L05: Query Processing & Tuning Queries

1. L05: Query Processing & Tuning Queries Query Processing -- Principles Tuning Queries

2. Query Processing & Optimization • Any high-level query (SQL) on a database must be processed, optimized and executed by the DBMS • The high-level query is scanned, and parsed to check for syntactic correctness • An internal representation of a query is created, which is either a query tree or a query graph • The DBMS then devises an execution strategy for retrieving the result of the query. (An execution strategy is a plan for executing the query by accessing the data, and storing the intermediate results) • The process of choosing one out of the many execution strategies is known as query optimization

3. Query Processor & Query Optimizer • A query processor is a module in the DBMS that performs the tasks to process, to optimize, and to generate execution strategy for a high-level query • Queries expressed in SQL can have multiple equivalent relational algebra query expressions. The query optimizer must select the optimal one

4. Database Query Processing

5. Relational Operations • We will consider how to implement: • Selection : Selects a subset of rows from relation. • Projection : Deletes unwanted columns from relation. • Join : Allows us to combine two relations. • Since each operation returns a relation, opreations can be composed!

6. SELECT * FROM Reserves R WHERE R.rname < ‘C%’ Simple Selections • Of the form • Size of result approximated as size of R * reduction factor; we will consider how to estimate reduction factors later. • With no index, unsorted: Must essentially scan the whole relation; cost is |R| (# pages in R). • With an index on selection attribute: Use index to find qualifying data entries, then retrieve corresponding data records. (Hash index useful only for equality selections.)

7. Using an Index for Selections • Cost depends on #qualifying tuples, and clustering. • Cost of finding qualifying data entries (typically small) plus cost of retrieving records (could be large w/o clustering). • In example, assuming uniform distribution of names, about 10% of tuples qualify (100 pages, 10000 tuples). With a clustered index, cost is little more than 100 I/Os; if unclustered, upto 10000 I/Os! • Important refinement for unclustered indexes: • 1. Find qualifying data entries. • 2. Sort the rid’s of the data records to be retrieved. • 3. Fetch rids in order. This ensures that each data page is looked at just once (though # of such pages likely to be higher than with clustering).

8. Selection - Summary • R.attr op value (R) • Size of result = R * selectivity • Processing methods • Table scan • Index scan • B+-tree, Hash index • Clustered vs. non-clustered • Clustered index: Good • Non-clustered index: • Good for low selectivity • Worse than scan for high selectivity SELECT * FROM Reserves R WHERE R.rname < ‘C%’

9. General Selection Conditions (day<8/9/94 AND rname=‘Paul’) OR bid=5 OR sid=3 • Such selection conditions are first converted to conjunctive normal form (CNF): (day<8/9/94 OR bid=5 OR sid=3 ) AND (rname=‘Paul’ OR bid=5 OR sid=3) • We only discuss the case with no ORs (a conjunction of terms of the form attr op value). • An index matches (a conjunction of) terms that involve only attributes in a prefix of the search key. • Index on <a, b, c> matches a=5 AND b= 3, but not b=3.

10. Two Approaches to General Selections • First approach: Find 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. • 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.

11. Intersection of Rids • Second approach (if we have 2 or more matching indexes that use Alternatives (2) or (3) for data entries): • Get sets of rids of data records using each matching index. • Then intersect these sets of rids (we’ll discuss intersection soon!) • Retrieve the records and apply any remaining terms. • Consider day<8/9/94 AND bid=5 AND sid=3. If we have a B+ tree index on day and an index on sid, both using Alternative (2), we can retrieve rids of records satisfying day<8/9/94 using the first, rids of recs satisfying sid=3 using the second, intersect, retrieve records and check bid=5.

12. The Projection Operation SELECTDISTINCT R.sid, R.bid FROM Reserves R • An approach based on sorting: • Modify Pass 0 of external sort to eliminate unwanted fields. Thus, runs of about 2B pages are produced, but tuples in runs are smaller than input tuples. (Size ratio depends on # and size of fields that are dropped.) • Modify merging passes to eliminate duplicates. Thus, number of result tuples smaller than input. (Difference depends on # of duplicates.) • Cost: In Pass 0, read original relation (size M), write out same number of smaller tuples. In merging passes, fewer tuples written out in each pass. Using Reserves example, 1000 input pages reduced to 250 in Pass 0 if size ratio is 0.25

13. Projection Based on Hashing • Partitioning phase: Read R using one input buffer. For each tuple, discard unwanted fields, apply hash function h1 to choose one of B-1 output buffers. • Result is B-1 partitions (of tuples with no unwanted fields). 2 tuples from different partitions guaranteed to be distinct. • Duplicate elimination phase: For each partition, read it and build an in-memory hash table, using hash fn h2 (<> h1) on all fields, while discarding duplicates. • If partition does not fit in memory, can apply hash-based projection algorithm recursively to this partition. • Cost: For partitioning, read R, write out each tuple, but with fewer fields. This is read in next phase.

