SCEP runs unscheduled full scans daily on some PC's
In Configuration Manager I've set daily quick scans, and once every thursday a full scan. This works perfectly for almost every client, except a few. They display the correct information in "Settings" (below), but instead the run full scans every
day. It seems to start between 01:00-02:00 at night.
The logs just tell me that yes indeed a scan was started, but not why.
Immediately when i change the policy in Configuration Manager and remove the full scan on thursdays, the daily full scans stop on the clients with the problems. Very strange behaviour..
Hi,
Have you checked the Eventlog?
Please check the Event Viewer -> Windows Logs -> System to see if the user who started full scans on Thusday and other dates are the same account.
Best Regards,
Joyce
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100% C: usage when SCEP is running a full scan
Hi,
I currently have an issue where a 2012 Server is filling its C: drive while running full scans.
Files are created in C:\Windows\Temp with the prefix TMP00000 and no extension. These seem to grow in size until they are filling the remaining free space (~40GB) and then once it hits ~100MB free it deletes the file.
Procmon tells me that the MsMpEng.exe is creating and writing to this file.
Can anyone tell me why it is creating these files and why it is trying to use all remaining free space? I cant see any configuration to allow me to change from C:\Windows\Temp (I can only assume its using the %temp% variable).
I notice that while the file is filling up and eating free space, the Item listed on the SCEP console is still changing so it doesn't appear to be getting stuck on a single file. I also note that there are no entries listed in the Event viewer either.
Thanks
AlexHi Alex,
Maybe you can use one of the following workarounds:
1. Disable Scan archive files.
2. Limit the archive size and depth. This can be only configured via GPO (ADMX).
3. Change the temp folder location.
http://blogs.msdn.com/b/astebner/archive/2006/01/15/513134.aspx
Thanks.
Please remember to click “Mark as Answer” on the post that helps you, and to click “Unmark as Answer” if a marked post does not actually answer your question. This can be beneficial to other community members reading the thread. -
Proxies Missing and Full scans not working
We had some major issues upgrading our Facilis shared storage to 'multi-user write' and in the process we lost all the proxy files/thumbnails for any video that was NOT stored in a archive device in FCSvr.
Now the Facilis is stabilized - but I am left will 20,000 plus video flies that I need to recreate proxies/thumnails for. Any of the video that was stored on archive devices in FCSvr still has all the proxy files/thumbnails intact.
So - I think I need to run a full scan on my edit-in-place device that has all the clips with missing proxy files? But this is not working. I keep getting:
"ERROR: E_INVAL pm_transcoder ..."
I can add new media to FCSvr with no problem - proxies and thumbnails get generated. But I can not seem to figure out how to make a full scan work. It tries three files - they all immediately fail and the full scan stops.
Any help would be greatly appreciated!In the process of making the changes you made, did any paths change? If the device paths changed at all, you could have trouble if you don't change the device in FCSvr too.
Also, Did you bring FCSvr down before you started the upgrade? If FCSvr was running and tried to access the device while it was not mounted, specifically if it tried to write to either the shared storage device or the proxy device while it was unavailable to FCSvr, it may have created a "phantom device" in the /Volumes folder. I've had this problem several times. Try this - using Terminal on the FCSvr machine, cd to /Volumes. Does everything look normal there? -
I've used Firefox for years with no problems. I've not installed any new extensions or programs between the time it worked great and when the problem began.
I click to start FF (shortcut, directly on program file, etc.) and I get an hourglass for a couple of seconds like it's starting, then nada. As mentioned, I see FF show up as a process in Task Manager for about 3 seconds, then disappear.
Solutions I've tried include: Uninstall/reinstall on the same drive and a separate drive, registry cleanup/optimization w/Norton Utilities, full scan for malware/viruses with Norton, stopping some autostartup programs that were installed and in use while FF was running fine.I've used Firefox for years with no problems. I've not installed any new extensions or programs between the time it worked great and when the problem began.
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I am running full upload daily now i want to load delta records how can i
Hi Gurus,
I am running full upload daily now i want to load delta records how can i do that could any body help recording this..
Regards
Kiran KumarHello Kiran,
First make sure that you datasource is delta enabled. If so create a new infopackage for Init without data transfer and execute it. Then create a InfoPackage for Delta.
Thanks
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How do you run a full virus scan
How do you run a full virus scan on the Imac
Take a look at this article:
http://en.wikipedia.org/wiki/Phishing
Running a virus scan would not help at all - someone got your email address and is now hoping you're dumb enough to use the links they provided and give them information so they can use your credit card or checking acount number, password, etc.
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The plan doesn't use the index but the cost of INDEX FULL SCAN looks better
Hi,
Well, I'm sure I miss the boat... and if the question is pretty tricky, the answer is probably :"You're stupid Greg!". Well anyway, you'll probably be interested in trying to answer it as I've spent some times on it without any result ! I use Oracle XE on Windows...
