# How the CBO works Jonathan Lewis www.jlcomp.demon.co.uk.

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How the CBO works Jonathan Lewis www.jlcomp.demon.co.uk

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Who am I Independent Consultant. 18+ years experience. Design, Strategy, Reviews, Briefings, Seminars, Tutorials, Trouble-shooting www.jlcomp.demon.co.uk

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Highlights A Puzzle Basic Costs Correcting Oracle's assumptions Oracle 9 learns (Join Mechanics - time permitting) Q and A

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 A puzzle (v8.1 - 4K) create table t1 as select trunc((rownum-1)/15)n1, trunc((rownum-1)/15)n2, rpad('x',100)v1 from all_objects where rownum <= 3000; create table t2 as select mod(rownum,200)n1, mod(rownum,200)n2, rpad('x',100)v1 from all_objects where rownum <= 3000; We construct two sets of data with identical content, although we do use two different mathematical methods to get 15 rows each for 200 different values.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 A puzzle - indexed create index t_i1 on t1(n1); create index t_i2 on t2(n1); analyze table t1 compute statistics; analyze table t2 compute statistics; We create indexes and generate statistics. In newer versions, we should use the dbms_stats package, not analyze. (Note- compute is often over-kill)

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 A puzzle - checking data USER_TABLES TABLE_NAME BLOCKS NUM_ROWS AVG_ROW_LEN ----------- ------ ---------- ----------- T1 96 3000 111 T2 96 3000 111 USER_TAB_COLUMNS TAB COL LOW_VALUE HIGH_VALUE NUM_DISTINCT --- ---- --------- ---------- ------------ T1 N1 80 C20264 200 T2 N1 80 C20264 200 We can check that statistics like number of rows, column values and column counts are identical. The data contents is identical across the two tables.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 A puzzle - the problem select * from t2 where n1 = 45; SELECT STATEMENT Optimizer=CHOOSE (Cost=15 Card=15) TABLE ACCESS (FULL) OF 'T2' (Cost=15 Card=15) select * from t1 where n1 = 45; SELECT STATEMENT Optimizer=CHOOSE (Cost=2 Card=15) TABLE ACCESS (BY INDEX ROWID) OF 'T1' (Cost=2 Card=15) INDEX (RANGE SCAN) OF 'T_I1' (NON-UNIQUE) (Cost=1 Card=15) We now run exactly the same query against the two sets of data - with autotrace switched on - and find that the execution plans are different.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 A puzzle - force it select /*+ index(t2) */ * from t2 where n1 = 45; SELECT STATEMENT Optimizer=CHOOSE (Cost=16 Card=15) TABLE ACCESS (BY INDEX ROWID) OF 'T2' (Cost=16 Card=15) INDEX (RANGE SCAN) OF 'T_I2' (NON-UNIQUE) (Cost=1 Card=15) select * from t2 where n1 = 45; SELECT STATEMENT Optimizer=CHOOSE (Cost=15 Card=15) TABLE ACCESS (FULL) OF 'T2' (Cost=15 Card=15) Why has Oracle ignored the index on T2 ? Put in the hint(s) to make it happen, and see if we get any clues. The cost of a tablescan is cheaper !

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 A puzzle - the detail select table_nametab, num_rowsnum_rows, avg_leaf_blocks_per_keyl_blocks, avg_data_blocks_per_key d_blocks, clustering_factorcl_fac from user_indexes; TAB NUM_ROWS L_BLOCKS D_BLOCKS CL_FAC ---- -------- -------- -------- ------ T1 3000 1 1 96 T2 3000 1 15 3000 Why is the tablescan cheaper ? We look at the data scattering, rather than the data content, and find the answer. The clustering is different.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 A puzzle - the difference The data on the left shows the effect of the trunc() function, the data on the right shows the mod() effect. The statistics describe the data perfectly.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 The arithmetic T2 by index one index block, 15 data blocks = 16 T2 by scan 96 blocks / 8 (multiblock read) = 12 (this is a first approximation) T1 by index one index block, one data block = 2 Silly assumption 1: Every logical request turns into a physical read. Silly assumption 2: A multiblock read is just as fast as a single block read.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Multiblock Read Actual Adjusted 96 / 6.589 = 14.57 (and add 1 in 9.2) The cost of a tablescan uses an 'adjusted' db_file_multiblock_read_count. Under 'traditional' costing the v9 cost is always one more than the v8 cost..

