win2000as routing

D

Dejan Rodiger

Hi,

Does Win 2000 Advanced Server have some bugs in routing?
I have Ethernet card and modem on the server and I have created Incoming
connections. It gives IP address to the calling PC.
I have created permanent route (or even non persistant one) to the calling
network with modem as a gateway. After the calling PC disconnects, and calls
again, my routing doesn't work. I can only ping remote IP address, but it is
actually localhost (because routing table is wrong).

For example:
Ethernet card IP address is 192.168.0.2 with default gateway on 192.168.0.1.
Modem IF is 0x1000000.
After remote PC connects, my modem has IP 192.168.1.1 and the remote has
192.168.1.2

Remote network has this parameters 192.168.2.0/24 and all PC's have gateway
192.168.2.1 (the PC with the modem)
Now I create route:
route add 192.168.2.0 netmask 255.255.255.0 192.168.1.2 IF 0x10000000

Now I can ping remote network (192.168.2.1-254)

After I disconnects and connects again, my routing table is all wrong.
It says that my route for network 192.168.2.0 has gateway 127.0.0.1
Even after remote PC connects again, it doesn't resolve the routing table.

For now the solution was:
route -f
route add 192.168.2.0 netmask 255.255.255.0 192.168.1.2 IF 0x10000000
But with this I don't have gateway for my ethernet card any more.

This is a bug!
 
B

Bill Grant

If it was a bug I think somebody else would have noticed it by now. It
has been in production for over six years.
 
B

Bill Grant

If you are trying to use this method to route between the two sites, you
are using the wrong approach. You do not need to configure routing manually.
Manual config fails because the interface does not exist until you make the
connection.

The correct procedure is to use RRAS and set up a demand-dial interface
to act as the connection point, not the default dialup interface. You link
the subnet route to this interface (using the new static route wizard). When
the remote site dials in, it connects to the demand-dial interface and the
subnet route becomes active. (It is stored in the registry). When the
connection is broken, the extra route is removed from the routing table.

Check the entries in help on lan to lan or router to router connections.
 
D

Dejan Rodiger

Bill Grant said the following on 03.03.2006 02:01:
If you are trying to use this method to route between the two sites, you
are using the wrong approach. You do not need to configure routing manually.
Manual config fails because the interface does not exist until you make the
connection.

The correct procedure is to use RRAS and set up a demand-dial interface
to act as the connection point, not the default dialup interface. You link
the subnet route to this interface (using the new static route wizard). When
the remote site dials in, it connects to the demand-dial interface and the
subnet route becomes active. (It is stored in the registry). When the
connection is broken, the extra route is removed from the routing table.

Check the entries in help on lan to lan or router to router connections.

Thank you Bill for your response, I will look at RRAS. I didn't want to
complicate with RRAS, since it was working OK (until this).
 
B

Bill Grant

For a full implementation of MSFT LAN to LAN routing you need a RRAS
server at both ends. This is what the documentation talks about. But if you
have a small site with no router or server you can use a workstation to set
up the connection.

You must have RRAS at the "answering" end so that you can use a
demand-dial interface to link the static route to. When you make the
connection you must use the name of the demand-dial interface as the
username, so that the connection uses the correct interface and sets up the
return route. Otherwise you still have the same problem as before (because
it connects to the default interface).
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Top