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

Local Native

Wow. I signed up just to say Thank You.

This process allows me to subnet in my head within 5-10 seconds on most questions. I have some trouble with how many hosts can be created, and how many subnets can be created questions ... is that just pure memorization of tables?

I'll post some questions if I still can't get a method around them.

Local Native

gouki2005 wrote: »

can you explain me both for example

What is the valid host range of the 7th subnet of address 10.0.0.0/14?

i use the method so 16-14 = 2
now 2^2= 4

so size of the blocks is 4 for so

10.0.0.0
10.4.0.0
10.8.0.0
10.12.0.0
10.16.0.0
10.20.0.0
10.24.0.0
10.28.0.0

so i thought the valid range was 10.25.0.0 to 10.26.0.0 it works for class C

You should be right, unless the question specifically states you can not use IP Subnet-Zero.

earweed

People keep asking why this isn't a sticky. It's probably because it get's referenced by so many people here at TE that it gets posted to at least weekly and maintains a high spot in the forum on its own.

alan2308

earweed wrote: »

People keep asking why this isn't a sticky. It's probably because it get's referenced by so many people here at TE that it gets posted to at least weekly and maintains a high spot in the forum on its own.

It is referenced in the CCNA FAQ sticky. Which everyone should read through.

And besides that, there's way too many epic threads here to sticky them all. The first 5 pages would all be stickies if we did that.

Local Native

Okay ... I'm having trouble answering questions like this:

You have a Class B network address and need 500 subnets with about 100 hosts per subnet? What subnet mask should be used?

Is this going to be pure memorization of a table listing how many subnets and hosts per mask, or is there a mathematical way to figure it out for each problem? I'd prefer the latter.

miller811

Local Native wrote: »

Okay ... I'm having trouble answering questions like this:

You have a Class B network address and need 500 subnets with about 100 hosts per subnet? What subnet mask should be used?

Is this going to be pure memorization of a table listing how many subnets and hosts per mask, or is there a mathematical way to figure it out for each problem? I'd prefer the latter.

it is simple math and the more you practice the easier it becomes.

a class b address is 255.255.0.0 or a /16 with 16 bits for the network and 16 bits for hosts

solve the first part of the question and it will lead you to the second part of the question.

since you need 500 subnets, first determine how many bits are required to reach 500 at a minimum

2/8 = 256 = /24
2/9 = 512 = /25

so your class B address would be given a mask of 255.255.255.128 or /25

this leave 128 bits-2 or 2/7 left for hosts....on each of the 512 subnets

Somnipotent

this thread really nailed down my subnetting confidence. thanks!!

Somnipotent

miller811 wrote: »

here is my favorite site to test your skills

IP Subnet Practice

click the new problem button and then solve

with practice, you will be able to subnet in your head

great site miller! subnettingquestions.com was getting kinda boring

ScytheX10

Edit: Nevermind

anobomski

Hi,

could someone explain how to use this method method when you are only given a required number of host and clients but no mask.

For example:
Needed usable subnets: 250
Network address: 109.0.0.0
figure out the Custom Subnet Mask

and:
A service provider has given you the Class C network range 209.50.1.0. break it into as many subnets as possible as long as there are at least 50 clients per network.

miller811 posted an explanation to a similar question (below)but i dont get it.

Quote:
Originally Posted by Local Native
Okay ... I'm having trouble answering questions like this:

You have a Class B network address and need 500 subnets with about 100 hosts per subnet? What subnet mask should be used?

ANS:
it is simple math and the more you practice the easier it becomes.

a class b address is 255.255.0.0 or a /16 with 16 bits for the network and 16 bits for hosts

solve the first part of the question and it will lead you to the second part of the question.

since you need 500 subnets, first determine how many bits are required to reach 500 at a minimum

2/8 = 256 = /24
2/9 = 512 = /25

so your class B address would be given a mask of 255.255.255.128 or /25

this leave 128 bits-2 or 2/7 left for hosts....on each of the 512 subnets

bermovick

It's a LOT easier when you're working to match the number of hosts. 100 hosts per subnet. The lowest bit that can encompass 100 hosts is 128: 10000000 (starting in the 4th octet and working left). That gives 8+8+8+1 network bits and 7 host bits = a mask of 255.255.255.128 or /25 in CIDR.

miller811

anobomski wrote: »

Hi,

could someone explain how to use this method method when you are only given a required number of host and clients but no mask.

