CIDR >Don't Totally Get It <

  • Thread starter Thread starter Wayne B.
  • Start date Start date
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Wayne B.

Would anyone out there happen to have a site that gives
examples on how to configure CIDR. I've been to many
sites that goes over it, but, not one gives >EXAMPLES< on
how to work it out. I'm taking the server cert. test in
the near future and I NEED to find out how to figure this
stuff out.

Such as ------>> 192.60.128.0/22 <<----- How would
one, go about figuring out the network's starting address
(which I believe is pretty self-explanatory) I think?,
and the network's ending address.

How would I work that out on paper if something
similiar to this was to show up on a test?

I hope that I have expressed my shortcoming clearly
enough. That's the best way that I can describe what I'm
not understanding.

Appreciate Any and All Constructive tutoring and help!
THANX N ADVANCE The Rookie/Wayne B.
 
Wayne,
For the given example 192.60.128.0/22, 22 stands for number of bits that are
used for network.
If you don't know the basics of IP address here is little info.
IP address contains 32 bits represented in four 8 bit numbers. So actually
the above address in binary form looks like this:
1100 0000 0011 1100 1000 00 00 0000 0000
| | |
--------------------------------- -------------
| |
Network address Host address

As I said earlier in this address first 22 bits represent network address,
so the starting address would be the same (192.60.128.0) and the ending
address would be the keeping the first 22 bits as they are and 1's for the
rest of the bits, which is
1100 0000 0011 1100 1000 00 11 1111 1111
| | |
--------------------------------- -------------
| |
Network address Host address

If you represent the above in dotted decimal notation it would be
192.60.131.255.
Hope this helps.

Thanks,
Rama Krishna Amaravadi
 
Thanks for the reply Rama. That's the way I understood
it also until I came across a question on a Transcender.
It's saying that a machine (m1) that has a mask of /18
would not be able to communicate with another machine (m2)
on a different subnet with the mask of /19 because it
would see that machine as local and try to send packets
directly to it.
Reason being, that their >18< leftmost bits are
identical, even though they have different masks.

So my question now is............ is it possible for 2
machines that have different masks to look at each other
as if they are on the same subnet, even though they
aren't??
THANX AGAIN, Wayne B.

-----Original Message-----
Wayne,
For the given example 192.60.128.0/22, 22 stands for number of bits that are
used for network.
If you don't know the basics of IP address here is little info.
IP address contains 32 bits represented in four 8 bit numbers. So actually
the above address in binary form looks like this:
1100 0000 0011 1100 1000 00 00 0000 0000
| | |
--------------------------------- -------------
Click to expand...
| |
Network address Host address

As I said earlier in this address first 22 bits represent network address,
so the starting address would be the same (192.60.128.0) and the ending
address would be the keeping the first 22 bits as they are and 1's for the
rest of the bits, which is
1100 0000 0011 1100 1000 00 11 1111 1111
| | |
--------------------------------- -------------
Click to expand...
| |
 
Thanks for the reply Rama. That's the way I understood
it also until I came across a question on a Transcender.
It's saying that a machine (m1) that has a mask of /18
would not be able to communicate with another machine (m2)
on a different subnet with the mask of /19 because it
would see that machine as local and try to send packets
directly to it.
Click to expand...
Reason being, that their >18< leftmost bits are
identical, even though they have different masks.

So my question now is............ is it possible for 2
machines that have different masks to look at each other
as if they are on the same subnet, even though they
aren't??
Click to expand...

Because the mask is used to make the "same subnet" or
"different subnet" decision.

The weird part is that SOMETIMES the wrong mask will
work -- it's too simplistic to say that such will always fail.

If my machines thinks you are on my subnet, I broadcast
an "arp" and when you don't answer (broadcasts don't
typically cross routers or subnets) then I figure you don't
exist.
 
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