Introduction to Networking
Variable Length Subnet Masking (VLSM)
What do you know?
Do you know how to convert decimal numbers to Base2
numbers and vice versa?
Do you know what a subnet is?
Do you know what VLSM stands for and what it is?
Do you know how to perform the VLSM process to
devise a logical network scheme?




***
2
Richard Hancock - 2010
Objectives
Be able to describe what subnetting is and what it’s
benefits are
Be able to define VLSM and describe what it does
Be able to describe the advantages of VLSM
Be able to perform VLSM operations on given IP
addresses




***
3
Richard Hancock - 2010
Subnetting

The process of dividing a network up into subnets and to
assign each subnet a valid network IP address, and the
hosts on that subnet valid IP addresses
Subnet 2
Subnet 1
Subnet 3
Subnet benefits
Makes larger networks more manageable
Reduces bandwidth consumption as a router must
forward packets between subnets
Can provide a level of security in the network
But most importantly, it allows you to preserve your IP
address allocation and use it more efficiently and
effectively




***
5
Richard Hancock - 2010
Types of subnetting
There are two types of subnetting:
 Classful
 Classless (VLSM)
Classful subnetting is used in older network protocols and has
various issues that reduce it’s effectiveness
Classful subnetting would not allow you to use Subnet Zero
Classless subnetting (VLSM) is a more efficient system to
preserve IP addresses and is used in modern routing protocols
Classless subnetting allows you to use Subnet Zero





***
6
Richard Hancock - 2010
Some rules
You cannot use the Network address or the Broadcast
address as a host address in either Classful or
Classless subnetting!
Once a subnet address is allocated to a subnet with
it’s subnet mask it cannot be used for subnetting again


***
7
Richard Hancock - 2010
Variable Length Subnet Masks
More than one subnet mask

172.80.32.0
172.80.32.1 – 39.254
172.80.40.0
172.80.40.1 – 47.254
172.80.24.0
172.80.24.1 – 31.254
172.80.8.0
172.80.8.1 – 15.254
172.80.16.0
172.80.16.1 – 23.254
8
Richard Hancock - 2010
We need
1.
2.
3.
An IP address to perform VLSM on
The number of segments we want to divide the major
network into
The number of hosts involved in each part of each
segment
***
9
Richard Hancock - 2010
Example using a Class C network address
192.168.1.0
60 hosts
No subnet mask
120 hosts
No subnet mask
30 hosts
No subnet mask
10
Richard Hancock - 2010
Process
1.
2.
3.
4.
5.
6.
11
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
Richard Hancock - 2010
Step 1
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
1.


How many bits would we need to make 120?
To accomodate120 hosts we need to use 7 bits from
the host portion of the address (27 - 2 = 126)

60 hosts
120 hosts
30 hosts
12
Richard Hancock - 2010
Step 2
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
11111111.1111111.1111111.1000000
2.


What is 11111111.11111111.11111111.10000000 expressed
in base10?
Converted to decimal (base10) we get 255.255.255.128

13
Richard Hancock - 2010
Step 3
3.

Write down the subnet addresses to fit the subnet mask
Now we need to find the subnet addresses that this subnet
mask will create



Therefore the subnets would be




Use the formula (256 - the subnet mask)
256 – 128 = 128
192.168.1.0 and
192.168.1.128
We can now assign 192.168.1.0/25 to accommodate the 120
segment
192.168.1.128 can be used for further subnetting for the other
two segments
14
Richard Hancock - 2010
So far...
60 hosts (62 in total)
No subnet mask
120 hosts (126 in total)
192.168.1.0/25
30 hosts (30 in total)
No subnet mask
15
Richard Hancock - 2010
Step 4
4.



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 192.168.1.128
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
1111111.1111111.1111111.11000000
Converted to decimal this will be?

16
255.255.255.192
Done
60 hosts
(62 in total)
30 hosts
(30 in total)
Richard Hancock - 2010
Step 5
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
5.


