Thanks guys.
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of Bret Hughes
Sent: Friday, August 11, 2000 7:33 PM
To: [EMAIL PROTECTED]
Subject: Re: IP Address Blocks
Roadrunner wrote:
> Hi,
>
> It looks to me that you own 64.184.117.0 + 64.184.117.28 and every
number in
> between on subnetmask 255.255.255.240.
>
> However I thought that the '0' isn't used 'cause it means
something like:
> "unknown value" or in this case unknown computer. But it could
also mean
> 'all computers'.
>
> anyway,
> When someone else gives you a completely different answer, please
take that
> one 'cause i've only installed a simple homenetwork.
>
> Remon
>
> (I don't know everything, but who does)
>
I dont believe this is correct but I will have to think about it
bear with me
as I do so in written form.
The notation given is really saying the same thing.
64.184.117.0/28 means your network starts at 64.184.117.0 and the
first 28 bits
are used for the network.
The netmask of 255.255.255.240 is an older notation for the /28.
255.255.255.240 in binary is three octets of all ones and the 240
is, in binary
if I am not mistaken, 11110000. So.... 3X8 + 4 = 28 bits used for
the
network. That leaves the remaing four bits for the host portion.
So what you have I believe is the numbers 64.184.117.0 through
64.184.117.15
yielding 14 usable addresses. The first and last are not usable as
host since
the first is the network address and the last is the subnet
broadcast.
Therefore you should have the use of 64.184.117.1 - 64.184.117.14 as
hosts.
I believe this is correct but one of the gurus will probably have
stepped in by
the time I get this written and sent.
Hope this helps. At least the network started at a 0 in the last
octet. I
really get a headache trying to visualize it if it is not.
Bret
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