UPDATE STATISTICS WITH FULL SCAN - High reads than normal
Hi
Update statistics when it runs with full sampling, reads high amount of data. my database size is 300GB and update stats is reading TBs of data from the data drive. Did anyone come across this wierd scenario?
Thanks
Vicky
Thanks
I have been running US on 1 TB database and did not notice high reading.... It does not block anyone but you may notice some performnace degradation...
Have you tried Ola's great utility
EXECUTE dbo.IndexOptimize @Databases = ‘USER_DATABASES’,
@FragmentationHigh_LOB = ‘INDEX_REBUILD_OFFLINE’,
@FragmentationHigh_NonLOB = ‘INDEX_REBUILD_ONLINE’,
@FragmentationMedium_LOB = ‘INDEX_REORGANIZE_STATISTICS_UPDATE’,
@FragmentationMedium_NonLOB = ‘INDEX_REORGANIZE_STATISTICS_UPDATE’,
@FragmentationLow_LOB = ‘NOTHING’,
@FragmentationLow_NonLOB = ‘NOTHING’,
@FragmentationLevel1 = 5,
@FragmentationLevel2 = 30,
@PageCountLevel = 1000
How does all this work?
Indexes with a fragmentation over 30% are rebuilt, online if possible.
Indexes with a fragmentation between 5% and 30% are reorganized and also have their statistics updated.
Index with a fragmentation under 5% (or a size of under 1000 pages) are ignored.
It is available on http://ola.hallengren.com.
Best Regards,Uri Dimant SQL Server MVP,
http://sqlblog.com/blogs/uri_dimant/
MS SQL optimization: MS SQL Development and Optimization
MS SQL Blog:
Large scale of database and data cleansing
Remote DBA Services:
Improves MS SQL Database Performance
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Regards>
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>
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Hello everybody,
Table MKPF
Statistics are current (|Changes| < 50 %)
New Method EH
New Sample Size ???
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Slow Query Using index. Fast with full table Scan.
Hi;
(Thanks for the links)
Here's my question correctly formated.
The query:
SELECT count(1)
from ehgeoconstru ec
where ec.TYPE='BAR'
AND ( ec.birthDate <= TO_DATE('2009-10-06 11:52:12', 'YYYY-MM-DD HH24:MI:SS') )
and deathdate is null
and substr(ec.strgfd, 1, length('[CIMText')) <> '[CIMText'Runs on 32 seconds!
Same query, but with one extra where clause:
SELECT count(1)
from ehgeoconstru ec
where ec.TYPE='BAR'
and ( (ec.contextVersion = 'REALWORLD') --- ADDED HERE
AND ( ec.birthDate <= TO_DATE('2009-10-06 11:52:12', 'YYYY-MM-DD HH24:MI:SS') ) )
and deathdate is null
and substr(ec.strgfd, 1, length('[CIMText')) <> '[CIMText'This runs in 400 seconds.
It should return data from one table, given the conditions.
