要考024的请注意了

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查看11 | 回复9 | 2006-4-16 22:44:45 | 显示全部楼层 |阅读模式
我昨天考试通过了024,是我第一回考试,由于考试费用太贵了1000元。所以我书看了三遍才敢来考试。由于准备充分我只使用了25分钟57/57满分通过了考试,我非常高兴。
在学习的过程中,我看过STS帮助复习。考前看了TESTKING,ITPASS两本资料。里面有大概4-8%的错题。我基本对所有的错误进行了纠正,希望能够帮助大家。
以下是TESTKING中的错误,29题之前应该也有一两个错误,我每记下来,应该很简单的。

QUESTION NO 29
Which three actions will cause queries to place a table’s blocks at the most-recently-used end of the LRU
list? (Choose three)
A. Creating a table with the CACHE option.
B. Querying the table by using a CACHE hint.
C. Ensuring the query performs a full table scan.
D. Defining the table without the option for caching.
E. Altering an existing table to set the CACHE option.
F. Ensuring the query does not retrieve data through index lookup.
G. Creating a separate database buffer cache to hold cached table.
Answer: A, C, E
corract a,b,e
You alter this behavior if you do the following:
? Create a table using the CACHE clause
? Alter a table using the CACHE clause
? Code the CACHE hint clause into a query
If you use one of these methods, the Oracle server places the table blocks at the most
recently used end of the LRU list. Use the CACHE clause when you create small
lookup tables used by many users. You may overcrowd the buffer cache if you have
too many cached tables.

QUESTION NO 36
Which two statements about row migration are true? (Choose two)
A. Row migration is caused by a PCTREE value set too low.
B. Row migration can be resolved using the ANALYZE command.
C. Row migration can be reduced by choosing a larger block size.
D. Row migration means that row pieces are stored in different blocks.
E. Queries that use an index to select migrated rows perform additional I/O
Answer: A, B
CORRECT:A,E

QUESTION NO 70
You determined that the values for REQUEST_FAILURES as seen from
V$SHARED_POOL_RESERVED is more than zero and always increasing. Which two actions would be
appropriate? (Choose two)
A. Decrease the value for LARGE_POOL_SIZE parameter.
B. Increase the value for LARGE_POOL_SIZE parameter.
C. Increase the value for SHARED_POOL_SIZE parameter.
D. Decrease the value for SHARED_POOL_SIZE parameter.
E. Increase the value for SHARED_POOL_RESERVED_SIZE parameter.
F. Decrease the value for SHARED_POOL_RESERVED_SIZE parameter.
Answer: C, F
CORRECT :C,E
The reserved pool is too small when the value for REQUEST_FAILURES is more
than zero and increasing. To resolve this, you can increase the value for the
SHARED_POOL_RESERVED_SIZE and SHARED_POOL_SIZE accordingly.
QUESTION NO 92
Which two statements regarding OLTP systems are true? (Choose two)
A. Use literals for optimally shared SQL rather than bind variables to keep the overhead of parsing to a
minimum.
B. To avoid the performance load of dynamic space allocation, allocate space explicitly so tables,
clusters and indexes.
C. B-tree indexing is preferred to bitmap indexing, because of locking issues affecting DML operations.
D. Use hash clusters especially on tables that are heavily inserted into, because of the use of space and
the number of blocks that need to be visited.
E. Use application code to enforce business rules instead of constraints, because constraints are
extremely expensive to process.
Answer: B, D
CORRECT:B,C
以下是ITPASS中的错题,和应该注意的题:
21. Which three statements regarding the SECONDS_IN_WAIT value for the Log Buffer Space event in the
V$SESSION_WAIT view are true?
A. A_SECONDS_IN_WAIT value close the zero is ideal.
B. A nonzero value in the SECONDS_IN_WAIT may include disk I/O contention on the redo log files.
C. The SECONDS_IN_WAIT value of the Log Buffer Space event indicates time spent waiting for space in
the redo log buffer.(*)
D. A nonzero value in the SECONDS_IN_WAIT may be an indication that the redo log buffers are too large
and log switches are not occurring fast enough.
correct:a
23. The cost-based optimizer can choose between a nested loops join and a sort-merge join operation. All tables are
analyzed, and the OPTIMIZER_MODE is set to FIRST_ROWS. Which execution plan will be the result?
A. the sort-merge join
B. the nested loops join (*)
C. This depends on some sort parameter values.
D. This depends on the number of rows in each table.
正确答案绝对是B,因为我选择B是满分;

