Ralph Shumaker wrote:
Then why would one use /27:
Because AT&T delegated that to you (indirectly).
AT&T is a /8:
they have authority over 12.XX.XX.XX
An ISP buying from AT&T may get a /16:
AT&T assigns them 12.169.0.0
The ISP can assign anything inside 12.169.X.X
An ISP buying from them may get a /24:
The previous ISP assigns 12.169.40.0
This ISP can assign anything inside 12.169.40.X
That ISP assigns you a /27:
You get 12.169.40.32
You can assign anything from
12.169.40.32 (all 0's in last 5 bits) to
12.169.40.63 (all 1's in last 5 bits)
12.169.40.36 is inside that range and is yours to assign.
(same host)
12.169.40.36
00001100 10101001 00101000 00100100 - 12.169.40.36 - host
11111111 11111111 11111111 11100000 - /27 netmask (if I got it)
----------------------------------- - bitwise AND the two together
00001100 10101001 00101000 00100000 - 12.169.40.32 ???
Since the last 5 bits will always AND to zeros, does this mean that
12.169.40.0 through 12.169.40.31 all belong to 12.169.40.36 /27?
12.169.40.32 to 12.169.40.63, actually.
And, yes, 12.169.40.36/27 means all of the addresses from 12.169.40.32
to 12.169.40.63 all belong to the same network.
Generally, by convention, we write that as 12.169.40.32/27. We set the
bits which would indicate the host to all 0's when we are actually
talking about the network/netmask.
-a
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