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Modern Database Systems - Lecture 01


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Slides for course "Modern Database Systems", taught at Aalto University during Spring 2016.

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Modern Database Systems - Lecture 01

  1. 1. Modern Database Systems Lecture 1 Aristides Gionis Michael Mathioudakis T.A.: Orestis Kostakis Spring 2016
  2. 2. logistics assignment will be up by Monday (you will receive email) due Feb 12th if you’re not registered... I will post material (slides and assignments) also at 2
  3. 3. in this lecture... review past material relational model and sql storage and indexing access cost analysis hash index b+ tree 3
  4. 4. relational model and SQL
  5. 5. relational model and sql what is the relational model? tabular representation of data why do we study it? supports simple and intuitive querying good for educational purposes most widely used 5
  6. 6. definitions relational database a set of relations relation example! schema name of relation + name and type of each field fields as columns instance a table with rows and columns 6
  7. 7. example relation: students cardinality (number of rows) = 3, degree (number of fields/columns) = 5 > can we have the same value twice in the same column? schema students(sid: integer, name: string, username: string, age: integer, gpa: real) sid name username age gpa 53666 Sam Jones jones 22 3.4 53688 Alice Smith smith 22 3.8 53650 Jon Edwards jon 23 2.4 7
  8. 8. querying major strength of relational model simple, intuitive, precise querying of data the DBMS is responsible for efficient evaluation Standard Query Language (SQL) the standard language for relational queries developed by IBM in the 1970s was standardized in 1986 latest standard in 2011 example! 8
  9. 9. example SQL query to find student records of age 23 SELECT * FROM students WHERE age=23 to find just names and usernames SELECT name, username FROM students WHERE age=23 sid name username age gpa 53666 Kate Jones jones 22 3.4 53688 Alice Smith smith 22 3.8 53650 Jon Edward jon 23 2.4 sid name username age gpa 53650 Jon Edward jon 23 2.4 name username Jon Edward jon 9
  10. 10. creating, altering, and destroying, relations in SQL CREATE TABLE students (sid CHAR(20), name CHAR(20), username CHAR(10), age INTEGER, gpa REAL); the type of each column is enforced by the DBMS DROP TABLE students; ALTER TABLE students ADD COLUMN firstYear integer; every tuple in the current instance is extended with a null value in the new column CREATE TABLE course (sid CHAR(20), points integer, grade CHAR(2)); destroy relation students (schema and instance) 10
  11. 11. adding and deleting tuples > what do the following statements do? INSERT INTO students(sid, name, username, age, gpa) VALUES (12345, “Kate Doe”, “kate”, 23, 4.0); DELETE FROM students WHERE name = ‘Jane Smith’; 11
  12. 12. candidate keys a set of fields is a candidate key (aka ‘key’) for a relation if... 1)  distinct tuples cannot have same values in all key fields, and 2)  this is not true for any subset of the key if only part (1) from above is true... we have a superkey possibly many candidate keys for a relation DBMS admin chooses one (1) of them as primary key an integrity constraint condition must be true for any instance of the database other integrity constraints? 12
  13. 13. candidate keys in SQL, use PRIMARY KEY to specify primary key UNIQUE to specify candidate keys example relation enrolled holds information about student enrollment to courses compare the following ‘create table’ statements use ICs carefully - they might forbid database instances that could arise in practice CREATE TABLE Enrolled (sid CHAR(20), cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid,cid)) CREATE TABLE Enrolled (sid CHAR(20) cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid), UNIQUE (cid, grade)) 13
  14. 14. storage and indexing 14
  15. 15. storage setting the DBMS uses disks as external storage to store relations into files of records disks retrieve random page at fixed cost cheaper to retrieve several consecutive pages than each by random access why? file organization method of arranging a file of records on external storage record: one row of a relation record is internally assigned a record id (rid) rid is sufficient to physically locate record (address) 15