1) Below is my query and its plan. You'll find the full schema generation script at the end of this email. Look at the cost (468) of the plan and the cost of the same query when you force the use of the index (116). Why is this ?
select count(distinct col5)
2 from demo
3 where col1 between 1 and 50000
4 and col2=col1
5 and col3=col1
6 and col4=col1;
Plan d'execution
Plan hash value: 2072716547
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
| 0 | SELECT STATEMENT | | 1 | 116 | 468 (2)| 00:00:06 |
| 1 | SORT GROUP BY | | 1 | 116 | | |
|* 2 | TABLE ACCESS FULL| DEMO | 1 | 116 | 468 (2)| 00:00:06 |
Predicate Information (identified by operation id):
2 - filter("COL2"="COL1" AND "COL3"="COL1" AND "COL4"="COL1" AND
"COL1"<=50000 AND "COL2"<=50000 AND "COL3"<=50000 AND "COL4"<=50000 AND
"COL1">=1 AND "COL2">=1 AND "COL3">=1 AND "COL4">=1)2) When I force the use of an index (with a Hint), You'll see the cost of the plan is 116 which is definitly better than the TABLE ACCESS FULL (468) :
SQL> l
1 select /*+ index(demo demo_idx)*/ count(distinct col5)
2 from demo
3 where col1 between 1 and 50000
4 and col2=col1
5 and col3=col1
6* and col4=col1
SQL> /
Plan d'execution
Plan hash value: 189561699
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
| 0 | SELECT STATEMENT | | 1 | 116 | 437 (2)| 00:00:06 |
| 1 | SORT GROUP BY | | 1 | 116 | | |
| 2 | TABLE ACCESS BY INDEX ROWID| DEMO | 1 | 116 | 437 (2)| 00:00:06 |
|* 3 | INDEX FULL SCAN | DEMO_IDX | 1 | | 436 (2)| 00:00:06 |
Predicate Information (identified by operation id):
3 - filter("COL2"="COL1" AND "COL3"="COL1" AND "COL4"="COL1" AND
"COL1"<=50000 AND "COL2"<=50000 AND "COL3"<=50000 AND "COL4"<=50000 AND
"COL1">=1 AND "COL2">=1 AND "COL3">=1 AND "COL4">=1)3) My question is why is plan1 used while plan2 should be considered better by the optimizer regarding the cost (to make the case even more complex, plan1 is actually more efficient but this is out of the scope of my question. I know that and I know why !).
You'll find a script to generate the structures and data below. I can send you the 10053 traces if you what to go furthermore. Take care the index is a REVERSE index (Don't know if query rewrite should be enabled in order to take advantage of this type of index but it is set to "true" (and "trusted") :
drop table demo;
create table demo (col1 number not null,
col2 number,
col3 number,
col4 number,
col5 varchar2(500));
begin
for i in 1..100000 loop
insert into demo values (i,i,i,i,'This column is used to raise the High Water Mark and '||
' the cost of an TABLE ACCESS FULL operation');
end loop;
end;
commit;
create index demo_idx on demo(col1,col2,col3,col4) reverse;
exec dbms_stats.gather_table_stats(USER, 'DEMO', cascade=>true, -
method_opt=>'FOR ALL COLUMNS SIZE 254', no_invalidate=>false) Any comments are welcome ! Best Regards,
Gregory
Message was edited by:
arkzoyd... I've added the "pre" tagsI suspect this has something to do with db_file_multiblock_read_count
After running provided creation statements by you I got following results:
SQL> show parameter multiblock
NAME TYPE VALUE
db_file_multiblock_read_count integer 16
SQL> set autotrace on
SQL> select count(distinct col5)
2 from demo
3 where col1 between 1 and 50000
4 and col2=col1
5 and col3=col1
6 and col4=col1
7 /
COUNT(DISTINCTCOL5)
1
Execution Plan
Plan hash value: 2072716547
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
| 0 | SELECT STATEMENT | | 1 | 116 | 375 (1)| 00:00:05 |
| 1 | SORT GROUP BY | | 1 | 116 | | |
|* 2 | TABLE ACCESS FULL| DEMO | 1 | 116 | 375 (1)| 00:00:05 |
Predicate Information (identified by operation id):
2 - filter("COL2"="COL1" AND "COL3"="COL1" AND "COL4"="COL1" AND
"COL1"<=50000 AND "COL2"<=50000 AND "COL3"<=50000 AND "COL4"<=5000
0 AND
"COL1">=1 AND "COL2">=1 AND "COL3">=1 AND "COL4">=1)
Statistics
196 recursive calls
0 db block gets
1734 consistent gets
850 physical reads
0 redo size
422 bytes sent via SQL*Net to client
385 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
7 sorts (memory)
0 sorts (disk)
1 rows processed
SQL> select /*+ index(demo demo_idx)*/ count(distinct col5)
2 from demo
3 where col1 between 1 and 50000
4 and col2=col1
5 and col3=col1
6 and col4=col1
7 /
COUNT(DISTINCTCOL5)
1
Execution Plan
Plan hash value: 189561699
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| T
ime |
| 0 | SELECT STATEMENT | | 1 | 116 | 431 (1)| 0
0:00:06 |
| 1 | SORT GROUP BY | | 1 | 116 | |
|
| 2 | TABLE ACCESS BY INDEX ROWID| DEMO | 1 | 116 | 431 (1)| 0
0:00:06 |
|* 3 | INDEX FULL SCAN | DEMO_IDX | 1 | | 430 (1)| 0
0:00:06 |
Predicate Information (identified by operation id):
3 - filter("COL2"="COL1" AND "COL3"="COL1" AND "COL4"="COL1" AND
"COL1"<=50000 AND "COL2"<=50000 AND "COL3"<=50000 AND "COL4"<=5000
0 AND
"COL1">=1 AND "COL2">=1 AND "COL3">=1 AND "COL4">=1)
Statistics
1 recursive calls
0 db block gets
50426 consistent gets
428 physical reads
0 redo size
422 bytes sent via SQL*Net to client
385 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
1 sorts (memory)
0 sorts (disk)
1 rows processedNow I modify multiblock_read_count and full scan cost is going up although anyway Oracle by default chooses full scan instead of index access.
