Another way to look at it is that you have one long contiguous line of
addresses
that you need to break up into different size groups that must also be
contiguous.
For example: 192.168.1.0 /24
Assume we need three networks (nets a&b) with 40 hosts, two networks
(nets c&d) with 25 hosts, and 3 networks (nets e,f,&g) with 12 hosts.
Our available subnet area is 192.168.1.0 - 192.168.1.255
Shown graphically (hope this doesn't get munged in transmission;
if so copy and paste into word using fixed width and 10).
.0
.255
/24:
|-------------------------------------------------------------------------------|
.128
/25:
|-------------------------------------|-----------------------------------------|
.64 .192
/26:
|-----------------|-------------------|-------------------|---------------------|
(a) ( b)
.160 .224
/27:
|-----------------|-------------------|---------|---------|----------|----------|
(c) (d)
.208 .240
/28:
|-----------------|-------------------|---------|---------|-----|----|----|-----|
(e)
(f) (g)
>From this we can see that Subnet blocks 192.168.1.0 & .64 are used for
nets a & b.
Blocks .128 & .160 are used for nets c & d, while blocks .192, 208 & 224
are used
for nets e, f, & g. This leaves one block, .240 for future use or to
further
subnet for /30's to address serial links.
By looking at it visually, there isn't any danger of overlapping
previously assigned
blocks. It also shows us where supernetting will occur so we can
properly assign
the blocks for easy aggregation. Since the blocks must be recombined
for supernetting
in the same manner they were subnetted, we can see that nets a & b could
be put on
one router and we would only advertise a /25 (192.168.1.0) upstream.
Likewise, blocks
.128 & .160 could be on a second router and advertised as a /26
(192.168.1.128).
We can also see that we couldn't put nets a-d on the same router and
advertise as a
/25 because they don't all come from the same /25 block. We would have
to use two
advertisements, a /25 & /26, if they were on the same router.
The same method can be used for address ranges that cross octet
boundries:
172.16.0.0 /16
0.0 255.255
/16: |---------------------------------------------------|
128.0
/17: |-----------------------|---------------------------|
and so on.....
I find showing it visually to my students makes understanding a lot
easier.
HTH,
Prof. Tom Lisa, CCAI
Community College of Southern Nevada
Cisco ATC/Regional Networking Academy
"Cunctando restituit rem"
John Neiberger wrote:
> The key is that you must completely unlearn classful thinking. Forget
> that
> you ever learned it. Completely ignore any prior classful subnet
> boundaries
> that you were forced to memorize. It's all just one big IP address
> space
> that you choose to carve up any way you like. As long as you do it
> correctly
> and don't have any overlap the subnetting scheme is up to you.
>
> Another helpful tip: don't ever use classful terminology any more!
> Don't say
> "Class A" to refer to an 8-bit prefix or subnet mask; don't say "Class
> C" to
> refer to a 24-bit mask, or /24. That will help move your brain away
> from
> that type of thinking.
>
> Think of your address space as a big pie, and each time you cut a
> segment in
> half you're adding one more bit to the subnet mask. Here's an example:
>
> You start with 10.20.30.0/24 (255.255.255.0) and we'll think of that
> as a
> whole pie. You don't need that many addresses in your subnet so you
> decide
> to break it up into smaller pieces. What do you do? Cut your pie in
> half
> (draw this out, it helps!).
>
> Your pie now has two halves and these represent two subnets with /25
> masks
> with no overlap. Let's say you want to further subnet one of those
> subnets.
> Cut it in half again! You now have a /25 and two /26s with no overlap.
> If
> you further cut one of those /26 subnets into two pieces you have two
> /27s.
> See how easy that is?
>
> Draw this out on paper and write down your subnet information as you
> go,
> like this:
>
> 10.20.30.0/24 (10.20.30.0-255) becomes
> 10.20.30.0/25 (10.20.30.0-127) and 10.20.30.128/25 (10.20.30.128-255)
> 10.20.30.128/25 further subnetted becomes 10.20.30.128/26
> (10.20.30.128-191)
> and 10.20.30.192/26 (10.20.30.192-255)
>
> And so on... practice it this way for a while and after a short time
> it
> will be second nature for you to subnet existing networks without
> accidentally overlapping them.
>
> HTH,
> John
>
> >>> Steven Aiello 9/9/03 12:03:06 PM >>>
> 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
> >>Store:
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> > **Please support GroupStudy by purchasing from the GroupStudy Store:
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> **Please support GroupStudy by purchasing from the GroupStudy Store:
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