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? *** 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 *** 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 1 Subnet 3 Subnet 2

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 *** 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 *** 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 *** Richard Hancock - 2010

Variable Length Subnet Masks 172.80.8.0 172.80.8.1 – 15.254 172.80.40.0 172.80.40.1 – 47.254 172.80.32.0 172.80.32.1 – 39.254 172.80.24.0 172.80.24.1 – 31.254 172.80.16.0 172.80.16.1 – 23.254 More than one subnet mask Richard Hancock - 2010

We need 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 *** 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 Richard Hancock - 2010

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 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 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) 120 hosts 60 hosts 30 hosts 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 What is 11111111.11111111.11111111.10000000 expressed in base10? Converted to decimal (base10) we get 255.255.255.128 Richard Hancock - 2010

Step 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 Use the formula (256 - the subnet mask) 256 – 128 = 128 Therefore the subnets would be 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 Richard Hancock - 2010

So far... 60 hosts (62 in total) No subnet mask 192.168.1.0/25 60 hosts (62 in total) No subnet mask 30 hosts (30 in total) Richard Hancock - 2010

Step 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? 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 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 Richard Hancock - 2010

So far... 60 hosts (62 in total) 192.168.1.128/26 192.168.1.0/25 60 hosts (62 in total) 192.168.1.128/26 30 hosts (30 in total) No subnet mask 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 25 – 2 = 30 hosts This will give us a subnet mask of 1111111.1111111.1111111.11100000 which is 255.255.255.224 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 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 Richard Hancock - 2010

Result! 192.168.1.0 60 hosts (62 in total) 192.168.1.128/26 192.168.1.0/25 60 hosts (62 in total) 192.168.1.128/26 30 hosts (30 in total) 192.168.1.192/27 192.168.1.0 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 hosts 25 – 2 =30 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 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 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. Remote E 30 hosts Remote F 30 hosts Remote G 30 hosts 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. 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. Richard Hancock - 2010

Exercise 192.168.3.0 30 hosts 6 hosts Backbone 126 hosts 6 hosts 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) 6 hosts 30 hosts 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) 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 Richard Hancock - 2010

Questions... ...are there any? Richard Hancock - 2010

End! Richard Hancock - 2010