The version of the database is Oracle9i Release 9.2.0.7.0
These are the parameters relevant to the optimizer:
SQL> show parameter optimizer
NAME TYPE VALUE
optimizer_dynamic_sampling integer 1
optimizer_features_enable string 9.2.0
optimizer_index_caching integer 99
optimizer_index_cost_adj integer 10
optimizer_max_permutations integer 2000
optimizer_mode string CHOOSE
SQL> Here is the output of EXPLAIN PLAN for the first fast query:
PLAN_TABLE_OUTPUT
| Id | Operation | Name | Rows | Bytes | Cost |
| 0 | SELECT STATEMENT | | | | |
| 1 | SORT AGGREGATE | | | | |
|* 2 | TABLE ACCESS FULL | EHCONS | | | |
Predicate Information (identified by operation id):
PLAN_TABLE_OUTPUT
2 - filter(SUBSTR("EC"."strgfd",1,8)<>'[CIMText' AND "EC"."DEATHDATE"
IS NULL AND "EC"."BIRTHDATE"<=TO_DATE('2009-10-06 11:52:12', 'yyyy
-mm-dd
hh24:mi:ss') AND "EC"."TYPE"='BAR')
Note: rule based optimizationHere is the output of EXPLAIN PLAN for the slow query:
PLAN_TABLE_OUTPUT
| |
| 1 | SORT AGGREGATE | | |
| |
|* 2 | TABLE ACCESS BY INDEX ROWID| ehgeoconstru | |
| |
|* 3 | INDEX RANGE SCAN | ehgeoconstru_VSN | |
| |
PLAN_TABLE_OUTPUT
Predicate Information (identified by operation id):
2 - filter(SUBSTR("EC"."strgfd",1,8)<>'[CIMText' AND "EC"."DEATHDATE" IS
NULL AND "EC"."TYPE"='BAR')
PLAN_TABLE_OUTPUT
3 - access("EC"."CONTEXTVERSION"='REALWORLD' AND "EC"."BIRTHDATE"<=TO_DATE('2
009-10-06
11:52:12', 'yyyy-mm-dd hh24:mi:ss'))
filter("EC"."BIRTHDATE"<=TO_DATE('2009-10-06 11:52:12', 'yyyy-mm-dd hh24:
mi:ss'))
Note: rule based optimizationThe TKPROF output for this slow statement is:
TKPROF: Release 9.2.0.7.0 - Production on Tue Nov 17 14:46:32 2009
Copyright (c) 1982, 2002, Oracle Corporation. All rights reserved.
Trace file: gen_ora_3120.trc
Sort options: prsela exeela fchela
count = number of times OCI procedure was executed
cpu = cpu time in seconds executing
elapsed = elapsed time in seconds executing
disk = number of physical reads of buffers from disk
query = number of buffers gotten for consistent read
current = number of buffers gotten in current mode (usually for update)
rows = number of rows processed by the fetch or execute call
SELECT count(1)
from ehgeoconstru ec
where ec.TYPE='BAR'
and ( (ec.contextVersion = 'REALWORLD')
AND ( ec.birthDate <= TO_DATE('2009-10-06 11:52:12', 'YYYY-MM-DD HH24:MI:SS') ) )
and deathdate is null
and substr(ec.strgfd, 1, length('[CIMText')) <> '[CIMText'
call count cpu elapsed disk query current rows
Parse 1 0.00 0.00 0 0 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 2 0.00 538.12 162221 1355323 0 1
total 4 0.00 538.12 162221 1355323 0 1
Misses in library cache during parse: 0
Optimizer goal: CHOOSE
Parsing user id: 153
Rows Row Source Operation
1 SORT AGGREGATE
27747 TABLE ACCESS BY INDEX ROWID OBJ#(73959)
2134955 INDEX RANGE SCAN OBJ#(73962) (object id 73962)
alter session set sql_trace=true
call count cpu elapsed disk query current rows
Parse 0 0.00 0.00 0 0 0 0
Execute 1 0.00 0.02 0 0 0 0
Fetch 0 0.00 0.00 0 0 0 0
total 1 0.00 0.02 0 0 0 0
Misses in library cache during parse: 0
Misses in library cache during execute: 1
Optimizer goal: CHOOSE
Parsing user id: 153
OVERALL TOTALS FOR ALL NON-RECURSIVE STATEMENTS
call count cpu elapsed disk query current rows
Parse 1 0.00 0.00 0 0 0 0
Execute 2 0.00 0.02 0 0 0 0
Fetch 2 0.00 538.12 162221 1355323 0 1
total 5 0.00 538.15 162221 1355323 0 1
Misses in library cache during parse: 0
Misses in library cache during execute: 1
OVERALL TOTALS FOR ALL RECURSIVE STATEMENTS
call count cpu elapsed disk query current rows
Parse 0 0.00 0.00 0 0 0 0
Execute 0 0.00 0.00 0 0 0 0
Fetch 0 0.00 0.00 0 0 0 0
total 0 0.00 0.00 0 0 0 0
Misses in library cache during parse: 0
2 user SQL statements in session.
