8 query

1. ECS-165A WQ'11 136 8. Query Processing Goals: Understand the basic concepts underlying the steps in query processing and optimization and estimating query processing cost; apply query optimization techniques; Contents: Overview Catalog Information for Cost Estimation Measures of Query Cost Selection Join Operations Other Operations Evaluation and Transformation of Expressions Query Processing Optimization Task: Find an ecient physical query plan (aka execution plan) for an SQL query Goal: Minimize the evaluation time for the query, i.e., compute query result as fast as possible Cost Factors: Disk accesses, read/write operations, [I/O, page transfer] (CPU time is typically ignored) Dept. of Computer Science UC Davis 8. Query Processing and Optimization

2. ECS-165A WQ'11 137 Basic Steps in Processing an SQL Query (System Catalogs) SQL Query Relational Algebra Expression Optimizer Statistics Query Result Evaluation Engine Execution Plan Data Files Parser Translator Parsing and Translating { Translate the query into its internal form (parse tree). This is then translated into an expression of the relational algebra. { Parser checks syntax, validates relations, attributes and access permissions Evaluation { The query execution engine takes a physical query plan (aka execution plan), executes the plan, and returns the result. Optimization: Find the cheapest execution plan for a query Dept. of Computer Science UC Davis 8. Query Processing and Optimization

3. ECS-165A WQ'11 138 A relational algebra expression may have many equivalent expressions, e.g., CName(Price5000((CUSTOMERS 1 ORDERS) 1 OFFERS)) CName((CUSTOMERS 1 ORDERS) 1 (Price5000(OFFERS))) Representation as logical query plan (a tree): o o CName Price 5000 CName Price 5000 OFFERS ORDERS CUSTOMERS ORDERS OFFERS CUSTOMERS Non-leaf nodes operations of relational algebra (with parameters); Leaf nodes relations A relational algebra expression can be evaluated in many ways. An annotated expression specifying detailed evaluation strategy is called the execution plan (includes, e.g., whether index is used, join algorithms, . . . ) Among all semantically equivalent expressions, the one with the least costly evaluation plan is chosen. Cost estimate of a plan is based on statistical information in the system catalogs. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

4. ECS-165A WQ'11 139 Catalog Information for Cost Estimation Information about relations and attributes: NR: number of tuples in the relation R. BR: number of blocks that contain tuples of the relation R. SR: size of a tuple of R. FR: blocking factor; number of tuples from R that

5. t into one block (FR = dNR=BRe) V(A; R): number of distinct values for attribute A in R. SC(A; R): selectivity of attribute A average number of tuples of R that satisfy an equality condition on A. SC(A; R) = NR=V(A; R). Information about indexes: HTI: number of levels in index I (B+-tree). LBI: number of blocks occupied by leaf nodes in index I (

6. rst-level blocks). ValI: number of distinct values for the search key. Some relevant tables in the Oracle system catalogs: USER TABLES USER TAB COLUMNS USER INDEXES NUM ROWS NUM DISTINCT BLEVEL BLOCKS LOW VALUE LEAF BLOCKS EMPTY BLOCKS HIGH VALUE DISTINCT KEYS AVG SPACE DENSITY AVG LEAF BLOCKS PER KEY CHAIN CNT NUM BUCKETS AVG ROW LEN LAST ANALYZED Dept. of Computer Science UC Davis 8. Query Processing and Optimization

7. ECS-165A WQ'11 140 Measures of Query Cost There are many possible ways to estimate cost, e.g., based on disk accesses, CPU time, or communication overhead. Disk access is the predominant cost (in terms of time); relatively easy to estimate; therefore, number of block transfers from/to disk is typically used as measure. { Simplifying assumption: each block transfer has the same cost. Cost of algorithm (e.g., for join or selection) depends on database buer size; more memory for DB buer reduces disk accesses. Thus DB buer size is a parameter for estimating cost. We refer to the cost estimate of algorithm S as cost(S). We do not consider cost of writing output to disk. Selection Operation A=a(R) where a is a constant value, A an attribute of R File Scan { search algorithms that locate and retrieve records that satisfy a selection condition S1 { Linear search cost(S1)= BR Dept. of Computer Science UC Davis 8. Query Processing and Optimization

