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Slide 1 4/21/2015 Lecture 8 Lecture 8: Schema Refinement and Normal Forms; Physical Design and Tuning Schema Refinement –Motivation –Anomalies, Redundancy.

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Presentation on theme: "Slide 1 4/21/2015 Lecture 8 Lecture 8: Schema Refinement and Normal Forms; Physical Design and Tuning Schema Refinement –Motivation –Anomalies, Redundancy."— Presentation transcript:

1 Slide 1 4/21/2015 Lecture 8 Lecture 8: Schema Refinement and Normal Forms; Physical Design and Tuning Schema Refinement –Motivation –Anomalies, Redundancy –Decomposition: a good solution –Keys and Functional Dependencies (FDs) –BCNF and Redundancy –Lossless Decompositions –Dependency Preserving Decompositions, Projections –Third Normal Form Physical Design –Performance and the workload –Choosing Indexes Identifying useful indexes, Too many indexes, How indexes are chosen –More Schema Refinement Denormalization, Vertical and Horizontal Decomposition Tuning the database and tuning queries

2 Slide 2 4/21/2015 Lecture 8 Learning Objectives LO8.1: Identify update, insertion and deletion anomalies LO8.2: Identify possible keys given an instance LO8.3: Identify possible functional dependencies in a relation LO8.4: Determine all keys in a schema LO8.5: Decompose a schema into BCNF schemas

3 Slide 3 4/21/2015 Lecture 8 Review We began the course with the life cycle of database applications First came Requirements Analysis from the customer We learned how to transform an RA into an ER diagram Then we transformed ER diagrams into relational schemas –and went on to implement the application by loading the data and writing SQL statements But different ER diagrams can lead to different relational schemas. This week we study which schemas are best.

4 Slide 4 4/21/2015 Lecture 8 What is Schema Refinement? Schema Refinement is the study of what should go where in a DBMS, or, which schemas are best to describe an application. For example, consider this schema Versus this one: Which schema do you think is best? Why? EID Name DeptID DeptName A01 Ali 12 Wing A12 Eric 10 Tail A13 Eric 12 Wing A03 Tyler 12 Wing EmpDept Emp EID Name DeptID A01 Ali 12 A12 Eric 10 A13 Eric 12 A03 Tyler 12 Dept DeptID DeptName 12 Wing 10 Tail

5 Slide 5 4/21/2015 Lecture 8 What’s wrong?* The first problem students usually identify with the EmpDept schema is that it combines two different ideas: employee information and department information. But what is wrong with this? 1.If we separated the two concepts we could save space. 2.Combining the two ideas leads to some bad anomalies. These two problems occur because DeptID determines DeptName, but DeptID is not a key. Let’s look into the anomalies further.

6 Slide 6 4/21/2015 Lecture 8 Anomalies, Redundancy* What anomalies are associated with EmpDept? Update Anomalies: Insertion Anomalies: Deletion Anomalies: EID Name DeptID DeptName A01 Ali 12 Wing A12 Eric 10 Tail A13 Eric 12 Wing A03 Tyler 12 Wing EmpDept

7 Slide 7 4/21/2015 Lecture 8 LO8.1:Practice Anomalies, Redundancies* Identify anomalies associated with this schema. Include update, insertion and deletion anomalies. EnrollStud(StudID, ClassID, Grade, ProfID, StudName) Why do these anomalies occur?

8 Slide 8 4/21/2015 Lecture 8 Decomposition: A good solution The intergalactic standard solution to the redundancy problem is to decompose redundant schemas, e.g., EmpDept becomes The secret to understanding when and how to decompose schemas is Functional Dependencies, a generalization of keys. When we say "X determines Y" we are stating a functional dependency. Emp EID Name DeptID A01 Ali 12 A12 Eric 10 A13 Eric 12 A03 Tyler 12 Dept DeptID DeptName 12 Wing 10 Tail

9 Slide 9 4/21/2015 Lecture 8 Review Keys Note that EID being a key* of EmpDept means that the values of EID are unique, and EID is minimal. Remember: you cannot determine keys from an instance, only from “natural” information or from a domain expert. Let’s practice keys by identifying possible keys in an instance. *sometimes called a candidate key EID Name DeptID DeptName A01 Ali 12 Wing A12 Eric 10 Tail A13 Eric 12 Wing A03 Tyler 12 Wing EmpDept

