Read consistency implementation in Oracle
In my next post , I will demonstrate that _db_block_max_cr_dba is a soft limit and CR clones more than its value can be created.
———————–
———————–
— Create a cluster with size = blocksize = 8k and index it
SQL> drop cluster index_cluster including tables;
create cluster index_cluster
( id number(3) )
size 8192;
1 33 5 5
2 34 5 5
3 33 5 5
4 33 5 5
5 33 5 5
6 32 5 5
7 34 5 5
8 34 5 5
9 33 5 5
10 32 5 5
…
90 34 5 5
91 34 5 5
92 30 5 5
93 34 5 5
94 34 5 5
95 34 5 5
96 34 5 5
97 33 5 5
98 33 5 5
99 34 5 5
100 34 5 5
vi cluster_factor.out
********************************************************************************
Here are the contents of the trace file:
In case of unorganized table, it can be seen that no. of blocks visited (3517) is approaches the number of rows (3400) in the table as rows for an id are scattered across a large no. of blocks.
SQL ID: 0npa78p7jkfa5
Plan Hash: 1120857569
SELECT *
FROM
UNORGANIZED WHERE ID = :B1
call count cpu elapsed disk query current rows
——- —— ——– ———- ———- ———- ———- ———-
Parse 1 0.00 0.00 0 0 0 0
Execute 100 0.00 0.00 0 0 0 0
Fetch 100 0.01 0.02 0 3517 0 3400
——- —— ——– ———- ———- ———- ———- ———-
total 201 0.01 0.02 0 3517 0 3400
*******************************************************************************
In case of single table index cluster,
Total I/O’s = I/O’s against the table + I/O’s against the table
i/O’s against the table = no. of table blocks across which various records for different id’s are stored
Since we saw earlier that records for each key value are scattered across 5 blocks,
I/O’s against the table = no. of distinct key values (id’s) * no. of blocks occupied by records for an id
= 100 * 5
= 500
Rest 100 I/O’s are made against against the index ( one I/O for each key value)
Hence total I/O’s = 100 + 500 = 600
********************************************************************************
SQL ID: 6qy378ww4729s
Plan Hash: 3651720007
SELECT *
FROM
INDEX_CLUSTER_TAB WHERE ID = :B1
call count cpu elapsed disk query current rows
——- —— ——– ———- ———- ———- ———- ———-
Parse 1 0.00 0.00 0 0 0 0
Execute 100 0.00 0.00 0 0 0 0
Fetch 100 0.00 0.00 0 600 0 3400
——- —— ——– ———- ———- ———- ———- ———-
total 201 0.00 0.00 0 600 0 3400
*******************************************************************************
In case of single table hash cluster, as index access is not needed,
Total I/O’s = I/O’s against the table
= no. of distinct key values (id’s) * no. of blocks occupied by records for an id
= 100 * 5
= 500
********************************************************************************
SQL ID: ctnu91v20p2x2
Plan Hash: 3860562250
SELECT *
FROM
HASH_CLUSTER_TAB WHERE ID = :B1
call count cpu elapsed disk query current rows
——- —— ——– ———- ———- ———- ———- ———-
Parse 1 0.00 0.00 0 0 0 0
Execute 100 0.00 0.00 0 0 0 0
Fetch 100 0.00 0.00 0 500 0 3400
——- —— ——– ———- ———- ———- ———- ———-
total 201 0.00 0.00 0 500 0 3400
*******************************************************************************
Summarizing the above results :
unorganized index_cluster_tab hash_cluster_tab
total CPU 0.01 0.00 0.00
elapsed 0.01 0.02 0.00
time
I/O’s 3517 600 500
Hence, it can be concluded that for exact match queries hash clusters give the best performance since least no. of I/O’s are made.
- Trace the query using range scan on three tables and verify that index cluster table gives the best performance .
– Let’s compare the statistics when rows for entire range of id’s are retrieved from the three tables
In case of unorganized table, it can be seen that Full table scan is done and
total I/O’s = Physical I/O’s + logical I/O’s
= 486 + 489 = 975
CPU usage = 0.01
total I/O’s = Physical I/O’s + logical I/O’s
= 385 + (501 + 1)
= 887
CPU usage = 0.01
total I/O’s = Physical I/O’s + logical I/O’s
= 503 + 506
= 1009
CPU usage = 0.04
total I/O’s 975 887 1009
CPU usage 0.01 0.01 0.04
Let’s compare clustering factor of indexes on the three tables.
Tables unorganized and index_cluster_tab already have index.
– Let’s create index on hash_cluster_tab and gather statistics .
SQL>create index hash_cluster_idx on hash_cluster_tab(id);
exec dbms_stats.gather_index_stats(USER, ‘HASH_CLUSTER_IDX’);
exec dbms_stats.gather_index_stats(USER, ‘INDEX_CLUSTER_IDX’);
– Find out clustering factor of the three tables.
SQL> select index_name, clustering_factor
from user_indexes
where index_name in (‘UNORGANIZED_IDX’, ‘INDEX_CLUSTER_IDX’, ‘HASH_CLUSTER_IDX’);
INDEX_NAME CLUSTERING_FACTOR
—————————— —————–
HASH_CLUSTER_IDX 500
INDEX_CLUSTER_IDX 100
UNORGANIZED_IDX 3311
– Note that
– clustering factor of index on unorganized table approaches no. of rows in the table (3400).
– clustering factor of index on hash_cluster_tab table = 500 . As entries for each id are spread across 5 blocks, 500 blocks need to be accessed to get all the rows and index is aware of this information.
– clustering factor of index on index_cluster_tab table = 100 as there are 100 entries (one for each id) in the index. Here also 500 table blocks need to be accessed to get all the rows but index contains information about only the first(or may be the last) data block for an id. Rest 4 blocks containing records for that id are chained to it and index does not have that information and clustering factor of an index is computed on the basis of the information available in the index. That’s why clustering factor in this case = no. of index entries.
*If too little space is reserved for each
key (small SIZE value), or if the cluster is created with too few hash keys (small HASHKEYS), then each key will split across multiple blocks negating the benefits of the cluster.When creating a hash cluster, it is important to choose the cluster key correctly and set the HASH IS, SIZE, and HASHKEYS parameters so that performance and space use are optimal.
* If too much space is reserved for each key (large SIZE value), or if the cluster is created with too many hash keys (large
HASHKEYS), then the cluster will contain thousands of empty blocks that slow down full table scans . A SIZE value much larger results in wasted space.
A properly sized hash cluster for a lookup table gives pretty much a SINGLE IO for a keyed lookup.
