Examining the Oracle Database 10053 Trace Event Dump File

When it comes to performance tuning, we can spend time on one or both ends of the problem. On the "before there is a problem" end, we (anyone who writes SQL) have the opportunity to write good or efficient SQL. The end result is a statement which is optimal, or nearly so, and for the most part, we've done our due diligence in terms of avoiding introducing a problem child SQL statement into the fray.

On the other end is where we get to practice our "Oracle CSI" skills and choose a tool which helps us solve the "why is this statement so bad?" issue. Common to both ends is the Cost Based Optimizer. The execution plan generated by the CBO is used to confirm several aspects of our statement of interest. Access method(s), or how Oracle intends on going about the task of getting rows is one key element within a plan. For example, if you were expecting an index to be used but the access method against that table reflects a full table scan, you know have some work to do regarding the index and perhaps some initialization parameters.

Another key area has to do with join methods. Aside from hints, we don't generally change what Oracle (the CBO) does. There's nothing we're coding that has much to do with which join method (typical methods being nested loop, hash, and merge sort) will be used. What influences the join method is the size of rowsets, and the CBO gets that from statistics. Looked at another way, Oracle naturally considers indexes once we put them in place. Oracle internally comes up with join methods and what we code indirectly influences how the two result sets are joined.

So, behind the scenes, what is the CBO doing when it comes to how it comes up with an execution plan? This is where the 10053 trace event comes into play. Other tools or settings show us WHAT the CBO comes up with; the 10053 setting tells us HOW the CBO came to its decision (the final execution plan).

For a relatively simple query, you might be amazed at all the work Oracle, via the CBO, goes through. Let's run a 10053 trace event and examine the contents of the trace file.

There are a couple of ways to start this trace event. A simple alter session command turns it on (and off).

ALTER SESSION SET EVENTS='10053 trace name context forever, level 1';
...your statement here...
ALTER SESSION SET EVENTS '10053 trace name context off';

Another method involves using ORADEBUG, which overall, has the advantage of outputting the name of the trace file, but has the disadvantage of having to connect as SYS. Using the ALTER SESSION option is more flexible as pretty much any user can do this.

conn / as sysdba
oradebug setmypid
oradebug unlimit
oradebug event 10053 trace name context forever, level 1
...your statement here...
oradebug event 10053 trace name context off
oradebug tracefile_name

Yet another method includes DBMS_SYSTEM and its SET_EV subprogram. This built-in doesn't have the best documentation in the world (or public support for that matter).

You have a choice of two levels with the 10053 trace event. Level 1 is more comprehensive than level 2. What is collected in the trace file includes:

Unlike "normal" tracing using SQL_TRACE, the trace file generated here is already formatted. Having the PLAN_TABLE in your schema (assuming this is done via ALTER SESSION by someone other than SYS) is needed as that is where, as always, the execution plan output is stored and 10053 tracing includes the output of the execution plan for your statement.

Let's take a fairly simple statement (as seen in various places throughout Oracle's documentation) and examine its 10053 output.

SELECT ch.channel_class, 
c.cust_city, 
t.calendar_quarter_desc, 
SUM(s.amount_sold) sales_amount 
FROM sh.sales s, 
sh.times t, 
sh.customers c, 
sh.channels ch 
WHERE s.time_id = t.time_id 
AND s.cust_id = c.cust_id 
AND s.channel_id = ch.channel_id 
AND c.cust_state_province = 'CA' 
AND ch.channel_desc in ('Internet','Catalog') 
AND t.calendar_quarter_desc IN ('1999-01','1999-02') 
GROUP BY ch.channel_class, c.cust_city, t.calendar_quarter_desc
ORDER by 1,2,3,4;

Using Oracle 11g, the output file can be found in the trace folder under this path (ORACLE_BASE for me is c:\apporacle):

$ORACLE_BASE\diag\rdbms\ora11g\ora11g\trace

This location corresponds to your background_dump_dest parameter. The trace file (for the above statement) has nearly 6,000 lines in it, and much of the output is similar blocks. You can clearly see how the optimizer tries different join orders, giving way to the idea that the order in which you list tables in the FROM clause is not necessarily the join order seen within the execution plan.

