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COIT20261 Network Routing And Switching 2

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Question:

1 – Address usage 

Consider the following classless address block:

154.78.177.3/27

List the addresses from this block that would be used as:

  1. the network address,
  2. the direct broadcast address, and
  3. the range available for hosts to use

Show the steps you took to arrive at your answers.

2– Allocating subnets from a block

A company has been granted a block of addresses which includes the address 138.77.216.5/24.   Answer the following questions, showing your calculations.

  1. Calculate the network address of this block and how many host addresses including special addresses this block can provide 
  2. Create the following 6 subnets for this company by calculating the subnet address for each subnet.  Answer this question by filling in the table in the Answer template.  Use CIDR format for the mask.
    1. 2 subnets with 32 addresses each
    2. 4 subnets with 16 addresses each 
  3. After some time, the company decides that it wants another subnet with 1,024 addresses.  Explain whether this can be allocated from the existing block.

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 in exploratory mode as suggested in some of the tutorial exercises.  Netstat and Tracert are included in Windows, while Wireshark is free to download and install.  Explore Wireshark, Netstat and Tracert, then complete this question.

  1. A wireshark scan has produced a packet capture, saved to a file named pcapngand available on the Unit website.  Download the file and open it in Wireshark, then answer these questions about the scan:

  2. Very briefly summarise in your own words the content in each of the three horizontal display windows in Wireshark

  3. In Frame 3, what brand of computer launched this scan and what was its IP address?  State where this information is found
  • Briefly explain the exchange event captured in frames 4 - 6 (.5 mark)
  1. Describe in your own words two specific network problems that a network administrator could use Wireshark for as a troubleshooting tool? 
  1. A ‘NETSTAT –aon’ command has given the output below (excerpted).   Briefly describe each column heading, and the states LISTENING, ESTABLISHED and CLOSE_WAIT. 

Active Connections

Proto  Local Address       Foreign Address      State         PID

TCP    10.0.0.58:139          0.0.0.0:0              LISTENING       

TCP    10.0.0.58:5040         0.0.0.0:0              LISTENING       7480

TCP    10.0.0.118:139         0.0.0.0:0              LISTENING       4

TCP    10.0.0.118:52450       52.63.165.133:443      ESTABLISHED     14080

TCP    10.0.0.118:52458       104.116.191.195:443    CLOSE_WAIT      8912

TCP    10.0.0.118:52791       40.100.151.2:443       ESTABLISHED     22400

TCP    10.0.0.118:52811       162.125.34.129:443     ESTABLISHED     4696

TCP    10.0.0.118:52820       34.232.224.128:443     CLOSE_WAIT      4696

TCP    10.0.0.118:52879       162.125.34.129:443     ESTABLISHED     4696

Do a TRACERT on your computer to www.google.com.  Paste the output to your assignment answer template and discuss the information being displayed.

 

Answer:

A 1:

a)

 

 

 

 

 

 

 

 

 

 

 

 

 

b)

 

 

 

 

 

 

 

 

 

c)

 

Here /27 is the subnet mask, ie) 255.255.255.224

 

Binary Format of Given Address is,

 

154.78.177.3         --->10011010.01001110.10110001.000 00011

 

255.255.255.224 --->11111111.11111111.11111111.111 00000

 

Taking AND operation between Given IP and Subnet Mask, The result is,

10011010.01001110.10110001.000 00000 (It is the network Address)

 

In Decimal form, the network Address is 154.78.177.0

 

From Subnet mask (binary form), Last five digit is 0. This denotes the number of host available. So 25 is 32

 

The number of IP available in this network is 32

 

Last address is the broadcast address

 

IP Range is 154.78.177.1 to 154.78.177.31

So Broadcast IP Address is 154.78.177.31

 

Usable IP range is IP Range is 154.78.177.1 to 154.78.177.30

 

We could not use the broadcast address as IP address

 

A 2:

a)

Here /24 is the subnet mask, ie) 255.255.255.0

 

Binary Format of Given Address is,

 

138.77.216.5   --->10001010.01001101.11011000.00000101

 

