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RE: Method R and CPU Time

From: MacGregor, Ian A. <ian_at_slac.stanford.edu>
Date: Tue, 6 Jul 2004 18:07:59 -0700
Message-ID: <26E3EC48949D134C94A1574B2C89466113A970@exchange2.slac.stanford.edu>


I have re-read Chaper 7 especially pg 153-154, a section entitled, "CPU Consumption Double-Counting". I expected the double counting to be negligible. The text does say USUALLY negligible. Thanks also for the multi-block read information. I did not notice this discrepancy on queries dominated by single-block read events. The multi-block reads in the example are not large, each is 8, 8k blocks or less, with very few or less. I don't consider these to be large reads.

 Ian

-----Original Message-----
From: Cary Millsap [mailto:cary.millsap_at_hotsos.com] Sent: Tuesday, July 06, 2004 1:58 PM
To: oracle-l_at_freelists.org
Subject: RE: Method R and CPU Time

Ian,

Some timed events like "SQL*Net message from client" (and "...to client" = and several others) are not double-counted within any "e" value. In the = book, I call these "between-call events". Other events that represent work = taking place within the context of a dbcall (which I call "within-call events") /are/ included within an "e" value.

But there is potential for significant double-counting between "ela" and = "c" values. I don't have a copy of the book handy (Optimizing Oracle Performance), but it's described in detail there (Chapter 7, I think). = The problem will occur most prominently when your application does large multi-block reads. Basically, the issue is that the time than an I/O = syscall spends consuming CPU is double-counted both in "c" for the dbcall and = "ela" for the read. This breaks the relationship e \approx c + \Sum ela. The larger the I/O size, the bigger the breakage.=20

Pictures in the book.

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

-----Original Message-----
From: oracle-l-bounce_at_freelists.org =
[mailto:oracle-l-bounce_at_freelists.org]
On Behalf Of MacGregor, Ian A.
Sent: Tuesday, July 06, 2004 3:48 PM
To: 'oracle-l_at_freelists.org'
Subject: 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=3D0, the statement had already been parsed before = this run.=20

SQL> select distinct misses from raw_ops_external;

    MISSES


         0

grep -i dep nlco_ora_2991.trc | wc

     604 1428 44874
grep -i dep=3D0 nlco_ora_2991.trc | wc

     604 1428 44874
grep -i dep=3D1 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.=20

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)=20

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 =3D 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]=20 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 =3D 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 =3D = Boston

-----Original Message-----
From: oracle-l-bounce_at_freelists.org =3D
[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 = =3D grep to extract the totals for all non-recursive SQL statements and then = =3D turned that output into an external table. I did the same thing with the =3D waits.
I had earlier checked the figures tkprof provides vs. the raw trace file = =3D for the same statement run once and found they agreed. I have not = summed =3D 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 =3D double-counting is not always incidental, but can be significant when = there are db file =3D I/O waits involved. If this is true, can one really use method R to =3D 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 =3D = 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
 =3D20

-----Original Message-----
From: Jonathan Lewis [mailto:jonathan_at_jlcomp.demon.co.uk]=3D20 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

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 =3D = different times with different bind variables, that is on average the elapsed time = =3D is 1.429 seconds per statement execution. Also because the report is =3D = based on 10 runs of a statement any discrepancies in the figuring of e, ela, = =3D 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 = =3D to 12:15 reported 630 seconds of CPU time. Again there are four CPU's, = =3D 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 = =3D CPU one needed to divide the reported CPU time by the number of processors = =3D 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 = =3D
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 =3D = 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 =3D = needs to collect signal data in real time and is employing direct mode inserts = =3D via OCI. Exactly how the partition sizes were decided, I don't know. = =3D Partition pruning is successful.

No one is complaining about the above response time, but it can vary =3D during the day due to machine load, and how much of the data is in = cache, at =3D times reaching unacceptable levels. Faster hardware is being = considered and =3D I'm trying to figure how much if any that would help by = figuring how much =3D 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



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Received on Tue Jul 06 2004 - 20:05:07 CDT

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