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Re: How to calculate ideal size of log_buffer parameter? [message #431769 is a reply to message #431736] |
Thu, 19 November 2009 07:06 |
halim
Messages: 100 Registered: September 2008
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Senior Member |
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Dears
Our database is performing slow from last few days .
we found that maximum time 4/5 redo log [out of 12(size 100mb per log file)] remain active.
at that time we found the below statestics....
What should we do now ?
SQL*Plus: Release 10.2.0.1.0 - Production on Thu Nov 19 19:48:24 2009
Copyright (c) 1982, 2005, Oracle. All rights reserved.
SQL> conn sys@live as sysdba
Enter password:
Connected.
SQL> SELECT SUM(value) "Redo Buffer Waits" FROM v$sysstat WHERE name = 'redo log space requests';
Redo Buffer Waits
-----------------
23555
SQL>
SQL> SELECT SUM(value) "Redo Buffer Waits" FROM v$sysstat WHERE name = 'redo log space wait time';
Redo Buffer Waits
-----------------
1155196
SQL> SELECT name,SUM(value) "Redo Buffer Waits" FROM v$sysstat WHERE name like '%redo log%'
2 group by name
3 /
NAME Redo Buffer Waits
---------------------------------------------------------------- -----------------------------
redo log space requests 23557
redo log space wait time 1155201
redo log switch interrupts 0
SQL>
SQL> select name, value
2 from v$parameter
3 where name in ('cpu_count','log_parallelism','log_buffer')
4 /
NAME VALUE
-------------------------------------------------------------------------------- ---------------------------------------------------------------------------------------
cpu_count 128
log_buffer 1253376
SQL> show sga
Total System Global Area 1.7952E+10 bytes
Fixed Size 2170576 bytes
Variable Size 8552915248 bytes
Database Buffers 9395240960 bytes
Redo Buffers 1294336 bytes
SQL> select
2 min(b.first_time - a.first_time) * 1440 * 60 seconds
3 from
4 sys.v_$instance i,
5 sys.v_$log_history a,
6 ( select
7 sequence#,
8 first_time
9 from
10 sys.v_$log
11 where
12 status = 'CURRENT'
13 union all
14 select
15 sequence#,
16 first_time
17 from
18 sys.v_$log_history
19 ) b
20 where
21 i.startup_time < a.first_time and
22 a.first_time < b.first_time and
23 a.sequence# + 1 = b.sequence#
24 /
SECONDS
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2
Database is in noarchive mode
and average session about 2500 .
Regards
Halim
[Updated on: Thu, 19 November 2009 07:17] Report message to a moderator
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