Subnet Calculation from a given IP range, using the classless Subnet mask. Calculating number of hosts in a subnet and number of subnets possible to create in a given IP range.
2. UNDERSTANDING SUBNETTING
• First let us take a short view at what are the subnets of classes
A, B, and C
• Class A subnet: 255.0.0.0 /8
which means 11111111.00000000.00000000.00000000
• Class B subnet: 255.255.0.0 /16
which means 11111111.11111111.00000000.00000000
• Class C subnet: 255.255.255.0 /24
which means 11111111.11111111.11111111.00000000
Whenever an IP is given (generally), it belongs to either of the three
i.e., Class A, B or C
But we need to attach multiple networks within the same IP range.
Thus we need to create subnets under the given IP.
3. CLASSLESS AND CLASSFUL
• In the previous slide you have seen that there are
three different subnet masks for Class A, B and C.
These masks are called Classful because they belong
to either class.
• A classless subnet mask can be created using subnet
of any class but a classless subnet does not belong to
any ‘Class’ (hence classless).
• A classless subnet exists only inside a private
network e.g., inside an office network or a college
network, etc.
4. CLASSLESS SUBNET MASK
• As mentioned previously, a classless subnet mask
is used to create a series of ‘subnets’ using one
network
Subnet 1
INTERNET
CLASS C IP RANGE
192.168.0.5/24
(/24 = 255.255.255.0)
Subnet 2
Subnet N
N subnets
5. EXAMPLE PROBLEM WITH SOLUTION
• Now with an example problem I will Illustrate the subnetting
process.
Problem
Given IP range: 172.168.0.2 /16
There are 14 departments in the office.
Create a network for each department and
determine how many PCs can be attached in
each department network
6. EXAMPLE PROBLEM WITH SOLUTION
From the given problem, there need to be 14 subnets.
The subnet mask is 255.255.0.0 (8+8=16 /16)
Which means the last two octets are usable.
(1 octet = 8bits
=> usable bits = 32-16 = 16 i.e., 2 octets)
Thus we need to create 14 subnets.
To create subnets the formula for number of bits used is = 2n-2 (where
n is the number of bits required)
Here the scenario is,
=> (2^n) – 2 = 14
=> n = log2(14+2) = 4 bits
7. EXAMPLE PROBLEM WITH SOLUTION
Thus, we need 4 bits to create 14 subnets.
These 4 bits will be the first 4 bits of the 1st usable Octet.
To create the Classless subnet mask, now we need to convert the first
4 bits of the 3rd Octet into 1’s. Hence the new Subnet (classless) will
look as follows: Network bits
Host bits
11111111.11111111.11110000.00000000
Which makes it 255.255.240.0 from 255.255.0.0 (/16)
i.e., prefix = /(8+8+4) = /20
From now on, subnet ID will be denoted with the 4 red bits only
Please note that the first and the last network are unusable as per
norms i.e., subnet 0000 and subnet 1111
Thus we are left with 0001 to 1110 usable networks i.e., 14 subnets
8. EXAMPLE PROBLEM WITH SOLUTION
Now we need to determine how many hosts can be attached to each subnet. This is very easy
and the formula remains the same.
From the last two octets we see that there are 4 network bits and (4+8) host bits.
Host bits = 12
No of hosts attachable = 2^12 – 2 = 4096 – 2
No. of hosts = 4094
The first and the last addresses are Network address and Broadcast address respectively.
Thus, first subnet will have the following:
subnet id = 0001
subnet mask
= 255.255.240.0
network address
= 172.168.16.0
broadcast address
= 172.168.31.255
first host
= 172.168.16.1
last host
= 172.168.31.254
[(00010000)b = (16)d]
[(00011111)b = (31)d]
9. EXAMPLE PROBLEM WITH SOLUTION
subnet id = 0010
subnet mask
network address
broadcast address
first host
last host
= 255.255.240.0
= 172.168.32.0
= 172.168.47.255
= 172.168.32.1
= 172.168.47.254
subnet id = 0011
subnet mask
network address
broadcast address
first host
last host
= 255.255.240.0
= 172.168.48.0
= 172.168.63.255
= 172.168.48.1
= 172.168.63.254
[(00010000)b = (16)d]
[(00011111)b = (31)d]
[(00010000)b = (16)d]
[(00011111)b = (31)d]
and so on…