Subnetting Made Easy

ricky8

subnetting is easy. you just have to memorize the increments.
128+64+32+16+8+4+2+1 = 128, 192, 224, 240, 248, 252, 254, 255 <---- MEMORIZE
now all you have to do is the grade 3 math. 256-increment # (for example 256-224 = 32)

EX:
network: 192.168.0.0 /27 (/27 = 255.255.255.224) 256- 224 = 32 (subnets will be 0, 32, 64, 96....224)
1st subnet: 192.168.0.0 - 192.168.0.31
2nd subnet: 192.168.0.32 - 192.168.0.63
3rd subnet: 192.168.0.64 - 192.168.0.95

etc...

hope this helps.

Noerc

Thanks for sharing, I havn't been great at subnetting but the more I read and practice it all comes together!

LinuxRacr

Anyone have a good method for figuring out which mask for each class will give you a given number of subnets when only the IP is available?

drkat

That kind of defeats the purpose of the subnetting... if given no subnet mask then you will use the default class mask for the range of IP's

then it's business as usual

I'm sure you can find a **** sheet somewhere given the range and available masks and the numbers it produces.

This comes to mind

http://www.aelius.com/njh/subnet_sheet.html

LinuxRacr

This is the question that spawned my earlier question:

You Have subnetted the 192.168.14.0 network into 8 subnets. What is the valid host range of the third subnet?

I had a mental block in solving it for some reason...

t_patrick

Awesome!

Carl_S_901

LinuxRacr wrote: »

This is the question that spawned my earlier question:

You Have subnetted the 192.168.14.0 network into 8 subnets. What is the valid host range of the third subnet?

I had a mental block in solving it for some reason...

Ok, we're not given a mask on this question so we will have to stick with the fact that it is a classful network. (i.e. Class C or /24)

One way to attack it is using the powers of 2. If we have a /24 we know that if we shift 1 bit to the right then we have twice as many networks (x 2). So /25 gives us two subnets. /26 gives us 4. /27 gives us 8.

We need 8 for the problem so /27 is our subnet mask. We can convert that to decimal by knowing that 32-27 = 5 and that 2^5 =32 and 256-32 = 224. So 255.255.255.224 is our decimal mask. 2^5 is also our block size or "magic number" so we can count up subnets:

1. 192.168.14.0
2. 192.168.14.32
3. 192.168.14.64
4. 192.168.14.96

That's enough... 192.168.14.64 is our third subnet. More details:

Subnet 192.168.14.64
Broadcast 192.168.14.95
Valid Hosts 192.168.14.65 - 192.168.14.94

This assumes that we are using the Zero Subnet. If "no ip subnet zero" is in play (not mentioned in this question) then the 3rd subnet would be 192.168.14.96 with a broadcast of 192.168.14.127 and valid hosts of 192.168.14.97 - 192.168.14.126.
Thanks


Carl S.

---

Check out my personal certification journey blog
Carl's Certification Journey | The road to getting certified can often be bumpy

drkat

You Have subnetted the 192.168.14.0 network into 8 subnets. What is the valid host range of the third subnet?

256/8 = 32 this is your block size.. then you count your block size

LinuxRacr

Thank you both! That makes it crystal clear!

The answer actually includes the zero subnet, so yes, Valid Hosts = 192.168.14.65 - 192.168.14.94.

LinuxRacr

Ok, I've been able to answer quite a few questions in my head now, but this is the latest one that is getting me:

How many subnets and hosts per subnet can you get from the 172.28.0.0/26 network.

The hosts per subnet part was easy:
Since we are past the /24 boundary, we subtract 26 from the next octet boundary of 32:
32-26=6
2^6=64
Hosts per subnet = 64-2=62.

It was the number of subnets that I was lost on.

Highlight with mouse to see answer below.
Answer: 1024 subnets and 62 hosts per subnet.

r_durant

LinuxRacr wrote: »

Ok, I've been able to answer quite a few questions in my head now, but this is the latest one that is getting me:

How many subnets and hosts per subnet can you get from the 172.28.0.0/26 network.

The hosts per subnet part was easy:
Since we are past the /24 boundary, we subtract 26 from the next octet boundary of 32:
32-26=6
2^6=64
Hosts per subnet = 64-2=62.

It was the number of subnets that I was lost on.

