h.lu/hkust l04: physical database design (2) introduction index selection partitioning &...
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H.Lu/HKUST
L04: Physical Database Design (2)
Introduction Index Selection Partitioning & Denormalization
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Tuning a Relational Schema
The choice of relational schema should be guided by the workload, in addition to redundancy issues: We may settle for a 3NF schema rather than BCNF. Workload may influence the choice we make in
decomposing a relation into 3NF or BCNF. We might denormalize (i.e., undo a decomposition step), or
we might add fields to a relation We may further decompose a BCNF schema! We might consider horizontal partitioning.
If such changes are made after a database is in use, called schema evolution; might want to mask some of these changes from applications by defining views.
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Example Schemas
We will concentrate on Contracts, denoted as CSJDPQV. The following ICs are given to hold: JPC, SD P, C is the primary key. What are the candidate keys for CSJDPQV? What normal form is this relation schema in?
Contracts (Cid, Sid, Jid, Did, Pid, Qty, Val)Depts (Did, Budget, Report)Suppliers (Sid, Address)Parts (Pid, Cost)Projects (Jid, Mgr)
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Denormalization Suppose that the following query is important:
Is the value of a contract less than the budget of the department?
To speed up this query, we might add a field budget B to Contracts. This introduces the FD DB wrt Contracts. Thus, Contracts is no longer in 3NF.
We might choose to modify Contracts thus if the query is sufficiently important, and we cannot obtain adequate performance otherwise (i.e., by adding indexes or by choosing an alternative 3NF schema.)
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Partitioning
Horizontal Partitioning: Distributing the rows of a table into several separate files Useful for situations where different users need
access to different rows Vertical Partitioning: Distributing the columns of a
table into several separate files Useful for situations where different users need
access to different columns The primary key must be repeated in each file
Combinations of Horizontal and VerticalPartitions often correspond with User Schemas (user views)
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Partitioning Advantages of Partitioning:
Records used together are grouped together Each partition can be optimized for performance Security, recovery Partitions stored on different disks: contention Take advantage of parallel processing capability
Disadvantages of Partitioning: Slow retrievals across partitions Complexity
Issues: Need to find suitable level Too little too much of irrelevant data access. Too much too much processing cost
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Horizontal Decompositions
Our definition of decomposition: Relation is replaced by a collection of relations that are projections. Most important case.
Sometimes, might want to replace relation by a collection of relations that are selections. Each new relation has same schema as the original,
but a subset of the rows. Collectively, new relations contain all rows of the
original. Typically, the new relations are disjoint.
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Horizontal Decompositions (Contd.)
Suppose that contracts with value > 10000 are subject to different rules. This means that queries on Contracts will often contain the condition val>10000.
One way to deal with this is to build a clustered B+ tree index on the val field of Contracts.
A second approach is to replace contracts by two new relations: LargeContracts and SmallContracts, with the same attributes (CSJDPQV). Performs like index on such queries, but no index overhead. Can build clustered indexes on other attributes, in addition!
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Masking Conceptual Schema Changes
The replacement of Contracts by LargeContracts and SmallContracts can be masked by the view.
However, queries with the condition val>10000 must be asked wrt LargeContracts for efficient execution: so users concerned with performance have to be aware of the change.
CREATE VIEW Contracts(cid, sid, jid, did, pid, qty, val)AS SELECT * FROM LargeContractsUNIONSELECT *FROM SmallContracts
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Decomposition of a BCNF Relation
Suppose that we choose { SDP, CSJDQV }. This is in BCNF, and there is no reason to decompose further (assuming that all known ICs are FDs).
However, suppose that these queries are important: Find the contracts held by supplier S. Find the contracts that department D is involved in.
Decomposing CSJDQV further into CS, CD and CJQV could speed up these queries. (Why?)
On the other hand, the following query is slower: Find the total value of all contracts held by supplier S.
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Vertical Partitioning
Vertical partitioning of a relation R produces partitions R1, R2, ..., Rm, each of which contains a subset of R's attributes as well as the primary key of R
The object of vertical partitioning is to reduce irrelevant attribute access, and thus irrelevant data access
``Optimal'' vertical partitioning minimizes the irrelevant data access for user applications
For a relation with m non-primary key attributes, the number of possible partitions is approximately equal to mm Hard to find an optimal solution Resort to heuristic approaches
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VP: Heuristic Approaches
Grouping: Assign each attribute to one fragment Join fragments until some criteria is satisfied
Splitting (our focus): Start with the original relation Generate partitions based on access behavior Closer to optimal; less overlapping fragments
Basic idea: Affinity of attributes A measure of closeness of these attributes
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Attribute Usage Matrices
Q = {q1 , q2, ..., qm} Set of user queries
R (A1, A2, ..., An) Relation R with n attributes
Usage matrix |Uij|m×n
Uij = 1 if attribute Aj is referenced by qi; Uij = 0 otherwise.
