Subnetting Made Easy

Hi all,

I've received an email from one of your members asking me to post up my technique for subnetting as links to external blogs are not allowed due to forum rules. I know that he benefited from it and he wishes to help out others so here goes:

First of all I need you to get rid of all of the negative thoughts surrounding subnetting. Put down all of the books that you have read about the subject and navigate away from other sites claiming to provide an easy way to subnet. This technique requires no charts, just simply the know-how to work with the powers of 2.

We need to start with the fundamentals of IP addressing. An IP address is made up of 32 bits, split into 4 octets (oct = 8, yes?). Some bits are reserved for identifying the network and the other bits are left to identify the host.

There are 3 main classes of IP address that we are concerned with.

Class A Range 0 - 127 in the first octet (0 and 127 are reserved)
Class B Range 128 - 191 in the first octet
Class C Range 192 - 223 in the first octet

Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.

NNNNNNNN . HHHHHHHH . HHHHHHHH . HHHHHHHH Class A
NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C

At each dot I like to think that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.

We will now look at typical questions that you may see on subnetting. More often than not they ask what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.

What subnet does 192.168.12.78/29 belong to?

You may wonder where to begin. Well to start with let's find the next boundary of this address.

Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.

We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-

192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
.............etc

Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.

What subnet does 172.16.116.4/19 sit on?

Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.

We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-

172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc

Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?

What subnet does 10.34.67.234/12 sit on?

Our mask is 12. Our next boundary is 16. Therefore 16 - 12 = 4. 2^4 = 16 which gives us our block size.

We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-

10.0.0.0
10.16.0.0
10.32.0.0
10.48.0.0
.............etc

Our address is 10.34.67.234 which must sit on the 10.32.0.0 subnet.

Hopefully the penny is starting to drop and you are slapping the side of your head realising that you were a fool to think that subnetting was hard. We will now change the type of question so that we have to give a particular host range of a subnet.

What is the valid host range of of the 4th subnet of 192.168.10.0/28?

Easy as pie! The block size is 16 since 32 - 28 = 4 and 2^4 = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.

192.168.10.0
192.168.10.16
192.168.10.32
192.168.10.48
192.168.10.64
.................etc

Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.

What is the valid host range of the 1st subnet of 172.16.0.0/17?

/17 tells us that the block size is 2^(24-17) = 2^7 = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-

172.16.0.0
172.16.128.0

The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).

What is the valid host range of the 7th subnet of address 10.0.0.0/14?

The block size is 4, from 16 - 14 = 2 then 22 = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.

The seventh subnet is 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remembering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255).

What if they give me the subnet mask in dotted decimal?

If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.

Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:

1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.
2. Subtract the number in that octet from 256 to get your block size e.g. above it is 256 - 248 = block size of 8.
3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).

Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.

One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.

What now?

Now it's time to go and pick up those books again and go straight to the practice questions, completely by-passing any of their techniques. Use my method and you will be laughing!

Happy subnetting!

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Comments

paulaz

Hey all. Hoping someone can help clear up what I sometimes get a little iffy on. With the following question What subnet does 10.34.67.234/12 sit on? The block size 16, so when I go to do my ranges, I just want to make sure the max is correct, that being the 3rd and 4th octet

10.0.0.0---10.15.254.255
10.16.0.0---10.31.254.255
10.32.0.0

10.47.254.255
10.48.0.0......etc

The answer is the 10.32.0.0 subnet. For that 3rd subnet, I put the max range of 10.47.254.255 but i don't exactly know why the third octet is 254. Is it correct that the subnet ID is 10.32.0.0 and the broadcast is 10.47.255.255? If so, I know that you cannot use the subnet ID and broadcast, but what is the "rule" that says the max third octet is 254? If I'm correct here, I'm only correct because I have seen it enough times but I don't why I'am right. Hopefully that makes sense.

therealone

Hi all, thanks for this awesome technique. It really helped me.
However, I stumbled upon a problem which I don't get it:

10.101.99.17 -- 255.255.254.0

So, we are in 3rd octet, right? 256-254 = 2, 2^2 = 4.

Incrementing third octet:
10.101.0.0
10.101.4.0
...
10.101.92.0
10.101.96.0
10.101.100.0

Therefore, 10.101.99.17 falls between 100 and 96 one, so it must be 10.101.96.0 the Network Address?
But it is wrong? Calculators say it must be 10.101.98.0?

