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| Author |
Message |
DC
Guest
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 Posted: Sun Feb 24, 2008 7:57 pm Post subject: CCNA, Calculating Subnets and ip subnet zero |
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Hi,
I know traditionally we didn't count the first and last subnet (all
zeroes and all ones). So a subnet mask of 255.255.255.192 would give
(2^2)-2, or 2, subnets while a mask of 255.255.255.224 would give
(2^3)-2, or 6, subnets.
So, for 192.168.201.0/26 the first subnet would be 192.168.201.64 and
the first host would be 192.168.201.65.
I know the reason for this, but it seemed rather silly and very wasteful
of a scarce resource.
Now, I'm given to believe that this traditional method is no longer used
(RFC1878 describes it as obsolete - http://tools.ietf.org/html/rfc1878).
Instead 255.255.255.192 would give 2^2, or 4, subnets and
255.255.255.224 would give 2^3, or 8, addresses.
I have no problem with this. My questions are:
.. Does anyone know if the latest CCNA exams assume the newer or the
older method (ie 2^s or (2^s)-2)?
.. If the newer method, are there any subnet exercises available on the
Internet? All the ones I have found seem to use the old method (and so
only allow 2 subnets for the mask 255.255.255.192).
Thanks |
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pk
Guest
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| Posted: Sun Feb 24, 2008 11:47 pm Post subject: Re: CCNA, Calculating Subnets and ip subnet zero |
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DC wrote:
| Quote: | Hi,
I know traditionally we didn't count the first and last subnet (all
zeroes and all ones). So a subnet mask of 255.255.255.192 would give
(2^2)-2, or 2, subnets while a mask of 255.255.255.224 would give
(2^3)-2, or 6, subnets.
|
Yes, assuming you are taking a /24 as a reference and you're using the
traditional (old) rule.
| Quote: | So, for 192.168.201.0/26 the first subnet would be 192.168.201.64 and
the first host would be 192.168.201.65.
I know the reason for this, but it seemed rather silly and very wasteful
of a scarce resource.
Now, I'm given to believe that this traditional method is no longer used
(RFC1878 describes it as obsolete - http://tools.ietf.org/html/rfc1878).
Instead 255.255.255.192 would give 2^2, or 4, subnets and
255.255.255.224 would give 2^3, or 8, addresses.
|
Correct. Moreover, nowadays almost all network devices think in CIDR terms,
so they are not aware of subnetting.
But nonetheless knowing subnetting is still important, because network
administrators can decide to subnet their address spaces, and thus they
must know how to do that.
| Quote: | I have no problem with this. My questions are:
. Does anyone know if the latest CCNA exams assume the newer or the
older method (ie 2^s or (2^s)-2)?
|
The new CCNA uses the new method, but only under certain conditions (ie,
classless routing protocol, VLSM, etc.). However, if no information is
given, it's generally safe to assume that the new rule can be used. For
more information, see the new CCNA prep material.
| Quote: | . If the newer method, are there any subnet exercises available on the
Internet? All the ones I have found seem to use the old method (and so
only allow 2 subnets for the mask 255.255.255.192).
|
Well, I think the main purpose of the exercises is to force you to do the
math and become familiar with it, rather than just providing an answer. So,
just use the old exercises and be aware that you have to add 2 to the
result shown by the exercise. |
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