CS586 Fall 2004

1
CS586 Fall 2004
Day 9

• Implementation of Database Management Systems
• Sid Kitchel

2
Plan of the Day

• Questions
• Introduction to Query Optimization
• Relational Algebra Equivalences
• Cost-based Optimization
• Tuning Query Performance

3
DB Logical Architecture
queries
Access Plan Executor
Query Execution engine
Parser
Operator Evaluator
Optimizer
Concurrency control
Access Methods
Transaction Manager
Recovery Manager
Buffer Manager
Lock Manager
Disk Manager
4
Getting Answers
SELECT cust_id, balance FROM accounts WHERE
cust_lname like Ander
Declarative Query Language
Query
Parse Tree
Internal Representation
Query Tree
Access Plan
Physical Operations
Manipulating data
5
Overview of Query Evaluation

• Plan Tree of R.A. ops, with choice of alg for
  each op.
• Each operator typically implemented using a
  pull interface when an operator is pulled
  for the next output tuples, it pulls on its
  inputs and computes them.
• Two main issues in query optimization
• For a given query, what plans are considered?
• Algorithm to search plan space for cheapest
  (estimated) plan.
• How is the cost of a plan estimated?
• Ideally Want to find best plan. Practically
  Avoid worst plans!
• We will study the System R approach. (System R is
  the early, pioneering implementation of an RDBMS
  by IBM Research Lab at San Jose.)

6
Relational Algebra Equivalences

• Allow us to choose different join orders and to
  push selections and projections ahead of joins.
• Selections
  (Cascade)

(Commute)

• Projections

(Cascade)
(Associative)

• Joins

R (S T) (R S) T
(Commute)
(R S) (S R)
R (S T) (T R) S

• Show that

7
More Equivalences

• A projection commutes with a selection that only
  uses attributes retained by the projection.
• Selection between attributes of the two arguments
  of a cross-product converts cross-product to a
  join.
• A selection on just attributes of R commutes with
  
• R S. (i.e., (R S) (R)
  S )
• Similarly, if a projection follows a join R
  S, we can push it by retaining only attributes
  of R (and S) that are needed for the join or are
  kept by the projection.

8
Some Common Techniques

• Algorithms for evaluating relational operators
  use some simple ideas extensively
• Indexing Can use WHERE conditions to retrieve
  small set of tuples (selections, joins)
• Iteration Sometimes, faster to scan all tuples
  even if there is an index. (And sometimes, we can
  scan the data entries in an index instead of the
  table itself.)
• Partitioning By using sorting or hashing, we can
  partition the input tuples and replace an
  expensive operation by similar operations on
  smaller inputs.

9
Statistics and Catalogs

• Luke, use the metadata!
• Need information about the relations and indexes
  involved. Catalogs typically contain at least
• tuples (NTuples) and pages (NPages) for each
  relation.
• distinct key values (NKeys) and NPages for each
  index.
• Index height, low/high key values (Low/High) for
  each tree index.
• Catalogs updated periodically.
• Updating whenever data changes is too expensive
  lots of approximation anyway, so slight
  inconsistency ok.
• More detailed information (e.g., histograms of
  the values in some field) are sometimes stored.

10
Access Paths

• An access path is a method of retrieving tuples
• File scan, or index that matches a selection (in
  the query)
• A tree index matches (a conjunction of) terms
  that involve only attributes in a prefix of the
  search key.
• E.g., Tree index on lta, b, cgt matches the
  selection a5 AND b3, and a5 AND bgt6, but not
  b3.
• A hash index matches (a conjunction of) terms
  that has a term attribute value for every
  attribute in the search key of the index.
• E.g., Hash index on lta, b, cgt matches a5 AND
  b3 AND c5 but it does not match b3, or a5
  AND b3, or agt5 AND b3 AND c5.

11
Highlights of System R Optimizer

• Impact
• Most widely used currently works well for lt 10
  joins.
• Cost estimation Approximate art at best.
• Statistics, maintained in system catalogs, used
  to estimate cost of operations and result sizes.
• Considers combination of CPU and I/O costs.
• Plan Space Too large, must be pruned.
• Only the space of left-deep plans is considered.
• Left-deep plans allow output of each operator to
  be pipelined into the next operator without
  storing it in a temporary relation.
• Cartesian products avoided.

