Re: Subnetting problem

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Exarctly. In original classful subnetting you had 2^n-2 "usable" subnets - 
you can't use the first or the last (what a waste). There probably are some 
routers out there that still cling to this, but I doubt you'll ever see one 
in a production environment (I haven't). So for practical purposes you can 
ignore the "-2". But if you're taking a test that is testing yor knowledge 
of classful subnets, the correct answer to your example is 6 subnets. In the 
real world you'd have 8.

"EEWashington" <EEWashington@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message 
news:FD476302-5616-4DD7-9F57-8FC7CCE5C1E5@xxxxxxxxxxxxxxxx
>I had a problem where I'm supposed to find subnets using the equation 2n-2
> with n being the number of bits I borrow from the original subnet mask. If 
> I
> wanted to borrow 3 bits, using traditional algebra it would be 2(3)-2 
> which
> equals 4 (subnets). But their example came up with 8 subnets so is that
> equation a misprint and should be 2 to the 3rd power (2x2x2 which equals 
> 8)?
> And if that is true then I'm assuming I can ignore the (-2) part of the
> equation
> -- 
> MCSA is just the beginning... 


.



Relevant Pages

  • Re: Subnetting problem
    ... >I had a problem where I'm supposed to find subnets using the equation 2n-2> with n being the number of bits I borrow from the original subnet mask. ... > wanted to borrow 3 bits, using traditional algebra it would be 2-2 ...
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  • Re: Subnetting problem
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