IP addressing and Subnetting:

To fully understand Subnetting, it is vital to know how many network and host bits are present in Class A, Class B, and Class C networks.

               

	First Octet (Range)	No. of Network Bits	No. of Host Bits	Default Mask
Class A	1 – 126	8	24	255.0.0.0
Class B	128 – 191	16	16	255.255.0.0
Class C	192 – 223	24	8	255.255.255.0

 

Network Address 17.1.1.1 broken down into network and host bits.

Dotted decimal	17	1	1	1
Octet in binary	0001001	00000001	00000001	00000001
Network / Host Bits	Network	Host	Host	Host

 

SUBNETTING:

Subnetting is simply a process of “borrowing” host bits in order to create the subnet portion of an address. Note in the following examples that the network portion of the address never changes. The subnet field always borrows from the host bits.

 

Why Use Subnetting?

Subnetting is a highly effective method of conserving  IP address. Consider a point to point ISDN connection with two host addresses, one on each side of the connection. Using an entire Class C address range for this network segment would be a waste of addresses. A default Class C network mask of 255.255.255.0 yields 254 usable host addresses, but only two are needed for this small network.

 

Subnetting allows the use of a “tighter” subnet mask that the default; that is, one that yields a smaller amount of network addresses. The benefit is that the address that would have been wasted are now still usable by other segments of the network.

 

 

Determining the number of valid subnets

To determine the number of valid subnets for a given number and mask, use this formula:
Number of subnets = (2 squared by the number of subnet bits) – 2

The number of subnet bits is determined by examining the default network mask for that class, and comparing it to the actual network mask. Taking network 172.20.20.0 255.255.255.0 as an example, the default mask for the class B network is 255.255.0.0. write out the default mask and the actual mask in binary:

	1st octet	2nd octet	3rd octet	4th octet
Default mask 255.255.0.0	1111111	1111111	0000000	0000000
Subnet mask 255.255.255.0	1111111	1111111	1111111	0000000
Type and number of bits	all network bits	all network bits	all subnet bits	all host bits

 

TRICK:

No. of subnet bits = network bits in (subnet mask) – network bits in (default subnet mask)
No. of hosts bits = 32 – no. of network bits in subnet mask (CIDR value)

Suppose:
x= no. of subnet bits     
y = no. of host bits

No. of valid subnet = 2x -2
No. of host per subnet = 2y – 2
Total no. of host = no. of valid subnet X no. of host per subnet

For example:
i) 172.16.15.0/24               (stated in prefix notation and /24 is called CIDR value)
x = 24 – 16 = 8                    
y= 32 – 24 = 8
No. of subnets = 28 – 2 = 254
No. of host per subnet= 28 – 2 = 254

ii) 10.16.15.0/24
x= 24 – 8 = 16
y= 32 – 24 = 8
No. of subnets = 28 – 2 = 254
No. of host per subnet = 28 – 2 = 254

iii) 192.168.1.64/27
x= 27 – 24 = 3
y= 32 – 27 = 5
No. of subnets = 23 – 2 = 6
No. of host per subnet = 25 – 2 = 30

 

Determining the subnet number (subnet address) of a given IP address and subnet mask
is accomplished by performing a Boolean AND operation.

First, the IP Address and its subnet mask will be converted to binary. The Boolean AND is simply a bit-by-bit comparison of the IP Address and the Subnet Mask.

1 + 1 = 1
0 + 1 = 0
0 + 0 = 0                                               
(if any value is 0 then result 0)

 

Consider Network Address 178.56.21.9/24. Convert the IP address and the subnet mask into binary, remembering that /24 is equivalent to 255.255.255.0

	1st octet	2nd octet	3rd octet	4th octet
IP Address 178.56.21.9	10110010	00111000	00010101	00001001
Subnet mask 255.255.255.0	1111111	1111111	1111111	00000000
AND Result	10110010	00111000	00010101	00000000
AND Result (decimal)	172	56	21	0

So, Subnet Address of 178.56.21.9/24 is 178.56.21.0/24

 

Example: What is the subnet address of 200.154.150.89/27

	1st octet	2nd octet	3rd octet	4th octet
IP Address 200.154.150.89	11001000	10011010	10010110	01011001
Subnet mask 255.255.255.224	1111111	1111111	1111111	11100000
AND Result	11001000	10011010	10010110	01000000
AND Result (decimal)	200	154	150	64

So, Subnet Address of 200.154.150.89/27 is 200.154.150.64/27

 

Host Range in a given subnet 200.154.150.64/27
Y = 32 – 27 = 5

Total no of valid host in a given subnet = 2y-2 = 30

First IP Address (in the subnet) (When all hosts bit are 0)	Valid Host Range (in the subnet)	Last IP Address (in the subnet) (When all host bits are 1)
200.154.150.64	200.154.150.65 to 200.154.150.94	200.154.150.95
200.154.150.96 (next n/w 64+32)

 

 

Meeting Stated Design Requirements:

Example:
Your network uses Class B 165.10.0.0. You need at least 150 subnets that have no more than 200 hosts apiece. What subnet mask should you use?

 

Solution:

No. of subnet ≥ 150
2X – 2 ≥ 150                          (so x = 8)

No. of hosts per subnet ≤ to 200
2Y – 2 ≤ 200                          (so x = 7)

As 165.10.0.0 is a Class B address with default mask 255.255.0.0. 16 bits are being used for network and leaving another 16 bits to be divided between subnet and host bits.

As no. of Hosts per subnet should not be more than 200, and No. of subnet can be more than 150. We are not going to use more than 7 bits for the hosts, it means we have 9 bits for the subnet.

So x = 9

so the subnet mask you should use is 255.255.255.128 or /25