10 Slide 10 4/21/2015 Lecture 8 LO8.2:Identify Possible Keys* Identify all possible Keys based on this instance: Time Flight Plane Origin Destination 9:57AM 157 abc SEA PDX 10:42AM 233 def PDX SEA 11:44AM 155 des ORD ATL 12:44PM 244 xdy ATL PDX 1:43PM 074 xyz SEA ATL 2:44PM 233 def PDX ATL 3:55PM 455 eff MSP SEA 5:44PM 120 ikk MSP PDX 7:55PM 233 abf CHI SEA

11 Slide 11 4/21/2015 Lecture 8 Functional Dependencies A key like EID has another property: If two rows have the same EID, then they have the same value of every other attribute. We say EID functionally determines all other attributes and write this Functional Dependency (FD): EID  Name, DeptID, DeptName Is Name  DeptID true? –No, because rows 2 and 3 have the same Name but not the same DeptID. EID Name DeptID DeptName A01 Ali 12 Wing A12 Eric 10 Tail A13 Eric 12 Wing A03 Tyler 12 Wing EmpDept

12 Slide 12 4/21/2015 Lecture 8 Functional Dependencies, ctd. Do you see any more FDs in EmpDept? –Yes, the FD DeptID  DeptName DEFINITION: If A and B are sets of attributes in a relation, we say that A (functionally) determines B, or A  B is a Functional Dependency (FD) if whenever two rows agree on A, they agree on B. In other words, the value of a row on A functionally determines its value on B. There are two special kinds of FDs: –Key FDs, X  A where X contains a key –Trivial FDs, such as Name  Name, or  Name,DeptID  DeptID EID Name DeptID DeptName A01 Ali 12 Wing A12 Eric 10 Tail A13 Eric 12 Wing A03 Tyler 12 Wing EmpDept

13 Slide 13 4/21/2015 Lecture 8 Identify (natural) FDs* What are the (natural) FDs in these relations? Identify the key FDs but ignore trivial FDs Customer(CustID, Address, City, Zip, State) EnrollStud(StudID, ClassID, Grade, ProfID, StudName, ProfName)

14 Slide 14 4/21/2015 Lecture 8 What are FDs? An FD is a generalization of the concept of key. FDs, like keys and foreign keys, are a kind of integrity constraint (IC). Like other ICs, FDs are part of a relation’s schema. For example, a schema might be: Assigned(EmpID Int, JobID Int, EmpName varchar(20), percent real, EmpID references…, JobID references…, PRIMARY KEY (EmpID, JobID)) FDs: EmpID  EmpName

15 Slide 15 4/21/2015 Lecture 8 How to determine FDs So far we have dealt with “natural” FDs. Sometimes it’s not clear what FDs apply in a relation, e.g., zip codes vs cities, or Supplier(Name, Address, Crating, Discount) – unclear what are the FDs. There are two ways to determine FDs –Infer them as “natural” FDs from your experience –You may be given them as part of the schema, by the instructor or by the customer. As with keys, you cannot determine FDs from an instance! –But you can tell if something is not an FD

16 Slide 16 4/21/2015 Lecture 8 LO8.3:Identify Possible FDs* Identify two possible non-key FDs based on this instance (identical to slide 10). Remember the possible keys for this instance are {Time}, {Plane, Dest}, {Origin, Dest} Time Flight Plane Origin Destination 9:57AM 157 abc SEA PDX 10:42AM 233 def PDX SEA 11:44AM 155 des ORD ATL 12:44PM 244 xdy ATL PDX 1:43PM 074 xyz SEA ATL 2:44PM 233 def PDX ATL 3:55PM 455 eff MSP SEA 5:44PM 120 ikk MSP PDX 7:55PM 233 abf CHI SEA

17 Slide 17 4/21/2015 Lecture 8 Reasoning about FDs EmpDept(EID, Name, DeptID, DeptName) Two natural FDs are EID  DeptID and DeptID  DeptName These two FDs imply the FD EID  DeptName –Because if two tuples agree on EID, then by the first FD they agree on DeptID, then by the second FD they agree on DeptName. The set of FDs implied by a given set F of FDs is called the closure of F and is denoted F +

18 Slide 18 4/21/2015 Lecture 8 Armstrong’s Axioms The closure of F can be computed using these axioms  Reflexivity: If X  Y, then X  Y  Augmentation: If X  Y, then XZ  YZ for any Z  Transitivity: If X  Y and Y  Z then X  Z Armstrong’s axioms are sound (they generate only FDs in F + when applied to FDs in F) and complete (repeated application of these axioms will generate all FDs in F + ).

