Subnetting of Class

Content
1. Subnetting of Class C
2. Subnetting of Class B
3. Subnetting of Class A

Subnetting of Class C
1. 194.168.5.0/24
Here 194 in 1st octet indicate that this is Class C IP address
/24 represents the subnet Mask i.e. 255.255.255.0
194.168.5 .0
Network bits Host Bits
No .of Host bits=8
2^8=256 Total IP address
2^8-2= 254 Total valid IP address

Now do subnetting of this IP address
1. 194.168.5.00000000
If we borrow 1 bit form Host bits to Network bits.
194.168.5.0 0000000
25 Network Bits 7 Host bits
● How many Subnetworks are created?
2^n where n is the no. of bits borrow from host by network.
So, 2^1= 2
Thus, 2 subnetwork are created by borrowing one bit from host to
network. Each subnetwork will contain 128 IP addresses in total

How many valid IP address will be there in one subnetwork?
2^n-2 where n indicate the no. of host bit left after borrowing a bit.
So, 2^7-2 = 128-2 = 126
There are 126 valid IP address in total.
● Subnet Mask of this IP address become: 192.168.5.0/25

2. 194.168.5.00000000
If we borrow 2 bit form Host bits to Network bits.
194.168.5.00 000000
26 Network Bits 6 Host bits
How many Subnetworks are created?
2^n where n is the no. of bits borrow from host by network.
So, 2^2= 4
Thus, 4 subnetwork are created by borrowing 2 bit from host to network.
Each subnetwork will contain 64 total IP addresses.

How many valid IP address will be there in one subnetwork?
2^n-2 where n indicate the no. of host bit left after borrowing a bit.
So, 2^6-2 = 64-2 = 62
There are 62 valid IP address in total.
● Subnet Mask of this IP address become: 192.168.5.0/26
Same way we can borrow 3,4 or so on to create subnetwork.
Bits can be borrow according to the no. of subnetwork we want to create.

Example: Find network IP, Broadcast IP,Subnet Mask and Host range of IP
address 200.200.100.0/24
Solution: It is a Class C address with subnet Mask:255.255.255.0
If we borrow 1 bit from host bits to network bit
200.200.100.0 0000000
25 Network Bits 7 Host bits
Subnetwork created: 2^n=2^1=2
Valid IP Address: 2^n-2 = 2^7-2=126

2 subnetwork created are
200.200.100.0/25 – subnet 1
200.200.100.128/25 – subnet 2
Subnet Mask: 255.255.255.128
(Last address of subnetwork is subnet mask of all other networks)

Subnetting of Class B
Subnetting of Class B is done when we require more IP than 255 IP’s in
Class C.
Class B contain 65536 IP addresses in total.
As Class B contain large no of IP’s so we count from least significant bits to
make the calculation easy.

1. 172.16.250.12/16
Here 172 in 1st octet indicate that this is Class B IP address
/16 represents the subnet Mask i.e. 255.255.0.0
172.16 .250.12
Network bits Host Bits
No .of Host bits=16
2^16=65536 Total IP address
2^16-2= 65534 Total valid IP address

Now do subnetting of this IP address:
1. 172.16.250.12
If we borrow 5 bit form Host bits to Network bits.
172.16.00000 000.00000000
21 Network Bits 11 Host bits
Subnetwork created: 2^n = 2^5 = 32
Valid IP Address: 2^n-2 = 2^11-2 = 2048-2 = 2046

32 subnetwork created are:
172.16.0.0/23 – subnet 1
172.16.8.0/23 – subnet 2
172.16.16.0/23 – subnet 3
……… ………
172.16.248.0/23 – subnet 128
Subnet Mask: 255.255.248.0
(Last address of subnetwork is subnet mask of all other networks)

Subnetting of Class A
1. 10.0.0.0/8
Here 10 in 1st octet indicate that this is Class A IP address
/8 represents the subnet Mask i.e. 255.0.0.0
10 .0.0.0
Network bits Host Bits
No .of Host bits=24
2^24=16777216 Total IP address

Now do subnetting of this IP address:
1. 10.0.0.0
If we borrow 11 bit form Host bits to Network bits.
10.00000000.000 00000.00000000
19 Network Bits 13 Host bits
Subnetwork created: 2^n = 2^11 = 2048
Valid IP Address: 2^n-2 = 2^13-2 = 4192-2 = 4190

2048 subnetwork created are:
10.0.0.0/19 – subnet 1
10.0.32.0/19 – subnet 2
10.0.64.0/19 – subnet 3
……… ………
10.0.224.0/19 – subnet n
10.1.0.0/19
10.1.32.0/19
……..
10.2.224.0./19
10.255.224.0 – Subnet 2048

Subnet Mask: 255.0.224.0
(Last address of subnetwork is subnet mask of all other networks)
Similarly we can find the host range as find in Class B and Class C

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