CS 587/410, Spring 2008

1. CS 587/410, Spring 2008 Relational Database Systems /Database Internals Len Shapiro, len at pdx.edu Syllabus: www.cs.pdx.edu/~len/587 • This work, and all other files in this series, are licensed under the Creative Commons Attribution License. To view a copy of this license, visit http://creativecommons.org/licenses or send a letter to Creative Commons, 171 2nd Street, Suite 300, San Francisco, California, 94105, USA. • These slides have benefitted from the content of slides developed by Raghu Ramakrishnan and Johannes Gehrke, whom I thank. • These slides are best viewed with “slide show”. To print the slides, consider using "File/Print/Properties/Pages per sheet", 4-up, instead of Power Point's "handouts" mode. • Slides will be online by 6PM each Monday evening. If not, email me!

2. Plan for today • 3x5 cards: Name, email, prerequisite • Review Syllabus • Personal Introductions • What is your background in Database? • What would you like to get out of the course? • Minibase Intro, Buffer Manager assignment • Lecture on Chapter 8 • Lecture on part of Chapter 9 needed for homework • Quit 10 minutes early to form teams • What is your background in Java?

3. Web Forms Applic. Front ends SQL interface Index Files System Catalog Data Files 8.1 Database Architecture (cf. pg. 20) Parser Plan Executor Optimizer Operator Evaluator Files & Access Methods Transaction Manager Recovery Manager Buffer Manager Lock Manager Disk Space Manager

4. Learning Objectives for Chapter 8 • Identify the input, output and function of each component in the architecture diagram • Be able to describe the relationships between different classifications of indexes • Alternatives, clustering, uniqueness, denseness, primary • For each cell in the table on the “cost of operations” slide , explain the underlying algorithm and justify the cost. • Definitions: Index only plans, composite keys • Given a workload, choose an optimal index configuration.

5. Table of Contents: Chapter 8 (*review) • 8.1 Data on external storage* • Database architecture • 8.2 File organizations and indexing* • Indexes • Alternatives for k* data entries • Index classification: clustered, dense, unique, primary • 8.3 Index search data structures* • Hash-based indexes • Tree indexes • 8.4 Comparing File Organizations • 8.5 Indexes and performance tuning*

6. 8.Storage 8.2 Alternative File Organizations Many 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. • Hashed files: Best for quick equality access. • RDBMSs typically use heapfiles plus indexes • A tree index using alternative 1 (see below) is similar to a sorted file, a hash index using alternative 1 is the same as a hash file.

7. How the Records Fall Heap Sorted Hashed Bucket 0 1 2 3 16 9 22 41 62 15 2 11 2 9 11 15 16 22 41 62 16 9 41 2 22 62 11 15

8. 8.Storage Indexes • An index on a file speeds up selections on the search key fields for the index. • Any subset of the fields, or a prefix of a field, 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). • An index contains a collection of data entries, and supports efficient retrieval of all data entries k*with a given key value k. • Given data entry k*, we can find record with key k in at most one disk I/O. (Details soon …)

9. Typical index architecture Search structure: given a key, it finds the data entry for that key • Used by modern DBMSs • What are typical search structures in CS? k* k* Data entries k* (Index File) (Data file) Data Records

10. 8.Storage Alternatives for Data Entry k*in Index • In a data entry k* we can store: • Data record with key value k, or • In this case the previous page is not accurate • <k, rid of data record with search key value k>, or • <k, list of rids of data records with search key k> • 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

11. 42 Red 30 Blue 10 Blue 14 Red 21 Blue INDEX 24 Green Alternative 1: k* holds the data . . .

12. 10 Blue 42 Red 30 Blue 21 Red 24 Blue INDEX 10 Green Alternative 2: One pointer per k*

13. 10 Blue 18 Red 30 Blue 21 Red 24 Blue INDEX Alternative 3: 1 or more pointers per k* 10 Green

14. 8.Storage 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 single-attribute index on a given collection of data records can use Alternative 1. (Otherwise, data records are duplicated, leading to redundant storage and potential inconsistency.) • Alternatives 2 and 3: • Alternative 3 more compact than Alternative 2, but leads to variable sized data entries even if search keys are of fixed length.

15. 8.Storage UNCLUSTERED Data entries (Data file) Data Records 8.2.1 Clustered vs. Unclustered Index • Suppose 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 recs is `close to’, but not identical to, the sort order.) Index entries CLUSTERED direct search for data entries Data entries (Index File) Data Records

16. 8.Storage 8.2.2 Index Classification • Primary vs. secondary: If search key contains primary key, then called primary index. • In practice, primary means alternative 1 • 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. • Alternative 1 implies clustered. • A file can be clustered on at most one single-attributesearch key. • Cost of retrieving data records through index varies greatly based on whether index is clustered or not!

