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

Ismaeljrp

Multiplication. Pay attention to the increment, and multiply to the closest subnet ID. You won't be seing any crazy high subnet numbers on the tests, at least that's what I hear.

How do you even list subnets ? I don't really get all those fancy stuff, I solve everything in my head with basic math. Since I already understand and can visualize the octets in binary format. I know that a 11110000 is a .240 or a /28 if it's in the last octet. I know the increment would be 16. 256-240=16. So if I see a 192.168.1.67 /28, considering the increment is 16, well I automatically know that 16 x 4 = 64. Which means that 192.168.1.67 is a valid host in the 192.168.1.64 subnet ID.

jsb515

anyone help me with this one? how are they getting a 117 subnet?

Which subnet does host 172.20.117.46 255.255.255.128 belong to?

here is how i did it..

128 64 32 16 8 4 2 1 || 128 192 224 240 248 252 254 255

/25

32-25=7

2^7=128

0.1-254 b255
128.0
128.1-254 b255
255.1-254 b255
256.0

Ltat42a

Remember your Classes of Addresses, A,B,&C. This is a Class B address. Default mask is 255.255.0.0.
The subnet mask they gave you is 255.255.255.128. 9 bits were used to create subnets, 7 bits left over for hosts. Subnetting takes place in the 4th octet in increments of 128.
Starting at 0, count like this-
172.20.117.0 (First subnet)
172.20.117.128 (Second subnet)

172.20.117.46 belongs to the first subnet.

hth

jsb515

ah ok i was thinking it was in the third octet but i forgot that you use the octet it borrows from. Thanks for the help.

jsb515

need some help again..

172.16.0.0 255.255.255.192

I do 32-26=6=2^6=64

here is where I believe I goof up.. the counting

since I borrowed in the fourth octet I start

172.16.0.64
172.16.0.128

but I got a feeling i'm counting this wrong..

When does this 0 change to 1 2 3 ect..?
..........\/
172.16.0.64

or do I count it like this

172.16.1.64 - 1.254
172.16.2.128 - 2.254

Ltat42a

Subnetting 172.16.0.0 with a mask of 255.255.255.192 (or /26), your subnetting takes place in the 4th octet.
The increments are 64, in the 4th octet. Starting at 0, your subnets would be:
172.16.0.0 - 172.16.0.63
172.16.0.64 - 172.16.0.127
172.16.0.128 - 172.16.0.191
172.168.0.192 - 172.168.0.255
172.16.1.0 - 172.16.1.63
etc
etc
etc

jsb515

192.168.4.255 /23

I thought this was a broadcast address at first glance but Boson said no it a host address.. so what would the broadcast of this address be?

atorven

192.168.5.255 - The increments are in 2 in the 3rd octet, so 0,2,4,6 e.t.c

BIGDub

I've been having a lot of problems with this lately. I've gotten some of it but I think I've been looking at too many subnetting methods and they are starting to conflict. I thought I was making headway last week. Now something has me confused. I have had a hit or miss record on subnettingquestions.com. I ran into one that makes me question the accuracy of their answers or their methodology in getting there.

Here is the question and the answer from the site:

Question: How many subnets and hosts per subnet can you get from the network 172.16.0.0 255.255.255.192?Answer: 1024 subnets and 62 hosts

Now the problem. I'm also studying the Todd Lammle CCENT Study Guide Copyright 2008 and there is an identical question in there, same IP address and subnet mask. However the book has the answer of 4 subnets and 16,382 hosts with 64 address block size making 62 available addresses per block.

