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Rebuilding Indexes and the Clustering Factor Quiz (One Of The Few) September 20, 2011

Posted by Richard Foote in Oracle Indexes, Clustering Factor, Index Rebuild, Quiz.
35 comments

Today’s question has been prompted by various recent comments regarding the Clustering Factor (CF) of an index and how to change the CF requires a reorg of the underlining table.

It used to be quite a common myth that if the CF of an index was greater that “X” or based on some nonsensical formula the CF was greater than “Y”, then rebuilding the index somehow made everything better again. I believe it’s now much more commonly accepted that rebuilding an index does not change the CF of an index in any way. By rebuilding an index, the actual order of the index entries remains unchanged as does the order of the rows within the table and so the resultant CF can’t possibly change.

Pick any index, no matter how bad the CF or badly fragmented the index or table, take fresh statistics and after rebuilding the index (and fresh statistics if compute statistics on the index isn’t performed), the CF will remain the same.

However, there are nearly always exceptions …

Give an example of when rebuilding an index will significantly change the actual CF of an index, assuming 100% accurate statistics are gathered before/after the index rebuild on a table with no DML.

There are actually two such examples that spring to mind :)

Big Tables, Sorts and Indexes Solution (Right On Mother) September 19, 2011

Posted by Richard Foote in CBO, Oracle Indexes, Quiz.
42 comments

My, what a clever lot we have reading this blog :)

Indeed, most of the work has already been done for me as we already have a couple of excellent demos in the list of comments.

The answer is Yes, the CBO will consider using the index and Yes, there may be a number of scenarios when the index will actually be selected by the CBO as it’s the genuinely cheaper option.

At the end of the day, it comes down to whether performing a Full Index Scan, one little block at a time but with the data coming out pre-sorted is cheaper than performing a multiblock Full Table Scan AND a subsequent sort of the data. It of course depends on various factors such as:

  • The most crucial of all, the Clustering Factor of the index. If the data is very well clustered according to the index, then the cost of reading the entire table via the index can be vastly reduced and the index access path becomes viable. A poor (or average) CF, and using the index is just too expensive. Radoslav Golian has an excellent example in the comments on when an index with an excellent CF is chosen by the CBO.
  • The cost of the sort. Sorts are really expensive, especially if Oracle is forced to go to disk and even more so if it has to perform a multi pass sort, so the more costly the sort, the more likely the index is the cheaper option.

An important point for the index to be considered is that it must have a NOT NULL constraint on the column(s), else the index is ignored as the CBO can’t guarantee all rows can referenced within the index.

The moral of this story is this. There is no selectivity by which an index is not considered by the CBO. An index can potentially select 100% of all rows, if doing so is the cheapest option available to the CBO.

I’ve discussed using an index to select 100% of all data before if anyone is interested.

New question coming soon !!

Big Tables, Sorts and Indexes Quiz (Candidate) September 14, 2011

Posted by Richard Foote in Oracle Indexes, Quiz.
27 comments

Following on from the previous quiz on Descending indexes. Simple scenario.

You have a huge table, 10 Million plus rows.

You have an index on a column with a NOT NULL constraint but there are various other columns in the table not included in the index.

You want to select all columns and all rows from the table so you don’t even have a  WHERE condition, but you want the data returned in the order of your indexed column, e.g.

SELECT * FROM really_big_table ORDER BY id;

So you have an index on the ID column.

Question: Is it possible for the CBO to use the index or for the CBO to even consider the index in the first place ? If so, how so ? If not, why not ?

Enjoy :)

 

UPDATE:

OK, all statistics, including system statistics need to be “accurate” and give the CBO correct details on the scenario (so no cheating with manipulating statistics).

No hints allowed and it’s the  ALL_ROWS optimizer_mode, as we want the best solution to retrieve every row from the 10M+ table. Again note, there is no WHERE condition in the query.

If it helps, the index can be unique and there may be as little as just 1 additional column in the table.

Descending Indexes Solution (Yellow Submarine) September 9, 2011

Posted by Richard Foote in Descending Indexes, Oracle Indexes, Quiz.
28 comments

Answers to the quiz on Descending Indexes and hopefully some useful dangers and tips on using them.

The answer to the first question is Yes, a “normal” Ascending Index can be used by the CBO to retrieve data in descending order and hence possibly avoid a sort. The reason being that leaf blocks in the index structure have effectively two pointers, one that points to the next leaf block in the index structure (except for the very last leaf block) and one that points to the previous block (except for the first leaf block). So the data in an index can be retrieved in either order.

The answer to the second question is Yes as well, a Descending Index can also be used to also retrieve data in either logical order as again all the leaf blocks have the two set of pointers.

That being the case, if an index has just the one column value, does it therefore make any difference which index one creates, ascending or descending ?

Hence my last question. The answer is maybe, as there are a number of fundamental differences in how each type of index is implemented.

Naturally, a little demo to illustrate :)

Let’s begin by creating a simple little table and a normal B-Tree index on an ID column, which has monotonically increasing values:

SQL> create table bowie (id number, name varchar2(30));

Table created.

SQL> create index bowie_id_i on bowie(id);

Index created.

SQL> insert into bowie select rownum, 'DAVID BOWIE' from dual connect by level <= 100000;

100000 rows created.

SQL> commit;

Commit complete.

Note the index is indeed a “Normal” B-Tree index and because the indexed values monotonically increase, all index leaf block splits are 90-10 splits resulting a perfectly compact, 100% utilised index structure:

SQL> select index_type from dba_indexes where index_name = 'BOWIE_ID_I';

INDEX_TYPE
---------------------------
NORMAL

SQL> analyze index bowie_id_i validate structure;

Index analyzed.

SQL> select lf_rows, lf_blks, pct_used from index_stats;

   LF_ROWS    LF_BLKS   PCT_USED
---------- ---------- ----------
    100000        199        100

Let’s now run a query to ensure the index is indeed used and that the sort can indeed be avoided. Note I’ve not actually collected any CBO statistics at this stage but I’m definitely using the CBO:

SQL> alter system set optimizer_mode='ALL_ROWS' scope=both;

System altered.

SQL> select * from bowie where id between 42 and 84 order by id desc;

43 rows selected.
Execution Plan
----------------------------------------------------------
Plan hash value: 2771731789

-------------------------------------------------------------------------------------------
| Id  | Operation                    | Name       | Rows  | Bytes | Cost (%CPU)| Time     |
-------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT             |            |    43 |  1290 |     1   (0)| 00:00:01 |
|   1 |  TABLE ACCESS BY INDEX ROWID | BOWIE      |    43 |  1290 |     1   (0)| 00:00:01 |
|*  2 |   INDEX RANGE SCAN DESCENDING| BOWIE_ID_I |    43 |       |     1   (0)| 00:00:01 |
-------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------

   2 - access("ID">=42 AND "ID"<=84)
       filter("ID">=42 AND "ID"<=84)

Note
-----
   - dynamic sampling used for this statement
Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          9  consistent gets
          0  physical reads
          0  redo size
       1092  bytes sent via SQL*Net to client
        418  bytes received via SQL*Net from client
          4  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
         43  rows processed

So the execution plan clearly shows the use of the index via an index range scan descending and that there are indeed no sort operations performed. There were no statistics gathered, so the CBO performed some dynamic sampling to determine a taste for the data.