For example:
Needed usable subnets: 250
Network address: 109.0.0.0
figure out the Custom Subnet Mask

and:
A service provider has given you the Class C network range 209.50.1.0. break it into as many subnets as possible as long as there are at least 50 clients per network.

miller811 posted an explanation to a similar question (below)but i dont get it.

Quote:
Originally Posted by Local Native
Okay ... I'm having trouble answering questions like this:

You have a Class B network address and need 500 subnets with about 100 hosts per subnet? What subnet mask should be used?

ANS:
it is simple math and the more you practice the easier it becomes.

a class b address is 255.255.0.0 or a /16 with 16 bits for the network and 16 bits for hosts

solve the first part of the question and it will lead you to the second part of the question.

since you need 500 subnets, first determine how many bits are required to reach 500 at a minimum

2/8 = 256 = /24
2/9 = 512 = /25

so your class B address would be given a mask of 255.255.255.128 or /25

this leave 128 bits-2 or 2/7 left for hosts....on each of the 512 subnets

always start with this chart to determine the class of address, and then determine the standard mask based on the address

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

For example:
Needed usable subnets: 250
Network address: 109.0.0.0
figure out the Custom Subnet Mask

currently the mask for this subnet would be 255.0.0.0 since it is a class A address

creating subnets from the current address you would start here.

255.128.0.0 would = 2 subnets 109.0.0.0, 109.128.0.0
255.192.0.0 would = 4 subnets 109.0.0.0, 109.64.0.0 x.128.0.0, x.192.0.0
255.224.0.0 would = 8 subnets 109.0.0.0 109.32.0.0 x.64.0.0 ... x.224.0.0
255.240.0.0 would = 16 subnets 109.0.0.0 - 109.240.0.0
255.248.0.0 would = 32 subnets 109.0.0.0 - 109.248.0.0
255.252.0.0 would =l 64 subnets 109.0.0.0 - 109.252.0.0
255.254.0.0 would =l 128 subnets 109.0.0.0 - 109.254.0.0
255.255.0.0 would =256 subnets 109.0.0.0 - 109.255.0.0

so the custom mask of 255.255.0.0 would meet the requirements of the 250 subnets.

and:
A service provider has given you the Class C network range 209.50.1.0. break it into as many subnets as possible as long as there are at least 50 clients per network.

so you know it is a class C address address which means the mask is 255.255.255.0

this allows for up to 254 hosts on with the stand mask. 256-2 = 254

so if you borrow the first bit the new mask would be
255.255.255.128 this would allow 2 subnets from the current address
209.50.1.0, 209.50.1.128 but this borrowing of 1 bit would equal 126 hosts per subnet, 128-2 = 126

255.255.255.192 would borrow 2 bits from the standard mask and would give you 62 hosts per subnet 64-2 = 62

255.255.255.224 would borrow 3 bits from the standard mask and would give you 30 hosts per subnets 32 -2 = 30

so the mask of 255.255.255.192 would meet the requirement

209.50.1.0 = first subnets with hosts 209.50.1.1 - 209.50.1.62
209.50.1.64 = 2nd subnet with hosts 209.50.1.65 - 209.50.1.126
209.50.1.128 = 3rd subnet with hosts 209.50.1.129 - 209.50.1.190
209.50.1.192 = 4th subnet with hosts 209.50.1.193 - 209.50.1.254

Hope that helps

aekash

Question: What is the last valid host on the subnetwork 10.57.240.0 255.255.240.0?
Answer: 10.57.255.254

Hi,

Some help needed please... I am unable to get my head around the above question. Could someone please explain how the answer is 10.57.255.254 ?