For this example, the CBO tries 21 permutations. Note that these are permutations, where order is important, not combinations, where order is not important.

Join order[1]: CHANNELS[CH]#0 TIMES[T]#1 CUSTOMERS[C]#2 SALES[S]#3
***********************
Best so far: Table#: 0 cost: 3.0015 card: 2.0000 bytes: 42
Table#: 1 cost: 37.2306 card: 362.0000 bytes: 13394
Table#: 2 cost: 146316.6073 card: 1233849.4565 bytes: 77732487
Table#: 3 cost: 152643.9004 card: 27500.9348 bytes: 2310084
***********************
Join order[2]: CHANNELS[CH]#0 TIMES[T]#1 SALES[S]#3 CUSTOMERS[C]#2
***********************
Best so far: Table#: 0 cost: 3.0015 card: 2.0000 bytes: 42
Table#: 1 cost: 37.2306 card: 362.0000 bytes: 13394
Table#: 3 cost: 536.1840 card: 56955.6791 bytes: 3303448
Table#: 2 cost: 946.4226 card: 27500.9348 bytes: 2310084
***********************
Join order[3]: CHANNELS[CH]#0 CUSTOMERS[C]#2 TIMES[T]#1 SALES[S]#3
Join order aborted: cost > best plan cost (--shown here once)
***********************
Join order[4]: CHANNELS[CH]#0 CUSTOMERS[C]#2 SALES[S]#3 TIMES[T]#1
Join order[5]: CHANNELS[CH]#0 SALES[S]#3 TIMES[T]#1 CUSTOMERS[C]#2
***********************
Best so far: Table#: 0 cost: 3.0015 card: 2.0000 bytes: 42
Table#: 3 cost: 501.9521 card: 459421.5000 bytes: 19295724
Table#: 1 cost: 522.8420 card: 56955.6791 bytes: 3303448
Table#: 2 cost: 933.0806 card: 27500.9348 bytes: 2310084
***********************
Join order[6]: CHANNELS[CH]#0 SALES[S]#3 CUSTOMERS[C]#2 TIMES[T]#1
Join order[7]: TIMES[T]#1 CHANNELS[CH]#0 CUSTOMERS[C]#2 SALES[S]#3
Join order[8]: TIMES[T]#1 CHANNELS[CH]#0 SALES[S]#3 CUSTOMERS[C]#2
Join order[9]: TIMES[T]#1 CUSTOMERS[C]#2 CHANNELS[CH]#0 SALES[S]#3
Join order[10]: TIMES[T]#1 SALES[S]#3 CHANNELS[CH]#0 CUSTOMERS[C]#2
***********************
Best so far: Table#: 1 cost: 18.1146 card: 181.0000 bytes: 2896
Table#: 3 cost: 517.0666 card: 113911.3582 bytes: 4214707
Table#: 0 cost: 521.1319 card: 56955.6791 bytes: 3303448
Table#: 2 cost: 931.3705 card: 27500.9348 bytes: 2310084
***********************
Join order[11]: TIMES[T]#1 SALES[S]#3 CUSTOMERS[C]#2 CHANNELS[CH]#0
***********************
Best so far: Table#: 1 cost: 18.1146 card: 181.0000 bytes: 2896
Table#: 3 cost: 517.0666 card: 113911.3582 bytes: 4214707
Table#: 2 cost: 923.7783 card: 55001.8696 bytes: 3465126
Table#: 0 cost: 931.3608 card: 27500.9348 bytes: 2310084
***********************
Join order[12]: CUSTOMERS[C]#2 CHANNELS[CH]#0 TIMES[T]#1 SALES[S]#3
Join order[13]: CUSTOMERS[C]#2 TIMES[T]#1 CHANNELS[CH]#0 SALES[S]#3
Join order[14]: CUSTOMERS[C]#2 SALES[S]#3 CHANNELS[CH]#0 TIMES[T]#1
Join order[15]: CUSTOMERS[C]#2 SALES[S]#3 TIMES[T]#1 CHANNELS[CH]#0
Join order[16]: SALES[S]#3 CHANNELS[CH]#0 TIMES[T]#1 CUSTOMERS[C]#2
Join order[17]: SALES[S]#3 CHANNELS[CH]#0 CUSTOMERS[C]#2 TIMES[T]#1
Join order[18]: SALES[S]#3 TIMES[T]#1 CHANNELS[CH]#0 CUSTOMERS[C]#2
Join order[19]: SALES[S]#3 TIMES[T]#1 CUSTOMERS[C]#2 CHANNELS[CH]#0
Join order[20]: SALES[S]#3 CUSTOMERS[C]#2 CHANNELS[CH]#0 TIMES[T]#1
Join order[21]: SALES[S]#3 CUSTOMERS[C]#2 TIMES[T]#1 CHANNELS[CH]#0

