Distributed Information Systems - Exercise 4

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1 WS 2004/2005 Distributed Information Systems - Exercise 4 Schema Fragmentation Date: Return: Horizontal Fragmentation Given the following relational database: Employees EmpNo EmpName Salary E1 Federer E2 Klum E3 Spears E4 Montoya 6000 E5 Beckham E6 Lopez ProjectAssignments EmpNo ProjNo MonthsSpent E1 P1 3 E2 P2 4 E2 P4 2 E3 P3 8 E4 P1 6 E5 P2 9 E6 P4 4.5 Projects ProjNo ProjName Location Budget P1 Training Geneva P2 Advertising Geneva P3 Management Munich P4 Customer Support Zurich Typical queries from applications are: At the controlling department located in Geneva: A1 SELECT P.ProjNo, P.Budget, PA.EmpNo, PA.MonthsSpent FROM Projects P, ProjectAssignments PA WHERE P.ProjNo = PA.ProjNo AND P.Budget < A2 SELECT PA.EmpNo, PA.ProjNo, PA.MonthsSpent FROM Employees E, ProjectAssignments PA WHERE E.EmpNo = PA.EmpNo AND E.Salary >

ProjectAssignments EmpNo ProjNo MonthsSpent E1 P1 3 E2 P2 4 E2 P4 2 E3 P3 8 E4 P1 6 E5 P2 9 E6 P4 4.

2 At the human resource department located in Munich: A3 SELECT * FROM Employees E A4 SELECT E.*, PA.MonthsSpent FROM Employees E, ProjectAssignments PA WHERE E.EmpNo = PA.EmpNo At the sales coordination department located in Zurich: A5 SELECT * FROM Projects WHERE Location!= Geneva a) For the relations Employees and Projects, determine minimal sets of simple predicates using the MinFrag algorithm. b) What are the corresponding fragments of the two relations Employees and Projects? c) How would you fragment ProjectAssignments horizontally? Give a short discussion of your decision. 2. Vertical Fragmentation Let Q = {Q1, Q2, Q3, Q4} be a set of queries, A = {A1, A2, A3, A4} a set of attributes, and S = {S1, S2, S3} a set of sites. The matrix in the Figure 1 describes the attribute usage values and the one in Figure 2 gives the application access frequencies. Assume that A1 is the primary key. Use the bond energy algorithm and the vertical partitioning algorithm to obtain a vertical fragmentation of the set of attributes in A. A1 Q Q Q Q Figure1: This matrix shows which attributes each query uses. S1 S2 S3 Q Q Q Q Figure 2: This matrix shows how often each site executes queries. 2

= Geneva a) For the relations Employees and Projects, determine minimal sets of simple predicates using the MinFrag algorithm.

3 WS 2004/2005 Distributed Information Systems - Exercise 4 - Solution Schema Fragmentation 1. Horizontal Fragmentation d) For the relations Employees and Projects, determine minimal sets of simple predicates using the MinFrag algorithm. For the table Projects the following simple predicates appear in the application queries: p1 = (Budget < ) p2 = (Location!= Geneva ) Using the MinFrag algorithm we get: Step 1: add p1 = (Budget < ) Step 2: add p2 = (Location!= Geneva ) fragments ok, p1 added not added as (p1 AND p2) and (p1 AND p2) don t create any new {Budget < } is the minimal complete set of predicates. For the table Employees there is only a single simple predicate. The minimal complete set contains therefore only one predicate: {Salary > 11500} e) What are the corresponding fragments of the two relations Employees and Projects? Employees, E.F1, E.F2 EmpNo EmpName Salary E1 Federer E2 Klum E3 Spears E4 Montoya 6000 E5 Beckham E6 Lopez

For the table Projects the following simple predicates appear in the application queries: p1 = (Budget < 110000) p2 = (Location!

4 Projects, P.F1, P.F2 ProjNo ProjName Location Budget P1 Training Geneva P2 Advertising Geneva P3 Management Munich P4 Customer Support Zurich f) How would you fragment ProjectAssignments horizontally? Give a short discussion of your decision. There are two possibilities to fragment ProjectAssignments horizontally: a) by deriving from Employees and b) by deriving from Projects. a) PA.E.F1, PA.E.F2 b) PA.P.F1, PA.P.F2 EmpNo ProjNo MonthsSpent EmpNo ProjNo MonthsSpent E1 P1 3 S1 E1 P1 3 E2 P2 4 S2 E2 P2 4 E2 P4 2 S3 E2 P4 2 E3 P3 8 S4 E3 P3 8 E4 P1 6 S5 E4 P1 6 E5 P2 9 S6 E5 P2 9 E6 P4 4.5 S7 E6 P4 4.5 Applications A1 & A2 in Geneva access S1, S2, S4, S5, S6, and S7. Application A4 in Munich accesses S1 to S7. Conclusion: All four fragments (PA.E.F1, PA.E.F2, PA.P.F1, and PA.P.F2) are accessed from Geneva and Munich. Therefore, we would not fragment ProjectAssignments and store it either in Geneva or Munich depending on the access frequencies of the applications. 4

horizontally? Give a short discussion of your decision. There are two possibilities to fragment ProjectAssignments horizontally: a) by deriving from Employees and b) by deriving from Projects. a) PA.