19 Slide 19 4/21/2015 Lecture 8 Determining Keys In order to determine if X is a key of a relation R, use this algorithm, which computes the attribute closure of X: AttClos = X; // Note: X is a set of attributes Repeat until there is no change  If there is an FD U  V with U  AttClos, then set AttClos = AttClos ∪ V  AttClos=R if and only if X is a key

20 Slide 20 4/21/2015 Lecture 8 LO8.4:Determining the keys of R* Given the schema: R(A,B,C,D,E) BC  A, DE  C. What are all the keys of this schema? Hint: any key must include A, BC or DE. Why?

21 Slide 21 4/21/2015 Lecture 8 Redundancy and FDs Consider the FDs in these examples: EmpDept(EID, Name, DeptID, DeptName) Assigned(EmpID, JobID, EmpName, percent) EnrollStud(StudID, ClassID, Grade) Remember that every non-key FD is associated with some redundancy, or anomalies, and vice-versa. Our game plan is to use non-key FDs to decompose any relation into a form that has no redudancy, a so-called normal form.

22 Slide 22 4/21/2015 Lecture 8 Boyce-Codd Normal Form (BCNF)* A relation is said to be in Boyce-Codd Normal Form if all its FDs are either trivial FDs or key FDs. Which of these relations is BCNF? EmpDept(EID, Name, DeptName) Assigned(EmpID, JobID, EmpName, percent) EnrollStud(StudID, ClassID, grade) Each BCNF relation with a single key looks like this Key Nonkey Attr1 Nonkey Attr2 Nonkey Attrk   

23 Slide 23 4/21/2015 Lecture 8 BCNF and Redundancy Theorem: BCNF relations have no redundancy. Proof: A relation has redundancy if there is an FD between two sets of attributes, say DeptID  DeptName, and there can be repeated entries of data for those attributes. For example, consider (12,Wing) in this example: But if the relation is BCNF, then the FD must be a key FD, and DeptID must be a key. Thus any pair such as (12,Wing) can appear only once. DeptID DeptName (Other attributes) 12 Wing 10 Tail 12 Wing

24 Slide 24 4/21/2015 Lecture 8 Decomposition into BCNF Here is an algorithm for decomposing an arbitrary relation R into a collection of BCNF relations: 1.If R is not in BCNF and X  A is a non-key FD, then decompose R into R  A and XA. 2.If R  A and/or XA is not in BCNF, recursively apply step 1.

25 Slide 25 4/21/2015 Lecture 8 Decomposing to BCNF* Given the schema EnrollStud(StudID, ClassID, Grade, ProfID, StudName) including its natural FDs, decompose it into BCNF relations.

26 Slide 26 4/21/2015 Lecture 8 LO8.5: Decomposing into BCNF* Given the schema MedsLabelDrug (Prescr#, CustID, Label, Drug), with FDs Prescr#  Label, Label  Drug decompose it into BCNF relations.

27 Slide 27 4/21/2015 Lecture 8 Where are we? We’ve accomplished a lot! –We began with a relational schema –We identified (redundancy, anomaly) problems with it –We learned how to use FDs to eliminate those problems with decompositions into BCNF. –Along the way, we learned a powerful tool: how to determine keys from FDs. There are two steps left –Showing that the BCNF decompositions do not lose information. –Discovering that they may lose FDs, and how to deal with that.

28 Slide 28 4/21/2015 Lecture 8 Lossless Decompositions Some decompositions lose information. Suppose we got carried away and further decomposed Enroll(StudID,ClassID,Grade) into StudGrade(StudID, Grade) and ClassGrade(ClassID, Grade) Here a row (123,B) in StudGrade means that student 123 got a B in some course, and (386,A) in ClassGrade means that some student got an A in course 386. But now we have no way of knowing which student got which grade in which class. This decomposition is lossy. It contains less information than the original schema. We want to generate only lossless decompositions when we design our databases.

29 Slide 29 4/21/2015 Lecture 8 Lossless Decompositions Definition:A decomposition of a schema R with FDs F, into attribute sets X and Y, is lossless with respect to F if for every instance r of R that satisfies F r =  X (r) ⋈  Y (r) In other words, we can recover r from the natural join of the decomposed versions of r.

