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PMIT-6102 Advanced Database Systems By- Jesmin Akhter Assistant Professor, IIT, Jahangirnagar University.

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Presentation on theme: "PMIT-6102 Advanced Database Systems By- Jesmin Akhter Assistant Professor, IIT, Jahangirnagar University."— Presentation transcript:

1 PMIT-6102 Advanced Database Systems By- Jesmin Akhter Assistant Professor, IIT, Jahangirnagar University

2 Lecture 05 Distributed Database Design

3 Outline Distributed Database Design  Distributed Design Issues o Fragmentation o Data Allocation Slide 3

4 Degree of Fragmentation Finding the suitable level of partitioning within this range To fragment to the level of individuals tuples or To fragment to the level of individuals attributes Relations (Not to fragment at all) finite number of alternatives

5 Correctness of Fragmentation Completeness  Decomposition of relation R into fragments R 1, R 2,..., R n is complete if and only if each data item in R can also be found in some R i  This property, which is identical to the lossless decomposition property of normalization  it ensures that the data in a global relation are mapped into fragments without any loss Slide 5

6 Correctness of Fragmentation Reconstruction  If relation R is decomposed into fragments F R ={R 1, R 2,..., R n }, then there should exist some relational operator ∇ such that R = ∇ Ri,  The reconstructability of the relation from its fragments ensures that constraints defined on the data in the form of dependencies are preserved. Slide 6

7 Correctness of Fragmentation Disjointness  If relation R is horizontally decomposed into fragments F R ={R 1, R 2,..., R n }, and data item d j is in R j, then d j should not be in any other fragment R k (k ≠ j ).  This criterion ensures that the horizontal fragments are disjoint.  If relation R is vertically decomposed, its primary key attributes are typically repeated in all its fragments (for reconstruction).  Therefore, in case of vertical partitioning, disjointness is defined only on the non-primary key attributes of a relation. Slide 7

8 Allocation Alternatives Non-replicated database  partitioned : each fragment resides at only one site Replicated database  fully replicated : each fragment at each site  partially replicated : each fragment at some of the sites Slide 8

9 Comparison of Replication Alternatives Slide 9

10 Information Requirements The information needed for distribution design can be divided into four categories:  Database information  Application information  Communication network information  Computer system information Slide 10

11 Fragmentation Horizontal Fragmentation (HF) Vertical Fragmentation (VF) Hybrid Fragmentation (HF) Slide 11

12 Horizontal Fragmentation (HF) Horizontal fragmentation partitions a relation along its tuples.  each fragment has a subset of the tuples of the relation. There are two versions of horizontal partitioning:  Primary horizontal fragmentation  Derived horizontal fragmentation Primary horizontal fragmentation of a relation is performed  using predicates that are defined on that relation. Derived horizontal fragmentation is the partitioning of a relation  results from predicates being defined on another relation. Slide 12

13 Information Requirements of HF Database Information  how the database relations are connected to one another, especially with joins.  In the relational model, these relationships are also depicted as relations.  cardinality of each relation: card(R) TITLE, SAL SKILL ENO, ENAME, TITLEPNO, PNAME, BUDGET, LOC ENO, PNO, RESP, DUR EMP PROJ ASG L1L1 L2L2 L3L3 Expression of Relationships Among Relations Using Links Slide 13

14 Information Requirements of HF 14 The direction of the link shows a one to many relationship. For each title there are multiple employees with their title. Thus there is a link between PAY and EMP relations.

