DBMS Internal Review. Outline Storage and Indexing Query Optimization.

DBMS InternalReview

Outline

• Storage and Indexing

• Query Optimization

Overview of Storage and Indexing

Data on External Storage• Disks: Can retrieve random page at fixed cost

– But reading several consecutive pages is much cheaper than reading them in random order

• Tapes: Can only read pages in sequence– Cheaper than disks; used for archival storage

• File organization: Method of arranging a file of records on external storage.– Record id (rid) is sufficient to physically locate record– Indexes are data structures that allow us to find the record

ids of records with given values in index search key fields

• Architecture: Buffer manager stages pages from external storage to main memory buffer pool. – File and index layers make calls to the buffer manager.

StorageManager

Buffer Manager

RecordID

File/Index

keyrecord

Record

Alternative File OrganizationsMany alternatives exist, each ideal for some situations, and not so good in others:– Heap (random order) files: Suitable when

typical access is a file scan retrieving all records.– Sorted Files: Best if records must be retrieved in

some order, or only a `range’ of records is needed.

– Indexes: Data structures to organize records via trees or hashing. • Like sorted files, they speed up searches for a subset

of records, based on values in certain (“search key”) fields

• Updates are much faster than in sorted files.

Indexes – Search Key

An index on a file speeds up selections on the search key fields for the index.

– Any subset of the fields of a relation can be the search key for an index on the relation.

– Search key is not the same as key (minimal set of fields that uniquely identify a record in a relation).

A B C D E

1

2

3

4

5

6

7

Index on AB

Search Key

Indexes – Data Entries An index contains a collection of data entries, and supports efficient retrieval of all data entries k* with a given key value k.

To locate (one or more) data records with search key value k– Search the index using k to find the

desired data entry k*– The data entry k* contains

information to locate (one or more) data records with search key value k

A B C D E

1

2

3

4

5

6

7

Search Key

Search key

Record ID

k

k*A data record

A data entry

Alternatives for Data Entry k* in Index

• Choice of alternative for data entries is orthogonal to the indexing technique used to locate data entries with a given key value k.– Examples of indexing techniques: B+ trees, hash-based structures– Typically, index contains auxiliary information that directs

searches to the desired data entries

Three alternatives:1. Actual data record (with search key

value k)2. <k, rid> pair, where rid is the record id

of data record with search key value k3. <k, rid-list> pair, where rid-list is a list of

rids of data records with search key k

Dataentries

Data Records

Key Indexingtechnique

Alternatives for Data Entries (Contd.)• Alternative 1:

– If this is used, index structure is a file organization for data records (instead of a Heap file or sorted file).

– At most one index on a given collection of data records can use Alternative 1. (Otherwise, data records are duplicated, leading to redundant storage and potential inconsistency.)

– If data records are very large, # of pages containing data records is high. Implies size of auxiliary information in the index is also large, typically.

DR

DR

DRDR

DR

DR DR DR

Key

DR Data records

Auxiliaryinformation

B-treeB-tree can be used to implement Alternative 1

Data records (instead of

data entries) stored intree node

The tree is relatively large

Variable sized data entries

Alternatives for Data Entries (Contd.)• Alternatives 2: Data entries <k, rid>,

typically much smaller than data records. So, better than Alternative 1 with large data records, especially if search keys are small. (Portion of index structure used to direct search, which depends on size of data entries, is much smaller than with Alternative 1.)

Dataentries

Data Records

Key

Alternative 3: Data entries <k, list-rid>, more compact than Alternative 2, but leads to variable sized data entries even if search keys are of fixed length.

Smaller than Alternative 1

Index Classification• Primary vs. secondary: If search key contains

primary key, then called primary index (alternative 1 is usually used to avoid one more I/O to bring in the matching data record).– Unique index: Search key contains a candidate key.

• Clustered vs. unclustered: If order of data records is the same as, or `close to’, order of data entries, then called clustered index.

Clustered IndexSuppose that Alternative (2) is used for data entries, and that the data records are stored in a Heap file.– To build clustered index, first sort the Heap file (with some

free space on each page for future inserts). – Overflow pages may be needed for inserts. (Thus, order of

data records is `close to’, but not identical to, the sort order.)

