Cost of hash join
- From: Naveen <new_dba_on_the_block@xxxxxxxxx>
- To: ORACLE-L@xxxxxxxxxxxxx
- Date: Tue, 7 Jun 2005 06:22:38 -0700 (PDT)
Hi All,
While going through the Oracle performance tuning
guide, I undestood the rationale behind the cost of
Sort-merge and nested loops joins.
But I failed to understand the cost associated to a
hash join. The guide says that the cost of hash join
is:
cost= (access cost of A * number of hash partitions of
B) + access cost of B.
Why is the access cost of A multiplied by no. of hash
partitions?
Is "A" read for each hash partition of B?
What I understand from the description of the hash
join in the tuning guide or from a few articles in
metalink is that Oracle reads "A", hashes the join key
and matches it with the corresponding hash partition
of "B" which is in memory.
So theoretically "A" should only be read once for all
the partitions in the memory.
Please let me know if I misread/misunderstood
something.
Naveen
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