I'm am calculating the above as follows:

1. The above IP address is a class A ( i.e. 255.0.0.0)
2. 240 is equal to four ones ( 128+64+32+16)
3. So 255(8 ones) +240 (4 ones) = 12 ones
4. The next boundry is 16, so 16-12 = 4
5. block size is 4
6 so, 10.57.0.0
10.57.4.0
10.57.8.0

7. Counting up in the block size of 4 does not give 10.57.255.254

Thanks,

earweed

aekash wrote: »

Question: What is the last valid host on the subnetwork 10.57.240.0 255.255.240.0?
Answer: 10.57.255.254

Hi,

Some help needed please... I am unable to get my head around the above question. Could someone please explain how the answer is 10.57.255.254 ?

I'm am calculating the above as follows:

1. The above IP address is a class A ( i.e. 255.0.0.0)
2. 240 is equal to four ones ( 128+64+32+16)
3. So 255(8 ones) +240 (4 ones) = 12 ones
4. The next boundry is 16, so 16-12 = 4
5. block size is 4
6 so, 10.57.0.0
10.57.4.0
10.57.8.0

7. Counting up in the block size of 4 does not give 10.57.255.254

Thanks,

When you said LAST valid host you automatically answered your question without requiring any work. Since 10.57.255.255 is the broadcast address the 10.57.255.254 is the LAST valid host.

aekash

earweed wrote: »

When you said LAST valid host you automatically answered your question without requiring any work. Since 10.57.255.255 is the broadcast address the 10.57.255.254 is the LAST valid host.

Thanks for getting back, much appriciated.

I understand the logic but how would counting up in the block size of 4 achive 255 ? Are you able to explain further ?

Are the steps I have used correct ?

earweed

The block size you're using is for subnets and you could find the last valid subnet that way. The last valid host on your last valid subnet would be 10.57.255.254.

aekash

earweed wrote: »

When you said LAST valid host you automatically answered your question without requiring any work. Since 10.57.255.255 is the broadcast address the 10.57.255.254 is the LAST valid host.

OK I think I have got it...

To calculate the block size would it be

(A)
1. The above IP address is a class A ( i.e. 255.0.0.0)
2. 240 is equal to four ones ( 128+64+32+16)
3. So 255(8 ones) +240 (4 ones) = 12 ones
4. The next boundry is 16, so 16-12 = 4
5. block size is 4

or (B)
1. The above IP address is a call A (i.e.255.0.0.0)
2 but the subnet mask which has been given is a 255.255.0.0
3. So 255 (8 ones) +255(8 ones) +240 (4 onnes) = 20 ones
4. The next boundry is 24, so 24-20 = 4
5. Block size is 4
6. 2 to the power of 4 = 16

Based on the block size of 16

172.57.0.0
172.57.16.
172.57.32
172.57.48
172.57.64
miss a few....
172.57.224
172.57.240
172.57.256

Host range betweeen 172.57.240 and 172.57.256 is:
241 -255 ?

Though I thought you cannot use the last broadcast address which is 255 or does this only apply to the last octect ?

Am I meant to subreact 4 from 16 or 24 in terms of the next boundry ? If the subnet mask is provided do I follow the subnet mask or the default subnet mask for the IP Address?

Thanks

lstudent

Hi all, I'm currently studying for CCNA. This post is in regards to the "broadcast subnet."

For example, take the following:

130.4.0.0 / 24 (255.255.255.0)

130.4.0.0 --> Classful network/subnet zero
130.4.1.0 --> First Nonzero subnet
.
.
.
130.4.255.0 --> which is the "broadcast subnet." Thus as are all addresses from 130.4.255.0 - 130.4.255.255 all "broadcast addresses"? or just 130.4.255.255 and 130.4.255.1 to 130.4.255.254 can be used as valid addresses? (This example follows the guidelines that the subnet number address and broadcast address are reserved.)

Please let me know if any of the above info or my questions need clarification. Thank you.

chmorin

lstudent wrote: »

Hi all, I'm currently studying for CCNA and working on finding all the give subnets given an address/mask.

For example, take the following:

130.4.0.0 / 24 (255.255.255.0)

130.4.0.0 --> Classful network/subnet zero
130.4.1.0 --> First Nonzero subnet
.
.
.
130.4.255.0 --> which is the "broadcast subnet." Thus as are all addresses from 130.4.255.0 - 130.4.255.255 all "broadcast addresses"? or just 130.4.255.255 and 130.4.255.1 to 130.4.255.254 can be used as valid addresses? (This example follows the guidelines that the subnet number address and broadcast address are reserved.)