Index entries

Data entries

direct search for

(Index File)

(Data file)

Data Records

data entries

CLUSTERED

Heap file

Only One Clustered Index

StudentID Age Income Phone

563121325 3212645418

572361672 4075493124

678563276 3219659332

698394250 4073459876

720357320 4077589092

734705862 3214551023

809435620 4077652364

903429554 4071245436

Sortedaccordingto SSN

Clus

tere

dSortedaccordingto phone#

Unclustered

Data file – sorted according to SSN

<Stu

dent

ID, r

id>

• A file can have only one clustered index & alternative 1 often used• Cost of retrieving data record through index varies greatly based on whether

index is clustered or not

Hash-based Index

h

e.g., h(key) = key mod N

key

Primarybucket pages

Overflow pages

2

0

N-1

h(key) = ID of hash bucket

An entry may be: a record (alternative 0), a <k, rid> (alternative 2), or a <k, list_rid> (alternative 3).

Hash-Based IndexesGood for equality selections.

• Index is a collection of buckets. Bucket = primary page plus zero or more overflow pages.

• Hashing function h: h is applied to the search key fields of r. h(r) = ID of bucket in which record r belongs. Hash on the key fields to determine the bucket(s) Scan the data entries in these buckets to find the

matching <key, rid> (i.e., alternative 2)

Use rid to locate the record r

• If Alternative (1) is used, the buckets contain the data records (instead of <key, rid> or <key, rid-list> pairs)

B+ Tree Indexes

Non-leaf Pages

Leaf Pages

P0 K1 P1 K2 P0 . . . Km Pm

Index entryA non-leaf page

Leaf pages contain data entries, and are chained (prev & next) Non-leaf pages contain index entries and direct searches:

B+ Tree Example

• Find 28*? 29*? All > 15* and < 30*• Insert/delete: Find data entry in leaf, then

change it. Need to adjust parent sometimes.– And change sometimes bubbles up the tree

2* 3*

Root

17

30

14* 16* 33* 34* 38* 39*

135

7*5* 8* 22* 24*

27

27* 29*

Entries <= 17 Entries > 17

Choice of Indexes

• 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?

Choice of Indexes (Contd.)• 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.→ Obviously, this implies that we must understand how

a DBMS evaluates queries and creates query evaluation plans!

• 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.

Index Selection Guidelines (1)

Attributes in WHERE clause are candidates for index keys.–Exact match condition suggests hash index.

SELECT E.dnoFROM Employees EWHERE E.num = 568429543

–Range query suggests tree index.Clustering is especially useful for range queries; can also help on equality queries if there are many duplicates.

SELECT E.dno SELECT E.nameFROM Employees E FROM Employees EWHERE E.age > 40 WHERE E.dno=123

Dept. 123 has many

employees

Employees older than 40

Index Selection Guidelines (2)Multi-attribute search keys should be considered when a WHERE clause contains several conditions

SELECT E.nameFROM Employees EWHERE E.age = 20 AND E.sal = 50000

– An index on <age,sal> would be better than an index on age or an index on sal

– Reason: The qualifying data entries are grouped together

1 2

Index Selection Guidelines (3)

• Indexes can sometimes enable index-only evaluation for important queries.

Example: use leaf pages of index on age to compute the average age

• 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.

On age

Examples of Clustered Indexes (1)

What is the selectivity of the condition ? If most employees are older than 30, a sequential

scan of the relation would do almost as well

SELECT E.dnoFROM Emp EWHERE E.age>30

B+ tree index on E.age can be used to get qualifying tuples

What if only 10% are older than 30 ? Unclustered index: Performing one I/O per

qualifying tuple could be more expensive than a sequential scan

Clustered index: This is a good option


SELECT E.dno, COUNT (*)FROM Emp EWHERE E.age>25GROUP BY E.dno

Clustered E.dno index may be better ! For each dno, count the

tuples with E.age > 25 No need to sort !

Using E.age index may be costly Retrieve tuples with

E.age > 25 Sort the tuples on dno,

and count number of tuples for each dno

On age

Good to know DBMS

internal


• If many employees collect stamps, a clustered index on E.hobby is helpful – If this query is important, we

should consider this index

SELECT E.dnoFROM Emp EWHERE E.hobby=Stamps

SELECT E.dnoFROM Emp EWHERE E.eid=328169455

• An unclustered index on E.eid is good enough for the second query since no two employees have the same E.eid.

