Re: Multi-home issue (network address as default gateway?)



On Wed, 19 Apr 2006 15:04:55 -0700, koonalex wrote:

I have been using the following configuration for a year or so now:

8 Ip's from ISP.
Westell Modem/Router handles only public wan ips. DLink Broadband router
handles internal NAT so its wan port is connected to the Westell and has
one of the 8 ip's on the Wan side. Server connected to Westell with 3
public IP's multihomed.

The block of 8 IP's is *.*.*.80 - *.*.*.87 (network address is *.80 and
broadcast address is 87). When the Westell connects it reports to have the
*.86 address as its Wan Ip. I have given *.81 to the DLink router which
NAT's an internal 192.168 network. The server is connected to the Westell
and has *.82, 83 and 84 allocated to a single NIC. When I set this up a
while back (using Debian) on a 2.6.8 kernel and prior using a 2.4 kernel I
set the default gateway for the default route to be the *.80 network
address.This all works fine and I can bind multiple services to the same
port number on the different ip addresses utilising the network address
(*.80) as the default gateway.

I now require more hdd space so purchased a Promise SATA card to connect
some additonal hdd. However, to get this to work I had to use a 2.6.15
kernel (Debian Etch/Testing). So I found an old HDD and thought to rebuild
the server. However, on this new install, I can't get my old interfaces
file to work. The kernel complains of the *.80 being an invalid default
gateway address - when I try to ping the address, it replies saying if I
want to do a broadcast ping - this suggests that it does actually
understand that the *.80 is the network address. I tried changing the
default gateway to the *.86 address (which is the modem's pppoa address)
but this does not seem to work. The internal NAT network via the Dlink
works fine.

I can't seem to understand why I can use the network address successfully
as a default gateway in 2.6.8 but not 2.6.15.

Can anyone shed some light - I am tearing my hair out trying to make this
work. The aim is just to get to use my block of ips successfully.

Hello.

according to your own saying :

Your ISP given IP address range is X.Y.Z.80 to X.Y.Z.87
and this give you 8 IP addresses.
another way is saying that you subnet mask is is 29 bits.

Now : On this little network of yours,
(More correctly a subnet of a bigger network)
there are addresses already reserved.

The first address of any subnet is an address reserved for human writing.
It is used for human by humans to describe the whole of your little net.

So X.Y.Z.80 is a reserved address describing your own little net, with
the hosts that may or may not yet be connected to your subnet.

The last address of your subnet is also reserved. It's the broadcast
address.

You can't give this IP to a host as any packet going to that host
will also be broadcast-ed (and listened) to all the other hosts on
that local network.

So out of your 8 "available" addresses only 6 can really be used.

Actually, since you want to connect a router interface to that, you're
left with only 5 addresses...


Way back then, in the old time, some IP stacks used the first address
of a subnet as a broadcast address. Sun's stack used to do this.
So I assume the code in the Linux kernel must have been changed...
Why ? Can't say...

Network and IP subnetting is no magic but can be complex to approach the
first few times...

I'll spare you the RFCs on that topic but you can seek further information
on the net on that.

Though no "Real" Network Engineers will say so, it's useful to have subnet
charts and a there are even some program to calculate your own number of
hosts on a such or such a size sub-netted network.

These exist for Windows or for Linux.

For Windows : Look on download.com or other shareware download site...

For Linux/Gnome : Look for GIP
http://dag.wieers.com/packages/gip/

For the Web inclined among us :
http://jodies.de/ipcalc

And last : A good page filled with informations and tables :
http://www.subnetonline.com/subnet/step3.html

Hope it help...
.



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