SQL> alter session set db_file_multiblock_read_count = 8;
Session altered.
SQL> select count(distinct col5)
2 from demo
3 where col1 between 1 and 50000
4 and col2=col1
5 and col3=col1
6 and col4=col1
7 /
COUNT(DISTINCTCOL5)
1
Execution Plan
Plan hash value: 2072716547
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
| 0 | SELECT STATEMENT | | 1 | 116 | 463 (1)| 00:00:06 |
| 1 | SORT GROUP BY | | 1 | 116 | | |
|* 2 | TABLE ACCESS FULL| DEMO | 1 | 116 | 463 (1)| 00:00:06 |
Predicate Information (identified by operation id):
2 - filter("COL2"="COL1" AND "COL3"="COL1" AND "COL4"="COL1" AND
"COL1"<=50000 AND "COL2"<=50000 AND "COL3"<=50000 AND "COL4"<=5000
0 AND
"COL1">=1 AND "COL2">=1 AND "COL3">=1 AND "COL4">=1)
Statistics
1 recursive calls
0 db block gets
1697 consistent gets
850 physical reads
0 redo size
422 bytes sent via SQL*Net to client
385 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
1 sorts (memory)
0 sorts (disk)
1 rows processed
SQL> select /*+ index(demo demo_idx)*/ count(distinct col5)
2 from demo
3 where col1 between 1 and 50000
4 and col2=col1
5 and col3=col1
6 and col4=col1
7 /
COUNT(DISTINCTCOL5)
1
Execution Plan
Plan hash value: 189561699
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| T
ime |
| 0 | SELECT STATEMENT | | 1 | 116 | 431 (1)| 0
0:00:06 |
| 1 | SORT GROUP BY | | 1 | 116 | |
|
| 2 | TABLE ACCESS BY INDEX ROWID| DEMO | 1 | 116 | 431 (1)| 0
0:00:06 |
|* 3 | INDEX FULL SCAN | DEMO_IDX | 1 | | 430 (1)| 0
0:00:06 |
Predicate Information (identified by operation id):
3 - filter("COL2"="COL1" AND "COL3"="COL1" AND "COL4"="COL1" AND
"COL1"<=50000 AND "COL2"<=50000 AND "COL3"<=50000 AND "COL4"<=5000
0 AND
"COL1">=1 AND "COL2">=1 AND "COL3">=1 AND "COL4">=1)
Statistics
1 recursive calls
0 db block gets
50426 consistent gets
0 physical reads
0 redo size
422 bytes sent via SQL*Net to client
385 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
1 sorts (memory)
0 sorts (disk)
1 rows processedSo I don't know what is the default value of dbfmbrc in XE and not gone too deep to understand how for example system statistics may change your situation.
Gints Plivna
http://www.gplivna.eu
P.S. BTW I used Oracle Database 10g Enterprise Edition Release 10.2.0.1.0.
Message was edited by:
gintsp
listened to Williams suggestion :) -
Why a table full scan when I've got the PK in the WHERE clause?
There is a very complex query that I need to optimize in an Oracle 10gR2 environment. I am deconstructing it into layers to see what is causing the first bottleneck. The innermost portion is fine, with an explain plan cost of 54. With a typical value for the bind variable, it returns 125 zero_id values. There are over 100,000 rows in table T_ONE in my test database, but my customer has over one million rows in their production instance.
