A brief overview of subneting IP Addressing (VLSM)

A brief overview of subneting IP Addressing (VLSM)

IP was created as a way to hide the complexity of physical addressing by creating a virtual addressing scheme that is independent of the underlying network. IP does not ensure that data is delivered to the
application in the appropriate order; that responsibility is left to upper-layer protocols such as TCP and UDP.

Note: IP is a connectionless, Network-layer protocol.

An IP address is 32 bits long. The bits can be broken down into four bytes. Each byte is expressed in decimal form and separated from other bytes by a dot (that is, x.x.x.x). This is called dotted-decimal format. Each bit within a byte carries a binary weight (starting from left to right) of 128, 64, 32, 16, 8, 4, 2, 1. If you add up these values, you get a range of 0–255 for each byte.

For example, one byte can be translated from binary format to decimal format as follows:

128 64 32 16 8 4 2 1

0 1 0 1 1 0 0 1 = 0 + 64 + 0 + 32 + 16 + 0 + 0 + 1 = 113

IP addressing has been broken down into five separate classes based on the number of maximum hosts required by the network.

IP Address Classes

Class A 0 Network Host

Class B 10 Network Host

Class C 110 Network Host

Class D 1110 Multicast Address

Class E 1111 Reserved

You can see from above figure that each address class contains a network portion and a host portion. The network portion identifies the data link that is in common with all the devices attached to that network. The host portion uniquely identifies an end device connected to the network.

IP Address Classes

Class	Decimal Value of First Byte	Purpose	Max, Hosts
Class A	0–127	Large organizations	16,777,214
Class B	128–191	Medium-sized Organizations	65,543
Class C	192–223	Small organizations	254
Class D	224–247	Multicast addresses	n/a
Class E	248–255	Experimental	n/a

Private IP Addresses

Private address space is not recognized by the Internet and can be used by anyone for use within a private network. Public address space, on the other hand, is a unique address that is assigned to a company. Within Classes A, B, and C the following ranges have been defined as private.

Starting Address Ending Address
10.0.0.0 10.255.255.255
172.16.0.0 172.31.255.255
192.168.0.0 192.168.255.255

Address Masks
The network mask is used in conjunction with an IP address to delineate the network portion of an IP address from the host portion. Each major network address within its designated class has a standard network mask:

Address Class Network Mask
Class A 255.0.0.0
Class B 255.255.0.0
Class C 255.255.255.0

A major network address can be further divided into smaller networks by using a technique called subneting. When a major network is subnetted, the address can be broken into three parts:

1. The network portion

2. The subnet portion

3. The host portion

When a network mask is varied into further subnets like this, it is commonly referred to as a Variable Length Subnet Mask (VLSM).

Cisco often represents the subnet mask by identifying the number of bits used as the mask. For example, 192.174.10.0/30 would represent network 192.174.10.0 255.255.255.252. The value of 30 represents the number of bits used for the network portion of the address; in binary format, 30 would be

255.255.255.252 = 11111111.11111111.11111111.11111100 = 30

Let’s look at another example. Given the following 170.130.0.0/21, what is the subnet mask?

21 = 11111111.11111111.11111100.00000000

The network address and mask are

170.130.0.0 255.255.248.0.

Let’s say that we want to determine the network address, the broadcast address, and the available addresses that
Correspond with the given IP address:

150.34.74.53 255.255.240.0

1. Convert the IP address and its address mask into binary format.

150.34.74.53 = 10010110 00100010 010 01010 00110101
255.255.240.0 = 11111111 11111111 11110000 00000000

2. Perform a logical AND between the IP address and the mask.
A logical AND is a digital math operation that compares two bits of data to each other. The result of the operation is as follows:

0 and 0 = 0
0 and 1 = 0
1 and 0 = 0
1 and 1 = 1

So,

Host Address: 10010110 00100010 01001010 00110101
Mask: 11111111 11111111 11110000 00000000
Logical AND Result: 10010110 00100010 01000000 00000000

3. Convert the results of the logical AND back into decimal format; this is the network address:

10010110 00100010 01000000 00000000 = 50.34.64.0

4. Calculate the broadcast address.
Remember that the network mask is used to delineate the network portion of an IP address from the host portion. Mask bits are set to 1 if the corresponding bit in the IP address should be considered part of the network address and 0 if part of the host address.

150.34.74.53 = 10010110 00100010 0100 1010 00110101
255.255.240.0 = 11111111 11111111 1111 0000 00000000
Network Bits Host Bits

To determine the broadcast address, we need to replace each bit available within the host portion of the IP address with a value of 1.

So, the broadcast address of the network for the host 150.34.74.53 is
150.34.79.255 = 10010110 00100010 0100 1111 11111111
Network Bits Host Bits

Summary:

Given the IP address and address mask: 150.34.74.53 255.255.240.0, we have determined the following:

Network Address = 150.34.64.0
Broadcast Address = 150.34.79.255
Available Addresses = 150.34.64.1–150.34.79.254 (for a total of 4,078 hosts)