COIT20261 Network Routing and Switching Term Assignment Sample

Sample COIT20261 Network Routing and Switching Term Assignment

Using this information, answer Question 1 and Question 2. Show your calculations for all sub-questions.

Question 1 – Information about the block

a) How many addresses are available in the block allocated to the company, including all special addresses?
b) What is the network address of the block?
c) What is the direct broadcast address of the block?

Question 2– Allocating subnets from the block

Create five contiguous subnets from the given block beginning with the first address of the block, as follows:

i. The first subnet with 1024 addresses
ii. A second subnet with 512 addresses
iii. A third subnet with 256 addresses
iv. Two (2) subnets with 126 addresses each

For each subnet, show its prefix in CIDR format, its subnet address, and its direct broadcast address. Organize your data clearly in a table.

Question 3 – Network Tools (Windows)

Often the best way to gain an initial familiarity with network tools is to simply use them at a basic level, exploring and looking up information as you go. Some common tools you can explore include Wireshark, ipconfig, tracert, netstat, ping and arp. All but Wireshark are included in Windows, Wireshark is free to download and install. Explore these tools by researching online and trying them out yourself on your computer, then answer the following questions, using your own words. Paste screenshots of your tryouts (not downloaded/copied from the Internet!) of each tool you have chosen to answer the questions below.
a) Assume that you want to use a command-line network tool to check if you have internet connection from your computer. From your desktop what command (tool) could you try first to find out if your internet connection is functioning OK or not?

Explain the tool and its output in determining your internet connectivity status.

b) You want to find out what IPv6 address your PC is currently configured with. What command could you use to discover this? What other information could you discover from using that command? Explain your answer.

c) Assume that you want to know which TCP ports are currently open in your computer. This information is quite useful for assignment help in checking if there are malicious services running on your system that have been introduced via malware. Which command would you use to discover this information? What other information could you discover from using that command? Explain your answer.

Question 4 -- TCP

Study a diagram of a TCP segment header (for example, Figure 9.23 in your textbook), paying special attention to the header fields, then list each field (field name) along with the value in decimal that would be in that field as per the information provided below. You must briefly explain your answer for each field.

• The segment is from an Internet Relay Chat (IRC) client to an Internet Relay Chat server
• A port number of 49,600 was assigned on the client side
• There were 20 bytes of options
• The server will be instructed not to send more than 500 bytes at any one time
• The previous segment from the server had an acknowledgement number of 12,400 and a sequence number of 8,300
• The TCP checksum was calculated to equal binary zero
• The control flag fields indicate states of: not urgent; not first (sync request) or final (termination request) segment; no bypass of buffer queues required; and not a RESET.

Solution

A) 2048 addresses are available in the block allocated to the company, including all special addresses. These special addresses are network and broadcast addresses. Given the IP address is 140.66.36.120/21 it shows that the first 21 bits of this address are reserved for the network address. Out of a total of 32 bits, (32-21)11 bits are left. So, 2^11 = 2048.

B) Network address = 140.66.32.0
Given IP address = 140.66.36.120/21
Network bits are 21
Binary format = 10001100. 01000010. 00100100. 01111000
Where the network bits are first 21 bits 10001100. 01000010. 00100
Network address = 10001100. 01000010. 00100000. 00000000
In decimal = 140.66.32.0

C) Direct broadcast address of the block
Given IP address = 140.66.36.120/21
The binary form of the given IP = 10001100. 01000010. 00100100. 01111000
In the direct broadcast address of any address block, network bits are all 1 and host bits are all 0.
Direct broadcast address = 11111111.11111111.11111000.00000000
In decimal = 255.255.248.0

2)
Question 2 – Allocating subnets from the block
Organise and write your answer neatly in a table and show calculations

The given IP address here is 140.66.36.120/21 which shows that 21 bits are used for the network part and the remaining bits 32-21=11 bits are used for the subnet and host part in the network.

After applying an AND operation with a subnet mask which is 255.255.252.0, the result generated is the address of the block is 140.66.36.0/23.
These 9 bits can be used to calculate the host - 2 11 = 2048.

By using VLSM (variable length subnet masking), each subnet will be defined with a different requirement (1024+512+256+128+128=2048). The CIDR routing can be done through 2 11=2048 bits, where 1024 bits are the most significant 0 and the remaining 1024 bits are the most significant bit 1.

