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- 1. Design Problem Distributed Database Systems Fall 2012 Design problem of distributed systems: Making decisions about the placement of data and programs across the sites of a computer Distributed Database Design network as well as possibly designing the network itself. SL02 In DDBMS, the distribution of applications involves Distribution of the DDBMS software Distribution of applications that run on the database Distribution of applications will not be considered in the following; Design problem instead the distribution of data is studied. Design strategies (top-down, bottom-up) Fragmentation (horizontal, vertical) Allocation and replication of fragments, optimality, heuristics DDBS12, SL02 1/60 ¨ M. Bohlen DDBS12, SL02 2/60 ¨ M. BohlenFramework of Distribution Design Strategies/1 Dimension for the analysis of distributed systems Level of sharing: no sharing, data sharing, data + program sharing Behavior of access patterns: static, dynamic Level of knowledge on access pattern behavior: no information, partial information, complete information Top-down approach Designing systems from scratch Homogeneous systems Bottom-up approach The databases already exist at a number of sites The databases should be connected to solve common tasks Distributed database design should be considered within this general framework. DDBS12, SL02 3/60 ¨ M. Bohlen DDBS12, SL02 4/60 ¨ M. Bohlen
- 2. Design Strategies/2 Design Strategies/3 Top-down design strategy Distribution design is the central part of the design in DDBMSs (the other tasks are similar to traditional databases). Objective: Design the LCSs by distributing the data over the sites. Two main aspects have to be designed carefully: Fragmentation: Relation may be divided into a number of sub-relations, which are distributed Allocation and replication: Each fragment is stored at site(s) with ”optimal” distribution Distribution design issues Why fragment at all? How to fragment? How much to fragment? How to test correctness? How to allocate? DDBS12, SL02 5/60 ¨ M. Bohlen DDBS12, SL02 6/60 ¨ M. BohlenDesign Strategies/4 Fragmentation/1 Bottom-up design strategy What is a reasonable unit of distribution? Relation or fragment of relation? Relations as unit of distribution: If the relation is not replicated, we get a high volume of remote data accesses. If the relation is replicated, we get unnecessary replications, which cause problems in executing updates and waste disk space Might be an OK solution, if queries need all the data in the relation and data stays only at the sites that use the data Fragments of relation as unit of distribution: Application views are usually subsets of relations Thus, locality of accesses of applications is deﬁned on subsets of relations Permits a number of transactions to execute concurrently, since they will access different portions of a relation Parallel execution of a single query (intra-query concurrency) However, semantic data control (especially integrity enforcement) is more difﬁcult ⇒ Fragments of relations are (usually) appropriate unit of distribution. DDBS12, SL02 7/60 ¨ M. Bohlen DDBS12, SL02 8/60 ¨ M. Bohlen
- 3. Fragmentation/2 Fragmentation/3 Types of Fragmentation Horizontal: partitions a relation along its tuples Fragmentation aims to improve: Vertical: partitions a relation along its attributes Mixed/hybrid: a combination of horizontal and vertical fragmentation Reliability Performance Balanced storage capacity and costs Communication costs Security The following information is used to decide fragmentation: Quantitative information: cardinality of relations, frequency of queries, (a) Horizontal Fragmentation site where query is run, selectivity of the queries, etc. Qualitative information: predicates in queries, types of access of data, read/write, etc. (b) Vertical Fragmentation (c) Mixed Fragmentation DDBS12, SL02 9/60 ¨ M. Bohlen DDBS12, SL02 10/60 ¨ M. BohlenFragmentation/4 Fragmentation/5 Example of database instance Example (contd.): Horizontal fragmentation of PROJ relation PROJ1: projects with budgets less than 200 000 PROJ2: projects with budgets greater than or equal to 200 000 Data DDBS12, SL02 11/60 ¨ M. Bohlen DDBS12, SL02 12/60 ¨ M. Bohlen