Btw, my broadcast address (next NA .100, therefore 100-1: 99, 10.101.99.255 is correct)

kpc21

Hi everyone,

@therealone, you are getting it wrong.

Raising to the power is something which you have to do when you have the netmask given in the CIDR notation (here it would be /23).

Then it's one bit missing to 24, so you raise 2^1 = 2 and you know the networks will change in the increments of 2.

See - you have n bits in the octet (the bits being zeros in the netmask) available for the addresses for your network. How many binary numbers can you compose using those n bits? You have n positions which can take 2 possible values: 0 or 1. So the result is: 2 times 2 times 2 etc. repeated n times, in other words, 2^n. And this is the reason why we increment by 2^n to get to the address of the next network. This is the reason why it works - from the mathematical perspective, explained in such a way that, I hope, everyone is able to understand it.

When you have the netmask given in the 255.255.254.0 notation, then it's enough to do the subtraction.

256-254 = 2.

If someone wants a mathematical explanation - I will add it at the end of this post.

So you have the networks:
10.101.0.0
10.101.2.0
10.101.4.0
and so on, the decimal numbers representing the third octet will be the multiples of 2, so simply all the even numbers from 0 to 254.

For your address - 10.101.99.17 - the network will therefore range from 10.101.98.0 to 10.101.99.25 - of course, those are the network and the broadcast address and the available addresses are from 10.101.98.1 to 10.101.99.254.

I promised an explanation, why the subtraction method works.

Let's think of an example, such one which will be clearer than the one discussed before.

A common netmask for point-to-point links: /30 or 255.255.255.252. Let's focus on the last octet. From the CIDR notation you easily see that the binary representation of the last octet is 11111100.

In binary you easily see all the possible network addresses possible to create with this mask (let's separate the network part and the host part with a space for clarity):
000000 00
000001 00
000010 00
000011 00
and so on, up to:
111111 00

They will be, generally, incremented by binary 100. Which is decimal 4.

How to easily get this 100 from the netmask?

When we subtract 11111100 from 11111111, we get simply 11. This is what you have in the host part of the broadcast address (of course if it wasn't the last octet, this would be a whole subnetwork containing not only the broadcast address, but also plenty of addresses available for our hosts - but it doesn't change anything for us, the last address in this subnet would be anyway the broadcast).

To move to the first address to the next network it's enough to add 1 to the broadcast address. We get the 100 by adding 1 to 11.

And this works for any netmask.

In decimal... our netmask has 252 in the octet which interests us. We subtract it from 255 (11111111) and add 1. Which is equivalent to subtracting from 256. 256-252 = 4, so we know that we increment by 4 to move to the next network address, that addresses separated by at least 4 will be certainly in different networks and that we have 4 addresses available in the network (two of which are the network address and the broadcast address, so actually we have 2 addresses available for the hosts, which is ideal for point-to-point links - by the way, actually it's possible to have a /31 netmask and use the "network" and "broadcast" addresses as hosts addresses on a point-to-point link, but it's out of the scope for the CCNA and if they ask if it's possible, even though they shouldn't, the correct answer will be most likely that it isn't possible).

cymatikata

therealone wrote: »

Hi all, thanks for this awesome technique. It really helped me.
However, I stumbled upon a problem which I don't get it:

10.101.99.17 -- 255.255.254.0

So, we are in 3rd octet, right? 256-254 = 2, 2^2 = 4.

Incrementing third octet:
10.101.0.0
10.101.4.0
...
10.101.92.0
10.101.96.0
10.101.100.0

Therefore, 10.101.99.17 falls between 100 and 96 one, so it must be 10.101.96.0 the Network Address?
But it is wrong? Calculators say it must be 10.101.98.0?

Btw, my broadcast address (next NA .100, therefore 100-1: 99, 10.101.99.255 is correct)

10.101.99.17 /23

/23 means we only care about the 3rd octet for starts.

So the so octets always repeat their ranges: 128 | 192 | 224 | 240 | 248 | 252 | 254 | 255 (17|18|19|20|21|22|23|24 bit positioning since we are talking the 3rd out of 4 octets).

23 = 254. Take that and subtract from 256. You get 2 and you start your networks on the 2's like this:

0
2
4
6
8
.
.
.254

that means 10.101.99.17 falls in the 98-99 network

10.101.98.1 through 10.101.99.254 is your usable range. 10.101.98.0 is your network and 10.101.99.255 is your broadcast.