8. ECS-165A WQ'11 141 Selection Operation (cont.) S2 { Binary search, i.e., the

9. le ordered based on attribute A (primary index) cost(S2) = dlog2(BR)e + SC(A; R) FR 1 { dlog2(BR)e cost to locate the

10. rst tuple using binary search { Second term blocks that contain records satisfying the selection. { If A is primary key, then SC(A; R) = 1, hence cost(S2) = dlog2(BR)e. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

11. ECS-165A WQ'11 142 Example (for Employee DB) { FEmployee = 10; V(Deptno; Employee) = 50 (dierent departments) { NEmployee = 10; 000 (Relation Employee has 10,000 tuples) { Assume selection Deptno=20(Employee) and Employee is sorted on search key Deptno : =) 10,000/50 = 200 tuples in Employee belong to Deptno 20; (assuming an equal distribution) 200/10 = 20 blocks for these tuples =) A binary search

12. nding the

13. rst block would require dlog2(1; 000)e = 10 block accesses Total cost of binary search is 10+20 block accesses (versus 1,000 for linear search and Employee not sorted by Deptno). Dept. of Computer Science UC Davis 8. Query Processing and Optimization

14. ECS-165A WQ'11 143 Index scan { search algorithms that use an index (here, a B+-tree); selection condition is on search key of index S3 { Primary index I for A, A primary key, equality A = a cost(S3) = HTI + 1 (only 1 tuple satis

15. es condition) S4 { Primary index I on non-key A equality A = a cost(S4) = HTI + SC(A; R) FR S5 { Non-primary (non-clustered) index on non-key A, equality A = a cost(S5) = HTI + SC(A; R) Worst case: each matching record resides in a dierent block. Example (Cont.): { Assume primary (B+-tree) index for attribute Deptno { 200/10=20 blocks accesses are required to read Employee tuples { If B+-tree index stores 20 pointers per (inner) node, then the B+-tree index must have between 3 and 5 leaf nodes and the entire tree has a depth of 2 =) a total of 22 blocks must be read. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

16. ECS-165A WQ'11 144 Selections Involving Comparisons Selections of the form Av(R) or Av(R) are implemented using a

17. le scan or binary search, or by using either a { S6 { A primary index on A, or { S7 { A secondary index on A (in this case, typically a linear

18. le scan may be cheaper; but this depends on the selectivity of A) Complex Selections General pattern: { Conjunction { 1^:::^n(R) { Disjunction { 1_:::_n(R) { Negation { :(R) The selectivity of a condition i is the probability that a tuple in the relation R satis

19. es i. If si is the number of tuples in R that satisfy i, then i's selectivity is estimated as si=NR. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

20. ECS-165A WQ'11 145 Join Operations There are several dierent algorithms that can be used to implement joins (natural-, equi-, condition-join) { Nested-Loop Join { Block Nested-Loop Join { Index Nested-Loop Join { Sort-Merge Join { Hash-Join Choice of a particular algorithm is based on cost estimate For this, join size estimates are required and in particular cost estimates for outer-level operations in a relational algebra expression. Example: Assume the query CUSTOMERS 1 ORDERS (with join attribute only being CName) { NCUSTOMERS = 5,000 tuples { FCUSTOMERS = 20, i.e., BCUSTOMERS = 5,000/20 = 250 blocks { NORDERS = 10,000 tuples { FORDERS = 25, i.e., BORDERS = 400 blocks { V(CName; ORDERS) = 2,500, meaning that in this relation, on average, each customer has four orders { Also assume that CName in ORDERS is a foreign key on CUSTOMERS Dept. of Computer Science UC Davis 8. Query Processing and Optimization

21. ECS-165A WQ'11 146 Estimating the Size of Joins The Cartesian product R S results in NR NS tuples; each tuple requires SR + SS bytes. If schema(R) schema(S) = primary key for R, then a tuple of S will match with at most one tuple from R. Therefore, the number of tuples in R1S is not greater than NS If schema(R) schema(S) = foreign key in S referencing R, then the number of tuples in R1S is exactly NS. Other cases are symmetric. In the example query CUSTOMERS 1 ORDERS, CName in ORDERS is a foreign key of CUSTOMERS; the result thus has exactly NORDERS = 10,000 tuples If schema(R) schema(S) = fAg is not a key for R or S; assume that every tuple in R produces tuples in R 1 S. Then the number of tuples in R 1 S is estimated to be: NR NS V(A; S) If the reverse is true, the estimate is NR NS V(A; R) and the lower of the two estimates is probably the more accurate one. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

