Physical Database Design I

1
Physical Database Design I
Simple queries no joins, no complex aggregate
functions

• About 25 of
• Chapter 20

Focus of this Lecture Designing Databases for
Simple Queries
2
Overview

• After ER design, schema refinement, and the
  definition of views, we have the conceptual and
  external schemas for our database.
• The next step is to choose storage stucture,
  indexes, make clustering decisions, and to refine
  the conceptual and external schemas (if
  necessary) to meet performance goals.
• We must begin by understanding the workload
• The most important queries and how often they
  arise.
• The most important updates and how often they
  arise.
• The desired performance for these queries and
  updates.

3
Decisions to Make

• How should relations be stored?
• What indexes should we create?
• Which relations should have indexes? What
  field(s) should be the search key? Should we
  build several indexes?
• For each index, what kind of an index should it
  be?
• Clustered? Hash/tree? Dynamic/static?
  Dense/sparse?
• Should we make changes to the relational database
  schema?
• Consider alternative normalized schemas?
  (Remember, there are many choices in decomposing
  into BCNF, etc.)
• Should we undo some decomposition steps and
  settle for a lower normal form?
  (Denormalization.)
• Horizontal partitioning, replication, views ...

4
Choice of Indexes

• One approach consider the most important queries
  in turn. Consider the best plan using the
  current indexes, and see if a better plan is
  possible with an additional index. If so, create
  it.
• Before creating an index, must also consider the
  impact on updates in the workload!
• Trade-off indexes can make queries go faster,
  updates slower. Require disk space, too.

5
Issues in Index Selection

• Attributes mentioned in a WHERE clause are
  candidates for index search keys.
• Exact match condition suggests hash index.
• Range query suggests tree index.
• Clustering is especially useful for range
  queries, although it can help on equality queries
  as well in the presence of duplicates.
• Try to choose indexes that benefit as many
  queries as possible. Since only one index can be
  clustered per relation, choose it based on
  important queries that would benefit the most
  from clustering.

6
Multi-Attribute Index Keys

• To retrieve Emp records with age30 AND sal4000,
  an index on ltage,salgt would be better than an
  index on age or an index on sal.
• Such indexes also called composite or
  concatenated indexes.
• Choice of index key orthogonal to clustering etc.
• If condition is 20ltagelt30 AND 3000ltsallt5000
• Clustered tree index on ltage,salgt or ltsal,agegt is
  best.
• If condition is age30 AND 3000ltsallt5000
• Clustered ltage,salgt index much better than
  ltsal,agegt index!
• Composite indexes are larger, updated more often.

7
Index-Only Plans

• A number of queries can be answered without
  retrieving any tuples the relations themselves!!

Relation Emp(ssn,dno,age,salary,)
Index Give the -of-employees for
each department
SELECT E.dno, COUNT() FROM Emp E GROUP BY
E.dno
ltE.dnogt
Index Give avg. salaries of 25-year
olds in 3000-5000 range
ltE. age,E.salgt
SELECT AVG(E.sal) FROM Emp E WHERE E.age25
AND E.sal BETWEEN 3000 AND 5000
8
Using Index Structures I

• Assume a relation student(ssn, name, age, gpa)
  is given that contains 100000 tuples which are
  stored in 1000 blocks (100 tuples fit into one
  block) using heap file organization.
  Additionally, an index on the age attribute
  (which is an integer field) has been created that
  takes 80 blocks of storage, and an index on gpa
  (which is a real number) has been created that
  takes 150 blocks of storage. Both index
  structures are implemented using static hashing,
  and you can assume that there are no overflow
  pages.  
• How many block accesses does the best
  implementation of the following queries take (you
  can either use the index if helpful or not use
  the index)? Give reasons for your answers!
• Remark Index on X means that the attributes
  belonging to X are used as the hash-key
• Q1) Give the age of all the students that are
  named Liu (assume that there are 23 Lius in
  the database) 2
• 1000 (reading the student relation sequentially)
•  
• Q2) Find all students of age 46 in the database
  (assume that there are 37
• students of that age in the database) 2
• 1(index) 37 (tuple block)
•  

9
Using Index Structures II

• Q3) Find the student with the highest GPA in the
  database (assume there single best student in
  the database) 3
• 150(index) 1 (tuple block)
•  
•  
• Q4) Give the ssn of all students whose gpa
  is between 3.4 and 3.6 (assume that there are 500
  students that match this condition). 2
• 150 (index) 500 (tuple block)
•  
• Q5) Delete all students whose age is equal to 53
  (there are 5 students of that age) 6
• Finding tuples to be deleted 1 (index) 5
  (tuple blocks) 6 reads of blocks
• Updating tuples 5 writes of block
• Updating age index 1 write of index block
• Updating gpa index 5 writes of index blocks
• Total 6 reads of blocks and 5 writes of blocks

10
Selecting Composite Index Structure

• Another Design Problem Assume we have a relation
  R(A,B,C,D) that has 1,000,000 tuples that are
  distributed over 1000 blocks. Moreover static
  hashing is used to implement index structures
  (assume no overflow pages and block are 100
  filled) and index pointers and the attributes
  A,B,C, D all require the same amount of storage.
  Each A value occurs 100 time times and each B
  value occurs 2000 times in the database. Assume
  the following query is given
• Select D
• from R
• where Avalue and Bvalue (returns 20
  tuples)
• Solutions
• Index on B does not help
• Index on A --- cost 1 100
• Index on A,B --- cost 1 20
• Index on A,B,D --- index size1000 index only
  scan does not help (hashed on A,B)
• Index of A and Index on B --- compute block
  pointers for each index there are 2000 pointers
  in the B index and 100 pointers in the A index
• Cost 1(finding pointers in Index A) 1 (finding
  pointers in index B) 1?(cost of computing the
  intersection of index pointer) 20 (cost of
  accessing the tuples of the relation)23
• Remark cost would be higher if number of index
  pointer to be merged would be larger)

11
Summary

• Indexes must be chosen to speed up important
  queries (and perhaps some updates!).
• Index maintenance overhead on updates to key
  fields.
• Choose indexes that can help many queries, if
  possible.
• Build indexes to support index-only strategies.
• Clustering is an important decision only one
  index on a given relation can be clustered!
• Order of fields in composite index key can be
  important.
• Static indexes may have to be periodically
  re-built.
• Statistics have to be periodically updated.