SELECT . . . WHERE cluster_key < . . . ;
References:
Related links:
Clustering Factor Demystified Part-I
Clustering Factor Demystified Part-II
Direct Read Enhancements in 11g
SQL Profile Demystified : Part – I
SQL Profile Demystified Part – II
Undocumented Parameters in Oracle 11g
———————–
SQL> select count(*) from p5;
SQL>select count(*) from p5;
SQL>select /*+ parallel (pdef 8) */ count(*) from pdef;
Related links :
Database Index
Automatic Degree Of Paralellism – Part-I
Oracle 11g : Automatic DOP – Parallel Threshold
Oracle 11g : Parallel Statement Queueing
Parallel_Adaptive_Multi_User
Parallel_Min_Percent
Parallel_Min_Servers
2. PARALLEL_MIN_TIME_THRESHOLD : Oracle 11gR2 will ascertain if the query’s estimated execution time is likely to run longer than the acceptable value (in seconds) for PARALLEL_MIN_TIME_THRESHOLD and, if sufficient resources for parallel execution exist right now, it will allow the query to execute; otherwise, it will delay its execution until sufficient resources exist. This helps prevent a single parallel query from consuming excessive resources at the cost of other non-parallelizable operations.The default of this parameter is 10 seconds.
3. PARALLEL_ADAPTIVE_MULTI_USER : If it is set to TRUE, then Oracle will adjust the degree of parallel based on the overall load on the system. When the system is more heavily loaded, then the degree of parallelism will be reduced.
4. If PARALLEL_IO_CAP is set to TRUE in 11g or higher, then Oracle will limit the Degree of Parallelism to that which the IO subsystem can support. The IO subsystem limits can be calculated by using the procedure DBMS_RESOURCE_MANAGER.CALIBRATE_IO.
5. PARALLEL_MAX_SERVERS : To avoid an arbitrary number of parallel processes to be running on a system, which may overload that system, the parameter parallel_max_servers provides a hard upper boundary. Regardless of any other setting, the degree of parallelism cannot exceed that which can be supported by PARALLEL_MAX_SERVERS. For most SQL statements, the number of servers required will be twice the Degree of Parallelism.
6. PARALLEL_DEGREE_LIMIT : It is the maximum DOP that can be used. It can take various values :
Note that even if you increase the parameter value, the automatically computed degree of parallelism may not increase, because the database may determine that the computed degree of parallelism is sufficient.The DOP that we run the statement with is the minimum value of the computed DOP (or ideal DOP) and that parallel_degree_limit parameter.
• A degree of parallelism can be specified at the table or index level by using the PARALLEL clause of CREATE TABLE, CREATE INDEX, ALTER TABLE or ALTER INDEX.
• The PARALLEL hint can be used to specify the degree of parallelism for a specific table within a query.
In my next article in this series , I will practically demonstrate oracle behaviour for various values of the parameter PARALLEL_DEGREE_POLICY i.e. MANUAL, LIMITED AND AUTO
References:
http://searchitchannel.techtarget.com/feature/Using-parallel-SQL-to-improve-Oracle-database-performance
http://docs.oracle.com/cd/E11882_01/server.112/e25523/parallel002.htm
http://kerryosborne.oracle-guy.com/2014/01/does-parallel_degree_limit-work-with-parallel_degree_policymanual/
http://www.databasejournal.com/features/oracle/oracle-parallel-processing-new-and-improved.html
http://www.pythian.com/blog/secrets-of-oracles-automatic-degree-of-parallelism/
http://hemantoracledba.blogspot.in/2015/02/parallel-execution-1-parallel-hint-and.html?utm_source=feedburner&utm_medium=email&utm_campaign=Feed:+blogspot/JcnHK+(Hemant%27s+Oracle+DBA+Blog)
https://hourim.wordpress.com/2015/02/24/parallel-query-broadcast-distribution-and-temp-space/
http://www.oracle.com/technetwork/articles/datawarehouse/twp-parallel-execution-fundamentals-133639.pdf
https://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:1657271217898
What Is Auto DOP?
Configuring And Controlling Auto DOP
https://blogs.oracle.com/datawarehousing/entry/parallel_degree_limit_parallel_max
In Memory Parallel execution in Oracle Database 11g R2
Parameter Changes For Parallel Execution In Oracel Database 12cR2
———————————————————————————————
Related links :
Database Index
Automatic Degree Of Paralellism – Part-II
Oracle 11g : Automatic DOP – Parallel Threshold
Oracle 11g : Parallel Statement Queueing
Parallel_Adaptive_Multi_User
Parallel_Min_Percent
Parallel_Min_Servers
– Using the rowid obtained above, access the table by rowid (i.e. one I/O to get the data)
whereas, the procedure followed in a hash cluster is :
– Hash the primary key to get the physical location of the record
– Perform single I/O to read the block.
Thus, hash clusters can reduce I/O on tables.
create cluster <cluster_name> (cluster_key <datatype>)
size <size_number> single table hashkeys <hash_keys_number> hash is
<expr>;
Let’s understand various clauses in create cluster command:
CLUSTER_KEY <datatype>
The cluster key should have a single column containing only integers.
Hash clusters having composite cluster keys or cluster keys made up of non integer columns use the internal hash function.
If a non-integer cluster key value is supplied and internal hash function is bypassed , the operation (INSERT or UPDATE statement) is rolled back and an error is returned.
SIZE <size_number>
Specifies the amount of space in bytes reserved in a block to store all rows having the same cluster key value or the same hash value. This space decides the maximum number of cluster or hash values stored in a data block.
SINGLE TABLE
SINGLE TABLE indicates that this hash cluster can contain only one
table.. However, you may drop the table and create another table in the
same cluster.
HASHKEYS <hash_keys_number>
Specify the HASHKEYS clause to create a hash cluster and specify the number of hash values for the hash cluster.The HASHKEYS value specifies and limits the number of unique hash values that can be generated by the hash function used by the cluster. (how many distinct values you anticipate for the cluster key over time)Oracle Database rounds up the HASHKEYS value to the nearest prime number to obtain the actual number of hash values. The minimum value for this parameter
is 2. If you omit the HASHKEYS parameter, the database creates an indexed cluster by default.
When you create a hash cluster, the database immediately allocates space for the cluster based on the values of the SIZE and HASHKEYS parameters.
It allocates a hash table to hold HASHKEY number of cluster keys of SIZE bytes each.
HASH IS <expr>
Specify an expression to be used as the hash function for the hash cluster. Must evaluate to a positive value
If you omit the HASH IS clause, then Oracle Database uses an internal hash function for the hash cluster.
—- Overview –
– create two single table hash clusters with
size = 8k , hashkeys = 4
cluster HSH_CLUSTER_ROWSIZ_8K and HSH_CLUSTER_ROWSIZ_4K
— Create table hash_cluster_tab_ROWSIZE_8k in cluster HSH_CLUSTER_ROWSIZ_8K with row size such that only one record fits one block
— Create table hash_cluster_tab_ROWSIZE_4k in cluster HSH_CLUSTER_ROWSIZ_4K with row size such that two records fit one block
— insert records for 5 distinct key values to both the tables
— Each record goes to a different block as hashkeys has been set to 5 (next prime no.) and each block can have records with only one hash key.
— Try to access record with a key value – single I/O
— Insert records for 6th key values although provision has made for 5 keys only and check that hash collision takes place as id = 6 is mapped to one of the already existing hash values (id=1)
— Add another record for id = 1 and check that multiple blocks containing rows with same hash value are chained together.