Let’s now change the optimizer_mode to CHOOSE, a common default setting (especially pre 11g, this example is run on a 10.2.0.4 database) and re-run the query:

SQL> select * from bowie where id between 42 and 84 order by id desc;

43 rows selected.
Execution Plan
----------------------------------------------------------
Plan hash value: 3062669298

---------------------------------------------------
| Id  | Operation                    | Name       |
---------------------------------------------------
|   0 | SELECT STATEMENT             |            |
|   1 |  SORT ORDER BY               |            |
|   2 |   TABLE ACCESS BY INDEX ROWID| BOWIE      |
|*  3 |    INDEX RANGE SCAN          | BOWIE_ID_I |
---------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   3 - access("ID">=42 AND "ID"<=84)

Note
-----
   - rule based optimizer used (consider using cbo)
Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          3  consistent gets
          0  physical reads
          0  redo size
       1092  bytes sent via SQL*Net to client
        418  bytes received via SQL*Net from client
          4  SQL*Net roundtrips to/from client
          1  sorts (memory)
          0  sorts (disk)
         43  rows processed

No statistics on the table now means the Rule Based Optimizer kicks in and although a sort operation is performed (as there’s no descending scan), Oracle at least used the index.

OK, let’s now run the exact same sequence of events, but this time using a Descending Index.

SQL> drop table bowie;

Table dropped.

SQL> create table bowie (id number, name varchar2(30));

Table created.

SQL> create index bowie_id_i on bowie(id desc);

Index created.

SQL> insert into bowie select rownum, 'DAVID BOWIE' from dual connect by level <= 100000;

100000 rows created.

SQL> commit;

Commit complete.

So it’s the exact same table and set of data. Let’s now look at the type of index created:

SQL> select index_type from dba_indexes where index_name = 'BOWIE_ID_I';

INDEX_TYPE
---------------------------
FUNCTION-BASED NORMAL

OK, Difference Number 1. A Descending Index is no ordinary “Normal” index, but is implemented as a  “Function-Based Normal” index instead. This means there’ll be a new hidden virtual column created behind the scenes and that the Rule Based Optimizer is going to have an issue here as it can’t cope with Function-based Indexes.

Let’s look at some Index_Stats:

SQL> analyze index bowie_id_i validate structure;

Index analyzed.
SQL> select lf_rows, lf_blks, pct_used from index_stats;

   LF_ROWS    LF_BLKS   PCT_USED
---------- ---------- ----------
    100000        426         50

Difference Number 2: This index is approximately double the size of the previous index and only half as efficient with its storage. Why ? Because as the data is now inserted in reverse logical order, the last index leaf block no longer receives the largest current index value and so 90-10 splits are not performed. As only 50-50 splits are performed, the index structure is left with 50% empty blocks which can not be reused. Unfortunately, a possible candidate for periodic index rebuilds …

Let’s now re-run the query using the CBO:

SQL> select * from bowie where id between 42 and 84 order by id desc;

43 rows selected.
Execution Plan
----------------------------------------------------------
Plan hash value: 3472402785

------------------------------------------------------------------------------------------
| Id  | Operation                   | Name       | Rows  | Bytes | Cost (%CPU)|Time     |
------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT            |            |   208 |  6240 |     1   (0)|00:00:01 |
|   1 |  TABLE ACCESS BY INDEX ROWID| BOWIE      |   208 |  6240 |     1   (0)|00:00:01 |
|*  2 |   INDEX RANGE SCAN          | BOWIE_ID_I |     1 |       |     1   (0)|00:00:01 |
------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------

   2 - access(SYS_OP_DESCEND("ID")>=HEXTORAW('3EAAFF')  AND
              SYS_OP_DESCEND("ID")<=HEXTORAW('3ED4FF') )
       filter(SYS_OP_UNDESCEND(SYS_OP_DESCEND("ID"))>=42 AND
              SYS_OP_UNDESCEND(SYS_OP_DESCEND("ID"))<=84)

Note
-----
   - dynamic sampling used for this statement
Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          9  consistent gets
          0  physical reads
          0  redo size
       1092  bytes sent via SQL*Net to client
        418  bytes received via SQL*Net from client
          4  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
         43  rows processed

Difference Number 3. Although the same execution plan with the same number of consistent gets is performed, the cardinality estimates are not as accurate and the SYS_OP_DESCEND and SYS_OP_UNDESCEND functions are used as access/filter conditions as they’re the functions implemented in the function-based index.

If we run the same query using the Rule Based Optimizer (remember, we “forgot” to collect statistics on the table):

SQL> alter system set optimizer_mode='CHOOSE' scope=both;

System altered.

SQL> select * from bowie where id between 42 and 84 order by id desc;

43 rows selected.
Execution Plan
----------------------------------------------------------
Plan hash value: 2027917145

------------------------------------
| Id  | Operation          | Name  |
------------------------------------
|   0 | SELECT STATEMENT   |       |
|   1 |  SORT ORDER BY     |       |
|*  2 |   TABLE ACCESS FULL| BOWIE |
------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   2 - filter("ID"<=84 AND "ID">=42)

Note
-----
   - rule based optimizer used (consider using cbo)
Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
        309  consistent gets
          0  physical reads
          0  redo size
       1092  bytes sent via SQL*Net to client
        418  bytes received via SQL*Net from client
          4  SQL*Net roundtrips to/from client
          1  sorts (memory)
          0  sorts (disk)
         43  rows processed

Difference Number 4. The Rule based Optimizer does not support Function-Based Indexes and so the index is now completely ignored. Oracle has no choice here but to perform the much more expensive Full Table Scan, when previously the ascending index was used.

A Descending Index can potentially be useful in a concatenated, multi-column index, in which the columns could be ordered in a combination of ascending/descending order that could in turn return the data in a required specific order, thereby negating the need for a potentially expensive sort operation.

However, with a single column index, one would need to question the need for making such an index descending …

Having fun :) Enjoy your weekend !!

Descending Indexes Quiz (Up On The Ladder) September 8, 2011

Posted by Richard Foote in Descending Indexes, Oracle Indexes, Quiz.
13 comments

OK, you won’t find the answer to these questions on my blog, so using my search facility won’t be of any help :)

Actually, it’s quite an easy one this, honest ;)

If you have a query such as:

SELECT * FROM bowie WHERE id BETWEEN 42 and 84 ORDER BY id DESC;

1) Can a default B-Tree index on the single ID column that stores data in Ascending order be used by the CBO to retrieve the data automatically in the required Descending order, without the need for a sort operation ?

