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

beach5563

This is a good method overall but I don't know if this is good for beginning subnetting. Seems like you may need to learn some basic stuff like the powers of 2 and maybe address ranges. I could be wrong but I think I will work on that stuff first so you can have an idea of whats going on underneath everything.

mohammadzzz

hi

Got this question on www.subnettingquestions.com,Please can any one explain this

what valid host range is the ip 172.21.88.41 255.255.252.0 a part of ?anserwer 172.21.88.1 to.172.21.91.254

my calculation

256-252 =4,block size 2^4=16 so

1st network 172.21.0.0
2nd network 172.21.16.0
3rd network 172.21.32.0
4th network 172.21.48.0
5th network 172.21.64.0
6th network 172.21.80.0
7th network 172.21.96.0

so my calculation say it is part of major 172.21.80.0 to be exact and goes to last limit of 172.21.95.254,but the answer from site says 172.21.88.1 to 172.21.91.254;can someone please explain where i have gone wrong ?Thanks

mohammadzzz

Thank you .It helps a lot

mella060 wrote: »

So assume the networks in your example are...

10.0.0.0/24
10.0.1.0/24
10.0.2.0/24
10.0.3.0/24

Step 1 - Re-list the subnet ID's (numbers) and prefix lenghs in decimal and calculate the subnet broadcast addresses.

10.0.0.0/24 - broadcast address is 10.0.0.255
10.0.1.0/24 - broadcast address is 10.0.1.255
10.0.2.0/24 - broadcast address is 10.0.2.255
10.0.3.0/24 - broadcast address is 10.0.3.255

Step 2 - Note the low and high end of the range of addresses for all combined subnets by noting the numerically lowest subnet ID and the numerically highest subnet broadcast address. So in your example, 10.0.0.0 would be the lowest subnet ID and 10.0.3.255 would be the highest broadcast address.

Step 3 - Pick the shortest prefix length of all subnets to be summarized (subordinate) subnets, and then subtract 1. So in your example, all the subnets have a prefix length of /24, so subtract 1 from 24...24-1 = 23. So calculate the range of addresses you would get with a prefix length of /23.

10.0.0.0/23 - 10.0.0.1 - 10.0.1.255...doesn't match all the addresses to be summarized. So subtract 1 from 23 and see if that includes all the subnets and their addresses to be summarized.

10.0.0.0/22 - range 10.0.0.1 - 10.0.3.255...This exactly matches the range of addresses shown above so you have found the subnet and mask to use in the best summary route: 10.0.0.0/22

Answer A 10.0.0.0/21 would also work but it is not the best summary route in this example. The best summary route is the one that includes all the subnets to be summarized, but with as few other addresses as possible.

xnx

mohammadzzz wrote: »

hi

Got this question on www.subnettingquestions.com,Please can any one explain this

what valid host range is the ip 172.21.88.41 255.255.252.0 a part of ?anserwer 172.21.88.1 to.172.21.91.254

my calculation

256-252 =4,block size 2^4=16 so

1st network 172.21.0.0
2nd network 172.21.16.0
3rd network 172.21.32.0
4th network 172.21.48.0
5th network 172.21.64.0
6th network 172.21.80.0
7th network 172.21.96.0

so my calculation say it is part of major 172.21.80.0 to be exact and goes to last limit of 172.21.95.254,but the answer from site says 172.21.88.1 to 172.21.91.254;can someone please explain where i have gone wrong ?Thanks

255.255.252 = /22 = Goes up in 4's on the 3rd octet

172.21.80.0
172.21.84.0
172.21.88.0

goat1974

I am new to Subnetting and taking a class a UMUC, I need to subnet 211.168.122.0/24 to 11 subnets and for subnets #1, #4, #5, and #11 I need to know The new subnet mask after the Subnetting, Subnet’s network address, Subnet’s broadcast address, Subnet’s range of available IP addresses and The calculations on how you get to the answers

mohammadzzz

Yes,thanks but i am still confused,if you follow the original method in this post,then i do not get the networks what you have mentioned.Please stipulate.