– Create 3 clusters with size = 2K, hashkeys = 4
HSH_CLUSTER_SIZ_2K_ROW_1K, HSH_CLUSTER_SIZ_2K_ROW_2K and HSH_CLUSTER_SIZ_2K_ROW_4K
— Create a table HSH_TAB_SIZ_2K_ROWSIZ_1K in cluster HSH_CLUSTER_SIZ_2K_ROW_1K with row size such that row size =1/2( specified size) = 1/2(2K) = 1K
— Create a table HSH_TAB_SIZ_2K_ROWSIZ_2K in cluster HSH_CLUSTER_SIZ_2K_ROW_2Kwith row size such that row size = specified size 2K
— Create a table HSH_TAB_SIZ_2K_ROWSIZ_4K in cluster HSH_CLUSTER_SIZ_2K_ROW_4K with row size such that row size = 2( specified size) = 2(2K) = 4K
— Insert records for 5 distinct key values in the three tables
— Check that each block contains records for a maximum of 3 hashkeys.
— Insert 3 records with id = 1 in all the 3 tables
— Add records for id’s = 6 to 9 i.e. four keys more than what we have defined the cluster for (5).
— Note that it takes time to add these records
— Check that there is collision for hash keys – Multiple key values correspond to the same hash key – the hash chain for a hash key becomes longer and contains records with different key values.
— Check that overallocation takes place i.e. a block holds rows for hashkeys more than it is expected to hold (3).
– Implementation –
– create two single table hash clusters with
size = 8k , hashkeys = 4
cluster HSH_CLUSTER_ROWSIZ_8K and HSH_CLUSTER_ROWSIZ_4K
– The cluster key column for the clusters is id. The column in table in this cluster does not have to be called ID, but it must be NUMBER(2), to match this definition.
– Also we have specified a SIZE 8K option which means that we expect about 8K bytes of data to be associated with each cluster key value. Oracle will use it to compute the maximum number of cluster keys that could fit per block. Here, maximum no. of hash keys per block = Block size/SIZE = 8K/8K = 1 i.e. each block can contain records having only one hash key.
– The value of HASHKEYS limits the number of unique hash values that can be generated by the hash function used for the cluster. Oracle rounds the number you specify for HASHKEYS to the nearest prime number (5 here as we have set HASHKEYS to 4 and 5 is the next prime number) i.e. for any cluster key value, the hash function generates a maximum of 5 values.
SQL>Create tablespace mssm datafile
‘/u01/app/oracle/oradata/orcl/mssm01.dbf’ size 100m
segment space management manual;
alter system set deferred_segment_creation=FALSE;
drop cluster HSH_CLUSTER_ROWSIZ_8K including tables;
create cluster HSH_CLUSTER_ROWSIZ_8K ( id number(2) )
size 8192 single table hash is id hashkeys 4 tablespace mssm;
drop cluster HSH_CLUSTER_ROWSIZ_4K including tables;
create cluster HSH_CLUSTER_ROWSIZ_4K ( id number(2) )
size 8192 single table hash is id hashkeys 4 tablespace mssm;
Let’s create tables in these clusters ..
– Create table
hash_cluster_tab_ROWSIZE_8k in cluster HSH_CLUSTER_ROWSIZ_8K with row
size such that only one record fits one block
drop table hash_cluster_tab_rowsize_8k purge;
CREATE TABLE HASH_CLUSTER_TAB_ROWSIZE_8K
(id number(2) ,
txt1 char(2000),
txt2 char(2000),
txt3 char(2000)
) CLUSTER HSH_CLUSTER_ROWSIZ_8K( id );
– Create table
hash_cluster_tab_ROWSIZE_4k in cluster HSH_CLUSTER_ROWSIZ_4K with row
size such that two records fit one block
drop table hash_cluster_tab_rowsize_4k purge;
CREATE TABLE hash_cluster_tab_rowsize_4k
(id number(2) ,
txt1 char(2000),
txt2 char(1000)) CLUSTER HSH_CLUSTER_ROWSIZ_4K( id );
– check the blocks assigned to the clusters –
SQL> col segment_name for a30
select segment_name, extent_id, block_id, blocks
from dba_extents where segment_name like ‘%HASH_CLUSTER%';
SEGMENT_NAME EXTENT_ID BLOCK_ID BLOCKS
—————————— ———- ———- ———-
HSH_CLUSTER_ROWSIZ_8K 0 128 8
HSH_CLUSTER_ROWSIZ_4K 0 136 8
– Insert 5 records each with a distinct
key value to both the tables
SQL>Begin
for i in 1..5 loop
insert into hash_cluster_tab_rowsize_8k values (i, ‘x’, ‘x’, ‘x’);
insert into hash_cluster_tab_rowsize_4k values (i, ‘x’, ‘x’);
end loop;
commit;
end;
/
Let’s find out the blocks the records have
gone to ..
– Each record goes to a different block as hashkeys has been set to 5 (next prime no.)
First block assigned to the cluster(128, 136) is used to store header info.
Records occupy blocks second block onwards
Note that record for id=5 goes to the first block as mod (id, hashkeys) decides the block no. for a key and mod (5,5) = 0
0 denotes the first block to accommodate records for the table
Note that records for id= 1 to 4 go to the subsequent blocks as mod (1,5) = 1, mod(2,5) = 2 …
In hash_cluster_tab_rowsize_4k, even though two rows can fit one block, still each row has gone to a different block as one block has been assigned to a hash value. and
SQL>select ‘HASH_CLUSTER_TABLE_8K’
TAB_NAME , id, dbms_rowid.rowid_block_number(t1.rowid) hash_cluster_block
from hash_cluster_tab_rowsize_8k t1
union
select ‘HASH_CLUSTER_TABLE_4K’, id,
dbms_rowid.rowid_block_number(t2.rowid) hash_cluster_block
from hash_cluster_tab_rowsize_4k t2
order by 1,2,3;
TAB_NAME
ID HASH_CLUSTER_BLOCK
——————— ———- ——————
HASH_CLUSTER_TABLE_4K
1
138
HASH_CLUSTER_TABLE_4K
2
139
HASH_CLUSTER_TABLE_4K
3
140
HASH_CLUSTER_TABLE_4K
4
141
HASH_CLUSTER_TABLE_4K
5
137
HASH_CLUSTER_TABLE_8K
1
130
HASH_CLUSTER_TABLE_8K
2
131
HASH_CLUSTER_TABLE_8K
3
132
HASH_CLUSTER_TABLE_8K
4
133
HASH_CLUSTER_TABLE_8K
5
129
– Try to access record with a key value –
single I/O is needed (cr=1) as hash access is used
SQL>conn / as sysdba
alter session set tracefile_identifier=’hash_cluster';
alter session set sql_trace=true;
select * from hash_cluster_tab_rowsize_8k where id = 1;
select * from hash_cluster_tab_rowsize_4k where id = 1;
alter session set sql_trace=false;
col value for a75
select value from v$diag_info where upper(name) like ‘%TRACE FILE%';
VALUE
—————————————————————————
/u01/app/oracle/diag/rdbms/orcl/orcl/trace/orcl_ora_14452_hash_cluster.trc
ho
rm hash_cluster.out
ho tkprof
/u01/app/oracle/diag/rdbms/orcl/orcl/trace/orcl_ora_14452_hash_cluster.trc
hash_cluster.out
ho vi hash_cluster.out
********************************************************************************
select * from
hash_cluster_tab_rowsize_8k where id = 1
call count
cpu elapsed
disk query
current rows
——- —— ——– ———- ———- ———-
———- ———-
Parse
1 0.00
0.00
0 2
0 0
Execute 1
0.00
0.00
0
0
0 0
Fetch
2 0.00
0.00
0 1
0 1
——- —— ——– ———- ———- ———-
———- ———-
total
4 0.00
0.01
0 3
0 1
Rows Row Source Operation
——- —————————————————
1 TABLE ACCESS HASH
HASH_CLUSTER_TAB_ROWSIZE_8K (cr=1 pr=0
pw=0 time=0 us)
********************************************************************************
select * from
hash_cluster_tab_rowsize_4k where id = 1
call count
cpu elapsed
disk query
current rows
——- —— ——– ———- ———- ———-
———- ———-
Parse
1 0.00
0.00
0 2
0 0
Execute 1 0.00
0.00
0
0
0 0
Fetch
2 0.00
0.00
0 1
0 1
——- —— ——– ———- ———- ———-
———- ———-
total
4 0.00 0.00
0
3
0 1
Rows Row Source Operation
——- —————————————————
1 TABLE ACCESS HASH
HASH_CLUSTER_TAB_ROWSIZE_4K (cr=1 pr=0
pw=0 time=0 us)
********************************************************************************
– Insert records for 6th key values although
provision has made for 5 keys only
SQL> insert into
hash_cluster_tab_rowsize_8k values (6, ‘x’, ‘x’, ‘x’);
insert into hash_cluster_tab_rowsize_4k values (6,
‘x’, ‘x’);
commit;
– Let’s find out the blocks the records have
gone to .
