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*REAL QUESTION* - 70-291: VLSM's

 
 
Matrixx333
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Posts: n/a
 
      04-30-2009
Hey guys,

I'm studying for the 70-291 and I'm having a hard time understanding
VLSM's. I think I have it now, but I just need someone to confirm my
understanding. Below is an example Class B default network that I
broke up using my understanding of VLSM. If someone could confirm if
it is correct, and if not, point out my mistakes and help to explain
where I went wrong, I would greatly appreciate it. Thanks!

Default Network
150.1.0.0/16 - original network

150.1.0.0/24 (255.255.255.0) makes 256 subnets with 256 hosts per
subnet

150.1.1.0/24 (1 subnet, 254 hosts, 1 net id, 1 broadcast)
150.1.2.0/24 (1 subnet, 254 hosts, 1 net id, 1 broadcast)
150.1.3.0/24 (1 subnet, 254 hosts, 1 net id, 1 broadcast)
150.1.4.0/24 (1 subnet, 254 hosts, 1 net id, 1 broadcast)
.....
.....
.....
150.1.56.0/30 (255.255.255.252) - 64 subnets with 4 addresses per
subnet (2 assignable, 1 net id, 1 broadcast)
150.1.56.4/30
150.1.56.8/30
150.1.56.12/30
150.1.56.16/30
....
....
....
150.1.56.248/30
150.1.56.252/30

150.1.57.0/30 (255.255.255.252) - 64 subnets with 4 addresses per
subnet (2 assignable, 1 net id, 1 broadcast)
150.1.57.4/30
150.1.57.8/30
150.1.57.12/30
150.1.57.16/30
....
....
....
150.1.57.248/30
150.1.57.252/30

150.1.58.0/30 (255.255.255.252) - 64 subnets with 4 addresses per
subnet (2 assignable, 1 net id, 1 broadcast)
150.1.58.4/30
150.1.58.8/30
150.1.58.12/30
150.1.58.16/30
....
....
....
150.1.58.248/30
150.1.58.252/30

150.1.59.0/24 (1 subnet, 254 hosts, 1 net id, 1 broadcast)
150.1.60.0/24 (1 subnet, 254 hosts, 1 net id, 1 broadcast)
.....
.....
.....
150.1.128.0/27 (255.255.255.224) - 8 subnets with 32 addresses per
subnet (30 assignable, 1 net id, 1 broadcast)
150.1.128.32/27
150.1.128.64/27
150.1.128.128/27
150.1.128.160/27
150.1.128.192/27
150.1.128.224/27
.....
.....
.....
150.1.254.0 (1 subnet, 254 hosts, 1 net id, 1 broadcast)
150.1.255.0 (1 subnet, 254 hosts, 1 net id, 1 broadcast)
 
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Lawrence Garvin [MVP]
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      04-30-2009
"Matrixx333" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Hey guys,
>
> I'm studying for the 70-291 and I'm having a hard time understanding
> VLSM's.


Variable Length Subnet Masks... yes... as if subnetting weren't complex
enough, then they came along and created this ability to break up a network
into multiple subnets with different subnet masks.

>I think I have it now, but I just need someone to confirm my
> understanding. Below is an example Class B default network that I
> broke up using my understanding of VLSM. If someone could confirm if
> it is correct, and if not, point out my mistakes and help to explain
> where I went wrong, I would greatly appreciate it.


Your example, while perhaps unnecessarily complex, does appear to correctly
capture the essence of VLSM.

Let me offer a simpler example, just to confirm the scenario, and in the
form you're likely to see the concept tested:

Imagine a single Class C network -- 192.168.1.0 -- with 172 computers.

We need six subnetworks, two with 50 computers, two with 30 computers, and
two with 6 computers.

Let's first set up the two 50-computer subnets. For this we'll need two /26
subnets (netmask 255.255.255.192). We'll use the following:
192.168.1.64/26 (192.168.1.65 - 192.168.1.126)
192.168.1.128/26 (192.168.1.129 - 192.168.1.190)

Then the two 30 computer subnets. For this we'll need two /27 subnets
(netmask 255.255.255.224). We'll use the following:
192.168.1.32/27 (192.168.1.33 - 192.168.1.62)
192.168.1.192/27 (192.168.1.193 - 192.168.1.222)

Finally, the two 6-computer subnets. For this we'll need two /29 subnets
(netmask 255.255.255.24. We'll use the following:
192.168.1.24/29 (192.168.1.25 - 192.168.1.30)
192.168.1.224/29 (192.168.1.225 - 192.168.1.230)

At this point we could also have used several other /29 subnets, including:
192.168.1.8/29
192.168.1.16/29
192.168.1.232/29
192.168.1.240/29

Notice also, although no longer strictly required, I've avoided the use of
the first and last subnet in the range (e.g. the subnet with network ID '0'
and the subnet with broadcast ID '255').

The end results of our configuration:

Net1: 192.168.1.24/29 (6 hosts)
Net2: 192.168.1.32/27 (30 hosts)
Net3: 192.168.1.64/26 (50 hosts; with 12 unused HostIDs)
Net4: 192.168.1.128/26 (50 hosts; with 12 unused HostIDs)
Net5: 192.168.1.192/27 (30 hosts)
Net6: 192.168.1.224/29 (6 hosts)

The key point being here is that we've divided up a specified network into
multiple subnets with a *variable* length subnet mask (/26, /27, or /29),
depending on the specific subnet.

Hope this helps.

--
Lawrence Garvin, M.S., MCITP:EA, MCDBA
Principal/CTO, Onsite Technology Solutions, Houston, Texas
Microsoft MVP - Software Distribution (2005-2009)

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

 
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Matrixx333
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      05-01-2009
> The end results of our configuration:
>
> Net1: 192.168.1.24/29 (6 hosts)
> Net2: 192.168.1.32/27 (30 hosts)
> Net3: 192.168.1.64/26 (50 hosts; with 12 unused HostIDs)
> Net4: 192.168.1.128/26 (50 hosts; with 12 unused HostIDs)
> Net5: 192.168.1.192/27 (30 hosts)
> Net6: 192.168.1.224/29 (6 hosts)
>
> The key point being here is that we've divided up a specified network into
> multiple subnets with a *variable* length subnet mask (/26, /27, or /29),
> depending on the specific subnet.
>
> Hope this helps.


Thank you for the time to reply...It has helped to solidify my
understanding and give me the confidence that I needed. So the ranges:

192.168.1.1 - 192.168.1.23 and
192.168.1.225 - 192.168.1.254

basically go unused until there is a need for them at which point they
themselves can be given a VLSM?

Thanks again!
 
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