2) Can a Descending B-Tree index on the ID column be used by the CBO to retrieve the data automatically in Ascending order, without the need for a sort operation ?

3) Depending on your answers above, what are the differences (if any) between the implementation of an Ascending and Descending index ?

Enjoy !!

Best Method To Select One Row From Small Table – Solution (Revolution 1) September 7, 2011

Posted by Richard Foote in Oracle Indexes, Small Indexes.
4 comments

OK, time for some answers, although of course regular readers of this blog will already know the answer :)

When selecting one row from the small table as in the quiz in my previous post, the correct order is as follows:

1) PK access of an Index Organized Table.

This option only requires just the 1 consistent get (as the IOT only consists of just 1 block) and this consistent get is a “cheaper” consistent get examination, which in turn only requires the 1 latch. So 1 consistent get and 1 latch. For more info on this, see: http://richardfoote.wordpress.com/2009/05/27/indexes-and-small-tables-part-vii-cluster-one/.

2) Use of Unique Index With a Heap Table.

This option only requires 2 consistent gets (one to read the index leaf block and one to read the table block). There can only be a maximum of one row when a single equality predicate is specified as the index is Unique, which means that 2 consistent gets is the maximum necessary (there may only be the 1 consistent get if there are either no rows to be returned or if all required columns can be found in the index). Additionally, because the index is Unique, both consistent gets are the cheaper, 1 latch consistent get examinations. So that’s 2 consistent gets and 2 latches. For more information, see: http://richardfoote.wordpress.com/2009/05/13/indexes-and-small-tables-part-v-its-no-game/.

3) Use of Non-Unique Index With a Heap Table

This option requires at most 3 consistent gets (one to read the leaf block, one to access the table and possibly one more to perform an additional fetch operation and checking the leaf block again in case of more rows, possibly necessary as the index is Non-Unique and there could be more than one row that matches an equality predicate). Unfortunately, as the index is Non-Unique, the consistent gets are the full-blown consistent gets which requires the buffer block to be pinned/unpinned via 2 latch calls. If the last value is specified, then the additional fetch may be unnecessary and if the index contains all the necessary columns, then just 1 consistent get would be necessary. But in the example provided, we’re looking at typically 3 consistent gets and 6 latches. For more information, see: http://richardfoote.wordpress.com/2009/05/05/indexes-and-small-tables-part-iv-treefingers/.

4) Use of a Full Table Scan

This option requires as a minimum 4 consistent gets (as Oracle needs to not only access the one block containing the 42 rows, but additionally the table segment header multiple times in order to determine the extent map, segment HWM etc.). These additional consistent gets are not generally an issue as these overheads are negligible for a typical FTS of a typical table. But in this example, with a tiny table, these consistent gets makes the difference. Note also that all 4 consistent gets require the block to be pinned/unpinned and so require 2 latch gets each. So that’s 4 consistent gets and 8 latches, minimum, even if all the rows can fit in 1 table block. On its own, no big deal, but if this small lookup table is accessed 10,000 times a minute, that’s potentially a lot of extra CPU and contention when performing a FTS over the above options. For more info, see: http://richardfoote.wordpress.com/2009/04/16/indexes-on-small-tables-part-i-one-of-the-few/ and the other articles on Indexes on Small Tables.

To those that got it right, well done. To those that didn’t, hopefully you’ve learnt something useful.

The moral of the story, no table is too small to potentially benefit from an index, even if the table is only one block in size :)

New question and discussion tomorrow.

Best Method To Select One Row From Small Table Quiz (Each Small Candle) September 5, 2011

Posted by Richard Foote in Oracle Indexes, Quiz, Small Indexes.
22 comments

Assume you have a tiny little table with just 42 rows (naturally) that all fit in one table block. Order the following options in order of “efficiency” (most efficient option first) when accessing just one of these rows:

1) Full Table Scan of Heap Table

2) PK access of an Index Organised Table

3) Index access of Heap Table via a Unique Index

4) Index access of Heap Table via a Non-Unique Index

If you think any of the options are the same, then you can order them as follows (example only):

1) Option 1

2) Option 2

2) Option 3

4) Option 4

Answer in the next few days …

UPDATE: just to clarify based on comments already made.

Yes, any index must visit the table as there are required columns within the table that are not stored in the index (this is implied by the above options). The table has only ever contained 42 rows and the are no additional table blocks below the table HWM (not that this really makes a difference to the answer). To keep it simple, the column being queried has a NOT NULL constraint (although it doesn’t really matter, except for when you want to put a PK constraint on it such as with the IOT option).

The Dark Side Of The Moon Immersion Box Set September 4, 2011

Posted by Richard Foote in Pink Floyd, Richard's Musings, The Dark Side of the Moon.
14 comments

It’s Father’s Day here today in Australia and because I’ve naturally been a really really good Dad all year, my family have given me a real treat for my present this year, the Immersion Box Set of the Pink Floyd classic, The Dark Side Of The Moon (although unfortunately, I have to wait a couple of weeks for it to get released until I can get my hands on it).

As the days of actually having a physical format for music (be it record or tape or CD or whatever) to hold and hug are fast disappearing in this age of digital downloads, Pink Floyd have decided to re-release their back catalogue in physical format one last time with some style.

All their albums are being re-released in new digitally remastered formats, but three of their very best albums (The Dark Side Of The Moon in late September, Wish You Were Here in November and The Wall in February 2012) get the special treatment with the release of Immersion Box Sets.

So what do you get in TDSOTM Immersion Box Set ?

The answer is heaps !!

In a very large box (naturally), you get:

Disc 1 , a CD containing a digitally remastered version of the album

Disc 2, a CD containing a previously unreleased live concert at Wembley dating back to 1974.

Disc 3, an audio DVD containing various 5.1 Surround Sound and Quadraphonic (as originally released in 1973) mixes of the album

Disc 4, a visual DVD containing various live performances, documentaries and all the original concert screen films (makes me want to go out and buy a circular TV !!)

Disc 5, a Blu-Ray containing both audio and video highlights of what I’ve listed already

Disc 6, a CD containing previously unreleased material, including demos and the various live sequences that didn’t quite make it onto the final album.

You also get a whole bunch of other goodies, including colour booklets, a photo book, Storm Thorgerson artwork and cards, a set of 9 coasters, a scarf (just in time for our Canberra summer), Pink Floyd marbles (of course) and replicas of various memorabilia.

I can’t wait !!

I’m almost pitying the neighbours already as I fully plan to sit in the middle of my surround sound system and play all this as it was intended. REALLY REALLY LOUD !!

If you want more details or you’re interested in buying this as well, simply click on the picture of the album artwork above.

