Department of Computer Science and Engineering, HKUST Slide Query Processing and Optimization Query Processing and Optimization

Department of Computer Science and Engineering, HKUST Slide 2 Introduction Users are expected to write “efficient” queries. But they do not always do that! –Users typically do not have enough information about the database to write efficient queries. E.g., no information on table size –Users would not know if a query is efficient or not without knowing how the DBMS’s query processor work DBMS’s job is to optimize the user’s query by: –Converting the query to an internal representation (tree or graph) –Evaluate the costs of several possible ways of executing the query and find the best one.

Department of Computer Science and Engineering, HKUST Slide 3 Steps in Query Processing SQL query Execution Plan Code Result Parse Tree Query Parsing Code Generation Query Optimization Runtime DB Processor Join ProjectEmployee Join Employee and Project using hash join, …...

Department of Computer Science and Engineering, HKUST Slide 4 File scan  scan all records of the file to find records that satisfy selection condition Binary search  when the file is sorted on attributes specified in the selection condition Index scan  using index to locate the qualified records –Primary index, single record retrieval  equality comparison on a primary key attribute with a primary index –Primary index, multiple records retrieval  comparison condition, etc. on a key field with primary index –Clustering index to retrieve multiple records –Secondary index to retrieve single or multiple records When would file scan be better than index scan? Select Operation

Department of Computer Science and Engineering, HKUST Slide 5 OP1 AND OP2 (e.g., EmpNo=123 AND Age=30)   Conjunctive selection: Evaluate the condition that has an index created (I.e., that can be evaluated very fast), get the qualified tuples and then check if these tuples satisfy the remaining conditions.   Conjunctive selection using composite index: if there is a composite index created on attributes involved in one or more conditions, then use the composite index to find the qualified tuples Complete Employee Records EmpNoAge Compositeindex   Conjunctive selection by intersection of record pointers: if secondary indexes are available, evaluate each condition and intersect the sets of record pointers obtained. Conjunctive Conditions

Department of Computer Science and Engineering, HKUST Slide 6   When there are more than one attribute with an index: – –use the one that costs least, and – –the one that returns the smallest number of qualified tuple   Disjunctive select conditions: OP1 or OP2 are much more costly:   potentially a large number of tuples will qualify   costly if any one of the condition doesn’t have an index created   selectivity of a condition  is the number of tuples that satisfy the condition divided by total number of tuples.   The smaller the selectivity, the fewer the number of tuples retrieved, and the higher the desirability of using that condition to retrieve the records. Conjunctive Conditions (cont.)

Department of Computer Science and Engineering, HKUST Slide 7 Join is one of the most time-consuming operations in query processing. Two-way join is a join of two relations, and there are many algorithms to evaluate the join. Multi-way join is a join of more than two relations; different orders of evaluating a multi-way join have different speeds We shall study methods for implementing two-way joins of form R A=B S Join Operation

Department of Computer Science and Engineering, HKUST Slide 8   Nested (inner-outer) Loop: For each record r in R (outer loop), retrieve every record s from S (inner loop) and check if r[A] = s[B]. R A=B S Join Algorithm: Nested (inner-outer) Loop for each tuple r in R do for each tuple s in S do if r.[A] = s[B] then output result end RS m tuples in R n tuples in S m*n checkings R and S can be reversed

Department of Computer Science and Engineering, HKUST Slide 9   If an index (or hash key) exists, say, on attribute B of S, should we put R in the outer loop or S? Why?   Records in the outer relation are accessed sequentially, an index on the outer relation doesn’t help;   Records in the inner relations are accessed randomly, so an index can retrieve all records in the inner relation that satisfy the join condition. When One Join Attributes is Indexed R S index on S for each tuple r in R do lookup r.[A] in S if found then output result end

Department of Computer Science and Engineering, HKUST Slide 10   Sort-merge join: if the records of R and S are sorted on the join attributes A and B, respectively, then the relations are scanned in say ascending order, matching the records that have same values for A and B. R A=B S Sort-Merge Join R and S are only scanned once. Even if the relations are not sorted, it is better to sort them first and do sort-merge join then doing double- loop join. if R and S are sorted, n + m if not sorted: n log(n) + m log(m) + m + n