- 4. Fragmentation/6 Correctness Rules of Fragmentation Example (contd.): Vertical fragmentation of PROJ relation PROJ1: information about project budgets Completeness PROJ2: information about project names and locations Decomposition of relation R into fragments R1 , R2 , . . . , Rn is complete iff each data item in R can also be found in some Ri . Reconstruction If relation R is decomposed into fragments R1 , R2 , . . . , Rn , then there should exist some relational operator that reconstructs R from its fragments, i.e., R = R1 . . . Rn Union to combine horizontal fragments Join to combine vertical fragments Disjointness If relation R is decomposed into fragments R1 , R2 , . . . , Rn and data item di appears in fragment Rj , then di should not appear in any other fragment Rk , k j (exception: primary key attribute for vertical fragmentation) For horizontal fragmentation, data item is a tuple For vertical fragmentation, data item is an attribute DDBS12, SL02 13/60 ¨ M. Bohlen DDBS12, SL02 14/60 ¨ M. BohlenIdea of Horizontal Fragmentation Information Requirements/1 Database information: Links between relations (a link models a 1:N relationship between Intuition behind horizontal fragmentation relations that are related to each other by an equality join) Every site should hold all information that is used to query the site The information at the site should be fragmented so the queries of the site run faster Horizontal fragmentation is deﬁned as selection operation, σp (R ) Example: σBUDGET <200K (PROJ ) σBUDGET ≥200K (PROJ ) Links (one to many) Cardinality of relations: card(R) DDBS12, SL02 15/60 ¨ M. Bohlen DDBS12, SL02 16/60 ¨ M. Bohlen
- 5. Information Requirements/2 02.1 Information Requirements/3 02.2Application information: Simple predicates used in queries Relation R [A1 , A2 , ..., An ] Ai θvi a simple predicate (θ ∈ {=, <, ≤, >, ≥, }, vi ∈ Di , Di is the Application information (cnt’d): domain of Ai ) Minterm selectivities For relation R we deﬁne Pr = {p1 , p2 , ..., pm } The number of tuples of the relation that would be accessed by a user Example: PNAME = ”Maintenance”, BUDGET < 200000 query, which is speciﬁed according to a given minterm predicate mi . Minterm predicates Access frequencies minterm predicates are conjunctions of simple predicates The frequency with which a user application qi accesses data. Relation R, Pr = {p1 , p2 , ..., pm } M = {m1 , m2 , ..., mr } such that M = {mi | mi = ∧pj ∈Pr pj ∗}, 1 ≤ j ≤ m, 1 ≤ i ≤ z where pj ∗ = pj or pj ∗ = ¬pj DDBS12, SL02 17/60 ¨ M. Bohlen DDBS12, SL02 18/60 ¨ M. BohlenHorizontal Fragmentation/1 Horizontal Fragmentation/2 A horizontal fragment Ri of relation R consists of all the tuples of R thatWe write Pr to denote the complete and minimal set of simple satisfy a predicate Fj :predicates generated from Pr: Rj = σFj (R ), 1 ≤ j ≤ w The set of predicates is complete if and only if any two tuples in the same fragment are referenced with the same probability by any where Fj is a selection formula, which is (preferably) a minterm predicate. application The set of predicates is minimal if and only if each simple predicate Computing the horizontal fragmentation: is relevant for determining the fragmentation and for each pair of 1. Determine Pr (80/20 rule) fragments there is at least one query that accesses the fragments 2. Compute Pr’ differently 3. Determine M 4. Minimize M (eliminating impossible minterms) DDBS12, SL02 19/60 ¨ M. Bohlen DDBS12, SL02 20/60 ¨ M. Bohlen