Access matrix |acci|
access frequency of qi
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VP - Matrices Examples Relation PROJ(PNO,PNAME,BUDGET,LOC), four
SQL queries sent to three sites: q1: SELECT BUDGET FROM PROJ WHERE PNO = val; q2: SELECT PNAME,BUDGET FROM PROJ; q3: SELECT PNAME FROM PROJWHERE LOC = val; q4: SELECT SUM(BUDGET) FROM PROJ WHERE
LOC=val;
15
5
25
3
acc
1100
1010
0110
0101
U
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Attribute Affinity Matrix
|affij|n×n : Affinity between two attributes Ai and Aj
affij = { k|Uki = 1 Ukj =1} acck
45
5
75
3
acc
783750
353545
755800
045045
aff
AA Matrix
U
1 0 1 0
0 1 1 0
0 1 0 1
0 0 1 1
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Bond Energy Clustering Algorithm
Determines groups of similar items (clusters of attributes with larger affinity values, and ones with smaller affinity values)
Final groupings are insensitive to the order in which items are presented to the algorithm
The computation time is O(n2 ) where n is the number of attributes
Secondary interrelationships between clustered attribute groups are identifiable
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Main Idea of BEA
Permute the attribute affinity matrix (AA) and generate a clustered affinity matrix (CA) to maximize the global affinity measure (AM)
where)],(),(
),(),()[,(
11
111 1
AAaffAAaff
AAaffAAaffAAaffAM
jiji
jijij
n
i
n
ji
0),(),(),(),(1100 AAiaffAAaffAAaffAAaff
ninij
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AM in Terms of Bond
Because the affinity matrix is symmetric,
, or
Let
then
AM = ∑[bond(Aj, Aj-1) + bond(Aj, Aj+1)]
)],(),()[,( 111 1
AAaffAAaffAAaffAM jijij
n
i
n
ji
)],(),(),(),([ 111 1
AAAAAAAA jijijij
n
i
n
ji affaffaffaffAM
n
zyzxzyx AAaffAAaffAAbond
1
),(),(),(
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Bond Energy Algorithm
Initialization : place and fix one of the columns of AA arbitrarily into CA
Iteration : Pick one of the remaining n i columns of AA and
place it in one of the i+1 positions in CA Choose the placement that makes greatest
contribution. Row ordering :
Change the placement of the rows accordingly
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Contribution of a Placement
Contribution of placing attribute Ak between Ai and Aj :
cont(Ai, Ak, Aj) = 2bond(Ai, Ak) + 2bond(Ak, Aj) – 2bond(Ai, Aj)
783750
353545
755800
045045
affbond(A1, A2) = 45*0+0*80+45*5+0*75=225bond(A1, A4) = 45*0+0*75+45*3+0*78=135bond(A4, A2) = 0*0+ 75*80+3*5+78*75=11865
If we place A4 between A1 and A2,
cont(A1, A4, A2 ) = 2bond(A1, A4) + 2bond(A4, A2) – 2bond(A1, A2) = 2*135 + 2*11865 - 2*225 = 23550
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BEA Example
cont(A0, A3, A1) = 8820cont(A1, A3, A2) = 10150cont(A2, A3, A4) = 1780
45
5
53
3
783750
353545
755800
045045
aff
45 0
0 80
45 5
0 75
A1 A2A3
45 45 0
0 5 80
45 53 5
0 3 75
45 45 0 0
0 5 80 75
45 53 5 3
0 3 75 78
A1 A3 A2
A1 A3 A2 A4
A1
A2
A3
A4
45 45 0 0
45 53 5 3
0 5 80 75
0 3 75 78
A1 A3 A2 A4
A1
A3
A2
A4
A1
A2
A3
A4
A1
A2
A3
A4
Two clusters: the upper left corner of the smaller affinity values, and the lower right corner of the larger affinity values
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BQ
OQ
VP Splitting
A1 A2 A3 … Ai Ai+1 An
A1 A2 A3 . Ai
Ai+1
An
TA
BA
Two attribute sets:TA : {A1, A2, ..., Ai} BA : {Ai+1, Ai+2, ..., An}
TQ
Three sets of apps:TQ : access TA only BQ : access BA only OQ: access both
The basic idea
Given a set of attributes {A1, A2, ..., An} and a set of applications, partition the attributes into two or more sets such that there are no (or minimal) applications that access more than one of the sets.
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VP Splitting Problem
Define: CTQ = total number of accesses to attributes by
applications that access only TA CBQ = total number of accesses to attributes by
applications that access only BA COQ = total number of accesses to attributes by
applications that access both TA & BA
Find a split point x (1≤x<n) which maximizes z
z = CTQ * CBQ COQ 2
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VP – The Splitting Algorithm
Input: Relation R, and CA, acc matrices Output: a set of fragments For each split point x (1≤x<n) , compute z
Choose the split point with the maximum z value and construct fragments
( )
i
iXQ
CXQq
qacc
XQ {TQ, BQ, OQ}
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VP: Splitting Example
45
5
75
3
acc
x 1 2 3
TA BA TQ BQ OQ CTQ CBQ COQ zA1 A3,2,4 Q2,3,4 Q1 0 83 45 -2025A1,3 A2,4 Q1 Q3 Q2,4 45 75 8 3311A1,3,2 A4 Q1,2 Q3,4 50 0 78 -6084
1100
1010
0110
0101
U
Partition: (A1,A3) (A2,A4)
45 45 0 0
45 53 5 3
0 5 80 75
0 3 75 78
A1 A3 A2 A4
A1
A3
A2
A4
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Complications in VP Partitioning Algorithm
Cluster forming in the middle of the CA matrix Shift a row up and a column left and apply the algorithm
to find the “best” partitioning point Do this for all possible shifts Cost O(n2)
More than two clusters M-way partitioning Try 1, 2, …, m-1 split points along the diagonal and try
to find the best point for each of these Cost O(2m)