WITH t_merged_id AS (SELECT DISTINCT zero_id FROM t_zero WHERE NVL(column2, zero_id) = :i_id)
SELECT t_one.one_id
FROM t_one
INNER JOIN t_two
ON t_one.column1 = t_two.two_id
INNER JOIN t_merged_id
ON t_two.column10 = t_merged_id.zero_id
UNION ALL
SELECT t_one.one_id
FROM t_one
INNER JOIN t_two
ON t_one.column2 = t_two.two_id
INNER JOIN t_merged_id
ON t_two.column10 = t_merged_id.zero_id
UNION ALL
SELECT t_one.one_id
FROM t_one
INNER JOIN t_three
ON t_one.column3 = t_three.three_id
INNER JOIN t_merged_id
ON t_three.column10 = t_merged_id.zero_id
UNION ALL
SELECT t_one.one_id
FROM t_one
INNER JOIN t_four
ON t_one.column4 = t_four.four_id
INNER JOIN t_two
ON t_four.column1 = t_two.two_id
INNER JOIN t_merged_id
ON t_two.two_id = t_merged_id.zero_id
UNION ALL
SELECT t_one.one_id
FROM t_one INNER JOIN t_merged_id ON t_one.column5 = t_merged_id.zero_id
UNION
SELECT t_one.one_id
FROM t_one INNER JOIN t_merged_id ON t_one.column6 = t_merged_id.zero_idHowever, the next step is to obtain a bunch of columns from T_ONE for each of those ONE_ID values. Adding that looks like the following, which causes a table full scan on T_ONE (and an explain plan cost over 1,500 for this query in my test system) and it takes far too long to return the results.
SELECT t_one.*
FROM t_one
INNER JOIN
(--This is the start of the query shown above
WITH t_merged_id AS (SELECT DISTINCT zero_id FROM t_zero WHERE NVL(column2, zero_id) = :i_id)
SELECT t_one.one_id
FROM t_one
INNER JOIN t_two
ON t_one.column1 = t_two.two_id
INNER JOIN t_merged_id
ON t_two.column10 = t_merged_id.zero_id
UNION ALL
SELECT t_one.one_id
FROM t_one
INNER JOIN t_two
ON t_one.column2 = t_two.two_id
INNER JOIN t_merged_id
ON t_two.column10 = t_merged_id.zero_id
UNION ALL
SELECT t_one.one_id
FROM t_one
INNER JOIN t_three
ON t_one.column3 = t_three.three_id
INNER JOIN t_merged_id
ON t_three.column10 = t_merged_id.zero_id
UNION ALL
SELECT t_one.one_id
FROM t_one
INNER JOIN t_four
ON t_one.column4 = t_four.four_id
INNER JOIN t_two
ON t_four.column1 = t_two.two_id
INNER JOIN t_merged_id
ON t_two.two_id = t_merged_id.zero_id
UNION ALL
SELECT t_one.one_id
FROM t_one INNER JOIN t_merged_id ON t_one.column5 = t_merged_id.zero_id
UNION
SELECT t_one.one_id
FROM t_one INNER JOIN t_merged_id ON t_one.column6 = t_merged_id.zero_id
--This is the end of the query shown above
) t_list
ON t_one.one_id = t_list.one_idMy question is, why wouldn’t Oracle use the existing index PK_T_ONE, which is keyed on T_ONE.ONE_ID? I tried refactoring the query using a “WHERE t_one.one_id IN” construct instead of the INNER JOIN but it didn’t make any difference. Neither did adding an index hint, which I hoped would force the use of the PK index.
Any ideas?I was able to completely resolve my problem, but I still want to understand why the original query wouldn't use an index.
(My solution was to move all the joins and where clauses from the query that wrapped the one we've been discussing and put them into each SELECT in the UNION, so there is no longer any inner subquery. So instead of trying to first get a list of ID values from the subquery, get the full records for the IDs from an outer query, and then joining to the outer query, I made SELECT in the UNION contain the full logic. This makes the query a lot more verbose, because all the joins and wheres are repeated six times, but it does use the index and returns in 0.04 seconds instead of over nine minutes in my test database.)
hoek wrote:
Values for optimizer_index_caching and optimizer_index_cost_adj are not the defaults. Any reasons for that?I am not a DBA and have no idea. However, I did a Google search and found this: http://decipherinfosys.wordpress.com/2007/02/13/optimizer_index_cost_adj-and-optimizer_index_caching/. Apparently Tom Kyte would approve more of our settings than the defaults.
hoek wrote:
Any chance to get a realistic dataset on your test server?Unfortunately, not for quite some time. The customer won't provide any real data, and generating data for testing is complex because of all the interrelationships. I have someone working on that. However, I was able to get back on the primary test server that has 136k records in the main table instead of only 2k. So far as I know, the Oracle configuration between the test server and the customer's server is the same. However, they have much more serious hardware that I do (more processors, more RAM, more platters). On the other hand, they have 10 times as much data.
hoek wrote:
Your second execution plan contains differents stats, they're not the 'common ones'. (E-rows etc.)The predicates are the same as the first. The 2nd plan was generated by the 10g-specific portion of Randolph's script using the command "select * from table(dbms_xplan.display_cursor(null, null, 'ALLSTATS LAST'));".
