That's the way I do it and it makes it possible to do everything in your
head, but when I explain it, it sounds like I'm trying to explain splitting
the atom. Some of my work colleagues have been admitted to secure units
after a lesson from me. It takes someone with a similarly warped mind to
understand me.
Is there a copyright on that text. Do you mind if I keep it to pass the
knowledge?
Gaz
""Tony Medeiros"" wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Paul,
> There are many ways to do this. I'll show you my technique that is fast
and
> works good.
>
> First, Take the number in the right most octet that isn't 255 and
subtract
> it from 256. In your case it's 248. This gives you 8.
>
> Now, 8 is the number that the subnets increment. The first subnet is
zero,
> then 8, then 16, then 24, etc.
>
> Now, take the number in you last octet and divide it by 8. In your case
> it's 184 divided by 8 with equals a nice round 23. Because there are no
> remainders in you division, that means you are given a valid network
number,
> not a host number on that network.
>
> Now, Because of your mask we know the networks increment by 8. So your
> network is 213.13.184.184. The next network is 213.13.184.172. Your
hosts
> are the numbers in between. BUT...
>
> Always remember your network number is the network identifier and can't be
> used for a host. And, the last number in your network is the broadcast
> address for your subnet.
>
> So, The network is 213.13.184.184 (can't use this for a host), The host
> address are 213.13.184.185, 186, 187, 188, 189, 170. The broadcast
address
> is 213.13.184.171. and the number of the next subnet is 213.13.184.172.
>
> Remember that you always loose 2 address for each subnet: one for the
> network number and one for the broadcast. A nice rule of thumb is that if
> the networks increment by 4, you have 2 usable host addresses. By 8, 6
> usable addresses. By 16, 14 usable address, etc
>
> If this still confuses you, write it out in binary. If you learn it in
> binary, you've learned it !!
>
> Tony M.
> #6172
>
>
> ----- Original Message -----
> From: paul
> To:
> Sent: Wednesday, July 11, 2001 3:11 AM
> Subject: calculating allowable hosts [7:11902]
>
>
> > Greetz,
> > Can anyone explain how you get the allowable host addresses from the
> > following 213.13.184.184 with 255.255.255.248?(or show where I can
figure
> > this out )
> > Thanx
Message Posted at:
http://www.groupstudy.com/form/read.php?f=7&i=12009&t=11902
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