I believe that Waleed's response is essentially correct.
The numbers still don't make sense, but ...
a) 250,000 blocks with the mbrc = 64. The t/s cost
(according to my observations on 9.0.1 and
8.1.7) should be based on an adjusted mbrc
of 15.5
250,000 / 15.5 = 16,000
b) Using an index to hit every block, Oracle is
able to determine from data clustering stats
that many consecutive index leaf values will
hit a single data block, therefore can estimate
the single read-count via index as 250,000 -
but you have told it to reduce this by a factor
of 1/100 - for a total of 2,500.
Unfortunately for the theory, both your costs
are 4,924 - so the T/S cost is close to the
traditional "table blocks / simple mbrc", and
index cost is too high be a factor of 2, which
may be due to some internal bitmap index
clustering fudge factor - some of the critical
bitmap index numbers apparently used to be
hard-coded constants. Possibly it just means
that your bitmap column has two values (and
of course I haven't allowed a count for the
number of leaf blocks in the index !).
Details notwithstanding - I suspect that calculations
like the above are the reason why Oracle decided
that a full indexed path was cheaper than a scan.
Jonathan Lewis
http://www.jlcomp.demon.co.uk
Author of:
Practical Oracle 8i: Building Efficient Databases
Next Seminar - Australia - July/August
http://www.jlcomp.demon.co.uk/seminar.html
Host to The Co-Operative Oracle Users' FAQ
http://www.jlcomp.demon.co.uk/faq/ind_faq.html
-----Original Message-----
To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
Date: 09 May 2002 22:16
choice
|Hi Johnathan,
|
|here is the skinny:
|
|db_file_multiblock_read_count = 64
|number of rows=15m
|blocks=251071
|empty_blocks=0
|db_block_size=16384
|
|total plan cost=4924
|tablescan cost = 4924
|
|Jack
|
|
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