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

oli356

I didn't have as many subnetting questions on my ICND1 today as I thought there would be. The ones I had as well I found very easy.

LinuxRacr

That's exactly the impression I got.

oli356

I did 2 practise exams through the networking acadamy and they had A LOT more in. Not complaining though.

drkat

The subnetting questions on the exam(s) arent really that hard.. if you're doing well on subnetting-questions then you can nail the CCNA - the random ones on the site are 10x tougher

oli356

drkat wrote: »

The subnetting questions on the exam(s) arent really that hard.. if you're doing well on subnetting-questions then you can nail the CCNA - the random ones on the site are 10x tougher

Most of mine were how many hosts can be used, we need 'x' hosts what subnet mask should we use? So easy yes. I've done ones where you have to find the network of the hosts and then the routers interface and find that network, takes a lot more time.

Does anyone else not like using the ANDing method? I feel like it would take me longer to work out the network address in binary than it would to just find what the network ID is going up in, for example 4 and then just figure it out from there..

chaddsuk

Hey

I actually made a video on youtube showing the magic number method for anyone that is struggling with this, you can find it at the following link if your interested.

Subnetting Made Easy - CCNA exam method - Magic number NEW - YouTube

I did make a subnetting video on youtube around a year or two ago thinking no one would watch it however 32,000 people watched it so I felt compelled to make another but this time without a bunch of typo's!

Hope it helps someone

beach5563

chaddsuk wrote: »

Hey

I actually made a video on youtube showing the magic number method for anyone that is struggling with this, you can find it at the following link if your interested.

Subnetting Made Easy - CCNA exam method - Magic number NEW - YouTube

I did make a subnetting video on youtube around a year or two ago thinking no one would watch it however 32,000 people watched it so I felt compelled to make another but this time without a bunch of typo's!

Hope it helps someone

I actually just took a quick look at both of your videos. I was wondering do you suggest learning all the binary stuff and basic IP addressing stuff first? It seems like it kind of starts at a more advanced level but I could be wrong. I'm thinking if I could get the subnetting stuff down real real good I may just go ahead and take the one test route. I have taken CCNA classes in the past and have had some training on switches and routers. I really like your videos though.

rukisama

Thank you so much for this method. Subnetting was something I understood in a kind of general way, but the knowledge of the nuts and bolts of it always seemed elusive. After using this method and practicing for about a day, though, I can solve any subnetting problem in my head in (at most) 30 seconds, and usually more like 10 seconds. I really appreciate it!

FloOz

wow this is great. subbed to thread for future refrence

DoubleD

chaddsuk wrote: »

I did a quick youtube video on the subnetting technique I use for exams which I found invaluable at the time (and still continue to use it). Of course I took the time to learn the "and or" binary method first however check out the video if you feel you need to speed things up. (Hope it helps someone)

Subnetting Made Easy - CCNA exam method - Magic number - YouTube

I like this thread and this site
Learn To Subnet: A Free, Lecture-Based Presentation on IP Addressing and Subnetting

thanks for the thread

dazl1212

Mods, could this be made a sticky please?

JustFred

I agree, making it sticky would help a lot of folks and also avoid all the other subnetting questions all over the place. You can ask every subnet question here.

dazl1212

JustFred wrote: »

I agree, making it sticky would help a lot of folks and also avoid all the other subnetting questions all over the place. You can ask every subnet question here.

This man knows

RouteMyPacket

I have to admit subnetting has been kinda funny to revisit now that I plan to sit a test. I like so many others just use a calculator in the real world but having gone through this in school years ago I remembered the basics of subnetting on paper and i'm trying to figure which way is easiest for me to get through the exam.

For me right now I have to write it out on paper to get the answers most of the time

RouteMyPacket

LinuxRacr wrote: »

All, I found this subnet calculator years ago to help with some sys admin work I was doing (since I wasn't interested then in being a network guy), and it is one of the best I've seen:

Network Calculators

You can calculate what you need, and then hit the explain button to break down the why, and it goes in-depth with the explanation.

Wow, very cool calc..thanks. I've used this one for years

Online IP CIDR / VLSM Supernet Calculator

MAC_Addy

RouteMyPacket wrote: »

Wow, very cool calc..thanks. I've used this one for years

Online IP CIDR / VLSM Supernet Calculator

I believe that using a calculator for subnetting practice to be a bit redundant. You need to be able to figure it out in your head easily without using a calculator for the exams. In the real world, this might be useful. But if you learn subnetting and actually understand it, then you won't ever need a calculator.

NetworkVeteran

MAC_Addy wrote: »

But if you learn subnetting and actually understand it, then you won't ever need a calculator.

Needing to use a calculator would've blocked me from obtaining my current job. Before relying on them, I would consider whether networking is one of your core skills, or it's just something you want to know a little bit about in support of your other skills.

(Similarly, while the argument has been made before we do not need to know basic multiplication and addition given calculators, your career growth may be challenged if your role involves lots of mathematics and you rely on a calculator for those.)

RouteMyPacket

MAC_Addy wrote: »

I believe that using a calculator for subnetting practice to be a bit redundant. You need to be able to figure it out in your head easily without using a calculator for the exams. In the real world, this might be useful. But if you learn subnetting and actually understand it, then you won't ever need a calculator.

