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

Spoonroom

gandalph wrote: »

Third octect would look like this

128|64|32|16|8|4|2|1
1|1|1|1|0|0|0|0

11110000 = 240

So how do you get to that?

miller811

Spoonroom wrote: »

Ok, I'm still having problems with these type of questions, where they give an ip and subnet mask, and ask for the host etc. It's been answered before, but I still don't get it.

Originally Posted by Nzastudios
What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
Answer: 10.5.176.1

How does 240 mean that you're borrowing the first 4 bits of the third octet???

the easiest way to figure those out is to do the following

determine the interesting octet. (the non 255. one)
in this case it is the third octet, which is 240
each octet totals 256 (0-255)
so 256-240 = 16
your block size is 16.

so with the address 10.5.178.10 you would start with the non interesting octets (255.255.0.0)
10.5.x.x
since your block size is 16 you would apply it to the base
10.5.0.0
10.5.16.0
10.5.32.0
10.5.48.0
10.5.64.0
10.5.80.0
10.5.96.0
10.5.112.0
10.5.128.0
10.5.144.0
10.5.160.0
10.5.176.0 (winner winner chicken dinner)
10.5.192.0

now you have determined the subnet it belongs in. the next ip address 10.5.176.1 is the first available host

the broadcast address is easily determined by taking the next subnet (10.5.192.0) and subtracting 1 = 10.5.191.255

the last available host would be one address below the broadcast
10.5.191.255 - 1 = 10.5.191.254

hope that helps

Spoonroom

miller811 wrote: »

the easiest way to figure those out is to do the following

determine the interesting octet. (the non 255. one)
in this case it is the third octet, which is 240
each octet totals 256 (0-255)
so 256-240 = 16
your block size is 16.

so with the address 10.5.178.10 you would start with the non interesting octets (255.255.0.0)
10.5.x.x
since your block size is 16 you would apply it to the base
10.5.0.0
10.5.16.0
10.5.32.0
10.5.48.0
10.5.64.0
10.5.80.0
10.5.96.0
10.5.112.0
10.5.128.0
10.5.144.0
10.5.160.0
10.5.176.0 (winner winner chicken dinner)
10.5.192.0

now you have determined the subnet it belongs in. the next ip address 10.5.176.1 is the first available host

the broadcast address is easily determined by taking the next subnet (10.5.192.0) and subtracting 1 = 10.5.191.255

the last available host would be one address below the broadcast
10.5.191.255 - 1 = 10.5.191.254

hope that helps

This made absolutely no sense when I read through it the first time, but then I tried another example, and it worked!

This is so cool!

Thx for the help, lemme go try some more examples...

Cisco Inferno

Could anyone seem to help me with these type of problems? i have no clue how to start. every other question takes me 4 seconds.
heres an example.
thanks

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0/28?
Answer: 4096 subnets and 14 hosts

miller811

Cisco Inferno wrote: »

Could anyone seem to help me with these type of problems? i have no clue how to start. every other question takes me 4 seconds.
heres an example.
thanks

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0/28?
Answer: 4096 subnets and 14 hosts

this is fairly easy to do. It is all based on the powers of 2....
see the chart below.

1 - 2
2 - 4
3 - 8
4 - 16
5 - 32
6 - 64
7 - 128
8 - 256
9 - 512
10 - 1024
11 - 2048
12 - 4096
13 - 8192
14 - 16384
15 - 32768
16 - 65536

So in your example the address given is a class B address. So under normal circumstances it is using a 16 bit mask. 255.255.0.0
In this example it is borrowing 12 of the possible bits for the subnets, and 4 bits for hosts.

so 2 to the 12 = 4096 = to the number of subnets
and then 4 bits are used for the number of hosts 2 to the 4 or 16... but you need to subtract 2 bits, one for the network address and one for the broadcast on each of the 4096 subnets......

Hope that helps.

Cisco Inferno

miller811 wrote: »

this is fairly easy to do. It is all based on the powers of 2....
see the chart below.

