Database Systems: Design, Implementation, & Management
Database Systems: Design, Implementation, & Management
12th Edition
ISBN: 9781305627482
Author: Carlos Coronel, Steven Morris
Publisher: Cengage Learning
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Chapter 6, Problem 4P

a.

Program Plan Intro

Normalization:

The process used to minimize data redundancy and dependency in a relational database is known as normalization. The database table is divided into two or more tables and defines the relationship between those tables.

Second normal form (2NF):

  • The value of all non-primary key attributes should be dependent on the primary key attribute.
    • If any attribute is depending on the partial primary key then it should determine the other attributes for an instance of the entity.
  • The partial dependencies should be removed from the data model.

Third normal form (3NF):

  • The value of any non-primary key attributes will not depend on any other non-primary key attributes.
    • If any non-primary key attributes depend on any other non-primary key attribute then it should be moved or deleted.
    • It is termed as transitive dependency.

Partial dependency:

A partial dependency exists at that time of an attributes depends only a part of primary key. This dependency is related with 1st normal form.

Transitive dependency:

A transitive dependency exists at that time of an attributes depends on another attribute which is not part of primary key.

Functional dependency:

An association between two attributes or two set of attributes in a same relational database table, which is having some constraints is known as functional dependency.

  • In a table one attribute is functionally dependent on another attribute to take one value.

a.

Expert Solution
Check Mark

Explanation of Solution

Construct the dependency diagram with all partial and transitive dependencies:

The relational schema for given STUDENT table is given below:

STUDENT(STU_NUM, STU_LNAME, STU_MAJOR, DEPT_CODE, DEPT_NAME,

DEPT_PHONE, ADVISOR_LNAME, ADVISOR_OFFICE, ADVISOR_BLDG, ADVISOR_PHONE, STU_GPA, STU_HOURS, STU_CLASS)

  • Here, “STU_NUM” indicates the primary key.

The representation of dependency diagram with all transitive dependencies is shown below:

Database Systems: Design, Implementation, & Management, Chapter 6, Problem 4P , additional homework tip  1

Explanation:

In the above dependency diagram,

  • The transitive dependency is,

    DEPT_CODE -> (DEPT_NAME, DEPT_PHONE, COLLEGE_NAME)

    ADV_OFFICE -> (ADV_BUILDINGS)

    STU_HOURS -> (STU_CLASS)

b.

Program Plan Intro

Normalization:

The process used to minimize data redundancy and dependency in a relational database is known as normalization. The database table is divided into two or more tables and defines the relationship between those tables.

Second normal form (2NF):

  • The value of all non-primary key attributes should be dependent on the primary key attribute.
    • If any attribute is depending on the partial primary key then it should determine the other attributes for an instance of the entity.
  • The partial dependencies should be removed from the data model.

Third normal form (3NF):

  • The value of any non-primary key attributes will not depend on any other non-primary key attributes.
    • If any non-primary key attributes depend on any other non-primary key attribute then it should be moved or deleted.
    • It is termed as transitive dependency.

Partial dependency:

A partial dependency exists at that time of an attributes depends only a part of primary key. This dependency is related with 1st normal form.

Transitive dependency:

A transitive dependency exists at that time of an attributes depends on another attribute which is not part of primary key.

Functional dependency:

An association between two attributes or two set of attributes in a same relational database table, which is having some constraints is known as functional dependency.

  • In a table one attribute is functionally dependent on another attribute to take one value.

b.

Expert Solution
Check Mark

Explanation of Solution

Construct the dependency diagram:

The new dependency diagram is represented by removing all transitive dependencies in STUDENT table.

First table:

The relational schema for first table is given below:

STUDENT(STU_NUM, STU_LNAME, STU_MAJOR, DEPT_CODE, ADVISOR_NUM STU_GPA, STU_HOURS, STU_CLASS)

  • Here, “STU_NUM” indicates the primary keys and “ADVISOR_NUM” indicates the foreign key.
  • The relation is in third normal form (2NF), since there is transitive dependency in this table.

The representation of dependency diagram in first table is shown below:

Database Systems: Design, Implementation, & Management, Chapter 6, Problem 4P , additional homework tip  2

Second table:

The relational schema for second table is given below:

MAJOR(MAJOR_CODE, DEPT_CODE, MAJOR_DESCRIPTION)

  • Here, “MAJOR_CODE” indicates the primary key.
  • The relation is in third normal form (3NF), since there is no transitive dependency and no repeated attributes.

The representation of dependency diagram in second table is shown below:

Database Systems: Design, Implementation, & Management, Chapter 6, Problem 4P , additional homework tip  3

Third table:

The relational schema for third table is given below:

BUILDING(BLDG_CODE, BLDG_NAME, BLDG_MANAGER)

  • Here, “BLDG_CODE” indicates the primary key.
  • The relation is in third normal form (3NF), since there is no transitive dependency and no repeated attributes.

The representation of dependency diagram in third table is shown below:

Database Systems: Design, Implementation, & Management, Chapter 6, Problem 4P , additional homework tip  4

Fourth table:

The relational schema for fourth table is given below:

DEPARTMENT(DEPT_NAME, DEPT_PHONE, COLL_CODE)

  • Here, “DEPT_NAME” indicates the primary key.
  • The relation is in third normal form (3NF) but it does not meet the BCNF, since there is no transitive dependency and no repeated attributes.

The representation of dependency diagram in fourth table is shown below:

Database Systems: Design, Implementation, & Management, Chapter 6, Problem 4P , additional homework tip  5

Fifth table:

The relational schema for fifth table is given below:

COLLEGE(COLL_CODE, COLL_NAME)

  • Here, “COLL_CODE” indicates the primary key.
  • The relation is in third normal form (3NF), since there is no transitive dependency and no repeated attributes.