  16. 16. alternative file organizations heap files random order suitable when typical access is a file scan to retrieve all records sorted files records are sorted - typically by column value(s) suitable if records must be retrieved by same order indexes data structures that allow organized access to records… ... via search keys - typically column value(s) updates are faster than in sorted files -- why? 16
  17. 17. data structures that allow us to find rids of records with specified column values any subset of the columns of a relation can be the search key for an index search key is not same as primary / candidate key indexes an index contains a collection of data entries supports efficient retrieval of data entries k* with a given key value k index entries data entries data records index file data file 17
  18. 18. types of data entries three alternatives 1. data record with key value k 2. (k, rid of data record with search key k) 3. (k, list of rids of data records with search key k) type of data entries is orthogonal to index structure example of index structure B+ trees or hash tables 18
  19. 19. data entries of type 1 index structure is a file organization for data records we just have an ‘index file’ index entries data records index file > how many indexes of a relation can be of type 1? 19
  20. 20. types of data entries - types 2 & 3 data entries typically much smaller than data records > why? index entries data entries data records index file data file type 3 is more compact than type 2 > why? 20
  21. 21. index classes primary vs secondary primary: if search key contains a primary key unique index: search key contains a candidate key clustered vs unclustered if order of data records is same as that of data entries makes big difference for some queries! > can alternative 1 indexes be unclustered? unclustered clustered 21
  22. 22. hash-based indexes retrieve records with exactly specified search-key values suitable for equality queries index is collection of buckets bucket = 1 or more disk pages hashing function h h(r) = bucket where record r belongs, based on its column values data entries are … ... type 1: the buckets contain data records ... type 2 or 3: the buckets contain (key, rid) or (key, rids) pairs 22
  23. 23. hash-based indexes Smith, 44, 3000 Jones, 40, 6003 Tracy, 44, 5004 Ashby, 25, 3000 Basu, 33, 4003 Kate, 29, 2007 Cass, 50, 5004 Basu, 33, 6003 age h1 relation employes(name CHAR(100), age INTEGER, salary INTEGER) 3000 3000 5004 5004 4003 2007 6003 6003 salaryh2 clustered (type 1) hash index on age unclustered (type 2) hash index on salary 23
  24. 24. leaf pages contain data entries, and are chained (prev & next) non-leaf pages have index entries; only used to direct searches P0 K 1 P 1 K 2 P 2 K m P m index entry b+ tree indexes non-leaf pages leaf pages (sorted by search key) 24
  25. 25. example b+ tree find 28*? 29*? all > 15* and < 30*? insert/delete find data entry in leaf, then update it need to adjust parent sometimes 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 note that data entries in leaf level are sorted
  26. 26. access-cost analysis 26
  27. 27. access-cost model ● relation students ○  B: number of data pages, R: number of records per page ● execute typical select-from-where query ○  D: (average) time to read or write one disk page SELECT * FROM students WHERE <...> ● estimate running time of query ○  ignore cpu costs ○  number of disk accesses (read/writes) is the bottleneck 27
  28. 28. file organizations heap file (random order; inserts at eof) sorted file, sorted on <age, gpa> clustered B+ tree file (type 1 data entries) on search key <age, gpa> heap file with unclustered B+ tree index on search key <age, gpa> heaf file with unclustered hash index on search key <age, gpa> 28
  29. 29. queries to compare insert record SELECT * FROM students SELECT * FROM students WHERE age = 22 and gpa = 4.0 SELECT * FROM student WHERE age >= 20 INSERT INTO STUDENTS (sid, name, username, age, gpa) VALUES (12345, “Michael”, “mike”, 32, 2.6) scan - fetch all records equality search range search 29