1 - 2
2 - 4
3 - 8
4 - 16
5 - 32
6 - 64
7 - 128
8 - 256
9 - 512
10 - 1024
11 - 2048
12 - 4096
13 - 8192
14 - 16384
15 - 32768
16 - 65536

So in your example the address given is a class B address. So under normal circumstances it is using a 16 bit mask. 255.255.0.0
In this example it is borrowing 12 of the possible bits for the subnets, and 4 bits for hosts.

so 2 to the 12 = 4096 = to the number of subnets
and then 4 bits are used for the number of hosts 2 to the 4 or 16... but you need to subtract 2 bits, one for the network address and one for the broadcast on each of the 4096 subnets......

Hope that helps.

aha the key is to use the additional bits on top of already using 255.255.0.0!

thanks. basically lammle's method learned in 30 seconds of reading this post instead of 3 days on a page

sthompson86

Thanks for this post. I agree with the poster above, out of all the ways I have learned sub netting this week, this way is by far the fastest.

I am horrible at math well not horrible, I just do not trust my self doing it in my head, so the less processes I have to do for sub netting the better, for I always write everything out.

Jas21

Hi

I joined to say thanks for this method - it has started to come together for me after struggling with subnetting using other methods for ages - (even Jeremy's CBT Nuggets method)

However, I am still struggling with these types of questions:

Question: You are designing a subnet mask for the 172.27.0.0 network. You want 30 subnets with up to 2000 hosts on each subnet. What subnet mask should you use?

Answer: 255.255.248.0

Is there an easy way of quickly knowing what mask(s) give you number of networks and hosts?

I also have problems subnetting in the 3rd octet when it's a class B address - this gets me every time - class C I have no problems with! For example:

Question: What is the last valid host on the subnetwork 10.194.176.0 255.255.240.0?

Answer: 10.194.191.254

Thanks again - great method and resource - Jason

EDIT - Actually I worked both these out from post#6 on this thread! I am buzzing!!

mattau

hi Jas. This guy helped me alot with how he does it. he uses a system called the magic number.

basically all you have to do is find the magic number and its all easy from there.
you should watch a have of his subnetting vids

YouTube - danscourses's Channel

What is the last valid host on the subnetwork 10.194.176.0 255.255.240.0?

the "important octet is the third octet as it is something other than 255.

what number bit is 240 ? it is the number 16 bit

thats your magic number!

so youre only interested in counting up in 16's in the THIRD octet.

in this case you can just say to yourself. 10 x 16 is 160. 160 + 16 is 176 + another 16 = 192

the subnet range is therefor 176 - 192

host range 177 - 191.254 with the broadcast being 191.255

mattau

same thing for this to

You are designing a subnet mask for the 172.27.0.0 network. You want 30 subnets with up to 2000 hosts on each subnet. What subnet mask should you use?

subnets you take the left bits and hosts are the right most bits

11111111 1111111 0000000 0000000 etc

you know if you want 30 subnets your closest option is the 32 bit as the next 1 up is 64 and that is to many.

just write out the third octet of zeros and treat each zero as 2 and go
0 x 0 = 4 x 0 = 8 x 0 = 16 x = 32

how many bits is that ? 6 bits from the left which is

255.255.248.0

as you borrowed 6 to the power of 2 bits in the THIRD octet

you will pick it up so quick in the vids trust me

Jas21

Mattau - thanks for your replies and link - thanks for taking the time

Jason

veritas_libertas

I'm thankful for this thread. I've read many ways of doing subnetting and this the only one that makes it really simple for me to understand.

CompUBug

Why isnt this thread a sticky?
They need to make this a sticky.

prsreek

hi ...
i am a CCNA beginner..
can anybody explain me how to solve the below given question.
Q) Find out the valid ip range.
172.16.0.0/16 for 300 hosts.