xnx wrote: »

255.255.252 = /22 = Goes up in 4's on the 3rd octet

172.21.80.0
172.21.84.0
172.21.88.0

EdTheLad

mohammadzzz wrote: »

my calculation

256-252 =4,block size 2^4=16 so

256-252=4, block size is 4 2^2 = 4

255 = 11111111
252 = 11111100

two 0's means a possible 4 combinations hence a block of 4!

mohammadzzz

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.254.0?

since its a class c,default mask is 255.255.255.0,block size is 256-254=2,but no idea what to do after this as 254/2=127 subnets and hosts are 2^8=256

answer is Answer: 128 subnets and 510 host,

but how.....any help will be apperciated.regards

EdTheLad

172 is class B.

mohammadzzz

hi

i am not too sure how to answer questions like this.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

any help

ednard

mohammadzzz wrote: »

hi

i am not too sure how to answer questions like this.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

any help

You need to write down the binary to get these figures, unless you're quick enough to do it in your head.

So 172 is a Class B network, so let's write down the binary for 172.21.0.0.

E.G. 1 - 11111111.11111111.00000000.00000000

The subnet they've given is 255.255.255.240, though, so we need to write this in binary now, and compare the two.

E.G. 2 - 11111111.11111111.11111111.11110000

So we've got the 16 network bits from the class B network (shown in E.G. 1), and now with the subnet they've given us (shown in E.G. 2), we've got another 12 bits for the subnet (as well as the 16 bits for networks), as well as 4 for the host part (shown in 0's). Therefore:

2^12 = the amounts of subnets we can make
2^4 (-2) = the amounts of hosts we can have (we -2 for the network address and broadcast address)

mohammadzzz

Thanks ednard.It helps.

mohammadzzz

Hi

i cant figure this one out...any explanation

A company needs 300 subnets and 50 hosts per network.We have 1 class b ip.What mask will we use

255.255.255.0

255.255.255.128

255.255.252.0

255.255.255.224

255.255.255.192

I have tried every thing i can think off,like for 50 hosts we need 2^6 bits-2=62 hosts

so 255.255.255.nn h h h h h h (so now we have 2+4+8+16+32+64=126),beyond that i am clueless how he drived value of .192 in last octet

255.255.255.126

any help please.

correct answer

255.255.255.128 & 255.255.255.192

MrPuzzlez

Remember! It's not cheating if you use your fingers! That's what I do every now and then.

hell911

thanks for this topic..

but i also used this video to learn, and.. it's was so easy!!

https://www.youtube.com/watch?v=KDig0lOO95M

Kenneth196

mohammadzzz wrote: »

Hi

i cant figure this one out...any explanation

A company needs 300 subnets and 50 hosts per network.We have 1 class b ip.What mask will we use

255.255.255.0

255.255.255.128

255.255.252.0

255.255.255.224

255.255.255.192

I have tried every thing i can think off,like for 50 hosts we need 2^6 bits-2=62 hosts

so 255.255.255.nn h h h h h h (so now we have 2+4+8+16+32+64=126),beyond that i am clueless how he drived value of .192 in last octet

255.255.255.126

any help please.

