query optimizer (chapter 9.0 - 9.6). optimization minimizes uses of resources by choosing best set...
TRANSCRIPT
Query Optimizer
(Chapter 9.0 - 9.6)
Optimization
• Minimizes uses of resources by choosing best set of alternative query access plans
• considers I/O cost, CPU cost
• gathers statistics - may become out of date (DB2 - RUNSTATS)
• selectivity of values - 1/domain - used to determine number of tuples of each values
Filter Factor - selectivity
• Fraction of rows with specified values(s) for specified attribute that result from the predicate restriction
• FF(c)= # records satisfying conditiontotal# of records in relation
• Estimate attribute with i distinct values as: ( |R|/i) / |R| = 1/col_cardinality
e.g. (10,000/2)/10,000 = 1/2
Filter Factor FF
• FF tells how many tuples satisfy predicate - hopefully only need to access those tuples + index
• Statistical assumptions - uniform distribution of column values, independent join distribution of values from any 2 columns
Assumptions
• Attribute values independent
• Conjunctive select (independent) C1 and C2 FF(C1) * FF(C2)
• e.g. 1/2 (gender) * 1/4 (class) = 1/8 freshman female in CS
Information for Optimization
1. SYSCOLUMNS col_name, table_name, #of values, High, Low
2. Cluster Ratio how well clustering property holds for rows with respect to a given index if 100% clustered - clustered with updates, becomes less clustered if clustering ratio 80% or more, use sequential
prefetch 3. Statistics on columns that deviate strongly from the
uniform assumption
Examples of FF
• if SQL statement specified: – col = const,
• DB2 assumes FF is 1/col_cardinality
– col between const1 and const2• DB2 assumes FF=(const2 - const1)/(High - Low)
• Predicates involving non-correlated subselects can be used for index retrieval but FF not predictable by simple formula
Explain Plan
• You can have access to query plan with
EXPLAIN PLAN statement for SQL_query in ORACLE
gives access type (index) col
Plans using Indexes
Can use an index if index matches select condition in where clause:
• A matching index scan - only have to access a limited number of contiguous leaf entries to access data
• Predicate screening – index entries to eliminate RIDs• Non-matching index scan – use index to identify RIDs• Index-only retrieval – don’t have to access data, RIDs• Multiple index retrieval – use >1 index to identify RIDs
Matching index scan
When is a matching index scan used? Assume a table T1 with multiple indexes on
columns C1, C2 and C3
1. Single where clause and (one) index matches Select * from T1
where C1=10 search B+-tree to leaf level for leftmost entry having specified values
useful for =, between
Index Scan used
2. If multiple where clauses and all '=' Select * from T1 where C1=10 and C2=5 and C3=1
a) if there is a separate index for each clause must choose one of the indexes b) if there is a composite index and a select
condition matches all index columns onlyhave to read contiguous leaf pages
FF = FF(P1) * FF(P2) * ...
Index Scan used
3. If all select conditions match composite index columns and some selects are a range Select * from T1 where C1=10 and C2
between 5 and 50
- not all entries on contiguous leaf pages
If must examine index entries to determine if in the result called predicate screening
Predicate screening
• discard RIDs based on values (for index)
• will access fewer tuples because RIDs used instead to eliminate potential tuples
Index Scan used
4. If select conditions match some index columns of composite index
Select * from T1 where C1=10 and C2=30 and C6=20
- a matching scan can be used if at least one of the columns in select is first column of index – must eliminate tuples with what indexes you can, then
examine the tuples
Rules for predicate matching
Decide how many attributes to match in a composite index after the first column, so can read in a small contiguous range of leaf entries in B+-tree to get RIDs
• Match first column of composite index then: – look at index columns from left to right – Match ends when no predicate found – If range (<=, like, between) for a column, match
terminates thereafter
If a range, easier to scan all entries for range - treat rest of entries as screening predicates
Non-matching index scan
• attributes in where clause don't include initial attribute of index
Select * from T1
where C2=30 and C3=15
search leaf entries of index and compare values for entries
must read in all leaf pages to find C2, C3 values e.g. 50 index pages vs 500,000 data pages
Index only retrieval
• elements retrieved in select clause are attributes of compose index
• don't need to access rows (actual data) Select C1, C3 from T1
where C1=5 and C3 between 2 and 5
Select count(*) from T1
Multiple Index Access
• If conjunctive conditions in where clause (and), can use >1 index – Extract RIDs from each index satisfying
matching predicate – Intersect lists of RIDs (and them) from each
index – Final list - satisfies all predicates indexed
Multiple Index Access
– If disjunctive conditions (or) Union the two lists of RIDs
Query optimizer rules for RIDs (DB2)
1. predicted active resulting RIDs must not be > 50% of RID pool
2. Limit to any single RID list the size of the RID memory pool (16M RIDs)
3. RID list cannot be generated by screening predicates
Rules cont’d
Optimizer determines diminishing returns using multiple index access
1. List indexes with matching predicates in where clause
2. Place indexes in order by increasing filter factor3. For successive indexes, extract RID list only if
reduced cost for final row returned e.g. no sense reading 100's of pages of a new index to get number of rows to only 1 tuple
Example with Multiple Indexes
Table prospects: 50M rows
Indexes: zipcode – 100,000 values
hobby – 100 values
age – 50 values
incomeclass – 10 values
Example with Multiple Indexes
Select name, stradr from prospectswhere zipcode between 02159 and 02658and age = 40 and hobby = ‘chess’ and incomeclass = 10;
FF in ascending order:1. FF(zipcode) = 500/100,000 = 1/2002. FF(hobby) = 1/1003. FF(age) = 1/504. FF(incomeclass) = 1/10
Example
(1) 50,000,000/200 = 250,000
(2) 250,000/100 = 2500
(3) 2500/50 = 50
(4) 50/10 = 5
How much time will this take? Is it cost effective to use all of these indexes?
see textbook Pg. 579