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

ricky8

subnetting is easy. you just have to memorize the increments.
128+64+32+16+8+4+2+1 = 128, 192, 224, 240, 248, 252, 254, 255 <---- MEMORIZE
now all you have to do is the grade 3 math. 256-increment # (for example 256-224 = 32)

EX:
network: 192.168.0.0 /27 (/27 = 255.255.255.224) 256- 224 = 32 (subnets will be 0, 32, 64, 96....224)
1st subnet: 192.168.0.0 - 192.168.0.31
2nd subnet: 192.168.0.32 - 192.168.0.63
3rd subnet: 192.168.0.64 - 192.168.0.95

etc...

hope this helps.

Noerc

Thanks for sharing, I havn't been great at subnetting but the more I read and practice it all comes together!

LinuxRacr

Anyone have a good method for figuring out which mask for each class will give you a given number of subnets when only the IP is available?

drkat

That kind of defeats the purpose of the subnetting... if given no subnet mask then you will use the default class mask for the range of IP's

then it's business as usual

I'm sure you can find a **** sheet somewhere given the range and available masks and the numbers it produces.

This comes to mind

http://www.aelius.com/njh/subnet_sheet.html

LinuxRacr

This is the question that spawned my earlier question:

You Have subnetted the 192.168.14.0 network into 8 subnets. What is the valid host range of the third subnet?

I had a mental block in solving it for some reason...

t_patrick

Awesome!

Carl_S_901

LinuxRacr wrote: »

This is the question that spawned my earlier question:

You Have subnetted the 192.168.14.0 network into 8 subnets. What is the valid host range of the third subnet?

I had a mental block in solving it for some reason...

Ok, we're not given a mask on this question so we will have to stick with the fact that it is a classful network. (i.e. Class C or /24)

One way to attack it is using the powers of 2. If we have a /24 we know that if we shift 1 bit to the right then we have twice as many networks (x 2). So /25 gives us two subnets. /26 gives us 4. /27 gives us 8.

We need 8 for the problem so /27 is our subnet mask. We can convert that to decimal by knowing that 32-27 = 5 and that 2^5 =32 and 256-32 = 224. So 255.255.255.224 is our decimal mask. 2^5 is also our block size or "magic number" so we can count up subnets:

1. 192.168.14.0
2. 192.168.14.32
3. 192.168.14.64
4. 192.168.14.96

That's enough... 192.168.14.64 is our third subnet. More details:

Subnet 192.168.14.64
Broadcast 192.168.14.95
Valid Hosts 192.168.14.65 - 192.168.14.94

This assumes that we are using the Zero Subnet. If "no ip subnet zero" is in play (not mentioned in this question) then the 3rd subnet would be 192.168.14.96 with a broadcast of 192.168.14.127 and valid hosts of 192.168.14.97 - 192.168.14.126.
Thanks

Carl S.

Check out my personal certification journey blog
Carl's Certification Journey | The road to getting certified can often be bumpy

drkat

You Have subnetted the 192.168.14.0 network into 8 subnets. What is the valid host range of the third subnet?

256/8 = 32 this is your block size.. then you count your block size

LinuxRacr

Thank you both! That makes it crystal clear!

The answer actually includes the zero subnet, so yes, Valid Hosts = 192.168.14.65 - 192.168.14.94.

LinuxRacr

Ok, I've been able to answer quite a few questions in my head now, but this is the latest one that is getting me:

How many subnets and hosts per subnet can you get from the 172.28.0.0/26 network.

The hosts per subnet part was easy:
Since we are past the /24 boundary, we subtract 26 from the next octet boundary of 32:
32-26=6
2^6=64
Hosts per subnet = 64-2=62.

It was the number of subnets that I was lost on.

Highlight with mouse to see answer below.
Answer: 1024 subnets and 62 hosts per subnet.

r_durant

LinuxRacr wrote: »

Ok, I've been able to answer quite a few questions in my head now, but this is the latest one that is getting me:

How many subnets and hosts per subnet can you get from the 172.28.0.0/26 network.

The hosts per subnet part was easy:
Since we are past the /24 boundary, we subtract 26 from the next octet boundary of 32:
32-26=6
2^6=64
Hosts per subnet = 64-2=62.

It was the number of subnets that I was lost on.

Highlight with mouse to see answer below.
Answer: 1024 subnets and 62 hosts per subnet.

