lecture 05: relational database design by er- & eer-to- relational mapping dr. dang tran khanh...

Lecture 05: Relational Database Design by ER- & EER-to-Relational

Mapping

Dr. Dang Tran Khanh

M.eng. Tran Minh QuangDepartment of Information Systems

Faculty of Computer Science & Engineering

Dr. Dang Tran Khanh ([email protected]), M.eng. Tran Minh Quang ([email protected]) 2

Outline

Main Phases of Database Design Conceptual Database Design Logical Database Design

– ER- & EER-to-Relational Mapping Exercises Reading Suggestion:

– [1]: Chapters 7, 12– [2]: Chapters 15, 16


Main Phases of Database Design

Three main phases– Conceptual database design– Logical database design– Physical database design

Detailed discussions: see [1] (chapter 12)– Six phases

A simplified diagram to illustrate the main phases of database design



Conceptual database design– The process of constructing a model of the data

used in an enterprise, independent of all physical considerations

Logical database design– The process of constructing a model of the data

used in an enterprise based on a specific data model (e.g. relational), but independent of a particular DBMS and other physical considerations



Physical database design– The process of producing a description of the

implementation of the database on secondary storage; it describes the base relations, file organizations, and indexes design used to achieve efficient access to the data, and any associated integrity constraints and security measures


Conceptual Database DesignSummarization

Read [1]: chapters 3, 12 for details To build a conceptual data model of the data

requirements of the enterprise– Model comprises entity types, relationship types,

attributes and attribute domains, primary and alternate keys, structural and integrity constraints



Step 1: Identify entity types Step 2: Identify relationship types Step 3: Identify and associate attributes with entity or

relationship types Step 4: Determine attribute domains Step 5: Determine candidate, primary, and alternate key

attributes Step 6: Consider use of enhanced modeling concepts (optional

step) Step 7: Check model for redundancy Step 8: Validate conceptual model against user transactions Step 9: Review conceptual data model with user



Step 1: Identify entity types– To identify the required entity types

Step 2: Identify relationship types– To identify the important relationships that exist between the

entity types Step 3: Identify and associate attributes with entity or

relationship types– To associate attributes with the appropriate entity or

relationship types and document the details of each attribute Step 4: Determine attribute domains

– To determine domains for the attributes in the data model and document the details of each domain



Step 5: Determine candidate, primary, and alternate key attributes

– To identify the candidate key(s) for each entity and if there is more than one candidate key, to choose one to be the primary key and the others as alternate keys

Step 6: Consider use of enhanced modeling concepts (optional step)

– To consider the use of enhanced modeling concepts, such as specialization/generalization, categories (union types)

Step 7: Check model for redundancy – To check for the presence of any redundancy in the model

and to remove any that does exist



Step 8: Validate conceptual model against user transactions

– To ensure that the conceptual model supports the required transactions

Step 9: Review conceptual data model with user– To review the conceptual data model with the user to

ensure that the model is a ‘true’ representation of the data requirements of the enterprise

Check with the previous ERD exercises !!

A simplified diagram to illustrate the main phases of database design


Logical Database Design

To translate the conceptual data model into a logical data model and then to validate this model to check that it is structurally correct using normalization and supports the required transactions


Logical Database Design

Logical database design for the relational model– Step 1: Derive relations for logical data model– Step 2: Validate relations using normalization– Step 3: Validate relations against user transactions– Step 4: Define integrity constraints– Step 5: Review logical data model with user– Step 6: Merge logical data models into global model

(optional step)– Step 7: Check for future growth

ER- & EER-to-Relational Mapping

The ERD for the COMPANY database

Result of mapping the COMPANY ER schema into a relational schema



ER-– Step 1: Mapping of Regular Entity Types– Step 2: Mapping of Weak Entity Types– Step 3: Mapping of Binary 1:1 Relationship Types– Step 4: Mapping of Binary 1:N Relationship Types– Step 5: Mapping of Binary M:N Relationship Types– Step 6: Mapping of Multivalued attributes– Step 7: Mapping of N-ary Relationship Types

EER-– Step 8: Options for Mapping Specialization or Generalization.– Step 9: Mapping of Union Types (Categories)


ER-to-Relational Mapping

Step 1: Mapping of Regular (strong) Entity Types– Entity --> Relation– Attribute of entity --> Attribute of relation– Primary key of entity --> Primary key of relation– Example: We create the relations EMPLOYEE,

