A seminar on Subnetting

The Institution of electronics and telecommunication engineers, Delhi centre.

Presented by: SanjayStream: CS

Guided by: Mr. Nibhesh kr. Das

What was that masked address?

To break the network down into pieces, each of which can be addressed separately.Controls network trafficReduces broadcastsOrganization of IP address space

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?)

Suppose we need 14 usable subnets, how many bits do we borrow?Remember, borrowing n bits give us:

2n – 2 subnetsTry 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)

Write it with the network octet in binary:

200.129.41.0000 0000 break here

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

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 ^ 1110on the first subnet: 200.129.41.30

Broadcast address 200.129.41.0001 ^ 1111for the first subnet: 200.129.41.31

Examples:Second usable 200.129.41.0010 ^ 0000

subnet address: 200.129.41.32Third usable 200.129.41.0011 ^ 0000

subnet address: 200.129.41.48Fourth usable 200.129.41.0100 ^ 0000

subnet address: 200.129.41.64...

Last usable 200.129.41.1110 ^ 0000subnet address: 200.129.41.224

The subnet mask (in binary) has:all ones in the network and subnet portion of the

addressall zeros in the host potion of the addressThe subnet mask for the previous example is:

255.255.255. 240255.255.255. 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.

Example (our subnet mask is 255.255.255.240)IP host address: 200.129. 41.23Last octet to binary: 200.129. 41.0001 0111AND subnet mask: 255.255.255.1111 0000 200.129. 41.0001 0000Subnet Address: 200.129. 41.16

So the host address 200.129. 41.23 is on the 200.129.41.16 subnet.

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.

Suppose we need 80 usable subnets, how many bits do we borrow?Remember, borrowing n bits give us:

2n – 2 subnetsTry 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)

Write it with the network octets in binary:

132.178.0000000 0.00000000

break here

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

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

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 ^ 0.00000000subnet address: 132.178.182.0

. . .

Last usable 132.178.1111110 ^ 0.00000000subnet address: 132.178.252.0

The subnet mask for this example is:255.255.254.0255.255.1111111 ^ 0.00000000

ANDing this mask with any valid host address on this network will always yield the subnet address.

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

Thank You….