Something is wrong. Even if the terminology were different you would still get the same numbers. It's either one of the two sources is wrong. It's also possible my interpretation is wrong but I can't see where it could be, there is a huge difference between 4 subnets and 1024 subnets.

atorven

Which answer do you agree with? Below is how I approached the question.
172.16.0.0 = Class B by default, 16bits would reserved for the network.
255.255.255.192 = 8 bits in the 3rd octet which would have been used for the host address have now been borrowed for the network portion, 2 bits have also been borrowed in the 4th octet so 8+2=10bits have been borrowed therefore 2^10=1024 subnets.
For the host part you should already know that 8 bits make up an octet, we have already borrowed 2 bits in that octet so we have 6 remain for our hosts therefore 64hosts but you have to subtract 2 hosts for the network address and broadcast address so we end up with 62 hosts per subnet.
HTH

sec_wiz

Hey guys,

Sorry I was unable to read over all pages of this thread so it may have been mentioned.

After reading the first post subnetting has finally clicked for me - in terms of finding subnet ranges and identifying the first and last host etc.

But how do I go using this method to determine how to create subnets as well as 'how many hosts and networks could you create if you had an ip of ______ and a mask of _______'?

thanks in advance!

jsb515

say i have a question like what ip class can 15 bits be borrowed to make subnets, how would you go about answering this one quickly?

ciscoman2012

jsb515 wrote: »

say i have a question like what ip class can 15 bits be borrowed to make subnets, how would you go about answering this one quickly?

Not sure if I'm thinking about this right but I would say Class B (the netmask would be 255.255.0.0)

255.255 would be the network portion and 0.0 would be the borrowed bits to subnet. For example 192.168.X.X the X's would be the subnetted portion.

oli356

Just feel like posting my experience today on the forum.

As part of my job I have to go on training for about 12 weeks this year. We do MTA exams, 3 of them. Each exam has a week of being taught/studying and then take the exam.
Though I have completed the exam I still have to attend the training and listen to the lessons, this time its the Networking exam.

So he started to explain IP addresses and went through the different classes (A, B, C) this was all fine. Then he got up to subnetting..

But it wasn't called subnetting anymore, he didn't even mention that word. It was called "CIDR IP". Basically if you are using a /8 /16 or /24 you have to write out the subnet mask e.g. 255.255.0.0
But if you want to "subnet" you have to use this "CIDR IP", so instead of writing out the subnet mask you HAVE to do the bit notation as the subnet mask and bit notation are nothing to do with each other. He didn't really explain any more than that, HOW you actually subnet.

He then tried to think of a bit notation for the IP address on the whiteboard, "I always forget what numbers I can use" I just spoke out and said 23.

Uh..... what!? Doesn't he know what CIDR notation is, what the /x represents, the binary - anything?
His explanation of VLANs yesterday didn't make any sense; he said kind of the opposite of what they are used for.

I was talking to one of my colleagues on WebEx at the time who is a double CCIE and told him what we had been told - he couldn't believe it!
/rant

pgnet

Hi,

Firstly thanks for this method, it’s by far been the easiest way I’ve been able to learn subnetting and I am able to answer most of the questions on subnettingquestions.com

The only one I am struggling with is the type ‘You are designing a subnet mask for the 192.168.123.0 network. You want 12 subnets with up to 11 hosts on each subnet. What subnet mask should you use?’

I seem to have a mental block with this, which is highlighted even more by trying the subnet game on cisco.com. On that game, which I urge you to try, the questions are similar.

One scenario is, there is a requirement for 2 subnets with 2 hosts per subnet.
You need to fill out the last octet values for each site in the following fields and are given the following which you cannot change:

Subnet mask 255.255.255
Router interface: 192.168.1
Broadcast address: 192.168.1
Network address: 192.168.1

To save me time with other questions I have been using the table:
128 64 32 16 8 4 2 1
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
128 192 224 240 248 252 254 255

Can you explain how I can work this out using the method originally posted and my table above?
Please do not write out binary methods etc of working this out, I just want the method from the original poster, Lordflasheart

I suspect there will be this type of question above in the exam so just need to get my head around it then should be sorted.