Highlight with mouse to see answer below.
Answer: 1024 subnets and 62 hosts per subnet.

172.28.0.0 is by default a class B subnet with 16 bits, the /26 gives you the ^10 bits required to subnet (/26 - /16 = 10)....2^10 = 1024

Hosts, as you rightfully said are 2^6-2 = 62

11111111.11111111.11111111.11000000

The subnet bits in red.

I'm not the 'greatest' at explaining stuff, but hope that helps

Carl_S_901

The only trick I would add here for doing them in your head:

2^10 is exactly right as r_durant has replied and we can further make the math a little easier in our head because we know that a full octet 2^8 is alway 256

therefore we can say that this is 2^2 * 256 OR 4 * 256 which is of course 1024

So basically, when I need to calculate the number of X bits that span across multiple octets (one of which will always be the full octet or it wouldn't span into another) then I figure out the number for the interesting octet (partial) and multiply.

This gets a bit more difficult to do in your head though with a Class A though.

For example: How many subnets of /28 in a Class A:

We know the classful boundary is /8 so we have 2^20 or 2 ^ 8 (256) * 2 ^ 8 (256) * 2 ^ 4 (16)

2^8 * 2^8 = 2^16 of course and is equal to 65536 so this would be 65536 * 16 which is pretty hard to pull off in your head.

Answer would be:

1,048,576 subnets

I doubt you would see anything like that on an exam. If you did you will need to grab the pen to do the final calculation

LinuxRacr

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.252.0?

Answer: 64 subnets and 1022 host <

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Highlight with mouse to see.



I worked this one out on paper, and this one seems wrong to me....

r_durant

LinuxRacr wrote: »

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.252.0?

Answer: 64 subnets and 1022 host <

---

Highlight with mouse to see.




I worked this one out on paper, and this one seems wrong to me....

Why do you think it's wrong?

You have 11111111.11111111.11111100.00000000

2^6 bits for the subnets which = 64
2^10-2 bits for the hosts which = 1022

LinuxRacr

I believe my lack of understanding coupled with the method in which I was trying to solve this lead to my conclusion... sigh...

Two steps forward, one step back...

r_durant

LinuxRacr wrote: »

I believe my lack of understanding coupled with the method in which I was trying to solve this lead to my conclusion... sigh...

Two steps forward, one step back...

IMO...The thing about subnetting is that each person will have a different method of working these problems out, in some cases people will use the same method. Just develop a method that you are comfortable with, master it and then practice it repeatedly. Also, have confidence in yourself...

LinuxRacr

Thanks for the tutorials, and I do have confidence in myself. I just hate setbacks when I seem to be moving forward. I just have to sure up this hole in my learning, and I'll be good to go.

Carl_S_901

I agree with r_durant in that there are a lot of different ways to do it.

Find what works best for you, get it down solid, be able to do it fast in your head, and then pretty much ignore reading about subnetting in the near future so as not to throw a wrench in your subnetting machine. (your brain)

Rhodan

Hi all,

Great thread! I'm stuck with this one:

You are designing a subnet mask for the 192.168.100.0 network. You want 3 subnets with up to 50 hosts on each subnet. What subnet mask should you use?

The answer given is 255.255.255.192

How do I work this out?

LinuxRacr

Rhodan wrote: »

Hi all,

Great thread! I'm stuck with this one:

You are designing a subnet mask for the 192.168.100.0 network. You want 3 subnets with up to 50 hosts on each subnet. What subnet mask should you use?

The answer given is 255.255.255.192

How do I work this out?

First let's find out our blocksize of each subnet:

256-192=64

So 64 is the blocksize of each subnet, and you get a total of 4 subnets.

192.168.100.0 - 192.168.100.63
192.168.100.64 - 192.168.100.127
192.168.100.128 - 192.168.100.191
192.168.100.192 - 192.168.100.255

With that IP, that's the best you can do. Remember with IP addresses, they are binary, so everything is a power of two, so you have to have 4 subnets. Three is more than two, so the next power of 2 is 4.

Carl_S_901

Rhodan,

The smallest block size that will support 50 hosts is 64.

Your mask is 255.255.255.192 (256-64) This is also known as a /26

You will have 4 subnets.

Couple of ways to do this:

1. If you have a block size of 64 then you are using 6 bits for hosts (2^6) which means you have 2 bits for network (2^2) beyond the mask of /24 you were given.

