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. _______________________________________________ rrd-users mailing list [email protected] https://lists.oetiker.ch/cgi-bin/listinfo/rrd-users
