You sure about that, Chuck? ;-)

2^n-2 = 8  ! a total of 8 subnets needed !
2^n = 10   ! add 2 to both sides !
n = 4      ! 2^4-2 = 14 !

128 64 32 16 8 4 2 1
 1   1  1  1 0 0 0 0

= 240, or answer A in the original post.

BJ



-------Original Message-------
From: The Long and Winding Road 
Sent: 12/05/02 09:48 AM
To: [EMAIL PROTECTED]
Subject: Re: VLSM Question [7:58569]

> you sure about that, Tom?


172.100.00000000.0
255.255.11100000.0
subnet bits = 11111.0
172.100.0.0 through 172.100.31.0 for /24's

these would be SUMMARIZED using the 224 mask in the third octet.

if you only want eight /24's, then the answer is

172.100.00000000.0
255.255.11111000.0
subnet bits =     111     eight subnets of /24
summarized as 172.100.0.0/21 ( 248 )



--
TANSTAAFL
"there ain't no such thing as a free lunch"




""Tom Lisa""  wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> If the test prep you are using is for the CCNA exam then "C" is the
> correct
> "Cisco" answer (the use of Class B/Class C terminology makes me think
> this is the case). This is because Cisco still insists, at the CCNA
> level, on
> computing subnets using the formula 2^n-2.  This assumes that subnet
> zero and the all ones subnet are unusable.  Therefore you have to create
> 16 subnets, resulting in 14 "usable" to get the required 8 subnets.
>
> In the "real" world, 255.255.224.0 is correct.
> BTW, what is the VLSM question here?
>
> HTH,
> Prof. Tom Lisa, CCAI
> Community College of Southern Nevada
> Cisco ATC/Regional Networking Academy
> "Cunctando restituit rem"
>
> Richard Burdette wrote:
>
>   A prep test I am using has a question for which I disagree with the
>   answer.
>   Here is the question
>
>   If I had a Class B address, what subnet mask would I use if I wanted
>   to
>   split it into 8 class C addresses?
>
>   a.255.255.240.0
>   b.255.255.255.0
>   c.255.255.248.0
>   d.255.255.254.0
>
>   The answer from the test is c.
>
>   I think the answer is not even listed; 255.255.224.0 because to add
>   eight
>   additional subnets we need 2^3=8 bits of subnet which equates to 224
>   of
>   mask.  Am I right or wrong?
>
>   Rich
>
>
>
>
>   [EMAIL PROTECTED]




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