This will definitely make my Recommendations Page :)

METHOD_OPT=> SIZE AUTO Quiz Solution (The Trickster) September 1, 2011

Posted by Richard Foote in CBO, Histograms, Oracle Indexes, Oracle Statistics.
16 comments

I was going to leave it for a few days but there have already been so many comments and discussions on all this, I thought I better write something up. In case anyone was wondering, yes I probably am driving my colleagues at work mad with my “Question of the Day” !!

Unfortunately, some might be disappointed at both Oracle and myself :)

Yes, I did kinda set things up to trick the unwary and yes, perhaps the answer isn’t what many are expecting.

The answer to my previous question of which column is going to have a histogram when using the METHOD_OPT  SIZE AUTO option is in fact Column 2. Well done to everyone who got it right.

Why ?

The simplest answer is because it’s the only column of the three that has 254 or less distinct values.

Here’s the key point. When using METHOD_OPT SIZE AUTO, every column with 254 or less distinct values that has been referenced within a predicate, will have a Frequency-based histogram. Each and every one of them, regardless of whether the data is actually skewed or not. So Column 2 with only 254 distinct values AND having previously been referenced in a predicate was guaranteed to have a histogram.

If a column has more than 254 distinct values, whether it then has a Height-Based histogram depends on how the data is skewed. If the data is perfectly evenly distributed, then it won’t have a histogram. Column 1, having sequenced based unique values will not meet the criteria and so not have a histogram.

Column 3 is interesting. Having inserted the outlier value, it now has 255 distinct values and so no longer qualifies for an automatic frequency based histogram. However, if all its values are evenly distributed, then it won’t qualify for a height based histogram either and Column 3 only has just the one outlier value, all other values are evenly distributed values. Unfortunately, Oracle doesn’t pick up on rare outlier values (even if you collect 100% statistics and it’s one of the low/high points of the column) and so will not generate a height-based histogram.

The only column that qualifies is Column 2.

A demo to illustrate. First, let’s create and populate our table:

SQL> create table bowie (id number, code1 number, code2 number);

Table created.

SQL> insert into bowie select rownum, mod(rownum,254), mod(rownum,254) from dual  connect by level <= 1000000;

1000000 rows created.

SQL> commit;

Commit complete.

Notice I’m using a MOD function to generate a perfectly even distribution of data. I’ve noticed a few examples (such as that by Charles Hooper in the comments of the Quiz posting), in which the DBMS_RANDOM function is used. Note this will almost certainly generate data with enough natural skewness on a 1M table with 254 random values that when the outlier 255th value is introduced, it will qualify for a height-based histogram. Very easy way to test and find out. Simply generate the 1M data with 255 random values and I suggest a height-based histogram is created regardless.

OK, I’ll run some SQL to generate sufficient workload to qualify the columns for automatic histograms:

SQL> select * from bowie where id = 42;
SQL> select * from bowie where code1 = 42;
SQL> select * from bowie where code2 = 42;

BTW, the difference between the SIZE AUTO and SIZE SKEWONLY options, is that AUTO requires previous workload to suggest a histogram might be relevant, SKEWONLY does not. 

If we were to collect statistics at this stage, we would notice that the second and third columns both have a Frequency-Based histogram as both columns only have 254 distinct values and so automatically qualify:

SQL> exec dbms_stats.gather_table_stats(ownname=>'BOWIE', tabname=>'BOWIE', estimate_percent=> null, cascade=>true);

PL/SQL procedure successfully completed.

SQL> select column_name, histogram from dba_tab_columns where table_name = 'BOWIE';

COLUMN_NAME                    HISTOGRAM
------------------------------ ---------------
ID                             NONE
CODE1                          FREQUENCY
CODE2                          FREQUENCY

If we were to run a query using the third column, notice how the cardinality estimates aren’t too bad in this example:

SQL> select * from bowie where code2 > 600;

no rows selected

Execution Plan
----------------------------------------------------------
Plan hash value: 1845943507

---------------------------------------------------------------------------
| Id  | Operation         | Name  | Rows  | Bytes | Cost (%CPU)| Time     |
---------------------------------------------------------------------------
|   0 | SELECT STATEMENT  |       |     1 |    13 |   660   (2)| 00:00:08 |
|*  1 |  TABLE ACCESS FULL| BOWIE |     1 |    13 |   660   (2)| 00:00:08 |
---------------------------------------------------------------------------

There are no rows that are greater than 600 and so an estimate of 1 isn’t too bad at all.

OK, let’s add in this one, tiny little row and collect fresh, <strong>100% accurate statistics</strong> (Note: the accurate statistics is very important as Niall’s examples has demonstrated):

&nbsp;

SQL> insert into bowie values (1000001, 42, 99999999);

1 row created.

SQL> commit;

Commit complete.

SQL> exec dbms_stats.gather_table_stats(ownname=>'BOWIE', tabname=>'BOWIE', estimate_percent=> null, cascade=>true);

PL/SQL procedure successfully completed.

SQL> select column_name, histogram from dba_tab_columns where table_name = 'BOWIE';

COLUMN_NAME                    HISTOGRAM
------------------------------ ---------------
ID                             NONE
CODE1                          FREQUENCY
CODE2                          NONE

Note that the third column now has 255 distinct values and so no longer qualifies for the automatic Frequency-Based histogram. As most of its data is perfectly evenly distributed with just the one outlier value, the column doesn’t qualify for a Height-based histogram either and so now has no histogram at all.

Note as I collected 100% accurate statistics, Oracle is definitely aware of this outlier value:

SQL> select column_name, low_value, high_value from dba_tab_columns where table_name='BOWIE' and column_name='CODE2';

COLUMN_NAME  LOW_VALUE  HIGH_VALUE
------------ ---------- ------------
CODE2        80         C464646464

SQL> var high_num number
SQL> exec dbms_stats.convert_raw_value('C464646464',:high_num);

PL/SQL procedure successfully completed.

SQL> print high_num

  HIGH_NUM
----------
  99999999

But it’s not enough for Oracle to automatically generate a histogram. Which is a shame really, because now we can have all sorts of problems:

SQL> select * from bowie where code2 > 600;
Execution Plan
----------------------------------------------------------
Plan hash value: 1845943507

---------------------------------------------------------------------------
| Id  | Operation         | Name  | Rows  | Bytes | Cost (%CPU)| Time     |
---------------------------------------------------------------------------
|   0 | SELECT STATEMENT  |       |   999K|    12M|   660   (2)| 00:00:08 |
|*  1 |  TABLE ACCESS FULL| BOWIE |   999K|    12M|   660   (2)| 00:00:08 |
---------------------------------------------------------------------------

When previously it had the cardinality estimates spot on, now they’re terrible (expecting not 1 row but 999,000 rows !!) because without a histogram, Oracle is assuming even distribution between its low and high point values.

I’m not a great fan of either the SIZE AUTO or SIZE SKEWONLY options ;)

Hope you’re enjoying these little quizzes, I’ll have another one for you all soon.

