The document discusses subnetting in communication networks, specifically focusing on fixed length subnetting, the division of large IP address blocks into smaller subnets, and the importance of subnet masks. It provides examples of how to divide specific address blocks into subnets, detailing the network and broadcast IDs, as well as exercises for practical understanding. Additionally, references for further reading on the topic are provided.

1. Communication
Networks
Network Layer Services
Fixed Length Subnetting

2. Subnets and Subnet Mask
• What are subnets?
• Dividing the large address space into smaller blocks of IP Address
• Why do we go for subnets?
• If the blocks of IP address are used as such millions of IP address are
wasted.
• Managing the Large block of IP Address is also difficult
• What are the types of subnets?
• Fixed Length Subnetting
• Variable Length Subnetting
• Subnet Mask
• Will be discussed few slides later
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3. Fixed Length Subnetting
• Lets consider an Organization and the entire organization has different
departments / Sections
I
II III
IVECE
EEE
CSE
MECH
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4. Fixed Length – Example 1
Divide the address block 200.1.2.0 into 2 subnets
Given address block is Class C → 24 bits for NID and 8 bits for HID
200.1.2.00000000
200.1.2.11111111
200.1.2.0
200.1.2.255 200.1.2.X0000000
X = 0 X = 1
Network ID : 200.1.2.0
Broadcast ID: 200.1.2.255
1 bit is required
to make 2 parts
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5. 200.1.2.X0000000
X = 0 X = 1
200.1.2.0_ _ _ _ _ _ _
Runs from
200.1.2.00000000
200.1.2.00000001
200.1.2.00000010
200.1.2.00000011
.
.
200.1.2.01111111
200.1.2.1_ _ _ _ _ _ _
Runs from
200.1.2.10000000
200.1.2.10000001
200.1.2.10000010
200.1.2.10000011
.
.
200.1.2.11111111
200.1.2.0
200.1.2.127
200.1.2.128
200.1.2.255
Network ID : 200.1.2.0
Broadcast ID: 200.1.2.127
Network ID : 200.1.2.128
Broadcast ID: 200.1.2.255
27 = 128
Address in Each Subnet
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6. Fixed Length – Example 2
Divide the address block 200.1.2.0 into 4 subnets
Given address block is Class C → 24 bits for NID and 8 bits for HID
200.1.2.00000000
200.1.2.11111111
200.1.2.0
200.1.2.255 200.1.2.XX000000
XX = 00
Network ID : 200.1.2.0
Broadcast ID: 200.1.2.255
XX = 01 XX = 10
XX = 11
2 bits are required
to make 4 parts
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7. 200.1.2.XX000000
XX = 00
XX = 01 XX = 10
XX = 11
200.1.2.00 _ _ _ _ _ _
Runs from
200.1.2.00000000
to
200.1.2.00111111
200.1.2.01 _ _ _ _ _ _
Runs from
200.1.2.01000000
to
200.1.2.01111111
200.1.2.11 _ _ _ _ _ _
Runs from
200.1.2.11000000
to
200.1.2.11111111
200.1.2.10 _ _ _ _ _ _
Runs from
200.1.2.10000000
to
200.1.2.10111111
200.1.2.0 → Network ID
200.1.2.63 → Broadcast ID
200.1.2.128 → Network ID
200.1.2.191 → Broadcast ID
200.1.2.192 → Network ID
200.1.2.255 → Broadcast ID
200.1.2.64 → Network ID
200.1.2.127 → Broadcast ID
26 = 64
Address in Each Subnet
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8. Fixed Length – Exercise 1
Divide the address block 192.168.9.0 into 5 networks
Given address block is Class C → 24 bits for NID and 8 bits for HID
192.168.9.00000000
192.168.9.11111111
192.168.9.0
192.168.9.255 192.168.9.XXX00000
Network ID : 192.168.9.0
Broadcast ID: 192.168.9.255
3 bits are required
to make 8 parts
000
001
010
011 100
101
110
111
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9. NET 1
192.168.9.00000000 192.168.9.0
192.168.9.00011111 192.168.9.31
NET 2
192.168.9.00100000 192.168.9.32
192.168.9.00111111 192.168.9.63
NET 3
192.168.9.01000000 192.168.9.64
192.168.9.01011111 192.168.9.95
NET 4
192.168.9.01100000 192.168.9.96
192.168.9.01111111 192.168.9.127
NET 5
192.168.9.10000000 192.168.9.128
192.168.9.10011111 192.168.9.159
NET 6
192.168.9.10100000 192.168.9.160
192.168.9.10111111 192.168.9.191
NET 7
192.168.9.11000000 192.168.9.192
192.168.9.11011111 192.168.9.223
NET 8
192.168.9.11100000 192.168.9.224
192.168.9.11111111 192.168.9.255

10. Fixed Length – Exercise 2
Divide the address space 172.16.0.0 into 3 networks
Given address block is Class B → 16 bits for NID and 16 bits for HID
172.16.00000000.00000000
To
172.16.11111111.11111111
172.16.XX000000.00000000
XX = 00
Network ID : 172.16.0.0
Broadcast ID: 172.16.255.255
XX = 01 XX = 10
XX = 11
2 bits are required
to make 4 parts
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11. NET 1
172.16.00000000.00000000 172.16.0.0
172.16.00111111.11111111 172.16.63.255
NET 2
172.16.01000000.00000000 172.16.64.0
172.16.01111111.11111111 172.16.127.255
NET 3
172.16.10000000.00000000 172.16.128.0
172.16.10111111.11111111 172.16.191.255
NET 4
172.16.11000000.00000000 172.16.192.0
172.16.11111111.11111111 172.16.255.255
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12. Fixed Length – Exercise 3
• Divide the address space 192.168.9.0 into suitable number of
networks so that each network can handle at least 10 hosts
In each subnet 2 address are not usable
so total no of address required is 12
2? Gives 12 address → This is not possible
24 Gives 16 address → This is the nearest possible value
192.168.9.XXXX_ _ _ _
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13. References:
• Behrouz A. Forouzan, ―Data communication and Networking, Fifth
Edition, Tata McGraw – Hill, 2013
• Larry L. Peterson, Bruce S. Davie, ―Computer Networks: A Systems
Approach, Fifth Edition, Morgan Kaufmann Publishers, 2011.
• Few online References (Will be Mentioned in the description Section)
Thank You…