0 internal SQL statements in session.
2 SQL statements in session.
Trace file: gen_ora_3120.trc
Trace file compatibility: 9.02.00
Sort options: prsela exeela fchela
2 sessions in tracefile.
2 user SQL statements in trace file.
0 internal SQL statements in trace file.
2 SQL statements in trace file.
2 unique SQL statements in trace file.
94 lines in trace file.Edited by: PauloSMO on 17/Nov/2009 4:21
Edited by: PauloSMO on 17/Nov/2009 7:07
Edited by: PauloSMO on 17/Nov/2009 7:38 - Changed title to be more correct.Although your optimizer_mode is choose, it appears that there are no statistics gathered on ehgeoconstru. The lack of cost estimate and estimated row counts from each step of the plan, and the "Note: rule based optimization" at the end of both plans would tend to confirm this.
Optimizer_mode choose means that if statistics are gathered then it will use the CBO, but if no statistics are present in any of the tables in the query, then the Rule Based Optimizer will be used. The RBO tends to be index happy at the best of times. I'm guessing that the index ehgeoconstru_VSN has contextversion as the leading column and also includes birthdate.
You can either gather statistics on the table (if all of the other tables have statistics) using dbms_stats.gather_table_stats, or hint the query to use a full scan instead of the index. Another alternative would be to apply a function or operation against the contextversion to preclude the use of the index. something like this:
SELECT COUNT(*)
FROM ehgeoconstru ec
WHERE ec.type='BAR' and
ec.contextVersion||'' = 'REALWORLD'
ec.birthDate <= TO_DATE('2009-10-06 11:52:12', 'YYYY-MM-DD HH24:MI:SS') and
deathdate is null and
SUBSTR(ec.strgfd, 1, LENGTH('[CIMText')) <> '[CIMText'or perhaps UPPER(ec.contextVersion) if that would not change the rows returned.
John -
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For information about the Other category in the Storage display, see this support article. If the Storage display seems to be inaccurate, try rebuilding the Spotlight index.
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sudo /Applications/GrandPerspective.app/Contents/MacOS/GrandPerspective
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☞ In the Finder, select Go ▹ Utilities from the menu bar, or press the key combination shift-command-U. The application is in the folder that opens.
☞ Open LaunchPad. Click Utilities, then Terminal in the icon grid.
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When you're done with the app, quit it and also quit Terminal. -
How oracle decide whetehr to use index or full scan (statistics)
Hi Guys,
Let say i have a index on a column.
The table and index statistics has been gathered. (without histograms).
Let say i perform a select * from table where a=5;
Oracle will perform a full scan.
But from which statistics it will be able to know indeed most of the column = 5? (histograms not used)
After analyzing, we get the below:
Table Statistics :
(NUM_ROWS)
(BLOCKS)
(EMPTY_BLOCKS)
(AVG_SPACE)
(CHAIN_COUNT)
(AVG_ROW_LEN)
Index Statistics :
(BLEVEL)
(LEAF_BLOCKS)
(DISTINCT_KEYS)
(AVG_LEAF_BLOCKS_PER_KEY)
(AVG_DATA_BLOCKS_PER_KEY)
(CLUSTERING_FACTOR)
thanks
Index Column (A)
======
1
1
2
2
5
5
5
5
5
5I have prepared some explanation and have not noticed that the topic has been marked as answered.
This my sentence is not completely true.
A column "without histograms" means that the column has only one bucket. More correct: even without histograms there are data in dba_tab_histograms which we can consider as one bucket for whole column. In fact these data are retrieved from hist_head$, not from histgrm$ as usual buckets.
Technically there is no any buckets without gathered histograms.
Let's create a table with skewed data distribution.