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Multiblock Read Count select * from t2 where n1 = 45; -- tablescan cost was 15 alter session set DB_FILE_MULTIBLOCK_READ_COUNT = 4; SELECT STATEMENT Optimizer=CHOOSE (Cost=16 Card=15) TABLE ACCESS (BY INDEX ROWID) OF 'T2' (Cost=16 Card=15) INDEX (RANGE SCAN) OF 'T_I2' (NON-UNIQUE) (Cost=1 Card=15) (tablescan cost would be 23) alter session set DB_FILE_MULTIBLOCK_READ_COUNT = 16; SELECT STATEMENT Optimizer=CHOOSE (Cost=10 Card=15) TABLE ACCESS (FULL) OF 'T2' (Cost=10 Card=15) We can affect access paths by changing the db_file_multiblock_read_count. But this is a bit of a drastic change on a production system.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Single-block adjustment select * from t2 where n1 = 45;-- index access cost was 16 alter session set OPTIMIZER_INDEX_COST_ADJ = 50; SELECT STATEMENT Optimizer=CHOOSE (Cost=8 Card=15) TABLE ACCESS (BY INDEX ROWID) OF 'T2' (Cost=8 Card=15) INDEX (RANGE SCAN) OF 'T_I2' (NON-UNIQUE) (Cost=1 Card=15) alter session set OPTIMIZER_INDEX_COST_ADJ = 25; SELECT STATEMENT Optimizer=CHOOSE (Cost=4 Card=15) TABLE ACCESS (BY INDEX ROWID) OF 'T2' (Cost=4 Card=15) INDEX (RANGE SCAN) OF 'T_I2' (NON-UNIQUE) (Cost=1 Card=15) ` Under Oracle 9 the numbers are slightly different. We can fix one of Oracle's silly assumptions - let it know that single block reads are cheaper (faster) than multiblock reads - by setting a percentage cost

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Join cost-adjustment select t2.n1, t1.n2 from t2,t1 where t2.n2 = 45 and t2.n1 = t1.n1; SELECT STATEMENT (Cost=31 Card=225) HASH JOIN (Cost=31 Card=25) TABLE ACCESS (FULL) OF T2 (Cost=15 Card=15) TABLE ACCESS (FULL) OF T1 (Cost=15 Card=3000) 31 = 15 + 15 + a bit We can even see the effect of this price fixing in joins. Unhinted, or unfixed, the optimizer chooses a hash join as the cheapest way to our two tables.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Hash Join (1) First (smaller) data set Hashed Second (larger) data set The first table is hashed in memory, the second table is used to probe the hash (build) table for matches. In simple cases the cost is easy to calculate.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Force a nested loop select /*+ ordered use_nl(t1) index(t1) */ t2.n1, t1.n2 fromt2,t1 wheret2.n2 = 45 andt2.n1 = t1.n1; NESTED LOOPS (Cost=45 Card=225) TABLE ACCESS (FULL) OF T2 (Cost=15, Card=15) TABLE ACCESS (BY ROWID) OF T1(Cost=2,Card=3000) INDEX(RANGE SCAN) OF T_I1(NON-UNIQUE)(Cost=1) As usual, to investigate why a plan is going wrong, we hint it to make it do what we want - and then look for clues in the resulting cost lines.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Forced NL cost alter session set OPTIMIZER_INDEX_COST_ADJ = 100; -- def NESTED LOOPS (Cost=45 Card=225) TABLE ACCESS(FULL) OF T2 (Cost=15, Card=15) TABLE ACCESS(BY ROWID) OF T1(Cost=2, Card=3000) INDEX(RANGE SCAN) OF T_I1(NON-UNIQUE)(Cost=1) T2- cost = 15 Estimated rows = 15 For each row from T2 we access T1 by complete key value T1-Cost per access = 2 Cost for 15 accesses = 15 x 2 = 30 Total cost of query = cost of T2 + total cost of T1 = 15 + 30 = 45 The nested loop algorithm is: for each row in the outer table, use the value in that row to access the inner table - hence the simple formula.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Nested Loop T1 The basic arithmetic of the nested loop join is visible in the picture. We do three indexed access into T2, but need the three driving rows from T1 first.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Nested Loops - recosted alter session set OPTIMIZER_INDEX_COST_ADJ = 50; NESTED LOOPS (Cost=30 Card=225) TABLE ACCESS(FULL) OF T2 (Cost=15, Card=15) TABLE ACCESS(BY ROWID) OF T1(Cost=1, Card=3000) INDEX(RANGE SCAN) OF T_I1(NON-UNIQUE)(Cost=1) T2- cost = 15 Estimated rows = 15 For each row from T2 we access T1 by complete key value T1-Cost per access = 2 x 50% = 1 Cost for 15 accesses = 15 x 1 = 15 Total cost of query = cost of T2 + total cost of T1 = 15 + 15 = 30 What happens to the cost when we tell Oracle that single block reads cost half as much as it would otherwise charge ?

