Subnetting in 293
RogueIT
In the question they ask what is the first subnet of 10.0.0.0/19.
I thought it would be 10.0.0.0 with the first usable address as 10.0.0.1,
but they give the answer as 10.0.32.1 to 10.0.63.254.
Why isn't 10.0.0.1 to 10.0.31.254 the first subnet?
Thanks,
RogueIT
JohnB
That is a Class A address, which has an 8 bit network mask.
With the example using 19 bits, having borrowed 11 bits from the host
portion of the address.
So the mask in dotted decimal format is 255.255.224.0
Which gives a subnet block size of 32 (256-224=32)
So the address ranges for each subnet would be:
10.0.0.0 - 10.0.31.255
10.0.32.0 - 10.0.63.255
10.0.64.0 - 10.0.95.255
10.0.96.0 - 10.0.127.255
and so on, up to....
10.0.224.0 10.0.255.255
With the first address being the subnet, and the last being the broadcast
address, for each range
So you're right, the *first* subnet is 10.0.0.0
The answer they give is the usable *range*, which is different than a subnet
number. And their range, is the second range.
"RogueIT" <Rog...@discussions.microsoft.com> wrote in message
news:EE612157-9D75-4E43...@microsoft.com...
Lawrence Garvin [MVP]
news:EE612157-9D75-4E43...@microsoft.com...
For the same reason the first subnet is not used in the other two examples:
192.168.214.0/29 -> First subnet 192.168.214.9
172.28.0.0/20 -> First subnet 172.28.16.1
so
10.0.0.1/19 -> First subnet 10.0.32.1
The problem is that applying the same, archaic, rule should also have
excluded the LAST subnet from that list as well.
This archaic rule had to do with the fact that the SubnetID portion of the
address also could not be all zeros or all ones.
Thus, 10.0.0.0/19 cannot be a SUBNET; 172.28.0.0/20 cannot be a SUBNET;
192.168.214.0/29 cannot be a SUBNET.
The question is severely misleading because it only partially applies the
rule. If the rule is to be applied properly, then
10.255.255.255/19 cannot be a subnet broadcast address, thus the subnet
10.255.224.1/19 should be invalid.
172.28.255.255/20 cannot be a subnet broadcast address; thus the subnet
172.28.240.1/20 should be invalid.
192.168.214.255/29 cannot be a subnet broadcast address; thus the subnet
192.168.214.248/29 should be invalid
But, as I noted, the rule is archaic. The explanation from the 70-291 book
(p2-28) is much clearer:
To determine the number of subnets available within an address space,
simply calculate the value of 2^y, where y equals the number of bits in the
subnet ID.... You do not usually need to subtract 2 from this total because
most modern routers ... can accept a subnet ID made up of all 1s or all 0s.
The 70-293 book has essentially the same statement, but because it did such
a poor job of explaining the *subnet* related portion of this rule, the
point is somewhat lost. From p2-27 of the 70-293 book:
Most routers and operating systems, including Windows Server 2003, now
enable you to use all zeros for a network or SUBNET identifier, but you must
be sure that all your equipment supports these values before you decide to
use them.
The fact is that Microsoft TCP/IP has always allowed all 1s or all 0s; in
fact, it was Microsoft TCP/IP that actually broke the rule to begin with and
permitted it to be done on Windows NT, when other *standard* equipment was
still conformant to the RFC that specified you could not do that.
For purposes of Microsoft Certification Exams -- your answer, that
10.0.0.1/19 is a valid subnet -- is correct.
btw, note also there are several items of errata in that question, all of
which are documented at
http://support.microsoft.com/kb/832375/en-us
--
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/profile/Lawrence.Garvin