On Wednesday, 11/11/2009 at 08:23 EST, "Frank M. Ramaekers"
<framaek...@ailife.com> wrote:
>
> In other words:
>
> VSWITCH ETH0
> 10.1.20.2 <-------> 10.1.20.20 <-----> Everything else
> z/VSE z/VM
> \
> \
> Everything else
I am going to chastise myself (yet again) for dealing with a configuration
problem without requiring the picture FIRST. Strange as it sounds, your
picture is not valid or you aren't understanding VSWITCHes.
Whenever a host touches a network of any sort (LAN or point-to-point), an
IP address comes into existence. (Special dispensation given for
unnumbered IP interfaces, which z/VM TCP/IP doesn't support.) If the two
networks are not bridged by something outside of the hosts, then they are,
by definition, separate networks. Every network gets its own subnet, even
if if contains only two hosts. And within Subnet A, you never assign the
IP addresses belonging to Subnet B. Never.
In your drawing, I detect two networks:
1. A network labeled "Everything else" (10.1.0.0/16)
2. A network that has no label and no assigned subnet
You cannot communicate on Network 2 until you assign it a subnet that is
not within 10.1.0.0/16, and then assign IP addresses within that new
subnet.
If you simply trying to attach VM TCP/IP and z/VSE to a VSWITCH, the
picture looks like
10.1.20.2 10.1.20.20
| ETH0 | ETH0 (vNICs)
--------------VSWITCH------------
OSA | The VSWITCH is
| an ethernet bridge
---Real network------------------ 10.1.0.0/16
They can talk to each other and out to the network.
Alan Altmark
z/VM Development
IBM Endicott
