[rrd-users] RRDtool computing logic help

rseiwert rob at vca.com
Tue May 6 01:11:28 CEST 2014


Since no one else replied, I'll take a swing at this. 

RRDTool will build it's buckets on the step boundary from the epoch. Since
one element is not exactly on the boundary RRD will interpolate the data
received. 

If you don't want RRD fudging your numbers you need to enter data exactly on
the step boundary. 

Using your example I put in data at 4 and 8 seconds and got

rrdtool updatev mydata6.rrd 1396419604:8 1396419608:14
return_value = 0
[1396419604]RRA[AVERAGE][1]DS[mem] = 8.0000000000e+00
[1396419608]RRA[AVERAGE][1]DS[mem] = 1.4000000000e+01




--
View this message in context: http://rrd-mailinglists.937164.n2.nabble.com/RRDtool-computing-logic-help-tp7581888p7581991.html
Sent from the RRDtool Users Mailinglist mailing list archive at Nabble.com.



More information about the rrd-users mailing list