– In hash_cluster_tab_rowsize_4k, id = 6 is hashed
to one of the existing hash values and record is placed in the same block (138)
as the earlier containing record with same hash key as two records can fit one
block. Hence block 138 becomes overflow block as it contains more hashvalues
than it is configured for.
Here id = 6 has been mappped to same hash
value as id = 1 and has occupied the same block as id = 1
– In hash_cluster_tab_rowsize_8k, since one row can
fit one block, Newly added rows have gone to a new block (147) as
the earlier blocks can’t accommodate new rows.
But it has again been mapped to one of the
already existing hash values and is chained to the blocks containing those key
values.
SQL>select ‘HASH_CLUSTER_TABLE_8K’
TAB_NAME , id, dbms_rowid.rowid_block_number(t1.rowid) hash_cluster_block
from hash_cluster_tab_rowsize_8k t1
union
select ‘HASH_CLUSTER_TABLE_4K’, id,
dbms_rowid.rowid_block_number(t2.rowid) hash_cluster_block
from hash_cluster_tab_rowsize_4k t2
order by 1,2,3;
TAB_NAME
ID HASH_CLUSTER_BLOCK
——————— ———- ——————
HASH_CLUSTER_TABLE_4K
1
138
HASH_CLUSTER_TABLE_4K
2
139
HASH_CLUSTER_TABLE_4K
3
140
HASH_CLUSTER_TABLE_4K
4
141
HASH_CLUSTER_TABLE_4K
5
137
HASH_CLUSTER_TABLE_4K
6
138
HASH_CLUSTER_TABLE_8K
1
130
HASH_CLUSTER_TABLE_8K
2
131
HASH_CLUSTER_TABLE_8K
3
132
HASH_CLUSTER_TABLE_8K
4
133
HASH_CLUSTER_TABLE_8K
5
129
HASH_CLUSTER_TABLE_8K
6
147
– Let’s add another record for id = 1
In hash_cluster_tab_rowsize_2k, although we have space for another row
in blocks containing id = 2,3,4,5 , the new record goes to a new block(142) as
one block can contain only one hashvalue. From now onwards, a block which
contains id = 1 may also contain an entry for id=6 and vice versa since both of
them hash to the same value.
In hash_cluster_tab_rowsize_8k also , the new record goes to a new
block (135) as a block can contain only one row. Presuming that id = 6 and 1
hash to the same value, blocks containing id = 1 (130 and 135) have been
chained to block containing id=6 (135).
From now onwards, when we search for id =1 or id = 6 we will have to scan a
larger no. of blocks.Hence, if actual no. of cluster keys exceeds the specified
value for HASHKEYS, the likelihood of a collision (two cluster
key values having the same hash value) increases and performance degrades.
SQL>
insert into
hash_cluster_tab_rowsize_8k values (1, ‘x’, ‘x’, ‘x’);
insert into
hash_cluster_tab_rowsize_4k values (1, ‘x’, ‘x’);
commit;
select ‘HASH_CLUSTER_TABLE_8K’
TAB_NAME , id, dbms_rowid.rowid_block_number(t1.rowid) hash_cluster_block
from hash_cluster_tab_rowsize_8k t1
union all
select ‘HASH_CLUSTER_TABLE_4K’, id,
dbms_rowid.rowid_block_number(t2.rowid) hash_cluster_block
from hash_cluster_tab_rowsize_4k t2
order by 1,2,3;
TAB_NAME
ID HASH_CLUSTER_BLOCK
——————— ———- ——————
HASH_CLUSTER_TABLE_4K 1
138
HASH_CLUSTER_TABLE_4K 1
142
HASH_CLUSTER_TABLE_4K
2
139
HASH_CLUSTER_TABLE_4K
3
140
HASH_CLUSTER_TABLE_4K
4
141
HASH_CLUSTER_TABLE_4K
5
137
HASH_CLUSTER_TABLE_4K 6
138
HASH_CLUSTER_TABLE_8K 1
130
HASH_CLUSTER_TABLE_8K 1
135
HASH_CLUSTER_TABLE_8K
2
131
HASH_CLUSTER_TABLE_8K
3
132
HASH_CLUSTER_TABLE_8K
4
133
HASH_CLUSTER_TABLE_8K
5
129
HASH_CLUSTER_TABLE_8K
6
134
Therefore, the distribution of rows in a hash cluster is directly
controlled by the value set for the HASHKEYS parameter. With a larger number of
hash keys for a given number of rows, the likelihood of
a collision (two cluster key values having the same hash value)
decreases. Minimizing the number of collisions is important because overflow
blocks (thus extra I/O) might be necessary to store rows with hash values that
collide.
Now let’s play around with SIZE clause...
The maximum number of hash keys assigned per data block is determined by the
SIZE parameter of the CREATE CLUSTER command. SIZE is an estimate of the total
amount of space in bytes required to store the average number of rows associated
with each hash value. For example, if the available free space per data block
is 1700 bytes and SIZE is set to 500 bytes, three hash keys (round(1700/500))
are assigned per data block.
Note: The importance of the SIZE parameter of hash clusters is analogous
to that of the SIZE parameter for index clusters. However, with index clusters,
SIZE applies to rows with the same cluster key value instead of the same hash
value.
Although the maximum number of hash key values per data block is determined
by SIZE, Oracle does not actually reserve space for each hash key value in the
block. For example, if SIZE determines that three hash key values are allowed
per block, this does not prevent rows for one hash key value from taking up all
of the available space in the block. If there are more rows for a given hash
key value than can fit in a single block, the block is chained, as necessary.
