Subnetting (2)
Last updated
Last updated
No. of subnets = 2^x, where x = no. of borrowed bits.
If you want 5 subnets, you'll need to use /27 mask.
/27 mask has 3 borrowed bits which gives 8 subnets (2^3).
Divide 192.168.255.0/24 into 5 subnets of equal size.
Find the number of subnets from the previous step Number of subnets
To find the subnets, consider the last octet.
You have 3 borrowed bits and 5 host bits.
The 3rd borrowed bit from the left corresponds to 32nd place in the octet.
Now, add 32 to the first address (first subnet) to get the next subnet and so on.
192.168.255.0/27, ..255.32/27, ..255.64/27, ..255.96/27, ..255.128/27
However there will be 3 other subnets as /27 prefix length allows upto 8 subnets.
What subnet does host 192.168.5.57/27 belong to?
Convert it into dotted decimal.
Its last octet is 00111001.
The first 3 are the borrowed bits since /27 mask.
To find the network or the subnet the host is in, change all the host bits to 0.
Convert back to dotted decimal. It leaves us with the last octet of 00100000 which equals 32.
This means the host belongs to the subnet 192.168.5.32/27.
Another example : 192.168.29.219/29 belongs to 192.168.29.216/29 subnet.
/25
2
126
/26
4
62
/27
8
30
/28
16
14
/29
32
6
/30
64
2
/31
128
0(2)
/32
256
0(1)
Each bit you borrow doubles the number of subnets.
Class B has more host bits than Class C, which means many more possible subnets can be created with a Class B network compared to Class C.
The process of subnetting Class A, B, C networks is EXACTLY the same.
Example 1
172.16.0.0/16 network. Create 80 subnets for your company's various LANs. What prefix length to use?
Use 2^x formula.
If you don't borrow any bit, you can't make a subnet. It's just one large /16 network.
If you borrow 1 bit, you can make 2 subnets, with a prefix of /17 and a mask of 255.255.128.0
Borrowing 7 bits gives us 128 subnets, with a mask of /23 and a mask of 255.255.254.0
To find the subnets, add the value of the last borrowed bit, which is 2 (because the 7th bit is 2nd place).
Some subnets are 192.168.0.0/23, 192.168.2.0/27, 192.168.4.0/27, 192.168.6.0/27 and so on.
Example 2
172.22.0.0/16 network. Divide the network into 500 subnets. Identify the prefix length to be used.
9 borrowed bits creates 512 subnets, so the prefix length would be /25, with a mask of 255.255.255.128
Every possible subnet in the 3rd octet (256) has a combination on 0 and 1 from the 1st bit of 4th octet, which makes 512 subnets.
Example 3
172.18.0.0/16 network. Company requires 250 subnets with same number of hosts per subnet. Find the prefix length to be used.
Borrowing 8 bits gives 256 subnets.
This also gives 8 host bits, which equals 254 hosts per subnet.
What subnet does host 172.25.217.192/21 belong to?
Write the address in binary.
Identify the borrowed bits.
Change the hosts to 0.
Convert back to dotted decimal.
You get 172.25.216.0/21
number of hosts in /21 subnet is 2046 and not 2044
172.30.0.0/16 network. 100 subnets with atleast 500 hosts per subnet. Find the prefix length.
Borrowing 7 bits gives 128 subnets and leaves us with 512 hosts (9 bits, 510 usable addresses).
So /23 prefix is used which has a mask of 255.255.254.0
172.21.111.201/20. Identify the subnet
Write in binary (last 2 octets : 01101111.11001001)
Change host bits to 0 (last 2 octets : 01100000.00000000)
Convert back to dotted decimal - 172.21.96.0/20
192.168.91.78/26. Find the broadcast address of the network of the address
Same prev steps
Change all host bits to 1 instead.
192.168.91.127
172.16.0.0/16 network. Divide into 4 subnets of equal size. Find network and broadcast address of the second subnet.
For 4 subnets, borrow 2 bits (2^2=4). So /18 prefix.
Network : 172.16.64.0/18
Broadcast : 172.16.127.255/18
172.30.0.0/16 network. Divide into subnets of 1000 hosts each. Find the number of subnets possible.
For 1000 hosts, 10 host bits are needed.
So 6 bits can be borrowed.
So 64 subnets (2^6).
Continued :