70-293

I'm studying for the 70-293 at the moment and I noticed that in my MS Press
book as well as in the videos I watch through a commercial portal, they
preach about using the "2^x -2" formula for calculating not only the maximum
number of usable hosts, but for the maximum usable number of subnets. It was
my understanding that most all modern NOSes and routers can use subnets
consisting of all 1s and all 0s.

Has anyone taken this test recently? Does it assume that you'll be applying
the older formula when calculating subnets or does it assume all subnets
will be usable?

Thanks,
Tyler Cobb

Tyler Cobb

> It was my understanding that most all modern NOSes and routers can use
subnets consisting of all 1s and all 0s.

By that, I take it you mean the 255.255.255.0 type of subnets, which are
commonly used for the 192.168.x.x ip range.

The idea behind this section is to get you to understand how wasteful this
type of subnetting is, and how to squeeze them for maximum efficiency.

Look at the typical home network installation. A router, two computers.
192.168.0.1 for the router, 192.168.0.2 and 192.168.0.3 for the computers. If
you did the default subnet mask of "all or nothing", you would use
255.255.255.0. This means the broadcast address would be 192.168.0.255 and
the ip range of 192.168.0.4-192.168.0.254 is available but unused. So it
would be better to use a subnet mask of 255.255.255.248, which gives you 6 ip
addresses.

Now in the real world, it doesn't matter what you use on a small home
network. But on a large one, it makes sense to route efficiently. That means
having some spare addresses, but not too many, and also limiting the amount
of broadcast traffic on any one segment.

So if a company has Site A with 200 computers, Site B with 100 and Sites C
and D with 50 each, you could

a) have one big network with 400 contiguous ip's
b) split it into 4 networks with 254 ip's each
c) split it into 4 networks, one with 254 ip's, one with 126 ip's and two
with 62 apiece

The simplest solution would be (a), but again, the broadcast traffic from
one computer would go to all 400.
With (b), it is better, limiting the broadcast traffic to stay within the
site it originated from. But the smaller sites will have many unused ip's.
But (c) reduces the size of each site to what is needed, limiting waste.

So in the above scenario:
Site A has 192.168.0.1-254 and 255.255.255.0
Site B has 192.168.1.1-126 and 255.255.255.128
Site C has 192.168.1.129-190 and 255.255.255.192
Site D has 192.168.1.193-254 and 255.255.255.192

as ip addresses and masks for the computers.

Remember, when making an ip range, the first number is the Subnet ID, and
the last is the broadcast address (hence the gaps at 127, 128, 191 and 192.
Also, 0 and 255 are typically ID and broadcast respectively, unless you get
into supernetting)

The reason to calculate the maximum number of subnets, is that once you get
into CIDR and supernetting, you need to make sure the range you use will
allow you to break it into enough subnets to do what you need.

Pete Jones, MCSE:Security MCSE:Messaging

That's not really what I was asking, but a great read, nonetheless. If ever
there were any other minor details that were left with any amount of
uncertainty, your post has cleared them up. Thanks for the reply!

I will try to clarify my original statement:

Using the classful IP addressing system, when you subnet a network, you have
to apply the formula 2^x-2 not only to the host addresses but also to the
subnet addresses. This is because the classful system cannot use a subnet
with a binary bitmask consisting of all zeroes or all ones. Therefore, in
classful IP subnetting, the first usable host address for 192.168.0.0/26
would actually be 192.168.0.65 and the last usable host address in the first
subnet being 126. This system is, as you can guess, antiquated.

In the modern classless addressing scheme, most operating systems and
network gear can deal with the first and last subnet in a subnetting range
and allow a network range to exist there and be utilized. The formula 2^x-2
still applies here but only to the hosts. Therefore, using the previous
example, in a classless system, the first usable host address for
192.168.0.0/26 would actually be 192.168.0.1 with the last usable host
address in the first subnet being 62 (with 0 being the network and 63 being
the broadcast).

I'm sure you already know all of this. I suppose I just wasn't clear in my
original post. Mainly this response is just for my own personal affirmation
as I continue to study for the MCSE.

So, here is my original question somewhat reworded:

Since they teach the classful method in my MS Press book and on the MS
videos in my portal I use, does anyone know if the actual 70-293 test will
assume classful or classless IP subnet calculations?

Thanks,
Tyler Cobb

Tyler Cobb

> Using the classful IP addressing system, when you subnet a network, you have
> to apply the formula 2^x-2 not only to the host addresses but also to the
> subnet addresses. This is because the classful system cannot use a subnet
> with a binary bitmask consisting of all zeroes or all ones. Therefore, in
> classful IP subnetting, the first usable host address for 192.168.0.0/26
> would actually be 192.168.0.65 and the last usable host address in the first
> subnet being 126. This system is, as you can guess, antiquated.

Almost. 192.168.0.65 is the first host address on the SECOND subnet.

> In the modern classless addressing scheme, most operating systems and
> network gear can deal with the first and last subnet in a subnetting range
> and allow a network range to exist there and be utilized. The formula 2^x-2
> still applies here but only to the hosts. Therefore, using the previous
> example, in a classless system, the first usable host address for
> 192.168.0.0/26 would actually be 192.168.0.1 with the last usable host
> address in the first subnet being 62 (with 0 being the network and 63 being
> the broadcast).