22. ECS-165A WQ'11 147 Size estimates for CUSTOMERS 1 ORDERS without using information about foreign keys: { V(CName; CUSTOMERS) = 5,000, and V(CName; ORDERS) = 2,500 { The two estimates are 5,000*10,000/2,500=20,000 and 5,000*10,000/5,000=10,000. We choose the lower estimate, which, in this case, is the same as our earlier computation using foreign key information. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

23. ECS-165A WQ'11 148 Nested-Loop Join Evaluate the condition join R 1C S for each tuple tR in R do begin for each tuple tS in S do begin check whether pair (tR; tS) satis

24. es join condition if they do, add tR tS to the result end end R is called the outer and S the inner relation of the join. Requires no indexes and can be used with any kind of join condition. Worst case: db buer can only hold one block of each relation =) BR + NR BS disk accesses Best case: both relations

25. t into db buer =) BR + BS disk accesses. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

26. ECS-165A WQ'11 149 An Improvement: Block Nested-Loop Join Evaluate the condition join R 1C S for each block BR of R do begin for each block BS of S do begin for each tuple tR in BR do for each tuple tS in BS do check whether pair (tR; tS) satis

27. es join condition if they do, add tR tS to the result end end end end Also requires no indexes and can be used with any kind of join condition. Worst case: db buer can only hold one block of each relation =) BR + BR BS disk accesses. Best case: both relations

28. t into db buer =) BR + BS disk accesses. If smaller relation completely

29. ts into db buer, use that as inner relation. Reduces the cost estimate to BR + BS disk accesses. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

30. ECS-165A WQ'11 150 Block Nested-Loop Join (cont.) Some improvements of block nested-loop algorithm { If equi-join attribute is the key on inner relation, stop inner loop with

31. rst match { Use M 2 disk blocks as blocking unit for outer relation, where M = db buer size in blocks; use remaining two blocks to buer inner relation and output. Reduces number of scans of inner relation greatly. { Scan inner loop forward and backward alternately, to make use of blocks remaining in buer (with LRU replacement strategy) { Use index on inner relation, if available . . . Dept. of Computer Science UC Davis 8. Query Processing and Optimization

32. ECS-165A WQ'11 151 Index Nested-Loop Join If an index is available on the inner loop's join attribute and join is an equi-join or natural join, more ecient index lookups can replace

33. le scans. It is even possible (reasonable) to construct index just to compute a join. For each tuple tR in the outer relation R, use the index to lookup tuples in S that satisfy join condition with tR Worst case: db buer has space for only one page of R and one page of the index associated with S: { BR disk accesses to read R, and for each tuple in R, perform index lookup on S. { Cost of the join: BR + NR c, where c is the cost of a single selection on S using the join condition. If indexes are available on both R and S, use the one with the fewer tuples as the outer relation. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

34. ECS-165A WQ'11 152 Example: { Compute CUSTOMERS 1 ORDERS, with CUSTOMERS as the outer relation. { Let ORDERS have a primary B+-tree index on the join- attribute CName, which contains 20 entries per index node { Since ORDERS has 10,000 tuples, the height of the tree is 4, and one more access is needed to

35. nd the actual data records (based on tuple identi

36. er). { Since NCUSTOMERS is 5,000, the total cost is 250 + 5000 5 = 25,250 disk accesses. { This cost is lower than the 100,250 accesses needed for a block nested-loop join. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

37. ECS-165A WQ'11 153 Sort-Merge Join Basic idea:

38. rst sort both relations on join attribute (if not already sorted this way) Join steps are similar to the merge stage in the external sort-merge algorithm (discussed later) Every pair with same value on join attribute must be matched. values of join attributes 1 1 2 2 2 3 3 3 4 5 5 Relation R Relation S If no repeated join attribute values, each tuple needs to be read only once. As a result, each block is read only once. Thus, the number of block accesses is BR + BS (plus the cost of sorting, if relations are unsorted). Worst case: all join attribute values are the same. Then the number of block accesses is BR + BR BS. If one relation is sorted and the other has a secondary B+-tree index on the join attribute, a hybrid merge-join is possible. The sorted relation is merged with the leaf node entries of the B+-tree. The result is sorted on the addresses (rids) of the unsorted relation's tuples, and then the addresses can be replaced by the actual tuples eciently. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