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Index Caching (NL only) Basic nested loop cost (hinted) NESTED LOOPS (Cost=45 Card=225) TABLE ACCESS (FULL) OF 'T2' (Cost=15, Card=15) TABLE ACCESS (BY INDEX ROWID) OF 'T1' (Cost=2, Card=3000) INDEX (RANGE SCAN) OF 'T_I1' (NON-UNIQUE) (Cost=1) alter session set OPTIMIZER_INDEX_CACHING = 100; NESTED LOOPS (Cost=30 Card=225) TABLE ACCESS (FULL) OF 'T2' (Cost=15, Card=15) TABLE ACCESS (BY INDEX ROWID) OF 'T1' (Cost=1, Card=3000) INDEX (RANGE SCAN) OF 'T_I1' (NON-UNIQUE) We can improve silly assumption 2 (every logical I/O is also a physical I/O). Index blocks are often cached. So tell the optimizer how good our cache is.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Simplifications What about the blevel ? What about multi-column indexes ? What about unbounded ranges ? What about unique indexes ? What about bitmap indexes ? This was a visually helpful introduction This walk-through is intended to give you a gut-feeling of how the optimizer. works. But there are plenty of special cases, and bits of funny arithmetic.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 An improved approximation: blevel + selectivity * leaf_blocks + selectivity * clustering_factor (see Wolfgang Breitling's paper to IOUG-A 2002) For equality on all index columns: avg_leaf_blocks_per_key  sel * leaf_blocks avg_data_blocks_per_key  sel * clustering_factor For multi column indexes, or when using a range scan, we need more precise arithmetic - but even our example was a special case of the general formula

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Adjusted cost: ( (blevel + selectivity * leaf_blocks) * (1 - optimizer_index_caching/100) + selectivity * clustering_factor ) * optimizer_index_cost_adj / 100 The formula that Wolfgang Breitling proposed has to be adjusted to handle the two 'fudge factor' parameters. This formula seems to be about right. Index bit Table bit

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 System Statistics (v9) dbms_stats.gather_system_stats('start') dbms_stats.gather_system_stats('stop') SNAME PNAME PVAL1 SYSSTATS_MAIN CPUSPEED 357 MHz SYSSTATS_MAIN SREADTIM 7.179 ms SYSSTATS_MAIN MREADTIM 18.559 ms SYSSTATS_MAIN MBRC 5 Single block is cheaper than multi. Used in t/s cost. But in version 9 you need the 'fudge factors' less (You could still use them as indicators of caching) - instead, you let Oracle learn about your hardware

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Conclusions Understand your data Data distribution is important Think about your parameters Help Oracle with the truth Use system statistics in v9

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Sort / Merge select count(t1.v1)ct_v1, count(t2.v2)ct_v2 from big1t1,big2t2 wheret2.n2 = t1.n1; SELECT STATEMENT (choose) Cost (963) SORT (aggregate) MERGE JOIN Cost (963 = 174 + 789) SORT (join) Cost (174) TABLE ACCESS (analyzed) T1 (full) Cost (23) SORT (join) Cost (789) TABLE ACCESS (analyzed) T2 (full) Cost (115) The cost of a sort-merge equi-join is typically the cost of acquiring each of the two data sets, plus the cost of making sure the two data sets are sorted.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Sort Merge Table 1 Sort Table 2 Sort Merge To next step - e.g. order by Once the two sets are in order, they can be shuffled together. The shuffling can be quick - the sorting may be the most expensive bit.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 In-memory sort PGA UGA (will be in SGA for Shared Servers (MTS)) sort_area_retained_size Sort_area_size - sort_area_retained_size To disc sort_area_retained_size In a merge join, even if the first sort completes in memory, it will still dump the excess over sort_area_retained_size to disc. (and so will the second sort)

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Big Sorts Sort Merge A one-pass sort. The data has been read, sorted, and dumped to disc in chunks, then re-read once to be merged into order, and dumped again.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Huge Sorts Sort Merge 1Merge 2 Merge 3 Multipass sort. After sorting the data in chunks, Oracle was unable to re-read the top of every chunk simultaneously, so we have multiple merge passes.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Hash Join (1) First (smaller) data set Hashed Second (larger) data set The first table is hashed in memory, the second table is used to probe the hash (build) table for matches. In simple cases the cost is easy to calculate.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Hash Join (2) Small data set Hashed and partitioned Bit mapped Dump to disc Big data set Dump to disc If the smaller data set cannot be hashed in memory, it partitioned, mapped, and partly dumped to disc. The larger data set is partitioned in the same way

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Hash Join (3) Small data set Hashed and partitioned Bit mapped Dump to disc Big data set Dump to disc And if things go really wrong (bad statistics) Oracle uses partitions which are too large - and the probe (secondary) partitions are re-read many times.

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Version 9 approach v\$sysstat Sorts, hashes, bitmap creates (v.9) workarea executions - optimal The job completed in memory - perfect. workarea executions - onepass The job required a dump to disk and single re-read. workarea executions - multipass Data was dumped to disc and re-read more than once. Under Oracle 9, you should be setting workarea_size_policy to true, and use the pga_aggregate_target to something big - the limit per user is 5% pga_aggregate_target= 500M worksize_area_policy = auto

© Jonathan Lewis 2001 - 2003 NoCOUG 2003 Conclusions 2 Sort joins have catastrophe points Hash joins have catastrophe points Work to avoid multi-pass pga_aggregate_target helps (v9)