>From what you say, I think you have it, but I'm not sure. Starting from the
bottom of a /24 subnet (Class C), you could have a /26 subnet, then two /27
subnets, then four /28 subnets, and finally another /26 subnet. Or you
could have two /28 subnets, one /27 subnet, one /26 subnet, followed by a
/25 subnet. The combination, and order, does not really matter, as long as
no IP addresses within the subnets overlap.
For instance, you couldn't have a /26 (64 addresses) followed by a /25 (128
addresses), followed by a /26 (64 addresses). Why? Because there can't be
any overlaps. The 64 would start at .0 and go to .63. The 128 would
start... Where? It can't start at .64, because that's in the middle of say
192.168.24.0/25 (which is 192.168.24.0-192.168.24.127). It would need to
start at .0 or .128. If it started at .128 then it would extend to .255, in
which case there wouldn't be room for the last /26 subnet. So, you re-order
them and use either a /26, /26, and /25, or /25, /26, and /26.
Remember, the whole classful/classless thing is routing protocol specific.
It has nothing to do with how hosts view IP addresses, or make "routing"
decisions (meaning whether to send it to a router or if the address is
local). The source code for a TCP/IP stack may look something like this:
# Assuming addresses/masks are 32-bit numbers, not dotted decimal
# string representations of addresses/masks.
# $ip_src is the IP address of the outgoing interface on the host
# $ip_dst is the IP address of the destination
# $ip_mask is the subnet mask on the outgoing interface
# $ip_gateway is the IP address of the default gateway
# check to see if destination address is in same subnet as our interface
if (($ip_src & $ip_mask) == ($ip_dst & $ip_mask)) {
# send directly to destination, possibly arping out first
} else {
# send to default gateway, $ip_gateway,
# possibly arping out first
}
There would obviously be more logic in there as you may have more than one
route and not a single default gateway, but the important point is that it
does not matter about the "classfulness" or "classlessness" of the subnet
mask. The host doesn't give a hoot. As long as the source and the
destination both agree whether they are in the same subnet or not everything
works fine. If they don't, you may need some ancient hack like proxy ARP,
but I don't know anyone in their right mind that would recommend
purposefully MIS-configuring a network so that it is required.
Fred Reimer - CCNA
Eclipsys Corporation, 200 Ashford Center North, Atlanta, GA 30338
Phone: 404-847-5177 Cell: 770-490-3071 Pager: 888-260-2050
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-----Original Message-----
From: Steven Aiello [mailto:[EMAIL PROTECTED]
Sent: Tuesday, September 09, 2003 1:21 PM
To: [EMAIL PROTECTED]
Subject: Re: Please Help - CIDR - How the bits work What I figured out
[7:75087]
I was stuck on the idea that you could ONLY re subnet a remaining piece
of a subnetwork. And not apply a mask to the whole span of the total
available network. You can (unless I'm incorrect here) you just have to
watch out for address over lap neer your subnetwork boundries.
I think I got it.
Man I love this news group!
Steve
Priscilla Oppenheimer wrote:
> Reimer, Fred wrote:
>
>>No offense, but this is CCNA material.
>
>
> Do they still teach classful for CCNA, though? Perhaps the only thing
that's
> hard for him is that 192.168.24.0 has a mask of 255.255.255.0 in a
classful
> system. Moving the prefix over to the left of that classful boundary isn't
> something they teach for CCNA yet. (They will soon. The new Networking
> Academy books teach it from the start now.)
>
> Priscilla
>
>
>>If you are going for
>>your CCNP, then
>>you should already have your CCNA and know the answer. But
>>anyway...
>>
>>If you need a network with 400 hosts, the smallest subnet would
>>have a /23
>>mask. So take the first part of your given network and assign
>>it to that:
>>
>>192.168.24.0/23 (192.168.24.0-192.168.25.255)
>>
>>Then you need one with 200 hosts. Well, that could fit within
>>a /24 subnet,
>>so assign the next available to that:
>>
>>192.168.26.0/24 (192.168.26.0-192.168.26.255)
>>
>>Now you only have 192.168.27.0/24 left from the original
>>192.168.24.0/23
>>(which covered 192.168.24.0-192.168.27.255). You need two
>>50's, so that
>>should fit within /26 subnets each. Assign them:
>>
>>192.168.27.0/26 (192.168.27.0-192.168.27.63)
>>192.168.27.64/26 (192.168.27.64-192.168.27.191)
>>
>>Finally, you need three subnets that can have two hosts each,
>>which would
>>fit within /30 subnets. So assign:
>>
>>192.168.27.192/30
>>192.168.27.196/30
>>192.168.27.200/30
>>
>>
>>Fred Reimer - CCNA
>>
>>
>>Eclipsys Corporation, 200 Ashford Center North, Atlanta, GA
>>30338
>>Phone: 404-847-5177 Cell: 770-490-3071 Pager: 888-260-2050
>>
>>
>>NOTICE; This email contains confidential or proprietary
>>information which
>>may be legally privileged. It is intended only for the named
>>recipient(s).
>>If an addressing or transmission error has misdirected the
>>email, please
>>notify the author by replying to this message. If you are not
>>the named
>>recipient, you are not authorized to use, disclose, distribute,
>>copy, print
>>or rely on this email, and should immediately delete it from
>>your computer.
>>
>>
>>-----Original Message-----
>>From: Steven Aiello [mailto:[EMAIL PROTECTED]
>>Sent: Tuesday, September 09, 2003 8:02 AM
>>To: [EMAIL PROTECTED]
>>Subject: Please Help - CIDR - How the bits work [7:75050]
>>
>>I just started my routing class for my CCNP. We are covering
>>CIDR. The
>>book is VEEEEEERY vague on how the bit patterns break down and
>>are used.
>>
>>
>>This was a problem posed in one of my CCNP labs
>>
>>I have network number
>>
>>192.168.24.0 / 22
>>
>>from this I need
>>networks with
>>
>>400 hosts
>>200 hosts
>>50 hosts
>>50 hosts
>>2 hosts (for serial int - no ip un-numbered allowed )
>>2 hosts
>>2 hosts
>>
>>Also no NATing
>>
>>Thanks all I really could use the help
>>
>>Steve
>>**Please support GroupStudy by purchasing from the GroupStudy
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> **Please support GroupStudy by purchasing from the GroupStudy Store:
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