Futura

prsreek wrote: »

hi ...
i am a CCNA beginner..
can anybody explain me how to solve the below given question.
Q) Find out the valid ip range.
172.16.0.0/16 for 300 hosts.

write out 300 in binary, thats how many host bits u need, the rest can be for subnets.

prsreek

Futura wrote: »

write out 300 in binary, thats how many host bits u need, the rest can be for subnets.

i dont get it..
plzz explain ..........

jdfriesen

prsreek wrote: »

hi ...
i am a CCNA beginner..
can anybody explain me how to solve the below given question.
Q) Find out the valid ip range.
172.16.0.0/16 for 300 hosts.

First, you've got to know your powers of 2. 2^8 is 256, which won't be enough, so you've got to go to 2^9 = 512, which gives you room for 510 hosts.

That means your network (assuming the zero subnet can be used, and that you're wanting the first valid address range) will be 172.16.0.0/23 (32-9 = 23).

Next, to figure out your ip range, you've got to calculate the block size. There are a few ways to do that, but the one that I tend to use is subtract the last non-zero subnet mask value from 256. In this case, the subnet mask is 255.255.254.0 (7 leading 1's in the octet = 254). So the block size is 2 (256-254). That means that your first two subnets will be:

172.16.0.0/23
172.16.2.0/23

Since the first and last address in the range are not valid to be assigned, your valid address range would be 172.16.0.1 - 172.16.1.254.

Jas21

Passed my ICND1 today and just wanted to say a massive thanks to Lord Flasheart (and all contributors to this thread) as this was the single biggest component of my finally having that 'Eureka' moment with subnetting!

From really not understanding the process a short while ago, I scored 100% on the "implement and IP addressing scheme and IP services to meet network requirements for a small branch office" section of the exam

Thanks loads everyone

-DeXteR-

Jas21 wrote: »

Passed my ICND1 today and just wanted to say a massive thanks to Lord Flasheart (and all contributors to this thread) as this was the single biggest component of my finally having that 'Eureka' moment with subnetting!

From really not understanding the process a short while ago, I scored 100% on the "implement and IP addressing scheme and IP services to meet network requirements for a small branch office" section of the exam

Thanks loads everyone

Congrats on the Pass

Jas21

Thanks!!

nimrod.sixty9

I cant figure out WTH Im doing wrong... Sometimes it works and sometimes it doesnt...

Heres one that it doesnt:

Question: How many subnets and hosts per subnet can you get from the network 172.26.0.0/27?

Answer: 2048 subnets and 30 hosts

Really? I got nowhere near this.

Can someone please explain?

@Jas21, Congrats!

jdfriesen

nimrod.sixty9 wrote: »

I cant figure out WTH Im doing wrong... Sometimes it works and sometimes it doesnt...

Heres one that it doesnt:

Question: How many subnets and hosts per subnet can you get from the network 172.26.0.0/27?

Answer: 2048 subnets and 30 hosts

Really? I got nowhere near this.

Can someone please explain?

@Jas21, Congrats!

The way I calculate the number of available subnets, is to subtract the CiDR value of the network class, from the CiDR value of the subnets, and then take 2 to the power of whatever the result is. In this case, it's a class B network, so the CiDR value is 16. 27 - 16 = 11, so to calculate the number of subnets, you can do 2^11 = 2048.

To calculate the number of hosts per network, calculate the block size, and subtract 2. In this case, the block size is 32 (/27 = 255.255.255.224, 256 - 224 = 32), so 32 - 2 = 30 hosts per subnet.

-DeXteR-

nimrod.sixty9 wrote: »

I cant figure out WTH Im doing wrong... Sometimes it works and sometimes it doesnt...

Heres one that it doesnt:

Question: How many subnets and hosts per subnet can you get from the network 172.26.0.0/27?

Answer: 2048 subnets and 30 hosts

Really? I got nowhere near this.

Can someone please explain?

@Jas21, Congrats!

since it is a class B address it has a default subnet mask of 255.255.0.0
In the given question /27 means 255.255.255.224 ,you have exactly 5 host bits and 11 subnet bits here
so 2^5 -2 = 30 hosts
similarly 2^11=2048 subnets
Hope that helped

nimrod.sixty9

Im confused, again. This says 8 16 24 and 32 are boundries. "The next boundry is..". So my math was done the same way. 27, the next boundry is 32. So I suppose thats where I messed up. So now I really dont know which one Im supposed to subtract.

