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Step 1: Binary IP Calculation

Step 2: Old-school IP Classes

Step 3: Subnet Masking

Step 4: Subnetting

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STEP 3: SUBNET MASKING

If you finished step 1 and 2, you now know that every IP address consists of 4 octets, 8 bits each and that every IP address,

depending on the Class, is separated into a network part and a hosts part.

When working with big network environments like the internet or major WAN's (Wide Area Networks), the standard type of

addressing can deplete the number of available IP addresses quickly. This is caused by i.e. the limitation of topologies, which

is the number of host per segment. The standard addressing of a Class A network, would allow for 126 networks of

16,777,214 hosts. This makes a total of 2,113,928,964 (2 billion) hosts. For a network like the internet, neither the number

of hosts or the number of networks is enough.

For this problem, subnetting can be a solution. Subnetting uses the available host bits and divides this part into a subnet part

and a hosts part. So, the IP address is split up in a network, subnet and host part. This is used for administrative ease, i.e.

by giving every department or geographical location it's own subnet. It also allows for mixing of different technologies such

as Token Ring and EtherNet. Another advantage is the reduction of network traffic by segmenting the network.

The bits / octet for determining the subnet are located at the following positions (Bold marks the spot):

Class Location of subnets octet

A W.X.Y.Z

B W.X.Y.Z

C W.X.Y.Z

Subnetting, or splitting the hosts part, is possible by defining a subnet mask. To start your way in determining the subnet

mask that is most suitable for you, determine the number of required subnets. Possible number of subnets are none, 2, 6,

14, 30, 62, 126 or 254. Convert the number of required subnets to b inary. For example:

30 subnets = 00011110

Now, count from right to left the numbers until you reach the last bit set to one. In this case that is five bits. To form a

subnet from this number, apply the number of bits on the subnet mask, from left to right. So, that would look like this:

5 bits from left to right = 11111000 binary = 248 in decimal.

By coverting the outcome to decimal value, you have determined the subnet mask bits. for a Class B IP address, to obtain 30

subnets, the subnet mask would be 255.255.248.0

A much faster way is to consult a list that shows you the right mask for the possible subnets, shown per Class, like the

following list:

Class A Table

Subnets Binary Mask Hosts

0 10000000 255.128.0.0 Invalid subnet

2 11000000 255.192.0.0 4,194,302

6 11100000 255.224.0.0 2,097,150

14 11110000 255.240.0.0 1,048,574

30 11111000 255.248.0.0 524,286

62 11111100 255.252.0.0 262,142

126 11111110 255.254.0.0 131,070

254 11111111 255.255.0.0 65,534

Class B Table

Subnets Binary Mask Hosts

0 10000000 255..255.128.0 Invalid subnet

2 11000000 255.255.192.0 16,382

6 11100000 255.255.224.0 8,190

14 11110000 255.255.240.0 4,094

30 11111000 255.255.248.0 2,046

62 11111100 255.255.252.0 1,022

126 11111110 255.255.254.0 510

254 11111111 255.255.255.0 254

Class C Table

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Subnets Binary Mask Hosts

0 10000000 255.255.255.128 Invalid subnet

2 11000000 255.255.255.192 62

6 11100000 255.255.255.224 30

14 11110000 255.255.255.240 14

30 11111000 255.255.255.248 6

62 11111100 255.255.255.252 2

126 11111110 255.255.255.254 Invalid subnet

254 11111111 255.255.255.255 Invalid subnet

The different Classes all contain one or more invalid subnet remarks. These options are invalid because when used they

would give all one's or all zero's. All one's or all zero's is not possible, because when ANDed, they would leave no subnet, or

no hosts. Try to AND one of the Invalid subnet options and see what happens.

It is a general rule that the network and the hosts ID can not be all zero, which would indicate the 'Local network', or cannot

all be one, which would indicate the broadcast address.

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