COIT20261 Network Routing and Switching (Term 1, 2018)

Running Head: COIT20261 Network Routing and Switching (Term 1, 2018) 1
Network Routing and Switching
Name
Tutor
Course
Date

COIT20261 Network Routing and Switching (Term 1, 2018) 2
Question 1 – Address usage (3 marks)
Consider the following classless address block:
154.78.177.3/27
List the addresses from this block that would be used as:
a) the network address,
b) the direct broadcast address, and
c) the range available for hosts to use
Following are the steps followed to arrive at the answer
Solution
154.78.177.3/27
The /27 in this address is an indication that the Ip address has its network address with a length
of 27 bits (the most significant bits). i.e. when we consider the ip address to be 32 bit long. So, in
my case scenario, 154.78.177.00000011 (the last 3 numbers have represented them as binary for
purposes of simplicity) up to 154.78.177.000 (8+8+8+3 = 27) is the network address and the
remaining 5 bits (00000) is for the host ip addresses (Fuller & Varadhan, 2013).
Note-
The network ID is simply the very first address of the total host ip’s while the direct broadcast
address is represented by the last address of total host ip’s (Fuller & Varadhan, 2013).

COIT20261 Network Routing and Switching (Term 1, 2018) 3
So the conclusion to this question is:
1. The network address becomes 154.78.177.0. This has been realized by setting all the 5 bits of
the host id to 0.
2. The direct broadcast address is 154.78.177.31. This has been realized by setting all the 5 bits
of the host to 1.
3. The range of available hosts is from 1 to 30 in the very last octet. I.e. from the 154.78.177.1 ip
address to 154.78.177.30 ip address (this is because the first and the last ip addresses are
reserved for the network address of the block and its direct broadcast address respectively)
(Huegen et al, 2011).
Question 2– Allocating subnets from a block (8 marks)
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.
a) Calculate the network address of this block and how many host addresses including
special addresses this block can provide (1 mark)
b) 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.
I. 2 subnets with 32 addresses each (2 marks)
II. 4 subnets with 16 addresses each (4 marks)
c) 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. (1 mark)

COIT20261 Network Routing and Switching (Term 1, 2018) 4
Solution
Subnetting is said to have taken place after extending the default subnet mask. Subnetting
cannot be performed having the default subnet mask and every class having its own default
subnet mask. To be able to know a subnetted subnet mask, the subnet mask is first written down
we first. Next on line is finding the host bits which have been borrowed in creating the subnets
and then convert them into decimal form. For instance in my question, I first find the subnet
mask of my given address 138.77.216.5/24? Since the address belongs to a class B address, and
class B addresses usually have default subnet masks of 255.255.0.0[ /16 in CIDR ] (Postel&
Mogul, 2015).
This means I have to borrow 8 bits from the host portion to be able to satisfy the
requirements for the address I have been given (/24=/16+8 bits). Bearing in mind that subnetting
proceeds from left to right, without skipping any network bit, the subnet mask in my given case
in binary form it becomes 11111111. 11111111.11111111.00000000. The first three octet
contains the default value so that its value in terms of decimal becomes 255.255.255. The 4th
octet is characterized by all its bits being off and therefore, the decimal representation is
0+0+0+0+0+0+0+0 =0. So my answer for subnet mask becomes 255.255.255.0 (Postel& Mogul,
2015).
Total number of subnets provided by the subnet mask
To get the total number of subnets which can be realized from a certain subnet mask the
formula applied is 2N, where N = the bits which are obtained from the host part to create the
subnets. In my question 138.77.216.5/24, N is 8. Examining the address keenly it is clear that the
address belongs to class B and the class B addresses have 255.255.0.0 [/16 in CIDR] as the

COIT20261 Network Routing and Switching (Term 1, 2018) 5
default subnet mask. From the given address the bits borrowed from the host are 24 - 16 = 8 host
bits. Now 28 = 256, so the answer becomes 256 (Postel& Mogul, 2015).
Clarification
My original network is a class B, so it has 16 bits in its default subnet mask, i borrowed 8 bits
from the host part (nnnnnnnn.nnnnnnnn.ssssssss.hhhhhhhh) of the original network, now if i do
2^8 i will get the total number of subnets that the network 138.77.x.x would have, 2^8 = 256.
Now to know how many subnets there are in the range of 138.77.216.x/24, i can take in
consideration the bits that have been borrowed from the host part only in the fourth octet which
are 0 bits, since the first three octets must match the address 138.77.216., so 2^0 = 1, i have 1
subnet that starts with 138.77.216.x.x and that is 138.77.216.0 (Schuler, 2013).
2 (a)
So the network address for this block of address is 138.77.216.0
2 (b)
i) 138.77.216.0 to 138.77.216.31 and
138.77.216.32 to 138.77.216.64
ii) 138.77.216.0 to138.77.216.15
138.77.216.16 to 138.77.216.31
138.77.216.32 to 138.77.216.47
138.77.216.48 to 138.77.216.64

COIT20261 Network Routing and Switching (Term 1, 2018) 6
2 (c)
From the above calculations, the total number of subnets is 1 (2^0) and the subnet is capable of
accommodating 256 hosts. This means the whole block of address will have 256 hosts. This
indicates that even if the company decides to add another subnet with 1024 addresses, there
would be no room for that.
Question 3 – Network Tools (Windows) (4 marks)
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.
a) A Wireshark scan has produced a packet capture, saved to a file named
wireshark_capture01.pcapng and available on the Unit website. Download the file and
open it in Wireshark, then answer these questions about the scan:
i. Very briefly summarise in your own words the content in each of the three horizontal
display windows in Wireshark (.5 mark)
ii. In Frame 3, what brand of computer launched this scan and what was its IP address?
State where this information is found (.5 mark)
iii. Briefly explain the exchange event captured in frames 4 - 6 (.5 mark)
iv. Describe in your own words two specific network problems that a network
administrator could use Wireshark for as a troubleshooting tool?
A (I)