14. Discussion of Projection • Sort-based approach is the standard; better handling of skew and result is sorted. • If an index on the relation contains all wanted attributes in its search key, can do index-only scan. • Apply projection techniques to data entries (much smaller!) • If an ordered (i.e., tree) index contains all wanted attributes as prefix of search key, can do even better: • Retrieve data entries in order (index-only scan), discard unwanted fields, compare adjacent tuples to check for duplicates.

15. Equality Joins With One Join Column SELECT * FROM Reserves R1, Sailors S1 WHERE R1.sid=S1.sid • In algebra: R S. Common! Must be carefully optimized. • R S is large; so, R S followed by a selection is inefficient. • Cost metric: # of I/Os. We will ignore output costs.

16. Notations • |R| = number of pages in outer table R • ||R|| = number of tuples in outer table R • |S| = number of pages in inner table S • ||S|| = number of tuples in inner table S • M = number of main memory pages allocated

17. Example of Join SELECT * FROM Sailors R, Reserve S WHERE R.sid=S.sid

18. Schema for Examples • 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. Sailors (sid: integer, sname: string, rating: integer, age: real) Reserves (sid: integer, bid: integer, day: dates, rname: string)

19. Simple Nested Loops Join foreach tuple r in R do foreach tuple s in S do if ri == sj then add <r, s> to result • For each tuple in the outer relation R, we scan the entire inner relation S. • Cost: |R| + ||R|| * |S| = 1000 + 100*1000*500 I/Os. • Page-oriented Nested Loops join: For each page of R, get each page of S, and write out matching pairs of tuples <r, s>, where r is in R-page and S is in S-page. • Cost: |R| + |R|*|S| = 1000 + 1000*500 • If smaller relation (S) is outer, cost = 500 + 500*1000

20. Block Nested Loops Join • Use one page as an input buffer for scanning the inner S, one page as the output buffer, and use all remaining pages to hold ``block’’ of outer R. • For each matching tuple r in R-block, s in S-page, add <r, s> to result. • Then read next R-block, scan S, etc. Read in block by block Hash Table for R (M-2)pages h( ) Input Buffer Output Buffer Read in page by page S R

21. Block Nested Loop Join • Scan inner table S per block of (M – 2) pages of R tuples • Each scan costs |S| pages • |R| / (M – 2) blocks of R tuples • |R| pages for outer table R • Total cost = |R| + |R| / (M – 2) * |S| pages • R should be the smaller table

22. Examples of Block Nested Loops • Cost: Scan of outer + #outer blocks * scan of inner • #outer blocks =no of pages of outer / blocksize • With Reserves (R) as outer, and 100 pages of R: • Cost of scanning R is 1000 I/Os; a total of 10 blocks. • Per block of R, we scan Sailors (S); 10*500 I/Os. • If space for just 90 pages of R, we would scan S 12 times. • With 100-page block of Sailors as outer: • Cost of scanning S is 500 I/Os; a total of 5 blocks. • Per block of S, we scan Reserves; 5*1000 I/Os. • With sequential reads considered, analysis changes: may be best to divide buffers evenly between R and S.

23. Index Nested Loops Join • Use one page as an input buffer for scanning the outer R, one page as the output buffer. • Use all remaining pages to hold index of inner S. • For each tuple r in R- page, probe index of S to find matches, s, add <r, s> to result. • Then read next R-page, repeat. Input Buffer Read in page by page Output Buffer Index for S S R

24. Index Nested Loop Join • Probe S index for matching S tuples per R tuple • Probe hash index: 1.2 I/Os • Probe B+ tree: 2-4 I/Os, plus retrieve matching S tuples: 1 I/O • For ||R|| tuples • |R| pages for outer table R • Total cost = |R| + ||R|| * index retrieval • Better than Block NL join only for small number of R tuples

25. Examples of Index Nested Loops • Hash-index (Alt. 2) on sid of Sailors (as inner): • Scan Reserves: 1000 page I/Os, 100*1000 tuples. • For each Reserves tuple: 1.2 I/Os to get data entry in index, plus 1 I/O to get (the exactly one) matching Sailors tuple. Total: 220,000 I/Os. • Hash-index (Alt. 2) on sid of Reserves (as inner): • Scan Sailors: 500 page I/Os, 80*500 tuples. • For each Sailors tuple: 1.2 I/Os to find index page with data entries, plus cost of retrieving matching Reserves tuples. Assuming uniform distribution, 2.5 reservations per sailor (100,000 / 40,000). Cost of retrieving them is 1 or 2.5 I/Os depending on whether the index is clustered.

