IP Address concept and subnetting - (Easy Way)

IP Address Concept and subnet mask

(Easy Way)

This three serial IP is open IP

172.16.x.x -../16

192.168.x.x - ../24

10.0.0.0 - ../8

Structure of IP, subnet mask, Network, Block size

Total number of bits is 32

Bits	subnet mask	Network	Block size
24	255.255.255.0	1	256
25	255.255.255.128	2	128
26	255.255.255.192	4	64
27	255.255.255.224	8	32
28	255.255.255.240	16	16
29	255.255.255.248	32	8
30	255.255.255.252	64	4
31	255.255.255.254	128	2

There are 4 sections of IP Address

In every section there are 8 bits that’s means in 4 section 8x4= 32 bits

In class c IP address there are 3 octet’s network and 1 Octet is host

As like 192.168.100. 1 (Normally the "/ " Means 24 if nothing is mentioned like /25 or /26 etc)
Here 192.168.100 is the Network and .1 is the host and its subnet mask is 255.255.255.0

When “/” will use after the host IP then like 192.168.100.1/25 then network will divided like below diagram.

Bit will be counted from the right side

1 is sign to enlighten the circuit of the bit means when /25 given the 1 in number 7 position become active then the system can be apply as like 2⁷=128

Ser No of bit - 7 6 5 4 3 2 1 0

Activation - 1 0 0 0 0 0 0 0

Logic - 2⁷2⁰2⁰ 2⁰2⁰2⁰2⁰2⁰

Total Bit - 128 0 0 0 0 0 0 0

Total = 128 (255.255.255.128)

When “/” will use after the host IP then like 192.168.100.1/26 then network will divided like below diagram.

Ser No of bit - 7 6 5 4 3 2 1 0

Activation - 1 0 0 0 0 0 0 0

Logic - 2⁷2⁰2⁰ 2⁰2⁰2⁰2⁰2⁰

Total Bit - 128 64 0 0 0 0 0 0

Total = 192 (255.255.255.192)

When “/” will use after the host IP then like 192.168.100.1/29 then what is its subnet mask, Network, hosts and block size

So, subnet mask will be

This number will be count as 29-24=5 so five circuits will become on

Which circuits will not become ON that will be still 0 figures

7 6 5 4 3 2 1 0

1 1 1 1 1 0 0 0

2⁷2⁶2⁵ 2⁴2³2⁰2⁰2⁰

128 64 32 16 8 0 0 0 Total = 248

So subnet mask is (255.255.255.248)

Now network will be

2^{n (Formula)}

2⁵=32 so network will be 32 (Here 2⁵ come from 29-24=5)

Now the block size is

2^{n (Formula)}

2^{3(Totally opposite site of network)}=8 (Here 2³come from 32-29=3)

Then the block will be like this way

0 -7 (Total 8 IP in a block)	First block
8 -15	Second block
16 – 23	Third block
24 – 31	Forth block
32 – 39	Fifth block
40 – 47	Sixth block
48 – 55	Seventh block
56 – 63	Eighth block
Continue……	Continue……
Continue……	Continue……
Continue……	Continue……
As the same way the IP will go to 32 separate network the last range will be the (248 -255)	And the Last block

And the valid hosts are

2ⁿ-2 (Logic)

2³-2

8-2

6 (that means here is 6 users)

How to make the IP address binary

As like we will make 203 to binary

Divided by	Divided to	Extras The binaries	Comments
2	203 – 1	1	All Odd number will be (minus)-1
2	101 – 1	1
2	50 -0	0
2	25 – 1	1
2	12 – 0	0
2	6 – 0	0
2	3 – 1	1
	1	1	Binary will count from below
The binary is	-	11001011	As the same way all binary can be find out