METHOD_OPT => SIZE AUTO Quiz (Automatic For The People) August 31, 2011

Posted by Richard Foote in Method_Opt Size AUTO, Oracle Indexes, Oracle Statistics.
40 comments

OK, a nice, easy, simple one today. No tricks, honest ;)

You have a table with 3 columns and lets say 1M rows.

Column 1 is effectively unique, so it has 1M distinct values. Let’s say 1 – 1000000, not that it really matters.

Column 2 has 254 distinct values, all evenly distributed so it has approximately the same number of rows for each value. Let’s say the values are 0-253 again it doesn’t really matter.

Column 3 is identical to Column 2, it also has 254 distinct values, all evenly distributed as well such that it also has approximately the same number of rows for each value. Let’s say the range of values are the same, 0-253, again it doesn’t really matter.

You have various queries in the database in which all 3 columns are referenced somewhere in WHERE conditions (eg. WHERE Column1 = 42).

You then insert just one row that has the following values based on our example: VALUES (1000001, 42, 99999999).

The key points here is that for Column1, it’s just another unique value, just 1 greater than the previous maximum value. Nothing special.

For Column2, it’s just another of the existing 254 values that doesn’t really change the even distribution of the data. Nothing special.

However, for Column 3, it’s not only a new value that didn’t previously exist (and so there’s just the one row with this value in the data, whereas all the other values correspond to roughly 1/254 of the rows) but it’s also a value that is way way way outside the normal range of existing values (nothing comes close to having a value of 99999999).

OK, we have the scenario, hopefully you can see where I going with this.

You decide to collect fresh statistics with DBMS_STATS, you want them to be completely accurate so you use a 100% sample size (or compute with estimate_percent=>null). But because you want to get with the AUTO program and make life easier for yourself, you decide to let Oracle work out which columns might require and benefit from a histogram by using METHOD_OPT=>’ FOR ALL COLUMNS SIZE AUTO’.

Now finally comes the question. Of the three columns, only one column will have a histogram. Which one and why is it so ?

If you understand how Oracle collects statistics, the answer will hopefully be obvious ;)

MIN / MAX Quiz Answer (One Shot) August 31, 2011

Posted by Richard Foote in CBO, Index Full Scan (Min/Max), MAX, MIN, Oracle Indexes.
9 comments

Not only are my regular blog readers a good deal better looking than the average person, but they’re quite a bit smarter as well :)

As most people have correctly identified, the answer I was after to my previous Min/Max Quiz is that Option 1 is indeed the odd one out, as it’s the only option that can’t make use of the Index Full Scan (Min/Max) access path.

If you’re after either the minimum or the maximum of a column value and the column is indexed, the CBO can potentially use the Index Full Scan (Min/Max), which simply navigates to the first OR last leaf block in the index structure looking for the Min or Max value in question. Oracle can of course navigate to the first (left-most) or last (right-most) leaf blocks very easily by simply following the associated first/last pointers in the Root/Branch structure of the index. All things being equal and providing there haven’t been any subsequent deletes to empty out the index entries from these leaf blocks, Oracle can very quickly determine the minimum or maximum value of the column.

However, the Index Full Scan (Min/Max) can only visit one side of the index, not both. Therefore, if you want both the minimum and the maximum column value, an Index Full Scan (Min/Max) is not viable and the CBO is forced to look for other alternatives. It sounds like such a trivial thing to implement but that’s how it goes. I do remember way back when Oracle9i was released and the introduction of the Index Skip Scan I thought perhaps Oracle might also soon introduce an index skip scan version of Min/Max (as it basically just needs to “skip” all the index leaf blocks in the “middle” of the index via another lookup of the index), but it was not to be.

So for a query such as in Option 1, if the column IS NULL and does not have a NOT NULL constraint, then:

  
SQL> select min(id), max(id) from muse;

   MIN(ID)    MAX(ID)
---------- ----------
         1    1000000
Execution Plan
----------------------------------------------------------
Plan hash value: 421245806

---------------------------------------------------------------------------
| Id  | Operation          | Name | Rows  | Bytes | Cost (%CPU)| Time     |
---------------------------------------------------------------------------
|   0 | SELECT STATEMENT   |      |     1 |     5 |  2125   (1)| 00:00:26 |
|   1 |  SORT AGGREGATE    |      |     1 |     5 |            |          |
|   2 |   TABLE ACCESS FULL| MUSE |  1000K|  4882K|  2125   (1)| 00:00:26 |
---------------------------------------------------------------------------

 
Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
       7795  consistent gets
       7788  physical reads
          0  redo size
        470  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
          1  rows processed

  

Then an (expensive) Full Table Scan is likely the way to go. However, if the column has a NOT NULL constraint and the index is indeed smaller than the parent table, then:

  
SQL> alter table muse modify id not null;

Table altered.

SQL> select min(id), max(id) from muse;

   MIN(ID)    MAX(ID)
---------- ----------
         1    1000000
Execution Plan
----------------------------------------------------------
Plan hash value: 1592024618

-----------------------------------------------------------------------------------
| Id  | Operation             | Name      | Rows  | Bytes | Cost (%CPU)| Time     |
-----------------------------------------------------------------------------------
|   0 | SELECT STATEMENT      |           |     1 |     5 |   611   (1)| 00:00:08 |
|   1 |  SORT AGGREGATE       |           |     1 |     5 |            |          |
|   2 |   INDEX FAST FULL SCAN| MUSE_ID_I |  1000K|  4882K|   611   (1)| 00:00:08 |
-----------------------------------------------------------------------------------

Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
       2242  consistent gets
          0  physical reads
          0  redo size
        470  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
          1  rows processed

  

Then an Index Fast Full Scan becomes viable.

All the other options I’ve used to return the Min/Max of the column all incorporate two separate SELECT clauses and so all can potentially use an Index Full Scan (Min/Max) access path for each distinct clause.

Be it when using a UNION operation:

  
SQL> select min(id) as "MIN(ID)/MAX(ID)" from muse union all select max(id) from muse;

 MIN(ID)/MAX(ID)
---------------
              1
        1000000
 
Execution Plan
----------------------------------------------------------
Plan hash value: 1370940131

-----------------------------------------------------------------------------------------
| Id  | Operation                   | Name      | Rows  | Bytes | Cost (%CPU)| Time     |
-----------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT            |           |     2 |    10 |     6  (50)| 00:00:01 |
|   1 |  UNION-ALL                  |           |       |       |            |          |
|   2 |   SORT AGGREGATE            |           |     1 |     5 |            |          |
|   3 |    INDEX FULL SCAN (MIN/MAX)| MUSE_ID_I |     1 |     5 |     3   (0)| 00:00:01 |
|   4 |   SORT AGGREGATE            |           |     1 |     5 |            |          |
|   5 |    INDEX FULL SCAN (MIN/MAX)| MUSE_ID_I |     1 |     5 |     3   (0)| 00:00:01 |
-----------------------------------------------------------------------------------------

Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          6  consistent gets
          0  physical reads
          0  redo size
        456  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
          2  rows processed

Although as pointed out in the comments, this does return 2 rows.