26. Sort-Merge Join • Sort R and S on the join column • Scan sorted R and S 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; do this 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. • R is scanned once; each S group is scanned once per matching R tuple. (Multiple scans of an S group are likely to find needed pages in buffer.)

27. External Sort ( K-Way Merge Sort)

28. Sort Merge Join • External-sort R: 2 |R| * (logM-1 |R|/M + 1) • Split R into |R|/M sorted runs each of size M: 2 |R| • Merge up to (M – 1) runs repeatedly • logM-1 |R|/M passes, each costing 2 |R| • External-sort S: 2 |S| * (logM-1 |S|/M + 1) • Merge matching tuples from sorted R and S: |R| + |S| • Total cost = 2 |R| * (logM-1 |R|/M + 1) + 2 |S| * (logM-1 |S|/M + 1) + |R| + |S| • If |R| < M*(M-1), cost = 5 * (|R| + |S|)

29. Example of Sort-Merge Join • Cost of merge is M+N, could be M*N (very unlikely!) • With 35, 100 or 300 buffer pages, both Reserves and Sailors can be sorted in 2 passes; total join cost: 7500. (BNL cost: 2500 to 15000 I/Os)

30. Hash Join – The Principle Ri If S Si for i j R

31. Original Relation Partitions OUTPUT Hash-Join 1 1 2 INPUT 2 • Partition Phase: Partition both relations using hash fn h1: R tuples in partition i will only match S tuples in partition i. hash function h1 . . . M-1 M-1 M main memory buffers Disk Disk Partitions of R & S Join Result Hash table for partition Ri (k < M -1 pages) Joining Phase: Read in a partition of R, hash it using h2 (<> h1!). Scan matching partition of S, search for matches. hash fn h2 h2 Output buffer Input buffer for Si M main memory buffers Disk Disk

32. Hash Join Algorithm • Partition phase: 2 (|R| + |S|) • Partition table R using hash function h1: 2 |R| • Partition table S using hash function h1: 2 |S| • R tuples in partition i will match only S tuples in partition i • R  (M – 1) partitions, each of size |R| / (M – 1) • Join phase: |R| + |S| • Read in a partition of R (|R| / (M – 1) < M -1) • Hash it using function h2 (<> h1!) • Scan corresponding S partition, search for matches • Total cost = 3 (|R| + |S|) pages • Condition: M > √f|R|, f ≈ 1.2 to account for hash table

33. General Join Conditions • Equalities over several attributes (e.g., R.sid=S.sid AND R.rname=S.sname): • For Index NL, build index on <sid, sname> (if S is inner); or use existing indexes on sid or sname. • For Sort-Merge and Hash Join, sort/partition on combination of the two join columns. • Inequality conditions (e.g., R.rname < S.sname): • For Index NL, need (clustered!) B+ tree index. • Range probes on inner; # matches likely to be much higher than for equality joins. • Hash Join, Sort Merge Join not applicable. • Block NL quite likely to be the best join method here.

34. Impact of Buffering • If several operations are executing concurrently, estimating the number of available buffer pages is guesswork. • Repeated access patterns interact with buffer replacement policy. • e.g., Inner relation is scanned repeatedly in Simple Nested Loop Join. With enough buffer pages to hold inner, replacement policy does not matter. Otherwise, MRU is best, LRU is worst (sequential flooding). • Does replacement policy matter for Block Nested Loops? • What about Index Nested Loops? Sort-Merge Join?

