Re: IP/Subnet question ,PLEASE!

From: Alon Brodski (abrodski_at_delete.012.net.il)
Date: 07/29/04 

Date: Thu, 29 Jul 2004 02:44:27 +0200

OK,let me try the last time the hard way :-)
4,449 hosts in Class B IP address:166.233.212.5
So instead of over 65,000 possible hosts I would have "ONLY" :-)===8,192 2
in power of 13
10100110111010011101010000000101
1111111111111111.11100000.00000000
so in dec my subnet mask would be here=255.255.224.0

"CZ" <CZ@no99spam.com> wrote in message
news:upgzfcMdEHA.4048@TK2MSFTNGP12.phx.gbl...
> >> I think the 255.255.192.0 is the right answer and not 255.255.206.0 but
> >> it
> would mean that we can only define a subnet much bigger than actual No.
> of hosts....ie.16,384 vs.12,506...
>
> Alon:
>
> 255.255.192.0 is the right answer (correct subnet mask).
>
> Impt concept:
> The subnet mask's decimal values are limited to the equivalent decimal
value
> of contiguous binary 1s from the left, or all binary 0s period.
>
> Therefore, the only possible subnet mask octect decimal values are:
> 0 for [0000 0000]
> 128 for [1000 0000]
> 192 for [1100 0000]
> 224 for [1110 0000]
> 240 for [1111 0000]
> 248 for [1111 1000]
> 252 for [1111 1100]
> 254 for [1111 1110]
> 255 for [1111 1111]
>
> >> Let's say we have a Class A IP address 84.94.158.20 and we have 12,506
> >> hosts
> on a subnet.So how would we know what subnet mask should be?
>
> Easy way:
> Use a subnet calculator and select a max # of hosts per subnet that is
> greater then the number you want, then use the related subnet mask that
the
> calculator gives you.
> I use the "IP Subnet Calculator" from Net3Group
>
> Hard way:
> We want 12,506 hosts in a single subnet.
> So, we want 2 to the nth power that allows more than 12,506 different
> numbers.
> 2 to the 14th = 16384.
>
> Host IDs are represented by contiguous binary 0s from the right in a
subnet
> mask:
> So, we want 14 contiguous 0s from the right.
> So, the subnet mask must be:
> 255.255.192.0 which is 8 0s in the right most octect, and 6 0s in the next
> octect for a total of 14 contiguous 0s.
>
>
>
>
>
>
>
>
>
> OP's post:
> Awhile ago I was asking some basic info about subnet masking.Now when I
know
> more,I'd like to ask deeper.
> I understand the idea behind the subnets ,but I still don't get the
> implementation part.
> Let's say we have a Class A IP address:84.94.158.20 and we have 12,506
hosts
> on a subnet.So how would we know what subnet mask should be?
> IP in bin is:
> 01010100.01011110.10011110.00010100 (I added dots for convenience only)
> 2 in a power of 14 is 16,384 and in a power of 13 is 8,192.But the 1st is
> too many and the 2nd is too few vs 12506 PC's that are given here.So I
> started playing with dec numbers and came to the closest number of 206 as
a
> 3rd dec octet of a subnet mask,ie 255.255.206.0 'cos 255-206=49 and 49
> times 256=12544 the highest integer that would give us more than 12506
> hosts in the subnet here.
> So the first thing is I don't know how then you define the EXACT number of
> hosts and 2nd if you translate 255.255.206.0 into bin=
> 01010100010111101001111000010100
> 11111111111111111100111000000000
> Doesn't look OK...but if I use powers example....2 in a power of 14...then
> it would be:
> 01010100010111101001111000010100
> 111111111111111111000000.00000000
> or 255.255.192.0
> I think the 255.255.192.0 is the right answer and not 255.255.206.0 but it
> would mean that we can only define a subnet much bigger than actual No.
> of hosts....ie.16,384 vs.12,506...
>
> So maybe anyone can explain it to me ?
>
>

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