VLSM Subnet Mask.

VLSM

VLSM is a Variable Length Subnet Mask. VLSM allows us to use a network multiple subnet mask.

Why VLSM is required?

VLSM is needed to handle IPs properly, is to minimize IP abuse.Because many clients of a single client need an IP of one range. So through VLSM I can give clients IP as per their requirement. Looking at an example we can easily explain.

Suppose is a new company. They need a certain number of IPs for their different departments. Their IP request is:

The management team will need 100 IPs.

Sales team will need 50 IPs.

The account team will require 25 IPs.

The IT team will need 5 IPs.

And our network is - 192.168.1.0

Before doing the Practical, I have reviewed some information before. For exiting the host number = the bits that are off is 2 ^ (total number) -2

For excluding network numbers = bits that will be extra on 2 ^ (total number) of bits= 24-last writ value for subnet ID release. 

Step-1: For 100 host.

When doing VLSM, it is best to take the highest number of IPs first. The result is easy to calculate. For example, the maximum number of IPs required is 100. So for 100 host IP we have to take 2 ^ 7 = 128-2 = 126. Then the subnet mux will be – 255.255.255.128 and the network will be – 192.168.1.0/25.

Step-2: For 50 host.

The second highest number of IPs is 50. Which is what the sales team needs. So for 50 host IP's we have to take (2 ^ 6) = 64-2 = 62. So since 6 bit is used for the host then the rest of the bits is (32-6) = 26.Again since Class C since 24 is fixed, extra bits are needed (26-24) = 2. According to the above data the value of the second bit is - 192. So our subnet mask is – 255.255.255.192. And our network will be – 192.168.1.128/26 because we had block size in the previous network – 128.

Step-3: For 25 host.

The third highest number of IPs required was 25. Which will be the rage for the accounts team. In total, we need to take 25 host IPs (2 ^ 5) = 32-2=30. So since there are 5 bits used for the host then the remaining bits are (32-5) = 27. Again since Class C, extra 24 bits are required (27-24) = 3. According to the above data, the value of 3rd bit is -224. So our subnet mask is – 255.255.255.224. And our network would be – 192.168.1.192/27 because we had a bit size in the previous network. Because 128 + 64 = 192 has been used.

Step-4: For 5 host.

Finally, the maximum number of IPs required is 5. Which is a must have for IT team members. So for 5 host IP we have to take (2 ^ 3) = 8-2 = 6. So since the 3 bit is used for the host then the rest of the bits is (32-3) = 29. Again since Class C has 20 fixed, extra bits are required (29-25) = 5. According to the above data, the value of the 5th bit is 248. So our subnet mask is 255.255.255.248. And our network would be -192.168.1.224/29 because our previous network was B-Squeeze-32. Because 192+32 = 224 have been used.