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

wallpaper_01

Unbelievable! Tried loads of techniques, trying to learn all the ways, I read this and it clicks instantly! I can answer every question within around 30 seconds now! Thanks so much!

mikroTik

How To Know Network Address of any Class From a Host Address?

So Simple..

A Host Address
192.168.0.2/24
192.168.0.2 Address in Binary
11000000.10101000.00000000.00000010
A Host Sub-net Mask
11111111.11111111.11111111.00000000
Keep Counting column by column and 1*0=0 and 1*1=1 ok ?
11000000.10101000.00000000.00000000

Now the above binary in decimal
192.168.0.0 (this is network address of 192.168.0.2)

that was an easy way you try by your self as much as hard you see to count any network address of any host.

mohammadzzz

Hi

using the above method i am still confused on the following from subnettingquestions.com

what is the broad cast address for 172.24.232.0 255.255.254.0

according to my calculations the block size is 24 where 256-232=24,we count into the third octet from zero to reach the above ip,but answer is 172.24.233.255,can somebody please help that how we drive to this answer.

fadhil

mohammadzzz wrote: »

Hi

using the above method i am still confused on the following from subnettingquestions.com

what is the broad cast address for 172.24.232.0 255.255.254.0

according to my calculations the block size is 24 where 256-232=24,we count into the third octet from zero to reach the above ip,but answer is 172.24.233.255,can somebody please help that how we drive to this answer.

always remember block size = 256 - subnet mask for 172.24.232.0 255.255.254.0, the block size is
256 -254=2.
to determine the broadcast address
since the block size is 2, the next network address will be 172.24.234.0 since the last address before next network address is the broadcast address. hence the broadcast address of 172.24.232.0 is 172.24.233.255

mohammadzzz

Zanzibar !

Thanks a looooooooooot brother,I do not have words to say how thankful i am with your help.....Thanks thanks

mella060

Another quick way to find the increment/block size is to work out the last bit position where the value is a 1. Remember that every subnet mask as a number of consecutive 1's. Also remember that in each octet you have the bit positions... 128 64 32 16 8 4 2 1

So in the above example 172.24.232.0 255.255.254.0, the subnet mask in binary would be...

11111111.11111111.11111110.00000000

Focusing on the 3rd octet, you get the following...128 62 32 16 8 4 2 1

So you can see that the value where the last 1 is is 2 which is your increment/block size.

IT69

This thread has got to be the best thing in the CCNA section, has helped so much....thanks so much to OP and others who contributed good info

gbdavidx

IT69 wrote: »

This thread has got to be the best thing in the CCNA section, has helped so much....thanks so much to OP and others who contributed good info

QFTquote for truth!! It has helped me understand subnetting

CCIE Wanna Be

thomas130 wrote: »

is there an easy to work this one out

class a network 10.0.0.0 create a subnet mask for the 600 subnets. THen Identify the 100th subnet

I can work out doing the subnet easy but is there an easier way to finding out the 100th subnet rather than writing each subnet down like this

10.0.0.0
10.0.64.0
10.0.128.0
10.0.192.0

It would take forever

First in order to get 600 subnets you have to borrow 10 bits, so the mask going forward will be /18. The math for this happens to work out perfectly for finding the 100th subnet. Each block in the 3rd octet will repeat the same four values with the 2nd octet incrementing by one. So 100 divided by 4 = 25. So the 100th subnet should be 10.25.192.0.

Please check my work, I did this in my head and I don't have access to subnet calculator to check my work.

chickenlicken09

Guys,

New to subnetting, i understand whats being done on the below question but why does it sit on the 192.168.12.72 subnet?
What am i missing? Its prob something really simple.

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.

pickarooney

This is the first, and so far only technique I've been able to understand and apply so first up thanks a million to the OP for posting it. I can do about 70% of the questions on subnettingquestions.com reasonably quickly but there is one type that consistently stumps me. If anyone can talk me through a method for working this out without resorting to writing down a list of 1s and 0s I'd be eternally grateful. I can't grasp the concept of borrowing bits which has been my downfall many times.

