comp 302 database systems - victoria · pdf filecomp 302 database systems time allowed: ......

EXAMINATIONS — 2007

MID-YEAR

COMP 302

Database Systems

Time allowed: 3 Hours

Instructions: Answer all questions.

Make sure that your answers are clear and to the point.

Calculators and printed foreign language dictionaries are allowed.

No reference material is allowed.

There are 180 marks on the exam.

Marks are shown for each question as a whole and also for their parts.

CONTENTS:

Question 1. Database Basics [20 marks]

Question 2. SQL and Relational Algebra [40 marks]

Question 3. Query Optimisation [20 marks]

Question 4. Enhanced Entity Relationship Data Model [25 marks]

Question 5. Mapping ER to Relational Data Model [20 marks]

Question 6. Functional Dependencies and Normalization [40 marks]

Question 7. Transaction Processing [15 marks]

Appendices:

A. SQL Reference

B. Student Database Schema

C. An Instance of the Student Database

D. EER Diagram

StudentId_____________

COMP302 continued 2

Question 1. Database Basics [20 marks]

a) [5 marks] Define the term database and identify its main features.

ANSWER

b) [2 marks] Define the term relational database schema (the schema of a whole relational

database).

ANSWER

c) [3 marks] Define the term relation schema key. List the properties of a relation schema key.

ANSWER


COMP302 continued 3

d) [5 marks] Define the term foreign key. List the properties of a foreign key.

ANSWER

e) [2 marks] What is a data manipulation language used for?

ANSWER

f) [3 marks] List the components (sublanguages) of the relational language SQL?

ANSWER


COMP302 continued 4

Question 2. SQL and Relational Algebra [40 marks]

Consider the Student database schema given below.

CREATE TABLE Student ( StudentId int PRIMARY KEY, Stud_Name varchar NOT NULL); CREATE TABLE Course ( CourseId char(7) PRIMARY KEY, Cour_Name varchar NOT NULL, NoOfPts int NOT NULL); CREATE TABLE Enrolled ( StudentId int NOT NULL REFERENCES Student, CourseId char(7) NOT NULL REFERENCES Course, Grade char(2), PRIMARY KEY (StudentId, CourseId)); CREATE TABLE Stud_Sport ( StudentId int NOT NULL REFERENCES Student, Sport_Name varchar NOT NULL, PRIMARY KEY (StudentId, Sport_Name));

Note: The same relational database schema is also given in the Appendix B to this exam paper. You

may find it convenient to tear it off and use when answering the following questions.

a) [14 marks] Write SQL queries for the database above to do the following:

i. [5 marks] Retrieve names of students that have enrolled the course with the name Database

Systems.

ANSWER


COMP302 continued 5

ii. [4 marks] Use either substring(string [from int] [for int]) function, or

LIKE expression, or SIMILAR TO operator to retrieve students whose name start with

capital letter S.

ANSWER

iii. [5 marks] Retrieve student id's and student names of students that passed at least one course

with a grade better than B+ (i.e. A-, A, or A+). Avoid duplicates.

ANSWER


COMP302 continued 6

b) [8 marks] Use relational algebra to write the following queries:

i. [5 marks] Retrieve names of students that have enrolled in Database Systems.

ANSWER

ii. [3 marks] Retrieve student id's and student names of students that passed at least one course

with a grade better than B+ (i.e. A-, A, or A+). Avoid duplicates.

ANSWER


COMP302 continued 7

(Spare Page for extra working)


COMP302 continued 9

i. [8 marks] Explain in English what the statement means.

ANSWER

ii. [2 marks] What will the statement return if executed against the database instance

given above?

ANSWER


COMP302 continued 10

d) [8 marks] Consider the following relational algebra expression

(StudentId, Stud_Name)FFFF(SUM NoOfPts)(Student*Enrolled*Course)

i. [5 marks] What would the relational algebra expression return if executed against the

database instance given above?

ANSWER

ii. [3 marks] Write an equivalent SQL statement.

ANSWER



Question 3. Query Optimisation [20 marks]

a) [10 marks] Consider the Student database schema given in Question 2 and Appendix B. Draw

the heuristic optimization tree that corresponds to the following SQL query:

SELECT Stud_Name FROM Student NATURAL JOIN Enrolled NATURAL JOIN Course WHERE Cour_Name = 'Database Systems';

ANSWER



b) [10 marks] Consider the SQL query

q2db=> SELECT s.StudentId, Stud_Name

FROM Student s

WHERE NOT EXISTS

((SELECT e.StudentId

FROM Enrolled e

WHERE s.Studentid = e.StudentId)

EXCEPT

(SELECT p.StudentId

FROM Stud_Sport p

WHERE Sport_Name = 'Squash' AND

s.Studentid = p.StudentId));

This is the same query you considered in Question 2 c). This query is not only logically complex,

but also very inefficient.

i. [5 marks] Justify the claim that that the query above is very inefficient.