5 2. Vertical Fragmentation Use the bond energy algorithm and the vertical partitioning algorithm to obtain a vertical fragmentation of the set of attributes in A. A1 Q Q Q Q S1 S2 S3 Sum Q Q Q Q Bond energy algorithm: A1 is the primary key and has to be in all fragments. Therefore, we don t have to consider it. We can calculate the following attribute affinity matrix: bond(a2, A3) = 45* *55 + 0*0 = 4500 bond(a2, A4) = 45*0 + 45*0 + 0*40 = 0 bond(a3, A4) = 45*0 + 55*0 + 0*40 = 0 Calculate the contribution of the column depending on its position: A4-A2-A3 cont(_, A4, A2) = bond(_, A4) + bond (A4, A2) - bond(_, A2) = 0 A2-A4-A3 cont(a2, A4, A3) = bond(a2, A4) + bond(a4, A3) - bond(a2, A3) = = A2-A3-A4 cont(a3, A4, _) = bond(a3, A4) + bond(a4, _) - bond(a3, _) = 0 Both A4-A2-A3 and A2-A3-A4 look good Q Q Q Q S1 S2 S3 Sum Q Q Q Q

Therefore, we don t have to consider it.

6 accesses(fragment 1: {A2}): 0 accesses(fragment 2: {A3, A4}): 50 accesses(fragment 1 AND fragment 2): 45 sq = accesses(fragment 1: {A2, A3}): 55 accesses(fragment 2: {A4}): 40 accesses(fragment 1 AND fragment 2): 0 sq = 2200 accesses(fragment 1: {A2, A4}): 40 accesses(fragment 2: {A3}): 10 accesses(fragment 1 AND fragment 2): 45 sq = The two partitions are therefore {A1, A4} and {A1, A2, A3}. 6

AND fragment 2): 0 sq = 2200 accesses(fragment 1: {A2, A4}): 40 accesses(fragment 2: {A3}): 10

7 The same calculation with all attributes: We can calculate the following attribute affinity matrix: A1 A A A We can randomly choose the first two columns. We choose A1, A2. Two determine the position of A3 first calculate the bond energy between two columns: bond(a1, A2) = 55* * * *0 = 2175 bond(a2, A3) = 15* * *55 + 0*0 = 4725 bond(a1, A3) = 55* * * *0 = 2325 Calculate the contribution of the column depending on its position: A3-A1-A2: cont(_, A3, A1) = bond(_, A3) + bond(a3, A1) - bond(_, A1) = = 2325 A1-A3-A2: cont(a1, A3, A2) = bond(a1, A3) + bond(a3, A2) - bond(a1, A2) = = 4875 A1-A2-A3: cont(a2, A3, A_) = bond(a2, A3) + bond(a3, A_) - bond(a2, A_) = = 4725 The order A1-A3-A2 leads to the highest bond energy. Same for A4: bond(a1, A4) = 55* *0 + 15*0 + 40*40 = 3800 bond(a2, A4) = 15* *0 + 45*0 + 0*40 = 600 bond(a3, A4) = 15* *0 + 55*0 + 0*40 = 600 A4-A1-A3-A2: cont(_, A4, A1) = bond(_, A4) + bond(a4, A1) bond(_, A1) = = 3800 A1-A4-A3-A2: cont(a1, A4, A3) = bond(a1, A4) + bond(a4, A3) bond (A1, A3) = = 2075 A1-A3-A4-A2: cont(a3, A4, A2) = bond(a3, A4) + bond(a4, A2) bond(a3, A2) = =

Two determine the position of A3 first calculate the bond energy between two columns: bond(a1, A2) = 55*15 + 15*45 + 15*45 + 40*0 = 2175 bond(a2, A3) = 15*15 + 45*45 + 45*55 + 0*0 = 4725 bond(a1, A3)

8 A1-A3-A2-A4: cont(a2, A4, A_) = bond(a2, A4) + bond(a4, A_) bond(a2, A_) = = 600 A4-A1-A3-A2 looks good A4 A1 A3 A2 A A A A A1 Q Q Q Q S1 S2 S3 Sum Q Q Q Q There are now several possibilities to split the table: split quality (sq) = accesses(fragment 1) * accesses(fragment 2) - accesses(fragment 1 AND fragment 2)^2 A4 A1 A3 A2 A A A A accesses(fragment 1): 0 accesses(fragment 2): 55 accesses(fragment 1 AND fragment 2): 40 sq = A4 A1 A3 A2 A A A A accesses(fragment 1): 40 accesses(fragment 2): 40 accesses(fragment 1 AND fragment 2): 15 sq = 1375 A4 A1 A3 A2 A A A A accesses(fragment 1): 50 accesses(fragment 2): 0 8

accesses(fragment 2) - accesses(fragment 1 AND fragment 2)^2 A4 A1 A3 A2 A4 40 40 0 0 A1 40 55 15 15 A3 0 15 55 45 A2 0 15 45 45 accesses(fragment 1): 0 accesses(fragment 2): 55 accesses(fragment 1

9 accesses(fragment 1 AND fragment 2): 45 sq = The two partitions are therefore {A1, A4} and {A1, A2, A3}. The primary key A1 has to be in all partitions. 9

Physical Database Design and Tuning

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