How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.255.192

It's a class B address whose default mask is 255.255.0.0 and the last octet has the mask, which is 64 less than 256 so I think this is the block size. I can't get past the next bit, working out the number of subnets and hosts. I also can't work out how to convert from digital format mask to slash format.

CCIE Wanna Be

pickarooney wrote: »

This is the first, and so far only technique I've been able to understand and apply so first up thanks a million to the OP for posting it. I can do about 70% of the questions on subnettingquestions.com reasonably quickly but there is one type that consistently stumps me. If anyone can talk me through a method for working this out without resorting to writing down a list of 1s and 0s I'd be eternally grateful. I can't grasp the concept of borrowing bits which has been my downfall many times.

How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.255.192

It's a class B address whose default mask is 255.255.0.0 and the last octet has the mask, which is 64 less than 256 so I think this is the block size. I can't get past the next bit, working out the number of subnets and hosts. I also can't work out how to convert from digital format mask to slash format.

Since this was originally a Class B the first 16 bits are reserved for the network address 172.19.xxxxxxxx.xxxxxxxx, where the x's represent the number of bit available to borrow. So with subnet mask 255.255.255.192, 8 bit were borrowed from octet #3 (hence the 255 in the mask), and an additional 2 bits were borrowed from the last octet (hence to 192 in the mask), so that gives us 10 bits for the subnet portion of the address, so in total if you were to subnet this class B address to the mask listed you would have 2^10 or 1024 possible subnets. To figure out the slash notation, any 255 in the mask represents 8 bits, in this case there are 3 so that = 24, and the two additional bit from the 4th octet to get /26 for subnet mask. In order to find the number of host subtract the number of reserved network bit from the total number of bits available which is 32. So 32-26 = 6, so bits are available for hosts, so using the formula 2^h-2 or 2^6 -2 or 62 hosts/network. Yes, 64 is your increment for your subnets.

First subnet is:
172.19.0.0

First broadcast:
172.19.0.63

valid Hosts:
172.19.0.1-172.19.0.62

Next subnet: 172.19.0.64, 172.19.0.128, 172.19.0.192, 172.19.1.0, 172.19.1.64..........

Since you also borrowed bits from the 3rd octet for the subnet, this value will increment as well, since mask is 255, the increment in this octet is one (256-255=1).

So doing the math, the last octet will repeat 0, 64, 128, 192, and the third octet will increment by by one from zero-255, so that 256 possible values in the 3rd octet and 4 in the last, so 256*4 =1024, which as you can see is the value we got for the number of available subnets.

If I have confused you more I apologize, otherwise I hope this helps.

pickarooney

I understand some of that... You lost me a little bit at borrowing bits, though. I can just about grasp that the change from default 255.255.0.0 involves borrowing taking 255 from the host portion and adding it to the third octet. I know that this is 8 bits just because it's a standard number. I've no idea how you get the additional 2 bits from the number 192 though.

It looks like it would take 10 minutes to work all this out, with moving from digital to slash and then working out the blocks, but I assume you added in this step specifically because I asked how to switch between them. I just about understand the 3x8 but again where does the 2 come from? If the number in the 3rd octet was anything other than 255 I don't understand how to work out how many are borrowed and as a result in the next step I don't get how to work out the number of subnets provided by this mask.

So I think there are two sticking points which, if I can understand them, will make everything clear
1. How to work out the number of bits borrowed from an octet if the number is not 255 or 0
2. How to count the number of subnets available from a mask other than 255

CCIE Wanna Be

pickarooney wrote: »

I understand some of that... You lost me a little bit at borrowing bits, though. I can just about grasp that the change from default 255.255.0.0 involves borrowing taking 255 from the host portion and adding it to the third octet. I know that this is 8 bits just because it's a standard number. I've no idea how you get the additional 2 bits from the number 192 though.

It looks like it would take 10 minutes to work all this out, with moving from digital to slash and then working out the blocks, but I assume you added in this step specifically because I asked how to switch between them. I just about understand the 3x8 but again where does the 2 come from? If the number in the 3rd octet was anything other than 255 I don't understand how to work out how many are borrowed and as a result in the next step I don't get how to work out the number of subnets provided by this mask.

