Oracle b tree index internals - rebuilding the thruth

Copyright: Richard Foote Consulting
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Oracle BOracle B--Tree Index Internals:Tree Index Internals:
Rebuilding The TruthRebuilding The Truth
Richard Foote

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ObjectivesObjectives
Dispel many myths associated with Oracle
B-Tree Indexes
Explain how to investigate index internals
Explain and prove how Oracle B-Tree
Indexes work
Explain when index rebuilds might be
appropriate

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““ExpertExpert”” quotes regarding Indexesquotes regarding Indexes
“Note that Oracle indexes will spawn to a fourth
level only in areas of the index where a massive
insert has occurred, such that 99% of the index
has three levels, but the index is reported as
having four levels. ” Don Burleson:
comp.databases.oracle.server newsgroup post dated
31st January 2003
“If the index clustering factor is high, an index
rebuild may be beneficial” Don Burleson: Inside
Oracle Indexing dated December 2003 at
www.DBAzine.com

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“The binary height increases mainly due to the size of the
table and the fact that the range of values in the indexed
columns is very narrow”. Richard Niemiec Oracle
Performance Tuning 1999.
“The index will be imbalanced if the growth is all on one
side such as when using sequence numbers as keys…
Reading the new entries will take longer” Richard Niemiec
Tuning for the Advanced DBA; Others will Require Oxygen 2001
“This tells us a lot about indexes, but what interests me is
the space the index is taking, what percentage of that is
really being used and what space is unusable because of
delete actions. Remember, that when rows are deleted, the
space is not re-used in the index.” John Wang: Resizing Your
Indexes When Every Byte Counts at www.DBAzine.com

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Index diagram showing an “unbalanced” Oracle index with leaf
nodes to the right of the index structure having more levels than
leaf nodes to the left. Mike Hordila: Setting Up An Automated Index
Rebuilding System at otn.oracle.com
“Deleted space is not reclaimed automatically unless there is an
exact match key inserted. This leads to index broadening and
increase in the indexes clustering factor. You need to reorganize
to reclaim white space. Generally rebuild index when the
clustering factor exceeds eight times the number of dirty blocks in
the base table, when the levels exceed two or when there are
excessive brown nodes in the index”” Mike Ault Advanced Oracle
Tuning Seminar at www.tusc.com/oracle/download/author_aultm.html

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Classic Oracle Index MythsClassic Oracle Index Myths
Oracle B-tree indexes can become “unbalanced” over time
and need to be rebuilt
Deleted space in an index is “deadwood” and over time
requires the index to be rebuilt
If an index reaches “x” number of levels, it becomes
inefficient and requires the index to be rebuilt
If an index has a poor clustering factor, the index needs to
be rebuilt
To improve performance, indexes need to be regularly
rebuilt

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Basic BBasic B--Tree IndexTree Index
Null – B1
Move – B2
Null – L1
Bolan – L2
Floyd – L3
Move – L4
Queen – L5
Ziggy – L6
ABBA rowid
ACDC rowid
Bolan rowid
Bowie rowid
Clash rowid
Floyd rowid
Kinks rowid
Lennon rowid
Move rowid
Police rowid
Pop rowid
Queen rowid
Reed rowid
Velvet rowid
Ziggy rowid
Root Block
Branch Blocks
Leaf Blocks

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Index FundamentalsIndex Fundamentals
Oracle’s implements a form of B*Tree index
Oracle’s B-Tree index is always balanced. Always.
Index entries must always be ordered.
An update consists of a delete and an insert
Each leaf block has pointers to next/previous blocks
Each leaf block contains the index entry with
corresponding rowid
Index scans use ‘sequential’, single block reads
(with exception of Fast Full Index Scan)

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TreedumpTreedump
alter session set events ‘immediate trace name treedump level 12345’;
where 12345 is the index object id
----- begin tree dump
branch: 0x8405dde 138436062 (0: nrow: 3, level: 3)
branch: 0xdc11022 230756386 (-1: nrow: 219, level: 2)
branch: 0x8405f15 138436373 (-1: nrow: 138, level: 1)
leaf: 0x8405ddf 138436063 (-1: nrow: 21 rrow: 21)
leaf: 0x8405de0 138436064 (0: nrow: 18 rrow: 13)
leaf: 0x8405de2 138436066 (1: nrow: 15 rrow: 15)
block type (branch or leaf) and corresponding rdba,
position within previous level block (starting at –1 except root starting at 0)
nrows: number of all index entries
rrows: number of current index entries
level : branch block level (leaf block implicitly 0)
Some versions of Oracle displays a block dump of all index leaf blocks
Treedumps perfectly highlight that indexes are always balanced and the
number of levels to all leaf blocks is always consistent

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Block DumpBlock Dump
To create formatted dumps of blocks:
alter system dump datafile 6 block 10;
alter system dump datafile 6 block min 5 block max 10
Creates the dump file in user_background_dest
To determine the datafile and block from a rba:
select DBMS_UTILITY.DATA_BLOCK_ADDRESS_FILE(138436069),
DBMS_UTILITY.DATA_BLOCK_ADDRESS_BLOCK(138436069)
from dual;