- 6. Horizontal Fragmentation/3 Horizontal Fragmentation/4 Example: Fragmentation of the PROJ relation Consider the following query: Find the name and budget of projects given their location. If access is only according to the location, the above set of The query is issued at all three locations predicates is complete Fragmentation based on LOC, using the set of predicates i.e., in each fragment PROJi each tuple has the same probability of {LOC = ‘Montreal ‘, LOC = ‘NewYork ‘, LOC = ‘Paris ‘} being accessed If there is a second query/application that accesses only those PROJ1 = σLOC =‘Montreal ‘ (PROJ ) PROJ2 = σLOC =‘NewYork ‘ (PROJ ) project tuples where the budget is less than $200K, the set of PNO PNAME BUDGET LOC PNO PNAME BUDGET LOC predicates is not complete. P2 DB Develop. 135000 New York P2 in PROJ2 has higher probability to be accessed P1 Instrument. 150000 Montreal P3 CAD/CAM 250000 New York PROJ3 = σLOC =‘Paris ‘ (PROJ ) PNO PNAME BUDGET LOC P4 Maintenance 310000 Paris DDBS12, SL02 21/60 ¨ M. Bohlen DDBS12, SL02 22/60 ¨ M. BohlenHorizontal Fragmentation/5 Horizontal Fragmentation/6 02.3 Example (contd.): Example (contd.): Now, PROJi , i = 1, 2, 3 will be split in two Add BUDGET < 200K and BUDGET ≥ 200K to the set of predicates fragments to make it complete. ⇒ {LOC = ‘Montreal ‘, LOC = ‘NewYork ‘, LOC = ‘Paris ‘, PROJ1 = σLOC =‘Montr ‘∧BDGT <200K (PROJ ) PROJ2 = σLOC =‘NY ‘∧BDGT <200K (PROJ ) BUDGET ≥ 200K , BUDGET < 200K } is a complete set PNO PNAME BUDGET LOC PNO PNAME BUDGET LOC Minterms to fragment the relation are given as follows: P1 Instrumnt 150000 Montreal P2 DB Devel 135000 New York PROJ3 = σLOC =‘Paris ‘∧BDGT ≥200K (PROJ ) PROJ2 = σLOC =‘NY ‘∧BDGT ≥200K (PROJ ) (LOC = ‘Montreal ‘) ∧ (BUDGET ≤ 200K ) PNO PNAME BUDGET LOC PNO PNAME BUDGET LOC (LOC = ‘Montreal ‘) ∧ (BUDGET > 200K ) P4 Maintenance 310000 Paris P3 CAD/CAM 250000 New York (LOC = ‘NewYork ‘) ∧ (BUDGET ≤ 200K ) (LOC = ‘NewYork ‘) ∧ (BUDGET > 200K ) Note that the following fragments are empty: (LOC = ‘Paris ‘) ∧ (BUDGET ≤ 200K ) σLOC =‘Paris ‘∧BDGT <200K (PROJ ) σLOC =‘Montreal ‘∧BDGT ≥200K (PROJ ) (LOC = ‘Paris ‘) ∧ (BUDGET > 200K ) DDBS12, SL02 23/60 ¨ M. Bohlen DDBS12, SL02 24/60 ¨ M. Bohlen
- 7. Derived Horizontal Fragmentation/1 02.5 Derived Horizontal Fragmentation/2 Deﬁned on a member (target) relation of a link according to a selection operation speciﬁed on its owner (source). Given a link L where owner (L ) = S and member (L ) = R, the Each link is an equijoin. derived horizontal fragments of R are deﬁned as Fragments of member relation can be deﬁned with a semijoin on fragments of owner relation. Ri = R |>< Si , 1 ≤ i ≤ w where w is the maximum number of fragments that will be deﬁned on R and Si = σFi (S ) where Fi is the formula according to which the primary horizontal fragment Si is deﬁned. DDBS12, SL02 25/60 ¨ M. Bohlen DDBS12, SL02 26/60 ¨ M. BohlenDerived Horizontal Fragmentation/3 02.7 Vertical Fragmentation/1 Objective of vertical fragmentation is to partition a relation into a set of smaller relations so that many of the applications will run on only one fragment. Given link L 1 where Vertical fragmentation of a relation R produces fragments owner (L 1) = PAY and R1 , R2 , . . . , each of which contains a subset of R’s attributes. member (L 1) = EMP Vertical fragmentation is deﬁned using the projection operation of and the relational algebra: PAY 1 = σSAL ≤30000 (PAY ) πA1 ,A2 ,...,An (R ) PAY 2 = σSAL >30000 (PAY ) we get EMP1 = EMP |>< PAY 1 Example: EMP2 = EMP |>< PAY 2 PROJ1 = πPNO ,BUDGET (PROJ ) PROJ2 = πPNO ,PNAME ,LOC (PROJ ) Vertical fragmentation has also been studied for (centralized) DBMS Smaller relations, and hence less page accesses e.g., MONET system DDBS12, SL02 27/60 ¨ M. Bohlen DDBS12, SL02 28/60 ¨ M. Bohlen