1

2

Indexes with Composite Search Keys • Composite Search Keys:

Search on a combination of fields.– Equality query: Every field

value is equal to a constant value, e.g. wrt <sal,age> index: age=20 and sal =75

– Range query: Some field value is not a constant, e.g., age =20; or age=20 and sal > 10

• Data entries in index sorted by search key to support range queries. Data entries in index

sorted by <sal,age>Data entriessorted by <sal>

sue 13 75

bob

cal

joe 12

10

20

8011

12

name age sal

<sal, age>

<age, sal> <age>

<sal>

12,20

12,10

11,80

13,75

20,12

10,12

75,13

80,11

11

12

12

13

10

20

75

80

Data recordssorted by name

Examples of composite key indexes

Data entries

Composite Search Keys Order of attributes is important for range queries

key<age,sal> ≠ key<sal,age>

To retrieve Emp records: – If condition is: age=30 AND 3000<sal<5000:

Clustered <age,sal> index much better than <sal,age> index!

Reason: The qualifying data entries are grouped together in <age,sal> index, but not in <sal,age> index

– If condition is: 20<age<30 AND 3000<sal<5000: Clustered tree index on <age,sal> or <sal,age> is best.

Composite indexes are larger, updated more often

Database Example

We use this database for the following query examples

dno mgr budgetDept

eid sal dno age phone

Emp

Foreign key

Index-Only Plans (1)A number of queries can be answered without retrieving any tuples from one or more of the relations involved if a suitable index is available.

SELECT D.mgrFROM Dept D, Emp EWHERE D.dno=E.dno

Find mangers of departments with at least one employee

Index on dno

If <E.dno> index is available, no need to retrieve Employees tuples, i.e., scan the E.dno data entries and pick up the corresponding Dept tuples


EmpNo need to perform join operation

Index-Only Plans (2)

eid sal dno age phoneEmp

Index on dno

SELECT E.dno, COUNT(*)FROM Emp EGROUP BY E.dno

If <E.dno> index is available, need only scan the data entries and count employees for each dno

Index-Only Plans (3)SELECT E.dno, MIN(E.sal)FROM Emp EGROUP BY E.dno

If <E.dno,E.sal> tree index is available, need only scan the data entries and compute MIN(E.sal) for each dno

Emp


Index on dno & sal


SELECT AVG(E.sal)FROM Emp EWHERE E.age=25 AND E.sal BETWEEN 300000 AND 500000

If <E.age,E.sal> or <E.sal,E.age> tree index is available, the average salary can be computed using only data entries in the index

Emp


Compute averagesalary of young executives


Using <age,dno> index• Use index find first data entry /w age = 30• Scan data entries with age = 30, and count

number of tuples for each dno (the departments are arranged continuously for age=30)

SELECT E.dno, COUNT (*)FROM Emp EWHERE E.age=30GROUP BY E.dno

Index-only plans are possible if the key is <dno,age> or we have a tree index with key <age,dno>

Using <dno,age> index• Scan all data entries• For each dno, count number of

tuples with age=30Do not scan

all ages→ Better !


Index only plans are possible – Using the <age,dno> index• For data entries with age > 30,

sort them on dno• Scan sorted data entries and

count number of data entries for each dno

– Using <dno,age> index • Scan data entries• For each dno, count number of

data entries with age > 30

SELECT E.dno, COUNT (*)FROM Emp EWHERE E.age>30GROUP BY E.dno

No sorting. Better !

Note: Sorting is not necessary if the department counters can fit in memory

Summary• Many alternative file organizations exist, each

appropriate in some situation.

• If selection queries are frequent, sorting the file or building an index is important.– Hash-based indexes only good for equality search.– Sorted files and tree-based indexes best for range

search; also good for equality search. (Files rarely kept sorted in practice; B+ tree index is better.)

• Index is a collection of data entries (<key, rid> pairs or <key, rid-list> pairs) plus a way to quickly find entries (using hashing or a search tree) with given key values.

Summary (Contd.)

• Data entries can be actual data records, <key, rid> pairs, or <key, rid-list> pairs.– Choice orthogonal to indexing technique used to

locate data entries with a given key value.

• Can have several indexes on a given file of data records, each with a different search key.

• Indexes can be classified as clustered vs. unclustered, and primary vs. secondary. Differences have important consequences for utility/performance.

Summary (Contd.)• 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?

• 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.

Overview of Query Optimization

Internal Query Representation

Query Plan: Tree of R.A. operations, with choice of algorithm for each operation.– 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.

Query results

Horizontal “filters”

Vertical “filters”

Overview of Query Evaluation

• 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!

Some Common TechniquesAlgorithms 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.)