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Block HeaderBlock Header
Start dump data blocks tsn: 39 file#: 2 minblk 467 maxblk 467
buffer tsn: 39 rdba: 0x008001d3 (2/467)
scn: 0x0000.043be543 seq: 0x02 flg: 0x00 tail: 0xe5430602
frmt: 0x02 chkval: 0x0000 type: 0x06=trans data
Block header dump: 0x008001d3
Object id on Block? Y
seg/obj: 0x7c74 csc: 0x00.43be543 itc: 1 flg: - typ: 2 - INDEX
fsl: 0 fnx: 0x0 ver: 0x01
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x000d.007.00007f63 0x03401186.067c.03 C--- 0 scn 0x0000.043be543
rdba - relative database block address of the branch block (file no/block no)
scn – system change number of the block when last changed
seq – number of block changes
type – block type
seg/obj – object id
typ – segment type (index)
Itl – Interested transaction Slots (default 2 for leaf blocks) including slot id,
transaction id, undo block address, flag and locking info and scn of transaction

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Common Index Header SectionCommon Index Header Section
header address 50794564=0x3071044
kdxcolev 1
KDXCOLEV Flags = - - -
kdxcolok 0
kdxcoopc 0x80: opcode=0: iot flags=--- is converted=Y
kdxconco 2
kdxcosdc 1
kdxconro 237
kdxcofbo 502=0x1f6
kdxcofeo 3120=0xc30
kdxcoavs 2618
kdxcolev: index level (0 represents leaf blocks)
kdxcolok: denotes whether structural block transaction is occurring
kdxcoopc: internal operation code
kdxconco: index column count
kdxcosdc: count of index structural changes involving block
kdxconro: number of index entries (does not include kdxbrlmc pointer)
kdxcofbo: offset to beginning of free space within block
kdxcofeo: offset to the end of free space (ie. first portion of block containing index data)
kdxcoavs: available space in block (effectively area between the two fields above)

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Branch Header SectionBranch Header Section
kdxbrlmc 8388627=0x800013
kdxbrsno 92
kdxbrbksz 8060
kdxbrlmc: block address if index value is less than the first (row#0) value
kdxbrsno: last index entry to be modified
kdxbrbksz: size of usable block space

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Leaf Header SectionLeaf Header Section
kdxlespl 0
kdxlende 0
kdxlenxt 8388628=0x800014
kdxleprv 8389306=0x8002ba
kdxledsz 0
kdxlebksz 8036
kdxlespl: bytes of uncommitted data at time of block split that have been cleaned out
kdxlende: number of deleted entries
kdxlenxt: pointer to the next leaf block in the index structure via corresponding rba
kdxleprv: pointer to the previous leaf block in the index structure via corresponding rba
kdxlebksz: usable block space (by default less than branch due to the additional ITL entry)

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Branch EntriesBranch Entries
row#0[4813] dba: 8388865=0x800101
col 0; len 12; (12): 41 6c 61 64 64 69 6e 20 53 61 6e 65
col 1; len 4; (4): 00 80 00 f6
row#1[4578] dba: 8389115=0x8001fb
col 0; len 12; (12): 41 6c 61 64 64 69 6e 20 53 61 6e 65
col 1; len 4; (4): 00 80 01 f1
Row number (starting at 0) [starting location in block] dba
Column number followed by column length followed by column value
Repeated for each indexed index
Repeated for each branch entry
Note: column value is abbreviated to smallest value that uniquely
defines path

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Leaf EntriesLeaf Entries
row#0[4616] flag: ----S, lock: 2
col 0; len 3; (3): 4c 6f 77
col 1; len 6; (6): 00 80 02 b5 00 6a
row number (starting at 0) followed by [starting location within block]
followed by various flags (locking information, deletion flag etc.)
index column number (starting at 0) followed by column length
followed by column value
repeated for each indexed column
repeated for each index entry

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Update of Index EntryUpdate of Index Entry
SQL> create table test_update (id number, name varchar2(10));
Table created.
SQL> create index test_update_idx on test_update (name);
Index created.
SQL> insert into test_update values (1, 'BOWIE');
1 row created.
SQL> commit;
Commit complete.
SQL> update test_update set name = 'ZIGGY' where id = 1;
1 row updated.
SQL> commit;
Commit complete.
SQL> select file_id, block_id from dba_extents where segment_name='TEST_UPDATE_IDX';
FILE_ID BLOCK_ID
---------- ---------------
12 83193
SQL> alter system dump datafile 12 block 83194;
System altered.
Note: add 1 to block_id else the segment header is dumped

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Block Dump After Index UpdateBlock Dump After Index Update
kdxconco 2
kdxcosdc 0
kdxconro 2
kdxcofbo 40=0x28
kdxcofeo 8006=0x1f46
kdxcoavs 7966
kdxlespl 0
kdxlende 1
kdxlenxt 0=0x0
kdxleprv 0=0x0
kdxledsz 0
kdxlebksz 8036
row#0[8021] flag: ---D-, lock: 2 => deleted index entry
col 0; len 5; (5): 42 4f 57 49 45
col 1; len 6; (6): 00 80 05 0a 00 00
row#1[8006] flag: -----, lock: 2
col 0; len 5; (5): 5a 49 47 47 59 => new index entry
col 1; len 6; (6): 00 80 05 0a 00 00

