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Home -> Community -> Mailing Lists -> Oracle-L -> Re: Method R and CPU Time
Just hand-waving and burbling out loud on this one, as I don't know how the 14.29 elapsed seconds is derived.
I note that you've got what seems to be a lot of CPU time recorded for what may be a relatively small amount of data handling.
Is it possible that the machine is running at high levels of CPU usage, so that you get a lot of:
Me running on CPU
Me still runnable but pre-emptively descheduled -
so Oracle won't be recording a wait event or CPU
time but real time is still passing
Me back at top of queue and running again
In this scenario, CPU and Wait time do not increase, but elapsed time increases. Of course, the tasks you are running, and the task other people are running concurrently would have to be very CPU intensive for the lost time to be such a large fraction of the wait time.
Regards
Jonathan Lewis
http://www.jlcomp.demon.co.uk/faq/ind_faq.html The Co-operative Oracle Users' FAQ
http://www.jlcomp.demon.co.uk/seminar.html Optimising Oracle Seminar - schedule updated May 1st
I'm having a hard time determining how much time is actually spent on CPU,
Consider the following
Runtime TRACE_ID EVENT WAITS
WAIT_SECS ELAPSED_SECS CPU_SECS
25-JUN-2004 12:05 nlco_ora_7279 db file sequential read 96
.27 14.29 13.07
25-JUN-2004 12:05 db file scattered read 378
7.14 14.29 13.07
25-JUN-2004 12:05 SQL*Net message from client 475
1.34 14.29 13.07
*************************** ----- -----
sum 949
8.75
The CPU_SECS + WAIT_SECS should equal elapsed secs at least approximately. I know there is some double counting between the file i/o waits and the CPU's but 13.07 + 8.75 = 21.82 and 21.82/14.29 * 100 = anout 153%. That's a lot more than the incidental double-counting error Cary speaks of in his book.
This particular machine has 4 CPU's and when looking at CPU usage at the system level one needs to take that into account. But what about for individual statements that are not parallelized; i.e., running against one and only one CPU, does one need to divide the CPU results by four on a four CPU machine.
Ian MacGregor
Stanford Linear Accelerator Center
iam_at_slac.stanford.edu
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