– Create 3 clusters with size = 2K, hashkeys
= 4
HSH_CLUSTER_SIZ_2K_ROW_1K, HSH_CLUSTER_SIZ_2K_ROW_2Kand
HSH_CLUSTER_SIZ_2K_ROW_4K
— Available free space is slightly less that blocksize of 8K
— No. of hashkeys that can be stored in the block < 4 (blocksize/size =
8k/2K)
= 3
SQL> drop cluster HSH_CLUSTER_SIZ_2K_ROW_1K
including tables;
create cluster HSH_CLUSTER_SIZ_2K_ROW_1K
( id number(2) )
size 2K single table hash is id hashkeys 4 tablespace mssm;
drop cluster
HSH_CLUSTER_SIZ_2K_ROW_2Kincluding tables;
create cluster hash_cluster_size_2k_row_2k
( id number(2) )
size 2K single table hash is id hashkeys 4 tablespace mssm;
drop cluster
HSH_CLUSTER_SIZ_2K_ROW_4K including tables;
create cluster HSH_CLUSTER_SIZ_2K_ROW_4K
( id number(2) )
size 2K single table hash is id hashkeys 4 tablespace mssm;
– Create a table HSH_TAB_SIZ_2K_ROWSIZ_1K in
cluster HSH_CLUSTER_SIZ_2K_ROW_1K with row size such that row size =1/2(
specified size) = 1/2(2K) = 1K
SQL>drop table HSH_TAB_SIZ_2K_ROWSIZ_1K
purge;
CREATE TABLE HSH_TAB_SIZ_2K_ROWSIZ_1K
(
id number(2) ,
txt1 char(400),
txt2 char(400)
)
CLUSTER HSH_CLUSTER_SIZ_2K_ROW_1K( id );
– Create a table HSH_TAB_SIZ_2K_ROWSIZ_2K in
cluster HSH_CLUSTER_SIZ_2K_ROW_2Kwith row size such that row size =
specified size 2K
SQL>drop table HSH_TAB_SIZ_2K_ROWSIZ_2K
purge;
CREATE TABLE HSH_TAB_SIZ_2K_ROWSIZ_2K
(
id number(2) ,
txt1 char(900),
txt2 char(900)
)
CLUSTER HASH_CLUSTER_SIZE_2K_row_2k( id );
– Create a table HSH_TAB_SIZ_2K_ROWSIZ_4K in
cluster HSH_CLUSTER_SIZ_2K_ROW_4K with row size such that row size = 2(
specified size) = 2(2K) = 4K
SQL>drop table HSH_TAB_SIZ_2K_ROWSIZ_4K
purge;
CREATE TABLE HSH_TAB_SIZ_2K_ROWSIZ_4K
(
id number(2) ,
txt1 char(1900),
txt2 char(1900)
)
CLUSTER HSH_CLUSTER_SIZ_2K_ROW_4K( id );
– Insert records for 5 distinct key values in
the three tables
SQL>Begin
for i in 1..5 loop
insert into
HSH_TAB_SIZ_2K_ROWSIZ_1K values (i, ‘x’, ‘x’);
insert into
HSH_TAB_SIZ_2K_ROWSIZ_2K values (i, ‘x’, ‘x’);
insert into
HSH_TAB_SIZ_2K_ROWSIZ_4K values (i, ‘x’, ‘x’);
end
loop;
commit;
end;
/
– Check that each block contains records for
a maximum of 3 hashkeys.
– For row size = 1K
8 rows can fit one block
Max no. of hash keys per block = 3
Max no. of records of distinct key values per block = 3
Records for id = 1,2 and 5 (3 hash keys) go to block 161
Records for id = 3 and 4 (only records left) go to block
162
– For row size = 2K
3 rows can fit one block
Max no. of hash keys per block = 3
Max no. of records of distinct key values per block = 3
Records for id = 1,2 and 5 (3 hash keys) go to block 169
Records for id = 3 and 4 (only records left) go to block
170
– For row size = 4K
1 row can fit one block
Max no. of hash keys per block = 3
Max no. of records possible per block = 1
Records for each id go to different block
SQL>select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t1.rowid) hash_cluster_block
from HSH_TAB_SIZ_2K_ROWSIZ_1K t1
union
select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t2.rowid) hash_cluster_block
from HSH_TAB_SIZ_2K_ROWSIZ_2K t2
union
select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t3.rowid) hash_cluster_block
from HSH_TAB_SIZ_2K_ROWSIZ_4K t3
order by 1,2,3;
TABLE_NAME
ID HASH_CLUSTER_BLOCK
————————————- ———- ——————
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K 1
161
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
2
161
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
3
162
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
4
162
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
5
161
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
1
169
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
2
169
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
3
170
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
4
170
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
5
169
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
177
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
2
179
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
3
178
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
4
180
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
5
181
— Let’s insert 3 records with id = 1 in
all the 3 tables
- For row size = 1K
No. of rows already there = 3 in block 161 containing id = 1
Since 8 rows can fit one block, all the 3 rows inserted go to
block 161
Now block 161 has 6 rows
i.e. although the maximum number of hash key values per data
block as determined by SIZE is 3, Oracle does not actually reserve space for
each hash key value in the block. Rows for one hash key value(id=1) can take up
all of the available space in the block.
– For row size = 2K
3 rows can fit one block
Block 169 containing id=1 already has 3 rows
Although 1 more row can fit in block 170 , newly added
rows for id=1 go to anothet block 172
i.e. when key value is inserted for the first time, its
blockmate keys are decided or let’s say that it is decided
records for which hash values will stay together. From then onwards, records
for its blockmate keys will always reside on the same block. Hence, once hash
values occupying a block have been grouped (1,2,5 and 3,4) further records will
occupy only the blocks along with their blockmates.
Block 172 is chained to block 169 as both contain id=1.
– For row size = 4K
1 row can fit one block
Each newly added record goes to new block
All the blocks containing id=1 (177, 184, 185, 186) will
be chained
SQL>select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t1.rowid) hash_cluster_block, count(*)
from HSH_TAB_SIZ_2K_ROWSIZ_1K t1
group by id,
dbms_rowid.rowid_block_number(t1.rowid)
union
select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t2.rowid) hash_cluster_block, count(*)
from HSH_TAB_SIZ_2K_ROWSIZ_2K t2
group by id,
dbms_rowid.rowid_block_number(t2.rowid)
union
select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t3.rowid) hash_cluster_block, count(*)
from HSH_TAB_SIZ_2K_ROWSIZ_4K t3
group by id,
dbms_rowid.rowid_block_number(t3.rowid)
order by 1,3,2;
TABLE_NAME
ID HASH_CLUSTER_BLOCK COUNT(*)
————————————- ———- ——————
———-
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
1
161 4
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
2
161 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
5
161 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
3
162 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
4
162 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
1
169 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
2
169 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
5
169 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
3
170 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
4
170 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
1
172 3
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
177 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
3
178 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
2
179 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
4
180 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
5
181 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
184 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
185 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
186 1
– Add records for id’s = 6 to 9 i.e. four
keys more than what we have defined the cluster for (5).