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Index StatisticsIndex Statistics
dba_indexes
– analyze command, or better still
– dbms_stats package
index_stats
– analyze index index_name validate structure;
– resource intensive, locking issues
v$segment_statistics (9.2)
– statistics_level = typical (or all)

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Statistics NotesStatistics Notes
blevel (dba_indexes) vs. height (index_stats)
blocks allocated, not necessarily used
lf_rows_len inclusive of row overheads (e.g. 12 bytes single
column index)
pct_used amount of space currently used in index structure:
(used_space/btree_space)*100. Note: index wide
Most index stats are inclusive of deleted entries:
– non-deleted rows = lf_rows – del_lf_rows
– pct_used by non-deleted rows = ((used_space – del_lf_rows_len) /
btree_space) * 100
And then there’s the clustering_factor

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Clustering FactorClustering Factor
A vital statistic used by the CBO
Determines the relative order of the table in relation to the
index
CF value corresponds to likely physical I/0s or blocks
visited during a full index scan (note same block could be
visited many times)
If the same block is read consecutively then Oracle
assumes only the 1 physical I/0 is necessary
The better the CF, the more efficient the access via the
corresponding index as less physical I/Os are likely
“Good” CF generally has value closer to blocks in table
“Bad” CF generally has a value closer to rows in table

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Index with Bad Clustering FactorIndex with Bad Clustering Factor

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Index with Good Clustering FactorIndex with Good Clustering Factor

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How To Improve The CFHow To Improve The CF
As some of the “expert” quotes suggest, rebuild index if
CP is poor is common advice
Unfortunately, as neither table nor index order changes,
the net effect is “disappointing”
To improve the CF, it’s the table that must be rebuilt
(and reordered)
If table has multiple indexes, careful consideration
needs to be given by which index to order table
Pre-fetch index reads improves poor CF performance
Rebuilding an index simply because it has a CF over a
certain threshold is futile and a silly myth

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Index Creation and PCTFREEIndex Creation and PCTFREE
When an index is created, Oracle reserves the pctfree
value as free space
pctfree has a default of 10% resulting in 10% of an
index remaining free after creation
Why ?
To reduce and delay the occurrence of index block splits
If there isn’t sufficient space in an index block for the
new entry, a block split is performed

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5050--50 Block Split50 Block Split
1
2
3
4
1
2
3
4
5

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5050--50 Leaf Block Block Steps50 Leaf Block Block Steps
An index block split is a relatively expensive operation:
1. Allocate new index block from index freelist
2. Redistribute block so the lower half (by volume) of index entries
remain in current block and move the other half into the new block
3. Insert the new index entry into appropriate leaf block
4. Update the previously full block such that its “next leaf block
pointer” (kdxlenxt) references the new block
5. Update the leaf block that was the right of the previously full block
such that its “previous leaf block pointer”(kdxleprv) also points to
the new block
6. Update the branch block that references the full block and add a
new entry to point to the new leaf block (effectively the lowest
value in the new leaf block)

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5050--50 Branch Block Splits50 Branch Block Splits
Insert operation is even more expensive if the
corresponding branch block is also full:
1. Allocate a new index block from the freelist
2. Redistribute the index entries in the branch block that is currently
full such that half of the branch entries (the greater values) are
placed in the new block
3. Insert the new branch entry into the appropriate branch block
4. Update the branch block in the level above and add a new entry
to point to the new branch block

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5050--50 Root Block Split50 Root Block Split
Root block is just a special case of a branch block:
1. Allocate two new blocks from the freelist
2. Redistributed the entries in the root block such that half the entries are
placed in one new block, the other half in the other block
3. Update the root block such that it now references the two new blocks
The root block is always physically the same block.
The root block split is the only time when the height of index increases
Therefore an index must always be balanced. Always !!
Suggestions that Oracle indexes become unbalanced are another silly
myth, made by those that don’t understand index block splits.

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Root Block Always The SameRoot Block Always The Same
SQL> create table same_root (id number, value varchar2(10));
Table created.
SQL> insert into same_root values (1, 'Bowie');
1 row created.
SQL> commit;
Commit complete.
SQL> create index same_root_idx on same_root(id);
Index created.
leaf: 0x800382 8389506 (0: nrow: 1 rrow: 1)
----- end tree dump
Then add enough rows to cause the index structure grow and root block to split….
branch: 0x800382 8389506 (0: nrow: 2, level: 1)
leaf: 0x800383 8389507 (-1: nrow: 540 rrow: 540)
leaf: 0x800384 8389508 (0: nrow: 460 rrow: 460)
----- end tree dump

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9090--10 Block Splits10 Block Splits
If the new insert index entry is the maximum
value, a 90-10 block split is performed
Reduces wastage of space for index with
monotonically increasing values
Rather than leaving behind ½ empty blocks, full
index blocks are generated
I prefer to call them 99-1 block splits as 90-10 is
misleading