39. ECS-165A WQ'11 154 Hash-Join { only applicable in case of equi-join or natural join { a hash function is used to partition tuples of both relations into sets that have the same hash value on the join attribute Partitioning Phase: 2 (BR + BS) block accesses Matching Phase: BR + BS block accesses (under the assumption that one partition of each relation

40. ts into the database buer) Cost Estimates for other Operations Sorting: If whole relation

41. ts into db buer ; quick-sort Or, build index on the relation, and use index to read relation in sorted order. Relation that does not

42. t into db buer;external sort-merge 1. Phase: Create runs by sorting portions of the relation in db buer 2. Phase: Read runs from disk and merge runs in sort order Dept. of Computer Science UC Davis 8. Query Processing and Optimization

43. ECS-165A WQ'11 155 Duplicate Elimination: Sorting: remove all but one copy of tuples having identical value(s) on projection attribute(s) Hashing: partition relation using hash function on projection attribute(s); then read partitions into buer and create in-memory hash index; tuple is only inserted into index if not already present Set Operations: Sorting or hashing Hashing: Partition both relations using the same hash function; use in-memory index for partitions Ri R [ S: if tuple in Ri or in Si, add tuple to result : if tuple in Ri and in Si, . . . : if tuple in Ri and not in Si, . . . Grouping and aggregation: Compute groups via sorting or hashing. Hashing: while groups (partitions) are built, compute partial aggregate values (for group attribute A, V(A,R) tuples to store values) Dept. of Computer Science UC Davis 8. Query Processing and Optimization

44. ECS-165A WQ'11 156 Evaluation of Expressions Strategy 1: materialization. Evaluate one operation at a time, starting at the lowest level. Use intermediate results materialized in temporary relations to evaluate next level operation(s). o Price 5000 CName CUSTOMERS ORDERS OFFERS First compute and store Price5000(OFFERS); then compute and store join of CUSTOMERS and ORDERS;

45. nally, join the two materialized relations and project on to CName. Strategy 2: pipelining. evaluate several operations simultaneously, and pass the result (tuple- or block-wise) on to the next operation. In the example above, once a tuple from OFFERS satisfying selection condition has been found, pass it on to the join. Similarly, don't store result of (

46. nal) join, but pass tuples directly to projection. Much cheaper than materialization, because temporary relations are not generated and stored on disk. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

47. ECS-165A WQ'11 157 Evaluation of Expressions (cont.) Pipelining is not always possible, e.g., for all operations that include sorting (blocking operation). Pipelining can be executed in either demand driven or producer driven fashion. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

48. ECS-165A WQ'11 158 Transformation of Relational Expressions Generating a query-evaluation plan for an expression of the relational algebra involves two steps: 1. generate logically equivalent expressions 2. annotate these evaluation plans by speci

49. c algorithms and access structures to get alternative query plans Use equivalence rules to transform a relational algebra expression into an equivalent one. Based on estimated cost, the most cost-eective annotated plan is selected for evaluation. The process is called cost-based query optimization. Equivalence of Expressions Result relations generated by two equivalent relational algebra expressions have the same set of attributes and contain the same set of tuples, although their attributes may be ordered dierently. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

50. ECS-165A WQ'11 159 Equivalence Rules (for expressions E, E1, E2, conditions Fi) Applying distribution and commutativity of relational algebra operations 1. F1(F2(E)) F1^F2(E) 2. F(E1 [[; ;] E2) F(E1) [[; ;] F(E2) 3. F(E1 E2) F0(F1(E1) F2(E2)); F F0 ^ F1 ^ F2, Fi contains only attributes of Ei; i = 1; 2. 4. A=B(E1 E2) E1 1 A=B E2 5. A(E1 [[; ;] E2)6 A(E1) [[; ;] A(E2) 6. A(E1 E2) A1(E1) A2(E2), with Ai = A f attributes in Eig; i = 1; 2. 7. E1 [[; ] E2 E2 [[; ] E1 (E1 [ E2) [ E3 E1 [ (E2 [ E3) (the analogous holds for ) 8. E1 E2 A1;A2(E2 E1) (E1 E2) E3 E1 (E2 E3) (E1 E2) E3 ((E1 E3) E2) 9. E1 1 E2 E2 1 E1 (E1 1 E2) 1 E3 E1 1 (E2 1 E3) The application of equivalence rules to a relational algebra expression is also sometimes called algebraic optimization. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