12
Cost Estimation

• For each plan considered, must estimate cost
• Must estimate cost of each operation in plan
  tree.
• Depends on input cardinalities.
• Weve already discussed how to estimate the cost
  of operations (sequential scan, index scan,
  joins, etc.)
• Must also estimate size of result for each
  operation in tree!
• Use information about the input relations.
• For selections and joins, assume independence of
  predicates.

13
Size Estimation and Reduction Factors
SELECT attribute list FROM relation list WHERE
term1 AND ... AND termk

• Consider a query block
• Maximum tuples in result is the product of the
  cardinalities of relations in the FROM clause.
• Reduction factor (RF) associated with each term
  reflects the impact of the term in reducing
  result size. Result cardinality Max tuples
  product of all RFs.
• Implicit assumption that terms are independent!
• Term colvalue has RF 1/NKeys(I), given index I
  on col
• Term col1col2 has RF 1/MAX(NKeys(I1), NKeys(I2))
• Term colgtvalue has RF (High(I)-value)/(High(I)-Low
  (I))

14
Schema for Examples
Sailors (sid integer, sname string, rating
integer, age real) Reserves (sid integer, bid
integer, day dates, rname string)

• Similar to old schema rname added for
  variations.
• Reserves
• Each tuple is 40 bytes long, 100 tuples per
  page, 1000 pages.
• Sailors
• Each tuple is 50 bytes long, 80 tuples per page,
  500 pages.

15
Motivating Example
RA Tree
SELECT S.sname FROM Reserves R, Sailors S WHERE
R.sidS.sid AND R.bid100 AND S.ratinggt5

• Cost 5005001000 I/Os
• By no means the worst plan!
• Misses several opportunities selections could
  have been pushed down earlier, no use is made
  of any available indexes, etc.
• Goal of optimization To find more efficient
  plans that compute the same answer.

Plan
16
Alternative Plans 1 (No Indexes)

• Main difference push down selects.
• With 5 buffers, cost of plan
• Scan Reserves (1000) write temp T1 (10 pages,
  if we have 100 boats, uniform distribution).
• Scan Sailors (500) write temp T2 (250 pages, if
  we have 10 ratings).
• Sort T1 (2210), sort T2 (23250), merge
  (10250)
• Total 3560 page I/Os.
• If we used BNL join, join cost 104250, total
  cost 2770.
• If we push projections, T1 has only sid, T2
  only sid and sname
• T1 fits in 3 pages, cost of BNL drops to under
  250 pages, total lt 2000.

17
Alternative Plans 2With Indexes
(On-the-fly)
sname
(On-the-fly)
rating gt 5

• With clustered index on bid of Reserves, we get
  100,000/100 1000 tuples on 1000/100 10
  pages.
• INL with pipelining (outer is not materialized).

(Index Nested Loops,
with pipelining )
sidsid
(Use hash
Sailors
bid100
index do
not write
result to
temp)
Reserves

• Projecting out unnecessary fields from outer
  doesnt help.

• Join column sid is a key for Sailors.
• At most one matching tuple, unclustered index on
  sid OK.

• Decision not to push ratinggt5 before the join
  is based on
• availability of sid index on Sailors.
• Cost Selection of Reserves tuples (10 I/Os)
  for each,
• must get matching Sailors tuple (10001.2)
  total 1210 I/Os.

18
Summary

• There are several alternative evaluation
  algorithms for each relational operator.
• A query is evaluated in steps
• 1) convert to a parse tree
• 2) convert to tree of R.A. operators and
• 3) convert to physical access plan by annotation,
  and
• 4) evaluating the annotated operators in the
  tree.
• What step was missed Query Optimization
• Before execution several alternative execution
  plans are considered and evaluated by the
  optimizer

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Summary (cont.)

• You must understand query optimization in order
  to fully understand the performance impact of a
  given database design (relations, indexes) on a
  workload (set of queries).
• Two parts to optimizing a query
• Consider a set of alternative plans.
• Must prune search space typically, left-deep
  plans only.
• Must estimate cost of each plan that is
  considered.
• Must estimate size of result and cost for each
  plan node.
• Key issues Statistics, indexes, operator
  implementations.

20
Tuning 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 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 are
  run.
• The most important updates and how often they are
  run.
• The desired performance for these queries and
  updates.

21
Decisions to Make

• 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?
• Should we make changes to the conceptual 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 ...

