Database Systems (資料庫系統) Lecture #6

(1)

Database Systems

( 資料庫系統 )

October 31, 2005

Lecture #6

(2)

Physical & Digital Space

Interaction

• Consider moving a pencil …

– Physical space movement mapping to digital space

– Digital space movement mapping to physical space

• Which one is more difficult?

• The Actuated Workbench (MIT media

lab)

(3)

Course Administration

• Assignment #2: due 11/2

• Assignment #3: post on the course homepage

– Due 11/8 (next Tuesday)

• Practicum Assignment #1:

– Post on the course homepage on 11/7

(4)

Reflection

• How to design a database?

– Conceptual design: ER Model

– Logical design: Relational Model

• How to ask questions on a database?

– Relational Algebra & SQLs

• What’s next?

– How to get fast access to records?

• File organizations & indexes

• What’s further next?

(5)

Overview of Storage &

Indexing

(6)

Outline

• Types of external storage devices • File organizations

– Questions:

• How to store table records on external storage device (e.g., a disk)?

• How to speed up access to needed records on a disk?

– Heap file, Sorted file

– Indexing data structures

• Tree-based indexing, Hash-based indexing • Comparison on file organizations

– Question: which one is better/worse in performance?

• Indexes and Performance

(7)

Data on External Storage

• External Storage: offer persistent

data storage

– Unlike physical memory, data saved on a persistent storage is not lost when the system shutdowns or crashes.

(8)

Types of External Storage

Devices

• Magnetic Disks: Can retrieve random page at fixed cost

– ~$1 per Gigabyte

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

• Tapes: Can only read pages in sequence

– $0.3 per Gigabyte

– Cheaper than disks; used for archival storage

• Other types of persistent storage devices:

– Optical storage (CD-R, CD-RW, DVD-R, DVD-RW) – Flash memory

(9)

Definition

• A record is a tuple or a row in a relation table.

– Fixed-size records or variable-size records

• A file is a collection of records.

– Store one table per file, or multiple tables in the s ame file

• A page is a fixed length block of data for disk I

/O.

– A file is consisted of pages.

– A data page also contains a collection of records. – Typical page sizes are 4 and 8 KB.

(10)

File Organization

• Method of arranging a file of records

on external storage.

– Record id (rid) is used to locate a record on a disk

• [page id, slot number]

– Indexes are data structures to efficiently search rids of given values

(11)

DB Storage and Indexing

• Layered Architecture

– Disk Space Manager allocates/de-allocates spaces on disks.

– Buffer manager moves pages between disks and main memory.

– File and index layers organize records on files, and manage the indexing data structure.

(12)

Alternative File

Organizations

• Many alternatives exist, each ideal for some situations, and not so good in others:

– Heap files: Records are unsorted. Suitable when typical

access is a file scan retrieving all records without any order.

• _{Fast update (insertions / deletions)}

– _{Sorted Files: Records are sorted. Best if records must be}

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

• _{Examples: employees are sorted by age.} • _{Slow update in comparison to heap file.}

– Indexes: Data structures to organize records via trees or

hashing.

• For example, create an index on employee age.

• Like sorted files, speed up searches for a subset of records that match

values in certain (“search key”) fields

(13)

Indexes

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

– _{Any subset of the attributes of a table can be the search}

key for an index on the relation.

– Search key does not have to be candidate key

• _{Example: employee age is not a candidate key.}

• An index file contains a collection of data entries (called k*).

– Quickly search an index to locate a data entry with a key value k.

• Example of a data entry: <age, rid>

– Can use the data entry to find the data record.

• Example of a data record: <name, age, salary>

– Can create multiple indexes on the same data records.

(14)

Indexing Example

Index Data Structure:

Index entries + Indexing method

Data entries

Data records

Search key value: find employees with age = 25 Index File (Small for efficient search) Data File (Large) Paul (k=25, Paul’s rid) (B+ Tree) (Hash) Mary

(15)

Alternatives for Data Entry k*

• Three alternatives for what to store in a data entry:

– (Alternative 1): Data record with key value k

• Example data record = data entry: <age, name, salary>

– (Alternative 2): <k, rid of data record with search key value k>

• Example data entry: <age, rid>

– (Alternative 3): <k, list of rids of data records with search key k>

• Example data entry: <age, rid_1, rid_2, …>

• Choice of alternative for data entries is independent of the indexing method.