17. Index Classification (Cont.) • Dense vs. Sparse: If there is at least one data entry per search key value (in some data record), then it’s dense. • Alternative 1 always leads to dense index. • Every sparse index is clustered! • Sparse indexes are smaller; however, some useful optimizations are based on dense indexes. Ashby, 25, 3000 22 Basu, 33, 4003 25 Bristow, 30, 2007 30 Ashby 33 Cass, 50, 5004 Cass Smith Daniels, 22, 6003 40 Jones, 40, 6003 44 44 Smith, 44, 3000 50 Tracy, 44, 5004 Sparse Index Dense Index on on Data File Name Age

18. Index Classification Clustered, sparse indexes are smaller; they work well for range searches and sorting. But…data blocks are difficult to keep sorted and some useful optimizations are based on dense indexes. Note: one file can have at most one clustered index - all of the additional indices must be unclustered. sparse dense clustered YES YES unclustered NO! YES

19. Smith,44,3000 Ashby,25,3000 Cass,50,5004 3000 4003 Jones,40,6003 Basu,33,4003 Daniels,22,6003 3000 2007 Bristow,29,2007 Tracy,44,5004 5004 6003 5004 6003 8.3.1 Hash-based Index Examples h(age)=00 h(sal)=00 age h(age)=01 sal h h h(sal)=11 h(age)=10 Employees file hashed on age Index on salary

20. 8.Storage 8.3.1 Hash-Based Indexes • Good for equality selections. • Index is a collection of buckets. • Bucket = primary page plus zero or moreoverflow pages. • Buckets contain data entries. • Hashing functionh: h(r) = bucket in which (data entry for) record r belongs. h looks at the search key fields of r. • No need for “index entries” in this scheme.

21. 8.Storage 8.3.2 B+ Tree Indexes Non-leaf Pages Leaf Pages (Sorted by search key) • Leaf pages containdata entries, and are chained (prev & next) • Non-leaf pages have index entries; only used to direct searches: index entry P K P K P P K m 0 1 2 1 m 2

22. 8.Storage Example B+ Tree Note how data entries in leaf level are sorted Root 17 • 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 Entries <= 17 Entries > 17 27 5 13 30 39* 2* 3* 5* 7* 8* 22* 24* 27* 29* 38* 33* 34* 14* 16*

23. 8.Storage 8.4 Comparing File Organizations • Assume an employee file is organized by the search key <age, salary>. • We will consider various file organizations, and operations like scan, equality, range, insert, delete. • For each file organization, we will compute the cost of each operation.

24. 8.Storage Comparing File Organizations • Heap files (random order; insert at eof) • Sorted files, sorted on <age, sal> • Clustered B+ tree file, Alternative (1), search key <age, sal> • Heap file with unclustered B + tree index on search key <age, sal> • Heap file with unclustered hash index on search key <age, sal>

25. 8.Storage Operations to Compare • Scan: Fetch all records from disk • Equality search • Range selection • Insert a record • Delete a record

26. 8.Storage Cost Model for Our Analysis We ignore CPU costs, for simplicity: • B: The number of data pages • R: Number of data records per page • F: Fanout, number of index records per page • D: (Average) time to read or write disk page • Measuring number of page I/O’s ignores gains of pre-fetching a sequence of pages; thus, even I/O cost is only approximated. • Average-case analysis; based on several simplistic assumptions. • Good enough to show the overall trends!

27. 8.Storage 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 record • Hash: No overflow buckets. • 80% page occupancy => File size = 1.25 data size • Tree: 67% occupancy (this is typical). • Implies file size = 1.5 data size

28. 8.Storage Assumptions (contd.) • Scans: • Leaf levels of a tree-index are chained. • Index data-entries plus actual file scanned for unclustered indexes. • Range searches: • We use tree indexes to restrict the set of data records fetched, but ignore hash indexes.

29. 8.Storage Cost of Operations • Several assumptions underlie these (rough) estimates!

30. 8.5 Indexes and Performance Tuning • One of a DBA’s most important duties is to deal with performance problems • One of the most effective methods for improving performance is to choose the best indexes for the given workload • Why not index everything? • What are the two costs of an index? • Part of a DBA’s job: choose indexes so the database will “run fast”. • What information does the DBA need, to do this?

31. 8.Storage 8.5.1 Understanding the Workload • For each query in the workload: • Which relations does it access? • Which attributes are retrieved? • Which attributes are involved in selection/join conditions? How selective are these conditions likely to be? • For each update in the workload: • Which attributes are involved in selection/join conditions? How selective are these conditions likely to be? • The type of update (INSERT/DELETE/UPDATE), and the attributes that are affected.