- 8. Vertical Fragmentation/2 Vertical Fragmentation/3 Two types of heuristics for vertical fragmentation exist: Vertical fragmentation is more complicated than horizontal Grouping: assign each attribute to one fragment, and at each step, fragmentation join some of the fragments until some criteria is satisﬁed. In horizontal partitioning: for n simple predicates, the number of Bottom-up approach possible minterms is 2n ; some of them can be ruled out by existing Splitting: starts with a relation and decides on beneﬁcial partitionings implications/constraints. based on the access behaviour of applications to the attributes. In vertical partitioning: for m non-primary key attributes, the number of possible fragments is equal to B (m) (= the mth Bell number), i.e., Top-down approach the number of partitions of a set with m members. Results in non-overlapping fragments For large numbers, B (m) ≈ mm (e.g., B (15) = 109 ) Optimal solution is probably closer to the full relation than to a set of small relations with only one attribute DDBS12, SL02 29/60 ¨ M. Bohlen DDBS12, SL02 30/60 ¨ M. BohlenVertical Fragmentation/4 Vertical Fragmentation/5 Given are the user queries/applications Q = (q1 , . . . , qq ) that will Application information: The major information required as input run on relation R (A1 , . . . , An ) for vertical fragmentation is related to applications (queries) Attribute Usage Matrix: Denotes which query uses which attribute: Since vertical fragmentation places in one fragment those attributes usually accessed together, there is a need for some measure that 1 iff qi uses Aj would deﬁne more precisely the notion of “togetherness”, i.e., how use (qi , Aj ) = 0 otherwise closely related the attributes are. This information is obtained from queries and collected in the Attribute Usage Matrix and Attribute Afﬁnity Matrix. The use (qi , •) vectors for each application are easy to deﬁne if the designer knows the applications that willl run on the DB (consider also the 80-20 rule) DDBS12, SL02 31/60 ¨ M. Bohlen DDBS12, SL02 32/60 ¨ M. Bohlen
- 9. Vertical Fragmentation/6 02.8 Vertical Fragmentation/7 Example: Consider relation PROJ (PNO , PNAME , BUDGET , LOC ) and queries: q1 = SELECT BUDGET FROM PROJ WHERE PNO=Value Attribute Afﬁnity Matrix: Denotes the frequency of two attributes Ai q2 = SELECT PNAME,BUDGET FROM PROJ and Aj with respect to a set of queries Q = (q1 , . . . , qn ): q3 = SELECT PNAME FROM PROJ WHERE LOC=Value aff (Ai , Aj ) = ( ref l (qk ) ∗ acc l (qk )) q4 = SELECT SUM(BUDGET) FROM PROJ WHERE LOC =Value use (q ,A )=1, k : use (qk ,Ai )=1 sites l k j Abbreviations: A1 = PNO , A2 = PNAME , A3 = BUDGET , A4 = LOC where Attribute Usage Matrix ref l (qk ) is the cost (= number of accesses to (Ai , Aj )) of query qK at site l acc l (qk ) is the frequency of query qk at site l DDBS12, SL02 33/60 ¨ M. Bohlen DDBS12, SL02 34/60 ¨ M. BohlenVertical Fragmentation/8 Vertical Fragmentation/9 Example (contd.): Let the cost of each query be ref l (qk ) = 1, and the frequency acc l (qk ) of the queries be as follows: Site1 Site2 Site3 acc1 (q1 ) = 15 acc2 (q1 ) = 20 acc3 (q1 ) = 10 Idea: Take the attribute afﬁnity matrix (AA) and reorganize the acc1 (q2 ) = 5 acc2 (q2 ) = 0 acc3 (q2 ) = 0 attribute orders to form clusters where the attributes in each cluster acc1 (q3 ) = 25 acc2 (q3 ) = 25 acc3 (q3 ) = 25 demonstrate high afﬁnity to one another. acc1 (q4 ) = 3 acc2 (q4 ) = 0 acc3 (q4 ) = 0 Bond energy algorithm (BEA) has been suggested for that purpose for several reasons: Attribute afﬁnity matrix aff (Ai , Aj ) = It is designed speciﬁcally to determine groups of similar items as opposed to a linear ordering of the items. The ﬁnal groupings are insensitive to the order in which items are presented. The computation time is reasonable (O (n2 ), where n is the number of attributes) e.g., aff (A1 , A3 ) = 1 =1 3=1 acc l (qk ) = k l acc 1 (q1 ) + acc 2 (q1 ) + acc 3 (q1 ) = 45 (q1 is the only query to access both A1 and A3 ) DDBS12, SL02 35/60 ¨ M. Bohlen DDBS12, SL02 36/60 ¨ M. Bohlen