A calculator for "practice" and for sitting the exam is not very good I agree. You can know subnetting all you want, in the real world you'll grab a calculator like everyone else.

NetworkVeteran

RouteMyPacket wrote: »

You can know subnetting all you want, in the real world you'll grab a calculator like everyone else.

Again, I would be ineligible for my present role if I needed to use a calculator. In the real world, there are many who have learned subnetting well enough to just know that a /24 will satisfy the need for a subnet with at least 250 IPs without having to whip out their laptop, go online, and use a subnetting calculator. I don't use such calculators.

(Similar, I do not use a calculator when I need to know what 8x8 is!)

I certainly don't deny there are many roles where that level of subnetting knowledge is not required. Some teams are expected to know less about networking. Those teams may have strengths in other areas, or they may simply be compensated less.

MickQ

In the real world I use my head. I use a subnet calc when I want to be really sure of something. If you don't exercise your brain, it'll go flabby.

SlayerX

How did Lordflasheart arrive at this final ip address 192.168.12.72? I understood all of it up until this point. Can anyone explain it to me in a simple way that is understandable?

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.

Excellent1

The address 192.168.12.72 is the subnet number for the address in question (192.168.12.7

. He is showing that because the mask is a /29, the block size is 8 (3 host bits, or 2 to the 3rd power- 2 x 2 x 2), the subnets increment by 8. That subnet encompasses all addresses within the .72 to .79 range, with 72 being the subnet and 79 being the broadcast address, with 73-78 being valid hosts for the .72 subnet.

Does this help or did I misunderstand your question?

allaboutjo

I would first like to say thanks for the post. The write up seems to be pretty straight forward. The only problem is, I am just getting my toes wet with subnetting and need a little guidance. A class I am taking has asked a specific question and I am having trouble deciphering it using the method.

We are using a Class C network of 204.15.5.0/24. Subnet the network in order to create five subnets. The largest subnet must support 28 host addresses. Is it possible with a class c network?

Is this the workup I should be looking at in order to complete the assignment?

Ismaeljrp

allaboutjo wrote: »

I would first like to say thanks for the post. The write up seems to be pretty straight forward. The only problem is, I am just getting my toes wet with subnetting and need a little guidance. A class I am taking has asked a specific question and I am having trouble deciphering it using the method.

We are using a Class C network of 204.15.5.0/24. Subnet the network in order to create five subnets. The largest subnet must support 28 host addresses. Is it possible with a class c network?

Is this the workup I should be looking at in order to complete the assignment?

Yes you can understand how to solve with this thread. I don't want to answer for you, so I need to ask you; do you understand the binary concepts related to addresses, and the borrowing of bits? That's really all you need to know, and how many hosts/networks you get from creating a new mask.

Here is a good video to help you with why we subnet, and about subnet masks.
Subnet Masks - Part 1 - CompTIA Network+ N10-004: 1.4 - YouTube

allaboutjo

Ismaeljrp wrote: »

Yes you can understand how to solve with this thread. I don't want to answer for you, so I need to ask you; do you understand the binary concepts related to addresses, and the borrowing of bits? That's really all you need to know, and how many hosts/networks you get from creating a new mask.

I know from general knowledge there are 8 bits in each byte. There is a total of 256 total (counting the 0) different masks. For class A, your ip would start in the range of 1-126. Class B range of 128-191 and class C range of 192-223. Since the number I am working with falls in the class C, the default subnet is going to be 255.255.255.0. What I am having a problem with is borrowing of bits.

allaboutjo

So if I am not mistaken, the answer is yes, I can create 5 subnets with the largest subnet supporting 28 host addresses. With the 255.255.255.0 subnet, when broken down to decimal, it would be 11111111.11111111.11111111.00000000. After creating the new subnets on the system, the decimal breakdown would be 11111111.11111111.11111111.11100000, which would allow me 8 networks with 30 hosts per network. Therefore by creating this breakdown, I have reached the minimum of 5 subnets (actually have 8 )and still meet the companies needs with having 30 hosts per network.

Did I figure this out right? I certainly hope so.....it sounds good in my head.

Ismaeljrp

allaboutjo wrote: »

30 hosts per network.

Did I figure this out right? I certainly hope so.....it sounds good in my head.

Good to go. With your new subnet mask of 255.255.255.224 you get 8 total subnets, and 32 - 2 hosts , 30. And you have it, remember, as long as you meet the minimum requirements, it doesn't matter if you have some IP's left over. Whenever you get asked for X amount of hosts per subnet, it just means " No less than X amount of hosts per subnet ".

jsb515

What about an address like this?

172.16.0.0 /24

i did it like this

32-24=8 then 2^8=256

cant count 256 so would I just start from 0?

Ismaeljrp

jsb515 wrote: »

What about an address like this?

172.16.0.0 /24

i did it like this

32-24=8 then 2^8=256

cant count 256 so would I just start from 0?

I'm not sure I get what you're asking. But I think what you mean is; if I have a /24 mask, that means I only have 8 host bits to work with? Yes start from zero, the zero network IS your network address and first subnet ID.

Just remember the diference between total addresses 256 in this case, and usable addresses 254 in this case per subnet.

jsb515

thanks for the response,

next question, is there a trick or simplistic way to get to subnets that are higher in count? instead of listing them all out?