  30. 30. cost analysis what is the estimated time for each query to run? under simplified model how many disk pages are accessed? time = #disk-accesses x D 30
  31. 31. cost analysis scan equality range insert heap sorted clustered unclustered b+ tree unclustered hash 31
  32. 32. heap file operation cost and explanation scan B; simply retrieve all pages equality search B in worst case; if we know that exactly one such record exists, the cost is 0.5B in expectation range search B; must retrieve all records insert 2; fetch and store back the last page of the file 32
  33. 33. sorted file operation cost and explanation scan B; simply retrieve all pages equality search log2B + #qualifying-pages; since the condition matches the index, we can find the page of the record with binary search that retrieves log2B pages; if more than one records qualify, retrieve sequentially #qualifying-pages after the first range search log2B + #qualifying-pages; as above, log2B pages are retrieved to find the first matching record, followed possibly by a number (#qualifying-pages) of pages with qualifying records insert log2B + B; find the position of the record in the file (log2B); then, read the second half of the file, insert the record, write the second half back (0.5B + 0.5B in expectation) 33
  34. 34. clustered b+ tree operation cost and explanation scan 1.5B; simply retrieve all record pages equality search logF1.5B + #qualifying-pages; find the first qualifying record and retrieve consecutive qualifying ones range search logF1.5B + #qualifying-pages; find the first qualifying record and retrieve consecutive qualifying ones insert logF1.5B + 1; search for record page (logF1.5B) and add record to it (1) assumptions: 2/3 = 67% occupancy of record pages, i.e. 1.5B record pages; fanout F 34
  35. 35. unclustered b+ tree operation cost and explanation scan B(R+0.15); scan the leaf level of the index (0.15B); for each data entry, fetch the page with the corresponding data record (6.7R x 0.15B = BR) equality search logF0.15B + #qualifying-records; locate the first data entry (logF0.15B) and do one disk access for every qualifying record (#qualifying-records) range search logF0.15B + #qualifying-records; locate the first data entry (logF0.15B) and do one disk access for every qualifying record (#qualifying-records) insert 3 + logF0.15B;insert at end of heap file (2), find page for data entry (logF0.15B) and update it (1) assumptions: the size of one data entry is 10% the size of one record; also, index pages have 2/3=67% occupancy; therefore, number of index leaf pages is 0.1*1.5B = 0.15B and number of data entries in one page are 10*0.67R = 6.7R 35
  36. 36. unclustered hash index operation cost and explanation scan B(R+0.125); retrieve pages that contain data entries (0.125B); for each data entry, fetch the page with the corresponding data record equality search 2; retrieve page with data entry (1) and page with data record (1) range search 0.125B + #qualifying-records; the hash index offers no help - scan index (0.125B) and retrieve pages of matching records; typically it’s better to scan entire heapfile (B) insert 4; insert record into heap file (1 read+1 write); insert record into hash index (1 read + 1 write) assumptions: the size of one data entry is 10% the size of one record; static hashing, no overflow pages (one bucket is one page); 4/5 = 80% occupancy; therefore , 0.1*1.25B = 0.125B pages for data entries and the number of data entries in a page is 10*0.8R = 8R 36
  37. 37. cost analysis scan equality range insert heap B B B 2 sorted B log2B + #qualifying- pages log2B + #qualifying- pages log2B + B clustered 1.5B logF1.5B + #qualifying- pages logF1.5B + #qualifying- pages logF1.5B + 1 unclustered b+ tree B(R+0.15) logF0.15B + #qualifying- records logF0.15B + #qualifying- records 3 + logF0.15B unclustered hash B(R+0.125) 2 0.125B + #qualifying- records 4 note we made several assumptions to obtain these numbers 37
  38. 38. the morale different queries have different cost for different file organizations > how would you use this analysis as a db admin? discuss 38