30 Slide 30 4/21/2015 Lecture 8 Example of a Lossless Decomposition EID Name DeptID DeptName A01 Ali 12 Wing A12 Eric 10 Tail A13 Eric 12 Wing A03 Tyler 12 Wing R=EmpDept = r X=EID,Name,DeptID EID Name DeptID A01 Ali 12 A12 Eric 10 A13 Eric 12 A03 Tyler 12 Y=DeptID,DeptName DeptID DeptName 12 Wing 10 Tail =X(r)=X(r) =Y(r)=Y(r) EID Name DeptID DeptName A01 Ali 12 Wing A12 Eric 10 Tail A13 Eric 12 Wing A03 Tyler 12 Wing r = =  X (r) ⋈  Y (r)

31 Slide 31 4/21/2015 Lecture 8 Example of a Lossy Decomposition StudID ClassID Grade 123 CS386 A 456 CS410 A R = Enroll = r StudID Grade 123 A 456 A X =StudID, Grade =X(r)=X(r) ClassID Grade CS386 A CS410 A Y =ClassID, Grade =Y(r)=Y(r) StudID ClassID Grade 123 CS386 A 123 CS410 A 456 CS410 A 456 CS386 A =  X (r) ⋈  Y (r) r  Note that the join has extra rows. This always happens in lossy decompositions

32 Slide 32 4/21/2015 Lecture 8 Producing only Lossless Decompositions In our design of database schemas we certainly want to produce only lossless decompositions. Fortunately this is easy to guarantee. Theorem: The decomposition of R with respect to FDs F into attribute sets R 1 and R 2 is lossless if and only if R 1  R 2 contains a key for either R 1 or R 2. Proof: Page 620 in the text. Corollary: The BCNF decomposition algorithm produces only lossless decompositions. Proof: In this case F includes the FD X  A and the decomposition is into R 1 =R  A and R 2 =XA. Then R 1  R 2 = X is a key for XA.

33 Slide 33 4/21/2015 Lecture 8 Where are we? In CS 3/586 we have learned how to transform A Requirements Analysis into an ER Diagram into a Relational Schema and to transform that losslessly into a BCNF schema. We recall from a previous picture that BCNF tables are particularly simple, so this looks like a perfect solution to a very general problem. But real schemas are not always BCNF. There is one more complexity to deal with.

34 Slide 34 4/21/2015 Lecture 8 Dependency Preserving Decompositions Decompositions should preserve FDs. FDs are business requirements that must be enforced. Consider an example: –Emp(Addr,City,State,Zip) ACS  Z, Z  S –Keys are ACS and ACZ. Consider the BCNF decomposition: (Address, City, Zip) ( Zip,State) –This is BCNF but it does not preserve ACS  Z –Consider the values ( 7315 SW84, Portland, 97223), ( 97223, OR), ( 7315 SW84, Portland, 00000), ( 00000, OR)

35 Slide 35 4/21/2015 Lecture 8 Third Normal Form Some schemas do not have a lossless, dependency preserving, decomposition into BCNF schemas. Because of this dilemma, researchers created another normal form called Third Normal Form (3NF), with the property that every schema has a lossless dependency preserving decomposition into 3NF schemas. A schema R with FDs F is in Third Normal Form if for every X  A in F, one of these is true: –X  A is a trivial FD (i.e., X contains A) –X  A is a key FD (i.e. X contains a key) –A is a part of some key for R Definition of BCNF! BCNF 3NF

36 Slide 36 4/21/2015 Lecture 8 Conclusion Almost all schemas in real life can be decomposed into BCNF schemas that preserve all FDs. In this case, life is wonderful. But every once in a while we get a schema like Emp(Addr,City,State,Zip) ACS  Z, Z  S Recall that its keys are ACS and ACZ. There is no decomposition into BCNF that preserves FDs! On the other hand, this schema is 3NF. Check it!

37 Slide 37 4/21/2015 Lecture 8 Conclusion, ctd. So in the rare case that we don’t have an ideal decomposition ( lossless, dependency preserving, into BCNF), rest assured that we can decompose into 3NF instead of BCNF and have lossless and dependency preservation. The proof of this assertion is in section

38 Slide 38 4/21/2015 Lecture 8 Physical Database Design Database development involves three steps 1.ER design 2.Schema refinement (normalization) and view definition  This defines the conceptual and external schemas 3.Physical Design  Choose indexes  More schema refinement  Consider denormalizing  Vertical and horizontal decomposition  Tuning the database and tuning queries  Deciding how the data will be stored on disks (omitted)

39 Slide 39 4/21/2015 Lecture 8 Performance and the Workload Note that ER design and normalization are logical concepts, while physical design is driven by performance needs. First the user tells you what information (logical) should be in the database, then s/he tells you how efficiently the database should perform (physical). We'll start the physical design process by learning how to choose indexes for a workload. We want to know: What Indexes might improve performance? What algorithms would they enable? What indexes are not useful together?