35. Summary of Join Operator • Simple nested loop: |R| + ||R|| * |S| • Block nested loop: |R| + |R| / (M – 2) * |S| • Index nested loop: |R| + ||R|| * index retrieval • Sort-merge: 2 |R| * (logM-1 |R|/M + 1) + 2 |S| * (logM-1 |S|/M + 1) + |R| + |S| • Hash Join: 3 * (|R| + |S|) • Condition: M > √f|R|

36. Overview of Query Processing Database Statistics Cost Model Query Optimizer Query Evaluator Parsed Query QEP Parser High Level Query Query Result

37. sname rating > 5 bid=100 sid=sid Sailors Reserves Query Optimization SELECT S.sname FROM Reserves R, Sailors S WHERE R.sid=S.sid AND R.bid=100 AND S.rating>5 • Given: An SQL query joining n tables • Dream: Map to most efficient plan • Reality: Avoid rotten plans • State of the art: • Most optimizers follow System R’s technique • Works fine up to about 10 joins

38. Many degrees of freedom Selection: scan versus (clustered, non-clustered) index Join: block nested loop, sort-merge, hash Relative order of the operators Exponential search space! Heuristics Push the selections down Push the projections down Delay Cartesian products System R: Only left-deep trees D C B A Complexity of Query Optimization

39. Equivalences in Relational Algebra • Selection: - cascade - commutative • Projection: - cascade • Join: - associative - commutative R (S T) (R S) T (R S) (S R)

40. Equivalences in Relational Algebra • 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 join 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

41. System R Optimizer • Find all plans for accessing each base table • For each table • Save cheapest unordered plan • Save cheapest plan for each interesting order • Discard all others • Try all ways of joining pairs of 1-table plans; save cheapest unordered + interesting ordered plans • Try all ways of joining 2-table with 1-table • Combine k-table with 1-table till you have full plan tree • At the top, to satisfy GROUP BY and ORDER BY • Use interesting ordered plan • Add a sort node to unordered plan

42. Source: Selinger et al, “Access Path Selection in a Relational Database Management System”

43. Note: Only branches for NL join are shown here. Additional branches for other join methods (e.g. sort-merge) are not shown. Source: Selinger et al, “Access Path Selection in a Relational Database Management System”

44. What is “Cheapest”? • Need information about the relations and indexes involved • Catalogstypically 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.

45. Estimating Result Size SELECT attribute list FROM relation list WHERE term1AND ... ANDtermk • 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 eachtermireflects the impact of the term in reducing result size • Term col=value has RF 1/NKeys(I) • Term col1=col2 has RF 1/MAX(NKeys(I1), NKeys(I2)) • Term col>value has RF (High(I)-value)/(High(I)-Low(I)) • Resultcardinality = Max # tuples * product of all RF’s. • Implicit assumption that terms are independent!

46. Cost Estimates for Single-Table Plans • Index I on primary key matches selection: • Cost is Height(I)+1 for a B+ tree, about 1.2 for hash index. • Clustered index I matching one or more selects: • (NPages(I)+NPages(R)) * product of RF’s of matching selects. • Non-clustered index I matching one or more selects: • (NPages(I)+NTuples(R)) * product of RF’s of matching selects. • Sequential scan of file: • NPages(R). Note:Typically, no duplicate elimination on projections! (Exception: Done on answers if user says DISTINCT.)

47. (On-the-fly) sname (Sort-Merge Join) sid=sid (Scan; (Scan; write to write to rating > 5 bid=100 temp T2) temp T1) Reserves Sailors Counting the Costs SELECT S.sname FROM Reserves R, Sailors S WHERE R.sid=S.sid AND R.bid=100 AND S.rating>5 • 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 (2*10*2), sort T2 (2*250*4), merge (10+250), total=2300 • Total: 4060 page I/Os • If we used BNL join, join cost = 10+4*250, 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 < 2000

48. Exercise • Reserves: 100,000 tuples, 100 tuples per page • With clustered index on bid of Reserves, we get 100,000/100 = 1000 tuples on 1000/100 = 10 pages • Join column sid is a key for Sailors - at most one matching tuple • Decision not to push rating>5 before the join is based on availability of sid index on Sailors • Cost: Selection of Reserves tuples (10 I/Os); for each tuple, must get matching Sailors tuple (1000*1.2); total 1210 I/Os (On-the-fly) sname (On-the-fly) rating > 5 (Index Nested Loops, with pipelining ) sid=sid (Use hash Index on sid) Sailors bid=100 (Use clustered index on sid) Reserves

49. L05: Query Processing & Tuning Queries Query Processing – Principles Tuning Queries

50. Avoid Redundant DISTINCT • DISTINCT usually entails a sort operation • Slow down query optimization because one more “interesting” order to consider • Remove if you know the result has no duplicates SELECT DISTINCT ssnum FROM Employee WHEREdept = ‘information systems’