15 Application Information  simple predicates : Given R[A 1, A 2, …, A n ], a simple predicate p j is p j : A i θValue where θ  {=,,≥,≠}, Value  D i and D i is the domain of A i. For relation R we define Pr = {p 1, p 2, …,p m } Example : PNAME = "Maintenance" BUDGET ≤ 200000 Information Requirements of HF Slide 15

16 Application Information  minterm predicates : Given R and Pr = {p 1, p 2, …,p m } define M = {m 1,m 2,…,m r } as M = { m i | m i =  p j  Pr  p j * }, 1≤j≤m, 1≤i≤z where p j * = p j or p j * = ¬(p j ). Information Requirements of HF Example m 1 : PNAME="Maintenance"  BUDGET≤200000 m 2 : NOT(PNAME="Maintenance")  BUDGET≤200000 m 3 : PNAME= "Maintenance"  NOT(BUDGET≤200000) m 4 : NOT(PNAME="Maintenance")  NOT(BUDGET≤200000) Slide 16

17 Application Information In terms of quantitative information about user applications, we need to have two sets of data.  minterm selectivities: sel(m i ) o The number of tuples of the relation that would be accessed by a user query which is specified according to a given minterm predicate m i.  access frequencies: acc(q i ) o The frequency with which a user application qi accesses data. o Access frequency for a minterm predicate can also be defined. Information Requirements of HF Slide 17

18 Primary Horizontal Fragmentation Definition : A primary horizontal fragmentation is defined by a selection operation on the owner relations of a database schema. Therefore, given relation R, its horizontal fragments are given by R i =  F i (R), 1 ≤ i ≤ w where F i is a selection formula used to obtain fragment R i. If F i is in conjunctive normal form, it is a minterm predicate (m i ).  A horizontal fragment R i of relation R consists of all the tuples of R which satisfy a minterm predicate m i.  Given a set of minterm predicates M, there are as many horizontal fragments of relation R as there are minterm predicates.  Set of horizontal fragments also referred to as minterm fragments. Slide 18

19 Primary Horizontal Fragmentation PROJ1 =  LOC=“Montreal” (PROJ) PROJ2 =  LOC=“New York” (PROJ) PROJ3 =  LOC=“Paris” (PROJ) Slide 19 Primary Horizontal Fragmentation of Relation PROJ

20 PHF – Algorithm Given:A relation R, the set of simple predicates Pr Output:The set of fragments of R = {R 1, R 2,…,R w } which obey the fragmentation rules. Preliminaries :  Pr should be complete  Pr should be minimal Slide 20

21 Completeness of Simple Predicates A set of simple predicates Pr is said to be complete if and only if there is an equal probability of access by every application to any tuple belonging to any minterm fragment that is defined according to Pr. Example :  Assume PROJ[PNO,PNAME,BUDGET,LOC] has two applications defined on it.  Find the budgets of projects at each location.(1)  Find projects with budgets less than $200000.(2) Slide 21

22 Completeness of Simple Predicates According to (1), Pr={LOC=“Montreal”,LOC=“New York”,LOC=“Paris”} which is not complete with respect to (2). Modify Pr ={LOC=“Montreal”,LOC=“New York”,LOC=“Paris”, BUDGET≤200000,BUDGET>200000} which is complete. Slide 22

23 Minimality of Simple Predicates If a predicate influences how fragmentation is performed, (i.e., causes a fragment f to be further fragmented into, say, f i and f j ) then there should be at least one application that accesses f i and f j differently. In other words, the simple predicate should be relevant in determining a fragmentation. If all the predicates of a set Pr are relevant, then Pr is minimal. P i is relevant if and only if  acc(m i ) the access frequency of a minterm m i. Slide 23

24 Minimality of Simple Predicates Example : Pr ={LOC=“Montreal”,LOC=“New York”, LOC=“Paris”, BUDGET≤200000,BUDGET>200000} is minimal (in addition to being complete). However, if we add PNAME = “Instrumentation” then Pr is not minimal. Slide 24

25 COM_MIN Algorithm Given:a relation R and a set of simple predicates Pr Output:a complete and minimal set of simple predicates Pr' for Pr Rule 1:a relation or fragment is partitioned into at least two parts which are accessed differently by at least one application. Slide 25

