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Re: Method R and CPU Time

From: Daniel Fink <Daniel.Fink_at_Sun.COM>
Date: Thu, 01 Jul 2004 14:59:57 -0600
Message-id: <40E47B4D.49E851F4@sun.com>


Ian,

I know this does not answer your question, but the fact that each event had the *exact* same CPU_SECs strikes me as odd. Are you sure this calculation/retrieval is correct?

As for CPU_SECS + WAIT_SECS = ELA_SECS, CPU is only incremented when the process is on the CPU and the CPU clock increments. It is theoretically possible that you could accumulate 0 cpu_seconds because the process was never on the CPU when the clock ticked. You could also have CPU seconds in the PARSE, EXEC and FETCH calls that may not be accounted for here.

Regards,
Daniel Fink

"MacGregor, Ian A." wrote:

> 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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Received on Thu Jul 01 2004 - 15:56:44 CDT

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