Or I could use Scalar Sub-Queries:

  
SQL> select (select min(id) from muse) "MIN(ID)", (select max(id) from muse) "MAX(ID)" from dual;

   MIN(ID)    MAX(ID)
---------- ----------
         1    1000000
Execution Plan
----------------------------------------------------------
Plan hash value: 2177063930

----------------------------------------------------------------------------------------
| Id  | Operation                  | Name      | Rows  | Bytes | Cost (%CPU)| Time     |
----------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT           |           |     1 |       |     2   (0)| 00:00:01 |
|   1 |  SORT AGGREGATE            |           |     1 |     5 |            |       |
|   2 |   INDEX FULL SCAN (MIN/MAX)| MUSE_ID_I |     1 |     5 |     3   (0)| 00:00:01 |
|   3 |  SORT AGGREGATE            |           |     1 |     5 |            |       |
|   4 |   INDEX FULL SCAN (MIN/MAX)| MUSE_ID_I |     1 |     5 |     3   (0)| 00:00:01 |
|   5 |  FAST DUAL                 |           |     1 |       |     2   (0)| 00:00:01 |
----------------------------------------------------------------------------------------

Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          6  consistent gets
          0  physical reads
          0  redo size
        468  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
          1  rows processed

  

Or indeed I could use a WITH clause:

  
SQL> with min_id as (select min(id) from muse), max_id as (select max(id) from muse) select * from min_id, max_id;

   MIN(ID)    MAX(ID)
---------- ----------
         1    1000000

 
Execution Plan
----------------------------------------------------------
Plan hash value: 3280440773

------------------------------------------------------------------------------------------
| Id  | Operation                    | Name      | Rows  | Bytes | Cost (%CPU)|Time     |
------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT             |           |     1 |    26 |     6   (0)|00:00:01 |
|   1 |  NESTED LOOPS                |           |     1 |    26 |     6   (0)|00:00:01 |
|   2 |   VIEW                       |           |     1 |    13 |     3   (0)|00:00:01 |
|   3 |    SORT AGGREGATE            |           |     1 |     5 |            |         |
|   4 |     INDEX FULL SCAN (MIN/MAX)| MUSE_ID_I |     1 |     5 |     3   (0)|00:00:01 |
|   5 |   VIEW                       |           |     1 |    13 |     3   (0)|00:00:01 |
|   6 |    SORT AGGREGATE            |           |     1 |     5 |            |         |
|   7 |     INDEX FULL SCAN (MIN/MAX)| MUSE_ID_I |     1 |     5 |     3   (0)|00:00:01 |
------------------------------------------------------------------------------------------

Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          6  consistent gets
          0  physical reads
          0  redo size
        470  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
          1  rows processed

They’re all subtly different but they all can make use of the Index Full Scan (Min/Max) scan for each separate SELECT clause and they all can perform the necessary resultant work with just 6 consistent gets in my example.

More quizzes to come …

MIN/MAX Quiz (Funtime) August 29, 2011

Posted by Richard Foote in MAX, MIN, Oracle Indexes.
8 comments

At my work I’ve recently introduced a little “Question of the Day” for my fellow DBAs, hopefully to pass on a few interesting little titbits of information and have a bit of fun along the way.

I was going to just write a blog article on the following topic of how Oracle deals with MIN and MAX queries, but decided to give you all the chance to have a bit of a think first via a little quiz. This was a question I asked recently, see what you think:

There are many ways one could write a query to determine the MIN and MAX of a column. That said, which of the following is the odd one out:

1) select min(id), max(id) from muse;

2) select min(id) as “MIN(ID)/MAX(ID)” from muse union all select max(id) from muse;

3) select (select min(id) from muse) “MIN(ID)”, (select max(id) from muse) “MAX(ID)” from dual;

4) with min_id as (select min(id) from muse), max_id as (select max(id) from muse) select * from min_id, max_id;

 

It’s a “big” table, there’s an index on the ID and the ID has a not null constraint (the constraint doesn’t actually matter to the answer of the question, but it can make a subtle difference).

Now, one could come up with lots of different possible answers (eg: option 4 is the only one that uses a WITH clause), however the answer I’m after is one that might be best associated with the common theme of this blog.

You don’t have to make a comment, you can just quietly ponder which option is somewhat different or maybe consider which one you may not want to use.

No more clues :)

More details in a few day’s time.

PS. Folk from my work and Jonathan Lewis may not answer as they already know the answer I’m after :)

BLEVEL 1 => BLEVEL 2 (Teenage Wildlife) August 23, 2011

Posted by Richard Foote in BLEVEL, CBO, Oracle Indexes.
4 comments

Jonathan Lewis recently wrote a really nice blog piece blevel=1 on the dangers of an index toggling between BLEVEL 1 and BLEVEL 2. I thought it would be useful to demonstrate this issue with a quick demo (Note: this example is on 11.2.0.1, with an 8K block size).
 
First, create a simple little table with 336,000 rows and an index on an ID number column:

  
SQL> create table major_tom (id number, code number, name varchar2(30));
 
Table created.
 
SQL> create index major_tom_i on major_tom(id);
 
Index created.
 
SQL> insert into major_tom select rownum, mod(rownum,100), 'GROUND CONTROL' from dual connect by level <=336000;
 
336000 rows created.
 
SQL> commit;
 
Commit complete.
 
SQL> exec dbms_stats.gather_table_stats(ownname=>null, tabname=> 'MAJOR_TOM', cascade=> true, estimate_percent=>null, method_opt=>'FOR ALL COLUMNS SIZE 1');
 
PL/SQL procedure successfully completed.
 
SQL> select blevel, leaf_blocks, clustering_factor from dba_indexes where index_name='MAJOR_TOM_I';
 
    BLEVEL LEAF_BLOCKS CLUSTERING_FACTOR
---------- ----------- -----------------
         1         671              1296

 
Note the index has 671 leaf blocks and a Blevel=1. This of course means the index basically consists of a Root Block, which in turn references all its 671 leaf blocks. Therefore to read a specific index entry, requires a read of the index root block followed by a read of the specific index leaf block. That’s 2 reads in total.
 
