IP SUBNETTING PART-1

IP Subnetting

Subnetting is a process of logical grouping of an IP network into two or more subnetworks.

Why subnetting is required?

1. Improved network performance due to less network congestion

2. Better Administration

3. Improved network security

4. Better utilization of IP address

Classful and Classless IP Address

In Classful IP Address, an IP address must use the default subnet mask for that IP address.For example, 120.10.0.0 is an IP address of class A and the default subnet mask for class A is 255.0.0.0. This subnet mask uses a total of 8 bits to define a network. It would be expressed in network notation as 120.10.0.0/8.

Class A IP address range : 1.0.0.0 to 126.255.255.255

Class B IP address range : 128.0.0.0 to 191.255.255.255

Class C IP address range : 192.0.0.0 to 223.255.255.255

                 Default Subnet Mask

Class	Octet
	1st	2nd	3rd	4th
A	255	0	0	0
	11111111	00000000	00000000	00000000
B	255	255	0	0
	11111111	11111111	00000000	00000000
C	255	255	255	0
	11111111	11111111	11111111	00000000

In Classless IP Address, an IP address allows to use Variable Length Subnet Mask (VLSM). As an example of a custom subnet mask, 210.10.20.0 is an IP address of class C and the default subnet mask for class C is 255.255.255.0. Suppose the custom subnet mask uses a total of 30 bits to define the subnetwork. It would be expressed in network notation as 210.10.20.0/30.

Convert Binary to Decimal
00000000	07+06+05+04+03+02+01+00	0+0+0+0+0+0+0+0	0
10000000	27+06+05+04+03+02+01+00	128+0+0+0+0+0+0+0	128
11000000	27+26+05+04+03+02+01+00	128+64+0+0+0+0+0+0	192
11100000	27+26+25+04+03+02+01+00	128+64+32+0+0+0+0+0	224
11110000	27+26+25+24+03+02+01+00	128+64+32+16+0+0+0+0	240
11111000	27+26+25+24+23+02+01+00	128+64+32+16+8+0+0+0	248
11111100	27+26+25+24+23+22+01+00	128+64+32+16+8+4+0+0	252
11111110	27+26+25+24+23+22+21+00	128+64+32+16+8+4+2+0	254
11111111	27+26+25+24+23+22+21+20	128+64+32+16+8+4+2+1	255

Example-1
210.10.20.56/30

Step 1: Identify the Class of the above IP Address.
             Here, 210.10.20.56 belongs to Class C

Step 2: Find the default subnet mask for that Class of IP address.
             Here, Default subnet mask for Class C is 255.255.255.0
                IP Address                    : 210.10.20.56
                Default Subnet Mask    : 255.255.255.0

Step 3: Identify the numbers of required network (1s) bits.
For the IP Address 210.10.20.56/30 the required network bits is 30.

Step 4: Find total number of sub networks.
Number of sub networks,  2X      
[x:numbers of network(1s)bits from host part of default sub netmask]
                                        =  26
                                        =  64

Step 5: Find the number of usable hosts in each sub networks.
Number of valid hosts per sub network, 2y - 2   
[y :numbers of host(0s)bits from new sub netmask]
                                                               = 22 - 2
                                                               = 4 - 2
                                                               = 2

Step 6: Find total number of IP address in each sub networks.
Block size of per subnetwork= 28 - new sub netmask
                                               =256 - 252
                                               =4

The blocks will be:

0,(0+4)=4,(4+4)=8,(8+4)=12,(12+4)=16,(16+4)=20,(20+4)=24,(24+4)=28,(28+4)=32,(32+4)=36, (36+4)=40,(40+4)=44,(44+4)=48,(48+4)=52,(52+4)=56,(56+4)=60,(60+4)=64,(64+4)=68,(68+4)=72,(72+4)=76,(76+4)=80,(80+4)=84,(84+4)=88,(88+4)=92,(92+4)=96,(96+4)=100,(100+4)=104,(104+4)=108,(108+4)=112,(112+4)=116,(116+4)=120,(120+4)=124,(124+4)=128, (128+4)=132,(132+4)=136,(136+4)=140,(140+4)=144,(144+4)=148,(148+4)=152,(152+4)=156, (156+4)=160,(160+4)=164,(164+4)=168,(168+4)=172,(172+4)=176,(176+4)=180,(180+4)=184, (184+4)=188,(188+4)=192,(192+4)=196,(196+4)=200,(200+4)=204,(204+4)=208,(208+4)=212, (212+4)=216,(216+4)=220,(220+4)=224,(224+4)=228,(228+4)=232,(232+4)=236,(236+4)=240, (240+4)=244,(244+4)=248,(248+4)=252,(252+4)=256.

1st Sub network
210.10.20.0---- Network IP address for the 1st sub network
210.10.20.1- - -Valid Host IP
210.10.20.2- - -Valid Host IP
210.10.20.3---- Broadcast IP address for the 1st sub network

2nd Sub network
210.10.20.4---- Network IP address for the 2nd sub network
210.10.20.5- - -Valid Host IP
210.10.20.6- - -Valid Host IP
210.10.20.7---- Broadcast IP address for the 2nd sub network

Example-2
Given address 192.10.20.19/28, find the valid host addresses on this subnet.

Step 1: Identify the Class of the above IP Address.
             Here, 192.10.20.19 belongs to Class C

Step 2: Find the default subnet mask for that Class of IP address.
             Here, Default subnet mask for Class C is 255.255.255.0
                IP Address                    : 192.10.20.19
                Default Subnet Mask    : 255.255.255.0

Step 3: Identify the numbers of required network (1s) bits.
For the IP Address 192.10.20.19/28 the required network bits is 28.

Default Subnet Mask: 255.255.255.0
                          =11111111.11111111.11111111.00000000
(By default class C uses 24 network bits.So, we need to borrow 4 network bits)
                          =11111111.11111111.11111111.11110000
                          =255.255.255.240

Step 4: Find total number of sub networks.
Number of sub networks,  2X      
[x:numbers of network(1s)bits from host part of default sub netmask]
                                        = 24
                                        = 16

Step 5: Find the number of usable hosts in each sub networks.
Number of valid hosts per sub network, 2y - 2   
[y :numbers of host(0s)bits from new sub netmask]
                                                              = 24 - 2
                                                              = 16 - 2
                                                              = 14

Step 6: Find total number of IP address in each sub networks.
Block size of per subnetwork= 28 - new sub netmask
                                               =256 - 240
                                               =16

The blocks will be:
0, (0+16)=16, (16+16)=32, (32+16)=48,- - - - 256.

Required Sub network
210.10.20.16---- Network IP address for the 3rd sub network
210.10.20.17- - - -
210.10.20.18       |
                  |         |
                  |         |  Valid host
210.10.20.29       |
210.10.20.30- - - -
210.10.20.31---- Broadcast IP address for the 3rd sub network