SQL> create table t as
2 select least(rownum,3) as val, '*' as pad
3 from dual
4 connect by level <= 1000000;
Table created
SQL> create index idx on t(val);
Index created
SQL> select val, count(*)
2 from t
3 group by val;
VAL COUNT(*)
1 1
2 1
3 999998So, we have table with very skewed data distribution.
Let's gather statistics without histograms.
SQL> exec dbms_stats.gather_table_stats( user, 'T', estimate_percent => 100, method_opt => 'for all columns size 1', cascade => true);
PL/SQL procedure successfully completed
SQL> select blocks, num_rows from dba_tab_statistics
2 where table_name = 'T';
BLOCKS NUM_ROWS
3106 1000000
SQL> select blevel, leaf_blocks, clustering_factor
2 from dba_ind_statistics t
3 where table_name = 'T'
4 and index_name = 'IDX';
BLEVEL LEAF_BLOCKS CLUSTERING_FACTOR
2 4017 3107
SQL> select column_name,
2 num_distinct,
3 density,
4 num_nulls,
5 low_value,
6 high_value
7 from dba_tab_col_statistics
8 where table_name = 'T'
9 and column_name = 'VAL';
COLUMN_NAME NUM_DISTINCT DENSITY NUM_NULLS LOW_VALUE HIGH_VALUE
VAL 3 0,33333333 0 C102 C104So, Oracle suggests that values between 1 and 3 (raw C102 and C104) are distributed uniform and the density of the distribution is 0.33.
Let's try to explain plan
SQL> explain plan for
2 select --+ no_cpu_costing
3 *
4 from t
5 where val = 1
6 ;
Explained
SQL> @plan
| Id | Operation | Name | Rows | Cost |
| 0 | SELECT STATEMENT | | 333K| 300 |
|* 1 | TABLE ACCESS FULL| T | 333K| 300 |
Predicate Information (identified by operation id):
1 - filter("VAL"=1)
Note
- cpu costing is off (consider enabling it)Below is an excerpt from trace 10053
BASE STATISTICAL INFORMATION
Table Stats::
Table: T Alias: T
#Rows: 1000000 #Blks: 3106 AvgRowLen: 5.00
Index Stats::
Index: IDX Col#: 1
LVLS: 2 #LB: 4017 #DK: 3 LB/K: 1339.00 DB/K: 1035.00 CLUF: 3107.00
SINGLE TABLE ACCESS PATH
BEGIN Single Table Cardinality Estimation
Column (#1): VAL(NUMBER)
AvgLen: 3.00 NDV: 3 Nulls: 0 Density: 0.33333 Min: 1 Max: 3
Table: T Alias: T
Card: Original: 1000000 Rounded: 333333 Computed: 333333.33 Non Adjusted: 333333.33
END Single Table Cardinality Estimation
Access Path: TableScan
Cost: 300.00 Resp: 300.00 Degree: 0
Cost_io: 300.00 Cost_cpu: 0
Resp_io: 300.00 Resp_cpu: 0
Access Path: index (AllEqRange)
Index: IDX
resc_io: 2377.00 resc_cpu: 0
ix_sel: 0.33333 ix_sel_with_filters: 0.33333
Cost: 2377.00 Resp: 2377.00 Degree: 1
Best:: AccessPath: TableScan
Cost: 300.00 Degree: 1 Resp: 300.00 Card: 333333.33 Bytes: 0Cost of FTS here is 300 and cost of Index Range Scan here is 2377.
I have disabled cpu costing, so selectivity does not affect the cost of FTS.
cost of Index Range Scan is calculated as
blevel + (leaf_blocks * selectivity + clustering_factor * selecivity) = 2 + (4017*0.33333 + 3107*0.33333) = 2377.
Oracle considers that it has to read 2 root/branch blocks of the index, 1339 leaf blocks of the index and 1036 blocks of the table.