Let’s run a query to return one row (note the ID column is effectively unique although I’ve only created a non-unique index):
 

SQL> select * from major_tom where id = 42;
 

Execution Plan
----------------------------------------------------------
Plan hash value: 4155681103
 
-------------------------------------------------------------------------------------------
| Id  | Operation                   | Name        | Rows  | Bytes | Cost (%CPU)| Time     |
-------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT            |             |     1 |    23 |     2   (0)| 00:00:01 |
|   1 |  TABLE ACCESS BY INDEX ROWID| MAJOR_TOM   |     1 |    23 |     2   (0)| 00:00:01 |
|*  2 |   INDEX RANGE SCAN          | MAJOR_TOM_I |     1 |       |     1   (0)| 00:00:01 |
-------------------------------------------------------------------------------------------
 

Predicate Information (identified by operation id):
---------------------------------------------------
 
   2 - access("ID"=42)
 

Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          4  consistent gets
          0  physical reads
          0  redo size
        531  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
          1  rows processed

 
 

Note: The cost of using the index is just 1 not 2 as perhaps expected. This is due to the CBO ignoring the Blevel in its calculations when the Blevel = 1.  As the index is relatively small, the CBO takes the approach that the root block is very likely already cached and so is not worth costing.
 
As the data is perfectly evenly distributed and effectively unique, the CBO has correctly estimated the number of returned rows as just 1. Therefore, the overall cost of the execution plan is just 2, 1 to read the leaf block and 1 to read the table block.
 
Notice that the number of consistent gets is 4. 1 to read the index root block, 1 for the index leaf block, 1 for the table block and as the index is non-unique, 1 for an additional fetch performed to check the index again that there are no further rows to be returned.
 
If we now create another table of 1M rows that will be used in a join operation:
 

 
SQL> create table ziggy (id number, code number, name varchar2(30));
 
Table created.
 
SQL> insert into ziggy select rownum, mod(rownum,10000), 'ZIGGY STARDUST' from dual connect by level <= 1000000;
 
1000000 rows created.
 
SQL> exec dbms_stats.gather_table_stats(ownname=>null, tabname=> 'ZIGGY', cascade=> true, estimate_percent=>null, method_opt=>'FOR ALL COLUMNS SIZE 1');
 
PL/SQL procedure successfully completed.

If we now join these 2 tables and select a moderate number of rows:
 

 
SQL> select * from ziggy z, major_tom m where z.id = m.id and z.code in (42, 4242);
 
68 rows selected.
 

Execution Plan
----------------------------------------------------------
Plan hash value: 2011771477
 
--------------------------------------------------------------------------------------------
| Id  | Operation                    | Name        | Rows  | Bytes | Cost (%CPU)| Time     |
--------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT             |             |   200 |  9400 |  1372   (2)| 00:00:17 |
|   1 |  NESTED LOOPS                |             |       |       |            |          |
|   2 |   NESTED LOOPS               |             |   200 |  9400 |  1372   (2)| 00:00:17 |
|*  3 |    TABLE ACCESS FULL         | ZIGGY       |   200 |  4800 |  1105   (2)| 00:00:14 |
|*  4 |    INDEX RANGE SCAN          | MAJOR_TOM_I |     1 |       |     1   (0)| 00:00:01 |
|   5 |   TABLE ACCESS BY INDEX ROWID| MAJOR_TOM   |     1 |    23 |     2   (0)| 00:00:01 |
--------------------------------------------------------------------------------------------
 

Predicate Information (identified by operation id):
---------------------------------------------------
 
   3 - filter("Z"."CODE"=42 OR "Z"."CODE"=4242)
   4 - access("Z"."ID"="M"."ID")
 

Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
       4175  consistent gets
       4024  physical reads
          0  redo size
       1950  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
         68  rows processed

The CBO goes for a Nested Loop join, primarily because the inner table is only accessed a relatively small number of times AND because the cost of doing so via the index is so damn cheap.
 
However, if we add just a few more rows and collect fresh statistics …
 

 
SQL> insert into major_tom select rownum+336000, mod(rownum,100), 'GROUND CONTROL' from dual connect by level <=500;
 
500 rows created.
 
SQL> commit;
 
Commit complete.
 
SQL> exec dbms_stats.gather_table_stats(ownname=>null, tabname=> 'MAJOR_TOM', cascade=> true, estimate_percent=>null, method_opt=>'FOR ALL COLUMNS SIZE 1');
 
PL/SQL procedure successfully completed.
 

SQL> select blevel, leaf_blocks, clustering_factor from dba_indexes where index_name='MAJOR_TOM_I';
 
    BLEVEL LEAF_BLOCKS CLUSTERING_FACTOR
---------- ----------- -----------------
         2         672              1298

The index has now toggled over to become a Blevel 2 index. We only added a handful of rows resulting in just the one additional index leaf block, but 672 is just too many to be referenced within the one index root block in this example. The root block has split, two new index branches have been created that now reference the leaf blocks and the root block now only references the two new branch blocks.
 
Overall, the changes are quite minor but the ramifications can be quite dramatic …
 
If we now select one row again:
 

 
SQL> select * from major_tom where id = 42;
 

Execution Plan
----------------------------------------------------------
Plan hash value: 4155681103
 
-------------------------------------------------------------------------------------------
| Id  | Operation                   | Name        | Rows  | Bytes | Cost (%CPU)| Time     |
-------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT            |             |     1 |    23 |     4   (0)| 00:00:01 |
|   1 |  TABLE ACCESS BY INDEX ROWID| MAJOR_TOM   |     1 |    23 |     4   (0)| 00:00:01 |
|*  2 |   INDEX RANGE SCAN          | MAJOR_TOM_I |     1 |       |     3   (0)| 00:00:01 |
-------------------------------------------------------------------------------------------
 

Predicate Information (identified by operation id):
---------------------------------------------------
 
   2 - access("ID"=42)
 

Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          5  consistent gets
          0  physical reads
          0  redo size
        531  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
          1  rows processed

The cost of the index has now jumped up by 2 from 1 to 3 with the overall costs up from 2 to 4, even though the index is practically the same size. As the Blevel is now 2, the CBO now includes the cost of the Blevel in its calculations. The cost associated with accessing the root block and a branch block all now count. Although overall the costs are still low, this increase actually represents a 100% increase in the use of the index for an equality search.
 
This increase can be significant if the index needs to be accessed multiple times. Let’s now re-run the join query:
 

 
SQL> select * from ziggy z, major_tom m where m.id = z.id and z.code in (42, 4242);
 
68 rows selected.
 

Execution Plan
----------------------------------------------------------
Plan hash value: 1928189744
 
--------------------------------------------------------------------------------
| Id  | Operation          | Name      | Rows  | Bytes | Cost (%CPU)| Time     |
--------------------------------------------------------------------------------
|   0 | SELECT STATEMENT   |           |   200 |  9400 |  1485   (2)| 00:00:18 |
|*  1 |  HASH JOIN         |           |   200 |  9400 |  1485   (2)| 00:00:18 |
|*  2 |   TABLE ACCESS FULL| ZIGGY     |   200 |  4800 |  1105   (2)| 00:00:14 |
|   3 |   TABLE ACCESS FULL| MAJOR_TOM |   336K|  7558K|   378   (1)| 00:00:05 |
--------------------------------------------------------------------------------
 

Predicate Information (identified by operation id):
---------------------------------------------------
 
   1 - access("M"."ID"="Z"."ID")
   2 - filter("Z"."CODE"=42 OR "Z"."CODE"=4242)
 

Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
       5366  consistent gets
       4024  physical reads
          0  redo size
       1964  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
         68  rows processed

Although it’s retrieving exactly the same data, the execution plan has changed significantly. The Nested Loop join is no longer as appealing to the CBO as the cost of accessing the inner table via the index has now effectively doubled. The CBO has now gone for a Hash Join, accessing both tables via Full Tables Scans. The overall cost of the Nested Loop plan was 1372, but this has increased to over 1485, the cost of the now so-called more efficient Hash Join plan.
 