- 10. Vertical Fragmentation/10 Vertical Fragmentation/11 Example (contd.): Cluster Afﬁnity Matrix CA after running BEA BEA: Input: AA matrix Output: Clustered AA matrix (CA) Permutation is done in such a way to maximize the following global afﬁnity mesaure (afﬁnity of Ai and Aj with their neighbors): n n Elements with similar values are grouped together, and two clusters can be identiﬁed AM = aff(Ai , Aj )[aff(Ai , Aj −1 ) + aff(Ai , Aj +1 ) + An additional partitioning algorithm is needed to identify the clusters i =1 j =1 in CA aff(Ai −1 , Aj ) + aff(Ai +1 , Aj )] Usually more clusters and more than one candidate partitioning, thus additional steps are needed to select the best clustering. The resulting fragmentation after partitioning (PNO is added in PROJ2 explicilty as key): PROJ1 = {PNO , BUDGET } PROJ2 = {PNO , PNAME , LOC } DDBS12, SL02 37/60 ¨ M. Bohlen DDBS12, SL02 38/60 ¨ M. BohlenBEA Algorithm/1 BEA Algorithm/2 In order to determine the best placement we deﬁne the contribution of a placement. Contribution of a placement: Input: The AA matrix Output: Clustered afﬁnity matrix CA which is a permutation of AA cont (Ai , Ak , Aj ) = 2 ∗ bond (Ai , Ak )+ Initialization: Place and ﬁx one of the columns of AA in CA (we 2 ∗ bond (Ak , Aj )− choose column 1 in our example) 2 ∗ bond (Ai , Aj ) Iteration: Place the remaining n-i columns in the remaining i+1 positions in the CA matrix. For each column choose the placement that makes the largest contribution to the global afﬁnity measure. Bond between a pair of attributes is deﬁned as: Row order: Order the rows according to the column ordering. bond (Ax , Ay ) = n aff (Az , Ax ) ∗ aff (Az , Ay ) z =1 DDBS12, SL02 39/60 ¨ M. Bohlen DDBS12, SL02 40/60 ¨ M. Bohlen
- 11. BEA Algorithm/3 BEA Algorithm/4 Ordering (0-3-1) :Consider the following AA matrix and the corresponding CA matrix where cont (A 0, A 3, A 1)A1 and A2 have been placed. = 2 ∗ bond (A 0, A 3) + 2 ∗ bond (A 3, A 1)–2 ∗ bond (A 0, A 1) A1 A2 A3 A4 A1 A2 = 2 ∗ 0 + 2 ∗ 4410 – 2 ∗ 0 = 8820 A1 45 0 45 0 A1 45 0 Ordering (1-3-2) : AA = A2 0 80 5 75 CA = A2 0 80 A3 45 5 53 3 A3 45 5 cont (A 1, A 3, A 2) A4 0 75 3 78 A4 0 75 = 2 ∗ bond (A 1, A 3) + 2 ∗ bond (A 3, A 2)–2 ∗ bond (A 1, A 2) = 2 ∗ 4410 + 2 ∗ 890 – 2 ∗ 225 = 10150bond (A3 , A1 ) = 45 ∗ 45 + 5 ∗ 0 + 53 ∗ 45 + 3 ∗ 0 = 4410 Ordering (2-3-4) :bond (A3 , A2 ) = 45 ∗ 0 + 5 ∗ 80 + 53 ∗ 5 + 3 ∗ 75 = 890 cont (A 2, A 3, A 4) = 2 ∗ bond (A 2, A 3) + 2 ∗ bond (A 3, A 4)–2 ∗ bond (A 2, A 4)The next step is to place A3 . = 1780 DDBS12, SL02 41/60 ¨ M. Bohlen DDBS12, SL02 42/60 ¨ M. BohlenBEA Algorithm/5 02.9 BEA Algorithm/6After adding A3 the CA matrix has the form The ﬁnal step is to divide a set of clustered attributes {A1 , ..., An } into A1 A3 A2 two sets {A1 , ..., Ai } and {Ai +1 , ..., An } such that the costs of queries 45 45 0 that access both sets are minimized. CA = 0 5 80 The appropriate split point on the diagonal must be determined: 45 53 5 0 3 75 A1 A3 A2 A4 A1 45 45 0 0After adding A4 and ordering rows the CA matrix has the form A3 45 53 5 3 A2 0 5 80 75 A1 A3 A2 A4 A4 0 3 75 78 A1 45 45 0 0 This is an optimization problem: the optimal point on the diagonal CA = A3 45 53 5 3 must be determined. A2 0 5 80 75 Note: Generalizations are needed to deal with A4 0 3 75 78 partitions located in the middle of the matrix multiple split points DDBS12, SL02 43/60 ¨ M. Bohlen DDBS12, SL02 44/60 ¨ M. Bohlen
