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

WickedAngel

Trifidw wrote: »

The IP address is a class b network that has a default prefix of /16 (255.255.0.0). If you subnet this into networks that provide 62 usable hosts you get 1024 subnets.

I think I see. For whatever reason, I had it in my head that I could calculate the # of subnets/hosts based on the CIDR designation without respect to IP address given.

If I'm understanding your post correctly, I just happened to get away with it for those others because the CIDR notations fell within the same octet as the default subnets of their associated IP addresses anyways.

rhone12

Hi Guys,

I'm just starting learning CISCO 1 - 4 modules and I'm mastering the best way of subnetting but there are still cases that I cannot answer subnetting questions correctly. Would you mind if you suggest me the easiest way? Also, can you please give me the steps on how to get the answer of this question below. I got this from subnettingquestions.com.

Question: How many subnets and hosts per subnet can you get from the network 172.21.0.0 255.255.255.240?
Answer: 4096 subnets and 14 hosts

* I always got 12 subnets and 14 hosts as my answer. Thank you!

lantech

Rhone,

I see how you got 14 hosts per subnet. But how did you get 12 subnets? I can't figure out how you managed that. Can you describe to us how you are figuring this out?

rhone12

Hi lantech,

I already checked it again and it's my mistake. Actually, I adapted LordFlasheart tips on how to do easy subnetting and it really works. I just forgot to raise 2 by 12 to get 4096 subnets.

This is how I got it: My IP is class B is also equal to /16 and my CIDR is /28 is also equal to 255.255.255.240. So, in LordFlasheart technique, 28 - 16 = 12. That's why I got 12 but my mistake, I forgot to raise 2 by 12 to get 4096.

I would like also to ask if my below technique is also applicable in getting Hosts and Subnets in any class of IPs w/o getting any errors. I learned it from my instructor and I don't know from whom he learned it also.

Here it goes: To get Host: 2 ^ (y) - 2 = hosts; where y is the nos. of zeros in the group depends on the class in the host group nearest from the network side starting from left. For example: /26 means 11111111.11111111.11111111.(11000000) = 6 zeros

To get Subnet: 2^(x)-2 = subnets; where y is the nos. of ones in the group depends on the class in the host group nearest from the network side starting from left. For example: /26 means 11111111.11111111.11111111.(11000000) = 2 ones.

Thank you!

knightrider56

Hi. I know its a dumb question but its not getting thru my head. If I have an address such as the following where the mask says /24. its right on the boundary for the first example post.
For an address such as this : 192.168.10.1/24 Would the subnet be 192.168.10.0 ?

or did I miss something in the explanation ?

for this one: 212.14.29.5/30 Would it be a subnet mask of 122.14.28.0 ? am i getting this or did i miss the boat?

veritas_libertas

knightrider56 wrote: »

Hi. I know its a dumb question but its not getting thru my head. If I have an address such as the following where the mask says /24. its right on the boundary for the first example post.
For an address such as this : 192.168.10.1/24 Would the subnet be 192.168.10.0 ?

You didn't miss anything. Don't think of it as a boundary, but rather as the Octet that you are in, which is what /24 would symbolize since:

8.16.24.32

In binary: 11111111.11111111.11111111.00000000

knightrider56 wrote: »

for this one: 212.14.29.5/30 Would it be a subnet mask of 122.14.28.0 ? am i getting this or did i miss the boat?

No it would be 212.14.29.4

The available host IPs are (212.14.29.5 - 212.14.29.6) Broadcast is 212.14.29.6

This is because 32-30=2 2^2=4

knightrider56

veritas_libertas wrote: »

You didn't miss anything. Don't think of it as a boundary, but rather as the Octet that you are in, which is what /24 would symbolize since:

8.16.24.32

In binary: 11111111.11111111.11111111.00000000

No it would be 212.14.29.4

The available host IPs are (212.14.29.5 - 212.14.29.6) Broadcast is 212.14.29.6

This is because 32-30=2 2^2=4

OK. But I still dont get this one.... 192.168.10.1/24 Would the subnet be 192.168.10.0 ?

What do you subtract from 24 to get your number to work with? or do you subtract anything ???

would 10.0.0.1/32 be subnet 10.0.0.0 ?

veritas_libertas

Yes 192.168.10.1 is the subnet.

There is nothing to subtract since /24 = a Class C network = 255.255.255.0 Since /24 is not being divided up, i.e. none of the eight host bits are being switched over to network bits 11111111.111111111.11111111.00000000 there is on the /24 network. 1-254 being the hosts with 0 being the network and 255 being the broadcast address.

Does that help?

drkat

I'm not sure here if we covered this but

Your boundary method is applied to the subnetting of the classful network. The hosts requirement is done based on the remaining 0's after you've subnetted your original network.

Quoting a previous post:

Question: What valid host range is the IP address 172.26.110.111/24 a part of?
Answer: 172.26.110.1 through to 172.26.110.254

How is this calculated? this is a class B n/w. I am confused.