2. Each time you move the mask a bit to the right you get twice as many subnets which are half the size. If you have a /24 and you borrow one bit and make it a /25 then you have two networks with 128 block size. If you move it one more to the right and make it /26 then you now have a block size of 64 and 4 networks.

Remember that usable hosts is always the size of the block MINUS 2 to account for the network and broadcast addresses.

Finally, if a question like this says you need 3 subnets that doesn't mean 3 and ONLY 3. It can be more so long as it meets the other requirements.

LinuxRacr

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.

LinuxRacr

Carl_S_901 wrote: »

Rhodan,

The smallest block size that will support 50 hosts is 64.

Your mask is 255.255.255.192 (256-64) This is also known as a /26

You will have 4 subnets.

Couple of ways to do this:

1. If you have a block size of 64 then you are using 6 bits for hosts (2^6) which means you have 2 bits for network (2^2) beyond the mask of /24 you were given.

2. Each time you move the mask a bit to the right you get twice as many subnets which are half the size. If you have a /24 and you borrow one bit and make it a /25 then you have two networks with 128 block size. If you move it one more to the right and make it /26 then you now have a block size of 64 and 4 networks.

Remember that usable hosts is always the size of the block MINUS 2 to account for the network and broadcast addresses.

Finally, if a question like this says you need 3 subnets that doesn't mean 3 and ONLY 3. It can be more so long as it meets the other requirements.

Thank you for the COMPLETE picture. I forgot that I should have approached this from the standpoint of NOT already having the answer....

Rhodan

Thanks for the quick replies, that makes sense.

Here's another type of question I want to check on:

How many subnets and hosts per subnet can you get from the network 172.22.0.0/25?

Answer given is 512 subnets and 126 hosts

This is how I work it out:
172.22.0.0 is a Class B network, so /16.

25 - 16 = 9 network bits
2^9 = 512 subnets

There are 7 host bits left from the /25, so (2^7) - 2 = 126 hosts.

Is this the correct way of working it out?

r_durant

Rhodan wrote: »

Thanks for the quick replies, that makes sense.

Here's another type of question I want to check on:

How many subnets and hosts per subnet can you get from the network 172.22.0.0/25?

Answer given is 512 subnets and 126 hosts

This is how I work it out:
172.22.0.0 is a Class B network, so /16.

25 - 16 = 9 network bits
2^9 = 512 subnets

There are 7 host bits left from the /25, so (2^7) - 2 = 126 hosts.

Is this the correct way of working it out?

That's correct!

elad123

Wickedkick wrote: »

Hi all,

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

-Wicked

Guys, I am reading through this but there is something that is not clicking for me, take the above for example if I am understanding this correctly:

24-22 = 2
2^2 = 4

so you increment the 3rd octet by 4

172.16.0.0
172.16.4.0
172.16.8.0

etc...

172.16.228.0
172.16.232.0


what is the shortcut to do that quickly instead of writing it all out?

r_durant

elad123 wrote: »

Guys, I am reading through this but there is something that is not clicking for me, take the above for example if I am understanding this correctly:

24-22 = 2
2^2 = 4

so you increment the 3rd octet by 4

172.16.0.0
172.16.4.0
172.16.8.0

etc...

172.16.228.0
172.16.232.0


what is the shortcut to do that quickly instead of writing it all out?

One shortcut you can use is to verify 228 is a multiple of 4 (228/4), if it is, then you know you're bang on. If it's not, then find the closest.

LinuxRacr

r_durant wrote: »

Why do you think it's wrong?

You have 11111111.11111111.11111100.00000000

2^6 bits for the subnets which = 64
2^10-2 bits for the hosts which = 1022

I'm still confused a little about where the 2^10 came from on this one..... Did it come from the 10 host bits (the zeros)?

oli356

LinuxRacr wrote: »

I'm still confused a little about where the 2^10 came from on this one..... Did it come from the 10 host bits (the zeros)?

Yes, to work out the number of hosts you do (2^x)-2, where x is the number of host bits. Host bits are the zeros, in this example there are 10 zeros accross the 2 octets. Of course you take away 2 because of the network ID and broadcast address.

LinuxRacr

Sure enough I had a question like that on my exam today...lol.