ANSWER

ii. [5 marks] Write a SQL query that will produce the same result as the query above, but that

run more efficiently. Explain why you think your query will run more efficiently.

ANSWER



Question 4. EER Data Model [25 marks]

Consider the EER diagram on the facing page. Redraw the diagram to include the following

structural constraints:

• A division operates many departments, but each department is operated by only one division.

Each division must have at least one department and each department must belong to a division.

• A department employs many employees, but there may be a department with no employees.

• Majority of employees called “dept_emps” are employed by a department. However, there are

also employees known as “free_lance_emps” who are not assigned to any department.

• A department may own many projects, but there may be a department with no projects. Each

project is existentially and identification dependent on only one department.

• Both dept_emps and free_lance_emps can work on many projects, but there are employees who

do not work on any project. A project has at lest one employee working on it.

• One of dept_emps ménages a department. Each department must have a manager.

• One of the employees runs a division. Each division must have someone running it.

Note: The same EER diagram is also given in the Appendix D to this exam paper. You may find it

convenient to tear it off and use when answering this question.

StudentId______________


Division Department

Employee

Project

Works_On

Operates

Employs Manages

Runs

Owns

StudentId______________


ANSWER

StudentId_______________________



StudentId_______________________


Question 5. Mapping ER to Relational Data Model [20 marks]

In this question you will be asked to map a number of ER diagrams into relational database schemas

(S, I ), where S is a set of relation schemas, and I is a set of referential integrity constraints. The ER

diagrams will contain only symbols for entity and relationship types, and relationship constraints.

Attributes will be given separately in the form

Entity_Type_Name (Attribute1,…, Attributen),

with entity type keys underlined, or

Relationship_Type_Name (Attribute1,…, Attributem),

where the set of a relationship attributes can be empty.

If an attribute A is multivalued, and its values are tuples of the form (b1,…, bn), it is denoted using

the following notation

A{(B1,…, Bn)},

where Bi, 1 ≤ i ≤ n, are attributes, and bi∈dom(Bi).

Represent your relation schema in the form

N(R, K),

where R is the set of attributes and K is the set of keys of the relation schema with the name N. To

indicate that a foreign key has to be not null, use the following notation

Null (Relation_Schema_Name, Attribute) = Not.

a) [5 marks] Map the following ER diagram into a relational database schema.

Lecturer Department Works_For

N 1

Lecturer (LecturerId, Lec_Name, Position)

Works_For ( )

Department (DeparmentId, Dep_Name)

StudentId_______________________


ANSWER

b) [7 marks] Map the following ER diagram into a relational database schema.

ANSWER

Student (StudentId, St_Name, Exams{(CourseId, Grade)})

Student

StudentId_______________________


c) [8 marks] Map the following ER diagram into a relational database schema. Include all

constraints you think should be satisfied.

ANSWER

Course(CourseId, Cour_Name, NoOfPts )

Prerequisite( )

Course Prerequsite is_a_prerequsite_of

has_prerequsites

N

N

StudentId_______________________



StudentId_______________________


Question 6. Functional Dependencies and Normalization [40 marks]

a) [5 marks] Consider a relation schema N (R, K), where R = {A, B, C} and K = {AB} (AB is the

primary key). Define a set of functional dependencies F under which N is in the first but not in

the second normal form.

ANSWER

b) [5 marks] Consider a relation schema N (R, K), where R = {A, B, C, D} and K = {AB} (AB is

the primary key). Define a set of functional dependencies F under which N is in the second but

not in the third normal form.

ANSWER

StudentId_______________________


c) [5 marks] Consider a relation schema N (R, K), where R = {A, B, C} and K = {AB} (AB is the

primary key). Define a set of functional dependencies F under which N is in the third but not in

the BCNF normal form.

ANSWER

d) [5 marks] Consider a relation schema N (R, K), where R = {A, B, C} and K = {AB} (AB is the

primary key). Define a set of functional dependencies F under which N is in the BCNF normal

form.

ANSWER

StudentId_______________________


e) [20 mark] Suppose you are given a relation schema N (R, K ), where R = {A, B, C, D} and K is

the set of relation schema keys. For each of the following sets of functional dependencies,

assuming those are the only functional dependencies that hold for N, do the following:

1. Identify all keys for N.

2. Identify the highest normal form that N satisfies (e.g. 1NF, 2NF, 3NF, BCNF).

3. If N is not in 3NF, transform it into a set of at least 3NF relation schemas that

preserve attributes and functional dependencies and have a lossless join property.

4. Check whether your decomposition satisfies BCNF normal form.

i. [10 marks] F = {A→B, BC→D, A→C}.