This is the result from running the script on the main test server:
NAME TYPE VALUE
_optimizer_cost_based_transformation string OFF
optimizer_dynamic_sampling integer 2
optimizer_features_enable string 10.2.0.4
optimizer_index_caching integer 95
optimizer_index_cost_adj integer 10
optimizer_mode string CHOOSE
optimizer_secure_view_merging boolean TRUE
db_file_multiblock_read_count integer 32
db_block_size integer 8192
cursor_sharing string FORCE
SNAME PNAME PVAL1 PVAL2
SYSSTATS_INFO STATUS COMPLETED
SYSSTATS_INFO DSTART 04-04-2008 07:02
SYSSTATS_INFO DSTOP 04-04-2008 07:02
SYSSTATS_INFO FLAGS 1
SYSSTATS_MAIN CPUSPEEDNW 646.57331
SYSSTATS_MAIN IOSEEKTIM 10
SYSSTATS_MAIN IOTFRSPEED 4096
SYSSTATS_MAIN SREADTIM
SYSSTATS_MAIN MREADTIM
SYSSTATS_MAIN CPUSPEED
SYSSTATS_MAIN MBRC
SYSSTATS_MAIN MAXTHR
SYSSTATS_MAIN SLAVETHR
SQL> SELECT st.*
2 FROM t_senttsk st INNER JOIN (WITH t_mrgdusr AS (SELECT DISTINCT usr_id
3 FROM t_usr
4 WHERE NVL(usr_mrgemstr, usr_id) = 10000002 /* i_payer_id */
5 )
6 SELECT t_senttsk.setk_id
7 FROM t_senttsk INNER JOIN t_mrgdusr
8 ON t_senttsk.setk_affn_memb = t_mrgdusr.usr_id
9 UNION
10 SELECT t_senttsk.setk_id
11 FROM t_senttsk INNER JOIN t_mrgdusr
12 ON t_senttsk.setk_ownr = t_mrgdusr.usr_id) t_affil
13 ON st.setk_id = t_affil.setk_id;
no rows selected
Elapsed: 00:13:14.54
Execution Plan
Plan hash value: 1241660758
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
| 0 | SELECT STATEMENT | | 169K| 64M| 1403 (3)| 00:00:17 |
| 1 | NESTED LOOPS | | 169K| 64M| 1403 (3)| 00:00:17 |
| 2 | TABLE ACCESS FULL | T_SENTTSK | 136K| 51M| 1400 (3)| 00:00:17 |
| 3 | VIEW | | 1 | 6 | 1 (0)| 00:00:01 |
| 4 | TEMP TABLE TRANSFORMATION | | | | | |
| 5 | LOAD AS SELECT | | | | | |
| 6 | TABLE ACCESS BY INDEX ROWID | T_USR | 1 | 8 | 1 (0)| 00:00:01 |
|* 7 | INDEX RANGE SCAN | IX_NVL_USR_MRGEMSTR_USR_ID | 1 | | 1 (0)| 00:00:01 |
| 8 | SORT UNIQUE | | | | | |
| 9 | UNION-ALL PARTITION | | | | | |
| 10 | NESTED LOOPS | | 1 | 25 | 3 (0)| 00:00:01 |
| 11 | TABLE ACCESS BY INDEX ROWID| T_SENTTSK | 1 | 12 | 1 (0)| 00:00:01 |
|* 12 | INDEX UNIQUE SCAN | PK_T_SENTTSK | 1 | | 1 (0)| 00:00:01 |
|* 13 | VIEW | | 1 | 13 | 2 (0)| 00:00:01 |
| 14 | TABLE ACCESS FULL | SYS_TEMP_0FD9D6608_399116CE | 1 | 6 | 2 (0)| 00:00:01 |
| 15 | NESTED LOOPS | | 1 | 22 | 3 (0)| 00:00:01 |
|* 16 | TABLE ACCESS BY INDEX ROWID| T_SENTTSK | 1 | 9 | 1 (0)| 00:00:01 |
|* 17 | INDEX UNIQUE SCAN | PK_T_SENTTSK | 1 | | 1 (0)| 00:00:01 |
|* 18 | VIEW | | 1 | 13 | 2 (0)| 00:00:01 |
| 19 | TABLE ACCESS FULL | SYS_TEMP_0FD9D6608_399116CE | 1 | 6 | 2 (0)| 00:00:01 |
Predicate Information (identified by operation id):
7 - access(NVL("USR_MRGEMSTR","USR_ID")=10000002)
12 - access("T_SENTTSK"."SETK_ID"="ST"."SETK_ID")
13 - filter("T_SENTTSK"."SETK_AFFN_MEMB"="T_MRGDUSR"."USR_ID")
16 - filter("T_SENTTSK"."SETK_OWNR" IS NOT NULL)
17 - access("T_SENTTSK"."SETK_ID"="ST"."SETK_ID")
18 - filter("T_SENTTSK"."SETK_OWNR"="T_MRGDUSR"."USR_ID")
Statistics
349 recursive calls
275041 db block gets
1239881 consistent gets
26 physical reads
52730252 redo size
3312 bytes sent via SQL*Net to client
240 bytes received via SQL*Net from client
1 SQL*Net roundtrips to/from client
