[mrtg] Re: Mathematical operation

[Paul C. Williamson]

It still needs to be in the form of OID1&OID2:public at a.b.c.d - OID1&OID2:public at a.b.c.d.

Do OID3&OID3:public at a.b.c.d - OID1&OID1:public at a.b.c.d * 10.  The blue and green will be the same then.  If you don't like that, try to find an OID that always responds with a 0 and insert that for either the first or second OID in each pair (so it would bet OID3&0:public - OID1&0:public)

Paul

>>> WEYMIENS Bruno FTRD/DMI/SOP <bruno.weymiens at rd.francetelecom.com> 09/03/01 07:39 AM >>>

Hi everyone,

I have 2 OIDs :
- .x.3:public at a.b.c.d and
- .x.1:public at a.b.c.d, which are 2 RTTs in 1/100th secondes, so I multiply
them by 10 to have millisecondes ;

For the moment being, I graph them as 2 separate targets, so I have 2 lines
as :
Target[Ping3]: .x.3&.x.3:public at a.b.c.d * 10
Target[Ping1]: .x.1&.x.1:public at a.b.c.d * 10
which runs pretty well.

now, i'd like to substract ping1 from ping3 ;
I looked at he FAQ and found :
"Target[ezwf]: 2:public at wellfleetA + 1:public at wellfleetA * 4:public at ciscoF"

I understood that MRTG needed 2 OIDs in 1 target to graph (that's why I have
smthg like "OID&OID:community at IP"), but in the FAQ response, I don't see how
to match 2 OIDs with the formula ; 
I tried smthg like :
"OID3 - OID1:community at IP&OID3 - OID1:community at IP", which gives us :

Target[Ping_arc2]: .x.3:public at a.b.c.d -
.x.1:public at a.b.c.d&.x.3:public at a.b.c.d - .x.1:public at a.b.c.d * 10,

but it didn't do well...

do I have to write smthg like :
"OID3:community at IP - OID1:community at IP", which gives :
Target[Ping_arc2]: .x.3:public at a.b.c.d - .x.1:public at a.b.c.d * 10 
?

thanks in advance

Bruno

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