department of computer science and engineering, hkust slide 1 13-14. query processing and...
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Department of Computer Science and Engineering, HKUST Slide 1
13-14. Query Processing and Optimization13-14. Query Processing and Optimization
Department of Computer Science and Engineering, HKUST Slide 2
IntroductionIntroduction
• 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 ProcessingSteps 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 Projectusing 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 recordsWhen would file scan be better than index scan?When would file scan be better than index scan?
Select OperationSelect 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 RecordsComplete Employee RecordsEmpNoEmpNo AgeAge012012 2525123123 3030
CompositeCompositeindexindex
Conjunctive selection by intersection of record pointers: if secondary indexes are available, evaluate each condition and intersect the sets of record pointers obtained.
Conjunctive ConditionsConjunctive 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.)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 OperationJoin 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) LoopJoin Algorithm: Nested (inner-outer) Loop
for each tuple r in Rdo for each tuple s in S
do if r.[A] = s[B] then output result
endend
00050005000200020004000400020002000200020001000100050005
000500050002000200020002000300030002000200050005
RRSS
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 IndexedWhen One Join Attributes is Indexed
00050005000200020004000400020002000200020001000100050005
RR
000500050002000200020002000300030002000200050005
SSindex on Sindex on S
for each tuple r in Rdo lookup r.[A] in S
if found then output resultend
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
00010001000200020002000200020002000400040005000500050005
000200020002000200020002000300030005000500050005
Sort-Merge JoinSort-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”.
00010001000200020002000200020002000400040005000500050005
000200020002000200020002000300030005000500050005
00010001 000200020002000200020002
00040004 0005000500050005
000200020002000200020002
00030003
0005000500050005
Hash Join MethodHash 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 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 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 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,Birthdate
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 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 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
DepartmentProject
The query optimizer transforms this canonical query into an efficient final query
Heuristics for Optimizing a 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
where ProjName=‘Aquarius’ AND
Project.ProjNo=WorksOn.ProjNo AND
Employee.EmpNo = WorksOn.EmpNo AND
Birthdate >‘DEC-31-1957’
WorksOn (EmpNo, ProjNo, Hours)
EmpName
ProjName=‘Aquarius’ AND Project.ProjNo=Project.ProjNo
AND Employee.EmpNo=WorksOn.EmpNo
AND Birthdate > ‘DEC-31-1957’
Project
WorksOnEmployee
Example Example
Department of Computer Science and Engineering, HKUST Slide 19
EmpName
ProjNo=ProjNo
Project
WorksOn
Employee
ProjName=‘Aquarius’
Birthdate > ‘dec-31-1957’
EmpNo=EmpNo
Example Example
Push all the conditions as far downthe tree as possible
Expensive due to largeExpensive due to largesize of Employeesize of Employee
Department of Computer Science and Engineering, HKUST Slide 20
Example Example
EmpName
EmpNo=EmpNo
Employee
WorksOn
Project
Birthdate > ‘dec-31-1957’
PNAME=‘Aquarius’
ProjNo=ProjNo
Rearrange join sequence accordingto estimates of relation sizes
Department of Computer Science and Engineering, HKUST Slide 21
Only need ProjNo attribute fromProject and WorksOn
Only need EmpNo attribute fromEmployee and WorksOn andEmpName from Employee
Example Example
Replace cross products and selectionsequence with a join operation EmpName
EmpNo= EmpNo
EmployeeWorksOn
Project
Birthdate > ‘dec-31-1957’
ProjName=‘Aquarius’
ProjNo= ProjNo
Department of Computer Science and Engineering, HKUST Slide 22
Example Example
Push projection as far down thequery tree as possible
LNAME
EmpNo = EmpNo
Employee
Birthdate > ‘dec-31-1957’
WorksOn
Project
ProjName=‘Aquarius’
ProjNo= ProjNo
EmpNo, EmpNameEmpNo
EmpNo, ProjNo ProjNo
Department of Computer Science and Engineering, HKUST Slide 23
1. Cascade of : A conjunctive selection condition can be broken up into a cascade (sequence) of individual operations:
• c1 AND c2 AND...AND cn(R) c1
(c2(...(cn
(R))..))