Dataentries

Data Records

Key

Some Common TechniquesAlgorithms 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.

Watch for these techniques as we discuss query evaluation!

Statistics and Catalogs• 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.

• More detailed information (e.g., histograms of the values in some field) are sometimes stored.

Avoid using unclustered index in

this range

Statistics and Catalogs• 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.

• More detailed information (e.g., histograms of the values in some field) are sometimes stored.

• Catalogs updated periodically.– Updating whenever data changes is too expensive; lots of

approximation anyway, so slight inconsistency ok.

Access Paths An access path is a method of retrieving tuples:

Examples: 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.

Example: Tree index on <a, b, c> matches the selection a=5 AND b=3, and a=5 AND b>6, but not b=3.

A hash index matches (a conjunction of) terms that has a term attribute = value for every attribute in the search key of the index.

Example: Hash index on <a, b, c> matches a=5 AND b=3 AND c=5; but it does not match b=3, or a=5 AND b=3, or a>5 AND b=3 AND c=5.

One Approach to SelectionsFind the most selective access path, retrieve tuples using it, and apply any remaining terms that don’t match the index:

– Most selective access path: An index or file scan that we estimate will require the fewest page I/Os.

– Terms that match this index reduce the number of tuples retrieved; other terms are used to discard some retrieved tuples, but do not affect number of tuples/pages fetched.

Database

Most selective access path

Matching Other terms

Fewest I/O’s

No additional

I/O’s

One Approach to Selections - Example

Consider day<8/9/94 AND bid=5 AND sid=3. – A B+-tree index on day can

be used; then, bid=5 and sid=3 must be checked for each retrieved tuple.

– Similarly, a hash index on <bid, sid> could be used; day<8/9/94 must then be checked.

Database

B+ tree on“day”

Check “bid” and “sid”

Fewest I/O’s

No additional

I/O’s

Using an Index for SelectionsCost depends on #qualifying tuples, and clustering.

Cost = Cost(finding qualifying data entries) + Cost(retrieving records)

Typically small

Could be large w/o clustering

Emp


Finding qualifying data entries

Retrieving matching data records

Using an Index for SelectionsCost depends on #qualifying tuples, and clustering.

Cost = Cost(finding qualifying data entries) + Cost(retrieving records)

Example: Assuming uniform distribution of names, about 5% of tuples qualify (say 100 pages, 10,000 tuples). – With a clustered index, cost is little more than 100

I/Os;– if unclustered, upto 10,000 I/Os!

SELECT *FROM Reserves RWHERE R.rname LIKE ‘C%’

Typically small

Could be large w/o clustering

Duplicates Elimination Using Sorting

• The expensive part is removing duplicates.SQL systems don’t remove duplicates unless the keyword DISTINCT is specified in a query.

• Sorting Approach: Sort on <sid, bid> and remove duplicates. (Can optimize this by dropping unwanted attributes while sorting.)

• Sorting Data Entries: If there is an index with both R.sid and R.bid in the search key, may be cheaper to sort data entries!

SELECT DISTINCT R.sid, R.bidFROM Reserves R

Duplicates Elimination Using HashingSELECT DISTINCT R.sid, R.bidFROM Reserves R

H1

Memory

B1B1B1B1B1B2What is the I/O cost ? Two rounds of I/O’s: Round 1 due to H1, and round 2 due to H2

H1 & H2 are distinct hash functions H2

B1.1 B1.2 B1.3 B1.4

Eliminate duplicates

One

buc

ket a

t a ti

me

Relation

Join: Index Nested Loops (1)

If there is an index on the join column of one relation (say S), can make it the inner and exploit the index.

Cost: M + ( (M*pR) cost of finding matching S tuples)

foreach tuple r in R doforeach tuple s in S where ri == sj do

add <r, s> to result

1

2

M

.

.

.

R 1

2

N

.

.

.

S

1 Index

pr tuples/page→

Fetch intomemory

R has (M p∙ R) tuples

Join: Index Nested Loops (2)• For each R tuple, cost of probing S index (i.e., finding data

entry) is about 1.2 for hash index, 2-4 for B+ tree.

Dataentries

Data Records

Key

2~4 levels

1.2 I/O

Join: Index Nested Loops (3)• For each R tuple, cost of probing S index (i.e., finding

data entry) is about 1.2 for hash index, 2-4 for B+ tree. • Cost of then finding S tuples (assuming Alt. (2) or (3)

for data entries) depends on clustering.– Clustered index: 1 I/O (typical), – Unclustered: up to 1 I/O per matching S tuple.