172.28.0.0 is by default a class B subnet with 16 bits, the /26 gives you the ^10 bits required to subnet (/26 - /16 = 10)....2^10 = 1024

Hosts, as you rightfully said are 2^6-2 = 62

11111111.11111111.11111111.11000000

The subnet bits in red.

I'm not the 'greatest' at explaining stuff, but hope that helps

Carl_S_901

The only trick I would add here for doing them in your head:

2^10 is exactly right as r_durant has replied and we can further make the math a little easier in our head because we know that a full octet 2^8 is alway 256

therefore we can say that this is 2^2 * 256 OR 4 * 256 which is of course 1024

So basically, when I need to calculate the number of X bits that span across multiple octets (one of which will always be the full octet or it wouldn't span into another) then I figure out the number for the interesting octet (partial) and multiply.

This gets a bit more difficult to do in your head though with a Class A though.

For example: How many subnets of /28 in a Class A:

We know the classful boundary is /8 so we have 2^20 or 2 ^ 8 (256) * 2 ^ 8 (256) * 2 ^ 4 (16)

2^8 * 2^8 = 2^16 of course and is equal to 65536 so this would be 65536 * 16 which is pretty hard to pull off in your head.

Answer would be:

1,048,576 subnets

I doubt you would see anything like that on an exam. If you did you will need to grab the pen to do the final calculation

LinuxRacr

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.252.0?

Answer: 64 subnets and 1022 host <

Highlight with mouse to see.

I worked this one out on paper, and this one seems wrong to me....

r_durant

LinuxRacr wrote: »

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.252.0?

Answer: 64 subnets and 1022 host <
Highlight with mouse to see.

I worked this one out on paper, and this one seems wrong to me....

Why do you think it's wrong?

You have 11111111.11111111.11111100.00000000

2^6 bits for the subnets which = 64
2^10-2 bits for the hosts which = 1022

LinuxRacr

I believe my lack of understanding coupled with the method in which I was trying to solve this lead to my conclusion... sigh...

Two steps forward, one step back...

r_durant

LinuxRacr wrote: »

I believe my lack of understanding coupled with the method in which I was trying to solve this lead to my conclusion... sigh...

Two steps forward, one step back...

IMO...The thing about subnetting is that each person will have a different method of working these problems out, in some cases people will use the same method. Just develop a method that you are comfortable with, master it and then practice it repeatedly. Also, have confidence in yourself...

LinuxRacr

Thanks for the tutorials, and I do have confidence in myself. I just hate setbacks when I seem to be moving forward. I just have to sure up this hole in my learning, and I'll be good to go.

Carl_S_901

I agree with r_durant in that there are a lot of different ways to do it.

Find what works best for you, get it down solid, be able to do it fast in your head, and then pretty much ignore reading about subnetting in the near future so as not to throw a wrench in your subnetting machine. (your brain)

Rhodan

Hi all,

Great thread! I'm stuck with this one:

You are designing a subnet mask for the 192.168.100.0 network. You want 3 subnets with up to 50 hosts on each subnet. What subnet mask should you use?

The answer given is 255.255.255.192

How do I work this out?

LinuxRacr

Rhodan wrote: »

Hi all,

Great thread! I'm stuck with this one:

You are designing a subnet mask for the 192.168.100.0 network. You want 3 subnets with up to 50 hosts on each subnet. What subnet mask should you use?

The answer given is 255.255.255.192

How do I work this out?

First let's find out our blocksize of each subnet:

256-192=64

So 64 is the blocksize of each subnet, and you get a total of 4 subnets.

192.168.100.0 - 192.168.100.63
192.168.100.64 - 192.168.100.127
192.168.100.128 - 192.168.100.191
192.168.100.192 - 192.168.100.255

With that IP, that's the best you can do. Remember with IP addresses, they are binary, so everything is a power of two, so you have to have 4 subnets. Three is more than two, so the next power of 2 is 4.

Carl_S_901

Rhodan,

The smallest block size that will support 50 hosts is 64.

Your mask is 255.255.255.192 (256-64) This is also known as a /26

You will have 4 subnets.

Couple of ways to do this:

1. If you have a block size of 64 then you are using 6 bits for hosts (2^6) which means you have 2 bits for network (2^2) beyond the mask of /24 you were given.

2. Each time you move the mask a bit to the right you get twice as many subnets which are half the size. If you have a /24 and you borrow one bit and make it a /25 then you have two networks with 128 block size. If you move it one more to the right and make it /26 then you now have a block size of 64 and 4 networks.

