COIT20261 Network Routing and Switching Written Assessment-1

COIT20261 Network Routing and Switching (Term 1, 2018)
Assignment item —Written Assessment-1
First Name:_________________________ Last Name:_________________________
Student ID: __________________________
Question Number Mark
allocated
Mark
earned
Question 1: (3 marks) 3
a) Given Address is 154.78.177.3 /27
First Part is Network or IP Address. It belongs to Class B network
Second Part is Subnet Mask (/27). Ie) 255.255.255.224
IPV4 consists four octets. Each octets has 0 to 255.
Four Octets Binary form
154.78.177.3 10011010.01001110.10110001.00000011
255.255.255.224 11111111.11111111.11111111.11100000
Take AND
operator
(0 AND 0 is 0
1 AND 0 is 0
0 AND 1 is 0
1 AND 1 is 1)
10011010.01001110.10110001.00000000
Last five binary digit is the Host Address
First three binary digit of the last octet is used to
find the number of subnet which can be
used/allocated in the given address. 23 =8,
Maximum eight subnet allowed in these given
address
So the network address is 154.78.177.0
1 mark
each
item,
total 3

COIT20261 Network Routing and Switching (Term 1, 2018)
Assignment item —Written Assessment-1
b)
c)
Four Octets Binary form
154.78.177.0
(Network
Address)
10011010.01001110.10110001.00000000
Replace last five
digit as 1. We get
the broadcast
address
10011010.01001110.10110001.00011111
Broad Cast Address is 154.78.177.31
The range of available host is the address between Network address and
Broadcast Address (exclusive)
Range of IP Address is 154.78.177.1 to 154.78.177.30
Question 2: (8 marks) 8
a)
Four Octets Binary form
138.77.216.5 10001010.01001101.11011000.00000101
255.255.255.0 11111111.11111111.11111111.00000000
Take AND
operator
(0 AND 0 is 0
1 AND 0 is 0
0 AND 1 is 0
1 AND 1 is 1)
10001010.01001101.11011000.00000000
Last eight binary digit is the Host Address.
So 28 - 1=256-1=255. The maximum number of
IP is 255
So the network address is 138.77.216.0
1

COIT20261 Network Routing and Switching (Term 1, 2018)
Assignment item —Written Assessment-1
b) Number of Address required is 32. i.e.) 25 is 32. Here 5 means the
number of Host. Total number of bit is 32 (four octets)
Subnet Mask is 32 – 5 = 27
6
c) Network allocation is the top down approach. We need to put the highest
IP required subnet in the top. In the last allocation, the highest number
of address is 32. But now, we need to add the new subnet which has
1024 address. We need to allocate the new subnet in the top. We should
redo the network allocation work
1
Subnet No. addresses Subnet address Mask /n
1 32 138.77.216.0 /27
2 32 138.77.216.32 /27
3 16 138.77.216.64 /28
4 16 138.77.216.80 /28
5 16 138.77.216.96 /28
6 16 138.77.216.112 /28
Subnet No. addresses Subnet address Mask /n
1 1024 138.77.216.0 /22
2 32 138.77.220.0 /27
3 32 138.77.220.32 /27
4 16 138.77.220.64 /28
5 16 138.77.220.80 /28
6 16 138.77.220.96 /28
7 16 138.77.220.112 /28