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9090--10 Splits10 Splits
1
2
3
4
1
2
3
4
5

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9090--10 Splits With 9i10 Splits With 9i
Spot the difference: Case 1
SQL> create table split_90_a (id number, value varchar2(10));
Table created.
SQL> create index split_90_a_idx on split_90_a(id);
Index created.
SQL> begin
2 for i in 1..10000 loop
3 insert into split_90_a values (i, 'Bowie');
4 end loop;
5 commit;
6 end;
7 /
PL/SQL procedure successfully completed.
SQL> analyze index split_90_a_idx validate structure;
Index analyzed.
SQL> select lf_blks, pct_used from index_stats;
LF_BLKS PCT_USED
---------- ----------
19 94

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9090--10 Splits With 9i10 Splits With 9i
Spot the difference: Case 2
SQL> create table split_90_a (id number, value varchar2(10));
Table created.
SQL> create index split_90_a_idx on split_90_a(id);
Index created.
SQL> begin
3 insert into split_90_a values (i, 'Bowie');
4 commit;
5 end loop;
6 end;
7 /
SQL> analyze index split_90_a_idx validate structure;
Index analyzed.
SQL> select lf_blks, pct_used from index_stats;
LF_BLKS PCT_USED
---------- ----------
36 51

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Deleted Index SpaceDeleted Index Space
When a delete (or update) is performed, Oracle marks the entry as deleted.
Relevant portions of a block dump:
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x000d.002.00007f5d 0x03400450.0671.01 CB-- 0 scn 0x0000.043be316
0x02 0x0014.029.00007e87 0x034005a3.0ac2.0e --U- 100 fsc 0x0960.043f9d45
kdxlende 100
kdxlenxt 8388865=0x800101
kdxleprv 0=0x0
kdxledsz 0
kdxlebksz 8036
row#0[4252] flag: ---D-, lock: 2
col 0; len 12; (12): 41 6c 61 64 64 69 6e 20 53 61 6e 65
col 1; len 6; (6): 00 80 00 0b 01 33
del_lf_rows and del_lf_rows_len in index_stats provide deletion stats

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Deleted SpaceDeleted Space –– Reused ?Reused ?
Create a table/index, insert 10 rows (indexed values 1,2,3,4,5,6,7,8,9,10),
commit, deleted 4 rows (2,4,6,8), commit;
Index_stats shows:
LF_ROWS DEL_LF_ROWS DEL_LF_ROWS_LEN USED_SPACE
-------------- --------------------- ----------------------------- -------------------
10 4 56 140
Treedump shows:
leaf: 0x8002ea 8389354 (0: nrow: 10 rrow: 6)
Block dump shows:
kdxconro 10
kdxlende 4
row#1[8012] flag: ---D-, lock: 2

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Deleted SpaceDeleted Space –– Reused ?Reused ?
Insert a new row (value 100) that is different and not within ranges of those
previously deleted, commit;
index_stats shows:
-------------- --------------------- ----------------------------- -------------------
7 0 0 98
Treedump shows:
Block dump shows:
kdxconro 7
kdxlende 0
All deleted index entries removed

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Deleted Index Space Is ReusedDeleted Index Space Is Reused
The previous example clearly illustrates that
any insert to a leaf block removes all
deleted entries
In randomly inserted indexes, deleted space
is not an issue as it will eventually be reused
But wait, there’s more …

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Deleted EntriesDeleted Entries –– Delayed Block CleanoutDelayed Block Cleanout
Same example as before:
create a table/index, insert values (1,2,3,4,5,6,7,8,9,10), commit;
delete 4 rows, values (2,4,6,8);
alter session set events ‘immediate trace name flush_cache’;
Index_stats shows:
-------------- --------------------- ----------------------------- -------------------
6 0 0 84
Treedump shows:
Block dump shows:
kdxconro 6
kdxlende 0
All deleted index entries removed

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Deleted Entries: DelayedDeleted Entries: Delayed
Block CleanoutBlock Cleanout
Long running transactions may result in dirty
blocks being flushed from memory before a
commit;
When subsequently accessed, delayed block
cleanout is performed
Delayed block cleanout results in all
corresponding deleted entries being cleaned out

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Deleted Leaf BlocksDeleted Leaf Blocks –– Reused ?Reused ?
Simple example to demonstrate if deleted leaf blocks are reused
SQL> create table test_empty_block (id number, value varchar2(10));
Table created.
SQL> begin
3 insert into test_empty_block values (i, 'Bowie');
4 end loop;
5 commit;
6 end;
7 /
SQL> create index test_empty_block_idx on test_empty_block (id);
Index created.

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SQL> delete test_empty_block where id between 1 and 9990;
9990 rows deleted.
SQL> commit;
Commit complete.
SQL> analyze index test_empty_block_idx validate structure;
Index analyzed.
SQL> select lf_blks, del_lf_rows from index_stats;
LF_BLKS DEL_LF_ROWS
---------- -----------
21 9990
Therefore all blocks except (probably) the last block holding the last 10
values are effectively empty.