The confusion comes from the formula - so when we were in high school the math teach would give you a formula to plug your numbers into - This really is no different.

172.26.110.111/24We have a class B address with 8 bits borrowed.. default class of /16 + 8 gets your /24

If we look at it from the very beginning... from the original network

172.16.0.0/16 has a mask of 255.255.0.0 - we add 8 bits on now we have a mask of 255.255.255.0 so now 172.16.0.0/16 became 172.16.0.0/24
So our network is now 172.16.0.0 1-254 and 255 for broadcast
172.16.1.0 is our next network
172.16.2.0 next and so forth

What we did was make 172.16.0.0 larger in terms of available networks (by adding the 8 bits) and smaller in terms of hosts as the case was reverse before subnetting it. This is how we conserve hosts and why subnetting was put in place to begin with so we dont waste 2^16 hosts in the original class B network.

knightrider56

veritas_libertas wrote: »

Yes 192.168.10.1 is the subnet.

There is nothing to subtract since /24 = a Class C network = 255.255.255.0 Since /24 is not being divided up, i.e. none of the eight host bits are being switched over to network bits 11111111.111111111.11111111.00000000 there is on the /24 network. 1-254 being the hosts with 0 being the network and 255 being the broadcast address.

Does that help?

I am trying to set up a router using FastEthernet0/0 to connect to another routers FastEthernet0/0

I tried using this command:

Router0(config-if)#ip address 192.168.10.1 192.168.10.0
Bad mask 0xC0A80A00 for address 192.168.10.1

and get this Bad mask error. Isnt this the subnet mask 192.168.10.0 for 192.168.10.1?

fadhil

kriscamaro68 wrote: »

Ok so I got this question on subnettingquestions.com and didn't see a way to answer it with your method: Question: You are designing a subnet mask for the 10.0.0.0 network. You want 3800 subnets with up to 3800 hosts on each subnet. What subnet mask should you use? Answer: 255.255.240.0 Here is also another one: Question: How many subnets and hosts per subnet can you get from the network 172.29.0.0 255.255.254.0? Answer: 128 subnets and 510 hosts Any way of explaining that out in your method. Thanks for that post I am finally starting to get subnetting.

always when you are doing these questions make sure,you meet the requirement of the customer. for the first question the requirement was 3800 subnet and 3800 host. for subnet make sure that you find power of 2,that is greater than or equal to number of subnet required 2^n>=3800. where n is the number of bits that have bee turn on. we get n equal to 12 that will be 4096 subnets in total. hence 12 will help to find the subnet mask and to check if number of host will be sufficient . this 12 is needed to set the left most bits of default mask(given mask) from zeros to ones, i.e to turn on the bits hence the subnet mask will be:- from default 255.0.0.0 in binaary 11111111.00000000.00000000.00000000 hence put those 12 bits in a subnet mask 11111111.11111111.11110000.00000000 in decimal 255.255.240.0 from this subnet will get 4094 hosts that is ( 2^f)-2 =4094 where f is the number of zeros from the subnet mask you found. 255.255.240.0. for question 2: for this question you didn't given number of subnets and hosts,just you just required to find them so to find this question,you have to work with subnet mask. from the subnet mask given 255.255.254.0 in binary numbers 11111111.11111111.11111110.00000000 so To number of host is =2^f -2 where f is number of zeros in a subnet mask. hence 2^9 -2= 510,the number of hosts. Easy to find number of subnet,we should look on class of the ip address given. the ip address is in class B, It's default subnet mask is 255.255.0.0. Hence from given subnet mask 255.255.254.0, It is seems like 7 bits have been turn on from default subnet mask. hence number of subnet will be 2^n, whrere n is the number of bits that have been turn on. 2^7=128. hence your answer are 128 number of subnets and 510 number of hosts

drkat

knightrider56 wrote: »

I am trying to set up a router using FastEthernet0/0 to connect to another routers FastEthernet0/0

I tried using this command:

Router0(config-if)#ip address 192.168.10.1 192.168.10.0
Bad mask 0xC0A80A00 for address 192.168.10.1

and get this Bad mask error. Isnt this the subnet mask 192.168.10.0 for 192.168.10.1?

No, your subnet is .0 but your mask is 255.255.255.0

veritas_libertas

knightrider56 wrote: »

I am trying to set up a router using FastEthernet0/0 to connect to another routers FastEthernet0/0

I tried using this command:

Router0(config-if)#ip address 192.168.10.1 192.168.10.0
Bad mask 0xC0A80A00 for address 192.168.10.1

and get this Bad mask error. Isnt this the subnet mask 192.168.10.0 for 192.168.10.1?

