[rrd-users] VDEF MINIMUM

[Alex van den Bogaerdt]

Are you saying that there is a discrepancy between what MIN and MAX
report, vs. what can be seen on the graph?

If so: are you sure you use the same (C)DEF to get min and max from?

> Hi Alex,
>
> Thanks for that, I'll study this some more.
>
> For now I'm doing it the "messy" way, i.e. fetch-ing all the data into an
> array, sorting it numerically ascending and then taking the first and last
> values of the array as minimum and maximum values plotted on the graph.
>
> As I say it's not as clean as using an rrdtool function directly to
> extract
> min/max, but it seems to give results which correlate exactly to the
> minimum
> and maximum points plotted on the graphs which is the main thing.
>
> Thanks,
> Oliver.
>
> -----Original Message-----
> From: Alex van den Bogaerdt
> Sent: Thursday, August 11, 2011 11:05 PM
> To: Oliver
> Cc: rrd-users at lists.oetiker.ch
> Subject: Re: [rrd-users] VDEF MINIMUM
>
> bottom paragraph of my reply, "... the other ..." is the relevant part
> then. You want to display a certain amount of RRA rows in less the amount
> of pixel columns, so RRDtool is going to do some on the fly consolidation.
>
> Try an end time which is n*60, try to display 400*60 seconds, do so in 400
> pixel columns.
> GPRINT last, min, average, max.
> Now try the same, except that you display 400*30 seconds.
>
> This part of the tool has had more than its fair share of problems in the
> past, I seem to recall an off by one error or two, and other problems.
> There could be differences between the various RRDtool versions, but all
> versions will have some form of on the fly consolidation, where multiple
> RRA rows will be combined into one pixel row.
>
> Once you understand the relatively easy n*30 vs. n*60 case, try to figure
> out what the tool does with n*40, n*50 and so on.
>
> GPRINTing the time component of the first and last sample on the graph may
> also be interesting.
>
> cheers,
> Alex
>
>
>> Hi Alex,
>>
>> Thank you for your detailed reply which has certainly helped me
>> understand
>> RRA calculations a bit better.
>>
>> However I am only using one RRA (I should have pointed this out in my
>> first
>> post):
>>
>> filename = "noise.rrd"
>> rrd_version = "0003"
>> step = 30
>> last_update = 1313098265
>> ds[noise].type = "GAUGE"
>> ds[noise].minimal_heartbeat = 60
>> ds[noise].min = NaN
>> ds[noise].max = NaN
>> ds[noise].last_ds = "9.4"
>> ds[noise].value = 5.6364270600e+01
>> ds[noise].unknown_sec = 0
>> rra[0].cf = "AVERAGE"
>> rra[0].rows = 5760
>> rra[0].cur_row = 4995
>> rra[0].pdp_per_row = 1
>> rra[0].xff = 5.0000000000e-01
>> rra[0].cdp_prep[0].value = NaN
>> rra[0].cdp_prep[0].unknown_datapoints = 0
>>
>> Does this influence your previous reply?
>>
>> Thanks,
>> Oliver.
>>
>> -----Original Message-----
>> From: Alex van den Bogaerdt
>> Sent: Tuesday, August 09, 2011 2:00 PM
>> To: rrd-users at lists.oetiker.ch
>> Subject: Re: [rrd-users] VDEF MINIMUM
>>
>> You are looking at averages. Your two tests use different RRAs. For
>> instance: one RRA consolidates 1 PDP into 1 CDP, the other consolidates
>> 3
>> PDPs into one CDP. Look at 10,20,30,40,50,60. One RRA has all 6 values,
>> the other RRA has average(10,20,30) equals 20, average(40,50,60) equals
>> 50.
>>
>> minimum(10,20,30,40,50,60) equals 10. minimum(20,50) equals 20. Your
>> test
>> would return 10 and 20.
>>
>> It should work much better if you look at minimums of minimums:
>>
>> The other RRA would contain min(10,20,30) equals 10, min(40,50,60)
>> equals
>> 40. Then min(10,20,30,40,50,60) and min(10,40) both return 10.
>>
>> There can be two places where consolidation takes place. One is when
>> data
>> is copied into RRAs, the other is when rrdtool graph needs data in a
>> resolution which is not available. If you take both into account, you
>> should be able to get predictable results.
>>
>> HTH
>> Alex
>>
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>>
>
>
>