Subnetting Made Easy

border5 wrote: »

One word Techexams.Net is Simply superb

That's more than one word

Siddarth Sajjan

Awasome simple and easy to understand concise and clear

NewManSoon

Can someone tell me why I was wrong.. why is there a block size of 63? I am confused..

NewManSoon wrote: »

Can someone tell me why I was wrong.. why is there a block size of 63? I am confused..

63 in binary = 00111111.
Using the prefix notation [red] (/18 or rather 255.255.192.0) covers 148, 250, and no bits afterwards.
148 .250 .00|010110 .01000010

The line marks the point where the bits in the third octect become hosts.

there's no bit in the 64th spot. 64th bit is in bold.

Converting the green host bits to all 0s, and converting the 3rd and 4th octect back to decimal will show the subnet number:
148.250.0.0.

The last valid host is (in bits):
148 .250 .00|111111 .11111110

Converting back to decimal to find the subnet:
148.250.63.254

NewManSoon

Roguetadhg wrote: »

63 in binary = 00111111.
Using the prefix notation [red] (/18 or rather 255.255.192.0) covers 148, 250, and no bits afterwards.
148 .250 .00|010110 .01000010

The line marks the point where the bits in the third octect become hosts.

there's no bit in the 64th spot. 64th bit is in bold.

Converting the green host bits to all 0s, and converting the 3rd and 4th octect back to decimal will show the subnet number:
148.250.0.0.

The last valid host is (in bits):
148 .250 .00|111111 .11111110

Converting back to decimal to find the subnet:
148.250.63.254

Thank you for your detailed reply. So , I can throw out any thing I have learned from this thread on figuring out the network of a given address? Basically, 192 is a 64 block size (256-192 = 64) OR next "boundry" method 24 - 18 = 6 , 2 to the power of 6 = 64. Maybe everything should just be done in binary

NewManSoon wrote: »

Thank you for your detailed reply. So , I can throw out any thing I have learned from this thread on figuring out the network of a given address? Basically, 192 is a 64 block size (256-192 = 64) OR next "boundry" method 24 - 18 = 6 , 2 to the power of 6 = 64. Maybe everything should just be done in binary

192 is a combination of the 128th, and 64th bit.

Subnet Mask Ends with: 128 | 192 | 224 | 240 | 248 | 252 | 254 | 255
Last subnet bit: ..........128 | 64 .|. 32 |.. 16 | ..8 .|.. 4 .| .2 ..|.. 1

After a while of looking at subnets, you'll learn those numbers like they're the back of your hand. I didn't believe it when I heard what I just typed either - but gosh darn it. I did.

Im using the periods are space holders, so ignore them.

NewManSoon

I am a complete idiot..I read the .66 as the third octet where .22 thinking .66 was in the .64 subnet!!

*backs out of thread slowly*

mvshijin

LordFlasheart wrote: »

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!

you great

lot of thanks

I do believe I have looked at so many different subnetting demo's that I have confused myself. I just looked at one not so long ago, and their method of finding the subnet range is to do 2^x-2 - now, before I thought that you only used the -2 when finding the host range. Is their method incorrect?

Excellent1

MAC_Addy wrote: »

I do believe I have looked at so many different subnetting demo's that I have confused myself. I just looked at one not so long ago, and their method of finding the subnet range is to do 2^x-2 - now, before I thought that you only used the -2 when finding the host range. Is their method incorrect?

It can be correct--it depends on whether or not the zero subnet is enabled. If it is not, the above is correct. If it is enabled, you leave off the "-2" from the network portion.

border5

My maths is not so good .every time i understand the topic but if someone change those digit and prefix then i get confused so could you please help me to remember the big monster of subnetting

Subnetting really isn't a big monster at all. Though, I can agree with your method of thinking. Before I got into subnetting I thought I was going to have a difficult time, and I did (hear me out on this). When I first started reading my books I actually skipped subnetting (Bad, bad idea). Then at the end of my studies I decided to read books, research on here (TE) and PRACTICE. Man, was I wrong about it being hard. All it takes is practice and some determination. Really, subnetting is a big fluffy bear once you grasp the concepts.

Here's a couple of charts that I have in PDF format that may help you along a little.

increments.pdf

subnetting chart.pdf

MAC_Addy wrote: »

Really, subnetting is a big fluffy bear once you grasp the concepts.

I felt inspired by this thread.

Motivational--Subnetting.jpg

Roguetadhg wrote: »

I felt inspired by this thread.

Haha, did you just make that? Or did I reference this without even knowing that it existed?

MAC_Addy wrote: »

Haha, did you just make that? Or did I reference this without even knowing that it existed?

The 6 layers are still in my photoshop

ritooraj

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.

ritooraj wrote: »

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.

/24 refers to 24 1's where a 1 means it matches the number so...

11111111.11111111.11111111.00000000 is the subnet mask. So the last octet can have any number you want in it (with the first and last being unusable.)

ritooraj

Trifidw wrote: »

/24 refers to 24 1's where a 1 means it matches the number so...

11111111.11111111.11111111.00000000 is the subnet mask. So the last octet can have any number you want in it (with the first and last being unusable.)

Thank you.
I was trying to apply the next boundary method to arrive at the answer but could not do it. if you can show me how it works with the next boundary method it would be really nice.

ritooraj wrote: »

Thank you.
I was trying to apply the next boundary method to arrive at the answer but could not do it. if you can show me how it works with the next boundary method it would be really nice.

Sorry, I am not familiar with the method. I do think you might benefit from reading about CIDR and possibly supernetting though.

lantech

ritooraj wrote: »

Thank you.
I was trying to apply the next boundary method to arrive at the answer but could not do it. if you can show me how it works with the next boundary method it would be really nice.

