COIT20261 Network Routing and Switching: Term 2, 2019 Assignment

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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
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Question 1: (3 marks) 3
a)
b)
c)
To obtain the network address we review the address 157.150.38.25/27
which has 5 host bits, with these number of bits , 32 is the subnet size,
which is the increment for each subnet
The final address is 157.150.38.30/27
The first subnet is 157.150.38.0/27
Directed broadcast address is the address after the last host address for
the subnet, 157.150.38.31/27 is the address.
The network has 27 bits while 5 bits are host bits applying the formula
2^5-(2) we have 30 addresses in the subnet
1 mark
each
item,
total 3

COIT20261 Network Routing and Switching Term 2, 2019
Assignment item —Written Assessment-1
Question 2: (8 marks) 8
<insert your table here>
Prefix Subnet size
172.38.216.120/29 8 Addresses
172.38.216.112/29 8 Addresses
172.38.216.96/28 16 Addresses
172.38.216.64/27 32 Addresses
172.38.216.32/27 32 Addresses
172.38.216.0/27 32 Addresses
6
Calculations: first address, total addresses, correct prefixes
The block granted to the company has the address as 172.38.216.22/25
which is a host address.
This represents 7 host bits with a subnet size of 128.
The network addresses are 172.38.216.0/25, and 172.38.216.128/25
The current block is part of the subnet 172.38.216.0/25
the 1st address in this block corresponds to 172.38.216.1/25
128 is the overall count of addresses.
To subnet the network address we use the number of host addresses
which is 32, 26 and 8. We borrow 2 bits from the host to create 4
subnets with 32 addresses each, next we use the next subnet borrowing
1 bit to create two networks of 16 addresses and Lastly I subnet the final
address to create two network with 8 host each as depicted in the table.
2
Question 3: (4 marks) 4

COIT20261 Network Routing and Switching Term 2, 2019
Assignment item —Written Assessment-1
3A
i
ii
Wireshark
Frame 5 (1 mark):
The packet in frame 5 is coming from server to client because the source
port is 443 and the destination port is 55090. The source port can only
be a server.
A HTTPS web based app was responsible for this ACK message
Frames 446 – 447 (1 mark):
Frame 446 and 447 are address resolution protocol request and reply
packets. In frame 446 the source with IP 10.0.0.22 sends an ARP request
to the IP 10.0.0.69 and get a reply in frame 447 having the MAC address
as 88:87:17:9d:a2:9b
2
3B Route print: 1

COIT20261 Network Routing and Switching Term 2, 2019
Assignment item —Written Assessment-1
Network destination: Basically, it is that subnet of a destination
networks we want to reach.
Netmask represents the subnet mask of each destination network.
Interface is the IP address by which we sent the traffic to destination
subnet.
Gateway: Basically, it is the IP address of the device to whom the traffic
would be directed.
The metric is a value that define the weight of a route
Default Route: Basically, it is a route of last resort, for forwarding the
traffic without an exact match that is 0.0.0.0 subnet and netmask 0.0.0.0
The all-link gateway traffic terminates on the PC.
3C IPCONFIG:
An interface is a port on through which traffic ingresses or egresses a
network device.
The Ethernet interface is disconnected and the
Wi-Fi-interface IP address is 192.168.8.101 its subnet mask is
255.255.255.0 and its gateway address is 192.168.8.1
1

COIT20261 Network Routing and Switching Term 2, 2019
Assignment item —Written Assessment-1
Total marks awarded 15 (max)
Less late penalties
Less plagiarism penalties
Total marks earned
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