Pay attention that selectivity is the major component of the cost of the Index Range Scan.
Let's try to gather histograms:
SQL> exec dbms_stats.gather_table_stats( user, 'T', estimate_percent => 100, method_opt => 'for columns val size 3', cascade => true);
PL/SQL procedure successfully completedIf you look at dba_tab_histograms you will see following
SQL> select endpoint_value,
2 endpoint_number
3 from dba_tab_histograms
4 where table_name = 'T'
5 and column_name = 'VAL'
6 ;
ENDPOINT_VALUE ENDPOINT_NUMBER
1 1
2 2
3 1000000ENDPOINT_VALUE is the column value (in number for any type of data) and ENDPOINT_NUMBER is cumulative number of rows.
Number of rows for any ENDPOINT_VALUE = ENDPOINT_NUMBER for this ENDPOINT_VALUE - ENDPOINT_NUMBER for the previous ENDPOINT_VALUE.
explain plan and 10053 trace of the same query:
| Id | Operation | Name | Rows | Cost |
| 0 | SELECT STATEMENT | | 1 | 4 |
| 1 | TABLE ACCESS BY INDEX ROWID| T | 1 | 4 |
|* 2 | INDEX RANGE SCAN | IDX | 1 | 3 |
Predicate Information (identified by operation id):
2 - access("VAL"=1)
Note
- cpu costing is off (consider enabling it)
BASE STATISTICAL INFORMATION
Table Stats::
Table: T Alias: T
#Rows: 1000000 #Blks: 3106 AvgRowLen: 5.00
Index Stats::
Index: IDX Col#: 1
LVLS: 2 #LB: 4017 #DK: 3 LB/K: 1339.00 DB/K: 1035.00 CLUF: 3107.00
SINGLE TABLE ACCESS PATH
BEGIN Single Table Cardinality Estimation
Column (#1): VAL(NUMBER)
AvgLen: 3.00 NDV: 3 Nulls: 0 Density: 5.0000e-07 Min: 1 Max: 3
Histogram: Freq #Bkts: 3 UncompBkts: 1000000 EndPtVals: 3
Table: T Alias: T
Card: Original: 1000000 Rounded: 1 Computed: 1.00 Non Adjusted: 1.00
END Single Table Cardinality Estimation
Access Path: TableScan
Cost: 300.00 Resp: 300.00 Degree: 0
Cost_io: 300.00 Cost_cpu: 0
Resp_io: 300.00 Resp_cpu: 0
Access Path: index (AllEqRange)
Index: IDX
resc_io: 4.00 resc_cpu: 0
ix_sel: 1.0000e-06 ix_sel_with_filters: 1.0000e-06
Cost: 4.00 Resp: 4.00 Degree: 1
Best:: AccessPath: IndexRange Index: IDX
Cost: 4.00 Degree: 1 Resp: 4.00 Card: 1.00 Bytes: 0Pay attention on selectivity, ix_sel: 1.0000e-06
Cost of the FTS is still the same = 300,
but cost of the Index Range Scan is 4 now: 2 root/branch blocks + 1 leaf block + 1 table block.
Thus, conclusion: histograms allows to calculate selectivity more accurate. The aim is to have more efficient execution plans.
Alexander Anokhin
http://alexanderanokhin.wordpress.com/ -
Do partition scans take longer than a full table scan on an unpartitioned table?
Hello there,
I have a range-partitioned table PART_TABLE which has 10 Million records and 10 partitions having 1 million records each. Partition is done based on a Column named ID which is a sequence from 1 to 10 million.
I created another table P2_BKP (doing a select * from part_table) which has the same dataset as that of PART_TABLE except that this table is not partitioned.
Now, I run a same query on both the tables to retrieve a range of data. Precisely I am trying to read only the data present in 5 partitions of the partitioned tables which theoretically requires less reads than when done on unpartitioned tables.
Yet, the query seems to take extra time on partitioned table than when run on unpartitioned table.Any specific reason why is this the case?