  39. 39. the morale know your workload what queries? how often? on what relations? what file organizations? what indexes would speed-up response times for your workload? hint: see WHERE clause for index key candidates why? what trade-offs will you face? hint: queries are faster but updates take time, index takes space we’ll see more complex cases in ‘query optimization’ 39
  40. 40. indexes with composite search keys composite search keys search on a combination of fields equality query every field value is equal to a constant e.g., age=20 and sal =75, wrt <sal,age> index range query some field value is not a constant e.g., age =20; or age=20 and sal > 10, wrt <sal,age> index data entries in index sorted by search key to support range queries (e.g., b+ trees) <sal, age> <age> <sal> data records sorted by name data entries sorted by <sal,age> data entries sorted by <sal> examples of composite key indexes 11,80 12,10 12,20 13,75 10,12 20,12 75,13 80,11 11 12 12 13 10 20 75 80 name age sal bob 12 10 cal 11 80 joe 12 20 sue 13 75 <age,sal> remember also composite indexes are larger, updated more often 40
  41. 41. composite search keys if condition is: 3000<sal<5000: <age,sal> index does not help! why? because the index does not match the selection condition index matches selection (condition ∧ ... ∧ ... ∧ condition) when: for hash index: only equality conditions for all fields for tree index: includes equality or range condition for a prefix of the search key 41
  42. 42. to retrieve employee records with age=30 AND sal=4000, an index on <age,sal> or <sal, age> would be better than an index on <age> or an index on <sal> if condition is: age=30 AND 3000<sal<5000: <age,sal> index much better than <sal,age> index! why? hint: allows us to allocate answer with contiguous data entries order can make a difference depending on the selectivity of each condition if condition is: 20<age<30 AND 3000<sal<5000: tree index on <age,sal> or <sal,age> make no difference if selectivity of each condition is the same composite search keys 42
  43. 43. index-only plans some queries can be answered without retrieving any data records if a suitable index is available example employees (name CHAR(100), depnum INTEGER, age INTEGER, salary INTEGER) SELECT depnum, COUNT(*) FROM employees GROUP BY depnum SELECT AVG(salary) FROM employees WHERE age=25 AND salary BETWEEN 3000 AND 5000 index on <depnum> b+ tree index on <age,salary> 43
  44. 44. index-only plans are possible with both <dno,age> or <age,dno> tree index <age, dno> is better why? SELECT E.dno, COUNT (*) FROM Emp E WHERE E.age=30 GROUP BY E.dno index-only plans 44
  45. 45. summary 45
  46. 46. summary ●  relational model and SQL ○  tabular representation ■  one record per row ■  schema determines names and types of columns ○  simple, intuitive querying language ■  statements to select records that satisfy a condition ■  specify columns to project ■  statements to insert and delete tuples 46 ●  storage ○  a DBMS might use different file organizations to store relations ○  heap file, sorted file, index ○  different queries have different access costs for different file organizations ○  having the right index can make a big difference in execution time ●  commonly used indexes ○  B+ tree and hash-based index
  47. 47. next b+ trees and hash-based index external sorting joins query optimization 47
  48. 48. references ●  “cowbook”, database management systems, by ramakrishnan and gehrke ●  “elmasri”, fundamentals of database systems, elmasri and navathe ●  other database textbooks ●  disk access analysis ○  cowbook, chapter 8 ●  b+ tree and hashing algorithms ○  elmasri ■  section 18.2: hash indexes ■  section 18.3.2: b+ trees ○  cowbook ■  chapters 10 and 11 48
  49. 49. credits slides based on material from database management systems, by ramakrishnan and gehrke 49
  50. 50. joins sid name username age gpa 53666 Sam Jones jones 22 3.4 53688 Alice Smith smith 22 3.8 53650 Jon Edwards jon 23 2.4 students sid points grade 53666 92 A 53688 35 D 53650 65 C course what does this compute? SELECT, C.grade FROM Students S,Course C WHERE S.sid = C.sid AND C.points > 60 C.grade Sam Jones A Jon Edwards C 50
  51. 51. index-only plans SELECT E.dno, COUNT (*) FROM Emp E WHERE E.age>30 GROUP BY E.dno what if we consider the second query? we’ll come back to this after external sorting