Introduction to Networking (Yarnfield)

Variable Length Subnet Masking (VLSM)

Objectives Define VLSM Describe the difference between classful

subnetting Describe the advantages of VLSM Be able to perform VLSM operations on give IP

addresses

Classful subnetting exercise 172.80.0.0 255.255.248.0 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?

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

172.80.8.0172.80.8.1 – 15.254

172.80.40.0172.80.40.1 – 47.254

172.80.32.0172.80.32.1 – 39.254

172.80.24.0172.80.24.1 – 31.254

172.80.16.0172.80.16.1 – 23.254

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

the network

We will... Create a number of subnet masks that suit

our needs more efficiently than a classful subnetting scheme could

Example using a Class C network address

120 hosts

60 hosts

30 hosts

192.168.1.0

Process1. Find the segment with the largest number of

hosts connected to it2. Find an appropriate subnet mask for the largest

segment3. Write down the subnet addresses to fit the

subnet mask4. Take one of the newly created subnet addresses

and apply a new subnet mask to it that is more appropriate

5. Write down the subnet addresses to fit the new subnet mask

6. Repeat from step 4 for smaller segments

Example continued1. 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 (27 - 2 = 126)

Example continued2. 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

Convert this to decimal and we get 255.255.255.128

Example continued3. 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 192.168.1.0 and

192.168.1.128 (remember we can now use subnet zero!)

We can now assign 192.168.1.0/25 to accommodate the 120 segment and have 192.168.1.128 to use for the other two segments

120 hosts (126 in total)192.168.1.0/25

60 hosts (62 in total)

30 hosts (30 in total)

Example continued4. 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 which is the same as 255.255.255.192

Example continued5. 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 192.168.1.128 and 192.168.1.192 We can now use 192.168.1.128/26 for the

segment with 60 hosts

120 hosts (126 in total)192.168.1.0/25

60 hosts (62 in total)192.168.1.128/26

30 hosts (30 in total)

Example continued4. 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

1111111.1111111.1111111.11100000 which is 255.255.255.224

Example continued5. 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 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

Result

120 hosts (126 in total)192.168.1.0/25

60 hosts (62 in total)192.168.1.128/26

30 hosts (30 in total)192.168.1.192/27

192.168.1.0

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

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

Remote E 30 hosts

Remote F 30 hosts

Remote G 30 hosts

Central

Exercise 192.168.3.0

Backbone126 hosts

6 hosts

6 hosts

6 hosts

30 hosts

30 hosts

30 hosts

Questions... ...are there any?