Doesn't the internet support only support high order bit subnetting
Don
----- Original Message -----
From: "John Neiberger" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Thursday, March 15, 2001 9:41 AM
Subject: Re: Discontiguous networks
> ouch! Please do not attempt this at home. Heck, please do not attempt
> this anywhere! :-)
>
> The binary mask for last octet would be 10100000. If you assume the use
> of subnet zero, your host address go from 1 to 31, then skip to 64-94.
> The next subnet in binary is 00100000 so the network address is
> 192.10.3.32 and host address are 33-63 and 95-126. This is painful,
> it's making my head hurt....
>
> Okay then, the next subnet in binary is 10000000, or .128 in decimal,
> and your host addresses are 129-159 and 192-223. The final subnet is
> 10100000, or .148 and the host addresses are 149-180 and 224-254.
>
> I think. That is really painful to think through and I was interrupted
> multiple times while trying to write this. Please forgive me if these
> numbers aren't correct.
>
> And always remember: Friends don't let friends use discontiguous
> subnet masks!
>
> >>> "Arthur Simplina" <[EMAIL PROTECTED]> 3/15/01 10:04:17 AM >>>
> I know that there was an earlier posting and a very good explanation on
>
> this. So, kindly bear with me.
>
> I am trying to figure out the subnets (and hosts) for this address:
>
> 192.10.3.0 with subnet mask 255.255.255.148.
>
> I am asking this out of curiosity and to learn how to go about this.
>
> Thanks for any help.
>
> Arthur
>
>
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