40 Slide 40 4/21/2015 Lecture 8 Example A B+ tree index on amount enables an index retrieval of tuples satisfying I.amount >1000. –But if there are many such tuples (the index is not selective) it may need to be clustered. An index on party enables an index retrieval of tuples satisfying C.party='IND'. –Again, selectivity matters. An index on C.commid or I.commid –enables an Index Nested Loop Join, but it might not be efficient if there are many tuples in the outer table. –Speeds up a Merge Sort join if one or both indexes are clustered Given an index on C.commid, an index on C.party is not useful and similarly for indexes on I.commid and I.amount. SELECT C.commname, I.donorname FROM comm C JOIN indiv I USING commid WHERE I.amount>1000 AND C.party='IND';

41 Slide 41 4/21/2015 Lecture 8 Too many indexes Why not declare all useful indexes? 1.The optimizer may not be able to support the plans you have in mind  Get to know your optimizer – use EXPLAIN 2.Indexes take up space  Though nowadays this is not a big problem 3.Indexes slow updates 4.Some indexes are not useful together 5.The optimizer will be slower because it has more choices 6.Indexes take time to create

42 Slide 42 4/21/2015 Lecture 8 Choosing indexes in the real world As illustrated on the previous two pages, choosing indexes is an extremely complex task. The big 3 commercial DBMSs provide utilities to do the work for you –Microsoft: AutoAdmin –DB2: Autonomic Computing –Oracle: Automatic Database Diagnostic Monitor

43 Slide 43 4/21/2015 Lecture 8 Automated Index Selection An algorithm for choosing indexes: –Input: schema, workload, performance requirements –Output: An index configuration whose cost (to execute the workload) is minimal. –Complexity: For a single table with 10 attributes, there are 30,240 different 5-attribute indexes. How do we choose among all those possibilities? –Consider only single- or two- attribute indexes. –Consider indexes only on relevant attributes –Still need to prune search space intelligently Computing the cost of a workload is very expensive – why?

44 Slide 44 4/21/2015 Lecture 8 More Schema Refinement We have studied one kind of schema refinement, namely normalizing a schema by decomposing it into 3NF or BCNF schemas. This is part of logical design. Physical design, driven by performance goals, includes other types of schema refinement, which we will study now. These include de-normalization (!), vertical decomposition and horizontal decomposition.

45 Slide 45 4/21/2015 Lecture 8 De-Normalization Recall the relation –CustState(CustID, Address, City, Zip, State) Here is its BCNF decomposition/Normalization –Cust(CustID,Address, City, Zip) State(Zip,State) Suppose we have done the normalization and the query SELECT C.CustID,C.Address, C.City, C.Zip, S.State FROM Cust C, State S WHERE C.Zip = S.Zip; is a frequent and important query in the company.

46 Slide 46 4/21/2015 Lecture 8 De-Normalization, ctd. The join query will be expensive, even if we declare indexes (which will be costly too). A possible solution is to denormalize the tables back to CustState. –Then the previous query will run much more quickly What are the disadvantages of denormalization? –Space wasted But space is cheap nowadays –Anomalies when data changes But zip codes and states are unlikely to change In real shops, denormalization is done to improve performance, even when data is likely to change.

47 Slide 47 4/21/2015 Lecture 8 Vertical Decomposition Consider the BCNF relation Emp(EID, Address, City, State, Wage, DeptID) Suppose that the HR department issues queries about EID, Address, City and State and the rest of the company issues queries about EID, Wage and DeptID. What is the advantage of storing the Emp information in these two relations? EmpHR(EID, Address, City, State) EmpComp(EID, Wage, DeptID) –All the queries will run faster because they process smaller tables. For obvious reasons this is called a vertical decomposition

48 Slide 48 4/21/2015 Lecture 8 Horizontal Decomposition Consider again the relation Emp(EID, Address, City, State, Wage, DeptID) Now suppose that most Emp queries are from the Washington or Oregon branches of the company, who issue queries about Washington or Oregon employees, respectively. Surely you see the advantage of storing the Emp information in two relations, EmpOR and EmpWA, consisting of OR and WA employees, respectively. Why is this called a horizontal decomposition?

