Subnetting with Jeremy Cioara method

adnankhq

Hi,

I like Jeremy's method for its simplicity as I dont enjoy mathematics much. I just stumbled upon a question which as appears to defy his logic of subnetting, ere's the question,

What subnet does host 172.28.188.66/27 belongs to?

This IP is of class B network and its original subnet mask is 255.255.0.0, that constitutes for 16 bits, and according to my understanding of his method ,I have to subtract 16 from 27 which is 11, and we only have 8 bits to play with, so how would you solve this particular question?

When I assumed it as a class C IP and used 24 rather then 16 it gave me the correct answer, where am confused is that the IP address is of class B and bit notation is of class C, how do you solve these kind of questions whith Jeremey's method?

Hope its not too confusing.

Thanks.

gorebrush

My method of solving -

Take a /27

25 26 27 28 29 30 31 32
1--1--1--0--0--0--0--0

If you just look at the fourth octet, you have 1, 2, 4, 8, 16 - these values for host bits.

Look at the next colum - 32.

32 - 1 = max number that can be added with the previous 5 bits, thus the subnets: -

172.28.188.0 - 31
172.28.188.32 - 63
172.28.188.64 - 95

Thus the answer is

172.28.188.64/27.

adnankhq

Thank your for your reply, but my question still stands unanswered, the given IP is of class B and shouldn't we start counting from 17, 18, 19, upto, 32?

adnankhq

or leme make it easier for you, what should we do when presented with a question which has IP from lets say class B and subnet from class C?

Jon_Cisco

I don't know what Jeremy method but I think the class B and C relate to how many sub networks you have.

So you would have 11 bits for your sub networks and 5 bits for you hosts.

Work in the octet that is being split but still count the other octets when determining the number of subnets.

I'm not sure if this makes sense but it seemed to be what you are asking about.

Jon

Bacon_00

So I learned using Jeremy's method, but I sort of adapted it to work better for me. To solve that problem, this is what happened in my head:

1. /27 is 3 bits more than /24, with /24 being the default class-C subnet mask that doesn't borrow anything from the 4th octet.
2. 3 bits from left to right into an octet is 128, 64, 32 -- these three bits are all part of the subnet, the rest is host.
3. I can't subtract 128 from 66, but I can subtract 64. That leaves 2 left, and I can't subtract 32 from 2. So that means that the remaining 2 are in the host portion, and the 64 is part of the subnet.
4. That means the subnet address is 172.28.188.64/27.

In step 1 I know that this is a class B address, but for answering the question in the problem it's sort of irrelevant. You can just treat it like a class C and, for me anyway, that makes it quicker and easier to solve.

xnx

This isn't aimed at anyone in particular, but don't over complicate it. /8, /16 or /24 are always going to be your starting points.

Lets say you have a mask of /27, you know that that's 3 more steps than /24, so count up: 128 + 64 + 32 = 224

256-224 = 32, now you know that that's your increment.

All you need to do now is find the closest network that's a multiple of 32, in this case it's 64.

172.28.188.64/27 is your subnet address.

gorebrush

xnx wrote: »

This isn't aimed at anyone in particular, but don't over complicate it. /8, /16 or /24 are always going to be your starting points.

Lets say you have a mask of /27, you know that that's 3 more steps than /24, so count up: 128 + 64 + 32 = 224

256-224 = 32, now you know that that's your increment.

All you need to do now is find the closest network that's a multiple of 32, in this case it's 64.

172.28.188.64/27 is your subnet address.

That was really what I aimed to explain, but admittedly did so rather poorly. Thanks.

roch_greg

I don't know the jeremy method either but here's my take.

172.28.188.66/27

where CDIR /27 equals 255.255.255.224 subnet mask

start by looking at the octet of interest in the subnet mask. In this case the 4th one (224).

Y, because as mentioned earlier that's the boundary where the split between networks/hosts occurs.

256 (magic #) - 224 = 32 as a incrementing value. Since this is a class B network (n.n.h.h) we need to account for the 3rd octet as well. So our networks are..

188.0 188.32 188.64 188.96.......

from here we can see that 188.64 ip address has to be on the 188.64 network because the next network 188.96 is out of range.

for broadcast addresses we have

188.31 188.63 188.95 ....

last hosts per net are.

188.30 188.62 188.94

and the first hosts are

188.1 188.33 188.65 188.97

Summing it all up

Network ID 188.0 188.32 188.64 188.96 188.128 .......
First Host 188.1 188.33 188.65 188.97
Last host 188.30 188.62 188.94 188.126
Broadcast 188.31 188.63 188.95 188.127

Summary:

you need to know what the CDIR notation equals as a subnet mask. It's really how many bits are on (1's) in the subnet mask.
Once you get the subnet mask the rest is cake.

Hope this helps.

vasyvasy

Here is how the calculus goes, all in your head:

1. Substract the current subnet mask from 32: 32-27=5

2. Calculate the increment for subnets: 2^5 = 32, this is the increment for calculating networks:

32-64-96-128-160-192-224-256, this gives

172.28.188.0-172.28.188.31
172.28.188.32-172.28.188.63
172.28.188.64-172.28.188.127
...

3. Identify the first subnet that can accomodate your original IP address
172.28.188.64/27

4. Have a beer


PS: This works when you consider the subnet mask in the appropriate byte, like this:
subnet mask: 0-7 -> calculate increment 8-subnetmask
subnet mask: 8-15 -> calculate increment 16-subnetmask
subnet mask: 16-23 -> calculate increment 24-subnetmask
subnet mask: 24-31 -> calculate increment 32-subnetmask

dpsmooth15

1.I am pretty freaking awesome at subnetting Ok back to you, there is more than one way to subnet, pick a technique that works for you, Jeremy has a method the "Network Doctor" on youtube has a method and at least 3 other people. Just learn one of them, …...and 90% of it can be done in your head.

Your Welcome,
Tech Exams Subnet King of Kings

ghoul

I think the problem you are having is that you are getting too caught up on the idea of "classful" addressing which isn't really relevant for this. I can completely understand why that would cause some confusion, though.

I'm not a master of subnetting, but I recommend Keith Barker's subnetting series on CBTNuggets. He has a very simplistic view of teaching it, and it's the only one that has really helped me out.

mella060

It is very simple. To find the increment, the last bit position in the subnet mask where the value is a 1 is your increment.

So for /27. In binary that is 11111111.11111111.11111111.11100000

Now, remember that each octet has the binary values 128 64 32 16 8 4 2 1

The last bit position where the value is a 1 is 32. That is your increment/block size.

So the subnets are...

172.28.188.0/27
172.28.188.32/27
172.28.188.64/27
172.28.188.96/27 and so on

So the subnet is 172.28.188.64/27

After you work with subnetting for a while this whole process takes around 5-10 seconds to do in your head.

discobiscuits

Hey dude think you might be confusing classful and classless networks?

Once you subnet and are out of /8 /16 and /24 i believe its basically classsles so ignore that.

As others have said look at which octet you are in and calc nets for /1,2,3,4,5,6,7,8 = 2,4,8,16,32,64,128,256

adnankhq

Wow, I just got caught up at work and couldn't respond to your comments, let me tell you, you guys are amazing! Seriously!

Thank you so very much to each of you!