– Indexing method takes a search key and finds the data entries matching the search key.

(16)

Alternatives for Data Entries

(Contd.)

• Alternative 1: data record with key

value k

– Data entries are also the data records. – At most one index on a given collection

of data records can use Alternative 1.

– If data records are very large, # of

pages containing data entries is high.

(17)

Alternatives for Data Entries

(Contd.)

• Alternative 2: <k, rid with key value k> • Alternative 3: <k, rid-list of data record

(s)>

– Data entries typically much smaller than data records.

• May lead to more efficient search than Alternative 1. Why?

– Alternative 3 more compact than Alternative 2,

• _{Lead to variable sized data entries (size of rid-li}

(18)

Index Classification

• Clustered vs. unclustered: If order of data records is the same as, or close to the

order of data entries, then it is called clustered index.

– Alternative 1 implies clustered; in practice, clustered also implies Alternative 1.

– One clustered index and multiple unclustered indexes

– Why is this important?

• _{Consider the cost of range search query: find all}

(19)

Clustered vs. Unclustered

Index

• Cost of retrieving data records through index varies great ly based on whether index is clustered or not!

• _{Examples: retrieve all the employees of ages 30~39.}

– _{What is the worst-case cost (# disk page I/Os) of clustered index?} – _{What is the worst-case cost of unclustered index?}

Cost = number of pages retrieved mr = # matched records; mp = # pages containing matched records

Index entries Data entries (Index File) (Data file) Data entries CLUSTERED UNCLUSTERED Ages = 30~39 Ages = 30~39

(20)

Hash-Based Indexes

• Good for equality selections.

– Data entries (key, rid) are grouped into buckets.

– Bucket = primary page plus zero or more overflow pages. – Hashing function h: h(r) = bucket in which record r

belongs. h looks at the search key fields of r.

– If Alternative (1) is used, the buckets contain the data records. H Smith, 44, 3000 500 4 -300 0 300 0 500 4 Jones, 40, 6003 Tracy, 44, 5004 Primary page Bucket A Salary Ashby, 25, 3000 Basu, 33, 4003 Bristow, 29, H(salary)=A H Age H(age)=00 H(age)=01

(21)

Hash-based Indexes (Cont)

• Search on key value:

– Apply key value to the hash function -> bucket number – Retrieve the primary page of the bucket. Search records

in the primary page. If not found, search the overflow pages.

– Cost of locating rids: # pages in bucket (small)

• Insert a record:

– Apply key value to the hash function -> bucket number – If all (primary & overflow) pages in that bucket are full,

allocate a new overflow page. – Cost: similar to search.

• Delete a record

(22)

B+ Tree Indexes

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

P0 _{K 1 P 1 K 2 P 2} K m P m

index entry

Non-leaf Pages

(23)

Example B+ Tree

• Find 7*, 29*? 15* < age < 30*

• Insert/delete: Find data entry in leaf, then change it. Need to adjust parent sometimes.

– And change sometimes bubbles up the tree (keep the tree balance) Root 17 30 13 5 27 2* 3* 5* 7* 8* _14*16* 22*24* 27*29* 33*34*38*39* Entries <= 17 Entries > 17

(24)

Cost Model for Our Analysis

• Ignore CPU costs, for simplicity.

• Measure disk I/O costs: number of page

I/O’s

– Ignores gains of pre-fetching a sequence of pages

• Cost analysis

– B: The number of data pages

(25)

Comparing File

Organizations

• Heap files (random order; insert at eof) • Sorted files, sorted on <age, salary> • Clustered B+ tree file, Alternative (1),

search key <age, salary>

• Heap file with unclustered B + tree index on search key <age, salary>

• Heap file with unclustered hash index on search key <age, salary>

(26)

Operations to Compare

• Scan: Fetch all records from disk • Equality search

– Example: find all employees with age = 23 and salary = 5000.

• Range selection

– Example: find all employees with age > 35.

• Insert a record

– Identify the page for inserting the record, fetch it, modify it, and write it back.

• Delete a record

(27)

Assumptions in Our Analysis

• Heap Files:

– Equality selection on key; exactly one

match.

• Sorted Files:

– Files compacted after deletions.