32. 8.Storage 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? • What kinds of indexes should we create? • Clustered? Hash/tree? • 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. • Also consider dropping indexes

33. 8.Storage 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! • Second approach: Use an Index Tuning Wizard [163]. • Inputs: A workload (it can also capture a workload) and an amount of disk storage for indexes • Output: an index configuration

34. 8.Storage 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. • 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. • 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.

35. 8.Storage Index-Only Plans SELECT E.dno, COUNT(*) FROM Emp E GROUP BY E.dno • 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. • For index-only strategies, clustering is not important! <E.dno> SELECT E.dno, MIN(E.sal) FROM Emp E GROUP BY E.dno <E.dno,E.sal> Tree index May help <E. age,E.sal> or <E.sal, E.age> SELECTAVG(E.sal) FROM Emp E WHERE E.age=25 AND E.sal BETWEEN 3000 AND 5000 Tree index!

36. 8.Storage Examples of Clustered Indexes SELECT E.dno FROM Emp E WHERE E.age>40 • B+ tree index on E.age can be used to get qualifying tuples. • How selective is the condition? • Is the index clustered? • Consider the GROUP BY query. • If many tuples have E.age > 10, using E.age index and sorting the retrieved tuples may be costly. • Clustered E.dno index may be better! • Equality queries and duplicates: • Clustering on E.hobby helps! 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

37. 8.Storage Indexes with Composite Search Keys Examples of composite key indexes using lexicographic order. • 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. 11,80 11 12 12,10 name age sal 12,20 12 13,75 bob 12 10 13 <age, sal> cal 11 80 <age> joe 12 20 10,12 sue 13 75 10 20 20,12 Data records sorted by name 75,13 75 80,11 80 <sal, age> <sal> Data entries in index sorted by <sal,age> Data entries sorted by <sal>

38. 8.Storage Composite Search Keys • To retrieve Emp records with age=30 ANDsal=4000, an index on <age,sal> would be better than an index on age or an index on sal. • Choice of index key orthogonal to clustering etc. • If condition is: 20<age<30 AND 3000<sal<5000: • Clustered tree index on <age,sal> or <sal,age> is best. • If condition is: age=30 AND 3000<sal<5000: • Clustered <age,sal> index much better than <sal,age> index! • Composite indexes are larger, updated more often.

39. Indexes in the real world • Types of indexes supported • Oracle, SQLServer and DB2 support only B+Tree indexes. Postgres supports hash indexes but does not recommend using them. • MySQL? • Everyone uses hash indexes for hash joins, but they are constructed on the fly.

40. Clustering in the real world • Oracle, SQL Server: • Declares a clustered index on any primary key. • It uses alternative 1. • DB2, Postgres • The user can define an index to be clustered. • The DBMS uses alternative 2. • At first the index will be clustered. • If you also specify a percentage of free space, it will place subsequent inserts/updates (Postgres:updates only) in sorted order so the index should stay clustered for a while. • It’s up to you to recluster the index if overflow chains get to be long. • MySQL

41. Text problems • Chapter 8, problems 1,3,5,7,9,11

42. Slides From Chapter 9 • Relevant to Buffer Management Assignment in Minibase • 9.3 Disk Space Management • 9.4 Buffer Management

43. 9.Disks 9.3 Disk Space Management • The lowest layer of DBMS software manages space on disk. • Higher levels call upon this layer to: • allocate/de-allocate a disk page • read/write a disk page • A request for a sequence of disk pages must be satisfied by allocating the pages sequentially on disk! Higher levels don’t need to know how this is done, or how free space on the disk is managed.

44. 9.Disks DB 9.4 Buffer Management in a DBMS Disk Page Requests from Higher Levels • Data must be in RAM for DBMS to operate on it! • Table of <frame#, pageid> pairs is maintained. BUFFER POOL disk page free frame MAIN MEMORY DISK choice of frame dictated by replacement policy

45. 9.Disks When a Disk Page is Requested ... • If requested disk page is not in buffer pool: • Choose a frame for replacement • If frame is dirty, write it to disk • Read requested page into chosen frame • Pin the page and return its address.

46. 9.Disks More on Buffer Management • Requestor of page must unpin it, and indicate whether page has been modified: • Dirty bit is used for this • What if requestor does not unpin the page? • Page in pool may be requested many times, • a pin count is used. A page is a candidate for replacement iff pin count = 0.

47. 9.Disks Buffer Replacement Policy • Frame is chosen for replacement by a replacement policy: • Least-recently-used (LRU), Clock, MRU etc. • Policy can have big impact on # of I/O’s; depends on the access pattern.