[mrtg] Re: 4 CPU monitor

Eric Brander Eric_Mailing_List at rednarb.com
Wed Aug 18 22:51:08 MEST 2004


Britt Tabor wrote:

> Thanks for everyone that replied to the post, I am planning on installing
> the RRDtool eventually but was able to solve it with an average of the 4
> cpu's with the following line...
>  
> 
> Target[server1.cpu]:
> (1.3.6.1.2.1.25.3.3.1.2.1&1.3.6.1.2.1.25.3.3.1.2.2:public at JGSPXPP01 / 2) +
> (1.3.6.1.2.1.25.3.3.1.2.3&1.3.6.1.2.1.25.3.3.1.2.4:public at JGSPXPP01 / 2)
> 
I think this will average processors 1 and 3 and graph the result on the 
input (green field), and average processors 2 and 4 and graph that 
result on the output (blue line). Please correct me if I'm wrong, list, 
as I am often that.

If this is indeed your desire, by all means go forth and multiply (or 
add and divide in this case).

If it were me, I'd either take an average of the 4 procs and graph that 
on one line, or I'd make 2 different graphs with no averaging. (I 
actually do the latter for my 4-proc system graphs).

This is how I would do 4 procs on one graph:

Target[server.proc.average]: (cpuoid1&cpuoid1:public$a.b.c.d + 
cpuoid2&cpuoid2:public$a.b.c.d + cpuoid3&cpuoid:public$a.b.c.d + 
cpuoid4&cpuoid:public$a.b.c.d) / 4
MaxBytes[server.proc.average]: 100
options[server.proc.average]: growright, gauge
withpeak[server.proc.average]: wmy
unscaled[server.proc.average]: dwmy
LegendI[server.proc.average]:
LegendO[server.proc.average]: Processor Utilization
Legend1[server.proc.average]:
Legend2[server.proc.average]:
Legend3[server.proc.average]:
Legend4[server.proc.average]: Peak Processor Utilization
Ylegend[server.proc.average]: % Usage
ShortLegend[server.proc.average]: %
PageTop[server.proc.average]: <H1>Average Processor Utilization</H1>

This will use both the input and output for the same data, making a neat 
looking green field with a blue line at the top, yet display only the 
blue line of the data below the graph. I use unscaled myself as I like 
to see at a glance the utilization. Without that parameter the graphs 
all look like they are having some outrageous peaking going on.

  I just wrote this up from memory so I hope I got everything.

HTH,

Eric Brander


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