Department of Computer Science and Engineering, HKUST Slide 11   Hash-join: R and S are both hashed to the same hash file based on the join attributes. Tuples in the same bucket are then “joined” Hash Join Method

Department of Computer Science and Engineering, HKUST Slide 12 Disk accesses are based on blocks, not individual tuples Main memory buffer can significantly reduce the number of disk accesses –Use the smaller relation in outer loop in nested loop method –Consider if 1 buffer is available, 2 buffers, m buffers When index is available, either the smaller relation or the one with large number of matching tuples should be used in the outer loop. If join attributes are not indexed, it may be faster to create the indexes on-the-fly (hash-join is close to generating a hash index on-the-fly) Sort-Merge is the most efficient; the relations are often sorted already Hash join is efficient if the hash file can be kept in the main memory Hints on Evaluating Joins

Department of Computer Science and Engineering, HKUST Slide 13 Give a relational algebra expression, how do we transform it to a more efficient one? Query Optimization Use the query tree as a tool to rearrange the operations of the relational algebra expression

Department of Computer Science and Engineering, HKUST Slide 14 A Query Tree Empolyee(EmpNo, EmpName, Address, Birthdate, DeptNo) Department (DeptNo, DeptName, MgrNo) Project (ProjNo, ProjName, ProjLocation, DeptNo) WorksOn(EmpNo, ProjNo, Hours)  ProjNo,DeptNo,EmpName,Address,Birthdat e MgrNo=EmpNo  ProjLocation=‘Stafford’ DeptNo=DeptNo Employee Department Project (3) (2) (1)

Department of Computer Science and Engineering, HKUST Slide 15 Structure and Execution of a Query Tree A query tree is a tree structure that corresponds to a relational algebra expression by representing the input relations as leaf nodes and the relational algebra operations as internal nodes of the tree An execution of the query tree consists of executing an internal node operation whenever its operands are available and then replacing that internal node by the relation that results from executing the operation

Department of Computer Science and Engineering, HKUST Slide 16 Heuristics for Optimizing a Query A query may have several equivalent query trees A query parser generates a standard canonical query tree from a SQL query tree –Cartesian products are first applied (FROM) –then the conditions (WHERE) –and finally projection (SELECT)

Department of Computer Science and Engineering, HKUST Slide 17  ProjNo,DeptNo,EmpName,Address,Birthdate  ProjLocation=‘Stafford’ AND MgrNo=EmpNo AND DeptNo=DeptNo, Employee   Department Project The query optimizer transforms this canonical query into an efficient final query Heuristics for Optimizing a Query select ProjNo, DeptNo, EmpName, Address, Birthdate from Project, Department, Employee where ProjLocation=‘Stafford’ and MrgNo=EmpNo and Department.DeptNo=Employee.DeptNo

Department of Computer Science and Engineering, HKUST Slide 18 Find the names of employees born after 1957 who work on a project named ‘Aquarius’ select EmpName from Employee, WorksOn, Project whereProjName=‘Aquarius’ AND Project.ProjNo=WorksOn.ProjNo AND Employee.EmpNo = WorksOn.EmpNo AND Birthdate >‘DEC ’ WorksOn (EmpNo, ProjNo, Hours)  EmpName  ProjName=‘Aquarius’ AND Project.ProjNo=Project.ProjNo AND Employee.EmpNo=WorksOn.EmpNo AND Birthdate > ‘DEC ’ Project   WorksOnEmployee Example

Department of Computer Science and Engineering, HKUST Slide 19  EmpName  ProjNo=ProjNo Project   WorksOn Employee  ProjName=‘Aquarius’  Birthdate > ‘dec ’  EmpNo=EmpNo Example Push all the conditions as far down the tree as possible Expensive due to large size of Employee

Department of Computer Science and Engineering, HKUST Slide 20 Example  EmpName  EmpNo=EmpNo Employee   WorksOn Project  Birthdate > ‘dec ’  PNAME=‘Aquarius’  ProjNo=ProjNo Rearrange join sequence according to estimates of relation sizes

