Cary, I hadn't gotten that far. Having finally read that section, it's all beginning to make much more sense. Interesting read on gettimeofday and getrusage and how they report time. Will definitely re-read a few times to get the full picture. Thanks for pointing me in the right direction. -- mohammed ________________________________ From: Cary Millsap <cary.millsap@xxxxxxxxxxxx> To: mkb125@xxxxxxxxx Cc: oracle-l@xxxxxxxxxxxxx Sent: Wed, October 28, 2009 5:44:18 PM Subject: Re: Extended SQL Trace Interpretation Question Mohammed, See pages 161–164 of the same book. It boils down to the fact that the c statistic comes from a less granular clock than the e statistic. Cary Millsap Method R Corporation http://method-r.com http://carymillsap.blogspot.com http://twitter.com/cary_millsap On Wed, Oct 28, 2009 at 3:32 PM, mkb <mkb125@xxxxxxxxx> wrote: > >I traced a very simple SQL statement (10046 level 12). Something like: > >>select id from issues > >>which returned 195 rows. > >>Here's a snippet from my trace file and on the PARSE line I get the following: > >>PARSING IN CURSOR #5 len=40 dep=0 uid=49 oct=3 lid=49 tim=1227288824370256 >>hv=1876028296 ad='777f6f40' >>select id from issues >>END OF STMT > >>PARSE >>#5:c=11998,e=11427,p=0,cr=0,cu=0,mis=1,r=0,dep=0,og=1,tim=1227288824370250 > >>BINDS #5: >>EXEC #5:c=0,e=82,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,tim=1227288824370419 > > >>My understanding was that the e ~ c + SUM ela (Optimizing Oracle Performance, >>Ch 5 p87, Millsap). So why is my e value less than my c value? Should it >>not be greater than or equal to my c value? > >>Database is 64-bit 10gR2 EE on RHAS 4u7. > >>Thanks > >>-- >>mohammed > > > > >>-- >//www.freelists.org/webpage/oracle-l > > >