General virtualization question

Bob_the_Goon

If you wanted a single server to host various virtual servers, each running within their own domains, would you need to link the virtualised server/s of each indiviudual domain to a SEPERATE physical network card?

I'm thinking of setting up a test environment with three seperate domains. I do not need the three domains to be able to see each other, however I would like the virtual servers to administer three seperate client PCs (not virtualised) connected to the host server via a switch.

Am I right by saying the the host server requires three separate NICs, each connected to their respective clients via a VLAN enabled switch?

This is probably a really simple/stupid question, but I'm new to this virtualization game.

Any help would be greatly appreciated!!

dynamik

Most virtualization packages allow you to setup virtual networks where you only link the VMs to each other. You wouldn't need a physical NIC for each VM.

Edit: Totally read your question wrong. You would however need a virtualization package that supports VLAN tagging. You could also add a multiport NIC if you just don't want to use a lot of slots on your motherboard. USB NICs could work in a test environment too if you simply don't have enough ports on your motherboard.

MentholMoose

If you want each domain on it's own broadcast domain, you don't need three NICs if your switch has VLAN support. You can setup a VLAN trunk with just one NIC connecting to one switchport. This may be unnecessary, though, since for many lab scenarios it is OK if multiple domains are on a single broadcast domain. Just configure them with different subnets and in most respects they will be separate.

For example, if you configure domain 1 machines on 10.0.1.0/24 and domain 2 machines on 10.0.2.0/24, they generally won't interfere with each other. The exception is that broadcast traffic would hit both domains, but this is not a problem unless you need broadcast traffic to only one domain (for example, configuring IP addresses with DHCP).