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

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


----- Original Message ----- 
From: "MacGregor, Ian A." <ian@xxxxxxxxxxxxxxxxx>
To: <oracle-l@xxxxxxxxxxxxx>
Sent: Thursday, July 01, 2004 9:42 PM
Subject: Method R and CPU Time


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@xxxxxxxxxxxxxxxxx



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