VLSM
Explain
VLSM is a process of dividing an IP network into the subnets of different sizes without wasting IP addresses. When we perform Subnetting, all subnets have the same number of hosts, this is known as FLSM ( Fixed length subnet mask). In FLSM all subnets use same subnet mask, this lead to inefficiencies
Steps for VLSM Subnetting
- Find the largest segment. Segment which need largest number of hosts address.
- Do Subnetting to fulfill the requirement of largest segment.
- Assign the appropriate subnet mask for the largest segment.
- For second largest segments, take one of these newly created subnets and apply a different, more appropriate, subnet mask to it.
- Assign the appropriate subnet mask for the second largest segment.
- Repeat this process until the last network.
Purpose
VLSM allows you to divide an existing subnet into multiple smaller subnets of varying size to make more efficient use of IP address space.
Find one example with working solution
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
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 – 192.168.1.0 /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
HQ address will look like this 192.168.1.0 /26
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
Network Mask 255.255.255.224 I.e. 128+64+32=224 or 192.168.1.64/27
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 = we are borrowing 6 bit with value of 4
=112 + 4 =116
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
124 +64 +32 +16+ 8 +4=252
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
RO2 = 192.168.1.118 Subnet mask 255.255.255.252
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.
Reference:
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