correct answer

255.255.255.128 & 255.255.255.192

Hello mohammadzzz,

For this particular problem, the key thing to note is that you have 1 class B IP address. Knowing that, the first 2 octets are notated as the network portion automatically. Those cannot be changed. If it was a class A IP address, only the first octet would be the network portion at a minimum, and if a class C IP, then the first three octets would be the network portion at a minimum. Now that's not to say that the other octets cant have network bits as well. i.e. Even though the first two octets of a class B IP address are network bits, the third octet can have all network bits also, which would mean that it has a subnet mask of 255.255.255.0
Sooooo.... we know that you have a class B IP address, you need atleast 300 subnets, and 50 hosts per subnet. The first two octets are automatically assigned as network bits so ignore those. Count on your hand(s), using the power of 2 until you get to 300 or over. So 2, 4, 8, 16, 32, 64, 128, 256, 512. How many fingers are you holding up? Should be 9. That means atleast 9 bits will have to be "borrowed" to have atleast 300 subnets. So knowing this, the first two octets are automatically assigned as network bits, and we also used 9 other bits for the network portion. Soooo... nnnnnnnn.nnnnnnnn.nnnnnnnn.nhhhhhhh - the first two octets are network bits, then we have 9 more network bits, and the rest are host.

Elijah09

Thanks, this was really helpful!

hell911

I have a problem, hope someone can help.

I have these values:

network address: 172.50.0.0
def subnet mask: 255.255.0.0
custom subnet mask: 255.255.224.0
total # of subnets: 8
number of usable subnets: 6
total # of host addresses: 8192
usable host address: 8190
# of bits borrowed: 3

if the question is, find first four subnets of this network. is the answer going to be...

172.50.0.0 - 172.50.0.255
172.50.1.0 - 172.50.1.255
172.50.2.0 - 172.50.2.255
172.50.3.0 - 172.50.3.255

Kenneth196

Assuming ip subnet zero is in play, then you are correct for the /16 subnet (or 255.255.0.0).

xnx

hell911 wrote: »

I have a problem, hope someone can help.

I have these values:

network address: 172.50.0.0
def subnet mask: 255.255.0.0
custom subnet mask: 255.255.224.0
total # of subnets: 8
number of usable subnets: 6
total # of host addresses: 8192
usable host address: 8190
# of bits borrowed: 3

if the question is, find first four subnets of this network. is the answer going to be...

172.50.0.0 - 172.50.0.255
172.50.1.0 - 172.50.1.255
172.50.2.0 - 172.50.2.255
172.50.3.0 - 172.50.3.255

172.50.0.0 /19 is:
172.50.0.0 - 172.50.31.255
172.50.32.0 - 172.50.63.255
172.50.64.0 - 172.50.95.255

etc etc..

hell911

why did the 3rd octet had a jump of 32? why not different number?

xnx

256-224 = 32

Also:

When a default mask is 255.255.0.0 that means it's a class B and has 16 bits on.

255.255.224.0 = 19 bits on (3 bits for subnet) > 2^3 = 8

Now you have 13 bits left for hosts > ((2^13)-2) = 8190

whiteje84

Thank You for this! It helped me a ton. Slow at first but I think it helped make things finally "click". Cheers!!

abyssinica

Thanks so much for this tutorial, it's fabulous, well-explained and easy to understand!

hmig89

Hi,

Thanks for posting this method.

It works however could up in the block sizes until I get up 128 or 200's take a lot of time. Is there a quicker way do this?

saraanand

I see such neat explanations here, just looking for a super like button.

hmig89

Hi,

This is by far the easiest method I have found, however I am struggling with the following question:
[h=2]What subnet mask would you use for the 172.18.0.0 network, such that you can get 30 subnets and 1090 hosts per subnet?[/h]
Can someone explain to me how I would work out this question using this method?

mikeybinec

Well, you're looking at a Class B IP range. The two octets on the right are where you will subnet. So, to get 32 networks "borrow" 5 bits 2^5 = 32 networks and what's left is your hosts per network 2^11 = 2048

XXXXXXXX.XXXXXXXX.NNNNNHHH.HHHHHHHH

Deathmage

I keep forgetting subnetting and then I look back on this post and I remember it. I dunno why like in 10 days I'll forget subnetting but as soon as I look at this I figure it out all over again. Maybe I just don't subnet enough at work for it to be solidified in my brain.

Anyone got some pointers on how to keep it sharp in your brain?

this post should really be made into a sticky....

Deathmage

Can this be made a sticky, spent 25 minutes looking for it....

every time I forget subnetting this page help re-plant the seed...