Answers To Question Number - 1

Answer to question number 1
Part a
The architecture provides a greater flexibility for the development teams of the
companies which would be helpful in updating any part of an application which is independent
from the other parts.
Part b
The ability changes a logical schema without changing the external schema is called
logical data independence. In the example provided a student can easily view their results in
MyInfo but cannot change in the Students record database.
Part c
The ability changes a physical schema without changing the application program is called
logical data independence. The developer would be able to add or update data which would not
require the program of MyInfo to be changed.
Answer to question number 2
Part a
Part i
The set of attribute which would be able to identify a tuple uniquely is known as the
super key. For example (Student Number, Student Name) and (Student Number, Unit code)
Part ii

Candidate key is a super key but not a primary key. For example, Student Number and
Unit code are the candidate keys.
Part iii
Primary is the unique identity attribute of a relation which is required for each tuple in a
relation. Student Number is the primary key in provided relation.
Part b
Part i
The primary key value of entity cannot be null as stated by the entity integrity constraint.
The Registration Number cannot be duplicated in MOTOR_VEHICLE.
Part ii
The primary key value of a table can be referenced in another table as per the referential
integrity constraints. The Registration Number in MOTOR_VEHICLE can be referenced in
MOTOR_VEHICLE_DRIVER.
Answer to question number 3
Part i
π Titles (Unit)⨝ σ(UnitOffering)Offering.Sememter = 1, Offering.Year = 2020
Part ii
π StudentName(Student)CourseCode = ICT285
Part iii

σ StudentName(Student)CourseCode = ICT285 ∧ CourseCode = ICT285
Part iv
σ StudentName(Student)CourseCode = ICT285 ∧ CourseCode = ICT285
Part v
π StudentName (Student) ⨝ σ(Enrollement)⨝ σ(UnitOffering)Offering.Sememter = 2,
Offering.Year = 2019
Answer to question number 4
Part i
Select * from Employee where empGender = ‘Female’;
Part ii
Select Employee.empName, Employee.empAddress from Employee inner join Department on
Employee.deptNo = Department.deptNo where Department.deptName = ‘IT’;
Part iii
Select Employee.empName, from Employee inner join Department on Employee.deptNo =
Department.deptNo inner join Project on Project.deptNo = Department.deptNo where
Project.projName = ‘Gigantic Project’;
Part iv