College of Computing and Informatics
Instructions:
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Pg. 01
Learning
Outcome(s):
Explain network
protocols,
including
Transport Control
Protocol / Internet
Protocol.
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Question One
2 Marks
Explain why
1. we do not need the router in inside the same Local Area Network (LAN)
A router is not required for direct communication between devices as long as they are
all connected to the same local area network and share the same IP address. In a Local
Area Network (LAN), all users are physically seated in the exact location, such as a
single building or educational institution (Froehlich & Hwang, 2021). A network switch
allows for the connection of many computers to form a larger LAN. The data packets
from the network's various computers and other devices are sent through this switch and
sent on their way. The MAC addresses of the gadgets are used to do this. Because
everything is connected to the same local area network (LAN) and has IP addresses from
the same subnet, no intermediary router is required for the devices to speak with one
another.
2. we need a router in the figure below
It is generally agreed that a router is required if two or more LANs are combined into a
single system. The router is essential networking equipment to facilitate connection
across several networks (Irei & Scarpati, 2021). Two LANs are joined using a router, as
shown in the accompanying diagram. The IP addresses and subnets each LAN uses are
entirely different from the others.
The router must operate as a gateway between devices on different LANs for them to
connect. If a device on LAN 1 (Alice site) interacts with a device on LAN 2 (Bob site),
it will send a data packet to its specified gateway. The IP address of the router interface
leading to LAN 1 reveals that this device is the gateway.
Pg. 02
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The router checks the IP address in the packet to determine where it needs to send the
data. The router then consults its routing database to determine which interfaces are
linked to LAN 2 and sends the packet via that path. The router performs this routing
function, enabling efficient communication between devices on separate LANs.
Pg. 03
Learning
Outcome(s):
Demonstrate
protocol
configuration,
network addressing
schemes,
and
analyze
packet
transmission
for
local and wide area
networks.
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3 Marks
Question Two
Find the class, Network ID, Host ID, and number of hosts of the following classful IP
addresses:
a. 130.34.54.12
The first byte of an IPv4 address indicates the address's class, which is as follows:
Figure 0.1:Class classification (Neha, 2021)
Based on the data, we know that address 130.34.54.12 belongs to Class B. IP
addresses are separated into network IDs and host IDs based on their class.
For Class B

The first two octets make up the Network ID, making 130.34 the appropriate
value.

The last two octets, or 54.12, make up the Host ID.
By design, the subnet mask for Class B networks is 255.255.0.0. Use this calculation
to determine how many hosts your Class B network might have.
𝑁𝑜 𝑜𝑓 ℎ𝑜𝑠𝑡 = 2ℎ𝑜𝑠𝑡 𝑏𝑖𝑡𝑠 − 2
In Class B addressing, the host address takes up 16 of the available 32 bits.
= 216 − 2
= 65,534 ℎ𝑜𝑠𝑡𝑠
Pg. 04
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b. 200.34.2.1
IPv4 address 200.34.2.1 falls under Class C according to the preceding table.IP
addresses are separated into network IDs and host IDs based on their class.
For Class C
 The Network ID begins with the first three octets, making 200.34.2 the whole
number.
 The value 1 represents the Host ID since it is the last octet.
By design, the subnet mask for Class C networks is 255.255.255.0. Use this
calculation to determine how many hosts your Class C network might have.
𝑁𝑜 𝑜𝑓 ℎ𝑜𝑠𝑡 = 2 ℎ𝑜𝑠𝑡 𝑏𝑖𝑡𝑠 − 2
Class C addresses have 8 bits set up for host identifiers.
= 28 − 2
= 254 ℎ𝑜𝑠𝑡𝑠
c. 01110111 11110011 10000111 11011101
The IP address supplied is in binary format. We change it in decimal form while each
octet represents an integer between 0 and 255.
01110111 = 119
11110011 = 243
10000111 = 135
11011101 = 221
So, the IP address is 119.243.135.221.
For Class A
 Since the Network ID occupies the first byte, the answer is 119.
Pg. 05
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 The remaining three octets make up the Host ID, making the full number 243.135.221.
By design, the subnet mask for Class A networks is 255.0.0.0. Use this calculation to
determine how many hosts your Class A network might have.
𝑁𝑜 𝑜𝑓 ℎ𝑜𝑠𝑡 = 2ℎ𝑜𝑠𝑡 𝑏𝑖𝑡𝑠 – 2
For Class A itt has 24 bits reserved for host addresses,
= 224 − 2
= 16,777,214 ℎ𝑜𝑠𝑡𝑠
Pg. 06
Learning
Outcome(s):
Demonstrate
protocol
configuration,
network addressing
schemes,
and
analyze
packet
transmission
for
local and wide area
networks.
