Variable Length Subnet Masking - VLSM - is a technique that allows network administrators to divide an IP address space to subnets of different sizes, unlike simple same-size Subnetting.
A Variable Length Subnet Mask (VLSM) in a way, means subnetting a subnet. To simplify futher, VLSM is the breaking down of IP addresses into subnets (multiple levels) and allocating it according to the individual need on a network. It can also be called a classless IP addressing. A classful addressing follows the general rule that has been proven to amount to IP address wastage.
Before you can understand VLSM, you have to be very familiar with IP address structure.
The best way you can learn how to subnet a subnet (VLSM) is with examples. Lets work with the diagram below:
Looking at the diagram, we have three LANs connected to each other with two WAN links.
The first thing to look out for is the number of subnets and number of hosts. In this case, an ISP allocated 192.168.1.0/24. Class C
HQ = 50 host
RO1 = 30 hosts
RO2 = 10 hosts
2 WAN links
HQ - 192.168.1.0 /26 Network address
HQ = 192.168.1.1 Gateway address
192.168.1.2, First usable address
192.168.1.62- Last usable address. Total address space -192.168.1.2 to 192.168.1.62
192.168.1.63 will be the broadcast address (remember to reserve the first and last address for the Network and Broadcast)
HQ Network Mask 255.255.255.192 - we got the 192 by adding the bit value from the left to the value we borrowed = 128+64=192
RO1 = 30 hosts
We are borrowing 3 bits with value of 32; this again is the closest we can get to the number of host needed.
RO1 address will start from 192.168.1.64 - Network address
Now we add the 32 to the 64 we borrowed earlier = 32+64 = 96
RO1 = 192.168.1.65 Gateway address
192.168.1.66 - First usable IP address
192.168.1.94 - Last usable IP address
192.168.1.95 Broadcast address – total address space – 192.168.1.66 –192.168.1. 94
RO2 = 192.168.1.96 Network address
We borrow 4 bits with the value of 16. That’s the closest we can go.
96+16= 112
So, 192.168.1.97- Gateway address
192.168.1.98 - First usable address
192.168.1.110 - Last usable address
192.168.1.111 broadcast
Total host address space – 192.168.1.98 to 192.168.1.110
Network Mask 255.255.255.240 or 192.168.1.96 /28
WAN links from HQ to RO1 Network address will be 192.168.1.112 /30 :
HQ se0/0 = 192.168.1.113
RO1 se0/0= 192.168.1.114
Mask for both links= 255.255.255.252 ( we got 252 by adding the bits value we borrowed i.e
WAN Link 2= 112+4=116
WAN Link from HQ to RO2 Network address = 192.168.1.116 /30
HQ = 192.168.1.117 subnet mask 255.255.255.252
The first thing to look out for is the number of subnets and number of hosts. In this case, an ISP allocated 192.168.1.0/24. Class C
HQ = 50 host
RO1 = 30 hosts
RO2 = 10 hosts
2 WAN links
We will try and subnet 192.168.1.0 /24 to sooth this network which allows a total number of 254 hosts I recommend you get familiar with this table below. I never leave home without it!
Lets begin with HQ with 50 hosts, using the table above:
We are borrowing 2 bits with value of 64. This is the closest we can get for 50 hosts.
HQ = 192.168.1.1 Gateway address
192.168.1.2, First usable address
192.168.1.62- Last usable address. Total address space -192.168.1.2 to 192.168.1.62
192.168.1.63 will be the broadcast address (remember to reserve the first and last address for the Network and Broadcast)
HQ Network Mask 255.255.255.192 - we got the 192 by adding the bit value from the left to the value we borrowed = 128+64=192
HQ address will look like this 192.168.1.0 /26
We are borrowing 3 bits with value of 32; this again is the closest we can get to the number of host needed.
RO1 address will start from 192.168.1.64 - Network address
Now we add the 32 to the 64 we borrowed earlier = 32+64 = 96
RO1 = 192.168.1.65 Gateway address
192.168.1.66 - First usable IP address
192.168.1.94 - Last usable IP address
192.168.1.95 Broadcast address – total address space – 192.168.1.66 –192.168.1. 94
Network Mask 255.255.255.224 I.e. 128+64+32=224 or 192.168.1.64/27
We borrow 4 bits with the value of 16. That’s the closest we can go.
96+16= 112
So, 192.168.1.97- Gateway address
192.168.1.98 - First usable address
192.168.1.110 - Last usable address
192.168.1.111 broadcast
Total host address space – 192.168.1.98 to 192.168.1.110
Network Mask 255.255.255.240 or 192.168.1.96 /28
WAN links = we are borrowing 6 bit with value of 4
=112 + 4 =116WAN links from HQ to RO1 Network address will be 192.168.1.112 /30 :
HQ se0/0 = 192.168.1.113
RO1 se0/0= 192.168.1.114
Mask for both links= 255.255.255.252 ( we got 252 by adding the bits value we borrowed i.e
124 +64 +32 +16+ 8 +4=252
WAN Link from HQ to RO2 Network address = 192.168.1.116 /30
HQ = 192.168.1.117 subnet mask 255.255.255.252
RO2 = 192.168.1.118 Subnet mask 255.255.255.252
Subnet Prefix / CIDR | Subnet mask | Usable IP address/hosts | Usable IP addresses + Network and Broadcast address |
/26 | 255.255.255.192 | 62 | 64 |
/27 | 255.255.255.224 | 30 | 32 |
/28 | 255.255.255.240 | 14 | 16 |
/29 | 255.255.255.248 | 6 | 8 |
/30 | 255.255.255.252 | 2 | 4 |
As I mentioned earlier, having this table will prove very helpful. For example, if you have a subnet with 50 hosts then you can easily see from the table that you will need a block size of 64. For a subnet of 30 hosts you will need a block size of 32.