- 12. BEA Algorithm/7 Correctness of Vertical Fragmentation Relation R is decomposed into fragments R1 , R2 , . . . , Rn AQ (qi ) = {Aj | use (qi , Aj ) = 1} attributes accessed by qi e.g., PROJ = {PNO , BUDGET , PNAME , LOC } into PROJ1 = {PNO , BUDGET } and PROJ2 = {PNO , PNAME , LOC } TQ = {qi | AQ (qi ) ⊆ TA } queries that access TA only BQ = {qi | AQ (qi ) ⊆ BA } queries that access BA only Completeness OQ = Q − {TQ ∪ BQ } queries that access TA and BA Guaranteed by the partitioning algortihm, which assigns each attribute in A to one partition CQ = q i ∈Q ∀Sj refj (qi ) ∗ accj (qi ) cost of all queries Reconstruction CTQ = refj (qi ) ∗ accj (qi ) cost of TQ queries Join to reconstruct vertical fragments qi ∈TQ ∀Sj R = R1 · · · Rn = PROJ1 PROJ2 CBQ = qi ∈BQ ∀Sj refj (qi ) ∗ accj (qi ) cost of BQ queries Disjointness COQ = qi ∈OQ ∀Sj refj (qi ) ∗ accj (qi ) cost of other queries Attributes have to be disjoint in VF. Two cases are distinguished: If tuple IDs are used, the fragments are really disjoint CTQ ∗ CBQ − COQ 2 maximize Otherwise, key attributes are replicated automatically by the system e.g., PNO in the above example DDBS12, SL02 45/60 ¨ M. Bohlen DDBS12, SL02 46/60 ¨ M. BohlenMixed Fragmentation Replication and Allocation In most cases simple horizontal or vertical fragmentation of a DB Replication: Which fragements shall be stored as multiple copies? schema will not be sufﬁcient to satisfy the requirements of the Complete Replication applications. Complete copy of the database is maintained in each site Mixed fragmentation (hybrid fragmentation): Consists of a Selective Replication horizontal fragment followed by a vertical fragmentation, or a vertical Selected fragments are replicated in some sites fragmentation followed by a horizontal fragmentation Allocation: On which sites to store the various fragments? Fragmentation is deﬁned using the selection and projection Centralized Consists of a single DB and DBMS stored at one site with users operations of relational algebra: distributed across the network Partitioned σp (πA1 ,...,An (R )) Database is partitioned into disjoint fragments, each fragment assigned πA1 ,...,An (σp (R )) to one site DDBS12, SL02 47/60 ¨ M. Bohlen DDBS12, SL02 48/60 ¨ M. Bohlen
- 13. Replication/1 Replication/2 Comparison of replication alternatives Replicated DB fully replicated: each fragment at each site partially replicated: each fragment at some of the sites Non-replicated DB (= partitioned DB) partitioned: each fragment resides at only one site Rule of thumb: read only queries If ≥ 1, then replication is advantageous, otherwise update queries replication may cause problems DDBS12, SL02 49/60 ¨ M. Bohlen DDBS12, SL02 50/60 ¨ M. BohlenFragment Allocation/1 Fragment Allocation/2 Fragment allocation problem Required information Given are: Database Information – fragments F = {F1 , F2 , ..., Fn } selectivity of fragments – network sites S = {S1 , S2 , ..., Sm } size of a fragment – and applications Q = {q1 , q2 , ..., ql } Application Information Find: the ”optimal” distribution of F to S RRij : number of read accesses of a query qi to a fragment Fj URij : number of update accesses of query qi to a fragment Fj Optimality uij : a matrix indicating which queries updates which fragments, Minimal cost rij : a similar matrix for retrievals originating site of each query Communication + storage + processing (read and update) Cost in terms of time (usually) Site Information Performance USCk : unit cost of storing data at a site Sk LPCk : cost of processing one unit of data at a site Sk Response time and/or throughput Network Information Constraints communication cost/frame between two sites Per site constraints (storage and processing) frame size DDBS12, SL02 51/60 ¨ M. Bohlen DDBS12, SL02 52/60 ¨ M. Bohlen
