Transcript Physical Database Design

L04: Physical Database Design (2)
Introduction
 Index Selection
 Partitioning & Denormalization

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Tuning a Relational Schema
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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
Contracts (Cid, Sid, Jid, Did, Pid, Qty, Val)
Depts (Did, Budget, Report)
Suppliers (Sid, Address)
Parts (Pid, Cost)
Projects (Jid, Mgr)

We will concentrate on Contracts, denoted as CSJDPQV. The
following ICs are given to hold: JPC, SD  P, C is the
primary key.
 What are the candidate keys for CSJDPQV?
 What normal form is this relation schema in?
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Denormalization
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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 DB 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 Vertical

Partitions often correspond with User Schemas (user views)
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Partitioning
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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.)
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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
CREATE VIEW Contracts(cid, sid, jid, did, pid, qty, val)
AS SELECT *
FROM LargeContracts
UNION
SELECT *
FROM SmallContracts
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.
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Decomposition of a BCNF Relation
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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
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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
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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
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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;
1
0
U 
0

0
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0 1 0
1 1 0
1 0 1

0 1 1
15 
5
acc   
 25
 
3
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Attribute Affinity Matrix

|affij|n×n : Affinity between two attributes Ai and Aj
affij = { k|Uki = 1  Ukj =1} acck
1
0
U
0

0
0 1 0
1 1 0
1 0 1

0 1 1
 45
5
acc   
 75
 
3
45 0 45 0 
 0 80 5 75

aff  
45 5 53 3 


0
75
3
78


AA Matrix
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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)
n
where
n
AM   aff ( Ai, A j )[ aff ( Ai , A j 1)  aff ( Ai , A j 1)
i 1 j 1
 aff ( Ai 1 , A j )  aff ( Ai 1 , A j )]
aff ( A0 , A j )  aff ( Ai , A0)  aff ( An 1 , Ai )  aff ( Ai , An 1)  0
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AM in Terms of Bond
Because the affinity matrix is symmetric,
n
n
AM   aff ( Ai, A j )[ aff ( Ai , A j 1)  aff ( Ai , A j 1)] , or
i 1 j 1
n
n
AM  [aff ( Ai, A j )aff ( Ai , A j 1)  aff ( Ai , A j )aff ( Ai , A j 1)]
i 1 j 1
Let
n
bond ( Ax , A y )   aff ( Az , Ax ) aff ( Az , A y )
z 1
then
AM = ∑[bond(Aj, Aj-1) + bond(Aj, Aj+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 ni 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)
45 0 45 0 
 0 80 5 75 bond(A1, A2) = 45*0+0*80+45*5+0*75=225
 bond(A1, A4) = 45*0+0*75+45*3+0*78=135
aff  
45 5 53 3  bond(A4, A2) = 0*0+ 75*80+3*5+78*75=11865


0
75
3
78


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
A1 A 2
A1
A2
A3
A4
45 0
 0 80

45 5

 0 75






A3
45
5
 
53
 
3
A1 A3 A2
A1
A2
A3
A4
45 45 0
 0 5 80

45 53 5

 0 3 75
cont(A0, A3, A1) = 8820
cont(A1, A3, A2) = 10150
cont(A2, A3, A4) = 1780






45 0 45 0 
 0 80 5 75

aff  
45 5 53 3 


0
75
3
78


A1 A3 A2 A4
A1
A2
A3
A4
45 45 0 0 
 0 5 80 75


45 53 5 3 


0
3
75
78


A1 A3 A2 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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A1
A3
A2
A4
45 45 0 0 
45 53 5 3 


 0 5 80 75


0
3
75
78


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VP Splitting
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.
A1 A2 A3 … Ai Ai+1
An Two attribute sets:
TA : {A1, A2, ..., Ai}
A1
TQ A
2
BA : {Ai+1, Ai+2, ..., An}
A3
TA
.
Three sets of apps:
OQ A
i
TQ : access TA only
Ai+1
BQ : access BA only
BA
BQ
An
OQ: access both
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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

CXQ 
 acc (q )
qi

 XQ
i
XQ  {TQ, BQ, OQ}
Choose the split point with the maximum z value and
construct fragments
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VP: Splitting Example
A1 A3 A2 A4
A1
A3
A2
A4
45 45 0 0 
45 53 5 3 


 0 5 80 75


 0 3 75 78
 45
5
acc   
 75
 
3
1
0
U 
0

0
0 1 0
1 1 0
1 0 1

0 1 1
TA
BA
TQ
BQ OQ CTQ CBQ COQ z
A3,2,4
Q2,3,4 Q1
0 83 45 -2025
1 A1
Q3
Q2,4 45 75
8 3311
2 A1,3 A2,4 Q1
3 A1,3,2 A4
Q1,2
Q3,4 50
0 78 -6084
x

Partition: (A1,A3) (A2,A4)
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Complications in VP Partitioning Algorithm
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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)
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