See above.

The point you make about the formula is correct about CIDR. It does allow
for a subnet consisting of a single IP, which means the subnet ID, broadcast
and host are all one and the same, although I can't think of any reasons to
do this. (Having said that, I fully expect people to chime in with scenarios
now.)

> Since they teach the classful method in my MS Press book and on the MS
> videos in my portal I use, does anyone know if the actual 70-293 test will
> assume classful or classless IP subnet calculations?

Stick with the Classful subnetting. Whilst CIDR is useful to know, I dont
believe it is part of the MCSE. What they are looking to test is that you can
take a range, and split it into smaller subnets.

Pete Jones, MCSE:Security MCSE:Messaging

"Pete Jones, MCSE:Security MCSE:Messaging"
< crosoft.com> wrote in
message news8958994-EE43-4D88-9C3D-...
>> Using the classful IP addressing system, when you subnet a network, you
>> have
>> to apply the formula 2^x-2 not only to the host addresses but also to the
>> subnet addresses. This is because the classful system cannot use a subnet
>> with a binary bitmask consisting of all zeroes or all ones. Therefore, in
>> classful IP subnetting, the first usable host address for 192.168.0.0/26
>> would actually be 192.168.0.65 and the last usable host address in the
>> first
>> subnet being 126. This system is, as you can guess, antiquated.
>
> Almost. 192.168.0.65 is the first host address on the SECOND subnet.

An ancient thread, no doubt, but deserves a reply, nonetheless. In the
strictest sense of subnetting, the technical limitation that precluded the
use of the .0 and .255 addresses for host addresses, also precluded the use
of the subnets containing the addresses x.x.x.0 and x.x.x.255 (all 0's or
all 1's in the subnet), thus Tyler's characterization of 192.168.0.65 as the
=first= available subnet is accurate under the original subnet
specifications.


>> In the modern classless addressing scheme, most operating systems and
>> network gear can deal with the first and last subnet in a subnetting
>> range
>> and allow a network range to exist there and be utilized. The formula
>> 2^x-2
>> still applies here but only to the hosts. Therefore, using the previous
>> example, in a classless system, the first usable host address for
>> 192.168.0.0/26 would actually be 192.168.0.1 with the last usable host
>> address in the first subnet being 62 (with 0 being the network and 63
>> being
>> the broadcast).

The great irony in this discussion is that it was =Microsoft= who first
started violating this fundamental rule, by actually programing the Windows
RIP implementation to actually allow the first and last subnets to be used.

So, yes, the 2^x-2 rule only applies to the number of hosts in a given
subnet, it no longer applies to the number of subnets available in a given
class.

Note also that this 2^x-2 rule has been pretty much obsoleted by the
availability of VLSM and CIDR, which now makes the available number of
subnets in any range fixed at n/4 where n is the total number of non-network
identifiers (e.g. in the network 192.168.0.0/28, there are 256 non-network
identifiers, and 64 maximum possible subnets (2 hosts each), or any
combination of subnets < 64.


> The point you make about the formula is correct about CIDR. It does allow
> for a subnet consisting of a single IP, which means the subnet ID,
> broadcast
> and host are all one and the same, although I can't think of any reasons
> to
> do this. (Having said that, I fully expect people to chime in with
> scenarios
> now.)

This is not true, CIDR would still not permit a subnet of a single IP,
because the limitation on the network and broadcast address always applies,
it's only the first and last subnet that's been made available. The smallest
possible subnet, even with CIDR, is as noted above -- 2 hosts. (e.g.
192.168.0.128/30). There would be *no* host address available in
192.168.0.128/31, because 192.168.0.128 would be the network address and
192.168.0.129 the broadcast address, leaving no host addresses available;
and 192.168.0.128/32 is simply a HOST address. (Take a look in the netmask
column of your 'netstat -r' or 'route print' commands.


>> Since they teach the classful method in my MS Press book and on the MS
>> videos in my portal I use, does anyone know if the actual 70-293 test
>> will
>> assume classful or classless IP subnet calculations?

> Stick with the Classful subnetting. Whilst CIDR is useful to know, I dont
> believe it is part of the MCSE. What they are looking to test is that you
> can
> take a range, and split it into smaller subnets.

You need to know *both*, but I believe Pete's point is valid; I don't think
VLSM or CIDR is directly tested on the cert exams, but they do use CIDR /xx
subnet notations.



--
Lawrence Garvin, M.S., MCITP, MCBMSP, MCTS(x4), MCP
Senior Data Architect, APQC, Houston, Texas
Microsoft MVP - Software Distribution (2005-200

MS WSUS Website: http://www.microsoft.com/wsus
My Websites: http://www.onsitechsolutions.com;
http://wsusinfo.onsitechsolutions.com
My MVP Profile: http://mvp.support.microsoft.com/pro...awrence.Garvin

Lawrence Garvin

Awesome! That's what I needed to know! Thanks, everyone!

Tyler Cobb