26 COM_MIN Algorithm  Initialization :  find a p i  Pr such that p i partitions R according to Rule 1  set Pr' = p i ; Pr  Pr – {p i } ; F  {f i }  Iteratively add predicates to Pr' until it is complete  find a p j  Pr such that p j partitions some f k defined according to minterm predicate over Pr' according to Rule 1  set Pr' = Pr'  {p j }; Pr  Pr – {p j }; F  F  {f j }  if  p k  Pr' which is nonrelevant then Pr'  Pr' – {p k } F  F – {f k } Slide 26

27 PHORIZONTAL Algorithm Makes use of COM_MIN to perform fragmentation. Input:a relation R and a set of simple predicates Pr Output:a set of minterm predicates M according to which relation R is to be fragmented  Pr'  COM_MIN (R,Pr)  determine the set M of minterm predicates  determine the set I of implications among p i  Pr‘  Iteratively eliminate the contradictory minterms from M  M  M-m i Slide 27

28 PHF – Example Two candidate relations : PAY and PROJ. Fragmentation of relation PAY  Application: Check the salary info and determine raise.  Employee records kept at two sites  application run at two sites  Simple predicates p 1 : SAL ≤ 30000 p 2 : SAL > 30000 Pr = {p 1,p 2 } which is complete and minimal Pr'=Pr  Minterm predicates m 1 : (SAL ≤ 30000) m 2 : NOT(SAL ≤ 30000)  (SAL > 30000) Slide 28

29 PHF – Example TITLE Mech. Eng. Programmer SAL 27000 24000 PAY 1 PAY 2 TITLE Elect. Eng. Syst. Anal. SAL 40000 34000 Slide 29

30 PHF – Example Fragmentation of relation PROJ  Applications: o Find the name and budget of projects given their location. – Issued at three sites o Access project information according to budget – one site accesses ≤200000 other accesses >200000  Simple predicates  For application (1) p 1 : LOC = “Montreal” p 2 : LOC = “New York” p 3 : LOC = “Paris”  For application (2) p 4 : BUDGET ≤ 200000 p 5 : BUDGET > 200000  Pr = Pr' = {p 1,p 2,p 3,p 4,p 5 } Slide 30

31 Fragmentation of relation PROJ continued  Minterm fragments left after elimination m 1 : (LOC = “Montreal”)  (BUDGET ≤ 200000) m 2 : (LOC = “Montreal”)  (BUDGET > 200000) m 3 : (LOC = “New York”)  (BUDGET ≤ 200000) m 4 : (LOC = “New York”)  (BUDGET > 200000) m 5 : (LOC = “Paris”)  (BUDGET ≤ 200000) m 6 : (LOC = “Paris”)  (BUDGET > 200000) PHF – Example Slide 31

32 PHF – Example PROJ 1 PNOPNAMEBUDGETLOC PNOPNAMEBUDGETLOC P1Instrumentation150000Montreal P2 Database Develop. 135000 New York PROJ 3 PROJ 4 PROJ 6 PNOPNAMEBUDGETLOC P3CAD/CAM250000New York PNOPNAMEBUDGETLOC MaintenanceP4310000Paris The result of the primary horizontal fragmentation of PROJ forms six fragments FPROJ = {PROJ1, PROJ2, PROJ3, PROJ4, PROJ5, PROJ} according to the minterm predicates M. Since fragments PROJ 2, and PROJ 5 are empty, they are not depicted in Figure Slide 32

33 Completeness  Since Pr' is complete and minimal, the selection predicates are complete Reconstruction  If relation R is fragmented into F R = {R 1,R 2,…,R r } R =   R i  F R R i Disjointness  Minterm predicates that form the basis of fragmentation should be mutually exclusive. PHF – Correctness Slide 33

34 Solved Problem Slide 34

35 Solved Problem Slide 35

36 Solved Problem Slide 36

37 Solved Problem Slide 37

38 Solved Problem Slide 38

39 Thank You Slide 39


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