Remember that usable hosts is always the size of the block MINUS 2 to account for the network and broadcast addresses.

Finally, if a question like this says you need 3 subnets that doesn't mean 3 and ONLY 3. It can be more so long as it meets the other requirements.

LinuxRacr

All, I found this subnet calculator years ago to help with some sys admin work I was doing (since I wasn't interested then in being a network guy), and it is one of the best I've seen:

Network Calculators

You can calculate what you need, and then hit the explain button to break down the why, and it goes in-depth with the explanation.

LinuxRacr

Carl_S_901 wrote: »

Rhodan,

The smallest block size that will support 50 hosts is 64.

Your mask is 255.255.255.192 (256-64) This is also known as a /26

You will have 4 subnets.

Couple of ways to do this:

1. If you have a block size of 64 then you are using 6 bits for hosts (2^6) which means you have 2 bits for network (2^2) beyond the mask of /24 you were given.

2. Each time you move the mask a bit to the right you get twice as many subnets which are half the size. If you have a /24 and you borrow one bit and make it a /25 then you have two networks with 128 block size. If you move it one more to the right and make it /26 then you now have a block size of 64 and 4 networks.

Remember that usable hosts is always the size of the block MINUS 2 to account for the network and broadcast addresses.

Finally, if a question like this says you need 3 subnets that doesn't mean 3 and ONLY 3. It can be more so long as it meets the other requirements.

Thank you for the COMPLETE picture. I forgot that I should have approached this from the standpoint of NOT already having the answer....

Rhodan

Thanks for the quick replies, that makes sense.

Here's another type of question I want to check on:

How many subnets and hosts per subnet can you get from the network 172.22.0.0/25?

Answer given is 512 subnets and 126 hosts

This is how I work it out:
172.22.0.0 is a Class B network, so /16.

25 - 16 = 9 network bits
2^9 = 512 subnets

There are 7 host bits left from the /25, so (2^7) - 2 = 126 hosts.

Is this the correct way of working it out?

r_durant

Rhodan wrote: »

Thanks for the quick replies, that makes sense.

Here's another type of question I want to check on:

How many subnets and hosts per subnet can you get from the network 172.22.0.0/25?

Answer given is 512 subnets and 126 hosts

This is how I work it out:
172.22.0.0 is a Class B network, so /16.

25 - 16 = 9 network bits
2^9 = 512 subnets

There are 7 host bits left from the /25, so (2^7) - 2 = 126 hosts.

Is this the correct way of working it out?

That's correct!

elad123

Wickedkick wrote: »

Hi all,

Question: What is the first valid host on the subnetwork that the node 172.16.228.120/22 belongs to?

and the actual answer to it is: Answer: 172.16.228.1

-Wicked

Guys, I am reading through this but there is something that is not clicking for me, take the above for example if I am understanding this correctly:

24-22 = 2
2^2 = 4

so you increment the 3rd octet by 4

172.16.0.0
172.16.4.0
172.16.8.0

etc...

172.16.228.0
172.16.232.0

what is the shortcut to do that quickly instead of writing it all out?

r_durant

elad123 wrote: »

Guys, I am reading through this but there is something that is not clicking for me, take the above for example if I am understanding this correctly:

24-22 = 2
2^2 = 4

so you increment the 3rd octet by 4

172.16.0.0
172.16.4.0
172.16.8.0

etc...

172.16.228.0
172.16.232.0

what is the shortcut to do that quickly instead of writing it all out?

One shortcut you can use is to verify 228 is a multiple of 4 (228/4), if it is, then you know you're bang on. If it's not, then find the closest.

LinuxRacr

r_durant wrote: »

Why do you think it's wrong?

You have 11111111.11111111.11111100.00000000

2^6 bits for the subnets which = 64
2^10-2 bits for the hosts which = 1022

I'm still confused a little about where the 2^10 came from on this one..... Did it come from the 10 host bits (the zeros)?

oli356

LinuxRacr wrote: »

I'm still confused a little about where the 2^10 came from on this one..... Did it come from the 10 host bits (the zeros)?

Yes, to work out the number of hosts you do (2^x)-2, where x is the number of host bits. Host bits are the zeros, in this example there are 10 zeros accross the 2 octets. Of course you take away 2 because of the network ID and broadcast address.

LinuxRacr

Sure enough I had a question like that on my exam today...lol.