modem utilization(ARC): 1.3.6.1.4.1.429.4.2.1.10.0
box temp (NMC): 1.3.6.1.4.1.429.1.2.2.5
-----Original Message-----
From: usr-tc-bounces+adam=semo.net(a)mailman.xmission.com
[mailto:usr-tc-bounces+adam=semo.net@mailman.xmission.com] On Behalf Of
Ben Winslow
Sent: Thursday, November 17, 2005 10:58 AM
To: Discussion "relating to the 3Com/US Robotics Total Control modem
systems.; totalcontrol(a)pagedata.com
Subject: Re: [USR-TC] mrtg ethernet traffic & free memory
On Thu, 2005-11-17 at 09:38 -0700, PD TC wrote:
> hello,
>
> We are new to the Total Controls and have been using Portmasters. We
> have been searching the manuals, the Internet and the maillist
> archives (through google searches) but have not found the answers for
> some questions.
>
> We are trying to get mrtg information on the ethernet traffic on the
> HiperARC card and the free memory on the Total Control. We saw a
> posting previously by Lewis Bergman in which he posted his .cfg
> information. We have tried implementing it and have been successful at
> retrieving the number of modems online and the temperature of the
> Total Control. We are not getting the ethernet traffic nor the free
> memory to come through.
The HiperARC and the NMC will have different IP addresses (and possibly
different SNMP communities) -- the ARC has the Ethernet statistics in
the expected place (e.g. mrtg's cfgmaker will have no problem finding
them), and the best way to get those numbers will be from the ARC
directly instead of the NMC (the ARC also has the number of connected
clients somewhere, I don't have the OID handy.)
The NMC will allow you to proxy your SNMP request to another one of the
cards (I can't remember the syntax off the top of my head, but it was
something like setting the community to slotid@yourcommunityname), but
it'll be more efficient to query the card directly unless there's
something that prevents you from doing so.
I can dig up the OID for connected users and probably the snmp proxy
syntax if you can't find them, but the info shouldn't be too difficult
to find with a lead.
Cheers,
--
Ben Winslow <rain(a)bluecherry.net>