136835 sorts (memory)
0 sorts (disk)
0 rows processed
SQL> SELECT /*+ gather_plan_statistics */ st.*
2 FROM t_senttsk st INNER JOIN (WITH t_mrgdusr AS (SELECT DISTINCT usr_id
3 FROM t_usr
4 WHERE NVL(usr_mrgemstr, usr_id) = 10000002 /* i_payer_id */
5 )
6 SELECT t_senttsk.setk_id
7 FROM t_senttsk INNER JOIN t_mrgdusr
8 ON t_senttsk.setk_affn_memb = t_mrgdusr.usr_id
9 UNION
10 SELECT t_senttsk.setk_id
11 FROM t_senttsk INNER JOIN t_mrgdusr
12 ON t_senttsk.setk_ownr = t_mrgdusr.usr_id) t_affil
13 ON st.setk_id = t_affil.setk_id;
no rows selected
Elapsed: 00:09:15.90
SQL>
SQL> select * from table(dbms_xplan.display_cursor(null, null, 'ALLSTATS LAST'));
PLAN_TABLE_OUTPUT
SQL_ID 2rc9d2c83a7ak, child number 0
SELECT /*+ gather_plan_statistics */ st.* FROM t_senttsk st INNER JOIN (WITH t_mrgdusr AS (SELECT DISTINCT usr_id
FROM t_usr WHERE NVL(usr_mrgemstr, usr_id) = :"SYS_B_0"
/* i_payer_id */ )
SELECT t_senttsk.setk_id FROM t_senttsk INNER JOIN t_mrgdusr
ON t_senttsk.setk_affn_memb = t_mrgdusr.usr_id UNION SELECT
t_senttsk.setk_id FROM t_senttsk INNER JOIN t_mrgdusr ON
t_senttsk.setk_ownr = t_mrgdusr.usr_id) t_affil ON st.setk_id = t_affil.setk_id
Plan hash value: 1065206678
| Id | Operation | Name | Starts | E-Rows | A-Rows | A-Time | Buffers | OMem | 1Mem | Used-Mem |
| 1 | NESTED LOOPS | | 1 | 169K| 0 |00:09:02.47 | 1514K| | | |
| 2 | TABLE ACCESS FULL | T_SENTTSK | 1 | 136K| 136K|00:00:01.64 | 7062 | | | |
| 3 | VIEW | | 136K| 1 | 0 |00:09:00.54 | 1507K| | | |
| 4 | TEMP TABLE TRANSFORMATION | | 136K| | 0 |00:09:00.12 | 1507K| | | |
| 5 | LOAD AS SELECT | | 136K| | 0 |00:08:24.31 | 548K| 1024 | 1024 | |
| 6 | TABLE ACCESS BY INDEX ROWID | T_USR | 136K| 1 | 0 |00:00:06.12 | 410K| | | |
|* 7 | INDEX RANGE SCAN | IX_NVL_USR_MRGEMSTR_USR_ID | 136K| 1 | 0 |00:00:05.41 | 410K| | | |
| 8 | SORT UNIQUE | | 136K| | 0 |00:00:19.10 | 822K| 1024 | 1024 | |
| 9 | UNION-ALL PARTITION | | 136K| | 0 |00:00:17.40 | 822K| | | |
| 10 | NESTED LOOPS | | 136K| 1 | 0 |00:00:08.02 | 411K| | | |
| 11 | TABLE ACCESS BY INDEX ROWID| T_SENTTSK | 136K| 1 | 136K|00:00:06.36 | 411K| | | |
|* 12 | INDEX UNIQUE SCAN | PK_T_SENTTSK | 136K| 1 | 136K|00:00:03.68 | 273K| | | |
|* 13 | VIEW | | 136K| 1 | 0 |00:00:01.03 | 0 | | | |
| 14 | TABLE ACCESS FULL | SYS_TEMP_0FD9D6609_399116CE | 136K| 1 | 0 |00:00:00.67 | 0 | | | |
| 15 | NESTED LOOPS | | 136K| 1 | 0 |00:00:06.54 | 411K| | | |
|* 16 | TABLE ACCESS BY INDEX ROWID| T_SENTTSK | 136K| 1 | 34256 |00:00:05.87 | 411K| | | |
|* 17 | INDEX UNIQUE SCAN | PK_T_SENTTSK | 136K| 1 | 136K|00:00:03.46 | 273K| | | |
|* 18 | VIEW | | 34256 | 1 | 0 |00:00:00.25 | 0 | | | |
| 19 | TABLE ACCESS FULL | SYS_TEMP_0FD9D6609_399116CE | 34256 | 1 | 0 |00:00:00.16 | 0 | | | |
Predicate Information (identified by operation id):
7 - access("T_USR"."SYS_NC00127$"=:SYS_B_0)
12 - access("T_SENTTSK"."SETK_ID"="ST"."SETK_ID")
13 - filter("T_SENTTSK"."SETK_AFFN_MEMB"="T_MRGDUSR"."USR_ID")
16 - filter("T_SENTTSK"."SETK_OWNR" IS NOT NULL)
17 - access("T_SENTTSK"."SETK_ID"="ST"."SETK_ID")
18 - filter("T_SENTTSK"."SETK_OWNR"="T_MRGDUSR"."USR_ID")
hoek wrote:Does rewriting 'the heart of the issue' into like below make any difference?
