Yet another subnetting topic =D

Sepher · #1 15-Sep-2008, 05:55 PM

Sorry to post another subnetting topic but I couldn't find an answer to my question.

Firstly I do know that what you need to know for the Net + exam is basic in regards to subnets. But I was curious to do some more research into Subnetting as I know its an important thing to know in Networking.

Through days of reading the internet and getting many headaches I'm finally starting to grasp the idea, well kind of haha.

So basically when you make a partial subnet, what does the host IP address become on that subnet?

Say you have a network for instance (class C) 192.168.1.x /24 where the subnet is 24 bits (the default 255.255.255.0) and you changed it too say 255.255.255.192 so that's /26 right?

What does an IP address on this subnet look like?! Because now the network part looks like this... (in Binary)

11111111.11111111.11111111.11 000000
-------------------------------- -------
Network Part Host Part (000000)

Does the IP address for this subnet become 192.168.192.x or something?

Does this make sense?

Thanks

PS. Oh and www.learntosubnet.com is a great site to learn this stuff

**hbroomhall** · #2 15-Sep-2008, 07:42 PM

There will be 4 possible subnets with the new mask:
192.168.1.0 - 192.168.1.63
192.168.1.64 - 192.168.1.127
192.168.1.128 - 192.168.1.191
192.168.1.192 - 192.168.1.255
(Assuming the base network is 192.168.1.0)
Usable hosts will, as usual, be these ranges less the network number and the broadcast address.

Note that the two extra network bits are in the 4th octet.

Harry.

Sepher · #3 15-Sep-2008, 08:47 PM

Complete edit of my post lol.

Ok Im starting to understand now I have read this site about it (http://www.troutman.org/tech/linux_guides/subnet.html)

For a subnet of 255.255.255.192 you can get four networks...

The first will be 192.168.50.0, broadcast 192.168.50.63, subnet mask 255.255.255.192
The second will be 192.168.50.64, broadcast 192.168.50.127, subnet mask 255.255.255.192
The third will be 192.168.50.128, broadcast 192.168.50.191, subnet mask 255.255.255.192
The fourth will be 192.168.50.193, broadcast 192.168.50.255, subnet mask 255.255.255.192

What is the purpose of splitting up into four sections? If they all use the same subnet?

Or even splitting it 16 ways in a 255.255.255.240 subnet?

And you cant use these addresses (if they are subnetted like this) in a default subnet (255.255.255.0) right?

**hbroomhall** · #4 15-Sep-2008, 10:22 PM

First - you do realize that you have shifted the base network address from 192.168.1.0 to 192.168.50.0? This will tend to cause confusion!

Next - the last line on your set of four is wrong - the network address is 192.168.50.192 not 192.168.50.193.

The reasons for sub-netting are many, but the usual one quoted is given as an example for an office:
You want to split sales, support, management and finance into separate networks so that people don't have automatic rights to see machines in other departments. So you subnet into 4 smaller networks.

Once you have sub-netted then you can't reuse IP addresses.

You *could*, of course, have two subnets of /26 and one of /25.

Harry.

Sepher · #5 16-Sep-2008, 12:35 AM

Ok I have finally cracked it, or so I think...

You've enlightened me alot to where I was going wrong and thank you for your help!

So heres what I've come up with...

Using a Class C IP address.... lets do a custom subnet

Code:

255.255.255.192 –  /26 – 11111111.11111111.11111111.11000000

Usable subnets

Code:

2^2 = 4 – 2 = 2 Subnet Masks (where 2 is the number of 1's)

Usable Hosts per subnet

Code:

2^6 = 64 hosts per Subnet – 62 usable hosts (where 6 is the number of 0's)

Total usable hosts

Code:

2 x 62 = 124 usable hosts

Total users left on unsubnetted subnet (255.255.255.0)

Code:

124 – 254 = 130 usable hosts on unsubnetted subnet (255.255.255.0)

And there you have it. Am I getting to obsessed

I don't understand why you cant use 2 of the available subnets, this is to do with some rule isn't it? I do understand that you cannot use the first or last IP address, broadcasting etc.

So does this add up?

I did not copy this from the net, well I did use a table so I could match the results. But it took me a while to get the same results and after looking at tons of web searches on subnetting I finally managed to write up my own summary of it and I got the same results as the table I was using.

Tinus1959 · #6 16-Sep-2008, 10:18 AM

In the past some network devices could not tell the differense between 192.168.50.0 /24 and 192.168.50.0 /26. They would simply see just the /24 unless it was clear that it was not a /24 network (because of the 192.168.50.64 address).

However keep in mind that the rule 2^x - 2 for the number of networks is not 'correct' anymore. Most 'modern' network devices do support the '0'-network, so today the rule would be 2^2 = 4 for a 255.255.255.192 network.
Just to make things complicated...

Sepher · #7 16-Sep-2008, 11:33 AM

Dam, so its now 4 subnets and not 2?

But are the rules still the same for the broadcast address of each subnet etc?

How complicated

**hbroomhall** · #8 16-Sep-2008, 12:23 PM

How subnetting is done has changed down the years.

The first major division you have to remember is between Classfull addressing and classless (CIDR).

The old classes are no longer used on the Internet, but there is a lot of kit still out there which uses classes. (Even Microsoft still uses it - if you put in a random IP address in the Windows GUI then the mask shown automatically will be that derived from the old class system. It is up to you to put the correct mask in.)

Even in the old class system the rules have changed. I believe that Cisco has a command that controls whether zeros are permissable in network addresses.

With CIDR the norm is that network addresses ending in zero are useable.

Harry.