TCP/IP Addressing & Subnetting

[micael]

Good post [webcarnage] !

[bimmer]

very Excellent and useful post.
keep it up WC ...

[Bimmersamster]

I understood all that, but now I have some questions:

In the first example

Lets say, for example, that you want 8 subnets and will be using a class C address.

You get 255.255.255.240 as the subnet mask...

so now we have 8 subnets, ok now waht about hosts? How many hosts can we have per subnet...

can u give an example of the whole 32 bit ip address and subnet for each of the 8 subnets?

I just want an example of how they look, to see if my calculations were correct...

thanks for the info WebCarnage, this could be of real help in my work.

[Bimmersamster]

if u want to have 8 subnets, you see that it would take 4 bits to make 8 in binary, so to get 8 subnets u would have 255.255.255.240...

but then again, to make 14 in binary it takes 4 bits....to make 10 or even 9 it takes 4 bits...so the 255.255.255.240 can be used to make 8 subnets or 14 subnets or 10 or 9?

clarify please!!!

[jared_c]

Nooooo.... Not subnetting!!!!! Ahhh.... I can feel my brain cells dying already.



I remember having this in class, and wanting to kill myself afterwards. Thanks for the tutorial though. I'm going to save it for future reference. It's a lot better than the one in my class book.

[nabylbt]

yeah subneting can be a pain when establishing vpns and configuring switches
but great post !

[[WebCarnage]]

Originially posted by: bimmersamster
Lets say, for example, that you want 8 subnets and will be using a class C address.
You get 255.255.255.240 as the subnet mask...
so now we have 8 subnets, ok now waht about hosts? How many hosts can we have per subnet...?

4 bits, subnet mask 255.255.255.240, 14 subnets, 14 hosts, net*hosts= 196

CIDR /28

clarify please!!!

Mate, I don't understand where your comming from. If your taking about bits and bytes, then listen up.

0     0     1     0     1     1     1     0
|     |     |      |     |     |     |      |
128  64   32      16      8     4     2     1

Now, that was a simple char, with a byte - or octet - consisting of 00101110. Now we add.

0 + 0 + 32 + 8 + 4 + 2 + 0 = 46

This happens because the 0 bits arn't added, while the 1's are.

Hope it helps