1

2

M

.

.

.

R

1 Index

1

2

N

.

.

.

S

pr tuples/page

Fetch intomemory

Join: Sort-Merge (R S)Sort R and S on the join column, then scan them to do a ``merge’’ (on join col.), and output result tuples.– Advance scan of R until

current R-tuple >= current S tuple, then – advance scan of S until

current S-tuple >= current R tuple; – Repeat the above steps until

current R tuple = current S tuple.– At this point, all R tuples with same

value in Ri (current R group) and all S tuples with same value in Sj (current S group) match; output <r, s> for all pairs of such tuples.

– Then resume scanning R and S.

i=j

SCAN

SCAN

SCAN SC

AN

RelationR RelationS

Often fits in one page

COST: R is scanned once; each S group is scanned once per matching R tuple.

An R group

Cost of Sort-Merge Join

Cost: M log M + N log N + (M+N)

Scan each relation once

The cost of scanning could be MN (very

unlikely)

SCAN

SCAN

SCAN SC

AN

RelationR RelationS

Often fits in one page Sort relation

R

Merger SortO

rigin

al re

latio

nIn-memory

local sorting, 2 rounds of I/O Merging two sorted lists

using 2 input buffers and 1 output buffer

Log28 or 3 passes (i.e., 6 rounds of I/Os)

Each pass requires 2 rounds of I/Os

Sorted relation

B1

B2 B3

Highlights of System R Optimizer• Impact:

– Most widely used currently; works well for < 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.O1

O2

O4

O3

R1 R2

R3

R4

R5

Left-deepplan

Query Block• An SQL query is parsed into a collection of query

blocks and then passed on to the query optimizer• A query block is an SQL query with no nesting

SELECT S.sid, MIN(R.day)FROM Sailors S, Reserves R, Boats BWHERE S.sid=R.sid AND R.bid-B.bid

AND B.colr=‘red’ AND S.rating = (SELECT MAX (S2.rating)

FROM Sailors S2 )GROUP BY S.sidHAVING COUNT(*)>1

SELECT S.sid, MIN(R.day)FROM Sailors S, Reserves R, Boats BWHERE S.sid=R.sid AND R.bid-B.bid

AND B.colr=‘red’ AND S.rating = Reference to nested blockGROUP BY S.sidHAVING COUNT(*)>1

SELECT MAX (S2.rating)FROM Sailors S2

The query optimizer considers each query block and chooses a query evaluation plan for that block.

Block 1

Block 2

Query ProcessorSQL Query

Plan Generator

Plan CostEstimator

Evaluationplan

Query Parser

Catalog Manager

Query Plan Evaluator

Cost Estimation

For each plan considered, must estimate cost:– Must estimate cost of each operation in plan tree.

• Depends on input cardinalities.• We’ve 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 (next page)

Size Estimation and Reduction Factors

• 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 RF’s Implicit assumption that terms are independent!

SELECT attribute listFROM relation listWHERE term1 AND ... AND termk

Consider a query block:

Size Estimation and Reduction Factors Result cardinality = Max # tuples product of all RF’s• Term col=value has RF 1/NKeys(I), given index I on col

EXAMPLE: If the number of distinct key values is 100, size(result) = 1% size(operand relation)

• Term col1=col2 has RF 1/MAX(NKeys(I1), NKeys(I2))OBSERVATION: – The smaller the numbers of distinct key values, the bigger the size of

the result due to more matches between the two operand relations. – If Nkeys(I1)=Nkeys(I2)=1, RF=1 and size(R1 R2) = size(R1R2)

• Term col>value has RF (High(I)-value)/(High(I)-Low(I))– RF is the percentage of the tuples that satisfy the term

High

LowValue

Matching area

Schema for ExamplesSailors (sid: integer, sname: string, rating: integer, age: real)Reserves (sid: integer, bid: integer, day: dates, rname: string)

.

.

.

Tuple is 50 bytes

80 tuples/page

500 pages

Sailors

.

.

.