DEPARTMENT, and PROJECT in the relational schema corresponding to the regular entities in the ER diagram. SSN, DNUMBER, and PNUMBER are the primary keys for the relations EMPLOYEE, DEPARTMENT, and PROJECT as shown


Strong Entity Types



Step 2: Mapping of Weak Entity Types– For each weak entity type W in the ER schema with owner entity type E,

create a relation R and include all simple attributes (or simple components of composite attributes) of W as attributes of R

– In addition, include as foreign key attributes of R the primary key attribute(s) of the relation(s) that correspond to the owner entity type(s)

– The primary key of R is the combination of the primary key(s) of the owner(s) and the partial key of the weak entity type W, if any

– Example: Create the relation DEPENDENT in this step to correspond to the weak entity type DEPENDENT. Include the primary key SSN of the EMPLOYEE relation as a foreign key attribute of DEPENDENT (renamed to ESSN)

The primary key of the DEPENDENT relation is the combination {ESSN, DEPENDENT_NAME} because DEPENDENT_NAME is the partial key of DEPENDENT

– Note: CASCADE option as implemented


Weak Entity Types Partial key

Owner’s PK

PK



ER-– Step 1: Mapping of Regular Entity TypesStep 1: Mapping of Regular Entity Types– Step 2: Mapping of Weak Entity TypesStep 2: Mapping of Weak Entity Types– Step 3: Mapping of Binary 1:1 Relationship Types– Step 4: Mapping of Binary 1:N Relationship Types– Step 5: Mapping of Binary M:N Relationship Types– Step 6: Mapping of Multivalued attributes– Step 7: Mapping of N-ary Relationship Types

Transformation of binary relationships - depends on functionality of relationship and membership class of participating entity types



Mandatory membership class– For two entity types E1 and E2: If E2 is a mandatory

member of an N:1 (or 1:1) relationship with E1, then the relation for E2 will include the prime attributes of E1 as a foreign key to represent the relationship

– For a 1:1 relationship: If the membership class for E1 and E2 are both mandatory, a foreign key can be used in either relation

– For an N:1 relationship: If the membership class of E2, which is at the N-side of the relationship, is optional (i.e. partial), then the above guideline is not applicable



Assume every module must be offered by a department, then the entity type MODULE is a mandatory member of the relationship OFFER. The relation for MODULE is:

MODULE(MDL-NUMBER, TITLE, TERM, ..., DNAME)

DEPARTMENT OFFER MODULE1 N


N:1 Relationships



Optional membership classes– If entity type E2 is an optional member of the N:1

relationship with entity type E1 (i.e. E2 is at the N-side of the relationship), then the relationship is usually represented by a new relation containing the prime attributes of E1 and E2, together with any attributes of the relationship. The key of the entity type at the N-side (i.e. E2) will become the key of the new relation

– If both entity types in a 1:1 relationship have the optional membership, a new relation is created which contains the prime attributes of both entity types, together with any attributes of the relationship. The prime attribute(s) of either entity type will be the key of the new relation



One possible representation of the relationship:BORROWER(BNUMBER, NAME, ADDRESS, ...)BOOK(ISBN, TITLE, ..., BNUMBER)

A better alternative:BORROWER(BNUMBER, NAME, ADDRESS, ...)BOOK(ISBN, TITLE, ...)ON_LOAN(ISBN, BNUMBER)

ON_LOAN BORROWER BOOK 1

N


1:N (both optional)


???

[1]: Step 4, p. 195, chapter 7



N:M binary relationships: – An N:M relationship is always represented by a new relation

which consists of the prime attributes of both participating entity types together with any attributes of the relationship

– The combination of the prime attributes will form the primary key of the new relation

Example: ENROL is an M:N relationship between STUDENT and MODULE. To represent the relationship, we have a new relation:

ENROL(SNUMBER, MDL-NUMBER, DATE)


M:N



ER-– Step 1: Mapping of Regular Entity TypesStep 1: Mapping of Regular Entity Types– Step 2: Mapping of Weak Entity TypesStep 2: Mapping of Weak Entity Types– Step 3: Mapping of Binary 1:1 Relationship TypesStep 3: Mapping of Binary 1:1 Relationship Types– Step 4: Mapping of Binary 1:N Relationship TypesStep 4: Mapping of Binary 1:N Relationship Types– Step 5: Mapping of Binary M:N Relationship Step 5: Mapping of Binary M:N Relationship

TypesTypes– Step 6: Mapping of Multivalued attributes– Step 7: Mapping of N-ary Relationship Types


Transformation of recursive/involuted relationships– Relationship among different instances of the same entity– The name(s) of the prime attribute(s) needs to be changed

to reflect the role each entity plays in the relationship

PERSON MARRY

1

1

EMPLOYEE SUPERVISE

N

1

PART COMPRISE

M

N



Example 1: 1:1 involuted relationship, in which the memberships for both entities are optional

PERSON(ID, NAME, ADDRESS, ...)