- 14. Fragment Allocation/3 Fragment Allocation/4 The total cost function has two components: storage and query We discuss an allocation model which attempts to processing. minimize the total cost of processing and storage meet certain response time restrictions TOC = STCjk + QPCi General Form: Sk ∈S Fj ∈F q i ∈Q min(Total Cost) Storage cost of fragment Fj at site Sk : subject to response time constraint STCjk = USCk ∗ size (Fj ) ∗ xij storage constraint processing constraint where USCk is the unit storage cost at site k Decision variable xij Query processing cost for a query qi is composed of two components: 1 if fragment Fi is stored at site Sj xij = composed of processing cost (PC) and transmission cost (TC) 0 otherwise QPCi = PCi + TCi DDBS12, SL02 53/60 ¨ M. Bohlen DDBS12, SL02 54/60 ¨ M. BohlenFragment Allocation/5 Fragment Allocation/6 Processing cost is a sum of three components: The transmission cost is composed of two components: access cost (AC), integrity contraint cost (IE), concurency control cost Cost of processing updates (TCU) and cost of processing retrievals (CC) (TCR) PCi = ACi + IEi + CCi TCi = TCUi + TCRi Access cost: Cost of updates: Inform all the sites that have replicas + a short conﬁrmation message ACi = (URij + RRij ) ∗ xij ∗ LPCk back sk ∈S Fj ∈F TCUi = uij ∗ (update message cost + acknowledgment cost) where LPCk is the unit process cost at site k Sk ∈S Fj ∈F Integrity and concurrency costs: Can be similarly computed, though depends on the speciﬁc constraints Retrieval cost: Send retrieval request to all sites that have a copy of fragments that are Note: ACi assumes that processing a query involves decomposing needed + sending back the results from these sites to the originating it into a set of subqueries, each of which works on a fragment, ..., site. This is a very simplistic model Does not take into consideration different query costs depending on TCRi = min (xjk ∗ (cost retrieval request + cost sending back result)) S k ∈S F j ∈F the operator or different algorithms that are applied DDBS12, SL02 55/60 ¨ M. Bohlen DDBS12, SL02 56/60 ¨ M. Bohlen
- 15. Fragment Allocation/7 Fragment Allocation/8 Solution Methods Modeling the constraints The complexity of this allocation model/problem is NP-complete Correspondence between the allocation problem and similar Response time constraint for a query qi problems in other areas execution time of qi ≤ max. allowable response time for qi Plant location problem in operations research Knapsack problem Storage constraints for a site Sk Network ﬂow problem Hence, solutions from these areas can be re-used storage requirement of Fj at Sk ≤ storage capacity of Sk Use different heuristics to reduce the search space F j ∈F Assume that all candidate partitionings have been determined together with their associated costs and beneﬁts in terms of query processing. Processing constraints for a site Sk The problem is then reduced to ﬁnd the optimal partitioning and placement for each relation processing load of qi at site Sk ≤ processing capacity ofSk Ignore replication at the ﬁrst step and ﬁnd an optimal non-replicated qi ∈Q solution Replication is then handeled in a second step on top of the previous non-replicated solution. DDBS12, SL02 57/60 ¨ M. Bohlen DDBS12, SL02 58/60 ¨ M. BohlenConclusion/1 Conclusion/2 Distributed design decides on the placement of (parts of the) data and programs across the sites of a computer network There are two key technical questions: fragmentation and allocation/replication of data Allocation/Replication of data Horizontal fragmentation is deﬁned via the selection operation Type of replication: no replication, partial replication, full replication σp (R ) Optimal allocation/replication modelled as a cost function under a set of constraints Rewrites the queries of each site in the conjunctive normal form and The complexity of the problem is NP-complete ﬁnds a minimal and complete set of conjunctions to determine Use of different heuristics to reduce the complexity fragmentation Vertical fragmentation via the projection operation πA (R ) Computes the attribute afﬁnity matrix and groups “similar” attributes together Mixed fragmentation is a combination of both approaches DDBS12, SL02 59/60 ¨ M. Bohlen DDBS12, SL02 60/60 ¨ M. Bohlen

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