49 Slide 49 4/21/2015 Lecture 8 Masking Decompositions with Views If someone in the company wants to issue a query about the old Emp relation, or if there is old software that uses the Emp relation, this is possible with the use of a view, for example CREATE VIEW Emp AS SELECT * FROM EmpOR UNION SELECT * FROM EmpWA;

50 Slide 50 4/21/2015 Lecture 8 Tuning the Database We have described the steps a DBA takes during initial physical design of a database, driven by performance requirements: choosing indexes, denormalization, and physical storage and refining schemas. These steps continue throughout the life of a database, because everything about the database changes: queries and their importance, schemas, and data. Changing the design of a database during the life of a database is called tuning. –Tuning also involves other steps such as updating statistics and reclustering tables. Tuning is driven by two kinds of information –Utilities that generate performance statistics E.g., disk usage, response times –User complaints Hopefully utilties will warn the DBA of problems before users complain.

51 Slide 51 4/21/2015 Lecture 8 Tuning Queries Sometimes a utility or a customer will identify a specific query as a problem (poor respose time and/or excessive use of resources). What should you do? The first step: is it the fault of the DBMS? –Check to see how much time/resources the DBMS is using vs the network, the OS, etc. The next step is to use EXPLAIN/SHOW PLAN, etc to find out what plan the optimizer is using to execute the query, then tune the query. There are various techniques to tune queries: –Rewrite the query to use existing indexes –Simplify the query, e.g., by eliminating DISTINCT, GROUP BY/HAVING clauses, or eliminating temporary relations –Flatten nested queries (already studied) –Alter the index configuration (already studied)

52 Slide 52 4/21/2015 Lecture 8 Rewriting a query to use existing indexes Consider the query SELECT E.EID FROM Emp E WHERE E.salary=1000 OR E.age=25; Suppose there are selective indexes on salary and age, but the optimizer is scanning the entire table. You could rewrite the query as a UNION SELECT E.EID FROM Emp E, Dept D WHERE E.salary=1000 UNION SELECT E.EID FROM Emp E, Dept D WHERE E.age=25;

53 Slide 53 4/21/2015 Lecture 8 Practice: Simplifying Queries* Can you simplify these queries? SELECT DISTINCT E.EID FROM Emp E WHERE E.salary > 1000; SELECT AVG(E.salary) FROM Emp E WHERE E.salary > 1000 GROUP BY E.age HAVING E.age=25;

54 Slide 54 4/21/2015 Lecture 8 Practice: Eliminate temp relations Usually (not always) an optimizer is more efficient without temporary relations. Can you combine these into one query? SELECT E.sal, D.dno INTO Temp FROM Emp E, Dept D WHERE E.dno=D.dno AND D.mgrname=‘Joe’; SELECT T.dno, AVG(T.sal) FROM Temp T GROUP BY T.dno;

55 Slide 55 4/21/2015 Lecture 8 LO8.1:Exercise* Identify anomalies associated with this schema. Include update, insertion and deletion anomalies. Assigned(EmpID, JobID, EmpName, percent) Why do these anomalies occur?

56 Slide 56 4/21/2015 Lecture 8 LO8.2: Exercise* Identify some possible keys based on this instance. Include only keys with one or two attributes: T W X Y Z s A 1 B 2 t X 5 X 4 u Z 9 Z 2 s A 2 B 1 r X 1 B 2

57 Slide 57 4/21/2015 Lecture 8 LO8.3: EXERCISE* Identify two possible non-key FDs based on this instance (identical to the previous slide): T W X Y Z s A 1 B 2 t X 5 X 4 u Z 9 Z 2 s A 2 B 1 r X 1 B 2

58 Slide 58 4/21/2015 Lecture 8 LO8.4: EXERCISE* Given the schema R(A,B,C,D,E) AB  D, CD  AE What are all the keys of this schema?

59 Slide 59 4/21/2015 Lecture 8 LO8.5: EXERCISE* Given the schema LoansBC(Branch#, Loan#, Amt, Assets, Cust#, CustName) including the FDs Branch#  Assets, Cust#  CustName, decompose it into BCNF relations.


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