Well, I was going to try and explain it but I'm not sure that I can so that you would understand. I just look at that address and subnet combination and know what network it's on.

That's a very easy subnetting question to get actually since it's a class B network with a /24 subnet mask. With a class B network address and a /24 subnet mask you should get to the point where you know that you would have 8 subnet bits which will equal 256 subnets with 256 ip addresses per subnet. Since you can't use the first and last ip address in a subnet mask that gives you 254 host addresses per subnet.

You also have to watch out for the zero subnet and broadcast subnet as well. Those might reduce your subnets by two. There are various factors that go into whether you can use the zero subnet and broadcast subnets and you do have to be aware of them and if those factors are relevant to a question on the exam.

If you need practice with subnetting you can go to subnettingquestions.com and it will generating subnetting scenerios for you and tell you what the right answer is.

ritooraj

Thank you, Lantech.

binargs

what i do is i do a quick "braindump" before the exam.

before the exam starts you have 15 minutes to read the instructions for the exam that we all done before.

during this time i make my self a nice little chart

128 192 224 240 248 252 254 255
25 26 27 28 29 30 31 32
17 18 19 20 21 22 23 24
128 64 32 16 8 4 2 1
127 63 31 15 7 3 1 0

and then do a number list increment of 4

0 4 8 12 16 20 24 .... etc

with this chart you can get through any subnet questions with a breeze, and you don't really have to fear the time clock anymore.

edit: the colums of those charts suppose to align but i guess this forum have a limit of 1 white space per character.

lantech

binargs wrote:

i make my self a nice little chart

128 192 224 240 248 252 254 255 - This would be the subnet number in decimal
25 26 27 28 29 30 31 32 - This is the mask for the last octet
17 18 19 20 21 22 23 24 - This is the mask for the second to last octet
128 64 32 16 8 4 2 1 - This is the decimal value of each binary location?
127 63 31 15 7 3 1 0 - What is this for?

and then do a number list increment of 4

0 4 8 12 16 20 24 .... etc - Why do this?

I understand most of the things on your chart there are just a couple of questions that I have.

lantech wrote: »

I understand most of the things on your chart there are just a couple of questions that I have.

127 63 31 15 7 3 1 0 is wildcard masks. I don't get the increments of 4 either.

binargs

lantech wrote: »

I understand most of the things on your chart there are just a couple of questions that I have.

the row with 128 64 32 means the number of host that particular /## subnet will hold. you will need to subtract 2 to factor in network id and bcast add.

127 row is wild card mask for acl, ospf and eigrp

number list of 4 will list pretty much all the the possibility that the network starting number. that list will also contain network number that consist 8, 16, 32, 64, 128. it makes the starting range for a subnet a breeze.

with that chart and nicely written aligned columns, subnet should be a treat on the cisco exams.

Wickedkick

Hi all,

I've been checking the forums from time to time and theres some really good tips on here, however I've just got to a question where either I'm being really stupid or don't fully understand this concept yet anyway here goes;

I got the following question:

Question: What is the first valid host on the subnetwork that the node 172.16.228.120/22 belongs to?

and the actual answer to it is: Answer: 172.16.228.1

The way I work it out I got the Answer of 172.16.228.4

Heres my working out:

/22 is in the 24 boundary so I take 22 from 24 which gives me 2, then I do 2^2 which gives me the answer of 4 so I believe if my workings are correct which they probably aren't the answer is 172.16.228.4

If someone could show me how to get to the actual answer and maybe explain (if you know) where I'm going wrong?

Thanks,

-Wicked

lantech

Wicked,

Which octect were you incrememting? It looks like from your answer you were incrementing the last octet which isn't correct. When you are subnetting you have to find the interesting octet. Which would be the the non 255 number. Which in this case is in the third octet. That is the number you will increment for your networks. The last octet would be used for getting your host addresses.

For your example, you rightly subtracted 24 - 22 to get 2 subnet bits. That two subnet bits means your going to increment the numbers in the third octet by 4 for each network range. With a /22 mask you have 10 host bits that gives you 1022 hosts per network. The calculation is 2^10 - 2

so your networks would look like this:

172.16.0.0 - 172.16.3.255
172.16.4.0 - 172.16.7.255
172.16.8.0 - 172.16.11.255

You would continue like that until you got to the 172.16.228.0 network. So your first host address in that network would be 172.16.228.1 and the broadcast address would be 172.16.31.255.

WickedAngel

Thanks to the methods posted here, I've pretty much got full control over most types of subnetting questions I've been facing short of the following.

Questions like this have given me no problem.

Question: How many subnets and hosts per subnet can you get from the network 172.22.0.0/23?

Answer: 128 subnets and 510 hosts

Question: How many subnets and hosts per subnet can you get from the network 172.17.0.0/20?

Answer: 16 subnets and 4094 hosts

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.248.0?

Answer: 32 subnets and 2046 hosts

Question: How many subnets and hosts per subnet can you get from the network 192.168.171.0/30?

Answer: 64 subnets and 2 hosts

However, this last question is different. I understand why there are 62 hosts available ((2^6)-2) but I have no idea how it was acceptable to repurpose the 8 bits from the preceding octet to get to the 1024 subnets indicated in the solution. In a /26 network, shouldn't there only be 2 bits available to be applied as network bits for the purposes of subnetting? The solution requires those 2 plus the preceding 8.

Question: How many subnets and hosts per subnet can you get from the network 172.18.0.0/26?

Answer: 1024 subnets and 62 hosts