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So the database is saying, If we were to read every row in COLOCATED via the index COLOCATED_PK from start to finish, we would perform 1,190 I/Os. However, if we did the same to DISORGANIZED, we would perform 99,930 I/Os against the table. The reason for the large difference is that as Oracle range scans through the index structure, if it discovers the next row in the index is on the same database block as the prior row, it does not perform another I/O to get the table block from the buffer cache. It already has a handle to one and just uses it. However, if the next row is not on the same block, then it will release that block and perform another I/O into the buffer cache to retrieve the next block to be processed. Hence the COLOCATED_PK index, as we range scan through it, will discover that the next row is almost always on the same block as the prior row. The DISORGANIZED_PK index will discover the opposite is true. In fact, we can actually see this measurement is very accurate. If we hint to the optimizer to use an index full scan to read the entire table and just count the number of non-null Y values, we can see exactly how many I/Os it will take to read the entire table via the index: select count(Y) from (select /*+ INDEX(COLOCATED COLOCATED_PK) */ * from colocated) call count ------- -----Parse 1 Execute 1 Fetch 2 ------- -----total 4 cpu elapsed disk query current -------- ---------- ---------- ---------- ---------0.00 0.00 0 0 0 0.00 0.00 0 0 0 0.07 0.07 0 1399 0 -------- ---------- ---------- ---------- ---------0.07 0.07 0 1399 0 rows ---------0 0 1 ---------1
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Rows Row Source Operation ------- --------------------------------------------------1 SORT AGGREGATE (cr=1399 pr=0 pw=0 time=0 us) 100000 TABLE ACCESS BY INDEX ROWID COLOCATED (cr=1399 pr=0 pw=0 time=237977... 100000 INDEX FULL SCAN COLOCATED_PK (cr=209 pr=0 pw=0 time=81712 us cost=... ******************************************************************************** select count(Y) from (select /*+ INDEX(DISORGANIZED DISORGANIZED_PK) */ * from disorganized) call count ------- -----Parse 1 Execute 1 Fetch 2 ------- -----total 4 Rows ------1 100000 100000 cpu elapsed disk query current -------- ---------- ---------- ---------- ---------0.00 0.00 0 0 0 0.00 0.00 0 0 0 0.27 0.27 0 100139 0 -------- ---------- ---------- ---------- ---------0.27 0.27 0 100139 0 rows ---------0 0 1 ---------1
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Row Source Operation --------------------------------------------------SORT AGGREGATE (cr=100139 pr=0 pw=0 time=0 us) TABLE ACCESS BY INDEX ROWID DISORGANIZED (cr=100139 pr=0 pw=0 time=... INDEX FULL SCAN DISORGANIZED_PK (cr=209 pr=0 pw=0 time=86955 us ...
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In both cases, the index needed to perform 209 logical I/Os (cr=209 in the Row Source Operation lines). If you subtract 209 from the total consistent reads and measure just the number of I/Os against the table, then you ll find that they are identical to the clustering factor for each respective index. The COLOCATED_PK is a classic the table is well ordered example, whereas the DISORGANIZED_PK is a classic the table is very randomly ordered example. It is interesting to see how this affects the optimizer now. If we attempt to retrieve 25,000 rows, Oracle will now choose a full table scan for both queries (retrieving
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25 percent of the rows via an index is not the optimal plan, even for the very ordered table). However, if we drop down to 10 percent (bear in mind that 10 percent is not a threshold value it is just a number less than 25 percent that caused an index range scan to happen in this case) of the table data: ops$tkyte%ORA11GR2> set autotrace traceonly explain ops$tkyte%ORA11GR2> select * from colocated where x between 20000 and 30000; Execution Plan ---------------------------------------------------------Plan hash value: 1550765370 --------------------------------------------------------------------------------| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| --------------------------------------------------------------------------------| 0 | SELECT STATEMENT | | 10002 | 791K| 142 (0)| | 1 | TABLE ACCESS BY INDEX ROWID| COLOCATED | 10002 | 791K| 142 (0)| |* 2 | INDEX RANGE SCAN | COLOCATED_PK | 10002 | | 22 (0)| --------------------------------------------------------------------------------Predicate Information (identified by operation id): --------------------------------------------------2 - access("X">=20000 AND "X"<=30000) ops$tkyte%ORA11GR2> select * from disorganized where x between 20000 and 30000; Execution Plan ---------------------------------------------------------Plan hash value: 2727546897 ---------------------------------------------------------------------------------| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | ---------------------------------------------------------------------------------| 0 | SELECT STATEMENT | | 10002 | 791K| 333 (1)| 00:00:04 | |* 1 | TABLE ACCESS FULL| DISORGANIZED | 10002 | 791K| 333 (1)| 00:00:04 | ---------------------------------------------------------------------------------Predicate Information (identified by operation id): --------------------------------------------------1 - filter("X"<=30000 AND "X">=20000) Here we have the same table structures the same indexes but different clustering factors. The optimizer in this case chose an index access plan for the COLOCATED table and a full scan access plan for the DISORGANIZED table. The key point to this discussion is that indexes are not always the appropriate access method. The optimizer may very well be correct in choosing to not use an index, as the preceding example demonstrates. Many factors influence the use of an index by the optimizer, including physical data layout. You might be tempted therefore to run out and try to rebuild all of your tables now to make all indexes have a good clustering factor, but that would be a waste of time in most cases. It will affect cases where you do index range scans of a large percentage of a table. Additionally, you must keep in mind that, in general, the table will have only one index with a good clustering factor! The rows in a table may be sorted in only one way. In the example just shown, if I had another index on the column Y it would be
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