Query execution is faster - as SYS user vs. application user in 11.1.07

In +11.1.0.7+ database, we have noticed execution of an application process as SYS user is faster than the application user. To quantify this execution time was 2 hours as application user and same process when executed as SYS user completed in 1 hour.
We ran tracing for this process and tkprof output showed follwoing:
OVERALL TOTALS FOR ALL RECURSIVE STATEMENTS:
_call          count      cpu       elapsed disk query current rows_
Execute 3705586 2440.49 *4294.54* 293602 21835223 23300886 3801288 <<< ==== SYS
Execute 3703285 6356.69 *8576.50* 303175 21785419 23378086 3801291 <<< ==== Application user
We have SR open with Oracle, but thought if some one knows about this issue would be great.
Edited by: dhyani on Aug 17, 2010 3:16 PM

What about the "Predicate Information" section of the execution plans - were those exactly the same? Did you use AUTOTRACE, EXPLAIN PLAN FOR, or the method that I showed to retrieve the execution plans? Where the SQL_ID and CHILD_NUMBERS displayed exactly the same? Is the STATISTIC_LEVEL set to TYPICAL for both sessions? You can check the setting in SQL*Plus like this:
SQL> SHOW PARAMETER STATISTICS_LEVEL
NAME                                 TYPE        VALUE
statistics_level                     string      typicalAs a demonstration, I will change the value at the session level and then change the setting back:
SQL> ALTER SESSION SET STATISTICS_LEVEL='ALL';
SQL> SHOW PARAMETER STATISTICS_LEVEL
NAME                                 TYPE        VALUE
statistics_level                     string      ALL
SQL> ALTER SESSION SET STATISTICS_LEVEL='TYPICAL';If the above does not show a difference, you will need to capture a 10046 trace at either level 8 or level 12 for both sessions and compare the wait events and the STAT lines in the 10046 trace to see where the additional time is being spent. Reference these two blog articles for the 10046 trace:
http://hoopercharles.wordpress.com/2009/12/01/10046-extended-sql-trace-interpretation/
http://hoopercharles.wordpress.com/2010/01/26/10046-extended-sql-trace-interpretation-2/
Keep in mind that the goal is to not change anything until we understand the problem better. Maybe, just maybe, someone created a table (or view) or two in the SYS schema (or used a table/view name that matched a SYS owned object), in a normal user's schema, or there are synonyms pointing to different tables. So, it probably would be a good idea to verify that the results are exactly the same when the SYS user executes the SQL statement and a normal user executes the SQL statement.
Charles Hooper
Co-author of "Expert Oracle Practices: Oracle Database Administration from the Oak Table"
http://hoopercharles.wordpress.com/
IT Manager/Oracle DBA
K&M Machine-Fabricating, Inc.

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    If you find item 2 is true then separating the select from the update will likely eliminate the 1555. You do this by building a driving table that has the keys of the rows to be updated or deleted. Then you use the driver to control accessing the target table.
    3- If the cause of the 1555 is or may be delayed block cleanout then select * from the target prior to running the long running query.
    Realistically you might need to increase the size of the undo tablespace to hold all the change data and the value of the undo_retention parameter to be longer than the job run time. Which brings up back to option 1. Tune every query in the process so that the job run time is reduced to optimal.
    HTH -- Mark D Powell --
    dear mark
    Thanks for the excellent advise..I found that the error is coming because of frequent commits..which is item 2 as u righly mentioned ..
    I think I need to keep a watch on the queries running , I was just trying to find the execution time for the queries..If there is any way to find the query duration without running a trace ..
    regards
    abhishek

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