Thanks for any help in advance

Ltat42a

pgnet wrote: »

One scenario is, there is a requirement for 2 subnets with 2 hosts per subnet.
You need to fill out the last octet values for each site in the following fields and are given the following which you cannot change:

Subnet mask 255.255.255
Router interface: 192.168.1
Broadcast address: 192.168.1
Network address: 192.168.1

The only subnet mask that will give you only 2 hosts (think WAN links) - 255.255.255.252

Now....fill in the rest

pgnet

I've tried attaching a screenshot but I can't in case I'm decribing it wrong. The answer I saw on the site below is

2 Areas = 255.255.255.128
1st room
Network 192.168.1.0
Broadcast 192.168.1.127
Router 192.168.1.1
2nd Room
Network 192.168.1.128
Broadcast 192.168.1.255
Router 192.168.1.129

The URL is https://learningnetwork.cisco.com/docs/DOC-1802

As above it asks me to fill out these fields:
Subnet mask 255.255.255
Router interface: 192.168.1
Broadcast address: 192.168.1
Network address: 192.168.1

And the work order (scenario question) is 'Four computers each in two areas that need there own subnet'.

Thought I'd got subnetting before seeing this

...

mackster

Hi everyone. This is a great post and it really makes subnetting easy.

Though I may not be able to see the wood for the trees on one particular question.

_______________________________
A RIB has the following:

Prefix | Next Hop
24.1.36.0/24 | 10.0.0.2
24.1.36.8/29 | 10.0.0.5
24.1.36.8/30 | 10.0.0.9

Given the avove RIB, what is the next hop for a packet destined for 24.1.36.14?
________________________________________

From my evaluation, I can see two possible hops, 10.0.0.5 and 10.0.0.9.

I think if someone can take me through thetheory on this, I can see what I am doing wrong.

tearofs

mackster wrote: »

Hi everyone. This is a great post and it really makes subnetting easy.

Though I may not be able to see the wood for the trees on one particular question.

_______________________________
A RIB has the following:

Prefix | Next Hop
24.1.36.0/24 | 10.0.0.2
24.1.36.8/29 | 10.0.0.5
24.1.36.8/30 | 10.0.0.9

Given the avove RIB, what is the next hop for a packet destined for 24.1.36.14?
________________________________________

From my evaluation, I can see two possible hops, 10.0.0.5 and 10.0.0.9.

I think if someone can take me through thetheory on this, I can see what I am doing wrong.

24.1.36.14 is within both 24.1.36.0/24 & 24.1.36.8/29 subnets, So I believe the possible hops are 10.0.0.2 and .5

boredgamelad

When a given destination IP matches more than one route, the router chooses the most specific route (aka the one with the longest network prefix, or highest CIDR number).

24.1.36.14 is within both the /24 and /29 subnets, but it would forward to 10.0.0.5 because /29 is more specific than /24.

iverson2k10

Here is one I don't get

what is the host range for 10.145.100.192 255.255.240.0?

I got 10.145.192.1 I would think its + 12 more for the range should be 192+12 = 204

10.145.192.1
10.145.204.254

I was wrong and the answer they got was 10.145.192.1 - 10.145.207.254. Even IP calculator confirms it.

beach5563

Just started using this method today, love it so far, was able to understand how to find which subnet an ip address belongs too and some of the valid host range concept of a particular subnet but there are a couple that kind of got me a little confused. I may just need more practice but one was asking about the valid host range for the 7th subnet of 10.0.0.0/14. So i understand about finding the boundary and getting the block size but the answer on Loardflashearts document is 10.24.0.1 to 10.27.255.254, so I guess the 255.254 stuff knocked me for a loop. Just trying to gigure out am I missing something that he already explained in the beginning, is there a certain part I need to go over again? Most of this stuff was coming along just fine til I got this one. There was another one like this too, I think it was a class B one though. Just wondering if I could get some input. Thanks guys

Ltat42a

iverson2k10 wrote: »

Here is one I don't get

what is the host range for 10.145.100.192 255.255.240.0?

I got 10.145.192.1 I would think its + 12 more for the range should be 192+12 = 204

10.145.192.1
10.145.204.254

I was wrong and the answer they got was 10.145.192.1 - 10.145.207.254. Even IP calculator confirms it.

