So I presume by joing the last two subnets I will be wasting two less IPs,
right. i.e one for the network 192.168.1.128 and one for the broadcast
192.168.1.255 giving me from 192.168.1.129 - 192.168.1.254 hosts. Other
wise I will be wasting 4 IPs for the last two subnets, right. So doing this
saves me two IPs and 192.168.1.192 and 192.168.1.191 will be regarded as
valid host ips which otherwise would be network and broadcast address for
the last two subnets. Please comment on this. Can a linux be used as a
bridge effectively ?
Thanks,
Irfan Akber
----------
> From: Keith Owens <[EMAIL PROTECTED]>
> To: Irfan Akber <[EMAIL PROTECTED]>
> Cc: [EMAIL PROTECTED]
> Subject: Re: Subnetting
> Date: Friday, January 01, 1999 12:05 AM
>
> On Thu, 31 Dec 1998 21:51:21 +0500,
> "Irfan Akber" <[EMAIL PROTECTED]> wrote:
> >What is the rule for supernetting 2-3 subnets to appear as one single
> >block.
> >e.g I have 4 subnets of 192.168.1.0 giving 4 subnets each with 62
usable
>
> A subnet must have n leading 1's in its netmask. Therefore it contains
> 2**(32-n) addresses. Grab ftp://ftp.ocs.com.au/ipcalc.pl.gz (Perl 5)
> or ipcalc.tcl.gz (Tcl/Tk 8).
>
> ipcalc.pl 192.168.1.0/26
>
> IP address 192 . 168 . 1 . 0 / 26 192.168.1.0/26
> Netmask bits 11111111 11111111 11111111 11000000
> Netmask bytes 255 . 255 . 255 . 192 255.255.255.192
> Address bits 11000000 10101000 00000001 00000000
> Network 192 . 168 . 1 . 0 192.168.1.0
> Broadcast 192 . 168 . 1 . 63 192.168.1.63
> First Host 192 . 168 . 1 . 1 192.168.1.1
> Last Host 192 . 168 . 1 . 62 192.168.1.62
> Total Hosts 62
> PTR 0.1.168.192.in-addr.arpa
> IP Address (hex) C0A80100
>
> ipcalc.pl 192.168.1.128/25
>
> IP address 192 . 168 . 1 . 128 / 25
192.168.1.128/25
> Netmask bits 11111111 11111111 11111111 10000000
> Netmask bytes 255 . 255 . 255 . 128
255.255.255.128
> Address bits 11000000 10101000 00000001 10000000
> Network 192 . 168 . 1 . 128 192.168.1.128
> Broadcast 192 . 168 . 1 . 255 192.168.1.255
> First Host 192 . 168 . 1 . 129 192.168.1.129
> Last Host 192 . 168 . 1 . 254 192.168.1.254
> Total Hosts 126
> PTR 128.1.168.192.in-addr.arpa
> IP Address (hex) C0A80180
>
> The best you can do is join subnets 3 and 4 together. Joing 2, 3 and 4
> would give 192 addresses which is not a power of two therefore you
> cannot join 2, 3, and 4.
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