So I think there are two sticking points which, if I can understand them, will make everything clear
1. How to work out the number of bits borrowed from an octet if the number is not 255 or 0
2. How to count the number of subnets available from a mask other than 255

Question 1:

Network and subnet bits are arranged from left to right in each octet. Each bit in the octet that represents the network is set to one and there can be no intervening zeros. This arrangement of bits in binary is called high order, because that are assigned form left to right.

So each bit in the octet represents the following values from left to right

>>>>>>>>>>>>>>>>>
Bit Position: 7 6 5 4 3 2 1 0
Decimal Value: 128 64 32 16 8 4 2 1

So counting from left to right borrowed bits will equal the additive decimal value

1 bit = 128 in binary 10000000
2 bits = 128 + 64 or 192 binary 11000000
3 bits = 128 + 64 + 32 or 224 binary 11100000
4 bits = 128 +64 +32 + 16 or 240 binary 11110000
5 bits = 128 + 64 + 32 + 16 + 8 or 248 binary 11111000
6 bits = 128 + 64 + 32 + 16 + 8 + 4 or 252 binary 11111100
7 bits = 128 + 64 + 32 + 16 + 8 + 4 + 2 or 254 binary 11111110
8 bits = 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 or 255 binary 11111111

Underlined binary values represent the bits borrowed for the network portion

Since the subnet mask represents which bits are reserved for the network and subnet portion of the address and there can be no intervening zeros, the values above are the only value that can be placed in the subnet mask. So if you see for instance a 192 in the 4th octet for the subnet mask, you know that 2 bits have been borrowed from the 4th octet.

Since the original mask was class be the first two octets will not change, in the new mask the third octet changed to 255, looking above that takes 8 bits, and the 4th octet changed to 192, looking above that takes 2 bits, so 8 (from third octet) + 2 (from fourth octet) = 10. You have 10 additional bits for subnets so the equation 2^# of borrowed subnet bits or 2^10 or 1024

Question 2:

To figure out the number of subnets available, you take 2 and raise it to the power of the total number of borrowed bits, in the example it was 10
So it would be 2^10 = 1024. Remember its the total number, even if you cross an octet like in the example above, we took all eight bits form the 3rd octet and an additional two from 4th octet.

It does help to right these out in binary, at least until you get good understanding of the underlying logic behind how a router is actually figuring out what is a network and what is host. I took a semester long course on Cisco routers, the first half of the semester before we touch any equipment was on IPv4 addressing and subnetting both using Constant Length Subnet Masks (which what is shown above) and Variable Length Subnet Masks. The key here is to fully understand the binary logic, this article may help Subnetwork - Wikipedia, the free encyclopedia

Again, If I have confused you more I apologize, otherwise I hope this helps.

boredgamelad

This is exactly why I advocate learning to subnet by hand first. The other methods don't help if you don't understand what's happening to the bits themselves.

CCIE Wanna Be

boredgamelad wrote: »

This is exactly why I advocate learning to subnet by hand first. The other methods don't help if you don't understand what's happening to the bits themselves.

The professor I had for my class showed us the easy way first and had us do a whole bunch of subnetting with CLSM and then had us try it on VLSM

without knowing the binary first

. A lot of puzzled faces in the class, but I quickly understood the point he was trying to make since he was a stickler on absolutely knowing underlying binary logic

pickarooney

It's slowly sinking in that I get the physical manifestation of the bit masking and I guess the main thing is to just learn these pairs of numbers off by heart in order to be able to solve questions reasonably quickly

1 bit = 128
2 bits = 192
3 bits = 224
4 bits = 240
5 bits = 248
6 bits = 252
7 bits = 254
8 bits = 255

CCIE Wanna Be

pickarooney wrote: »

It's slowly sinking in that I get the physical manifestation of the bit masking and I guess the main thing is to just learn these pairs of numbers off by heart in order to be able to solve questions reasonably quickly

1 bit = 128
2 bits = 192
3 bits = 224
4 bits = 240
5 bits = 248
6 bits = 252
7 bits = 254
8 bits = 255

Indeed, but also remember that these are high-ordered bits (meaning they are assigned and added from left to right within each octet).

mohammadzzz

Question: How many subnets and hosts per subnet can you get from the network 172.17.0.0 255.255.254.0?

my calculation is 22 bits for networks and 8 for hosts but answer do not match up,any help please

Answer: 128 subnets and 510 hosts

boredgamelad

There is a maximum of 32 bits... 22 + 8 doesn't equal 32, so one or both of those numbers has to be wrong.