256 – 192 = 64
Therefore the new subnet addresses would be



192.168.1.128 and
192.168.1.192
We can now use 192.168.1.128/26 for the segment with
60 hosts
We have 192.168.1.192 left over to further subnet


17
Richard Hancock - 2010
So far...
60 hosts (62 in total)
192.168.1.128/26
120 hosts (126 in total)
192.168.1.0/25
30 hosts (30 in total)
No subnet mask
18
Richard Hancock - 2010
Step 4 is repeated
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
4.




25 – 2 = 30 hosts
This will give us a subnet mask of
1111111.1111111.1111111.11100000 which is
255.255.255.224

19
Richard Hancock - 2010
Step 5 is repeated
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
5.


256 – 224 = 32
Therefore the new subnet addresses would be
192.168.1.192 and 192.168.1.224
We can now use 192.168.1.192/27 for the segment with
30 hosts
We still have the new 192.168.1.224 subnet which could
be used for future growth



20
Richard Hancock - 2010
Result!
192.168.1.0
60 hosts (62 in total)
192.168.1.128/26
120 hosts (126 in total)
192.168.1.0/25
30 hosts (30 in total)
192.168.1.192/27
21
Richard Hancock - 2010
Exercise 1



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 E 30 hosts
Remote F 30 hosts
Remote G 30 hosts
22
Richard Hancock - 2010
Exercise 1 solution
Remote hosts
 25 – 2 =30 hosts
 11111111.11111111.1111111.11100000 (255.255.255.224)
 256 – 224 = 32







192.168.2.0/27 (assigned to segment)
192.168.2.32/27 (assigned to segment)
192.168.2.64/27 (assigned to segment)
192.168.2.96/27 (assigned to segment)
192.168.2.128/27 (assigned to segment)
192.168.2.160/27 (assigned to segment)
192.168.2.192/27 (assigned to segment)

192.168.2.224/27 (left for further subnetting)

Eight subnets created.
First seven give to remote sites; eighth subnet re-subnetted to
accommodate the P to P links.

23
Richard Hancock - 2010
Exercise 1 solution continued
P to P links
 22 – 2 = 2 hosts
 11111111.11111111.11111111.11111100 (255.255.255.252)
 256 – 252 = 4







192.168.2.224/30 (assigned to segment)
192.168.2.228/30 (assigned to segment)
192.168.2.232/30 (assigned to segment)
192.168.2.236/30 (assigned to segment)
192.168.2.240/30 (assigned to segment)
192.168.2.244/30 (assigned to segment)
192.168.2.248/30 (assigned to segment)

192.168.2.252/30 (expansion)

Eight subnets created supporting 2 IP addresses
Only seven subnets are needed, leaving one left over for expansion.

24
Richard Hancock - 2010
Exercise

192.168.3.0
30 hosts
6 hosts
Backbone
126 hosts
6 hosts
30 hosts
6 hosts
30 hosts
25
Richard Hancock - 2010
Exercise 2 solution
Backbone
27 – 2 = 126
11111111.1111111.11111111.10000000 (255.255.255.128)
256 – 128 = 128
192.168.3.0 /25(assigned to backbone)
192.168.3.128/25
30 Hosts
25 – 2 = 30
11111111.11111111.11111111.11100000 (255.255.255.224)
256 – 224 = 32
192.168.3.128/27 (assigned to segment)
192.168.3.160/27 (assigned to segment)
192.168.3.192/27 (assigned to segment)
192.168.3.224/27
6 hosts
23 – 2 =6
11111111.11111111.11111111.11111000 (255.255.255.248)
256 – 248 = 8
192.168.3.224/29 (assigned to segment)
192.168.3.232/29 (assigned to segment)
192.168.3.240/29 (assigned to segment)
192.168.3.248/29 (expansion)
26
Richard Hancock - 2010
Summary





Classless subnetting (VLSM) is used in most networks and uses
modern routing protocols
Subnetting is all about
 Preserving IP addresses
 Making large networks more manageable (logically)
 Preserving bandwidth
 Providing a level of security
To determine the number of hosts a subnet can support use the
formula 2n – 2
Always start the VLSM process with the segment with the largest
amount of hosts to accommodate
You cannot use the subnet address or broadcast address as a host
address
27
Richard Hancock - 2010
Questions...

...are there any?
28
Richard Hancock - 2010
End!
29
Richard Hancock - 2010