MARRY(HUSBAND-ID, WIFE_ID, DATE_OF_MARRIAGE)



Example 2: 1:M involuted relationship. – If the relationship is mandatory or almost mandatory:

EMPLOYEE(ID, ENAME, ..., SUPERVISOR_ID)– If the relationship is optional:

EMPLOYEE(ID, ENAME, ...)SUPERVISE(ID, START_DATE, ..., SUPERVISOR_ID)

Example 3: N:M involuted relationshipPART(PNUMBER, DESCRIPTION, ...)

COMPRISE( MAJOR-PNUMBER, MINOR-PNUMBER, QUANTITY)



ER-– Step 1: Mapping of Regular Entity TypesStep 1: Mapping of Regular Entity Types– Step 2: Mapping of Weak Entity TypesStep 2: Mapping of Weak Entity Types– Step 3: Mapping of Binary 1:1 Relationship TypesStep 3: Mapping of Binary 1:1 Relationship Types– Step 4: Mapping of Binary 1:N Relationship TypesStep 4: Mapping of Binary 1:N Relationship Types– Step 5: Mapping of Binary M:N Relationship TypesStep 5: Mapping of Binary M:N Relationship Types– Step 6: Mapping of Multivalued attributes– Step 7: Mapping of N-ary Relationship Types




Step 6: Mapping of Multivalued attributes– For each multivalued attribute A, create a new relation R. This

relation R will include an attribute corresponding to A, plus the primary key attribute K-as a foreign key in R-of the relation that represents the entity type or relationship type that has A as an attribute

– The primary key of R is the combination of A and K. If the multivalued attribute is composite, we include its simple components

Example: The relation DEPT_LOCATIONS is created. The attribute DLOCATION represents the multivalued attribute LOCATIONS of DEPARTMENT, while DNUMBER-as foreign key-represents the primary key of the DEPARTMENT relation. The primary key of R is the combination of {DNUMBER, DLOCATION}


Multivalued Attr.



Step 7: Mapping of N-ary Relationship Types– For each n-ary relationship type R, where n>2, create a

new relationship S to represent R– Include as foreign key attributes in S the primary keys of the

relations that represent the participating entity types– Also include any simple attributes of the n-ary relationship

type (or simple components of composite attributes) as attributes of S

Example: The relationship type SUPPY in the ER below. This can be mapped to the relation SUPPLY shown in the relational schema, whose primary key is the combination of the three foreign keys {SNAME, PARTNO, PROJNAME}

ER-to-Relational MappingFIGURE 4.11 Ternary relationship types(a) The SUPPLY relationship

Note: if the cardinality constraint on any of the entity types E participating in the relationship is 1, the PK should not include the FK attributes that reference the relation E’ corresponding to E (see section 4.7 [1])


ER-to-Relational MappingSummary of Mapping Constructs & Constraints

Correspondence between ER and Relational Models

ER Model Relational ModelEntity type “Entity” relation1:1 or 1:N relationship type Foreign key (or “relationship” relation)M:N relationship type “Relationship” relation and two foreign keysn-ary relationship type “Relationship” relation and n foreign keysSimple attribute AttributeComposite attribute Set of simple component attributesMultivalued attribute Relation and foreign keyValue set DomainKey attribute Primary (or secondary) key



ER-– Step 1: Mapping of Regular Entity TypesStep 1: Mapping of Regular Entity Types– Step 2: Mapping of Weak Entity TypesStep 2: Mapping of Weak Entity Types– Step 3: Mapping of Binary 1:1 Relationship TypesStep 3: Mapping of Binary 1:1 Relationship Types– Step 4: Mapping of Binary 1:N Relationship TypesStep 4: Mapping of Binary 1:N Relationship Types– Step 5: Mapping of Binary M:N Relationship TypesStep 5: Mapping of Binary M:N Relationship Types– Step 6: Mapping of Multivalued attributesStep 6: Mapping of Multivalued attributes– Step 7: Mapping of N-ary Relationship TypesStep 7: Mapping of N-ary Relationship Types



EER-to-Relational Mapping

Step8: Options for Mapping Specialization or Generalization.