172.17.0.0 / 255.255.254.0 = 172.17.0.0/23

Default class B: /16

23 - 16 = 7

2^7 = 128 subnets.

32 - 23 = 9

(2 ^ 9) - 2 = 510 hosts.

dazl1212

Say you have a 192.168.1.0 network 255.255.255.192 and you subnet it could you not use say 192.168.1.16 as the network address?
I'm 99% sure its a no but I just want to double check.
It is the only thing I am not sure of.

dazl1212

I just want to thank the OP for this thread. It has helped me immensely and I very nearly have subnetting down to less than minute

CCIE Wanna Be

dazl1212 wrote: »

Say you have a 192.168.1.0 network 255.255.255.192 and you subnet it could you not use say 192.168.1.16 as the network address?
I'm 99% sure its a no but I just want to double check.
It is the only thing I am not sure of.

In your example, the increment for the networks is 64, assuming you can use ip-subnet zero, your networks are:

network=192.168.1.0 valid host =192.168.1.1 to 192.168.1.62 broadcast=192.168.1.63

and so on.........

192.168.1.64
192.168.1.128
192.168.1.192

The ip address 192.168.1.16 is a valid host in the 192.168.1.0 network. So to answer your question directly, no you could not use this as a network address with the given subnet mask.

dazl1212

Thanks CCIEwannabe

CCIE Wanna Be

dazl1212 wrote: »

Thanks CCIEwannabe

you're welcome

mohammadzzz

hi

just wondering how do you find a correct subnet mask from just IP,lets say that we have two ip address given which are 172.16.1.126 for a node with 84 hosts in a vlan 1 and second is 172.16.1.129 for node in vlan2 with 114 hosts,connected by a switch,how do we find what subnet mask is used.

your options are

a-subnet mask used 255.255.255.192
b-subnet mask used 255.255.255.128
c-subnet mask used 255.255.255.133

answer is b:255.255.255.128

also i found this question and still do not know how to answer this

Q-which two subnets will be included in the summarized address of 172.31.80.0/20

options are

a-172.31.17.4/30
b-172.31.51.61/30
c-172.31.64.0/18
d-172.31.80.0/22
e-172.31.92.0/22
f-172.31.192.0/18

answer d and e

cant figure out how this is correct ????i tried route summarization and was unable to answer a similar question.Any help will be apperciated

dazl1212

Honestly no Idea

Ltat42a

Definitely need to know your block sizes or range multiples for questions like these. What subnet mask will allow you have 84 or 114 hosts in 1 subnet (or vlan)? 255.255.255.128. This mask gives you 2 subnets [0 & 128], with 126 available host IP's in each subnet.

For the summary question...what is the subnet range for that IP & Mask?
172.31.80.0/20

a /20 = 255.255.240.0. This meanss you're subnetting the Class B address in the third octet in range multiples of 16.
So...starting at 172.16.80.0, go up 16 subnets - 172.16.96.0 is the next subnet.
Anything in the range of 172.16.80.0 - 172.16.95.255 would be your answer.

hth

JDex

Really helpful!

Digitaljunkie

Hi folks

I'm having a little trouble well to be honest banging my head of the wall with questions such as the example below, please help!

Question: What is the broadcast address of the network 172.20.85.192/26?
Answer: 172.20.85.255

Using LordFlasheart's method i'm comming up short every time on these type of questions. Where am I going wrong?

I started with /26 so 26 - 24 = 2
So 2^2 = 4 Block size ?
So the network is 190, 191, 192,193 as this is the block of 4. I thought that 193 ie. 172.20.85.193 was the broadcast.

Is it that i'm miss reading / understanding the method and or the question ie. network and subnetwork?

Dj