If you have indexes that are on the boundary of increasing from a blevel=1 to a blevel=2, execution plans can potentially change significantly based on the differences in how indexes get costed. This can be especially troublesome when such indexes get regularly rebuilt as they may toggle between Blevel 1 and 2 based on associated space savings and can sometimes result in unpredictable performance depending on when new statistics get collected.
 
I liken it to a child growing up from being a young kid to a teenager. It may only be a difference of a year or so but boy, can the differences be dramatic !!

How To Become An Oracle Expert (The Scientist) August 13, 2011

Posted by Richard Foote in InSync11, Sydney Oracle Meetup.
7 comments

I’ve been invited by the Sydney Oracle Meetup Group to participate in an open discussion on “How To Become An Oracle Expert” next Tuesday, 16 August 2011 at 5:30pm.

There’s a great lineup, including:

Chris Muir (Oracle ACE Director, Australia)
–  Connor McDonald (Oracle ACE Director, Australia)
–  Craig Shallahamer (Oracle ACE Director, US)
–  Guy Harrison (Oracle ACE, Australia)
–  Marcelle Kratochvil (Oracle ACE Director, Australia)
–  Richard Foote (Oracle ACE Director, Australia)
–  Thomas Kyte (Oracle, US)
–  Tim Hall (Oracle ACE Director, UK)

For anyone in Sydney next week attending the InSync11 Conference, this is a great opportunity to meet and chat with some rather clever folks involved in the Oracle community. The venue has changed due to demand (there are currently 66 people down to attend), so if you’re interested, I  strongly suggest you create an account and enroll while you still can.

Hope to catch up with some of you at either InSync11 or at this event next week :)

 

Indexing A Column With Just One Distinct Value (All The Madmen) August 10, 2011

Posted by Richard Foote in CBO, Oracle Indexes.
11 comments

When one thinks about a column that might benefit from being indexed, one generally considers columns with lots of different values so that the selectivity of the column is such that relatively few rows get selected, making the index appealing to the cost based optimizer.
 
There are of course many exceptions to this generalisation and there are times when an index is the perfect and most efficient method when selecting 100% of all data, even if it involves a normal index range scan reading every row out of the table.
 
There are also times when a column that has very few distinct values should be indexed, even with a standard B-tree index.
 
In this example, I’ll index a column that only has just the 1 single value and the CBO will choose to use the index gladly.
 
Just going to create my standard little table and create an index on the CODE column:
 

 
SQL> create table bowie (id number, code number, name varchar2(50));
 
Table created.
 
SQL> create index bowie_code_i on bowie(code);
 
Index created.

 

 

I’ll now load the table with 1 million rows but all the CODE column will remain unpopulated and consist only of NULLS:
 

 
SQL> insert into bowie select rownum, null, 'Ziggy Stardust and the Spiders From Mars' from dual connect by level <= 1000000;
 
1000000 rows created.
 
SQL> commit;
 
Commit complete.

 

 

I’ll only populate one in every 10,000 rows that a CODE value of 42 (of course):
 

 
SQL> update bowie set code = 42 where mod(id,10000) = 0;
 
100 rows updated.
 
SQL> commit;
 
Commit complete.

 

 

Let’s collect accurate statistics, however note I’m not collecting histograms even though there are relatively few rows that have a CODE value of 42:
 

 
SQL> exec dbms_stats.gather_table_stats(ownname=>null, tabname=>'BOWIE', cascade=> true, estimate_percent=>null, method_opt=>'FOR ALL COLUMNS SIZE 1');
 
PL/SQL procedure successfully completed.

 

 

Indeed, the column only has the one value (42) which occurs relatively infrequently in my data, most rows however remain as NULL:
 

 
SQL> select code, count(*) from bowie group by code;
 
      CODE   COUNT(*)
---------- ----------
               999900
        42        100

 

 

Note that the index statistics clearly shows it only has the 1 distinct value. Remember, NULLS are not indexed by default in B-Tree indexes and as a result the index is tiny with just the one leaf block:
 

 
SQL> select blevel, leaf_blocks, distinct_keys from dba_indexes where index_name='BOWIE_CODE_I';
 
    BLEVEL LEAF_BLOCKS DISTINCT_KEYS
---------- ----------- -------------
         0           1             1

 

 

Note also that the column statistics clearly highlight there’s just the one distinct value, however it also records that there are many rows (999900) that are NULL:

 
 

 
SQL> select column_name, num_distinct, num_nulls from dba_tab_columns where table_name = 'BOWIE' and column_name = 'CODE';
 
COLUMN_NAME  NUM_DISTINCT  NUM_NULLS
------------ ------------ ----------
CODE                    1     999900

 

 

Therefore, Oracle with accurate statistics and without requiring any histograms has all the information it needs to know when selecting rows that contain the one and only distinct value of our CODE column, that it will only actually be selecting 100 rows out of the 1 million rows in the table.

 
 

 
SQL> select * from bowie where code = 42;
 
100 rows selected.
 

Execution Plan
----------------------------------------------------------
Plan hash value: 1602289932
 
--------------------------------------------------------------------------------------------
| Id  | Operation                   | Name         | Rows  | Bytes | Cost (%CPU)| Time     |
--------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT            |              |   100 |  4700 |   101   (0)| 00:00:02 |
|   1 |  TABLE ACCESS BY INDEX ROWID| BOWIE        |   100 |  4700 |   101   (0)| 00:00:02 |
|*  2 |   INDEX RANGE SCAN          | BOWIE_CODE_I |   100 |       |     1   (0)| 00:00:01 |
--------------------------------------------------------------------------------------------
 

Predicate Information (identified by operation id):
---------------------------------------------------
 
   2 - access("CODE"=42)
 

Statistics
----------------------------------------------------------
          1  recursive calls
          0  db block gets
        102  consistent gets
          0  physical reads
          0  redo size
       1423  bytes sent via SQL*Net to client
        395  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          0  sorts (memory)
          0  sorts (disk)
        100  rows processed

 

 

The CBO has the rows estimate/cardinality spot on (100) and has decided that the index is indeed the most efficient access path to select the 100 rows of interest.

With flag and bollean type columns, it might be worth some consideration simply storing the one value when one value is relatively rare, such that the resultant index can take advantage of not having to store any of the corresponding NULL values.
 
Even columns with just the one distinct value can potentially benefit from being indexed …

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