— Note that it takes time to add these records
SQL>Begin
for i in 6..9 loop
insert into
HSH_TAB_SIZ_2K_ROWSIZ_1K values (i, ‘x’, ‘x’);
insert into
HSH_TAB_SIZ_2K_ROWSIZ_2K values (i, ‘x’, ‘x’);
insert into
HSH_TAB_SIZ_2K_ROWSIZ_4K values (i, ‘x’, ‘x’);
end
loop;
commit;
end;
/
– Check that there is collision for hash keys
- Multiple key values correspond to the same hash key – the
hash chain for a hash key becomes longer and contains records with different
key values.
– Check that overallocation takes place i.e.
a block holds rows for hashkeys more than it is expected to hold (3).
– For row size = 1K
In block 161, no. of rows already there = 6 containing id
= 1
Since 8 rows can fit one block, rows for id = 6 and 7 go to block
161 i.e. now block 161 contains records for id = 1,2,5,6,7
i.e. it contains 2 hashkeys more than it has configured to hold
. This is called overallocation and block 161 is called overflow block.
From now onwards whenever a record is added with any of id =
1,2,5,6,7, it will be chained to block 161 .
All the blocks chained to block 161 will be searched whenever any of
these id’s is searched.
– For row size = 2K
3 rows can fit one block
Block 169 containing id=1 already has 3 rows with id =
1,2,5
Block 172 has 3 rows with id = 1 – It is already full –
it is chained to block 169
Block 170 has 2 rows with id = 3,4
Record for id=8 goes to block 170 i.e. from now onwards all
the blocks containing of id=3,4,8 will be chained to block 170.
All the blocks chained to block 170 will be
searched whenever any of these id’s (3,4,8) is searched.
Rows with id’s 7 and 9 go to block 173 . This block
will be chained with block 172 or 170
– For row size = 4K
1 row can fit one block
Each newly added record goes to new block
Blocks with id’s > 5 will be chained with any of the
earlier blocks with matching hash value.
e.g. mod(1,5) = mod (6,5) = 1
Hence block containing id = 6 i.e. block 183 will
chain with the all the blocks containing id = 1 (177, 184, 185, 186)
i.e. to search for id = 6, 5 blocks will be searched.
– With a larger number of hash keys for a given number of rows, the
likelihood of a collision (two cluster key values having the same
hash value) decreases.
SQL>select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t1.rowid) hash_cluster_block, count(*)
from HSH_TAB_SIZ_2K_ROWSIZ_1K t1
group by id,
dbms_rowid.rowid_block_number(t1.rowid)
union
select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t2.rowid) hash_cluster_block, count(*)
from HSH_TAB_SIZ_2K_ROWSIZ_2K t2
group by id,
dbms_rowid.rowid_block_number(t2.rowid)
union
select
‘HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K’ TABLE_NAME, id,
dbms_rowid.rowid_block_number(t3.rowid) hash_cluster_block, count(*)
from HSH_TAB_SIZ_2K_ROWSIZ_4K t3
group by id,
dbms_rowid.rowid_block_number(t3.rowid)
order by 1,3,2;
TABLE_NAME
ID HASH_CLUSTER_BLOCK COUNT(*)
————————————- ———- ——————
———-
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
1
161 4
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
2
161 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
5
161 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
6
161 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
7
161 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
3
162 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
4
162 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
8
162 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_1K
9
162 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
1
169 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
2
169 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
5
169 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
3
170 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
4
170 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
8
170 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
6
171 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
1
172 3
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
7
173 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_2K
9
173 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
177 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
3
178 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
2
179 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
4
180 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
5
181 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
9
182 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
6
183 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
184 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
185 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
1
186 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
7
187 1
HASH_CLUSTER_TABLE_SIZE_2K_ROWSIZE_4K
8
188 1
Specifying a User-Defined Hash Function
You can also specify any SQL expression as the hash function for a hash
cluster. If your cluster key values are not evenly distributed among the
cluster, you should consider creating your own hash function that more
efficiently distributes cluster rows among the hash values.
For example, if you have a hash cluster containing employee information and
the cluster key is the employee’s home area code, it is likely that many
employees will hash to the same hash value. To alleviate this problem, you can
place the following expression in the HASH IS clause of the CREATE CLUSTER
command:
MOD((emp.home_area_code + emp.home_prefix + emp.home_suffix), 101)
The expression takes the area code column and adds the phone prefix and
suffix columns, divides by the number of hash values (in this case 101), and
then uses the remainder as the hash value. The result is cluster rows more
evenly distributed among the various hash values.
CREATE CLUSTER address
(postal_code NUMBER, country_id CHAR(2))
HASHKEYS 20
HASH IS MOD(postal_code + country_id, 101);
Summary:
— Hash Cluster tables are appropriate for data that is read frequently via
an equality comparison on the key. If an index scan is used for a key
value, as more no. of users search for the same record, they hit the same index
block which becomes the “hot block” leading to more contention for
the cache buffers chains (cbc) latch. Replacing indexed tables with hash cluster
tables in this case can resolve the problem of contention for CBC latches.
— you cannot range scan a table in a hash cluster without adding a
conventional index to the table. In an index cluster, the query for range of values will be able to make use of the cluster key index to find these rows. In a hash cluster, this query would result in a full table scan unless you had an index on the key column. Only exact equality searches (including in lists and subqueries) may be made on the hash key without using an index that supports range scans.
— When you create a hash cluster table, you must determine in advance, the number of hash keys your table will ever have. The number of HASHKEYs in a hash cluster is a fixed size. You cannot change the size of the hash table without a rebuild of the cluster. This does not in any way limit the amount of data you can store in this cluster; it simply limits the number of unique hash keys that can be generated for this cluster. That may affect performance due to unintended hash collisions if the value was set too low. Getting the size of the HASHKEYs and SIZE parameters right is crucial to avoid a rebuild.
— With a hash cluster, the tables will start out big and will take longer
to create, as Oracle must initialize each block, an action that normally takes
place as data is added to the table. They have the potential to have data in
their first block and their last block, with nothing in between. Full scanning
a virtually empty hash cluster will take as long as full scanning a full hash cluster.
• The hash cluster is allocated right from the beginning. Oracle will take
your HASHKEYS/ trunc(blocksize/SIZE) and allocate and format that space right away. As soon as the first table is put in that cluster, any full scan will hit every allocated block. This is different from every other table in this respect.
• Updates to hash cluster tables do not introduce significant overhead,
unless you update the HASHKEY, which would not be a good idea, as it would
cause the row to migrate
. Hash clusters allocate all the storage for all the hash buckets when the cluster is created, so they may waste space.. Full scans on single table hash clusters will cost as much as they would in a heap table.
In my post Clustering Factor Demystified : Part – III, I have demonstrated the use of single table index and hash clusters to improve the clustering factor of an unorganized table.