select a.*
from foo a
where exists ( select null
from bar b
where a.foo_pk_id = b.foo_pk_id
and b.some_col = :bind_var
The UNION in the subquery seems to make that difficult. -
Storer going very slow on full scans
The storer service on my root server has started taking a lot of time to
process full scan .str files (2 - 5 min each). It use to process them in a
few seconds. This is causing the rolled up zip files to back up in
the "\ZENworks\Inv\ScanDir\DBDir\" directory. Our root server is a new
dual processor P4 with 4GB RAM and RAID5. While the storer is running, the
server usually reads around 25% utilization. I've not changed anything
recently on the server and I've tried rebooting. Any ideas as to why it
may have slowed down so much?On Thu, 27 Jan 2005 18:59:05 GMT, [email protected] wrote:
> I did try some different settings in the storerdebug.properties file but
> it didn't seem to make any difference. I set it back to default.
that's not the right file..
you configure it in the mgmtdbs.ncf file
-c should be around 75 MB to 10k wks... otherwise 175 MB till 35
-gn should be perhaps 100 if more than 40k wks..
also make sure you have enough ram and that you are using only zfd 4.x
wks... you did install the agent on all wks?
Marcus Breiden
Please change -- to - to mail me.
The content of this mail is my private and personal opinion.
http://www.edu-magic.net -
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/ -
Index vs Full Scan - Parallelism
Hi everyone, i have an problem my scenario is a plataform Linux x64 bits in RAC 10gR2 (10.2.0.5) and we are migrating to RAC 11gR2 (11.2.0.3.3) in AIX 7.1 and my query: Update movimientos set ESTADOENCOLADO = :"SYS_B_0" where tipoproceso = :"SYS_B_1" and ESTADOENCOLADO = :"SYS_B_2"; in my Oracle 10g uses an index in the table MOVIMIENTOS and in the Oracle 11g the optimizer takes a FULL SCAN.
In both database server the statistics are updated with estimate 100% even the system statistics, fixed tables and data dictionary are taken its statistics.
I review the cost in both plans in Oracle 11g (with a full scan and other using a hint to index) and the full scan plan has less cost that the index. My db_file_multiblock_read_count is not set up, but in the Oracle 10g i see a value of 16 and in the 11g i see a value dinamic of 192. Understand that it too helps an uses a full scan.
The problem is that the query taken in Oracle 10g 20 minutes, but in the Oracle 11g with full scan 1.4 h whereas with a hint takes too 20 minutes.
I use the advisor tuning and no recommendations.
So basicaly i fix it modifily my parameter optimizer_index_cost_adj to 80 and it works well.
My doubt is why Oracle11gR2 says me that a full scan is faster than take an index when in the real live it is not working in this mode. I need some posible reasons please.
Thank you very much if somebody can help me with some tip or idea.
Regards>
My doubt is why Oracle11gR2 says me that a full scan is faster than take an index when in the real live it is not working in this mode. I need some posible reasons please.
>
Well if you can't figure it out by looking at the plans what makes you think anyone can figure it out without looking at them?
We have no idea what the plans contain. Post the plans, record counts for the table(s) involved, record counts for any filter predicates and the other information needed. See the FAQ for how to post a tuning request and the information needed. -
I am having problems gettting my mail to work ... Mac telling me that my start up disk id full and to delete some files to free up space ... have done this but when i open mail from the dock it just brings up the colour wheel and nothing happening ... it shows when you right click on the mail icon that " application not responding"
If your hard drive is getting full, you need to free up, at least, 20 GBs of space on your iMac's hard drive.
If your Mac is running a fairly recent version of OS X, here are some general guidelines.
Follow some of my tips for cleaning out, deleting and archiving data from your Mac's internal hard drive.
Have you emptied your iMac's Trash icon in the Dock?
If you use iPhoto, iPhoto has its own trash that needs to be emptied, also.
If you use Apple Mail app, Apple Mail also has its own trash area that needs to be emptied, too!
Other things you can do to gain space.
Delete any old or no longer needed emails and/or archive to disc, flash drives or external hard drive, older emails you want to save.
Look through your Documents folder and delete any type of old useless type files like "Read Me" type files.
Again, archive to disc, flash drives, ext. hard drives or delete any old documents you no longer use or immediately need.
Download an app called OnyX for your version of OS X.
When you install and launch it, let it run the automatic ans S.M,A.R.T. tests, then go to the cleaning and maintenance tabs and run the tabs that have the ability to clean out all web browser cache files, web browser histories, system cache files, delete old error log files.
Typically, iTunes and iPhoto libraries are the biggest users of HD space.
move these files/data off of your internal drive to the external hard drive and deleted off of the internal hard drive.
If you have any other large folders of personal data or projects, these should be archived or moved, also, to the optical discs, flash drives or external hard drive and then either archived to disc and/or deleted off your internal hard drive.