Figure 1 Diagram to show CIDR and subnetting

First subnet - The 140.66.0010 0100. 0000 0000
Hence its subnet address becomes
140.66.36.0 and as we can see that the subnet bit will remain the same in all host parts hence its prefix is 22 (the first 22 bits are common in all hosts)
For broadcast address - all 1’s in the host part over here host bits are the last 8 bits hence
It converts to 140.66.0010 0100. 1111 1111
Which gives 140.66.36.255/24
So, the subnet mask for 1024 addresses subnet = 255.255.252.0

Second Subnet - Now as we can see from the figure the 2nd subnet needs a 512 host for that many addresses, we need 9 bits for a host which will leave 2 it for the subnet, for subnet identification it uses 10 bits
Hence,
140.66.0010 0101. 0000 0000 which gives
140.66.37.0/23 (prefix 23 because the first 23 bits will remain the same in all host parts)
For broadcast address put 1 in all host part
140.66.0010 0101. 0111 1111
Which gives 140.66.37.127/23

So, the subnet mask for 512 addresses subnet = 255.255.254.0
The third subnet - Now the third subnet needs address 256 which will require 8 bits for the host part and that will leave 3 bits for the subnet, for its subnet we will use bit 110
140.66.0010 0101. 1000 0000 which gives
140.66.37.128/26 (prefix 24 because the first 24 bits will remain the same in all host parts)
For broadcast address put 1 in all host part
140.66.0010 0101. 1011 1111
Which gives 140.66.37.191/24

So, the subnet mask for 256 addresses subnet = 255.255.255.0
The fourth subnet - Now the fourth subnet needs address 128 which will require 7 bits for the host part and that will leave 4 bits for the subnet, for its subnet we will use bit 1110
140.66.0010 0101. 1100 0000 which gives
140.66.37.192/25 (prefix 25 because the first 25 bits will remain the same in all host parts)
For broadcast address put 1 in all host part
140.66.0010 0101. 1101 1111 which gives 140.66.37.223/25
So, the subnet mask for 128 addresses subnet = 255.255.255.218

3)

Question 3 – Network tools - Windows (5 marks)

The command tool ping will be used. A ping is a tool for checking end-to-end connectivity between two endpoints and verifying internet connectivity.
Type “cmd” to bring up the Command Prompt.
Open the Command Prompt.
Type the “ping” command.
Type the IP address to ping.
Review the results displayed
Another tool that can be used here is ‘netstat -a’ to check active connections.

Figure 2 Result of ping command that shows successful pinging means an internet connection is active

The command to get the IPV6 address of the PC is "ipconfig /all"
This command is used to show IPv4 and IPv6 addresses with subnet masks and addresses. It can also be used to show DHCP and DNS settings. It also displays the current network that can be TCP/IP without using any kind of parameters.

Figure 3 Result of ipconfig/all command

Other information includes the drivers like media state, connection type of DNS, hostname, physical address, and status of DHCP enabled or not.
Netstat command is used to know which TCP ports are currently open in the computer.

Figure 4 Output of netstat -q command

Figure 5 Output of netstat -s command

Other than these, different information can be collected by using this command like active ports, current connection, segments received, send and retransmit, and failed attempt in connection.

4)

Question 4 – TCP (5 marks)

The first Row holds the source port address and Destination port Address which is 16 bits
The second Row covers the Sequence Number which is 16 bits
The third row comprises the Acknowledgement Number which is 16 bits
The fourth Row encompasses Header Length (4 bits) and (6 bits) reserved bits with 6 functions

Six Unique bits for six functions: They are

1. First One - URG - When this is 1 means - it is urgent and When this is 0 means - the segment is not urgent
2. Second One - ACK - Whether the segment contains valid Acknowledgement or not
3. Third One - PSH - Which means the request for Push
4. Fourth One - RST Which is for resetting the segment
5. Fifth One - SYN - Synchronize the segment with the following segments.
6. The Sixth One - FIN - Is for Terminating the Connection.

Fourth Row comprehends the window size that includes segment Length which is used to carry Data
Fifth Row consist of

Checksum – It is used to check the error occurrence in the segment

Urgent Pointer - This is for telling that the segment is very urgent. So, the TCP will give priority to this segment.

The sixth Row encompasses residual information of the segment and data will be added to this place.

- The segment from Internet Relay Chat client to Internet Relay Chat server is used for chat communication over the internet through different channels and access to the servers.

- A port number of 49,600 was assigned on the client side which is used for TCP/UDP connection. The Transmission Control Protocol is used for connection-oriented infrastructure and establishing a connection between the client and server side, and exchanging data.

- There were 20 bytes of options, then, the maximum header length will be 20B+60B = 80 B and the minimum will be 20B+20B=40B.

- The server will be instructed not to send more than 500 bytes at any one time which is more than the receiving speed. In this case, if the packet is too large from the server end, then the receiver client does not accept it. This packet will be divided into multiple segments and then sent.

- The previous segment from the server had an acknowledgment number of 12,400 and a sequence number of 8,300 where 8300 is the streaming rate of data from sending and receiving TCP whereas 12400 is the next sequence number which is the receiver for the destination node.

- The TCP checksum was calculated to equal binary zero which can be calculated by the sender and it represents that an error occurred during the transmission process.

- The control flag fields indicate states of
- not urgent – URG=0
- not first (sync request) or final (termination request) segment – ACK=0
- no bypass of buffer queues required – SYN=0
- not a RESET – RST=0

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Sample COIT20261 Network Routing and Switching Term Assignment

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