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RE: Method R and CPU Time

From: Cary Millsap <>
Date: Tue, 6 Jul 2004 11:22:33 -0500
Message-ID: <012001c46375$75c4ed20$6701a8c0@CVMLAP02>

Tkprof double-counts /horribly/. It begins with how it handles the = so-called
"idle events."

Cary Millsap
Hotsos Enterprises, Ltd.
* Nullius in verba *

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-----Original Message-----
From: =

On Behalf Of MacGregor, Ian A.
Sent: Saturday, July 03, 2004 2:31 PM
To: ''
Subject: RE: Method R and CPU Time

Thanks, these figures are from tkprof on a database. I used = grep
to extract the totals for all non-recursive SQL statements and then = turned
that output into an external table. I did the same thing with the = waits.
I had earlier checked the figures tkprof provides vs. the raw trace file = for
the same statement run once and found they agreed. I have not summed = the
waits from the raw trace file upon which tkprof was run, but the CPU and elapsed times do agree between the raw trace file and the tkprof output.

As CPU_SECS + Wait_SECS > 1.5 * Elapsed_SECS, it appears the = double-counting
is not always incidental, but can be significant when there are db file = I/O
waits involved. If this is true, can one really use method R to = evaluate
how a hardware upgrade will affect performance.

Remember, the figures are from tkprof which Cary states can get things wrong, and the wait times totals have not been confirmed as accurate = from
the raw trace data; hence, the question is premature. Has anyone had difficulty separating time spent waiting for file i/o vs. CPU time?

Ian MacGregor
Stanford Linear Accelerator Center

-----Original Message-----
From: Jonathan Lewis []=20 Sent: Saturday, July 03, 2004 3:03 AM
Subject: Re: Method R and CPU Time

Notes in-line.


Jonathan Lewis The Co-operative Oracle Users' FAQ Optimising Oracle Seminar - schedule updated May 1st

The figures represent totals from running the same statement 10 = different
times with different bind variables, that is on average the elapsed time = is
1.429 seconds per statement execution. Also because the report is = based
on 10 runs of a statement any discrepancies in the figuring of e, ela, = or c
are magnified.

[jpl] Not necessarily, though you may know it to be true in your case.
[jpl] In the general case, 10 runs would be more likely to flatten out
[jpl] minimise descrepancies.

The statements ran starting at 12:05 PM on Jun 25. Statspack from noon = to
12:15 reported 630 seconds of CPU time. Again there are four CPU's, = the
machine was not overloaded.

My original question had to do as to why "sum(ela)" + "c" was over 1.5 times as high as "e", and whether for a statement running on a single = CPU
one needed to divide the reported CPU time by the number of processors = on
the machine just as one would when looking at total CPU time across the entire machine. If I do that, then ela + c < e, but the error is much = much

[jpl] Without knowing what tools you are using to produce
[jpl] the numbers, and where they are coming from, and what
[jpl] actually is happening in the code, it is not possible to give
[jpl] a guaranteed answer to that question. But if you are just
[jpl] reading v$mystat and v$session_event for the session, and
[jpl] parts of the query are parallelised, you need to know that
[jpl] PX slave stats are summed back to the QC, but PX slave
[jpl] waits are not. So any attempt to add ela to c to get
[jpl] elapsed time would be misguided.
[jpl] On the other hand, you didn't mention any PX Deq wait
[jpl] time, and I assumed from the reference to ela and c that
[jpl] you are processing a 10046 trace file - so the simple answer
[jpl] to your original question is no - you don't need to divide
[jpl] the c figure by the number of processors.

There are things outside of disk waits and CPU times which need to be researched. Such as why submit 10 different requests for 10 different signals. The requests themselves union a daily partioned table with indexes and a non-indexed live table holding a single calendar days = worth of
data partitioned every 10 minutes. The non-indexed table is the one reporting the scattered read waits. The table is not indexed as it = needs to
collect signal data in real time and is employing direct mode inserts = via
OCI. Exactly how the partition sizes were decided, I don't know. = Partition
pruning is successful.

No one is complaining about the above response time, but it can vary = during
the day due to machine load, and how much of the data is in cache, at = times
reaching unacceptable levels. Faster hardware is being considered and = I'm
trying to figure how much if any that would help by figuring how much = time
is actually spent on CPU for these queries vs. waits for physical I/O.

Ian MacGregor
Stanford Linear Accelerator Center

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Received on Tue Jul 06 2004 - 11:22:06 CDT

Original text of this message