COIT20261 Network Routing and Switching Assignment 2022

COIT20261 Network Routing and Switching Term 2, 2019
Assignment item —Written Assessment-1
ANSWER TEMPLATE ASSIGNMENT ONE
Type your answers in the spaces provided
Marking criteria: Your answers will be marked based on technical correctness,
completeness, clarity and relevance. Questions that ask you to show your working or
calculations or the steps you took to arrive at your answers, may have marks deducted if
such information is not provided. If a question requires you to submit a graphic (e.g. a
screenshot or a diagram), the graphic must have sufficient resolution to show all its
details clearly and be of a reasonable size for normal reader viewing, with all or any text
within the graphic being legible and readable, in order to be marked.
First Name:_________________________ Last Name:____________________________
Student ID: __________________________
Your Answers Mark
allocated
Mark
earned
Question 1: (3 marks) 3
a)
b)
c)
For the classless address block 157.150.38.25/27
The network address is - 157.150.38.0 and the last usable address is
157.150.38.30
The direct broadcast address is - 157.150.38.31
30 address are available as the host address for the given address block.
1 mark
each
item,
total 3
Question 2: (8 marks) 8
Subnet Name No. of Address Subnet Address Mask
A 32 172.38.216.0 /26
B 32 172.38.216.64 /26
C 32 172.38.216.128 /26
6

COIT20261 Network Routing and Switching Term 2, 2019
Assignment item —Written Assessment-1
D 16 172.38.216.192 /27
E 8 172.38.216.224 /28
F 8 172.38.216.240 /28
Calculations: first address, total addresses, correct prefixes
Network Address = 172.38.216.22/25
Binary notation of the address block
10101100. 100110. 11011000. 00010110
Binary notation of the net mask
11111111. 11111111. 11111111. 10000000
Decimal notation of the wild card mask
00000000.00000000.00000000 .01111111
Network
10101100. 00100110. 11011000.00010110
= 172.38.216.22/25
172.38.216.1 is the first assignable IP address for the
address block
It can include 126 address including the broadcast address
The correct prefix for the address block is /24
Total IP address needed is 144 and /25 can allocate 126 IP
address and thus the subnet fails. Thus /24 is selected that is
capable of allocating 254 IP address.
2
Question 3: (4 marks) 4
3A
i
ii
Wireshark
Frame 5 (1 mark):
The packet is from client to server and it has been found from analyse
section Follow > TCP stream option in wireshark.
Secured web application using HTTPS is responsible for this ACK
message. This is because the port 443 is used as the source port for the
frame and it is used for HTTPS.
Frames 446 – 447 (1 mark):
2