Class A:
0.0.0.0 - 127.255.255.255 default subnet mask: 255.0.0.0 (or /8, 8 network bits and 24 host bits)
Class B:
128.0.0.0 - 191.255.255.255 default subnet mask: 255.255.0.0 (or /16, 16 network bits and 16 host bits)
Class C:
192.0.0.0 - 223.255.255.255 default subnet mask: 255.255.255.0 (or /24, 24 network bits and 8 host bits)
Class D: (Used for multicasting, not used for IP addressing)
224.0.0.0 - 239.255.255.255
Class E: (Unused)
240.0.0.0 - 255.255.255.255
Subnetting based on the number of networks:
Example A:
Say that our assigned network is 213.213.213.0 and we need to divide it into 5 networks.
Let's consult a table for our decimal to binary conversion to find what 5 is in binary. We work with 8 bits and we start from 0, so we work from 20 to 27:
That means we need 3 bits to represent number 5 (22 is the 3rd bit, we start from 0). That also means we need to borrow 3 host bits for our network.
That means that in reality we will have 23=8 networks.
Class C subnet mask: 255.255.255.0 (/24) or in binary: 11111111 11111111 11111111 00000000
Our subnet mask will be the default mask with the number of host bits that we are required to borrow. In our case, we need to borrow 3 host bits and move them as network bits. Note that we always borrow the first available network bits, so:
Our subnet mask: 11111111 11111111 11111111 11100000
In other words : 255 255 255 224 (/27, since we have 27 network bits enabled)
So now, the number of our hosts will be the remaining host bits. In our case 32.
The number of usable hosts in our network though will be the remaining host bits minus the network address and the broadcast address, so 2 to the power of our remaining host bits minus 2.
So, in our case (25)-2=32-2=30. We will have 30 hosts in 8 networks.
A faster way to find this number is with what Cisco refers to as the interesting part. The interesting part is
the part of the subnet mask which is not 255. We subtract this number from 256 and that will be our IP address increment.
This in our case is the fourth part of the IP address. So, In our case, 256-224=32.
Another way to find our IP address increment is to find out the least number where "1" was put in our subnet mask (in our case that was the third place from the end) which corresponds to "32" if we do a lookup on our binary table.
So our first IP allocation will be 213.213.213.32. Just write the increments and fill the spaces in-between.
213.213.213.0
213.213.213.32
213.213.213.64
213.213.213.96
etc.
So our network is:
213.213.213.0 - 213.213.213.31
213.213.213.32 - 213.213.213.63
213.213.213.64 - 213.213.213.95
etc.
.....
Example B:
We are given the network 130.130.0.0 and we want to divide it into 900 networks.
The sum of our 8-bit table (128 to 1) amounts to 255 (obviously). 8 bits are not enough. We need more entries:
No need to calculate any more, we are sure that this is the correct bit, we go over after that.
So we need to borrow 10 host bits. 210=1024 networks.
This is a Class B network so its network mask looks like this:
11111111 11111111 00000000 00000000
So borrowing 10 host bits, our netmask looks like this:
11111111 11111111 11111111 11000000 (/26)
255 255 255 192
256-192=64. So we have 64 hosts per network, 62 usable ones and our IP allocation
increments go like so:
130.130.130.0
130.130.130.64
130.130.130.128
etc.
Our IP ranges are:
130.130.130.0 - 130.130.130.63
130.130.130.64 - 130.130.130.127
etc.
.....
Example C:
We have the network 10.0.0.0 and we want to split it into 500 networks.
Let's consult a table for our decimal to binary conversion to find how 500 is represented:
The sum of our 8-bit table (128 to 1) amounts to 255 (obviously). 8 bits are not enough. We need one more entry:
That means that we need to borrow 9 host bits.
This is a class A address and its network mask looks like this:
11111111 00000000 00000000 00000000
So adding 9 bits makes our netmask like so:
11111111 11111111 10000000 00000000 (/17) or:
255 255 128 0
29=512, so we have 512 networks.
256-128=128, so we have 128 hosts in total and 126 usable hosts per network,so our IP allocation increments and our ranges are the following:
10.0.0.0 - 10.0.127.255
10.0.128.0 - 10.0.255.255
10.1.0.0 - 10.1.127.255
10.1.128.0 - 10.1.255.255
10.2.0.0 - 10.2.127.255
10.2.128.0 - 10.2.255.255
.......
Now that we've gone through the basics and learned how to subnet, we need to note:
It's always a good idea to subtract one when trying to subnet based on networks just in case our number is on a binary calculation boundary (e.g. 128,64,32,16,8,4,2,1).
Let's go over an example again with our new method:
We have the network 192.168.0.0 and we need to divide it into 16 networks.
We need to subtract one so instead of calculating for 16, we calculate for 16-1=15:
We need 4 bits, so 24; that means we will have 16 networks.
Class C so:
11111111 11111111 11111111 11110000
255 255 255 240 or /28 in CIDR notation.
And 24=16 IP addresses and 14 usable hosts. Our IP allocation is arranged by 256-240=16 increments.
192.168.0.0 - 192.168.0.15
192.168.0.16 - 192.168.0.31
192.168.0.32 - 192.168.0.63
etc.
If we had done the calculation without subtracting one, we would have provisioned for 32 networks.