Convert each specialization with m subclasses {S1, S2,….,Sm} and generalized superclass C, where the attributes of C are {k,a1,…an} and k is the (primary) key, into relational schemas using one of the four following options:

Option 8A: Multiple relations-Superclass and subclasses. Create a relation L for C with attributes Attrs(L) = {k,a1,…an} and PK(L) = k. Create a

relation Li for each subclass Si, 1 <= i <= m, with the attributesAttrs(Li) = {k} U {attributes of Si} and PK(Li)=k. This option works for any specialization (total or partial, disjoint of over-lapping).

Option 8B: Multiple relations-Subclass relations only Create a relation Li for each subclass Si, 1 <= i <= m, with the attributes Attr(Li) = {attributes

of Si} U {k,a1…,an} and PK(Li) = k. This option only works for a specialization whose subclasses are total (every entity in the superclass must belong to (at least) one of the subclasses)



Option 8C: Single relation with one type attribute

Create a single relation L with attributes Attrs(L) = {k,a1,…an} U {attributes of S1} U…U {attributes of Sm} U {t} and PK(L) = k. The attribute t is called a type (or discriminating) attribute that indicates the subclass to which each tuple belongs

Option 8D: Single relation with multiple type attributes Create a single relation schema L with attributes Attrs(L) = {k,a1,…an} U

{attributes of S1} U…U {attributes of Sm} U {t1, t2,…,tm} and PK(L) = k. Each ti, 1 <= i <= m, is a Boolean type attribute indicating whether a tuple belongs to the subclass Si

Option 8A is preferred !!

Example: Option 8A

Example: Option 8B

Tonnage

Example: Option 8C

EngType

Serving as the type attribute

Example: Option 8D

Boolean type attributes



Mapping of Shared Subclasses (Multiple Inheritance)

– A shared subclass, such as STUDENT_ASSISTANT, is a subclass of several classes, indicating multiple inheritance. These classes must all have the same key attribute; otherwise, the shared subclass would be modeled as a category.

– We can apply any of the options discussed in Step 8 to a shared subclass, subject to the restriction discussed in Step 8 of the mapping algorithm. Below both 8C and 8D are used for the shared class STUDENT_ASSISTANT

Example: Mapping of Shared Subclasses

Major

Course

Example: Mapping of Shared Subclasses



Step 9: Mapping of Union Types (Categories).

– For mapping a category whose defining superclasses have different keys, it is customary to specify a new key attribute, called a surrogate key, when creating a relation to correspond to the category.

– In the example below we can create a relation OWNER to correspond to the OWNER category and include any attributes of the category in this relation. The primary key of the OWNER relation is the surrogate key, which we called OwnerId

– We also include the surrogate key attribute OwnerId as FK in each relation corresponding to a superclass of the category in order to specify the correspondence in values between the surrogate key and the PK of each superclass

Example

CYear

OwnerId

Exercises (in-class)

Map the above ERD into relational schemas.Justify your choice of mapping options.


Summary

3 Main Phases of Database Design: An Overview Conceptual Database Design: A Summarization Logical Database Design

– ER- & EER-to-Relational Mapping Exercises & homework Reading Suggestion & homework: do not forget !! Next Lecture: (students’ presentation)

– Relational Algebra & Relational Calculus– [1]: Chapter 6


Exercises

Homework: – [1]: Exercise 7.5, p. 204– Map all ERDs of the previous exercises to

corresponding relational database schemas


Exercises

Homework:

Give a database scheme of a company as– customer(C-Id, name, addr, city)– product(P-Id, name, unit, description)– buying(C-id, P-id, date, quantity, price)– selling(C-id, P-id, date, quantity, price)Answer following questions using SQL


Exercises

Homework:1. Display all customers2. Display all customers but the information to be displayed

are customer id and customer name only3. Similar to question 2 but customers to be selected are in

HCM city only4. Display all the products sold on 10/03/20075. Display all the customers who supplied product P01 on

10/03/20076. Similar to question 5 but the information to be displayed

includes C-id, C-name, P-id, P-name.


Exercises

Homework:7. Display customers who buy the products supplied by

customer whose name is “Intel”

8. Display product which are concurrently bought and sold on 10/03/2007

9. Display customers who paid the greatest amount of money for buying product on 10/03/2007

10. Display customers who got the greatest amount of money by selling product to the company in the year 2006


Q&A

lecture 05: relational database design by er- & eer-to- relational mapping dr. dang tran khanh...

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