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Subnet Masking Made Simple

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Assumptions:

It is assumed that the reader understands IP addressing and how to recognize Class A/B/C addresses.

Additionally, it is assumed that the reader is familiar with converting to and from decimal to binary.

Why subnet:

Subnetting allows a network administer to segment their network into smaller networks. An example of

this would be to have IP phones on a different network than the data traffic in order to help reduce theamount of broadcast messages to the phone.

Lets do it:

We will take a private Class B address and subnet it. Here is our address: 172.20.0.0 255.255.0.0

An administrator can borrow the last 2 octets (host) to subnet. Lets say that it has been decided that the

customer wants 8 networks. They have IP phones, printer, servers, and so on that the company wants ondifferent networks.

1. Since the administrator wants 8 networks, we need to borrow enough bits to equal 8. Nowremember this is binary, so we need 3 bits (2 to the power of 3 equals 8. Additionally, you can

count in binary (2, 4, 8= 3 bits).

2. Plot those 3 bits from left to right as shown. Remember, we are borrowing our bits from the left.This is why we are plotting them from the most significant bit.

128 64 32 16 8 4 2 1

1 1 1

3. Now add 128+64+32 to get our subnet mask. 255.255.224.04. Now that we have our subnet mask we need to plot out our addresses. Highlight the farthest bit

on the right and that will be the value we use to add.

128 64 32 16 8 4 2 11 1 1

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Subnet Masking Made Simple

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The value here is 32, so our addresses for172.20.0.0255.255.224.0 are as follows:

Start ofNetwork

Start ofAddresses

End ofAddresses

BroadcastAddress

172.20.0.0 172.20.0.1 172.20.31.254 172.20.31.255

172.20.32.0 172.20.32.1 172.20.63.254 172.20.63.255

172.20.64.0 172.20.64.1 172.20.95.254 172.20.95.255

172.20.96.0 172.20.96.1 172.20.127.254 172.20.127.255

172.20.128.0 172.20.128.1 172.20.159.254 172.20.159.255

172.20.160.0 172.20.160.1 172.20.191.254 172.20.191.255

172.20.192.0 172.20.192.1 172.20.223.254 172.20.223.255

172.20.224.0 172.20.224.1 172.20.255.254 172.20.255.255

There we have our 8 networks. Keep in mind that we would have the same results if the network

administrator would have wanted 7 networks. We would still have borrowed 3 bits since 2 bits would

equal 4 networks.

Lets do it again:

Now lets take a Class C address. The method is the same, but the outcome looks a little different. Hereis our address: 192.168.1.0 255.255.255.0

An administrator can borrow the last octet (host) to subnet. Lets say that it has been decided that the

customer wants 11 networks. They have IP phones, printer, servers, and so on that the company wantson different networks.

5. Since the administrator wants 11 networks, we need to borrow enough bits to equal 11 (or closeto it). Now remember this is binary, so we need 4 bits (2 to the power of 4 equals 16.

Additionally, you can count in binary (2, 4, 8, 16= 4 bits). We cannot use 3 bits because we

want 11 and 8 is not enough. Even though we only want 11 networks 4 bits will have to do andthe administrator will have additional networks for growth.

6. Plot those 4 bits from left to right as shown. Remember, we are borrowing our bits from the left.This is why we are plotting them from the most significant bit.

128 64 32 16 8 4 2 1

1 1 1 1

7. Now add 128+64+32+16 to get our subnet mask. 255.255.255.2408. Now that we have our subnet mask we need to plot out our addresses. Highlight the farthest bit

on the right and that will be the value we use to add.

128 64 32 16 8 4 2 1

1 1 1 1

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The value here is 16, so our addresses for192.168.1.0255.255.255.240 are as follows:

Start of

Network

Start of

Addresses

End of

Addresses

Broadcast

Address192.168.1.0 192.168.1.1 192.168.1.14 192.168.1.15

192.168.1.16 192.168.1.17 192.168.1.30 192.168.1.31

192.168.1.32 192.168.1.33 192.168.1.46 192.168.1.47

192.168.1.48 192.168.1.49 192.168.1.62 192.168.1.63

192.168.1.64 192.168.1.65 192.168.1.78 192.168.1.79

192.168.1.80 192.168.1.81 192.168.1.94 192.168.1.95

192.168.1.96 192.168.1.97 192.168.1.110 192.168.1.111

192.168.1.112 192.168.1.113 192.168.1.126 192.168.1.127

192.168.1.128 192.168.1.129 192.168.1.142 192.168.1.143

192.168.1.144 192.168.1.145 192.168.1.158 192.168.1.159

192.168.1.160 192.168.1.161 192.168.1.174 192.168.1.175

192.168.1.176 192.168.1.177 192.168.1.190 192.168.1.191

192.168.1.192 192.168.1.193 192.168.1.206 192.168.1.207

192.168.1.208 192.168.1.209 192.168.1.222 192.168.1.223

192.168.1.224 192.168.1.225 192.168.1.238 192.168.1.239

192.168.1.240 192.168.1.241 192.168.1.254 192.168.1.255

There we have our 16 networks.

Additional notes:

The only subnet masks available are ending in: 128, 192, 224, 240, 248, 252, and 254. This can beconfirmed by adding up the binary values.

128 64 32 16 8 4 2 1

1 1 1 1 1 1 1

128

+64192

+32224

+16

240+8

248

+4

252+2

254