255.255.255.0 --->11111111.11111111.11111111.000 00000

 

Taking AND operation between Given IP and Subnet Mask, The result is,

10001010.01001101.11011000 .00000000 (It is the network Address

 

Network Address is 138.77.216.0

 

From Subnet mask (binary form), Last eight digit is 0. This denotes the number of host available. So 28 is 256

 

The number of IP available in this network is 255

 

Last address is the broadcast address

 

IP Range is 138.77.216.1 to 138.77.216.255

So Broadcast IP Address is 138.77.216.255

 

Usable IP Range is 138.77.216.1 to 138.77.216.254

 

b)

Subnet

No. addresses

Subnet address

Mask /n

1

32

138.77.216.0

255.255.255.224 (/27)

2

32

138.77.216.32

255.255.255.224 (/27)

3

16

138.77.216.64

255.255.255.240 (/28)

4

16

138.77.216.80

255.255.255.240 (/28)

5

16

138.77.216.96

255.255.255.240 (/28)

6

16

138.77.216.112

255.255.255.240 (/28)

 

 

c)

No, we could not allocate 1024 address from the given address

 

Given broadcast address is /24

 

The maximum IP Address of this broadcast is 255

 

If you want to set the 1024 address, we need to change the subnet mask

 

Another one, we need to design the network address using the high order of number of IP required

 

Subnet

No. addresses

Subnet address

Mask /n

1

1024

138.77.216.0

255.255.252.0 (/22)

2

32

138.77.220.0

255.255.255.224 (/27)

3

32

138.77.220.32

255.255.255.224 (/27)

4

16

138.77.220.64

255.255.255.240 (/28)

5

16

138.77.220.80

255.255.255.240 (/28)

6

16

138.77.220.96

255.255.255.240 (/28)

7

16

138.77.220.112

255.255.255.240 (/28)

 

 

A 3:

a)

i)

ii)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

iii)

 

 

 

 

 

 

iv)


First column, it denotes the frame 3.

Second column, it is the running time which the packet is captured from the starting.

 

Third Column: It denotes Source Address. It displays the current local system’s IP address

 

Fourth Column : Destination Address. It displays the remote systems’ IP Address

 

Fifth Column: Protocol, it displays the protocol name which is used to establish/communicate between client and server

 

 

Destination System is HewlettP_24:ea:fa

Source System is Cisco_3f:9a:7f

The above information is found from packet details panel (Ethernet II)

 

Destination IP is 138.77.216.88

Source IP is 104.224.42.136

The above information is found from packet details panel (Internet Protocol Version 4)

 

Frame 4 : Server responds the client by sending “Server hello”, along with random value. TLS version 2 protocol used.

 

Frame 5: It is a handshake protocol message.

After Server hello, Server sends its certificate to the client for authentication. Server may request a certificate from the client

Frame 6: Server key exchanged Server sends the “Server hello done”.

If the server has requested a certificate from the client, the client sends it

 

Wire shark tool is used to monitor or view the network packet information. These packet information is used to check or find the vulnerabilities of network.

Easy to filter or search the specific keyword using Wireshark.

b)

 

Column

Description

Proto

Protocol name such as TCP or UDP

Local Address

Working System Interface IP address

Foreign Address

Remote or gateway Address

State

It is a connection state between client and remote system

PID

Process ID, it denotes the OS kernel process number

 

LISTENING – This state indicates that remote system is ready to accept the connect

 

Line 1: Client system (10.0.0.58) is finding the required system connection. 0.0.0.0 means that system is listening on all the IP addresses assigned to the current system

 

ESTABLISHED – the remote system received the SYN (synchronize) signal from the local (client). This connection session is established

 

Line 4: Client System is established the connection of the remote system (52.63.165.133) of the port 443

 

CLOSE_WAIT – the remote system (server) received the first FIN signal from the local (client). This connection between client and server is in the process of being closed

 

c)

 

 

 

 

 

First line, It finds the default gateway (hop 1)

 

Twelfth line, it finds the google address which is 216.58.199.196

 

Number of Hop is 12 which is used to find the google address

 

 
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