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Home -> Community -> Mailing Lists -> Oracle-L -> RE: Method R and CPU Time
Thank you very much for responding. I decided to do away with tkprof and calculate the nubers myself. I chose another example where the statement is taking a longer time to run
Here is what I got from using the tkprof method
RUN_DATE TRACE_ID EVENT WAITS WAIT_SECS ELAPSED_SECS CPU_SECS
----------------------------------------------------------------------------------------------------------------
26-JUN-2004 03:07 nlco_ora_2991 db file sequential read 1165 4.9 170.87 142.1 26-JUN-2004 03:07 db file scattered read 6756 147.41 170.87 142.1 26-JUN-2004 03:07 SQL*Net message from client 550 1.46 170.87 142.1 *************** ---------- ---------- sum 8471 153.77 And here are the waits fropm the raw trace file from the above EVENT SUM(MICROSECONDS)/1000000
------------------------------ -------------------------
SQL*Net message from client 1.469555 SQL*Net message to client .002067 db file scattered read 147.412026 db file sequential read 4.906271 latch free .000009 ------------------------- sum 153.789928
And here are the elapsed tike and CPU details from the raw trace file
OPERA CPU_TIME ELAPSED_TIME
----- ---------- ------------
EXEC .03 .033659
FETCH 142.07 170.824816
PARSE 0 .013718
---------- ------------
sum 142.1 170.872193
All cursors are at dep=0, the statement had already been parsed before this run.
SQL> select distinct misses from raw_ops_external;
MISSES
0
grep -i dep nlco_ora_2991.trc | wc
604 1428 44874
grep -i dep=0 nlco_ora_2991.trc | wc
604 1428 44874
grep -i dep=1 nlco_ora_2991.trc | wc
0 0 0
What I did was to "grep -I wait" and write the results to a file. I then again used the external table feature and used sum and compute to produce the report. I then did the same thing with the parse, exec, and fetch lines. The report agrees favorably with the tkprof output.
Here are the external table definitions
SQL> describe raw_waits_external
Name Null? Type ----------------------------------------- -------- ---------------------------- TRACE_ID VARCHAR2(15) EVENT VARCHAR2(30) MICROSECONDS NUMBER(9) P1 NUMBER(15) P2 NUMBER(10) P3 NUMBER(2) SQL> describe raw_ops_external Name Null? Type
----------------------------------------- -------- ----------------------------
TRACE_ID VARCHAR2(15) OPERATION VARCHAR2(5) CPU NUMBER(10) ELAPSED NUMBER(10) PHYSICAL_R NUMBER(6) CONSISTENT_R NUMBER(6) CURRENT_R NUMBER(6) MISSES NUMBER(2) ROW_COUNT NUMBER(3) DEPTH NUMBER(2) OPT_GOAL NUMBER(2) TIM NUMBER(15)
Perhaps I'm making the same mistake as tkprof as our figures agree. Perhaps I have to throw out some of the lines from the trace files despite all depths being = 0 and there being no library cache misses. Any suggestion as to where the double-counting is occuring.
I'm ready to change the premise that one cannot always separate time spent on CPU and time spent waiting for file I/O via method R, because the overlap between the two may be significant not incidental to an assertion.
One thing I did not mention is that the trace was done at level 12.
Ian MacGregor
Stanford Linear Accelerator Center
ian_at_SLAC.Stanford.edu
-----Original Message-----
From: Cary Millsap [mailto:cary.millsap_at_hotsos.com]
Sent: Tuesday, July 06, 2004 9:23 AM
To: oracle-l_at_freelists.org
Subject: RE: Method R and CPU Time
Tkprof double-counts /horribly/. It begins with how it handles the = so-called "idle events."
Cary Millsap
Hotsos Enterprises, Ltd.
http://www.hotsos.com
* Nullius in verba *
Upcoming events:
- Performance Diagnosis 101: 6/22 Pittsburgh, 7/20 Cleveland, 8/10 = Boston - SQL Optimization 101: 5/24 San Diego, 6/14 Chicago, 6/28 Denver - Hotsos Symposium 2005: March 6-10 Dallas - Visit www.hotsos.com for schedule details...
-----Original Message-----
From: oracle-l-bounce_at_freelists.org =
[mailto:oracle-l-bounce_at_freelists.org]
On Behalf Of MacGregor, Ian A.
Sent: Saturday, July 03, 2004 2:31 PM
To: 'oracle-l_at_freelists.org'
Subject: RE: Method R and CPU Time
Thanks, these figures are from tkprof on a 9.2.0.4 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
ian_at_SLAC.Stanford.edu
=20
-----Original Message-----
From: Jonathan Lewis [mailto:jonathan_at_jlcomp.demon.co.uk]=20
Sent: Saturday, July 03, 2004 3:03 AM
To: oracle-l_at_freelists.org
Subject: Re: Method R and CPU Time
Notes in-line.
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
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 anomalies [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
less.
[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
ian_at_SLAC.Stanfod.edu
-- Archives are at http://www.freelists.org/archives/oracle-l/ FAQ is at http://www.freelists.org/help/fom-serve/cache/1.htmlReceived on Tue Jul 06 2004 - 15:44:31 CDT
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