22
Index Selection for Joins

• When considering a join condition
• Hash index on inner is very good for Index Nested
  Loops. (May not be available.)
• Should be clustered if join column is not key for
  inner, and inner tuples need to be retrieved.
• Clustered B tree on join column(s) good for
  Sort-Merge and for loop index probe joins.
• Bitmap index for joins or row pointer
  intersections

23
Example 1
SELECT E.ename, D.mgr FROM Emp E, Dept D WHERE
D.dnameToy AND E.dnoD.dno

• Hash index on D.dname supports Toy selection.
• Given this, index on D.dno is not needed.
• Hash index on E.dno allows us to get matching
  (inner) Emp tuples for each selected (outer) Dept
  tuple.
• What if WHERE included ... AND E.age 25
  ?
• Could retrieve Emp tuples using index on E.age,
  then join with Dept tuples satisfying dname
  selection. Comparable to strategy that used
  E.dno index.
• So, if E.age index is already created, this query
  provides much less motivation for adding an E.dno
  index.

24
SELECT E.ename, D.mgr FROM Emp E, Dept D WHERE
E.sal BETWEEN 10000 AND 20000 AND
E.hobbyStamps AND E.dnoD.dno
Example 2

• Clearly, Emp should be the outer relation.
• Suggests that we build a hash index on D.dno.
• What index should we build on Emp?
• B tree on E.sal could be used, OR an index on
  E.hobby could be used. Only one of these is
  needed, and which is better depends upon the
  selectivity of the conditions.
• As a rule of thumb, equality selections more
  selective than range selections.
• As both examples indicate, our choice of indexes
  is guided by the plan(s) that we expect an
  optimizer to consider for a query. Have to
  understand optimizers!

25
Tuning the Conceptual Schema

• The choice of conceptual schema should be guided
  by the workload, in addition to redundancy
  issues
• We may settle for a 3NF schema rather than BCNF.
• Workload may influence the choice we make in
  decomposing a relation into 3NF or BCNF.
• We may further decompose a BCNF schema!
• We might denormalize (i.e., undo a decomposition
  step), or we might add fields to a relation.
• We might consider horizontal decompositions.
• If such changes are made after a database is in
  use, called schema evolution might want to mask
  some of these changes from applications by
  defining views.

26
Example Schemas
Contracts (Cid, Sid, Jid, Did, Pid, Qty,
Val) Depts (Did, Budget, Report) Suppliers (Sid,
Address) Parts (Pid, Cost) Projects (Jid, Mgr)

• We will concentrate on Contracts, denoted as
  CSJDPQV. The following ICs are given to hold
  
• JP C, SD P, C is the
  primary key.
• What are the candidate keys for CSJDPQV?
• What normal form is this relation schema in?

27
Settling for 3NF vs BCNF

• CSJDPQV can be decomposed into SDP and CSJDQV,
  and both relations are in BCNF. (Which FD
  suggests that we do this?)
• Lossless decomposition, but not
  dependency-preserving.
• Adding CJP makes it dependency-preserving as
  well.
• Suppose that this query is very important
• Find the number of copies Q of part P ordered in
  contract C.
• Requires a join on the decomposed schema, but can
  be answered by a scan of the original relation
  CSJDPQV.
• Could lead us to settle for the 3NF schema
  CSJDPQV.

28
Denormalization

• Suppose that the following query is important
• Is the value of a contract less than the budget
  of the department?
• To speed up this query, we might add a field
  budget B to Contracts.
• This introduces the FD D B wrt Contracts.
• Thus, Contracts is no longer in 3NF.
• We might choose to modify Contracts thus if the
  query is sufficiently important, and we cannot
  obtain adequate performance otherwise (i.e., by
  adding indexes or by choosing an alternative 3NF
  schema.)

29
Choice of Decompositions

• There are 2 ways to decompose CSJDPQV into BCNF
• SDP and CSJDQV lossless-join but not
  dep-preserving.
• SDP, CSJDQV and CJP dep-preserving as well.
• The difference between these is really the cost
  of enforcing the FD JP C.
• 2nd decomposition Index on JP on relation CJP.
• 1st

CREATE ASSERTION CheckDep CHECK ( NOT
EXISTS ( SELECT FROM PartInfo P,
ContractInfo C WHERE P.sidC.sid AND
P.didC.did GROUP BY C.jid, P.pid HAVING COUNT
(C.cid) gt 1 ))
30
Choice of Decompositions (Cont.)