The best (lowest cost) permutation is join order #11, which starts with TIMES, joins to SALES, then joined to CUSTOMERS, and finally joined to CHANNELS. What does the execution plan look like?

PLAN_TABLE_OUTPUT
--------------------------------------------------------------
| Id | Operation | Name | Rows |
--------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1237 |
| 1 | SORT ORDER BY | | 1237 |
| 2 | HASH GROUP BY | | 1237 |
|* 3 | HASH JOIN | | 27501 |
|* 4 | TABLE ACCESS FULL | CHANNELS | 2 |
|* 5 | HASH JOIN | | 55002 |
|* 6 | TABLE ACCESS FULL | CUSTOMERS | 3408 |
|* 7 | HASH JOIN | | 113K|
| 8 | PART JOIN FILTER CREATE | :BF0000 | 181 |
|* 9 | TABLE ACCESS FULL | TIMES | 181 |
| 10 | PARTITION RANGE JOIN-FILTER| | 918K|
| 11 | TABLE ACCESS FULL | SALES | 918K|

Reading the lines (RSO, or row source operations), we see exactly that progression: TIMES-SALES-CUSTOMERS-CHANNELS.

Another line in the 10053 trace file shows that 21 permutations were tried and up to 2000 would have been considered. Since four tables results in 24 permutations (4!=4*3*2*1), why were three permutations skipped? In my example, table permutations 1230, 2031 and 2130 were not considered and there is nothing in the trace file to show why these were omitted (there is no obvious message about 2-0-3-1 being pruned because anything starting with 2-0 is guaranteed to be higher than what 2-0-1-3 had as a cost).

The trace file includes an extensive listing of all default hidden and unhidden parameters that are in place, plus a long list of bug fixes (enabled or not). Techniques such as join eliminations, query transformations, common subexpression elimination, self-join conversions, predicate move-around, and a slew of other approaches are detailed in the file.

A side benefit of running a trace is the output showing all of the statistics for a table and its associated indexes. The "BASE STATISTICAL INFORMATION" (section title in the file) for the customers table shows the following:

Table Stats::
Table: CUSTOMERS Alias: C
#Rows: 55500 #Blks: 1486 AvgRowLen: 181.00
Index Stats::
Index: CUSTOMERS_GENDER_BIX Col#: 4
LVLS: 1 #LB: 3 #DK: 2 LB/K: 1.00 DB/K: 2.00 CLUF: 5.00
Index: CUSTOMERS_MARITAL_BIX Col#: 6
LVLS: 1 #LB: 5 #DK: 11 LB/K: 1.00 DB/K: 1.00 CLUF: 18.00
Index: CUSTOMERS_PK Col#: 1
LVLS: 1 #LB: 115 #DK: 55500 LB/K: 1.00 DB/K: 1.00 CLUF: 54405.00
Index: CUSTOMERS_YOB_BIX Col#: 5
LVLS: 1 #LB: 19 #DK: 75 LB/K: 1.00 DB/K: 1.00 CLUF: 75.00

The output here can be correlated against DBA_TABLES and DBA_INDEXES if need be (DB/K, for example, corresponds to average data blocks per key).