I don't mean to be insulting, but are you using a book to guide you through studying? If not you are only going to get confused and frustrated. We are more than glad to help, and I don't mean to make you feel otherwise.

fadhil

you didn't write the subnet mask,that's why an error occurred. the subnet mask looks like 255.225.255.xxxxxxxxx for class C ip address . you were suppose to configure like 198.168.10.1 255.255.255.xxxxxxxxx

Tfoote01

For a very long time I have worked on mastering subnetting. I have tried many techniques and this method is ninja fast! I took notes, wrote out powers, masks, and any notes I needed and I can't tell you how blown away I am. I can burn through subnetting questions now. It is laughable now. A giant click happened with practicing and really poring over mistakes I made. I feel confident that I have subnetting 90% mastered. I need more practice with VLSM. Examples with multiple subnets and multiple host requirements are what I need to practive. I really feel confident about subnetting now! dear lord! It's like a giant dawning!

fadhil

The question is long but it is simple and has many approaches to reach to the answer/conclusion. you can use subneting only or subnetting and VLSM to solve the problem because both ways meet the requirements of a customer.
thank u.

FLEOHB

Is this site a good place to practice this method of subnetting? If not, know of any better ones?

MAC_Addy

FLEOHB wrote: »

Is this site a good place to practice this method of subnetting? If not, know of any better ones?

Yes! Many people on here use it including myself. Brilliant place to practice!

FLEOHB

Sweet, thanks MAC_Addy

LinuxRacr

Awesome! I have something similar from howtonetwork.net in the form of an iPhone app.

LinuxRacr

Having trouble understanding the following question:

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

Answer: 192.168.14.65 - 192.168.14.94

j-man

LinuxRacr wrote: »

Having trouble understanding the following question:

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

Answer: 192.168.14.65 - 192.168.14.94

First think about how many bits you need to create 8 subnets. 3 bits will do that. So that means your increment will be 32.

1st subnet 0 - 31 - valid host range 1 through 30
2nd subnet 32 - 63 - valid host range 33 through 62
3rd subnet 64 - 95 - valid host range 65 through 94

LinuxRacr

j-man wrote: »

First think about how many bits you need to create 8 subnets. 3 bits will do that. So that means your increment will be 32.

1st subnet 0 - 31 - valid host range 1 through 30
2nd subnet 32 - 63 - valid host range 33 through 62
3rd subnet 64 - 95 - valid host range 65 through 94

The red text above introduced a new concept to me to try to understand or reverse-engineer...

If the increment is 32, then that means that 2^5 is the blocksize, or increment. That means to get that, since this is a class C address default mask, we had to subtract 27 from 32 boundary to get the 5 to plug into 2^5. So by this logic, the CIDR is /27. The three bits you are referring to are the extra bits added past the /24 CIDR boundary.

Now my question is how did you figure out that 3 bits would work so quickly?

j-man

The question stated that we subnetted that class C network into 8 subnets. 2^3 = 8

The resulting subnet mask for that network would be 255.255.255.224 or in binary

11111111.11111111.11111111.11100000 The bits with 1 are the number of subnets created with this class C mask and the 0 bits are the number of hosts - 2 (2^5 - 2) or 30 because remember that the subnet number and broadcast are not usuable for hosts.

jamesp1983

Excellent technique. I have always enjoyed the dotted decimal method (256 - value present in interesting octet). Once you understand subnetting keep practicing it. You need to be fast and accurate at this.

jamesp1983

Tfoote01 wrote: »

For a very long time I have worked on mastering subnetting. I have tried many techniques and this method is ninja fast! I took notes, wrote out powers, masks, and any notes I needed and I can't tell you how blown away I am. I can burn through subnetting questions now. It is laughable now. A giant click happened with practicing and really poring over mistakes I made. I feel confident that I have subnetting 90% mastered. I need more practice with VLSM. Examples with multiple subnets and multiple host requirements are what I need to practive. I really feel confident about subnetting now! dear lord! It's like a giant dawning!

Congrats! It is an excellent feeling once you get it.

LinuxRacr

Thanks for the explanation. This is adding to the angles of attack in my subnetting arsenal.

radix

Having trouble how to answer these question from Lammle's book

You have a Class B network and need 29 subnets. What is your mask?
Answer: 255.255.248.0

is there a way to solve this or do we need to memorize the netmask?

MAC_Addy

radix wrote: »

is there a way to solve this or do we need to memorize the netmask?

You can use my chart if you want.

But just remember that Class B you can count from:

255.255.0.0 = 1 subnet
255.255.128.0 = 2 subnets
255.255.192.0 = 4 subnets
255.255.224.0 = 8 subnets
255.255.240.0 = 16 subnets
255.255.248.0 = 32 subnets
255.255.252.0 = 64 subnets
255.255.254.0 = 128 subnets

That's how I figure out how many subnets for each class.

subnetting chart.pdf

radix

MAC_Addy wrote: »

You can use my chart if you want.

But just remember that Class B you can count from:

255.255.0.0 = 1 subnet
255.255.128.0 = 2 subnets
255.255.192.0 = 4 subnets
255.255.224.0 = 8 subnets
255.255.240.0 = 16 subnets
255.255.248.0 = 32 subnets
255.255.252.0 = 64 subnets
255.255.254.0 = 128 subnets

That's how I figure out how many subnets for each class.

oooh thanks will be a great help