[PLUG] ARP black magic question

[Zot O'Connor]

On Tue, 2003-08-26 at 12:45, Eric Harrison wrote:
> It appears I am in the need of being hit upside the head with the
> clue stick ;-)
> 
> 
> One particularly complex load-balancing setup we have is in need of
> a little arp black magic: I need server A to answer arp requests on
> behalf of server B.
> 
> That is, when some random host sends out an arp query "who has serverb?",
> Server A needs to respond "serverb is at 01:02:03:04:05:06", where 
> "01:02:03:04:05:06" is the real MAC address for server B. 
> 
> I'm not having much luck figuring out how to do this and all of the
> docs I can find cover how to hijack an IP via arp & proxy arp. Neither
> of these fit my scenario. 
> 
> Has anyone here setup something like this up? Given how much trouble I'm
> having I assume there is an embarrasingly simple solution ;-)
> 

yes I have.

It all backwards, an odd number of times.

The question is, do you want to proxy an IP or a mac address?

Lets assume outside and inside of a gateway.

Address 192.168.90.45 is inside

192.168.90.46 is the gateway's nic on the inside (eth1)
192.168.90.47 is the gateway's nic on the outside. (eth0)
192.168.90.1 is the real default gateway which is on the outside.

So we want to make 192.168.90.45 look like it available to the outside

we have eth1 for the inside, and eth0 for the outside

To publish to outside that .45 is to be "answered" by eth0
        arp -Ds 192.168.90.45  eth1 pub -i eth0

To publish to the inside that .1 is to answered by eth1
        arp -Ds 192.168.90.1  eth0 pub -i eth1

You should be able to do this with MAC Address as well.

Word of warning, this works real well under Linux, but other Unix-like
OSes including routers do not always support pub.



> -Eric
> 
> 
> 
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-- 
Zot O'Connor

http://www.ZotConsulting.com
http://www.WhiteKnightHackers.com