Router based VPN through single Ethernet port or easy on/off

1

I've recently moved out of the USA. Before my move I signed up for a VPN service.

I've been searching for a way to enable network-level VPN for all my devices through my router. It seem easy enough through a router that supports OpenVPN or through flashing tomato on a compatible router.

Having said that, I don't need to be VPN'd into the US all the time and the solutions I've found seem to be all or nothing.

Questions

  • Is there a way to enable VPN only on a single Ethernet port on a router (any router...I'm willing to purchase one). Doing this would allow a second VPN'd wireless connection or a hardwire VPN connection.

  • If VPN cannot be enabled for a single port, does a router exist that would allow easy switching on/off of VPN capabilities?

James Hill

Posted 2014-06-16T17:26:15.530

Reputation: 979

Answers

1

Some routers (especially enterprise-grade or Linux-based) support routing decisions based on source IP address (source based routing). Essentially you get multiple routing tables, and based on the source IP of the packet, the router will choose a routing table. You wouldn't even need a specific port on the router, you can just use an IP from your subnet to be routed over the VPN (statically assigned to a host).

If you want a different port, you could assign a different VLAN and subnet to that port, and use source based routing on the entire subnet.

At least Cisco IOS, Juniper, Linux, Vyatta and DD-WRT support this feature.

mtak

Posted 2014-06-16T17:26:15.530

Reputation: 11 805

Based on your new question I would assume this is the answer you are looking for. Could you please mark it as an answer so it can be useful for posterity as well?

– mtak – 2014-06-17T08:38:34.797

1Actually, it's not quite the answer I'm looking for. To get there, I would have had to modify my question so much that your answer wouldn't have fit (totally my fault, not yours). I thought it would be best to post a new question instead. I'm going to mark this as correct and hope for a more detailed answer to my new question. – James Hill – 2014-06-17T08:41:12.790