer online games, satellite cast for an live event.
5. How does TCP establish a connection? Explain the three-way handshake?
Host A sends a TCP SYNchronize packet to Host B
Host B receives A's SYN
Host B sends a SYNchronize-ACKnowledgement
Host A receives B's SYN-ACK
Host A sends ACKnowledge
Host B receives ACK.
TCP socket connection is ESTABLISHED.
Ref - 3 way handshake
6. Explain briefly different classes of networks? What is a subnet and why do networks need them?
Class A from N.H.H.H from 1 to 127
Class B from N.N.H.H from 128 to 127+64 = 191
Class C from N.N.N.H from 192 to 191+32 = 223
Dividing a network or logical partition or distribution of the network to organise and control the network flow is called subnet.
7. What is a subnet mask? And how is it related to different class of networks?
8. How does dynamic addressing work?
9. What benefits and problems does dynamic addressing provide?
10. What is address resolution (translation)? Explain how a DNS server does name resolution.
11. How is address resolution performed for network layer addresses?
12. How is address resolution performed for data link layer addresses?
Part 1: Wireshark. TCP - Transmission Control Protocol
In this practical part we would like to demonstrate how the basic functions of the TCP, namely,
• Linking to the Application Layer
• Segmenting
• Session management
are implemented in practical situations. We also need to know the structure of the TCP segment that you can conveniently access from http://en.wikipedia.org/wiki/Transmission_Control_Protocol#TCP_segment_structure
Download again from Moodle the Wireshark HTTP demo file WiresharkHTTPdemo090402.pcap and place it in a known folder, say, desktop. Invoke wireshark from the folder Start\Programs\Developments and open the demo file.
1. Linking to the Application Layer:
We start at the familiar frame 35. After selecting this frame in the top pane, in the middle pane select and expand the “Transmission Control Protocol” line.
a. Note the values of the source and destination ports.
b. Go to the frame 38 and again record the values of the source and destination port.
c. Describe the meaning of these values.
2. Segmenting
Start with the frame 42 (get Andrew’s photo)
a. How many TCP segments it takes to transmit the photo?
#Line no 112 to 43 = 71 and then ACK are 23. 71-23 = 48 -1 = 47
b. What is the total size of the photo?
66694
c. Read first about the “continuous ARQ” and then answer how many times the transmission of the photo segments have been acknowledged.
23
d. Give the numbers of the acknowledgment frames.
46,49,52,55,58,61,64,67,70,73,76,79,82,85,88,91,94,97,100,103,106,109,112
2-2
3. Session management
a. Identify from which packet or frame the start of TCP session between the
client (192.6168.1.2) and Server (130.194.64.145) is established. List the steps
required for session layer to identify the start and establishment of a session
between the communicating peers.
ip.src == 130.194.64.145 and ip.dst == 192.168.1.2 starts from line number 33
ip.src == 192.168.1.2 and ip.dst == 130.194.64.145 starts from line number 30
Wireshark Queried
3 WAYS HANDSHAKE
b. Identify from which packet or frame the ending of TCP session between the
client (192.6168.1.2) and Server (130.194.64.145) is finished. List the steps
required for session layer to identify the termination and finishing of a session
between the communicating peers.
3 WAYS HANDSHAKE -ACK AND SYN 30,33,34 STEPS PROCESS
4 WAYS ENDING WITH THE KEYOWRD - FIN
4. The length of the Ethernet frame
Verify that the Ethernet II frame does not have any information about its length.
Instead, it is dependent on the information supplied by the IP header:
8000 - IPV4
86DD - IPV6
TOTAL LENGHT - UNDER IP + 14 = 54 IPV4
TOTAL LENGHT - UNDER IP + 14 + 40 = 54 IPV6
1. What does the transport layer do?
Links Application to the network
2. What does the network layer do?
Routing and addressing
3. Compare and contrast the types of addresses used in a network.
Application layer address (www.google.com)Protocol Http
Network layer address (546.45.23.14 four bytes - protocol IP)
Data link layer address (6 bytes hexa decimal addressing 00-55-d4-gf-gg-55) Ethernet addressing
4. What is the function of Transport layer protocols TCP and UDP, Distinguish TCP from UDP protocol?
TCP - A typical TCP segment has 24 beyts or 192 bit header of control information which has information about the sender / receiver 's application layer. The message in receivers end tells in which application program it should be sent to. It also says which application was used at senders end to generate the message. TCP header also has a seq which helps to assemble tcp segments and complete the message. This makes sure no packet gets missed or lost. Requires connection ,Example web surfing.
UDP - TCP / IP has a second type of transport layer protocol which is called User datagram Protocol. It gets used for sending small messages which dont require to be segmented. It holds it has only 4 fields (8 bytes) plus application layer packet. It doesnt check for sequence so occasionally data gets lost. Connection less protocol. Example - IPTV, multilayer online games, satellite cast for an live event.
5. How does TCP establish a connection? Explain the three-way handshake?
Host A sends a TCP SYNchronize packet to Host B
Host B receives A's SYN
Host B sends a SYNchronize-ACKnowledgement
Host A receives B's SYN-ACK
Host A sends ACKnowledge
Host B receives ACK.
TCP socket connection is ESTABLISHED.
Ref - 3 way handshake
6. Explain briefly different classes of networks? What is a subnet and why do networks need them?
Class A from N.H.H.H from 1 to 127
Class B from N.N.H.H from 128 to 127+64 = 191
Class C from N.N.N.H from 192 to 191+32 = 223
Dividing a network or logical partition or distribution of the network to organise and control the network flow is called subnet.
