My contrarian $.02 :)
>Typically the first and last subnet are not used,
This might be true, but
>toss out 176 and 191,
is a non-sequitur :)
The 0 and -1 subnet restriction only apply to classful considerations.
We're already out of classful thinking here, because we were given a
non-/16 block (a /20) out of the Class B network, 172.16.0.0, and are
extending the prefix 4 bits to /24.
The only subnet addresses that would be considered problematic in our
general area would be 172.16.0.0/24 and 172.16.255.0/24 (classful
subnets 0 and -1).
Of course, neither of those prefixes fall within the block that we were
given, so it's not *our* problem :)
>this leaves 177 through 190, each with a 24 bit mask.
Thus, *all* "subnets" 172.16.[176-191].0/24 are valid.
I.e., no host or router would object to being given an address
172.16.176.1/24
or 172.16.191.1/24
and 172.16.191.255/24
would not be an all-subnets broadcast (just a simple directed
broadcast).
Thus, if additional choices had been:
G)��� 172.16.191.0/24
H)��� 172.16.176.0/24
Then the answer would have been
C, F, G, H
-------------------------------------------------
Tks��� ��� | <mailto:[EMAIL PROTECTED]>
BV��� ���� | <mailto:[EMAIL PROTECTED]>
Sr. Technical�Consultant,� SBM, A Gates/Arrow Co.
Vox 770-623-3430�����������11455 Lakefield Dr.
Fax 770-623-3429���������� Duluth, GA 30097-1511
=================================================
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of
Ed Moss
Sent: Saturday, January 27, 2001 10:50 PM
To: [EMAIL PROTECTED]
Subject: Re: Subnet question
The 20 bit prefix extends four bits into the third octet (176).
176 in binary is 10110000, so with the mask the address ends at 1011.
You want to use the next four bits for subnetting (last four 0's)
This gives the range of 10110000 (176) through 10111111 (191)
providing 16 subnets with 256 addresses in each subnet.
Typically the first and last subnet are not used, toss out 176 and 191,
this leaves 177 through 190, each with a 24 bit mask. (We started with
20 bits, and we added four bits for our own subnets).
Looking at the possible answers, the following fall in this range.
C) 172.16.183.0/24
F) 172.16.190.0/24
Ed
ORIGINAL:
----------------
Can anyone please explain to me how to derive the answer of this
question?
A company has been assigned a subnet of 172.16.176.0/20, and wants the
next four available bits to create 14 subents, each containing an equal
number of hosts.� Which of the following could represent one of these
subnets?
A)��� 172.16.255.0/24
B)��� 172.16.193.0/24
C)��� 172.16.183.0/24
D)��� 172.16.16.0/24
E)��� 172.16.0.0/24
F)��� 172.16.190.0/24
Answer is C and F
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