Tuple is 40 bytes

100 tuples/page

1000 pages

Reserves

Plan Generation – Relational Algebra

SELECT S.snameFROM Reserves R, Sailors SWHERE R.sid=S.sid AND R.bid=100 AND S.rating>5

sname( sid=sid bid=100 rating>5 (RS))

= sname( bid=100 rating>5 ( sid=sid (RS)))

= sname( bid=100 rating>5 (R S))

FROM clause WHERE clause SELECT clause

Plan Generation - Execution Plan


Reserves Sailors

sid=sid

bid=100 rating > 5

sname RA Tree

Reserves Sailors

sid=sid

bid=100 rating > 5

sname

(Simple Nested Loops)

(On-the-fly)

(On-the-fly)

Plan

sname( bid=100 rating>5 (R S))

On-the-fly Save I/O

Motivating Example

• Cost: 500 + 5001000 I/Os• By no means the worst plan! • Misses several opportunities:

selections could have been `pushed’ earlier, no use is made of any available indexes, etc.

• Goal of optimization: To find more efficient plans that compute the same answer.


Reserves Sailors

sid=sid

bid=100 rating > 5

snameRA Tree

Reserves Sailors

sid=sid

bid=100 rating > 5

sname

(Simple Nested Loops)

(On-the-fly)

(On-the-fly)Plan

(1,000 pages) (500 pages)

Alternative Plan 1 (No Indexes)

• Main difference: push selects• 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)– With 5 buffer pages, can do 4-way merges

Sort T1 (2+2)10, sort T2 (2+6)250, merge (10+250)– Total cost: 4,060 page I/Os

Reserves Sailors

sid=sid

bid=100

sname(On-the-fly)

rating > 5

(Scan;write to temp T1)

(Scan;write totemp T2)

(Sort-Merge Join)

Push selects before join

(1,000 pages) (500 pages)

Need log4(250/5) or 3 passesOne pass for local sort, each requiring 2 rounds of I/Os

Scan T1 & T2


If we used Block Nested Loop join, join cost = 10+4250, total cost = 2770.

Reserves Sailors

sid=sid

bid=100

sname(On-the-fly)

rating > 5



(Block Nested Loop Join)

Scan T2 for every 3 pages of T1

→ Since T1 has 10 pages, need to scan T2

four times, i.e.,

T1 has 10 pages

T1 T2

JOIN

MemoryBring in T1 3 pages at

a time

Need to scan T2 4 times


Reserves Sailors

sid=sid

bid=100

sname(On-the-fly)

rating > 5



(Block Nested Loop Join)

T1 T2

JOIN

Memory

T1 fits in 3 pages

Need to scan the

smaller T2 only once

If we `push’ projections, T1 has only sid, T2 only sid and sname:– T1 fits in 3 pages, need to scan T2

only once– cost of BNL drops to under 250

pages, total < 2000.

Alternative Plan 2 (with Indexes)• Reserves has 1000 pages and 100 tuples

per page → 100,000 tuples• There are 100 boats, equally popular• With clustered index on bid of Reserves,

we get 1000 tuples (= 100,000/100) on 10 pages (= 1,000/100).

• INL with pipelining (outer is not materialized).

Reserves

Sailors

sid=sid

bid=100

sname(On-the-fly)

rating > 5

(Use hashindex; donot writeresult to temp)

(Index Nested Loops,with pipelining )

(On-the-fly)

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

• Cost: Selection of Reserves tuples (10 I/Os); for each of 1,000 Reserve tuples, must get matching Sailors tuple (10001.2); total 1210 I/Os (plus 1,000 more I/Os if Sailors stored as alternative 2)

(unclusteredhash Index on sid OK)

Not materialized

Alternative Plan 2 (continued …)

Reserves

Sailors

sid=sid

bid=100

sname(On-the-fly)

rating > 5

(Use hash index; do not write result to temp)


(On-the-fly)

(unclusteredIndex on sid OK)

Decision not to push “rating>5” before the join is based on availability of sid index on Sailors

bid, sidProjecting out unnecessary

fields from outer relation (i.e.,

Reserves) doesn’t help because we

want to use the hash index Reserves

Sailors

sid=sid

bid=100

sname(On-the-fly)

rating > 5


(On-the-fly)

(unclusteredIndex on sid OK)

bid, sid

Projecting out unnecessary fields here doesn’t help

because we want to prob Sailors without

writing result to temp

Summary• There are several alternative evaluation algorithms for each

relational operator.• A query is evaluated by converting it to a tree of operators and

evaluating the operators in the tree.• 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.

DBMS Internal Review. Outline Storage and Indexing Query Optimization.

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Transcript of DBMS Internal Review. Outline Storage and Indexing Query Optimization.