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Now reinsert a similar volume but after the last current values
SQL> begin
2 for i in 20000..30000 loop
3 insert into test_empty_block values (i, 'Bowie');
4 end loop;
5 commit;
6 end;
7 /
SQL> analyze index test_empty_block_idx validate structure;
Index analyzed.
SQL> select lf_blks, del_lf_rows from index_stats;
LF_BLKS DEL_LF_ROWS
---------- -----------
21 0
Note all empty blocks have been reused and deleted rows cleanout.

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Empty Blocks Not UnlinkedEmpty Blocks Not Unlinked
Following select statement was executed after the 9990 deletions in previous example
SQL> select /*+ index test_empty_blocks */ * from test_empty_blocks
where id between 1 and 100000;
10 rows selected.
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE
1 0 TABLE ACCESS (BY INDEX ROWID) OF 'TEST_EMPTY_BLOCKS'
2 1 INDEX (RANGE SCAN) OF 'TEST_EMPTY_BLOCKS_IDX' (NON-UNIQUE)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
28 consistent gets
0 physical reads
0 redo size
549 bytes sent via SQL*Net to client
499 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
10 rows processed

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Conclusion: Deleted SpaceConclusion: Deleted Space
Deleted space is cleaned out by subsequent writes
Deleted space is cleaned out by delayed block
cleanout
Fully emptied blocks are placed on freelist and
recycled (although remain in the index structure)
Suggestions that deleted space can never be reused
are wrong and yet another silly myth

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Index FragmentationIndex Fragmentation
Although deleted index space is generally reusable,
there can be wasted space:
– Bug with 90-10 split algorithm in 9i (previously discussed)
– Too high PCTFREE
– Permanent table shrinkage
– Monotonically increasing index values and deletions
– Deletes or Updates that dramatically reduce occurrence
of specific index values
– Large volume of identical index entries

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Too High PCTFREEToo High PCTFREE
create index bowie_idx on bowie (id) pctfree 95;

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Permanent Table ShrinkagePermanent Table Shrinkage
But note that the table would look as follows:
Therefore it’s the table rather than just the indexes that should be rebuilt.
Current entries
Deleted entries

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Monotonically increasingMonotonically increasing
index values and deletionsindex values and deletions
• As previously discussed, fully deleted blocks are recycled
• Therefore it’s sparse deletions with monotonically increasing entries
Current entries
Deleted entries

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Deletes/Updates reduceDeletes/Updates reduce
occurrence of index valuesoccurrence of index values
• Similar to previous example but a range of values permanently deleted
• Again, sparse deletions as fully deleted leaf blocks are recycled

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Large Volume Of Identical ValuesLarge Volume Of Identical Values
Null – L1
AAA – L2
AAA – L3
BBB – L4
BBB – L5
BBB – L6
BBB – L7
AAA rowid
AAA rowid
AAA rowid
AAA rowid
AAA rowid
AAA rowid
BBB rowid
BBB rowid
BBB rowid
BBB rowid
BBB rowid
BBB rowid
BBB rowid
BBB rowid
BBB rowid
BBB rowid
CCC rowid
CCC rowid
To insert a new value for AAA, Oracle looks to the branch block and
performs the following logic:
1. Is AAA less than AAA. No, it is not less than AAA.
2. Is AAA greater or equal to AAA and less than AAA. No it is not less than AAA
3. Is AAA greater or equal to AAA and less than BBB. Yes, therefore the entry goes
into L3.

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Large Volume Of Identical ValuesLarge Volume Of Identical Values
There are a number of issues with this behaviour:
1. The only leaf blocks that can be inserted into are:
• L3 for values of AAA
• L7 for values of BBB or CCC
i.e. the last leaf blocks containing a specific value
2. All other leaf blocks are “isolated” in that they
cannot be considered by subsequent inserts
(assuming only current values)
3. The isolated blocks are ½ empty due to 50-50
block splits

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Index RebuildsIndex Rebuilds
• As discussed, most indexes are efficient at allocating and reusing space
• Randomly inserted indexes operate at average 25% free space
• Monotonically increasing indexes operate at close to 0% free space
• Deleted space is generally reusable
• Only in specific scenarios could unused space be expected to be higher
and remain unusable
• So when may index rebuilds be necessary ?

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Index RebuildsIndex Rebuilds
• An index rebuild should only be considered under the following
general guideline:
“The benefits of rebuilding the index are greater than the
overall costs of performing such a rebuild”
• Another way to look at this:
“If there are no measurable performance benefits of
performing an index rebuild, why bother ?”
• Another important point:
“If after a rebuild, the index soon reverts back to it’s previous
state, again why bother ?”
• The basic problem with index rebuilds improving performance is
that generally, the ratio of index blocks visited to table blocks
visited is relatively small.