27. What is the main reason for a row overflow area when creating index-organized tables?
A. avoid row chaining and migration
B. keep the b-tree structure densely clustered
C. speed up full table scans and fast full index scans
D. improve performance when the index-organized table is clustered *
correct:b
Of course, the primary key column must always be stored in the IOT index blocks as a
basis for search. But, you can choose to keep nonkey values in a separate area, the row
overflow area, so that the B-tree itself remains densely clustered.

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千问 | 2006-4-16 22:44:45 | 显示全部楼层
我将我学习过的TESTKING和ITPASSUPLOAD上来,不知有没有这个许可
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千问 | 2006-4-16 22:44:45 | 显示全部楼层
TESTKING
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千问 | 2006-4-16 22:44:45 | 显示全部楼层
thanks!!













:
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千问 | 2006-4-16 22:44:45 | 显示全部楼层
21. Which three statements regarding the SECONDS_IN_WAIT value for the Log Buffer Space event in the
V$SESSION_WAIT view are true?
A. A_SECONDS_IN_WAIT value close the zero is ideal.
B. A nonzero value in the SECONDS_IN_WAIT may include disk I/O contention on the redo log files.
C. The SECONDS_IN_WAIT value of the Log Buffer Space event indicates time spent waiting for space in
the redo log buffer.(*)
D. A nonzero value in the SECONDS_IN_WAIT may be an indication that the redo log buffers are too large
and log switches are not occurring fast enough.
correct:a
这到题是不是abc吗?
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千问 | 2006-4-16 22:44:45 | 显示全部楼层
SORRY ,我没有仔细看题。不过我考试中没有出现这道题
上面的题中,A,B,C应该是对的,我从书中找到了答案如下:
V$SESSION_WAIT View (continued)
Columns (continued)
? P3RAW: Third additional parameter value in hexadecimal
? WAIT_TIME:
? SECONDS_IN_WAIT: Number of seconds the event waited
? STATE: WAITING, WAITED UNKNOWN TIME, WAITED SHORT TIME (less
than one hundredth of a second), WAITED KNOWN TIME (the value is stored in
the WAIT_TIME column)
Note: Not all of the parameter columns are used for all events.
TIMED_STATISTICS Initialization Parameter
Set the TIMED_STATISTICS parameter to TRUE to retrieve values in the
WAIT_TIME column. It is a dynamic initialization parameter.
Value Explanation
> 0
= 0
= -1
= -2
The session’s last wait time
The session is currently waiting
The value was less than 1/100 of a second
The system cannot provide timing
information

Dynamic Views
? The V$SESSION_WAIT view indicates through the Log Buffer Space event if
there are any waits for space in the log buffer because the session is writing data
into the log buffer faster than LGWR can write it out.
SQL> select sid, event, seconds_in_wait, state
2 from v$session_wait
3 where event = ’log buffer space%’;
SID EVENT SECONDS_IN_WAIT STATE
------ --------------------------------- --------------- ------
5 log buffer space 110 WAITING
The SECONDS_IN_WAIT value of the Log Buffer Space event indicates the time
spent waiting for space in the redo log buffer because the log switch does not occur.
This is an indication that the buffers are being filled up faster than LGWR is writing
and may also indicate disk I/O contention on the redo log files.
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千问 | 2006-4-16 22:44:45 | 显示全部楼层
我也正要考024,看书还没看完,非常感谢楼上的,对我很有用。
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千问 | 2006-4-16 22:44:45 | 显示全部楼层
千言万语汇成一句话
谢谢
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千问 | 2006-4-16 22:44:45 | 显示全部楼层
谢谢你
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千问 | 2006-4-16 22:44:45 | 显示全部楼层
我马上也要考这门课啦非常感激!!
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