[racktables-users] Re: How to handle virtual machine's
- From: Denis Ovsienko <pilot@xxxxxxxxxx>
- To: racktables-users@xxxxxxxxxxxxx
- Date: Thu, 5 Feb 2009 19:11:26 +0300
> Help me to understand this please.
>
> On a xen-host there is for example the following interfaces:
>
> xenbr0 xenbr1 - which I understand as Xen bridge interfaces.
> peth0 peth1
> vif0.0, vif0.1, vif1.0, vif1.1
>
>
> So which of those do I configure in Racktables for the host?
>
> On the guest I have eth0 and eth1 as IPv4 configured. Then, according
> to what you said above I must use one veth-port on the guest. So I
> define such a port. Does it matter what I call it?
The default naming convention is the one, which is used in the real
world. Xen documentation states:
"Xen creates, by default, seven pair of "connected virtual ethernet
interfaces" for use by dom0. Think of them as two ethernet interfaces
connected by an internal crossover ethernet cable. veth0 is connected
to vif0.0, veth1 is connected to vif0.1, etc, up to veth7 -> vif0.7.
You can use them by configuring IP and MAC addresses on the veth# end,
then attaching the vif0.# end to a bridge."
For me, that sounds exactly as "eight" pairs, but in the Xen world
things may count differently. The manual tries to get into some odd
logic then (I used to take Xen sceptically), which can be safely
dropped with no sense lost. Anyway, the only meaningful mapping
required seems to be:
vif0.0 - dom0 eth0
vif1.0 - dom1 eth0
vif2.0 - dom2 eth0
vif3.0 - dom3 eth0
...and so on. I could be wrong, but I just don't care now.
That's basically all, that is necessary to show, which virtual machine
is executed by which real hardware. So after creating the "server"
objects you add a set of "vif" plugs to one and one "eth" plug to each
of others, all of these typed "veth/xen", so they can be connected
together later.
> How do I link it with the interface on the host?
The same standard way, which has been used to "connect" two ports
together. From the Ports tab through a pop-up window.
> What if another host has exactly the same interfaces as this one?
It doesn't matter, I guess.
--
Denis Ovsienko
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