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Index vs. Table Block VisitsIndex vs. Table Block Visits
Let’s look first at a theoretical example.
Index structure before rebuild – pct_used only 50%:
• Height = 3
• Branch blocks = 50 (+ 1 branch block for the root)
• Index Leaf blocks = 20,000
Index structure after rebuild – pct_used now 100%:
• Height = 3
• Branch blocks = 25 (+1 branch block for root)
• Index Leaf Blocks = 10,000
Table structure:
• Blocks = 100,000
• Rows = 1,000,000

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Example 1 – Single row select on unique index
Cost before rebuild = 1 root + 1 branch + 1 leaf + 1 table = 4 LIOs
Cost after rebuild = 1 root + 1 branch + 1 leaf + 1 table = 4 LIOs
Net benefit = 0% Note CF has no effect in this example
Example 2 – Range scan on 100 selected rows (0.01% selectivity)
Before Cost with worst CF = 1 rt + 1 br + 0.0001*20000 (2 leaf) + 100 table = 104 LIOs
After Cost with worst = 1 rt + 1 br + 0.0001*10000 (1 leaf) + 100 table = 103 LIOs
Net benefit = 1 LIO or 0.96%
Before Cost with best CF = 1 rt + 1 br + 0.0001*20000 (2 leaf) + 0.0001*100000 (10
table) = 14 LIOs
After Cost with best CF = 1 rt + 1 br + 0.0001*10000 (1 leaf) + 10 table = 13 LIOs
Net benefit = 1 LIO or 7.14%

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Example 3 – range scan on 10000 selected rows (1% selectivity)
Before cost (worst CF) = 1 rt + 1 br + 0.01*20000 (200 lf) + 10000 table = 10202 LIOs
After cost (worst CF) = 1 rt + 1 br + 0.01*10000 (100 lf) + 10000 table = 10102 LIOs
Net benefit = 100 LIOs or 0.98%
Before cost (best CF) = 1 rt + 1 br + 0.01*20000 (200 lf) + 0.01*100000 (1000 tbl) =
1202 LIOs
After cost (best CF) = 1 rt + 1 br + 0.01*10000 (100 lf) + 1000 table = 1102 LIOs
Net benefit = 100 LIOs 8.32%
Example 4 – range scan on 100000 select rows (10% selectivity)
Before cost (worst CF) = 1 rt + 1 br + 0.1*20000 (2000 lf) + 100000 (tbl) = 102002 LIOs
After cost (worst CF) = 1 rt + 1 br + 0.1*10000 (1000 lf) + 100000 tbl = 101002 LIOs
Before cost (best CF) = 1 rt + 1 br + 0.1*20000 (2000 lf) + 0.1*100000 (10000 tbl) =
12002 LIOs
After cost (best CF) = 1 rt + 1 br + 0.1*10000 (1000 lf) + 10000 tbl = 11002 LIOs

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Example 5 – Fast Full Index Scan (100% selectivity) assuming average 10
effective multiblock reads
Cost before rebuild = (1 root + 50 branch + 20000 leaf) / 10 = 2006 I/Os
Cost after rebuild = (1 root + 25 branch + 10000 leaf) / 10 = 1003 I/O
Net benefit = 1003 LIOs or 50%

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Index vs. Table Block Visit:Index vs. Table Block Visit:
ConclusionsConclusions
• If an index accesses a ‘small’ % of rows, index fragmentation is
unlikely to be an issue
• As an index accesses a ‘larger’ % of rows, the number of Index
LIOs increases but the ratio of index reads to table reads remains
constant
• Therefore caching characteristics of index becomes crucial as the
size of index and % of rows accessed increases
• The clustering factor of the index is an important variable in the
performance of an index and the possible effects of index
fragmentation as it impacts the ratio of index/table blocks accessed
• Index Fast Full Scans are likely to be most impacted by index
fragmentation as access costs are directly proportional to index size

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Index Rebuild: Case Study 1Index Rebuild: Case Study 1
Non ASSM, 8K block size tablespace
Index created with a perfect CF
Indexed columns represents just over 10% of
table columns
Test impact of differing index fragmentation
on differing cardinality queries

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create table test_case (id number, pad char(50), name1 char(50), name2
char(50), name3 char(50), name4 char(50), name5 char(50), name6
char(50), name7 char(50), name8 char(50), name9 char(50));
begin
For i in 1..1000000 loop
insert into test_case values (i, '**************************************************',
'David Bowie', 'Ziggy Stardust', 'Major Tom', 'Thin White Duke', 'Aladdin
Sane', 'David Jones', 'John', 'Sally', 'Jack');
End loop;
End;
/
create index test_case_idx on test_case(id, pad) pctfree 50;

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SQL> select * from test_case where id = 1000; -- select 1 row
SQL> select * from test_case where id between 100 and 199; -- select 100 rows
SQL> select * from test_case where id between 2000 and 2999; -- select 1,000 rows
SQL> select /*+ index(test_case) */ * from test_case where id between 300000 and
399999; -- select 100,000 rows
SQL> select /*+ index(test_case) */ id from test_case where id between 1 and 1000000;
-- select 1,000,000 rows
SQL> select /*+ index_ffs(test_case) */ id, pad from test_case where id between 1 and
1000000; -- select 1,000,000 rows via a Fast Full Index Scan
Note: Statements run several times to reduce parsing and caching differences

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The table had a total of 1,000,000 rows in 76,870 blocks while the index
had a clustering factor of 76,869 (i.e. perfect).
Index recreated with differing pctfree values and tests rerun.
HEIGHT BR_BLKS LF_BLKS PCT_USED
0 pctfree 3 14 8,264 100
25 pctfree 3 18 11,110 75
50 pctfree 3 27 16,947 49
75 pctfree 3 55 35715 24

