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Slides on Query Optimization - Advanced Database Systems | EECS 584, Study notes of Database Management Systems (DBMS)

Material Type: Notes; Professor: Lefevre; Class: Adv Dbase Sys; Subject: Electrical Engineering And Computer Science; University: University of Michigan - Ann Arbor; Term: Spring 2008;

Typology: Study notes

Pre 2010

Uploaded on 09/02/2009

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10/6/08 1
Query Optimization
Based on:
Selinger et al.
Access Path Selection in a
Relational Database Management System
, 1979
Chaudhuri.
An Overview of Query Optimization
in Relational Systems
, 1998
Presentation by: Kristen LeFevre
10/6/08 2
Relational Query Processing
Two Key Components:
Query Execution Engine
Implements a set of physical operators
Physical operators combine to form a query
execution plan
Query Optimizer
Generates the input for the execution engine
Chooses a good plan!
10/6/08 3
Query Execution Plan
Operator Interface: open(), getNext(),
close()
Intermediate Results (multiple ops):
Pipelined: Tuples resulting from one operator
fed directly into the next
Materialized: Create a temporary table to store
intermediate results
pf3
pf4
pf5

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10/6/08 1

Query Optimization

Based on:

Selinger et al. Access Path Selection in a

Relational Database Management System, 1979

Chaudhuri. An Overview of Query Optimization

in Relational Systems, 1998

Presentation by: Kristen LeFevre

10/6/08 2

Relational Query Processing

 Two Key Components:

 Query Execution Engine

 Implements a set of physical operators

 Physical operators combine to form a query

execution plan

 Query Optimizer

 Generates the input for the execution engine

 Chooses a good plan!

10/6/08 3

Query Execution Plan

 Operator Interface: open(), getNext(),

close()

 Intermediate Results (multiple ops):

 Pipelined: Tuples resulting from one operator

fed directly into the next

 Materialized: Create a temporary table to store

intermediate results

10/6/08 4

Example

Sort Clustered Index Scan B

Table Scan A

Merge-Join

(A.x=B.x)

Index Scan C

Index Nested Loop

(A.x=C.x)

Often many different plans that produce the same result!! 10/6/08 5

Example -- Alternative

Table Scan A Table Scan C

Nested Loop

(A.x=C.x)

Table Scan B

Nested Loop

(C.x=B.x)

10/6/08 6

Query Optimizer

 Given parsed representation of SQL query,

produce an efficient execution plan

 Key aspects of optimizer:

 Search space (Which plans do we consider?)

 Cost estimation technique to evaluate each

plan in the search space

 Plan enumeration algorithm

10/6/08 10

System R Statistics

 (Stored in the system catalogs)

 TCARD(T) : # pages in table T

 NCARD(T) : # tuples in table T

 P(T) : holdover from when a page could

hold tuples from different relations

 ICARD(I) : # distinct keys in index

 NINDX(I) : # pages in index

10/6/08 11

Step 1: Selectivity Estimation

 Using statistics, assign selectivity factor F for

each boolean predicate

 Very, very rough estimate!

 attr = val

 F = 1/ICARD(I) if suitable index exists

 1/10 otherwise

 val1 < attr < val

 F = (val2-val1) / (highkey - lowkey) if index exists

 1/4 otherwise

 exp1 AND exp

 F(exp1) * F(exp2)

 exp1 OR exp

 F(exp1) + F(exp2) - F(exp1) * F(exp2)

10/6/08 12

Alternatives

 Histograms over a single column

 Multidimensional histograms

 Sampling

Age

Frequency

Came about after System R

10/6/08 13 Step 2: Single-Table Cost Estimation  For each relation, calculate the cost of scanning the relation using each suitable index and table scan  Cost = #I/Os + W * RSI Calls

tuples RSS returns

Example: SELECT * FROM Employees WHERE Name = ‘Bob’ AND Salary = 50000 Assume: Clustered index on Name Unclustered index on Salary

Three Alternatives:

(estimate cost of each)

1. Use Name index

2. Use Salary index

3. Table scan

10/6/08 14 Step 3: Join Method and Ordering  System R: Only consider left-deep join trees  Used to restrict the search space  Left-deep plans can be fully pipelined.  Intermediate results not written to temporary files.  Not all left-deep trees are fully pipelined (e.g., SM join). A^ B C D A B C D A B C D outer (^) inner

Linear Tree: at least 1 child in every join node is a base relation

10/6/08 15 Enumeration of Left-Deep Plans

 Decide:

 Join order  Join method for each join

 Enumerated using N passes (if N relations joined):

 Pass 1: Find best 1-relation plan for each relation.  Pass 2: Find best way to join result of each 1-relation plan (as outer) to another relation. (All 2-relation plans.)  Pass N: Find best way to join result of a (N-1)-relation plan (as outer) to the N’th relation. (All N-relation plans.)

 For each subset of relations, retain only:

 Cheapest plan overall, plus  Cheapest plan for each interesting order of the tuples.