51. ECS-165A WQ'11 160 Examples: Selection: { Find the name of all customers who have ordered a product for more than $5,000 from a supplier located in Davis. CName(SAddress like 0%Davis%0 ^ Price5000 (CUSTOMERS 1 (ORDERS 1 (OFFERS 1 SUPPLIERS)))) Perform selection as early as possible (but take existing indexes on relations into account) CName(CUSTOMERS 1 (ORDERS 1 (Price5000(OFFERS) 1 (SAddress like 0%Davis%0(SUPPLIERS))))) Projection: { CName;account(CUSTOMERS 1 Prodname=0CDROM0(ORDERS)) Reduce the size of argument relation in join CName;account(CUSTOMERS 1 CName(Prodname=0CDROM0(ORDERS))) Projection should not be shifted before selections, because minimizing the number of tuples in general leads to more ecient plans than reducing the size of tuples. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

52. ECS-165A WQ'11 161 Join Ordering For relations R1, R2, R3, (R1 1 R2) 1 R3 R1 1 (R2 1 R3) If (R2 1 R3) is quite large and (R1 1 R2) is small, we choose (R1 1 R2) 1 R3 so that a smaller temporary relation is computed and materialized Example: List the name of all customers who have ordered a product from a supplier located in Davis. CName(SAddress like 0%Davis%0 (SUPPLIERS 1 ORDERS 1 CUSTOMERS)) ORDERS 1 CUSTOMERS is likely to be a large relation. Because it is likely that only a small fraction of suppliers are from Davis, we compute the join SAddress like 0%Davis%0(SUPPLIERS 1 ORDERS)

53. rst. Summary of Algebraic Optimization Rules 1. Perform selection as early as possible 2. Replace Cartesian Product by join whenever possible 3. Project out useless attributes early. 4. If there are several joins, perform most restrictive join

54. rst Dept. of Computer Science UC Davis 8. Query Processing and Optimization

55. ECS-165A WQ'11 162 Evaluation Plan An evaluation plan for a query exactly de

56. nes what algorithm is used for each operation, which access structures are used (tables, indexes, clusters), and how the execution of the operations is coordinated. Example of Annotated Evaluation Plan Query: List the name of all customers who have ordered a product that costs more than $5,000. Assume that for both CUSTOMERS and ORDERS an index on CName exists: I1(CName; CUSTOMERS); I2(CName; ORDERS). CName (sort to remove duplicates) o get tuples for tids of I index−nested loop join block nested−loop join Price 5000 I1 (CName, CUSTOMERS) 2 I (CName, ORDERS) ORDERS 2 pipeline pipeline full table scan Dept. of Computer Science UC Davis 8. Query Processing and Optimization

57. ECS-165A WQ'11 163 Choice of an Evaluation Plan Must consider interaction of evaluation techniques when choosing evaluation plan: choosing the algorithm with the least cost for each operation independently may not yield the best overall algorithm. Practical query optimizers incorporate elements of the following two optimization approaches: { Cost-based: enumerate all the plans and choose the best plan in a cost-based fashion. { Rule-based: Use rules (heuristics) to choose plan. Remarks on cost-based optimization: { Finding a join order for R1 1 R2 1 : : : 1 Rn: n! dierent left-deep join orders For example, for n = 9, the number is 362880. ; use of dynamic programming techniques Heuristic (or rule-based) optimization transforms a given query tree by using a set of rules that typically (but not in all cases) improve execution performance: { Perform selection early (reduces number of tuples) { Perform projection early (reduces number of attributes) { Perform most restrictive selection and join operations before other similar operations. Dept. of Computer Science UC Davis 8. Query Processing and Optimization

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