Please let me know if any of the above info or my questions need clarification. Thank you.

Either I missed something in my CCNA classes or you are barking up the wrong tree.

There are really three types of addresses. Network, host, and broadcast. The first address in a subnet is the network address, and the last is the broadcast address. All inbetween is the host addreses.

Using your example:
130.4.0.0/24 is the network address, with 130.4.0.255 as the broadcast address.
130.4.1.0/24 is the network address, with 130.4.1.255 as the broadcast address
.
.
.
130.4.255.0 is the network address, with 130.4.255.255 as the broadcast address.

The broadcast address storms to all host addresses in the subnet, where as the network address is a unique identifier for the subnet.

I have never heard of a broadcast subnet.

To answer your question, 130.4.255/24 is a usable subnet for hosts, excluding of course the network and broadcast address.

lstudent

That answers my question, its probably just a general term for the last subnet in a list of subnets for a given address/mask. p.389 of CCENT/CCNA ICND1 uses the term "Broadcast Subnet" in chart 12-37 as the last subnet when listing all the subnets for a given address/mask.

Thanks!

chmorin

lstudent wrote: »

That answers my question, its probably just a general term for the last subnet in a list of subnets for a given address/mask. p.389 of CCENT/CCNA ICND1 uses the term "Broadcast Subnet" in chart 12-37 as the last subnet when listing all the subnets for a given address/mask.

Thanks!

That is probably an accurate observation.

Gargamel

Morty3 wrote: »

Very much alike Jeremy from CBT nuggets way. Find the last 1 in the mask, that one is your increment. If the last one is a 8, the nets also hop with and 8.

Hi all
I have a little problem with cbtnuggets method to calculate...for example
10.0.0.0 255.0.0.0
and I have to divide in 4 networks
can someone try to calculate and give me a feedback how to do it (in cbtnugets method)...
I need 4 networks:

00000100 is 4 (So it takes 3 bits for number 4)

so next step is to add those bits in subnet mask to get increment and new sub-net mask:

11111111.11100000.00000000.00000000
what makes
255.224.0.0 and increment of 32(last bit in mask)

but subnet calculator gives me this:

Am I making a obvious mistake somewhere...please help..Im gona explode

Sorry guys, found it at the end of lecture, I noticed before that 4 and 32 ect. dont give wright result...so
2, 4, 8 , 16, 32 , 64, 128 give off calculations!!!! Those are exceptions!!!

mikej412

Gargamel wrote: »

00000100 is 4 (So it takes 3 bits for number 4)

?

That is the binary number 4 -- but 3 bits gives you 8 possible bit choices/options/networks -- 000 001 010 011 100 101 110 111 (0 through 7)

1 bit gives you 2 choices (or 2 networks) -- 0 or 1 (0 through 1)

2 bits gives you 4 choices (or 4 networks) -- 00 01 10 11 (0 through 3)

storma

Just like to say thanks to lordflasheart

could any help to me undetstand 23 bit mask
I have been using lordflasheart process but still cant get 23bit mask

the bit which is confusing is where you draw the boundry 24 - 23 =1 this is very a get confused

please help

Thanks storma

binaryhat

Hosts howto:

max = /32
Determining number of host’s:

2^(number of zeroes in the subnet mask)-2

if /26, 32-26=6

2^6-2=62 total hosts

8, 16, 24, 32 boundries

Subnet howto:

1) determine octet position from netmask
2) Take netmask subtract from boundry position
3) do 2^boundry difference = x, block size
4) count up by block size value
5) stop when block size value is close to ip address.

What subnet does 192.168.12.78/29 belong to?

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.

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

Our address is 192.168.12.78 so it must sit on the
192.168.12.72 subnet.

MD75028

Glad I found this post, it really helped me understand.

laidbackfreak

This should be made a sticky.

Spoonroom

I've just gone through the first page of this post now and already understand subnetting a lot better than from going through a lot of other books and tutorials. Thanks a lot!

Spoonroom

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.

Nzastudios wrote: »

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

240 mask means you are borrowing the first 4 bits of the third octet.

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

gandalph

Third octect would look like this

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

11110000 = 240