Testing home network

So I was reading the subnetting section ( www.firewall.cx/ip-subnetting-routing.php ) and thought I would test subnetting/dual ip's on my network and to see if I could get them to all communicate...it looked easier than its turning out...

Heres the setup

Belkin router:
Internal interface is 192.168.2.1/24

XP machine1:
address 192.168.2.5/24
GW: 192.168.2.1

XP machine2:
address 192.168.2.35/24
GW: 192.168.2.36

on my Debian box my /etc/network/interfaces file looks like:
auto eth0
address 192.168.2.10
netmask 255.255.255.224
network 192.168.2.0
broadcast 192.168.2.31
gateway 192.168.2.1

auto eth0:0
address 192.168.2.36
netmask 255.255.255.224
network 192.168.2.32
broadcast 192.168.2.63

The problem as I see is that XP machine2 can ping only 192.168.2.36, but cant get to the rest of the network/internet.
Can someone tell me why this isnt working? Thanks for the help.

Assuming you have XP1 and the debian box plugged into the router and XP2 plugged into your debian box, eth0:0 looks like an alias to me but Im not familiar with linux too much. Try changing the netmask of eth0 to 255.255.255.0 so its on the same network as XP1 and the netmask of XP to /27(255.255.255.224).

let us know how that works out for you.

Oops...i said the netmask on the xp boxes was 255.255.255.0 but i meant 255.255.255.224. So that would put eth0 on the same network as XP1. With that said, I cant plug xp2 into my debian box because It only has 1 NIC. Sorry about that misunderstanding. Thanks for the help

I'm not sure that your router will work for a setup like that. Other than that 192.168.2.0/27 is an invalid subnet address. Try using a different subnet for XP1 and eth0.

Why is 192.168.2.0/27 in invalid subnet address? (27 = 255.255.255.224, right???)

My router seems to be working fine with that address. I think the problem is something I'm missing in my setup on my debian box. Unless of course I'm wrong about the 27 subnet thing...

192.168.2.0 = 11000000.10101000.00000010.00000000 in binary. Which is the default class C network address. 192.168.2.224 is also an invalid subnet address because the broadcast address will be 255.255.255.255. That is why the total number of useable networks or hosts is 2^x - 2. The first and last are unusable.