——————————–
References :
http://www.relationaldbdesign.com/extended-database-features/module3/oracle-hash-cluster.php
http://docs.oracle.com/cd/B19306_01/server.102/b14231/hash.htm
http://simpleoracledba.blogspot.in/2009/05/oracle-single-table-hash-clusters-are.html
http://jonathanlewis.wordpress.com/2009/10/05/hash-clusters-2/
http://www.iselfschooling.com/mc4articles/mc4cluster.htm
http://dbaforums.org/oracle/index.php?showtopic=18947
http://www.orafaq.com/tuningguide/bad%20hash%20cluster.html
http://www.riddle.ru/mirrors/oracledocs/server/scn73/ch507.html
http://www.dba-oracle.com/t_hash_tables_clusters.htm
http://docs.oracle.com/cd/E11882_01/server.112/e25789/tablecls.htm
http://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:707440500346653259
http://stackoverflow.com/questions/2714099/oracle-hash-cluster-overflow-blocks
http://www.orafaq.com/forum/t/141142/2/
https://forums.oracle.com/forums/thread.jspa?threadID=2512488&start=0&tstart=0
———————————————————————————————————–
Related links :
Clustering Factor Demystified : Part – I
Clustering Factor Demystified : Part – II
Clustering Factor Demystified : Part – III
——————-
old 6: where dbarfil = &file_no and
As I discussed in my earlier post on parent and child cursors, multiple child cursors may be created for the same parent cursor if bind variables have different values.
In this post I will discuss about the parameter CURSOR_SHARING which controls the sharing of child cursors if bind variables have different values.
The parameter CURSOR_SHARING can take 3 values :
Let’s see the impact of different values :
CURSOR_SHARING = EXACT
– In this case when the same statement is issued with different literals, multiple parent cursors will be created.
— create a test table with
1 record with id1 = id2 = 1
1000 records with id1 = id2 = 2
2000 records with id1 = id2= 3
— create an index on the table
HR> drop table test purge;
create table test (id1 number, id2 number, txt char(1000));
insert into test values (1,1, ‘one’);
begin
for i in 1..1000 loop
insert into test values (2,2, ‘two’);
insert into test values (3,3, ‘three’);
end loop;
end;
/
insert into test select * from test where id1=3;
commit;
create index test_idx1 on test(id1);
create index test_idx2 on test(id2);
select id1,id2, count(*)
from test
group by id1,id2;
CURSOR_SHARING=EXACT
— Flush the shared pool
Set cursor_sharing=exact
SYS>alter system set CURSOR_SHARING=’EXACT';
alter system flush shared_pool;
sho parameter CURSOR_SHARING
NAME TYPE VALUE
—– —- —–
cursor_sharing string EXACT
— Issue identical statements with different values of literals
HR>conn hr/hr
select count(*) from test where id1=1;
select count(*) from test where id1=2;
select count(*) from test where id1=3;
— Check that the 3 parent cursors have been created
— Note that there is one record for each statement in v$sqlarea as one parent cursor is created for each sql statement since each of these statements differ in their text.
SYS>col sql_text for a30 word_wrapped
SELECT SQL_TEXT , SQL_ID, VERSION_COUNT, HASH_VALUE,PLAN_HASH_VALUE
FROM V$SQLAREA
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID VERSION_COUNT HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ————- ———- —————
select count(*) from test 1n09m564gh0q3 1 2297955011 4192825871
where id1=3
select count(*) from test 20nhaap8uxf7s 1 1370405112 3507950989
where id1=2
select count(*) from test bavqx2mw26wg0 1 4163072480 3507950989
where id1=1
— Note that 3 child cursors have been created for the 3 statements
SYS>col child_number for 99
SELECT SQL_TEXT, SQL_ID, CHILD_NUMBER CHILD#, HASH_VALUE,
PLAN_HASH_VALUE
FROM V$SQL
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID CHILD# HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ———- ———- —————
select count(*) from test 1n09m564gh0q3 0 2297955011 4192825871
where id1=3
select count(*) from test 20nhaap8uxf7s 0 1370405112 3507950989
where id1=2
select count(*) from test bavqx2mw26wg0 0 4163072480 3507950989
where id1=1
— We can see that in all 6 cursors have been created :
– 3 parent cursors and
– 3 child cursors
Each of the cursor occupies memory. Parent cursors contain sql text whereas
child cursor contains execution plan, execution statistics and execution
environment. If we replace literal with a bind variable, all the 3 statements
will be identical and hence only parent cursor needs to be created. Multiple
child cursors can be created for different values of the bind variables.
That’s what CURSOR_SHARING=SIMILAR does. It replaces literals in the otherwise
identical SQL statements with bind variables and only one parent cursor is
created.
If histogram on a column is created with only one bucket,i.e. it does not know about the skew
in data, only one child cursor will be created.
If histogram is created on a column with >1 buckets i.e. it knows about skew in data in that
column, it will create one child cursor for each statement even of the execution plan is same.
Thus CURSOR_SHARING=SIMILAR reduces the no. parent cursors.
If there is skew in data
If histogram on the column containing skewed data is there
multiple child cursors may be created – one for each value of the bind variable
else (histogram is not available)
only one child cursor will be created.
else (Data is not skewed)
only one child cursor will be created.
Now, since there is identical skewed data in id1 and id2 , we will create histogram on id1
with one bucket and on id2 with 4 buckets and see the difference.
CURSOR_SHARING=SIMILAR WITHOUT HISTOGRAM
— create histogram only on id1 with one bucket so that optimizer does not
know about the skew —
HR>exec dbms_stats.gather_table_stats(OWNNAME => ‘HR’,-
TABNAME => ‘TEST’,-
ESTIMATE_PERCENT =>null,-
METHOD_OPT => ‘FOR COLUMNS SIZE 1 ID1′);
— Set cursor_sharing = similar —
— Flush the shared pool
SYS>alter system set CURSOR_SHARING=’SIMILAR';
alter system flush shared_pool;
sho parameter CURSOR_SHARING
— Issue identical statements with different values of literals for the column on which histogram is not there (id1)
HR>conn hr/hr
select count(*) from test where id1=1;
select count(*) from test where id1=2;
select count(*) from test where id1=3;
— Check that the only 1 parent cursor has been created and literal has been replaced by bind variable. ( 1 record in v$SQLAREA)
.There is only one child cursor (version_count=1) since the optimizer does not know about skew in data
SYS>col sql_text for a30 word_wrapped
SELECT SQL_TEXT , SQL_ID, VERSION_COUNT, HASH_VALUE,PLAN_HASH_VALUE
FROM V$SQLAREA
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID VERSION_COUNT HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ————- ———- —————
select count(*) from test 07tpk6bm7j4qm 1 3866661587 3507950989
where id1=:”SYS_B_0″
— Note there is only one child cursor created i.e. same execution plan will be used for different values of the bind variable
SYS>col child_number for 99
SELECT SQL_TEXT, SQL_ID, CHILD_NUMBER CHILD#, HASH_VALUE,
PLAN_HASH_VALUE
FROM V$SQL
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID CHILD# HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ———- ———- —————
select count(*) from test 07tpk6bm7j4qm 0 3866661587 3507950989
where id1=:”SYS_B_0″
CURSOR_SHARING=SIMILAR WITH HISTOGRAM
Parent
— create histogram on id2 with 4 buckets so that optimizer knows about the skew in data —
HR>exec dbms_stats.gather_table_stats(OWNNAME => ‘HR’,-
TABNAME => ‘TEST’,-
ESTIMATE_PERCENT =>null,-
CASCADE => TRUE,-
METHOD_OPT => ‘FOR COLUMNS SIZE 4 ID2′);
— Issue identical statements with different values of literals for the column on which histogram is there (id2)
SYS>alter system flush shared_pool;
HR>conn hr/hr
select count(*) from test where id2=1;
select count(*) from test where id2=2;
select count(*) from test where id2=3;
— Check that the only 1 parent cursor has been created and literal has been replaced by bind variable. ( 1 record in v$SQLAREA)
.There are 3 child cursors (version_count=3)
SYS>col sql_text for a30 word_wrapped
SELECT SQL_TEXT , SQL_ID, VERSION_COUNT, HASH_VALUE,PLAN_HASH_VALUE
FROM V$SQLAREA
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID VERSION_COUNT HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ————- ———- ————–
select count(*) from test 3tcujqmqnqs8t 3 3981140249 2432738936
where id2=:”SYS_B_0″
— Note that 3 child cursors have been created as optimizer realizes that data is skewed and different execution plans will be more efficient for different values of the bind variable.