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Case Study 1 % of Table 0 pctfree 25 pctfree 50 pctfree 75 pctfree
1 row 0.001% 0:0.01 0:0.01 0:0.01 0:0.01
100 rows 0.01% 0:0.01 0:0.01 0:0.01 0:0.01
1,000 rows 0.1% 0:0.01 0:0.01 0:0.01 0:0.01
10,000 rows 1% 0:0.03 0:0.03 0:0.03 0:0.03
50,000 rows 5% 0:10:05 0:12.06 0:17.00 0:18:06
100,000 rows 10% 0:22:06 0:23.08 0:27.09 0:33.06
1,000,000 rows 100% 3:57:09 4:43:03 5:12.05 5:29:08
1,000,000 FFS 100% 0:18.06 0:19:02 0:24.01 0:36.06

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Case Study 1: CommentsCase Study 1: Comments
• All indexes resulted in identical executions plans
• No difference with statements that access < 10,000 rows
• Differences emerge between 10000 and 50000 due to caching
restrictions (25,000 approximate point of some differences)
• 50000 rows marks point where index hint required to force use
of hint, therefore index issues somewhat redundant
• General exception Index Fast Full Scan where performance is
most effected and directly proportional is index size
• Summary: queries up to 25,000 rows (2.5%) little to no
difference, 25,000 – 50,000 some differences emerged, 50,000+
index not used anyway

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Similar to case 1 but importantly with a
much worse CF
Also size of index designed to increase
index height when poorly fragmented
Non-Unique values results in less efficient
branch entries management as second pad
column required

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begin
insert into test_case2 values (0, '**************************************************', 'David Bowie', …);
for a in 1..100 loop
for b in 1..10 loop
for c in 1..10 loop
for d in 1..5 loop
for d in 1..2 loop
for e in 1..10 loop
end loop;
end loop;
end loop;
end loop;
end loop;
end loop;
commit;
end;
/

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SQL> select * from test_case2 where id = 0; -- 1 row
SQL> select * from test_case2 where id = 1; -- 100 rows
SQL> select * from test_case2 where id = 2; -- 1,000 rows
SQL> select * from test_case2 where id = 3; -- 10,000 rows
SQL> select /*+ index (test_case2) */ * from test_case2 where id = 4; -- 50,000 rows
SQL> select /*+ index (test_case2) */ * from test_case2 where id = 5; -- 100,000 rows
SQL> select /*+ index (test_case2) */ * from test_case2 where id = 6; -- 1,000,000 rows
SQL> select /*+ ffs_index (test_case2) */ id, pad from test_case2 where id = 6; --
1,000,000 rows
Note: Statements run several times to reduce parsing and caching differences.
May not necessarily be appropriate with first execution times being relevant.

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The table had a total of 1,161,101 rows in 82,938 blocks while the index
had a clustering factor of 226,965 (i.e. much worse than case 1).
HEIGHT BR_BLKS LF_BLKS PCT_USED
0 pctfree 3 79 9,440 100
25 pctfree 3 107 12,760 75
50 pctfree 4 163 19,352 49
75 pctfree 4 346 41,468 24

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Test Case 2 % of Table 0%
pctfree
25% pctfree 50% pctfree 75% pctfree
1 row 0.000086% 0:0.01 0:0.01 0:0.01 0:0.01
100 rows 0.0086% 0:0.01 0:0.01 0:0.01 0:0.01
1,000 rows 0.086% 0:0.01 0:0.01 0:0.01 0:0.01
10,000 rows 0.86% 0:27:02 0:27.06 0:29.03 0:35.03
50,000 rows 4.31% 0:41:00 0:42.03 0:53.07 1:03.04
100,000 rows 8.61% 0:52.03 1:01:04 1:20.08 2:08.02
1,000,000 rows 86.12%
4:01.07 4:57:02 5:54:00 6:12:02
1,000.000 FFS 86.12% 0:18.01 0:21:09 0:23.07 0:36.05

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Case Study 2: CommentsCase Study 2: Comments
• Index is clearly less efficient and generates slower execution times
for statements > 1,000 rows
• All indexes resulted in identical executions plans
• No difference with statements that access < 1,000 rows
• Differences emerge with >= 10,000 rows although only significantly
so for pct_free >= 50%
• Because of the poorer CF, 10,000 rows marks the boundary where
index is automatically used by the CBO
• Again, Fast Full Index Scan performance directly proportional to
index size
• Summary: index only an issue for a very narrow cases with between
10,000 – 50,000 rows and 50%+ pct_free

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Index SelectivityIndex Selectivity ““ZonesZones””
Green Zone: Index fragmentation makes no difference
because of low LIOs and high caching characteristics making
index rebuilds pointless. Most OLTP queries belong here.
Orange Zone: Selectivity generates significant index I/Os
and index caching is reduced. Likelihood increases closer to
index appropriateness boundary. If ratio of index/table reads
impacted, some performance degradation possible.
Red Zone: Selectivity so high that index rarely used by CBO,
therefore index fragmentation generally not an issue.
Exception Index Fast Full Scan execution plans.