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применения Heading 1 к тексту, который должен
здесь отображаться.
3 Marks
Question Three
Find the Number of bits for Host ID, Network ID, sub-networks, and Maximum number
of sub-networks of the following classless address:
a. 190.107.16.17/18
Network ID
Subnet masks are denoted with the forward slash notation (/18). This number specifies
the size of the network ID in bits. There are 18 bits involved here.
Host ID:
𝐻𝑜𝑠𝑡 𝐼𝐷 = 𝑇𝑜𝑡𝑎𝑙 𝐵𝑖𝑡𝑠 − 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝐵𝑖𝑡𝑠
= 32 − 18
= 14 𝑏𝑖𝑡𝑠
Sub Network:
This address has Class B from the classification table so it has 16 bit network portion.
𝑁𝑜 𝑜𝑓 𝑠𝑢𝑏𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 = 2𝑏𝑖𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑢𝑏𝑛𝑒𝑡 𝑚𝑎𝑠𝑘 − 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑙𝑎𝑠𝑠 𝑛𝑒𝑡𝑤𝑜𝑟𝑘 𝑏𝑖𝑡𝑠
= 218−16 = 4
Maximum no of subnets
The number of possible subnets is limited by how many bits may be borrowed from
the host part for subnetting. We may use (32 - 18 - 2) = 12 bits for further subnetting
if we set aside 2 bits for hosts.
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑏𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 = 212 = 4096
Pg. 07
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b. 70.25.33.24/29
Network ID:
Subnet masks are denoted with the forward slash notation (/29). This number specifies
the size of the network ID in bits. There are 29 bits involved here.
Host ID:
𝐻𝑜𝑠𝑡 𝐼𝐷 = 𝑇𝑜𝑡𝑎𝑙 𝐵𝑖𝑡𝑠 − 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝐵𝑖𝑡𝑠
= 32 − 29
= 3 𝑏𝑖𝑡𝑠
Number of sub-networks:
This address has Class A from the classification table so it has 8 bit network portion.
𝑁𝑜 𝑜𝑓 𝑠𝑢𝑏𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 = 2 𝑏𝑖𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑢𝑏𝑛𝑒𝑡 𝑚𝑎𝑠𝑘 − 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑙𝑎𝑠𝑠 𝑛𝑒𝑡𝑤𝑜𝑟𝑘 𝑏𝑖𝑡𝑠
= 229−8
= 221 = 2,097,152
Maximum number of sub-networks:
The number of possible subnets is limited by how many bits may be borrowed from the
host part for subnetting. We may use (32 - 29 - 2) = 1 bits for further subnetting if we set
aside 2 bits for hosts.
𝑇ℎ𝑒 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑏 𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 = 21 = 2
Pg. 08
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c. 200.235.1.5/28
Network ID:
Subnet masks are denoted with the forward slash notation (/28). This number specifies
the size of the network ID in bits. There are 28 bits involved here.
Host ID:
𝐻𝑜𝑠𝑡 𝐼𝐷 = 𝑇𝑜𝑡𝑎𝑙 𝐵𝑖𝑡𝑠 − 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝐵𝑖𝑡𝑠
= 32 − 28
= 4 𝑏𝑖𝑡𝑠
Number of sub-networks:
This address has Class C from the classification table so it has 24 bit network portion.
𝑁𝑜 𝑜𝑓 𝑠𝑢𝑏𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 = 2 𝑏𝑖𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑢𝑏𝑛𝑒𝑡 𝑚𝑎𝑠𝑘 − 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑙𝑎𝑠𝑠 𝑛𝑒𝑡𝑤𝑜𝑟𝑘 𝑏𝑖𝑡𝑠
= 228−24
= 24 = 16
Maximum number of sub-networks:
The number of possible subnets is limited by how many bits may be borrowed from the
host part for subnetting. We may use (32 - 28 - 2) = 2 bits for further subnetting if we set
aside 2 bits for hosts.
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑏𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 = 22 = 4
Reference:
Froehlich, A., & Hwang, D. (2021, February 12). What is a lan? - definition from
whatis.com. Networking.
https://www.techtarget.com/searchnetworking/definition/local-area-networkLAN
Irei, A., & Scarpati, J. (2021, October 6). What is a router? definition from
searchnetworking. Networking.
https://www.techtarget.com/searchnetworking/definition/router
Neha, T. (2021, September 21). What is classful addressing in ipv4? Subnetting,
Supernetting & Disadvantages. Binary Terms. https://binaryterms.com/classfuladdressing-in-ipv4.html