Good Luck! -
Helix Fans in keyboard portion ALWAYS running at full speed
Hi,
I was wondering if anyone from Lenovo support (or a fellow user) could recommend a Lenovo sponsored solution for the problem regarding the fan in the keyboard tray always running at full speed on the Helix? I know there are a bunch of 3rd party fan control solutions out there, but some users have cautioned that they cause unexpected touch screen problems.
When I use my Helix to remote login to other computers, the fans in the keyboard dock slow down and stop, and the noise difference is incredible. It would be so much better if the Helix was always that quite...
Any solutions?
Thanks!bgiles9091,
We've had fan feedback for a while in the forum on helix. All of the samples here perform as expected, so to revisit the issue we may need a sample from a customer.
I would also ensure you are at the latest BIOS and power management driver levels, and have installed the firmware update on your keyboard dock.
Thanks,
Mark
ThinkPads: S30, T43, X60t, X1, W700ds, IdeaPad Y710, IdeaCentre: A300, IdeaPad K1
Mark Hopkins
Program Manager, Lenovo Social Media (Services)
twitter @lenovoforums
English Community Deutsche Community Comunidad en Español Русскоязычное Сообщество -
I need to run a clean scan report on my os x mac purchased in 2006.
Can anyone help me identify a software that will do this?
Thanks Madhumannclean scan report
What do you mean by this? What results are you looking for? List some of the results you would see in the report.
Are you saying you want to check the hd for problems?
Verify a disk
As an alternative, you can verify that the filesystem on the disk is correct. You will not be able to repair the file system.
I suggest that you use disk utility to verify that your startup disk is OK.
Macintosh-HD -> Applications -> Utilities -> Disk Utility
Start up disk utility. On the left pane view, you will see a list of all your disks.
Click on your startup disk.
Click on the First Aid Tab.
Click on verify.
Hopefully your disk will verify. If not, you have to boot from your installation DVD and run Disk Utility from there to attempt to repair your file-system.
Robert -
We have a set of oracle clients running on T5220 zones that need some help
Greetings all -
We have a set of oracle clients running on T5220 zones that need some help.
If, for example, I execute the query "select (all) from dba_objects", trace output reports 90% of the elapsed time spent under "SQLNet message from client".
Here are OS details for the clients: Solaris 10 5/08 s10s_u5wos_10 SPARC
Running "uname -a" from the client gives:
(SunOS bmc-ste-app 5.10 Generic_127127-11 sun4v sparc SUNW,SPARC-Enterprise-T5220)
Here are OS details for the dataserver: Enterprise Linux Enterprise Linux Server release 5.2
Running "uname -a" from the dataserver gives:
(2.6.18-92.1.6.0.2.el5 #1 SMP Thu Jun 26 17:44:55 EDT 2008 x86_64 x86_64 x86_64 GNU/Linux)
The RDBMS is 10.2.0.4
Again, please note, there is no application in the picture here. We are just running a simple catalog query (select * from dba_objects) from sqlplus. Why the wait on something like this? The wait also occurs when a different query is used (select (all) from SYS.COL$ where rownum < 50000). And it doesn't matter if I used the full client or the instant client. I still get the same high-wait.
We had thought - maybe this is a network thing - so we put a test client on the same subnet as the dataserver. This helped - but not enough. The wait is still way too high even with firewalls taken out of the equation.
We've also looked at arraysize, SDU, kernel settings. And we've spent time going over tkprof, truss and sqlnet-tracing.
Has anyone ever had to solve this HIGH WAIT ON CLIENT issue? Is there a work around or some tweak I'm missing.
Is anyone configured like we are (linux dataserver, solaris clients)? If so - did you see anything like this?
tia -
Jim
Edited by: jim1768 on Mar 31, 2010 1:47 PM
Edited by: jim1768 on Mar 31, 2010 2:12 PM
Edited by: jim1768 on Mar 31, 2010 2:13 PMHello,
We have the exact same issue. Did you ever solve this issue? We have a t5220 and have just upgraded our 11.5.10.2 11i system on it from 9.2.0.8 32-bit sparc to 11.2.0.1 64-bit sparc. Things should be faster but arent, and our consultant has tracked it down to high wait times when the apps tier using forms connects to the database tier on the same box. So even though the t5220 is both server and client, there is something about the client sql connection.through TNS that his having trouble. Thanks for any information. We've also created an S/R with Oracle.
Note: I am well aware of the other issues with the CMT server series but this particular issue seems independent of the regular / known issues and remains a mystery to us. Other known issues with the CMT servers for SPARC:
Metalink Note 781763.1 (Migration from fast single threaded CPU machine to CMT UltraSPARC T1 & T2)
http://blogs.sun.com/glennf/resource/Optimizing_Oracle_CMT_v1.pdf
http://blogs.sun.com/glennf/tags/throughput
http://blogs.sun.com/glennf/entry/getting_past_go_with_sparc
http://www.oracle.com/apps_benchmark/doc/E-Bus-11i-PAY_ORA_SUN-T5220.pdf (this paper has some oracle init settings at the end. The kernel settings have been included in the OS upgrade)
http://blogs.sun.com/mandalika/entry/siebel_on_sun_cmt_hardware
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