The representation of dependency diagram in fifth table is shown below:

Database Systems: Design, Implementation, & Management, Chapter 6, Problem 4P , additional homework tip  6

Sixth table:

The relational schema for sixth table is given below:

ADVISOR(ADV_NUM, ADV_LASTNAME, ADV_OFFICE, ADV_BUILDING, ADV_PHONE)

  • Here, “ADV_NUM” indicates the primary key.
  • The dotted transitive dependency line specifies that dependency is interpretation.
  • The relation is in third normal form (2NF), since there is transitive dependency in this table.

The representation of dependency diagram in sixth table is shown below:

Database Systems: Design, Implementation, & Management, Chapter 6, Problem 4P , additional homework tip  7

c.

Explanation of Solution

The representation of Crow’s Foot Entity Relational Diagram (ERD) is shown below:

The following data model shows the solution for the given question.

Database Systems: Design, Implementation, & Management, Chapter 6, Problem 4P , additional homework tip  8

Explanation:

  • In the above data model, the college owns more than one department.
    • The “COLLEGE” entity contains the “COLL_CODE” and “COLL_NAME” attributes.
      • The primary key of “COLL_CODE” entity is “INV_NUM”.
    • The “DEPARTMENT” entity contains the “DEPT_CODE”, “DEPT_NAME”, “DEPT_PHONE” and “COLL_CODE” attributes.
      • The primary of this entity is “DEPT_CODE”.
      • The foreign key of this entity is “COLL_CODE”.
  • The department offers more than one major.
    • The “MAJOR” entity contains the “MAJOR_CODE”, “MAJOR_NAME”, and “DEPT_CODE” attributes.
      • The primary key of “MAJOR_CODE” for this entity.
      • The foreign key of this entity is “VEND_CODE”.
  • The major attracts more than one student.
    • The “STUDENT” entity contains the “STU_NUM”, “STU_LNAME”, “STU_CLASS”, “STU_HOURS”, “STU_GPA”, “MAJOR_CODE”, and “ADV_NUM” attributes.
      • The primary key of “STU_NUM” entity.
      • The foreign key of this entity is “MAJOR_CODE” and “ADV_NUM”.
  • The building houses more than one advisor.
    • The “ADVISOR” entity contains the “ADV_NUM”, “DEPT_CODE”, “BLDG_CODE”, “ADV_LNAME”, “ADV_OFFICE”, and “ADV_PHONE” attributes.
      • The primary key of “ADV_NUM” entity.
      • The foreign key of this entity is “DEPT_CODE” and “BLDG_CODE”.
    • The “BUILDING” entity contains the “BLDG_CODE”, “BLDG_NAME” and “BLDG_MANAGER” attributes.
      • The primary key of “BLDG_CODE” entity.
  • The advisor advises more than one student.

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Using the STUDENT table structure shown in Table P6.4, do the following:a. Write the relational schema and draw its dependency diagram. Identify all dependencies, including all transitive dependencies.b. Write the relational schema and draw the dependency diagram to meet the 3NF requirements to the greatest practical extent possible. If you believe that practical considerations dictate using a 2NF structure, explain why your decision to retain 2NF is appropriate. If necessary, add or modify attributes to create appropriate determinants and to adhere to the naming conventions.c. Using the results of Problem 4, draw the Crow's Foot ERD.Note: Although the completed student hours (STU_HOURS) do determine the student classification (STU_ClASS), this dependency is not as obvious as you might initially assume it to be. For example, a student is considered a junior if the student has completed between 61 and 90 credit hours.To upload and submit your assignment, click the Choose File button…
Using the STUDENT table structure shown in Table P6.4, do the following:a. Write the relational schema and draw its dependency diagram. Identify all dependencies, including all transitive dependencies.b. Write the relational schema and draw the dependency diagram to meet the 3NF requirements to the greatest practical extent possible. If you believe that practical considerations dictate using a 2NF structure, explain why your decision to retain 2NF is appropriate. If necessary, add or modify attributes to create appropriate determinants and to adhere to the naming conventions.c. Using the results of Problem 4, draw the Crow's Foot ERD.Note: Although the completed student hours (STU_HOURS) do determine the student classification (STU_ClASS), this dependency is not as obvious as you might initially assume it to be. For example, a student is considered a junior if the student has completed between 61 and 90 credit hours.
Using the STUDENT table structure shown in Table 1, do the following: a. Write the relational schema and draw its dependency diagram. Identify all dependencies, including all transitive dependencies. b. Write the relational schema and draw the dependency diagram to meet the 3NF requirements to the greatest practical extent possible. If you believe that practical considerations dictate using a 2NF structure, explain why your decision to retain 2NF is appropriate. If necessary, add or modify attributes to create appropriate determinants and to adhere to the naming conventions. c. Using the results of Problem b, draw the Crow's Foot ERD. TABLE 1: STUDENT INFO ATTRIBUTE SAMPLE SAMPLE SAMPLE SAMPLE SAMPLE NAME VALUE VALUE VALUE VALUE VALUE STU_NUM 211343 200128 199876 198648 223456 STU LNAME Stephanos Smith. Jones Ortiz McKulski STU_MAJOR Accounting Accounting Marketing Marketing Statistics DEPT CODE ACCT АССT MKTG MKTG ΜΑΤΗ DEPT NAΜΕ Accounting Accounting Marketing Marketing Mathematics…
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