That answer doesn't seem right.
10.145.100.192 with a mask of 255.255.240.0. You're subnetting in the 3rd octet with a block size (or range size) of 16.
10.145.100.192 belongs to the network 10.145.96.0. The valid host range would be 10.145.96.0 - 10.145.111.254.
The next network would be 10.145.112.0.

boredgamelad

beach5563 wrote: »

one was asking about the valid host range for the 7th subnet of 10.0.0.0/14. So i understand about finding the boundary and getting the block size but the answer on Loardflashearts document is 10.24.0.1 to 10.27.255.254, so I guess the 255.254 stuff knocked me for a loop.

It's important to remember how to count IP addresses both up and down. A refresher: if you count IP addresses from 0.0.0.0 you've got 0.0.0.0, 0.0.0.1, 0.0.0.2, and so on until you hit 0.0.0.255. When you add 1 to 255, you have exceeded the value for the fourth octet (you can't have 0.0.0.256). So the previous octet increments by 1, making the next IP address 0.0.1.0. Likewise, if we count down one from 0.0.1.0, the next lowest IP address is 0.0.0.255. Keeping that in mind, let's look at your example.

You've been given 10.0.0.0/14, or 10.0.0.0 / 255.252.0.0. Our interesting octet is the second and 256-252 gives us an increment of 4:

10.0.0.0 < 1st subnet
10.4.0.0 < 2nd subnet
10.8.0.0 < etc.
10.12.0.0
10.16.0.0
10.20.0.0
10.24.0.0 < 7th subnet
10.28.0.0
[...]

10.24.0.0 is the 7th subnet. We know from our increment that 10.28.0.0 is the 8th subnet. So, we can determine the range for the 10.24.0.0 subnet by figuring out what IP address comes before 10.28.0.0. In this case, it's 10.27.255.255. The range of IPs that belong to the 10.24.0.0 network are 10.24.0.0 through 10.27.255.255.

Since the question is asking about valid hosts, we can't stop there. Knowing that the first IP address is reserved for the subnet ID (10.24.0.0) and the last IP address is reserved for broadcasts (10.27.255.255), that makes all other IP addresses in that subnet valid hosts. The valid host range for any subnet is going to be (subnet ID + 1) through (broadcast ID - 1). In other words, 10.24.0.1 (10.24.0.0 + 1) through 10.27.255.254 (10.27.255.255 - 1).

Remember, once you've found the subnet ID:

First valid host = subnet ID + 1
Last valid host = broadcast ID -1
Broadcast address = next subnet ID -1

Hope this helps.

beach5563

@boredgamlad, thanks so much, I'm going to practice this stuff again today, I really appreciate that. Man I was so close yesterday but by the end of the weekend I should have this down. I may study some basic ip addressing stuff today too maybe.

richypc

For the network 192.0.2.0/23, which option is a valid IP address that can be assigned to a host?

A. 192.0.2.0
B. 192.0.2.255
C. 192.0.3.255
D. 192.0.4.0

Answer: B

Can anyone explain this one, having trouble. The others I do just fine with.

New2IT

Using the method described in this thread:

192.0.2.0/23
23 is in the 3^rd octet, so 24-23 = 1, then you take 2 to the 1^st power = 2
So 2 is your increment in the 3^rd octet

192.0.0.0
192.0.2.0
192.0.4.0

So in the subnet of 192.0.2.0 you have:
192.0.2.0 network address (answer A)
192.0.2.1 – 192.0.3.254 host range, 192.0.2.255 is within the host range (answer

192.0.3.255 broadcast address (answer C)

192.0.4.0 is the next subnet network address (answer D)

I am new to this so I hope it makes sense.

boredgamelad

I was just typing up an answer when I saw New2IT's post. You've got it exactly right.

lmoworld

Awesome Thread.

richypc

Yeah I see, I was trying hard to do in my head like I have with the rest. Just easier and better to write a few of them out sometimes.