[mrtg] Re: Adding two OIDs together.

[Sverre Munthe]

To take the average CPU load on 4 cup's on our SQL server I take
(OID1&OID1 + OID2&OID2 + OID3&OID3 + OID4&OID4) / 4

Without the parentheses you will add the first 3 and then add 25% of the
4th.  :)

Or this one who shows the temp probe on the back of the server and the
average of the temp probes on the 4 cup's

(1.3.6.1.4.1.674.10892.1.700.20.1.6.1.7&1.3.6.1.4.1.674.10892.1.700.20.1
.6.1.1:public at 10.47.2.2 +
1.3.6.1.4.1.674.10892.1.700.20.1.6.1.7&1.3.6.1.4.1.674.10892.1.700.20.1.
6.1.2:public at 10.47.2.2 +
1.3.6.1.4.1.674.10892.1.700.20.1.6.1.7&1.3.6.1.4.1.674.10892.1.700.20.1.
6.1.3:public at 10.47.2.2 +
1.3.6.1.4.1.674.10892.1.700.20.1.6.1.7&1.3.6.1.4.1.674.10892.1.700.20.1.
6.1.4:public at 10.47.2.2) / 40

I had a real hard time before I understood I had to use the parentheses
:)

Sverre


-----Original Message-----
From: Scott Schappell [mailto:archon at silvertree.org] 
Sent: 15. oktober 2003 21:52
To: mrtg at list.ee.ethz.ch
Subject: [mrtg] Re: Adding two OIDs together.

PAUL WILLIAMSON wrote:
> OID1&OID2 + OID3&OID4 = OID1 + OID3   and OID2 + OID4
> 
> I thought it said that in the docs somewhere...
> 
> Paul
It doesn't, I actually had the exact same question and after 
experimenting with it I came to the answer you provided, Paul.

Also any mathematical expression is applied to all counters.

target1&target2 / 4 would divide both targets by 4. I tried grouping by 
() and {} but MRTG explodes in a frenzy of log activity :).

HTH
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