Yet another benefit of running this trace event is that you can see what Oracle comes up with for a stored outline.

Outline Data:
/*+
BEGIN_OUTLINE_DATA
IGNORE_OPTIM_EMBEDDED_HINTS
OPTIMIZER_FEATURES_ENABLE('11.2.0.1')
DB_VERSION('11.2.0.1')
ALL_ROWS
OUTLINE_LEAF(@"SEL$1")
FULL(@"SEL$1" "T"@"SEL$1")
FULL(@"SEL$1" "S"@"SEL$1")
FULL(@"SEL$1" "C"@"SEL$1")
FULL(@"SEL$1" "CH"@"SEL$1")
LEADING(@"SEL$1" "T"@"SEL$1" "S"@"SEL$1" "C"@"SEL$1" "CH"@"SEL$1")
USE_HASH(@"SEL$1" "S"@"SEL$1")
USE_HASH(@"SEL$1" "C"@"SEL$1")
USE_HASH(@"SEL$1" "CH"@"SEL$1")
PX_JOIN_FILTER(@"SEL$1" "S"@"SEL$1")
SWAP_JOIN_INPUTS(@"SEL$1" "C"@"SEL$1")
SWAP_JOIN_INPUTS(@"SEL$1" "CH"@"SEL$1")
USE_HASH_AGGREGATION(@"SEL$1")
END_OUTLINE_DATA
*/

In a SQL*Plus session, create a stored outline and query USER_OUTLINE_HINTS. You should expect to see the same output.

CREATE OUTLINE star_query
FOR CATEGORY test_outlines ON
;
 
SELECT join_pos, hint
FROM user_outline_hints
WHERE name = 'STAR_QUERY';
 
JOIN_POS HINT
---------- ------------------------------------------------------------
0 USE_HASH_AGGREGATION(@"SEL$1")
0 SWAP_JOIN_INPUTS(@"SEL$1" "CH"@"SEL$1")
0 SWAP_JOIN_INPUTS(@"SEL$1" "C"@"SEL$1")
0 PX_JOIN_FILTER(@"SEL$1" "S"@"SEL$1")
0 USE_HASH(@"SEL$1" "CH"@"SEL$1")
0 USE_HASH(@"SEL$1" "C"@"SEL$1")
0 USE_HASH(@"SEL$1" "S"@"SEL$1")
0 LEADING(@"SEL$1" "T"@"SEL$1" "S"@"SEL$1" "C"@"SEL$1" "CH"@"SEL$1")
4 FULL(@"SEL$1" "CH"@"SEL$1")
3 FULL(@"SEL$1" "C"@"SEL$1")
2 FULL(@"SEL$1" "S"@"SEL$1")
1 FULL(@"SEL$1" "T"@"SEL$1")
0 OUTLINE_LEAF(@"SEL$1")
0 ALL_ROWS
0 DB_VERSION('11.2.0.1')
0 OPTIMIZER_FEATURES_ENABLE('11.2.0.1')
0 IGNORE_OPTIM_EMBEDDED_HINTS

In Closing

Exploring the dump file from the 10053 trace event will give you a better understanding of what the optimizer is doing while it is coming up with the best plan it can (based on the number of permutations). The plan you see is probably the best possible plan, given certain conditions. One is good statistics. Another, and not quite as obvious, is what session or system parameters are set to. If you noticed the name of the outline (STAR_QUERY), does the execution plan look like the results of a star query? The answer is no (but Oracle can elect to use a "normal" plan if the cost is best), but for certainty, we know that a star query transformation was not considered because we did not see any star-related lines in the outline. One line in the outline would have been this:

OPT_PARAM('star_transformation_enabled' 'true')

Another tell-tale would have been the use of bitmap indexes in the execution plan. The overall point of this is that the CBO returned the best plan, given what it had to work with. It still may not have been the best plan overall because not all of the prerequisites to have a better plan were enabled.

Additional Resources

Oracle Wiki: HOW TO trace the CBO working out the execution path (event 10053
Oracle : Understanding Optimization

» See All Articles by Columnist Steve Callan