How did you figure "5 host bits and 11 subnet bits"? Would either of you like to break down the problem for me?

Thanks for the reponses!

jdfriesen

nimrod.sixty9 wrote: »

Im confused, again. This says 8 16 24 and 32 are boundries. "The next boundry is..". So my math was done the same way. 27, the next boundry is 32. So I suppose thats where I messed up. So now I really dont know which one Im supposed to subtract.

How did you figure "5 host bits and 11 subnet bits"? Would either of you like to break down the problem for me?

Thanks for the reponses!

There are 32 bits total, and 27 are excluded from the host portion, leaving 32-27=5 bits for the host section. Given that it is a class B network, which is /16, there are 27-16=11 bits for the subnet portion.

j-man

nimrod.sixty9 wrote: »

Question: How many subnets and hosts per subnet can you get from the network 172.26.0.0/27?

Answer: 2048 subnets and 30 hosts

Really? I got nowhere near this.

Can someone please explain?

It might help if you diagram it out. (N=Network, S=Subnet, H=Host)

172.26.0.0/27

The 27 indicates the network and subnet. Everything left over are hosts.

It's a class B address so you know that the first 16 bits are for the network and cannot be touched so it will look like the following:

NNNNNNNN.NNNNNNNN.

Now consider the /27. 16 out of the 27 are network which leaves 11 as the subnet number.

NNNNNNNN.NNNNNNNN.SSSSSSSS.SSS

Fill in the last remaining because they are the hosts

NNNNNNNN.NNNNNNNN.SSSSSSSS.SSSHHHHH

Now do the math
S^11 ( 2^11 ) = 2048 Subnets
H^5-2 ( 2^ 5 - 2 ) = 30 Hosts

To figure out the subnet if you want to convert from / notation to decimal remember to add from left to right (or memorize the following)

128, 192, 224, 240, 248, 252, 254, 255

In this case it's a /27 so there are 3 S bits assigned so it's a 224 subnet or
255.255.255.224

I was having a heck of a time until I read through this thread, did both binary and decimal methods and learned some shortcuts. Writing it out in binary really helped me because then I could start visualizing it.

FYI: This is based off what I read in the Odom ICND1 book. I don't want to be accused of plagiarism.

geek4god

j-man wrote: »

It might help if you diagram it out. (N=Network, S=Subnet, H=Host)

Now do the math
S^11 ( 2^11 ) = 2048 Subnets
H^5-2 ( 2^ 5 - 2 ) = 30 Hosts

Any tips on doing the 2 to the power of X math? Or just memorize them? I got 2^9 down do I just need to memorize the rest and if so what is the max power I need?

Cisco Inferno

geek4god wrote: »

Any tips on doing the 2 to the power of X math? Or just memorize them? I got 2^9 down do I just need to memorize the rest and if so what is the max power I need?

just keep doubling.
2,4,8,16,32,64,128.... youll memorize it soon.

j-man

geek4god wrote: »

Any tips on doing the 2 to the power of X math? Or just memorize them? I got 2^9 down do I just need to memorize the rest and if so what is the max power I need?

Nope. Exactly what Cisco Inferno said.

IMHO you should at least have 2^13 or 2^14 ready at will.

This is one of the things I always jot down before any practice exams. I'll start at ^13 and go to at least ^22

-DeXteR-

nimrod.sixty9 wrote: »

Im confused, again. This says 8 16 24 and 32 are boundries. "The next boundry is..". So my math was done the same way. 27, the next boundry is 32. So I suppose thats where I messed up. So now I really dont know which one Im supposed to subtract.

How did you figure "5 host bits and 11 subnet bits"? Would either of you like to break down the problem for me?

Thanks for the reponses!

Think the above posts will clear it for you hopefully ! Btw which book you referring to .Imo todd lammle is good for subnetting for starters