Variable Length Subnet Mask (VLSM)

In previous subnetting lessons here and here all our subnets had a “fixed size”. Each subnet had the same size. For example we took a class C network 192.168.1.0 and divided it 4 blocks:

Is this really an efficient way of creating subnets? Let’s say I would have the following requirements:

• One subnet for 12 hosts.
• One subnet for 44 hosts.
• One subnet for 2 hosts (point-to-point links are a good example where you only need 2 IP host addresses).
• One subnet for 24 hosts.

I have 4 subnets so it’s no problem, but I’m still wasting a lot of IP addresses. If we use a block of 64 for our subnet where I only need 2 IP addresses I’m throwing 62 IP addresses away.

Now you might think why we could care about this because we are using a private network address (192.168.1.0) and we have plenty of space. This is true, but what about the Internet? We don’t want to throw away valuable public IP addresses.

Let’s say I want to subnet my 192.168.1.0 network in the most efficient way, let’s take another look at the requirements I just showed you:

• One subnet for 12 hosts.
• One subnet for 44 hosts.
• One subnet for 2 hosts.
• One subnet for 24 hosts.

What kind of subnets would we need to fit in these hosts? Let’s see:

• 12 hosts, the smallest subnet would be a block of 16.
• 44 hosts, the smallest subnet would be a block of 64.
• 2 hosts, the smallest subnet would be a block of 4.
• 24 hosts, the smallest subnet would be a block of 32.

Let’s create the subnets. We take our block of “256”:

And chop it into the blocks we just specified:

We just saved ourselves some valuable IP addresses, now the next thing to do is answer the following questions:

• What are the network addresses?
• What are the broadcast addresses?
• What is the subnet mask?
• What are the usable host IP addresses?

Let’s answer these questions. We’ll start with the network addresses:

• Subnet 1: (size of 64)
  • network address: 192.168.1.0
• Subnet 2: (size of 32)
  • network address: 192.168.1.64
• Subnet 3: (size of 16)
  • network address: 192.168.1.96
• Subnet 4: (size of 4)
  • network address: 192.168.1.112
• Subnet 5: (this is where the free space starts)
  • network address: 192.168.1.116

Now we can fill in the broadcast addresses:

• Subnet 1: (size of 64)
  • network address: 192.168.1.0
  • broadcast address: 192.168.1.63
• Subnet 2: (size of 32)
  • network address: 192.168.1.64
  • broadcast address: 192.168.1.95
• Subnet 3: (size of 16)
  • network address: 192.168.1.96
  • broadcast address: 192.168.1.111
• Subnet 4: (size of 4)
  • network address: 192.168.1.112
  • broadcast address: 192.168.1.115

Because we have different subnet sizes, we need to calculate the subnet mask for each subnet. To find the subnet mask you can use this trick:

256 – subnet size = subnet mask

• Subnet 1: 256 – 64 = 192 so the subnet mask is 255.255.255.192
• Subnet 2: 256 – 32 = 224 so the subnet mask is 255.255.255.224
• Subnet 3: 256 – 16 = 240 so the subnet mask is 255.255.255.240
• Subnet 4: 256 – 4 = 252 so the subnet mask is 255.255.255.252

The only thing left to do is fill in the usable host IP addresses:

• Subnet 1: (size of 64)
  • network address: 192.168.1.0
  • first host: 192.168.1.1
  • last host: 192.168.1.62
  • broadcast address: 192.168.1.63
• Subnet 2: (size of 32)
  • network address: 192.168.1.64
  • first host: 192.168.1.65
  • last host: 192.168.1.94
  • broadcast address: 192.168.1.95
• Subnet 3: (size of 16)
  • network address: 192.168.1.96
  • first host: 192.168.1.97
  • last host: 192.168.1.110
  • broadcast address: 192.168.1.111
• Subnet 4: (size of 4)
  • network address: 192.168.1.112
  • first host: 192.168.1.113
  • last host: 192.168.1.114
  • broadcast address: 192.168.1.115

Here we go, we just subnetted our 192.168.1.0 /24 by using VLSM.

Let’s try another example but this time we use a Class B 172.16.0.0 network with different requirements:

• One subnet for 340 hosts.
• One subnet for 250 hosts.
• One subnet for 31 hosts.
• One subnet for 20 hosts.
• One subnet for 8 hosts.

To solve this question first we need to determine the “block” that we require: