COIT20261 Network Routing and SwitchingTerm 1, 2019 - Written Assessment-1

COIT20261 Network Routing and SwitchingTerm 1, 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: __________________________
Questions Mark
allocated
Mark
earned
Question 1: (3 marks) 3
a)
b)
c)
/26 converted to decimal becomes 255.255.255.192.
256-192=64, possible subnets are 0, 64, 128 ...
215.150.68.98 lies in sub 64 ie 64-128, therefore;
Network address: 215.150.68.64/26
This is given by next subnet minus 2 ie 128-2=126
Last usable address: 215.150.68.126/26
The broadcast address is given by next subnet minus 1 ie 128-1=127
Direct broadcast address :215.150.68.127
Number of available addresses =2n-2 where n is the number of
zeroes ie 26-2 =62
Available as host addresses : 62
1 mark
each
item,
total 3

COIT20261 Network Routing and SwitchingTerm 1, 2019
Assignment item —Written Assessment-1
Question 2: (8 marks) 8
Show the correct number of addresses for each subnet in the table for b)
in the second column labelled “No. addresses”.
2
Subnet No. addresses Subnet address Mask /n
1 126 172.38.0.0 /25
2 126 172.38.0.128 /25
3 62 172.38.1.0 /26
4 62 172.38.1.64 /26
5 30 172.38.1.128 /27
6 30 172.38.1.160 /27
Your Calculations:
To calculate the number of addresses, we use the formula 2n-2 = >
number of hosts where n=number of zeroes
Subnet 1 – 2n-2=>100
=2n =>102
=2n =128 ; we picked 128 because it is the nearest higher divisor of 2
N=7, ie 27= 128, therefore number of addresses = 128-2
=126
Note that the same formular is applied for subnet 2,3,4,5 and 6.
6
Question 3: (4 marks) 4
a)
i) Frame 4: Ethernet interface was used. By right clicking the packet and
click on filter capture, eth is displayed in the search area indicating that
this is an Ethernet interface.
2
1 mark
each
item,
total 2