Well, your numbers are a little off, but the general idea is sound. With
the way the question is worded, who knows exactly what the correct mask is.
Other than the obvious, maybe we can assume that the questioner is adding 10
additional bits to the native classful mask of 255.255.0.0 for a class B
network. This would produce a 255.255.255.192 mask.
Now for the math.
With 192 in the last octet, this leaves 6 bits for hosts
2^6 = 64 - 2 (don't forget this part!) = 62 usable host addresses
You also now have 10 bits for subnetting (additional to the native classful
mask)
So, 2^10 = 1024 - 2 (don't forget this part here either!) = 1022 usable
subnets
Notice we can use the same simple formula ( 2^n-2) to find both the number
of usable host addresses and subnets.
One other thing to be aware of: each subnet ID, which is the first address
in every subnet, always has an even number in the subnetted octet. In other
words, the first "usable" subnet in the previously mentioned example should
be 172.16.0.64. This is just a chracteristic of the process of
subnetting....something about always starting to count with a 0 instead of a
1.... ;-}
172.16.0.0 - 172.16.0.63
172.16.0.64 - 172.16.0.127
172.16.0.128 - 172.16.0.191
172.16.0.192 - 172.16.0.255
etc., etc., etc........ you get the idea
The website I think is a great resource for those just getting into TCP/IP
is:
http://www.learntosubnet.com/
Check it out!
BTW: it is true that many legacy device/software won't recognize the first
or last subnet. The first subnet, also called the "zero subnet", can be
configured for use on Cisco devices with the "ip subnet-zero" command.
Although not every vendor supports doing this, I would guess that a vast
majority of them do. It was only a short time ago when this was not the
case. I can remember when NT 4.0 would not support this. It probably does
now, with service packs and all, but it didn't always.
Rik
"Joshua Beining" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Based on the subnet mask (255.255.192.0) subtract 256 from 192 which is
64.
> Therefore 64 is your first subnet. To get the second subnet, add 64 to
the
> first subnet. To get the third add 64 to the second subnet and so on.
> Continue in this fashion until you reach 192. Remember that you cannot
use
> the ranges 172.16.0.1 - 172.16.0.62 and 172.16.255.193 -
172.16.255.254
> (network and broadcast respectively) unless your router is configured to
do
> so.
>
> Subnet Host Range
> 1 172.16.0.65 - 172.16.0.126
> 2 172.16.0.129 - 172.16.0.190
> 3 172.16.0.193 - 172.16.0.254
> 4 172.16.1.1 - 172.16.1.62
> 5 172.16.1.65 - 172.16.1.126
> 6 172.16.1.129 - 172.16.1.190
> 7 172.16.1.193 - 172.16.1.254
> 8 172.16.2.1 - 172.16.2.62
> 9 172.16.2.65 - 172.16.2.126
> 10 172.16.2.129 - 172.16.2.190
> .
> .
> .
> .
> .
> .
>
> 1015 172.16.253.193 - 172.16.253.254
> 1016 172.16.254.1 - 172.16.254.62
> 1017 172.16.254.65 - 172.16.254.126
> 1018 172.16.254.129 - 172.16.254.190
> 1019 172.16.254.193 - 172.16.254.254
> 1020 172.16.255.1 - 172.16.255.62
> 1021 172.16.255.65 - 172.16.255.126
> 1022 172.16.255.129 - 172.16.255.190
>
> -Joshua
> -----Original Message-----
> From: Lowell Sharrah [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, March 22, 2001 8:14 AM
> To: [EMAIL PROTECTED]
> Subject: RE: difficult ccna question
>
>
> how do you know where the first subnet begins?
>
> >>> Joshua Beining <[EMAIL PROTECTED]> 03/22/01 10:34AM >>>
> Just remember the following fomulas:
>
> (2^# of masked bits) - 2 = Total # of subnets
> (2^# of unmasked bits - 2 = Total # of hosts
>
> Based on this the correct answer is A.
>
> -----Original Message-----
> From: George [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, March 22, 2001 12:16 AM
> To: [EMAIL PROTECTED]
> Subject: difficult ccna question
>
>
> if you have a class B network with a 10-bit subnet mask, how many subnet
and
> how many hosts do you have?
>
> a. 1022 subnets, 62 hosts
> b. 62 subnets, 8190 hosts
> c. 8190 subnets, 254 hosts
> d. 254 subnets , 126 hosts
>
>
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