1. Introduction to Networking (Yarnfield)
Variable Length Subnet Masking (VLSM)

2. Objectives Define VLSM
Describe the difference between classful subnetting
Describe the advantages of VLSM
Be able to perform VLSM operations on give IP addresses

3. Classful subnetting exercise
Find
The first five subnet addresses
First host, last host and broadcast of each subnet
Default gateway
How many subnets can be made?
How many hosts per subnet?

4. VLSM defined More than one subnet mask
Using classful subnetting wastes IP addresses
Why? – – – – –

5. We need An IP address to perform VLSM on
The number of hosts involved in each part of the network

6. We will...
Create a number of subnet masks that suit our needs more efficiently than a classful subnetting scheme could

7. Example using a Class C network address
60 hosts
120 hosts
30 hosts

8. Process
Find the segment with the largest number of hosts connected to it
Find an appropriate subnet mask for the largest segment
Write down the subnet addresses to fit the subnet mask
Take one of the newly created subnet addresses and apply a new subnet mask to it that is more appropriate
Write down the subnet addresses to fit the new subnet mask
Repeat from step 4 for smaller segments

9. Example continued
Find the segment with the largest number of hosts connected to it
In the example the largest segment has 120 hosts connected so we must start with this segment
To accomodate120 hosts we need to use 7 bits from the host portion of the address ( = 126)

10. Example continued
Find an appropriate subnet mask for the largest segment
If we have borrowed 7 bits for our hosts the subnet mask (in binary) will be
Convert this to decimal and we get

11. Example continued
Write down the subnet addresses to fit the subnet mask
Now we need to find the subnet addresses that this subnet mask will create
256 – 128 = 128
Therefore the subnets would be and (remember we can now use subnet zero!)
We can now assign /25 to accommodate the 120 segment and have to use for the other two segments

12. 120 hosts (126 in total)
/25
60 hosts (62 in total)
30 hosts (30 in total)

13. Example continued
Take one of the newly created subnet addresses and apply a new subnet mask to it that is more appropriate
We still have two segments to deal with and we have a new subnet address to work with of
We must start with the larger segment, which has 60 hosts
To accommodate 60 hosts we need to borrow 6 bits from the host portion of the given IP address
26 – 2 = 62 hosts
This will give us a subnet mask of which is the same as

14. Example continued
Write down the subnet addresses to fit the new subnet mask
Now we need to find the subnet addresses that this subnet mask will create
256 – 192 = 64
Therefore the new subnet addresses would be and
We can now use /26 for the segment with 60 hosts

15. 120 hosts (126 in total)
/25
60 hosts (62 in total)
/26
30 hosts (30 in total)

16. Example continued
Take one of the newly created subnet addresses and apply a new subnet mask to it that is more appropriate
We still have the segment with 30 hosts to deal with
We work this out in the same way as before
To accommodate 30 hosts we need to borrow 5 bits from the host portion of the IP address
25 – 2 = 30 hosts
This will give us a subnet mask of which is

17. Example continued
Write down the subnet addresses to fit the new subnet mask
Now we need to find the subnet addresses that this subnet mask will create
256 – 224 = 32
Therefore the new subnet addresses would be and
We can now use /27 for the segment with 30 hosts
We still have the new subnet which could be used for future growth

18. Result 192.168.1.0 60 hosts (62 in total) 192.168.1.128/26
/25
60 hosts (62 in total)
/26
30 hosts (30 in total)
/27

19. Summary
To determine the number of hosts a subnet can support use the formula 2n – 2
Always start the process with the segment with the largest amount of hosts to accommodate
Classless subnetting deals with the hosts as opposed to classful subnetting which deals more with subnets

20. Exercise 192.168.2.0/24 7 remote sites, 30 hosts each P to P links
between routers
Remote A 30 hosts
Remote B 30 hosts
Remote C 30 hosts
Remote D 30 hosts
Central
Remote hosts
25 – 2 =30
( )
256 – 224 = 32
/27 (assigned to segment)
/27 (assigned to segment)
/27 (assigned to segment)
/27 (assigned to segment)
/27 (assigned to segment)
/27 (assigned to segment)
/27 (assigned to segment)
/27
Eight subnets created. First seven give to remote sites, eight subnet re-subnetted to accommodate the P to P links.
P to P links
22 – 2 = 2
( )
256 – 252 = 4
/30 (assigned to segment)
/30 (assigned to segment)
/30 (assigned to segment)
/30 (assigned to segment)
/30 (assigned to segment)
/30 (assigned to segment)
/30 (assigned to segment)
/30 (expansion)
Eight subnets created supporting 2 IP addresses. Only seven subnets are needed, leaving one left over for expansion.
Remote E 30 hosts
Remote F 30 hosts
Remote G 30 hosts

21. Exercise 192.168.3.0 30 hosts 6 hosts Backbone 126 hosts 6 hosts
27 – 2 = 126
( )
256 – 128 = 128
/25(assigned to backbone)
/25
30 Hosts
25 – 2 = 30
( )
256 – 224 = 32
/27 (assigned to segment)
/27 (assigned to segment)
/27 (assigned to segment)
/27
6 hosts
23 – 2 =6
( )
256 – 248 = 8
/29 (assigned to segment)
/29 (assigned to segment)
/29 (assigned to segment)
/29 (expansion)
6 hosts
30 hosts

22. Questions...
...are there any?