COIT20261 Network Routing and Switching Term 1, 2019 - Written Assessment-1

COIT20261 Network Routing and Switching Term 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)
Major network 215.150.68.98/26. The /26 means that 2 bits is borrowed
from the fourth octet. If two bits borrowed then 215.150.68.0 network
will have four subnets with 215.150.68.0, 215.150.68.64,
215.150.68.128 and 215.150.68.192 IP address. Therefore, the
215.150.68.98 has the network address 215.150.68.64 and the last
address is 215.150.68.126.
The direct broadcast address of a network is the last IP Address. This
last IP address of the network is 215.150.68.127.
Total available address as host address are 62. This is because,
215.150.68.64 is network address and 215.150.68.127 is broadcast
address.
1 mark
each
item,
total 3
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

COIT20261 Network Routing and Switching Term 1, 2019
Assignment item —Written Assessment-1
Subnet No. addresses Subnet address Mask /n
1 126 172.38.216.0 /25
2 126 172.38.216.128 /25
3 62 172.38.217.0 /26
4 62 172.38.217.64 /26
5 30 172.38.217.128 /27
6 30 172.38.217.160 /27
Your Calculations:
172.38.216.7/22 is the major network. The network address is
172.38.216.0 and the broadcast address is 172.38.219.255. If 100 IP
address is required then /25 mask is considered. /25 mask has 128
total IP address among which 126 addresses are available for
assigning. Therefore, for first two subnet /25 mask is selected.
The /26 mask has 62 assignable IP addresses. This fulfills the
requirement of 50 IP Addresses in subnet three and four. The
fourth and fifth subnet requires 25 addresses. /27 mask has 30
assignable IP addresses in the subnet. Therefore, the subnets
mentioned in the above table fulfills the subnet requirement.
6
Question 3: (4 marks) 4
a)
i)
ii)
Frame 4: The packet was captured in Interface id: 0 (\Device\Netgroup
Packet Filter_{2D7BDAD1-F89D-47B5-A4DB-62406675BB67}). The
Transmission Control Protocol’s transport layer protocol was used in
this transmission. This transport layer ensures end-to-end and error free
delivery of packets.
Frame 22406: The purpose of this message is ping to another device.
The source indicate it could receive 1548328 bytes on the next packet.
This is because the ACK has the value of 1548328.
2
1 mark
each
item,
total 2
for 3a.