Subnetting allows a network administrator to break a single network into multiple subnets. It is done by borrowing bits from the host portion of the IP address to create a subnet portion. This increases the number of available subnets while reducing the number of available hosts per subnet. The document provides examples of subnetting default class C and class B network addresses, showing how to determine the number of bits to borrow to meet subnet and host requirements, and how subnet masks are used to identify the subnet of a given host IP address.

2. Why Subnet?
• To break the network down into pieces, each of
which can be addressed separately.
• Controls network traffic
• Reduces broadcasts
• Can provide low level security with access lists on the
router
• Organization of IP address space

3. Subnetting a Default Class C
Network Address: 200.129.41.0
• Default Class C address is divided into network and
host portions as follows:
N . N . N . H
• To subnet we “borrow” bits from the host portion
of the address (8 bits for Class C)
N . N . N . x x x x x x x x
• Borrowing n bits yields 2n
– 2 subnets.
• Leaving n bits yields 2n
– 2 hosts.
• For a class C, we can borrow from 2 to 6 bits.
• Why not 1 bit? (How many usable subnets?)
• Why not 7 bits? (How many usable hosts?)

4. Subnetting a Default Class C
Network Address: 200.129.41.0
• Suppose we need 14 usable subnets, how many
bits do we borrow?
• Remember, borrowing n bits give us:
• 2n
– 2 subnets
• Try borrowing 3 bits (n = 3):
• 23
– 2 = 8 – 2
= 6 usable subnets (not enough)
• Try borrowing 4 bits
• 24
– 2 = 16 – 2
= 14 usable subnets (enough)

5. Subnetting a Default Class C
Network Address: 200.129.41.0
• Write it with the network octet in binary:
200.129.41.0000 0000
breakhere
• Borrowing 4 bits yields 14 usable subnets
• How many usable hosts per subnet?
• Same formula as subnets (2n
– 2)
• 4 host bits (n = 4)
• 24
– 2 = 16 – 2
= 14 usable hosts per subnet
subnet bits host bits

6. Subnetting a Default Class C
Network Address: 200.129.41.0
• Examples:
• First usable 200.129.41.0001 ^ 0000
subnet address: 200.129.41.16
• First usable host 200.129.41.0001 ^ 0001
on the first subnet: 200.129.41.17
• Second usable host 200.129.41.0001 ^ 0010
on the first subnet: 200.129.41.18
.
.
.
• Last usable host 200.129.41.0001 ^ 1110
on the first subnet: 200.129.41.30
• Broadcast address 200.129.41.0001 ^ 1111
for the first subnet: 200.129.41.31

7. Subnetting a Default Class C
Network Address: 200.129.41.0
• Examples:
• Second usable 200.129.41.0010 ^ 0000
subnet address: 200.129.41.32
• Third usable 200.129.41.0011 ^ 0000
subnet address: 200.129.41.48
• Fourth usable 200.129.41.0100 ^ 0000
subnet address: 200.129.41.64
.
.
.
• Last usable 200.129.41.1110 ^ 0000
subnet address: 200.129.41.224

8. The Subnet Mask: How the
Router Determines the Subnet
• The subnet mask (in binary) has:
• all ones in the network and subnet portion of the
address
• all zeros in the host potion of the address
• The subnet mask for the previous example is:
255.255.255. 240
255.255.255. 1111^ 0000 (128 + 64 + 32 + 16 =240)
• ANDing this mask with any valid host address on the
network will always yield the subnet address for that
host, where if both bits are 1 you get a 1 otherwise its a 0.

9. The Subnet Mask: How the
Router Determines the Subnet
• Example (our subnet mask is 255.255.255.240)
IP host address: 200.129. 41.23
Last octet to binary: 200.129. 41.0001 0111
AND subnet mask: 255.255.255.1111 0000
200.129. 41.0001 0000
Subnet Address: 200.129. 41.16
So the host address 200.129. 41.23 is on the
200.129.41.16 subnet.

10. Subnetting a Default Class B
Network Address: 132.178.0.0
• Default Class B address is divided into network and
host portions as follows:
N . N . H . H
• To subnet we “borrow” bits from the host portion
of the address (16 bits for Class B)
N . N . x x x x x x x x . x x x x x x x x
• For a class B, we can borrow from 2 to 14 bits.

11. Subnetting a Default Class B
Network Address: 132.178.0.0
• Suppose we need 80 usable subnets, how many
bits do we borrow?
• Remember, borrowing n bits give us:
• 2n
– 2 subnets
• Try borrowing 6 bits (n = 6):
• 26
– 2 = 64 – 2
= 62 usable subnets (not enough)
• Try borrowing 7 bits
• 27
– 2 = 128 – 2
= 126 usable subnets (enough)

12. Subnetting a Default Class B
Network Address: 132.178.0.0
• Write it with the network octets in binary:
132.178.0000000 0.00000000
breakhere
• Borrowing 7 bits yields 126 usable subnets
• How many usable hosts per subnet?
• Same formula as subnets (2n
– 2)
• 9 host bits (n = 9)
• 29
– 2 = 512 – 2
= 510 usable hosts per subnet
subnet bits host bits

13. Subnetting a Default Class B
Network Address: 132.178.0.0
• Examples:
• First usable 132.178.0000001 ^ 0.00000000
subnet address: 132.178.2.0
• First usable host 132.178.0000001 ^ 0.00000001
on the first subnet: 132.178.2.1
• Second usable host 132.178.0000001 ^ 0.00000010
on the first subnet: 132.178.2.2
.
.
.
• Last usable host 132.178.0000001 ^ 1.11111110
on the first subnet: 132.178.3.254
• Broadcast address 132.178.0000001 ^1.11111111
for the first subnet: 132.178.3.255

14. Subnetting a Default Class B
Network Address: 132.178.0.0
• Examples:
• Second usable 132.178.0000010 ^ 0.00000000
subnet address: 132.178.4.0
• Third usable 132.178.0000011 ^ 0.00000000
subnet address: 132.178.6.0
.
.
.
• Ninety-first usable 132.178.1011011 ^ 0.00000000
subnet address: 132.178.182.0
.
.
.
• Last usable 132.178.1111110 ^ 0.00000000
subnet address: 132.178.252.0

15. Subnetting a Default Class B
Network Address: 132.178.0.0
• The subnet mask for this example is:
255.255.254.0
255.255.1111111 ^ 0.00000000
• ANDing this mask with any valid host address on
this network will always yield the subnet address.
• Where if both bits are 1 you get a 1 otherwise its a
0.

16. Subnetting a Default Class B
Network Address: 132.178.0.0
• Example:
IP host address: 132.178.119.112
Last octets to binary: 132.178.0111011 ^ 1.01110000
AND subnet mask: 255.255.1111111 ^ 0.00000000
132.178.0111011 ^ 0.00000000
Subnet Address: 132.178.118.0