• Indexes:

– Alt (2), (3): data entry size = 10% size of data record

(28)

Heap Files

B: The number of data pages R: Number of records per page

• Scan: B

• Search with equality selection: ½ B • Search with range selection: B

• Insert: 2

– New record is inserted at the end of the file. Read/write out last page.

• Delete a record: search cost + 1 (no compacting) • Delete a record (with rid): 2

– search cost = 1

(29)

Sorted Files

B: The number of data pages R: Number of records per page

• Scan: B

• Search with equality selection: log2(B)

– Binary search

• Search with range selection: log2(B) + # pages of matched records

– 14<x<60: Search for first matching record / page, then find all the qualifying records / pages in sequential order.

• Insert: search cost + B

– Find the right position/page (search), make space for the inserting record by shifting all subsequent records by one slot, then insert.

• Delete: search cost + B

– Search the record, delete it, shift all subsequent records by one slot.

(30)

Clustered B+ Tree File

B: The number of data pages R: Number of records per page • Scan: 1.5 B

• Search with equality selection: logF(1.5 B)

– F is number of children per B+ tree node

• Search with range selection: logF(1.5 B) + # pages of

matched records

– Search for first matching record (page), then find all the qualifying records / pages in sequential order.

Un-occupancy in clustered file ~ 0.5 B

Index entries

Data entries = Data records CLUSTERED

(31)

Clustered B+ Tree File

• Insert: search cost + 1

– Find the right position (page), insert + write out the modified page. No need to shift records -> empty data entries in modified page.

• Delete: search cost + 1

– Search record, delete record, and write back modified page. No need to shift.

Index entries

Data entries = Data records CLUSTERED

(32)

Heap File with Unclustered

Tree Index

• Scan (in-order): 0.1 B + RB

– Unorder scan: B

• Search with equality selection: logF(0.1 B) + 1

– Search for data entry takes logF(0.1B) pages + one read on data

record page.

Data/index entry size ~ 0.1 data record size

UNCLUSTERED

(33)

Heap File with Unclustered

Tree Index

• Search with range selection: logF(0.1B) + # pages

of matched records

– Search for first matching data entry, then find all the qualifying entries in sequential order. But each data entry may point to a data record on a different data page.

• Insert: search cost + 3

– One read/write to heap file page + search + one write to data entry page.

(34)

Heap File with Unclustered

Hash Index

• Scan (in-order): 0.125 B + RB = (0.125 + R) B

– Unorder scan: B

• Search with equality selection: 2

– Hash function + read data entry page + read data record page • Search with range selection: B

– Hash function is useless, do unorder scan. Hash: No overflow buckets.

80% page occupancy => # data entry pages = 0.125B

H Smith, 44, 3000 500 4 -300 0 300 0 500 4 Jones, 40, 6003 Tracy, 44, 5004 Primary page Bucket A Salary Ashby, 25, 3000 Basu, 33, 4003 Bristow, 29, H(salary)=A

(35)

Heap File with Unclustered

Hash Index

• Insert: 4

– Read & Write Heap file page + Read & Write data entry page

• Delete: 4

Hash: No overflow buckets. 80% page occupancy => # data entry pages = 0.125B

H Smith, 44, 3000 500 4 -300 0 300 0 500 4 Jones, 40, 6003 Tracy, 44, 5004 Primary page Bucket A Salary Ashby, 25, 3000 Basu, 33, 4003 Bristow, 29, H(salary)=A

(36)

36

Cost of Operations



No one file organizations is uniformly superior

(a) S c a n (b) Equalit y (c) Range (d)

Insert (e) delete

(1) Heap B 0.5B B 2 Search

+ 1 (2) Sorted B _log₂_{(B) log}₂_{(B) +}

#matche s Searc h + B Search + B (3) Clustered Tree Index 1.5 B _log_F_(1.5 B) log_F(1.5B ) + #matche s Searc h + 1 Search + 1 (4) Uncluster ed Tree index B(R+ 0.1) 1+log_(0.1B)F log_F(0.1 B) + #mat ches Searc h + 3 Search + 3 (5) Uncluster ed Hash B(R+ 0.125 ) 2 B 4 4

(37)

General Guidelines

• An index supports efficient retrieval of data entries satisfying a selection

condition:

– Two types of selections: equality and range – Hash-based indexing is only optimized for

equality selection, useless for range selection. – Tree-based indexing is better for both.

– Tree-based clustering index is best for range selection.