• The following ICs were given to hold
  JP C, SD P, C is the
  primary key.
• Suppose that, in addition, a given supplier
  always charges the same price. If we decide that
  we want to decompose CSJDPQV into BCNF, we now
  have a third choice
• Begin by decomposing it into SPQV and CSJDPQ.
• Then, decompose CSJDPQ (not in 3NF) into SDP,
  CSJDQ.
• This gives us the lossless-join decomp SPQV,
  SDP, CSJDQ.
• To preserve JP C, we can add CJP, as
  before.
• Choice SPQV, SDP, CSJDQ or SDP, CSJDQV
  ?

31
Decomposition of a BCNF Relation

• Suppose that we choose SDP, CSJDQV . This is
  in BCNF, and there is no reason to decompose
  further (assuming that all known ICs are FDs).
• However, suppose that these queries are
  important
• Find the contracts held by supplier S.
• Find the contracts that department D is involved
  in.
• Decomposing CSJDQV further into CS, CD and CJQV
  could speed up these queries. (Why?)
• On the other hand, the following query is slower
• Find the total value of all contracts held by
  supplier S.

32
Horizontal Decompositions

• Our definition of decomposition Relation is
  replaced by a collection of relations that are
  projections. Most important case.
• Sometimes, might want to replace relation by a
  collection of relations that are selections.
• Each new relation has same schema as the
  original, but a subset of the rows.
• Collectively, new relations contain all rows of
  the original. Typically, the new relations are
  disjoint.
• Oracle calls this partitioning Informix calls
  it fragmentation

33
Horizontal Decompositions (Cont.)

• Suppose that contracts with value gt 10000 are
  subject to different rules. This means that
  queries on Contracts will often contain the
  condition val gt 10000.
• One way to deal with this is to build a clustered
  B tree index on the val field of Contracts.
• A second approach is to replace contracts by two
  new relations LargeContracts and
  SmallContracts, with the same attributes
  (CSJDPQV).
• Performs like index on such queries, but no index
  overhead.
• Can build clustered indexes on other attributes,
  in addition!

34
Masking Conceptual Schema Changes
CREATE VIEW Contracts(cid, sid, jid, did, pid,
qty, val) AS SELECT FROM
LargeContracts UNION SELECT FROM
SmallContracts

• The replacement of Contracts by LargeContracts
  and SmallContracts can be masked by the view.
• However, queries with the condition val gt 10000
  must be asked wrt LargeContracts for efficient
  execution so users concerned with performance
  have to be aware of the change.

35
Automatic Horizontal Partitioning

• Changing the schema by splitting tables into
  multiple sub-tables is pretty drastic
• The view on the previous page is also potentially
  expensive
• Many systems build horizontal partitioning into
  the physical storage of a tables data
• Logically a single table exists, but it is stored
  on multiple disk mount points so that an
  individual partition or fragment is grouped
  together
• This allows parallel I/O of the table and it
  allows a smart optimizer to do fragment
  elimination

36
Tuning Queries and Views

• If a query runs slower than expected, check if an
  index needs to be re-built, or if statistics are
  too old.
• Sometimes, the DBMS may not be executing the plan
  you had in mind. Common areas of weakness
• Selections involving arithmetic or string
  expressions.
• Selections involving OR conditions.
• Lack of evaluation features like index-only
  strategies or certain join methods or poor size
  estimation.
• Selections involving null values.
• Check the plan that is being used using the
  explain plan tool! Then adjust the choice of
  indexes or rewrite the query/view.

37
Tuning I / O

• Disk accesses and I/O are expensivewe do not
  want to make them worse by creating disk
  contention
• This occurs when multiple objects or tablespaces
  are on the same mount point
• Operations such as scan can be compromised by
  multiple users needing data from the same disk
  spindle
• Certain tasks such as logging are inherently
  intensive
• So one aspect of system tuning is to have an
  adequate number of mount points and try to spread
  tablespaces across them with especial attention
  to keeping very active objects separate.