0.0.0.0 - 127.255.255.255 default subnet mask: 255.0.0.0 (or /8, 8 network bits and 24 host bits)
Class B:
128.0.0.0 - 191.255.255.255 default subnet mask: 255.255.0.0 (or /16, 16 network bits and 16 host bits)
Class C:
192.0.0.0 - 223.255.255.255 default subnet mask: 255.255.255.0 (or /24, 24 network bits and 8 host bits)
Class D: (Used for multicasting, not used for IP addressing)
224.0.0.0 - 239.255.255.255
Class E: (Unused)
240.0.0.0 - 255.255.255.255
Subnetting based on the number of networks:
Example A:
Say that our assigned network is 213.213.213.0 and we need to divide it into 5 networks.
Let's consult a table for our decimal to binary conversion to find what 5 is in binary. We work with 8 bits and we start from 0, so we work from 20 to 27:
| 27 | 26 | 25 | 24 | 23 | 22 | 21 | 20 |
|---|---|---|---|---|---|---|---|
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 |
That means we need 3 bits to represent number 5 (22 is the 3rd bit, we start from 0). That also means we need to borrow 3 host bits for our network.
That means that in reality we will have 23=8 networks.
Class C subnet mask: 255.255.255.0 (/24) or in binary: 11111111 11111111 11111111 00000000
Our subnet mask will be the default mask with the number of host bits that we are required to borrow. In our case, we need to borrow 3 host bits and move them as network bits. Note that we always borrow the first available network bits, so:
Our subnet mask: 11111111 11111111 11111111 11100000
In other words : 255 255 255 224 (/27, since we have 27 network bits enabled)
So now, the number of our hosts will be the remaining host bits. In our case 32.
The number of usable hosts in our network though will be the remaining host bits minus the network address and the broadcast address, so 2 to the power of our remaining host bits minus 2.
So, in our case (25)-2=32-2=30. We will have 30 hosts in 8 networks.
A faster way to find this number is with what Cisco refers to as the interesting part. The interesting part is
the part of the subnet mask which is not 255. We subtract this number from 256 and that will be our IP address increment.
This in our case is the fourth part of the IP address. So, In our case, 256-224=32.
Another way to find our IP address increment is to find out the least number where "1" was put in our subnet mask (in our case that was the third place from the end) which corresponds to "32" if we do a lookup on our binary table.
So our first IP allocation will be 213.213.213.32. Just write the increments and fill the spaces in-between.
213.213.213.0
213.213.213.32
213.213.213.64
213.213.213.96
etc.
So our network is:
213.213.213.0 - 213.213.213.31
213.213.213.32 - 213.213.213.63
213.213.213.64 - 213.213.213.95
etc.
.....
Example B:
We are given the network 130.130.0.0 and we want to divide it into 900 networks.
The sum of our 8-bit table (128 to 1) amounts to 255 (obviously). 8 bits are not enough. We need more entries:
| 1024 | 512 | ................. |
|---|---|---|
| 0 | 1 | ................. |
No need to calculate any more, we are sure that this is the correct bit, we go over after that.
So we need to borrow 10 host bits. 210=1024 networks.
This is a Class B network so its network mask looks like this:
11111111 11111111 00000000 00000000
So borrowing 10 host bits, our netmask looks like this:
11111111 11111111 11111111 11000000 (/26)
255 255 255 192
256-192=64. So we have 64 hosts per network, 62 usable ones and our IP allocation
increments go like so:
130.130.130.0
130.130.130.64
130.130.130.128
etc.
Our IP ranges are:
130.130.130.0 - 130.130.130.63
130.130.130.64 - 130.130.130.127
etc.
.....
Example C:
We have the network 10.0.0.0 and we want to split it into 500 networks.
Let's consult a table for our decimal to binary conversion to find how 500 is represented:
The sum of our 8-bit table (128 to 1) amounts to 255 (obviously). 8 bits are not enough. We need one more entry:
| 256 | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 |
That means that we need to borrow 9 host bits.
This is a class A address and its network mask looks like this:
11111111 00000000 00000000 00000000
So adding 9 bits makes our netmask like so:
11111111 11111111 10000000 00000000 (/17) or:
255 255 128 0
29=512, so we have 512 networks.
256-128=128, so we have 128 hosts in total and 126 usable hosts per network,so our IP allocation increments and our ranges are the following:
10.0.0.0 - 10.0.127.255
10.0.128.0 - 10.0.255.255
10.1.0.0 - 10.1.127.255
10.1.128.0 - 10.1.255.255
10.2.0.0 - 10.2.127.255
10.2.128.0 - 10.2.255.255
.......
Now that we've gone through the basics and learned how to subnet, we need to note:
It's always a good idea to subtract one when trying to subnet based on networks just in case our number is on a binary calculation boundary (e.g. 128,64,32,16,8,4,2,1).
Let's go over an example again with our new method:
We have the network 192.168.0.0 and we need to divide it into 16 networks.
We need to subtract one so instead of calculating for 16, we calculate for 16-1=15:
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
We need 4 bits, so 24; that means we will have 16 networks.
Class C so:
11111111 11111111 11111111 11110000
255 255 255 240 or /28 in CIDR notation.
And 24=16 IP addresses and 14 usable hosts. Our IP allocation is arranged by 256-240=16 increments.
192.168.0.0 - 192.168.0.15
192.168.0.16 - 192.168.0.31
192.168.0.32 - 192.168.0.63
etc.
If we had done the calculation without subtracting one, we would have provisioned for 32 networks.
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