2. Commutativity of :c1
(c2(R)) c2
(c1(R))
3. Cascade of :
• List1(List2
(... (Listn(R))... )) List1
(R)
if List1 is included in List2…Listn; result is null if List1 is not in any of List2…Listn
Transformation RulesTransformation Rules
Department of Computer Science and Engineering, HKUST Slide 24
4. 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.)Transformation Rules (Cont.)
Department of Computer Science and Engineering, HKUST Slide 25
7. Commuting with JOIN: if List can be separated into List1 and List2 involving only attributes from R and S, respectively, and the join condition c involves only attributes in List:
• List(R c S) (List1(R) c List2
(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.)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 OptimizationHeuristic 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.)Heuristic Algebraic Optimization (Cont.)
Department of Computer Science and Engineering, HKUST Slide 28
Estimation of the Size of JoinsEstimation of the Size of Joins
• The Cartesian product r s contains nrns tuples; each tupleoccupies sr + ss 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 EstimationExample 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 5000.
• Data: R = Customer, S = Depositorcustomer = 10,000
fcustomer = 25
bcustomer = 10000/25 = 400
depositor = 5,000
fdepositor = 50
bdepositor = 5000/50 = 100
Department of Computer Science and Engineering, HKUST Slide 30
Estimation of the size of JoinsEstimation 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 JoinsEstimation of the size of Joins
• Compute the size estimates for depositor customer without using information about foreign keys: customer = 10,000
depositor = 5,000V(customer-name, depositor ) = 2500 V(customer-name, customer ) = 10000
– 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)Nested-Loop Join (Tuple-Based)
• Compute the theta join, r s for each tuple tr in r do begin
for each tuple ts in s do begin test pair (tr, ts) to see if they satisfy the join condition
if they do, add tr · ts 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 ns tuples in
the inner relation (s)
– Cost is nr x ns
Department of Computer Science and Engineering, HKUST Slide 33
Cost of Nested-Loop JoinCost of Nested-Loop Join
• If there is enough memory to hold only one block of each relation, the estimated cost is nr * bs + br disk accesses
• If the smaller relation fits entirely in memory, use it as the inner relation. This reduces the cost estimate to br + bs disk accesses.– br + bs is the minimum possible cost to read R and S once– Putting both relations in memory won’t reduce the cost further
br disk accesses toload R into buffer
RS
For each tuple in r, S has to be
read into buffer, bs 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)
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
br disk accesses toload R into buffer
RS
For each tuple in r, S has to be
read into buffer, bs 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)
Nested-Loop Join with Buffers (Still Tuple Based)
br disk accesses toload R into buffer
RS
• At this point, the first r-tuple has been compared to 6 s-tuples • The next step begins with the 2nd 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 = nr * bs + br disk accesses
Department of Computer Science and Engineering, HKUST Slide 36
Rewriting the Nested-Loop JoinRewriting the Nested-Loop Join
• To make use of the buffer efficiently, the algorithm has to be buffer-aware
• for each block Br in r do beginfor each block Bs in s do begin Do all tuples in Br and Bs: Br Bs
end end
RS
•Total cost = br * bs + br disk accesses
Department of Computer Science and Engineering, HKUST Slide 37
RS
• To make use of the buffer efficiently, the algorithm has to be rewritten
• for each block Br in r do beginfor each block Bs in s do begin Do all tuples in Br and Bs: Br Bs
end end
• Total cost = br * bs + br disk accesses
Rewriting the Nested-Loop JoinRewriting the Nested-Loop Join
Department of Computer Science and Engineering, HKUST Slide 38
RS
• To make use of the buffer efficiently, the algorithm has to be rewritten
• for each block Br in r do beginfor each block Bs in s do begin Do all tuples in Br and Bs: Br Bs
end end
• Total cost = br * bs + br disk accesses
Rewriting the Nested-Loop JoinRewriting the Nested-Loop Join