38
Tuning Queries and Views

• If a query runs slower than expected, check if an
  index needs to be re-built, or if statistics are
  too old.
• Sometimes, the DBMS may not be executing the plan
  you had in mind. Common areas of weakness
• Selections involving null values.
• Selections involving arithmetic or string
  expressions.
• Selections involving OR conditions.
• Lack of evaluation features like index-only
  strategies or certain join methods or poor size
  estimation.
• Check the plan that is being used! Then adjust
  the choice of indexes or rewrite the query/view.

39
Rewriting SQL Queries

• Complicated by interaction of
• NULLs, duplicates, aggregation, subqueries.
• Guideline Use only one query block, if
  possible.

SELECT DISTINCT FROM Sailors S WHERE S.sname
IN (SELECT Y.sname FROM YoungSailors Y)
SELECT DISTINCT S. FROM Sailors S,
YoungSailors Y WHERE S.sname Y.sname

SELECT FROM Sailors S WHERE S.sname
IN (SELECT DISTINCT Y.sname FROM
YoungSailors Y)
SELECT S. FROM Sailors S, YoungSailors
Y WHERE S.sname Y.sname

40
The Notorious COUNT Bug

• What happens when Employee is empty??

41
Summary on Unnesting Queries

• DISTINCT at top level Can ignore duplicates.
• Can sometimes infer DISTINCT at top level! (e.g.
  subquery clause matches at most one row)
• DISTINCT in subquery w/o DISTINCT at top Hard to
  convert.
• Subqueries inside OR Hard to convert.
• ALL subqueries Hard to convert.
• EXISTS and ANY are just like IN.
• Aggregates in subqueries Tricky.
• Good news Some systems now rewrite under the
  covers (e.g. DB2).

42
More Guidelines for Query Tuning

• Minimize the use of DISTINCT dont need it if
  duplicates are acceptable, or if answer contains
  a key.
• Minimize the use of GROUP BY and HAVING

SELECT MIN (E.age) FROM Employee E GROUP BY
E.dno HAVING E.dno102
SELECT MIN (E.age) FROM Employee E WHERE
E.dno102

• Consider DBMS use of index when writing
  arithmetic expressions E.age2D.age will
  benefit from index on E.age, but might not
  benefit from index on D.age!

43
Guidelines for Query Tuning (Cont.)
SELECT INTO Temp FROM Emp E, Dept D WHERE
E.dnoD.dno AND D.mgrnameJoe

• Avoid using intermediate
  relations

SELECT E.dno, AVG(E.sal) FROM Emp E, Dept
D WHERE E.dnoD.dno AND D.mgrnameJoe GROUP
BY E.dno
and
SELECT T.dno, AVG(T.sal) FROM Temp T GROUP BY
T.dno
vs.

• Does not materialize the intermediate table Temp.
• If there is a dense B tree index on ltdno, salgt,
  an index-only plan can be used to avoid
  retrieving Emp rows in the second query!

44
Summary

• Database design consists of several tasks
  requirements analysis, conceptual design, schema
  refinement, physical design and tuning.
• In general, have to go back and forth between
  these tasks to refine a database design, and
  decisions in one task can influence the choices
  in another task.
• Understanding the nature of the workload for the
  application, and the performance goals, is
  essential to developing a good design.
• What are the important queries and updates? What
  attributes/relations are involved?

45
Summary (Cont.)

• The conceptual schema should be refined by
  considering performance criteria and workload
• May choose 3NF or lower normal form over BCNF.
• May choose among alternative decompositions into
  BCNF (or 3NF) based upon the workload.
• May denormalize, or undo some decompositions.
• May decompose a BCNF relation further!
• May choose a horizontal decomposition of a
  relation.
• Importance of dependency-preservation based upon
  the dependency to be preserved, and the cost of
  the IC check.
• Can add a relation to ensure dep-preservation
  (for 3NF, not BCNF!) or else, can check
  dependency using a join.

46
Summary (Cont.)

• Over time, indexes have to be fine-tuned
  (dropped, created, re-built, re-analyzed ...)
  for performance.
• Should determine the plan used by the system, and
  adjust the choice of indexes appropriately.
• Make sure that the statistics stored in the
  system catalogs is up-to-datere-analyze all
  objects on a regular basis
• So, may have to rewrite the query/view
• Avoid nested queries, temporary relations,
  complex conditions, and operations like DISTINCT
  and GROUP BY.
• Tune the physical disk storage to minimize disk
  contention.
• Archive old dataonly keep what you need for
  actual operations