RE: Method R and CPU Time
- From: "MacGregor, Ian A." <ian@xxxxxxxxxxxxxxxxx>
- To: "'oracle-l@xxxxxxxxxxxxx'" <oracle-l@xxxxxxxxxxxxx>
- Date: Tue, 6 Jul 2004 18:07:59 -0700
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@xxxxxxxxxx]
Sent: Tuesday, July 06, 2004 1:58 PM
To: oracle-l@xxxxxxxxxxxxx
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
- 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@xxxxxxxxxxxxx =
[mailto:oracle-l-bounce@xxxxxxxxxxxxx]
On Behalf Of MacGregor, Ian A.
Sent: Tuesday, July 06, 2004 3:48 PM
To: 'oracle-l@xxxxxxxxxxxxx'
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@xxxxxxxxxxxxxxxxx
-----Original Message-----
From: Cary Millsap [mailto:cary.millsap@xxxxxxxxxx]=20
Sent: Tuesday, July 06, 2004 9:23 AM
To: oracle-l@xxxxxxxxxxxxx
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
- 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@xxxxxxxxxxxxx =3D
[mailto:oracle-l-bounce@xxxxxxxxxxxxx]
On Behalf Of MacGregor, Ian A.
Sent: Saturday, July 03, 2004 2:31 PM
To: 'oracle-l@xxxxxxxxxxxxx'
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@xxxxxxxxxxxxxxxxx
=3D20
-----Original Message-----
From: Jonathan Lewis [mailto:jonathan@xxxxxxxxxxxxxxxxxx]=3D20
Sent: Saturday, July 03, 2004 3:03 AM
To: oracle-l@xxxxxxxxxxxxx
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
----- Original Message -----
From: "MacGregor, Ian A." <ian@xxxxxxxxxxxxxxxxx>
To: <oracle-l@xxxxxxxxxxxxx>
Sent: Friday, July 02, 2004 3:42 PM
Subject: RE: Method R and CPU Time
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@xxxxxxxxxxxxxxxx
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