Below is the query I am trying to run on both the tables and their corresponding Explain Plans.
QUERY A
=========
select * from P2_BKP where id<5000000;
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
| 0 | SELECT STATEMENT | | 6573K| 720M| 12152 (2)| 00:02:26 |
|* 1 | TABLE ACCESS FULL| P2_BKP | 6573K| 720M| 12152 (2)| 00:02:26 |
QUERY B
========
select * from part_table where id<5000000;
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | Pstart| Pstop |
| 0 | SELECT STATEMENT | | 3983K| 436M| 22181 (73)| 00:04:27 | | |
| 1 | PARTITION RANGE ITERATOR| | 3983K| 436M| 22181 (73)| 00:04:27 | 1 | 5 |
|* 2 | TABLE ACCESS FULL | PART_TABLE | 3983K| 436M| 22181 (73)| 00:04:27 | 1 | 5 |at the risk of bringing unnecessary confusion into the discussion: I think there is a situation in 11g in which a Full Table Scan on a non partitioned table can be faster than the FTS on a corresponding partitioned table: if the size of the non partitioned table reaches a certain threshold (I think it's: blocks > _small_table_threshold * 5) the runtime engine may decide to use a serial direct path read to access the data. If the single partitions don't pass the threshold the engine will use the conventional path.
Here is a small example for my assertion:
-- I create a simple partitioned table
-- and a corresponding non-partitioned table
-- with 1M rows
drop table tab_part;
create table tab_part (
col_part number
, padding varchar2(100)
partition by list (col_part)
partition P00 values (0)
, partition P01 values (1)
, partition P02 values (2)
, partition P03 values (3)
, partition P04 values (4)
, partition P05 values (5)
, partition P06 values (6)
, partition P07 values (7)
, partition P08 values (8)
, partition P09 values (9)
insert into tab_part
select mod(rownum, 10)
, lpad('*', 100, '*')
from dual
connect by level <= 1000000;
exec dbms_stats.gather_table_stats(user, 'tab_part')
drop table tab_nopart;
create table tab_nopart
as
select *
from tab_part;
exec dbms_stats.gather_table_stats(user, 'tab_nopart')
-- my _small_table_threshold is 1777 and the partitions
-- have a size of ca. 1600 blocks while the non-partitioned table
-- contains 15360 blocks
-- I have to flush the buffer cache since
-- the direct path access is only used
-- if there are few blocks already in the cache
alter system flush buffer_cache;
-- the execution plans are not really exciting
| Id | Operation | Name | Rows | Cost (%CPU)| Time | Pstart| Pstop |
| 0 | SELECT STATEMENT | | 1 | 8089 (0)| 00:00:41 | | |
| 1 | SORT AGGREGATE | | 1 | | | | |
| 2 | PARTITION LIST ALL| | 1000K| 8089 (0)| 00:00:41 | 1 | 10 |
| 3 | TABLE ACCESS FULL| TAB_PART | 1000K| 8089 (0)| 00:00:41 | 1 | 10 |
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
| 0 | SELECT STATEMENT | | 1 | 7659 (0)| 00:00:39 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| TAB_NOPART | 1000K| 7659 (0)| 00:00:39 |
But on my PC the FTS on the non-partitioned table is faster than the FTS on the partitions (1sec to 3 sec.) and v$sesstat shows the reason for this difference:
-- non partitioned table
NAME DIFF
table scan rows gotten 1000000
file io wait time 15313
session logical reads 15156
physical reads 15153
consistent gets direct 15152
physical reads direct 15152
DB time 95
-- partitioned table
NAME DIFF
file io wait time 2746493
table scan rows gotten 1000000
session logical reads 15558
physical reads 15518
physical reads cache prefetch 15202
DB time 295
(maybe my choose of counters is questionable)
So it's possible to get slower access for an FTS on a partitioned table under special conditions.
Regards
Martin -
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