subnetting

[bogoreh]

hello evryone
i am a student of Networking and will like to have your help to understand subnetting clearly. first let me know you that i know the classes (A B C D) even very well but my issue is i can do well the subnetting and have tried to understant from so many video but they are just confusing me more and more so please can any one of you explain to me thks advance. secondly let me know what i don't understand let said you have a network ID which 192.168.30.0 and you need 8 subnet so how are you going to do.

[Alberto Bucur]

Hello,
Isn't hard to calculate:
192.168.30.0 with mask 255.255.255.0 has 256 IPs with
192.168.30.0 net ID and 192.168.30.255 brodcast adress.
This is equivalent with notation 192.168.30.0/24. 24 represent the number of bits used in mask.
255.255.255.0 - 11111111.11111111.11111111.00000000 that means 24 bits.

Lets split the network.
If i want to split in 2 subnetwork i need to use 25 bits in mask
255.255.255.128 - 11111111.11111111.11111111.10000000
If i want 4 subnets i need 26 bits
255.255.255.192 - 11111111.11111111.11111111.11000000
(192 = 128 + 64)
and so on.

Now, you need 8 subnets.
For that i need 27 bits in mask
255.255.255.224 - 11111111.11111111.11111111.11100000
(last octet in mask is 128+64+32 = 224)
The mask is set and need to calculate the IPs for each subnet.
256/8 = 32 adress per subnetwork.
First subnet will be:
192.168.30.0 - 192.168.30.31 (that are first 32 adresses = 0 + 31)
192.168.30.0 is the network ID
192.168.30.31 is broadcasting address
192.168.30.1 is the first usable adress
192.168.30.30 is the last usable adress
Second subnet:
192.168.30.32 - 192.168.30.63 (second interval 32 + 31)
192.168.30.32 is the network ID
192.168.30.63 is broadcasting address
192.168.30.33 is the first usable adress
192.168.30.62 is the last usable adress
And so on...
3 subnet:
192.168.30.64 - 192.168.30.95
4 subnet:
192.168.30.96 - 192.168.30.127
5 subnet:
192.168.30.128 - 192.168.30.159
6 subnet:
192.168.30.160 - 192.168.30.191
7 subnet:
192.168.30.192 - 192.168.30.223
8 subnet(last):
192.168.30.224 - 192.168.30.255
The mask for all this subnets are 255.255.255.224.

Now, lets split last subnet in other 2 subnets.
We have in this case 192.168.30.224/27.
We need a more bit in mask to make this split.(28 bits)
The mask will be:
255.255.255.240 - 11111111.11111111.11111111.11110000
(240 = 128+64+32+16)
192.168.30.224/27 have 32 adresses
Lets split in 2 subnets:
Each subnet have 32/2 = 16 adresses
First subnet will be:
192.168.30.224 - 192.168.30.239 (224 + 15)
The second
192.168.30.240 - 192.168.30.255 (240 + 15)

I hope understand me.

[bogoreh]

first thanks to answer me but how do you know that you need 27 bits?

[Alberto Bucur]

The number of bits is given by how many bits of mask are 1.
Your initial network have mask.
255.255.255.0 - 11111111.11111111.11111111.00000000 - that means 24 bits.

255.255.255.128 - 11111111.11111111.11111111.10000000 - that means 25 bits
Your initial network is splited in 2 subnets (2^1)
255.255.255.192 - 11111111.11111111.11111111.11000000 - that means 26 bits
Your initial network is splited in 4 subnets (2^2)
255.255.255.224 - 11111111.11111111.11111111.11100000 - that means 27 bits
Your initial network is splited in 8 subnets (2^3)
255.255.255.240 - 11111111.11111111.11111111.11110000 - that means 28 bits
Your initial network is splited in 16 subnets (2^4)
And so on...
How you see, each bit copleted with 1 means a power of 2.
The supernet of our initial network have mask
255.255.254.0 - 11111111.11111111.11111110.00000000 - that means 23 bits.
and 192.168.30.0 subnet is a part of
192.168.30.0 - 192.168.31.255 subnet with 512 adresses (510 for hosts)
That are writed 192.168.30.0/23

If i want to obtain our initial subnet 192.168.30.0 - 192.168.30.255 must split the supernet in 2. That means I need to complete with 1 one more bit in mask.
255.255.255.0 - 11111111.11111111.11111111.00000000 - that means 24 bits.
Now we have 2 subnet
192.168.30.0/24
192.168.31.0/24

[bogoreh]

thanks a lot this anwers is very helpful and i really understands o if now i try to sum u i have to use the 2^n to get bits that i have borrow do i?

[Alberto Bucur]

2^n give you number of subnets
n number of bits from mask borrowed.

[bogoreh]

good by how can i get the increment ?

[Alberto Bucur]

Just read carefully what i write before and make some exercises.