[rrd-users] RRD smoothing - please help

Nuno Pereira nunopereira at interacesso.pt
Tue Nov 11 12:32:06 CET 2008

Hi Joghan,


As far as I can understand, your problem is this: “last_update = 1226319614”.

As you have a step of 100, and the rrdtool stores the values at multiples of the step, when you send a value in a timestamp that isn’t a multiple of your step (100 in your case), it estimates the values at that time and stores it.


You can try and set the update time of rrdupdate to be a multiple of 100 (the one previous to the one that you are at that moment). But try to use a value retrieved by that “rounded” time, e.g., instead of change the update time, change the retrieve time to something closer to the multiple of 100 (but use the multiple of 100 as the update time).


I hope am right and this helps you.



Nuno Pereira


De: rrd-users-bounces at lists.oetiker.ch [mailto:rrd-users-bounces at lists.oetiker.ch] Em nome de johan thoren
Enviada: terça-feira, 11 de Novembro de 2008 08:13
Para: William R. Lorenz
Cc: rrd-users at lists.oetiker.ch
Assunto: Re: [rrd-users] RRD smoothing - please help


Here is the RRD info. Look specifically at rra[0].cdp_prep[1].value = 2.2500000000e+00
Im possitive that Im only sending integer values to RRD. And them are sent every 100 seconds. 


filename = "99.rrd"
rrd_version = "0003"
step = 100
last_update = 1226319614
ds[sPHI].type = "GAUGE"
ds[sPHI].minimal_heartbeat = 100
ds[sPHI].min = 0.0000000000e+00
ds[sPHI].max = 1.8760000000e+03
ds[sPHI].last_ds = "UNKN"
ds[sPHI].value = 0.0000000000e+00
ds[sPHI].unknown_sec = 0
ds[sPHO].type = "GAUGE"
ds[sPHO].minimal_heartbeat = 100
ds[sPHO].min = 0.0000000000e+00
ds[sPHO].max = 1.8760000000e+03
ds[sPHO].last_ds = "UNKN"
ds[sPHO].value = 4.2000000000e+01
ds[sPHO].unknown_sec = 0
ds[sARI].type = "GAUGE"
ds[sARI].minimal_heartbeat = 100
ds[sARI].min = 0.0000000000e+00
ds[sARI].max = 1.8760000000e+03
ds[sARI].last_ds = "UNKN"
ds[sARI].value = 0.0000000000e+00
ds[sARI].unknown_sec = 0
ds[sARO].type = "GAUGE"
ds[sARO].minimal_heartbeat = 100
ds[sARO].min = 0.0000000000e+00
ds[sARO].max = 1.8760000000e+03
ds[sARO].last_ds = "UNKN"
ds[sARO].value = 0.0000000000e+00
ds[sARO].unknown_sec = 0
rra[0].cf = "MAX"
rra[0].rows = 1000
rra[0].pdp_per_row = 3
rra[0].xff = 0.0000000000e+00
rra[0].cdp_prep[0].value = 0.0000000000e+00
rra[0].cdp_prep[0].unknown_datapoints = 0
rra[0].cdp_prep[1].value = 2.2500000000e+00
rra[0].cdp_prep[1].unknown_datapoints = 0
rra[0].cdp_prep[2].value = 0.0000000000e+00
rra[0].cdp_prep[2].unknown_datapoints = 0
rra[0].cdp_prep[3].value = 0.0000000000e+00
rra[0].cdp_prep[3].unknown_datapoints = 0




--- Den tis 2008-11-11 skrev William R. Lorenz <wrl at express.org>:

Från: William R. Lorenz <wrl at express.org>
Ämne: Re: [rrd-users] RRD smoothing - please help
Till: "johan thoren" <johan_thoren at yahoo.se>
Kopia: rrd-users at lists.oetiker.ch
Datum: tisdag 11 november 2008 08.26

Hej Johan,
On Tue, 11 Nov 2008, johan thoren wrote:
> William, thanks for your reply!  I looking to accomplish a five minute 
> graph with integer values like below. The problem is that I cant get rid 
> of the double X.XX values that is produced by RRD.
If you'd like to get rid of any double X.XX values and convert them to 
integers, have you considered a Perl or C or Bash or ZSH int/floor/ceil 
type of function? :-)  Perhaps there's some intermediate steps that could 
also happen between the data returned by rrdtool and your application? 
It should be rather trivial to convert all those mathematical values? :-)
> no.
> 4.                            ____
> 3.                     ____
> 2.       ____
> 1.____       ____
> 0.    300. 600. 900. 1200 time
With regards to your previous email (as you write -- "my goal is to store 
data and ONLY display the maximum value"), max values are not necessarily 
integer values.  Are you also able to share more about your overall goals?
As a disclaimer, I'm sure there's others on this mailing list that know
lot more than I do about such things, and I defer to their wise expertise.
Hope this may help, at least a bit.   Take care,
William R. Lorenz



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