7. What is a subnet mask? And how is it related to different class of networks?
8. How does dynamic addressing work?
9. What benefits and problems does dynamic addressing provide?
10. What is address resolution (translation)? Explain how a DNS server does name resolution.
11. How is address resolution performed for network layer addresses?
12. How is address resolution performed for data link layer addresses?
Thursday, March 27, 2014
Introduction to Data Communication
Topic: Introduction to Data Communications
Review Questions
Use these questions to review the important material from the Topic Notes and Chapter 1 of the prescribed textbook.
The review should take place in your own time. Where possible you should work with your colleagues.
You can discuss your answers with your tutor in the week 2 tutorial.
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1. How do LANs differ from MANs, WANs, and BNs?
2. What is a circuit?
3. What is a client?
4. What is a server?
5. Why are network layers important?
6. Describe the seven layers in the OSI network model and what they do.
7. Describe the five layers in the Internet network model and what they do.
8. Explain how a message is transmitted from one computer to another using layers.
9. Describe the three stages of standardization.
10. How are Internet standards developed?
11. Describe two important data communications standards-making bodies. How do they differ?
12. What is the purpose of a data communication standard?
Lab Preparation
To prepare for the lab in week 2:
1. Via the Web find out about:
a. The hexadecimal number system
b. The ASCII character set
2. Convert the decimal number 90 to a hexadecimal number
3. Convert the hexadecimal number determined in 2 above, to an 8 bit binary byte and then determine what ASCII character this byte represents.
Review Questions
Use these questions to review the important material from the Topic Notes and Chapter 1 of the prescribed textbook.
The review should take place in your own time. Where possible you should work with your colleagues.
You can discuss your answers with your tutor in the week 2 tutorial.
------------------------------------------------------------------------------------------------------------------------------------------------
1. How do LANs differ from MANs, WANs, and BNs?
2. What is a circuit?
3. What is a client?
4. What is a server?
5. Why are network layers important?
6. Describe the seven layers in the OSI network model and what they do.
7. Describe the five layers in the Internet network model and what they do.
8. Explain how a message is transmitted from one computer to another using layers.
9. Describe the three stages of standardization.
10. How are Internet standards developed?
11. Describe two important data communications standards-making bodies. How do they differ?
12. What is the purpose of a data communication standard?
Lab Preparation
To prepare for the lab in week 2:
1. Via the Web find out about:
a. The hexadecimal number system
b. The ASCII character set
2. Convert the decimal number 90 to a hexadecimal number
3. Convert the hexadecimal number determined in 2 above, to an 8 bit binary byte and then determine what ASCII character this byte represents.
Application Layer
Topic: Application Layer
Review Questions
Use these questions to review the important material from the Topic Notes and chapter2 of the prescribed textbook.
The review should take place in your own time. Where possible you should work with your colleagues.
You can discuss your answers with your tutor in the week 3 tutorial.
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1. What are the different types of application architectures?
2. Describe the four basic functions of an application software package.
3. What are the advantages and disadvantages of host-based networks versus client-server networks?
4. What is middleware and what does it do?
5. Suppose your organization was contemplating switching from a host-based architecture to client-server. What problems would you foresee?
6. Compare and contrast two-tiered, three-tiered, and n-tiered client server architectures. What are the technical differences and what advantages and disadvantages do each offer?
7. How does a thin client differ from a thick client?
8. For what is HTTP used? What are its major parts?
9. Describe how a Web browser and Web server work together to send a web page to a user.
10. Describe how mail user agents and message transfer agents work together to transfer mail messages.
11. What roles do SMTP, POP, and IMAP play in sending and receiving e-mail on the Internet?
12. What is FTP and why is it useful?
13. What is Telnet and why is it useful?
Lab Preparation
To prepare for the lab in week 3:
1. Study the structure of the HTTP request and response packets, refer Lecture – Week 2: Application Layer slides 23 to 27, chapter 2 of prescribed textbook and look at other source material on Web.
2. Check out Rex Swain’s HTTP Viewer at www.rexswain.com/httpview.html
Review Questions
Use these questions to review the important material from the Topic Notes and chapter2 of the prescribed textbook.
The review should take place in your own time. Where possible you should work with your colleagues.
You can discuss your answers with your tutor in the week 3 tutorial.
------------------------------------------------------------------------------------------------------------------------------------------------
1. What are the different types of application architectures?
2. Describe the four basic functions of an application software package.
3. What are the advantages and disadvantages of host-based networks versus client-server networks?
4. What is middleware and what does it do?
5. Suppose your organization was contemplating switching from a host-based architecture to client-server. What problems would you foresee?
6. Compare and contrast two-tiered, three-tiered, and n-tiered client server architectures. What are the technical differences and what advantages and disadvantages do each offer?
7. How does a thin client differ from a thick client?
8. For what is HTTP used? What are its major parts?
9. Describe how a Web browser and Web server work together to send a web page to a user.
10. Describe how mail user agents and message transfer agents work together to transfer mail messages.
11. What roles do SMTP, POP, and IMAP play in sending and receiving e-mail on the Internet?
12. What is FTP and why is it useful?
13. What is Telnet and why is it useful?
Lab Preparation
To prepare for the lab in week 3:
1. Study the structure of the HTTP request and response packets, refer Lecture – Week 2: Application Layer slides 23 to 27, chapter 2 of prescribed textbook and look at other source material on Web.
2. Check out Rex Swain’s HTTP Viewer at www.rexswain.com/httpview.html
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