— 2 children have same execution plan (PLAN_HASH_VALUE) (for id=2 and 3 (Full table scan )
SYS>col child_number for 99
SELECT SQL_TEXT, SQL_ID, CHILD_NUMBER CHILD#, HASH_VALUE,
PLAN_HASH_VALUE
FROM V$SQL
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID CHILD# HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ———- ———- ————–
select count(*) from test 3tcujqmqnqs8t 0 3981140249 2432738936
where id2=:”SYS_B_0″
select count(*) from test 3tcujqmqnqs8t 1 3981140249 2432738936
where id2=:”SYS_B_0″
select count(*) from test 3tcujqmqnqs8t 2 3981140249 1489241381
where id2=:”SYS_B_0″
Hence, it can be seen that setting CURSOR_SHARING=SIMILAR
– replaces literals with bind variables in otherwise identical sql statements
Ideally we would like one child cursor to be created if execution plan is same for different values of the bind variable.
Setting CURSOR_SHARING=FORCE IN 11G does precisely this but only if the optimizer is
aware about the skew in the data. Let’s see:
CURSOR_SHARING=FORCE IN 11G WITHOUT HISTOGRAM
Parent
alter system flush shared_pool;
HR>conn hr/hr
select count(*) from test where id1=1;
select count(*) from test where id1=2;
select count(*) from test where id1=3;
— Note that only one parent cursor is created
One child cursor has been created (version_count=1)
SYS>col sql_text for a30 word_wrapped
SELECT SQL_TEXT , SQL_ID, VERSION_COUNT, HASH_VALUE,PLAN_HASH_VALUE
FROM V$SQLAREA
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID VERSION_COUNT HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ————- ———- ————–
select count(*) from test 07tpk6bm7j4qm 1 3866661587 3507950989
where id1=:”SYS_B_0″
— Note that 1 child cursor has been created
SYS>col child_number for 99
SELECT SQL_TEXT, SQL_ID, CHILD_NUMBER CHILD#, HASH_VALUE,
PLAN_HASH_VALUE
FROM V$SQL
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID CHILD# HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ———- ———- —————
select count(*) from test 07tpk6bm7j4qm 0 3866661587 3507950989
where id1=:”SYS_B_0″
CURSOR_SHARING=FORCE IN 11G WITH HISTOGRAM
Parent
|
+—+—-+
| |
Child Child
– Flush the shared pool and issue query using the column with histogram on it so that optimizer is aware of the skew.
SYS> alter system flush shared_pool;
HR>conn hr/hr
select count(*) from test where id2=1;
select count(*) from test where id2=2;
select count(*) from test where id2=3;
— Note that only one parent cursor is created
Two child cursors have been created (version_count=2)
SYS>col sql_text for a30 word_wrapped
SELECT SQL_TEXT , SQL_ID, VERSION_COUNT, HASH_VALUE,PLAN_HASH_VALUE
FROM V$SQLAREA
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID VERSION_COUNT HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ————- ———- —————
select count(*) from test 3tcujqmqnqs8t 2 3981140249 2432738936
where id2=:”SYS_B_0″
— Note that 2 child cursors have been created and each child has a distinct execution plan (PLAN_HASH_VALUE)
SYS>col child_number for 99
SELECT SQL_TEXT, SQL_ID, CHILD_NUMBER CHILD#, HASH_VALUE,
PLAN_HASH_VALUE
FROM V$SQL
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID CHILD# HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ———- ———- —————
select count(*) from test 3tcujqmqnqs8t 0 3981140249 2432738936
where id2=:”SYS_B_0″
select count(*) from test 3tcujqmqnqs8t 1 3981140249 1489241381
where id2=:”SYS_B_0″
Hence, setting CURSOR_SHARING=FORCE in 11g will use the same child cursor if execution plan is same for different values of the bind variables which means saving in memory in the shared pool and saving in the time for scanning the hash chains in the library cache . This new feature of 11g is called ADAPTIVE CURSOR SHARING.
Note: The behaviour of CURSOR_SHARING=FORCE in 11g is different from 9i/10g. Earlier, it would peek the value of the bind variable during the first execution and decide on the eexcution plan. On subsequent execution of the same statement with different values of the bind variable, it would reuse the same plan irrespective of the skew in the data.
CURSOR_SHARING=FORCE IN 10G WITH/WITHOUT HISTOGRAM
Parent
|
SYS> alter system set optimizer_features_enable=’10.2.0.3′;
— Flush the shared pool and issue query using the column with histogram on
it so that optimizer is aware of the skew.
SYS> alter system flush shared_pool;
HR>conn hr/hr
select count(*) from test where id2=1;
select count(*) from test where id2=2;
select count(*) from test where id2=3;
— Note that only one parent cursor is created
Only child cursor has been created (version_count=1)
SYS>col sql_text for a30 word_wrapped
SELECT SQL_TEXT , SQL_ID, VERSION_COUNT, HASH_VALUE,PLAN_HASH_VALUE
FROM V$SQLAREA
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID VERSION_COUNT HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ————- ———- —————
select count(*) from test 3tcujqmqnqs8t 1 3981140249 2432738936
where id2=:”SYS_B_0″
— Note that 1 child cursor has been created
SYS>col child_number for 99
SELECT SQL_TEXT, SQL_ID, CHILD_NUMBER CHILD#, HASH_VALUE, PLAN_HASH_VALUE
FROM V$SQL
WHERE LOWER(SQL_TEXT) LIKE ‘select count(*) from test%’
AND LOWER(SQL_TEXT) NOT LIKE ‘%HASH%';
SQL_TEXT SQL_ID CHILD# HASH_VALUE PLAN_HASH_VALUE
—————————— ————- ———- ———- —————
select count(*) from test 3tcujqmqnqs8t 0 3981140249 2432738936
where id2=:”SYS_B_0″
– cleanup –
SYS>alter system set optimizer_features_enable=’11.2.0.1′;
drop table hr.test purge;
CONCLUSION:
CURSOR_SHARING = EXACT
– Causes maximum memory usage in library cache as two cursors – one parent and one child cursor are created for each distinct value of the bind variable.
– Gives best performance as optimizer creates different execution plan for each value of the bind variable.