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High Selectivity: Which Indexes ?High Selectivity: Which Indexes ?
With selectivity crucial, how to find candidate indexes ?
Oracle9i, the v$sql_plan view provides useful info:
select hash_value, object_name, cardinality, operation, options
from v$sql_plan
where operation = 'INDEX' and object_owner = 'BOWIE' and cardinality > 10000
order by cardinality;
HASH_VALUE OBJECT_NAME
---------- ----------------------------------------------------------------
CARDINALITY OPERATION OPTIONS
----------- ------------------------------ ------------------------------
2768360068 TEST_EMPTY_ASSM_IDX
10011 INDEX FAST FULL SCAN
Note: SQL efficiency still of paramount importance !!

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Index RebuildIndex Rebuild –– Inserts ?Inserts ?
Impact on subsequent inserts needs to be considered
Simple demo with index rebuilt with pctfree = 0:
SQL> create table test_insert_0 (id number, value varchar2(10));
Table created.
SQL> begin
3 insert into test_insert_0 values (i, 'Bowie');
4 end loop;
5 end;
6 /
SQL> create index test_insert_0_idx on test_insert_0(id) pctfree 0;
Index created.

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Now insert 10% of rows evenly across the index.
SQL> begin
3 insert into test_insert_0 values (i*10, 'Bowie');
4 end loop;
5 end;
6 /
Elapsed: 00:00:24.03
SQL> analyze index test_insert_0_idx validate structure;
Index analyzed.
SQL> select pct_used from index_stats;
PCT_USED
----------
55
So by adding approximately 10% of data, the pct_used has plummeted to
only 55%. It kind of makes the index rebuild a little pointless !!

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Repeat same test but with an index rebuilt with pctfree = 10
SQL> create index test_insert_10_idx on test_insert_10(id) pctfree 10;
Index created.
SQL> begin
3 insert into test_insert_10 values (i*10, 'Bowie');
4 end loop;
5 end;
6 /
Elapsed: 00:00:15.04 ===> faster than the previous pctfree 0 example.
SQL> analyze index test_insert_10_idx validate structure;
Index analyzed.
SQL> select pct_used from index_stats;
PCT_USED
----------
99
And we notice that the pct_used value remains much higher.

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Index RebuildIndex Rebuild –– Inserts ConclusionInserts Conclusion
Be very careful of pct_free value with rebuilds
Ensure there is sufficient free space to avoid
imminent block splits
Can impact subsequent insert performance
Can lead to large amounts of unused space due to
block splits making the rebuild pointless

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Index HeightIndex Height –– Rebuild Factor ?Rebuild Factor ?
The simple answer is no
Most index rebuilds do not result in a height reduction
If the pct_used is high, rebuild is pointless
If index creeps over height boundary, rebuild is still
pointless as:
– Instead of reading root, now read root + branch resulting in
just 1 additional cached I/O
– Index eventually will grow anyways
Rebuilding an index purely because of its height is yet
another myth

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Conditions for RebuildsConditions for Rebuilds
Large free space (generally 50%+), which indexes rarely reach, and
Large selectivity, which most index accesses never reach, and
Response times are adversely affected, which rarely are.
Note requirement of some free space anyways to avoid insert and
subsequent free space issues
Benefit of rebuild based on various dependencies which include:
– Size of index
– Clustering Factor
– Caching characteristics
– Frequency of index accesses
– Selectivity (cardinality) of index accesses
– Range of selectivity (random or specific range)
– Efficiency of dependent SQL
– Fragmentation characteristics (does it effect portion of index frequently used)
– I/O characteristics of index (serve contention or I/O bottlenecks)
– The list goes on and on ….

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Other Rebuild Issues To ConsiderOther Rebuild Issues To Consider
More efficient index structures can reduce stress
on buffer cache. Harder to formulate but
requires consideration
If you have the resources and you have the
appropriate maintenance window, then the cost
vs. benefit equation more favorable to rebuild
– Benefit maybe low but perhaps so is the relative cost
Rebuild or Coalesce ?

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Index CoalesceIndex Coalesce
More efficient, less resource intensive, less
locking issues than rebuild option
Can significantly reduce number of leaf blocks in
some scenarios
Requires sum of free space to exceed 50% +
pctfree in consecutive leaf blocks
However, as generally needs excessive 50%+
freespace for rebuild to be effective
Does not reduce index height
alter index bowie_idx coalesce;

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Index CoalesceIndex Coalesce
1 2 3 4 5 6 7 8 9
1,2,3 4 6,7 8,95

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SummarySummary
The vast majority of indexes do not require rebuilding
Oracle B-tree indexes can become “unbalanced” and
need to be rebuilt is a myth
Deleted space in an index is “deadwood” and over time
requires the index to be rebuilt is a myth
If an index reaches “x” number of levels, it becomes
inefficient and requires the index to be rebuilt is a myth
If an index has a poor clustering factor, the index needs
to be rebuilt is a myth
To improve performance, indexes need to be regularly
rebuilt is a myth

Oracle b tree index internals - rebuilding the thruth

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