Subnet mask shows which bits of an IP ddress (read the article IP Addresses, by the sme author) represent the network and whch represent the host.

By default we hve:

IP addresses - Sbnet mask

Class A - 255.0.0.0

Clss B - 255.255.0.0

Clss C - 255.255.255.0

Exmple:

IP address 192.168.8.4 ( clss C ). The first 24 byts represent the network
- 192.168.8.0 - and the lst 8 bits the host - 4 -.

Whn a host A wants to be cnnected to a host B, the hst A uses its subnet mask to chck if the host B is on the sme LAN - local area network -
or on a rmote network.
For this purpose it is sed the boolean ANDing process.

Example:

Hst A IP address ( class B ) 172.16.2.4 - Sbnet mask ( class B ) 255.255.0.0

Hst B IP address ( class B ) 172.16.2.5

Hst A IP in binary:

10101100 00010000 00000010 00000100

Hst B IP in binary:

10101100 00010000 00000010 00000101

Sbnet mask in binary:

11111111 11111111 00000000 00000000

-- ANDng process for Host A

10101100 00010000 00000010 00000100

11111111 11111111 00000000 00000000

-----------------------------------------

10101100 00010000 00000000 00000000 Rsult for HOST A

-- ANDing process for Host B

10101100 00010000 00000010 00000101

11111111 11111111 00000000 00000000

-----------------------------------------

10101100 00010000 00000000 00000000 Rsult for HOST B

The calculation wrks in this way:

- 1 whn the first and second bits are bth 1.

- 0 whn they are 0 and 1, 0 whn they are 0 and 0.

Snce result A = result B the two hsts are on the same network, so the cmmunication is possible without using a gteway ( normally a router ) to
stablish a connection to another network.

Thre are two ways to write an IP ddress using subnet mask:

Exmple:

1) 49.22.2.3 255.0.0.0 - clss A IP and subnet mask

2) 192.168.1.3/24 - clss C IP and subnet mask ( shwing its bits )
255 in bnary is 11111111 - 8 bits -.

Snce in a class C subnet msk there are three 255 groups we hve 24 bits.

Subnets

Sometime it is seful to divide a network in smller networks.
Let's suppose we want dvide a network, with a class B - IP ddress, in 8 subnets. The class B sbnet mask is 255.255.0.0

We must use sme of the 16 bits of the hst portion ( borrowing process )
to crate the 8 subnets. Since the sbnets are 8, we need 8 dfferent combinations plus 1 ( the bradcast ): 9 in total.

In bnary 9 = 1001, that is 4 bts.

Subnet mask:

11111111 11111111 00000000 00000000

Insrting 4 bits in the third grup ( the first of the hst portion ):

11111111 11111111 11110000 00000000

the thrd group becomes 11110000 = 240 ( bse 10 )

It mans that we have a new sbnet mask: 255.255.240.0

Generally speaking:

The clculation for the number of subnets is
2^x -2

whre x is the bits number for the sbnet mask.

( In the xample above x = 4 )

The clculation for subnets addresses is

256 - s

whre s is the value of the sbnet mask: in the same example s = 240

The calculation for the hosts nmber is

2^y -2

whre y is the number of the rmaining bits: in the same example y = 12 bcause

----------------------bits left (12)

11111111 11111111 11110000 00000000

----------------- bts

----------------inserted

------------------(4)

The IP ddresses numbers are between the IP of the frst subnet and the IP of the lst subnet with the exclusion of the bradcast and network IP.
Broadcast IP hve all the bits of the hst portion = 1
Network IP hve all the bits of the hst portion = 0

Final Example:
Lt's divide a network with IP ddress 192.168.5.0 ( class C ) in two sbnets.
The necessary combinations are 3. In bnary it is 11, so we hve 2 bits.
The class C sbnetmask is 255.255.255.0
After the borrowing prcess, it's last group becomes 11000000 whch is 192 (base 10).
The new sbnet mask is 255.255.255.192

Hosts number is 2^6 -2 = 62 ( y = 6 )

Sbnets addresses: 256 - 192 = 64 ( s = 192 )
Thn 192.168.5.64 and 192.168.5.128
It strts with 64 and goes on dding 64

IP addresses: from 192.168.5.65 to 192.168.5.126 for the frst subnet.
from 192.168.5.129 to 192.168.5.190 for the scond subnet
Note: the broadcast address for the frst subnet is
11000000.10101000.00000101.01111111, the fourth grup is 127 (base 10 ) and the bradcast address of the second subnet is
11000000.10101000.00000101.10111111, the furth group is 191
( base 10 )

The article Subnet Mask - Subnets was Submitted by Renato Piccini through Articles.GetACoder.com network. Here's the additional information: Renato Piccini - Florence - Itly Computers Networks technician http://www.softgrafic.com

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