Department of Computer Science and Engineering, HKUST Slide 21 Only need ProjNo attribute from Project and WorksOn Only need EmpNo attribute from Employee and WorksOn and EmpName from Employee Example Replace cross products and selection sequence with a join operation  EmpName EmpNo= EmpNo Employee WorksOn Project  Birthdate > ‘dec ’  ProjName=‘Aquarius’ ProjNo= ProjNo

Department of Computer Science and Engineering, HKUST Slide 22 Example Push projection as far down the query tree as possible  LNAME EmpNo = EmpNo Employee  Birthdate > ‘dec ’ WorksOn Project  ProjName=‘Aquarius’ ProjNo= ProjNo  EmpNo, EmpName  EmpNo  EmpNo, ProjNo  ProjNo

Department of Computer Science and Engineering, HKUST Slide Cascade of  : A conjunctive selection condition can be broken up into a cascade (sequence) of individual  operations:  c 1 AND c 2 AND...AND c n (R)   c 1 (  c 2 (...(  c n (R))..)) 2. Commutativity of  :  c 1 (  c 2 (R))   c 2 (  c 1 (R)) 3. Cascade of  :  List 1 (  List 2 (... (  List n (R))... ))   List 1 (R) if List 1 is included in List 2 …List n ; result is null if List 1 is not in any of List 2 …List n Transformation Rules

Department of Computer Science and Engineering, HKUST Slide Commuting  with  : if the projection list List1 involves only attributes that are in condition c  List1 (  c (R))   c (  List1 (R)) 5. Commutivity of JOIN or  : R S  S R 6. Commuting  with JOIN: if all the attributes in the selection condition c involve only the attributes of one of the relations being joined, say, R  c (R S)  (  c (R)) S Transformation Rules (Cont.)

Department of Computer Science and Engineering, HKUST Slide 25 7.Commuting  with JOIN: if List can be separated into List 1 and List 2 involving only attributes from R and S, respectively, and the join condition c involves only attributes in List:  List (R c S)  (  List 1 (R) c  List 2 (S)) 8.Commuting set operations:  and  are commutative 9.JOIN, , ,  are associative 10.  distributes over , ,   c (R  S)   c (R)   c (S) 11.  distributes over   List (R  S)  (  List (R)   List (S)) Transformation Rules (Cont.)

Department of Computer Science and Engineering, HKUST Slide 26   Use rule 1 to break up any  operation with conjunctive conditions into a sequence of  operations   Use rules 2, 4, 6, and 10 concerning commutativity of  with other operations to move each  operation as far down the query tree as possible based on the attributes in the  operations   Use rule 9 concerning associativity of binary operations to rearrange the leaf nodes of the tree so that the leaf node relations with the most restrictive  operations are executed Heuristic Algebraic Optimization

Department of Computer Science and Engineering, HKUST Slide 27   Combine sequences of Cartesian product and  operation representing a join condition into single JOIN operations   Use rules 3, 4, 7, and 11 concerning the cascading of  and commuting with other operations, break down a  and move the projection attributes down the tree as far as possible   Identify subtrees that represent groups of operations that can be executed by a single algorithm (select/join followed by project) Heuristic Algebraic Optimization (Cont.)

Department of Computer Science and Engineering, HKUST Slide 28 Estimation of the Size of Joins The Cartesian product r s contains n r n s tuples; each tuple occupies s r + s s bytes. If R  S = , then r s is the same as r x s. If R  S is a key for R, then a tuple of s will join with at most one tuple from r; therefore, the number of tuples in r s is no greater than the number of tuples in s. If R  S in S is a foreign key in S referencing R, then the number of tuples in r s is exactly the same as the number of tuples in s. The case for R  S being a foreign key referencing S is symmetric. R S Matching tuples

Department of Computer Science and Engineering, HKUST Slide 29 Example of Size Estimation In the example query depositor customer, customer-name in depositor is a foreign key of customer; hence, the result has exactly  depositor tuples, which is Data: R = Customer, S = Depositor  customer = 10,000 f customer = 25 b customer = 10000/25 = 400  depositor = 5,000 f depositor = 50 b depositor = 5000/50 = 100