(38)

General Guidelines (Cont)

• Clustered index can be more expensive than unclustered index:

– When inserting a new record into a full page, shift existing records into other pages → change data entries for these records → expensive.

(39)

Understanding the Workload

• How to decide the best indexing for a table? – Need to understand the workload

• For each query in the workload:

– Which tables does it access? – Which fields are retrieved?

– Which fields are involved in selection/join conditions?

How selective are these conditions likely to be? • For each update in the workload:

– Which fields are involved in selection/join conditions? – How selective are these conditions likely to be?

– The type of update (INSERT/DELETE/UPDATE), and

(40)

Choice of Indexes

• What indexes should we create?

– Which tables 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 or unclustered? – Hash index or Tree index?

(41)

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!

– For now, we discuss simple 1-table queries.

• Before creating an index, must also consider the impact on updates in the workload!

– Trade-off: Indexes can make queries go faster, updates

slower (because also have to update the indexes). Indexes also require disk space, too.

(42)

Index Selection Guidelines

• Attributes in WHERE clause are candidates for index keys.

– Exact match condition suggests hash index. – Range query suggests tree index.

• Clustering is especially useful for range queries; can also help on equality queries if there are many duplicates.

• Multi-attribute search keys should be considered when a

WHERE clause contains several conditions.

– Order of attributes is important for range queries.

– _{Such indexes can sometimes enable index-only strategies for} important queries.

• For index-only strategies, clustering is not important!

• 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.

(43)

Examples of Clustered

Indexes

• B+ tree index on E.age can b

e used to get qualifying rec ords.

– How selective is the condition ? (selective means % of qualif ied records)

– _{Is this index useful?}

• Consider the GROUP BY query.

– _Is_E.age_{index good? Why not?} – _Clustered_E.dno_{index may be b}

etter.

• Equality queries and duplica tes:

– _{Unclustering is bad in case of}

many qualified records.

– _{Clustering on}_E.hobby_helps!

SELECT E.dno FROM Emp E

WHERE E.age>40

SELECT E.dno, COUNT (*) FROM Emp E WHERE E.age>10 GROUP BY E.dno SELECT E.dno FROM Emp E WHERE E.hobby=Stamps

(44)

Composite Search Keys

• Search on a combination of

fields.

• Which index can be applied?

– Equality query: Every field v alue is equal to a constant v alue.

• age=12 and sal =10

– Range query:_{Some field value}

is not a constant.

• age =13; or sal=10 and age > 5

– The order of fields in compos ite key is important!

• <sal, age>: data entries are so rted by sal first, then age.

sue 13 75 bob cal joe 12 10 20 80 11 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 records sorted by name

Data entries in index Data entries Examples of composite key

(45)

Composite Search Keys

• To retrieve Emp records with age=30 AND sal=4000,

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

• If condition is: 20<age<30 AND 3000<sal<5000:

– _{Clustered tree index on <}age,sal_{> or <}sal,age_>.

• If condition is: age=30 AND 3000<sal<5000:

– _{Clustered <}_age,sal_{> index much better than <}_sal,age_{> index. W}

hy?

– _{For <age, sal>, find the first data entry with (age=30, sal=3}

000) and the qualified entries are likely to be qualified. H owever, for <sal, age>, find the first data entry with (sal=3 000, age=anything), subsequent entries can have any ages.

– The order of fields in composite key is important!

(46)

Index-Only Plans

• A number of queries can be

answered without retrieving any

records from one or more of the

relations involved if a suitable index is available.

<E.dno>

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

GROUP BY E.dno

<E.dno,E.sal> Tree Index SELECT E.dno, MIN(E.sal) FROM Emp E

(47)

Index-Only Evaluation

(Contd.)

• Index-only

evaluations are

possible if the key is

<dno,age> or a

tree index with key

<age,dno> • Consider selective-ness of condition vs. cost of sorting – Selective: <age, dno>

– Not selective: <dno,

age>

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

WHERE E.age=30

GROUP BY E.dno

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

WHERE E.age>30

(48)

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.

• Index is a collection of data entries plus a

way to quickly find entries with given key values.

(49)

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. Differences have important consequences for utility/performance.

(50)

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

fields/relations are involved?

• Indexes must be chosen to speed up important queries (and perhaps some updates!).

– Index maintenance overhead on updates to key fields. – 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