Department of Computer Science and Engineering, HKUST Slide 30 Estimation of the size of Joins If R  S = {A} is not a key for R or S. If we assume that every tuple t in R produces tuples in R S, number of tuples in R S is estimated to be:  r  s V(A, s) If the reverse is true, the estimates obtained will be:  r  s V(A, r) The lower of these two estimates is probably the more accurate one. Number of distinct values of A in s R S  s V(A, s)

Department of Computer Science and Engineering, HKUST Slide 31 Estimation of the size of Joins Compute the size estimates for depositor customer without using information about foreign keys: –  customer = 10,000  depositor = 5,000 V(customer-name, depositor ) = 2500 V(customer-name, customer ) = –The two estimates are 5000 * 10000/2500 = 20,000 and 5000 * 10000/10000 = 5000 –We choose the lower estimate, which, in this case, is the same as our earlier computation using foreign keys. There are 5,000 tuples in depositor relation but has only 2,500 distinct depositors, so every depositor has two accounts Customer-name is unique

Department of Computer Science and Engineering, HKUST Slide 32 Nested-Loop Join (Tuple-Based) Compute the theta join, r  s for each tuple t r in r do begin for each tuple t s in s do begin test pair (t r, t s ) to see if they satisfy the join condition  if they do, add t r · t s to the result. End end r is called the outer relation and s the inner relation of the join. Requires no indices and can be used with any kind of join condition. Expensive since it examines every pair of tuples in the two relations. –For each tuple in the outer relation (r), loop through all n s tuples in the inner relation (s) –Cost is n r x n s

Department of Computer Science and Engineering, HKUST Slide 33 Cost of Nested-Loop Join If there is enough memory to hold only one block of each relation, the estimated cost is n r * b s + b r disk accesses If the smaller relation fits entirely in memory, use it as the inner relation. This reduces the cost estimate to b r + b s disk accesses. –b r + b s is the minimum possible cost to read R and S once –Putting both relations in memory won’t reduce the cost further b r disk accesses to load R into buffer R S For each tuple in r, S has to be read into buffer, b s disk accesses no. of bocks in rno. of bocks in s

Department of Computer Science and Engineering, HKUST Slide 34 Nested-Loop Join with Buffers (Still Tuple Based) The algorithm is the same as in the previous slide –Tuples are fetched and compared one by one according to the double loop –OS or DBMS fetches a tuple from buffer if it is already there b r disk accesses to load R into buffer R S For each tuple in r, S has to be read into buffer, b s disk accesses At this point, one block of r is read, and the first r-tuple has been compared to 3 s-tuples (1 block of s)

Department of Computer Science and Engineering, HKUST Slide 35 Nested-Loop Join with Buffers (Still Tuple Based) b r disk accesses to load R into buffer R S At this point, the first r-tuple has been compared to 6 s-tuples The next step begins with the 2 nd tuple in r’s buffer; no access to r on disk is needed; however, the s-tuples have to be read from disk again Total cost = n r * b s + b r disk accesses

Department of Computer Science and Engineering, HKUST Slide 36 Rewriting the Nested-Loop Join To make use of the buffer efficiently, the algorithm has to be buffer-aware for each block B r in r do begin for each block B s in s do begin Do all tuples in B r and B s : B r  B s end end R S Total cost = b r * b s + b r disk accesses

Department of Computer Science and Engineering, HKUST Slide 37 R S To make use of the buffer efficiently, the algorithm has to be rewritten for each block B r in r do begin for each block B s in s do begin Do all tuples in Br and Bs: Br  Bs end end Total cost = b r * b s + b r disk accesses Rewriting the Nested-Loop Join

Department of Computer Science and Engineering, HKUST Slide 38 R S To make use of the buffer efficiently, the algorithm has to be rewritten for each block B r in r do begin for each block B s in s do begin Do all tuples in Br and Bs: Br  Bs end end Total cost = b r * b s + b r disk accesses Rewriting the Nested-Loop Join