ANSWER

ii. [10 marks] F = {B→C, D→A}.

ANSWER

StudentId_______________________



StudentId_______________________


Question 7. Transaction Processing and Concurrency Control [15 marks]

Consider the code fragments of a Java JDBC program on the facing page. Note that “…” after a

comment is in place of the code that implements the action of the comment.

a) [3 marks] Which commands determine transaction boundaries?

ANSWER

b) [3 marks] Which command performs actual locking in the book table and what is locked?

ANSWER

c) [3 marks] What will be the content of the res object if the command res.next(); returns

true?

ANSWER

d) [3 marks] Why is a try catch pair of blocks needed inside the catch block at the end of

the code?

ANSWER

e) [3 marks] What is the task of the con.setAutoCommit(false) command?

ANSWER

StudentId_______________________

COMP302 27 end

. . .

Connection con;

int j_isbn;

try {

// Establish a connection with a database

. . .

con.setAutoCommit(false);

// Do some processing

. . .

PreparedStatement retBook = con.prepareStatement("SELECT * FROM book WHERE isbn=? FOR UPDATE");

// Obtain data for the variable j_isbn

. . .

retBook.setInt(1, j_isbn);

ResultSet res = retBook.executeQuery();

if (res.next()) {

// Do some processing

. . .

// Do some database updating

. . .

}

con.commit();

con.setAutoCommit(true);

}

catch (SQLException ex) {

try {

con.rollback();

con.setAutoCommit(true);

}

catch (SQLException e) {

}

return "\nSQLException: " + ex.getMessage();

}

// Close connection and exit

. . .

*********

COMP302 Appendix 28

APPENDIX A

Simplified PostgreSQL documentation:

CREATE TABLE

CREATE TABLE table_name (

{ column_name data_type [ DEFAULT default_expr ] [ column_constraint [, ... ] ]

| table_constraint } [, ... ]

)

where column_constraint is:

[ CONSTRAINT constraint_name ]

{ NOT NULL | NULL | UNIQUE | PRIMARY KEY | CHECK (expression) |

REFERENCES reftable [ ( refcolumn ) ] [ ON DELETE action ] [ ON UPDATE action ] }

table_constraint is:

[ CONSTRAINT constraint_name ]

{ UNIQUE ( column_name [, ... ] ) |

PRIMARY KEY ( column_name [, ... ] ) |

CHECK ( expression ) |

FOREIGN KEY ( column_name [, ... ] ) REFERENCES reftable [ ( refcolumn [, ... ] ) ]

[ ON DELETE action ] [ ON UPDATE action ] }

and action is one of RESTRICT, CASCADE, SET NULL, or SET DEFAULT

SELECT

SELECT [ ALL | DISTINCT ]

* | expression [ AS output_name ] [, ...]

[ FROM from_item [, ...] ]

[ WHERE condition ]

[ GROUP BY expression [, ...] ]

[ HAVING condition [, ...] ]

[ { UNION | INTERSECT | EXCEPT } [ ALL ] select ]

[ ORDER BY expression [ ASC | DESC | USING operator ] [, ...] ]

[ FOR UPDATE [ OF tablename [, ...] ] ]

where from_item can be:

[ ONLY ] table_name [ * ] [ [ AS ] alias [ ( column_alias_list ) ] ] |

( select ) [ AS ] alias [ ( column_alias_list ) ] |

from_item [ NATURAL ] [ join_type ] JOIN from_item [ ON join_condition | USING ( join_column_list ) ]

and join_type can be:

INNER |

LEFT [ OUTER ] |

RIGHT [ OUTER ] |

FULL [ OUTER ] |

CROSS

For INNER (the default) and OUTER join types, exactly one of NATURAL, ON join_condition, or

USING ( join_column_list ) must appear. For CROSS JOIN, none of these items may appear.

CREATE VIEW

CREATE VIEW view [ ( column name list ) ] AS SELECT query

Some Data Types

integer, int, smallint

character[n], char[n], character varying[n], varchar[n], varchar

numeric, numeric[precision], numeric[precision, scale], real, double

boolean, date,

Note: [ xxx ] means xxx is optional, { xxx | yyy } means xxx or yyy.

COMP302 Appendix 29

APPENDIX B

Student Database Schema

CREATE TABLE Student ( StudentId int PRIMARY KEY, Stud_Name varchar NOT NULL); CREATE TABLE Course ( CourseId char(7) PRIMARY KEY, Cour_Name varchar NOT NULL, NoOfPts int NOT NULL); CREATE TABLE Enrolled ( StudentId int NOT NULL REFERENCES Student, CourseId char(7) NOT NULL REFERENCES Course, Grade char(2), PRIMARY KEY (StudentId, CourseId)); CREATE TABLE Stud_Sport ( StudentId int NOT NULL REFERENCES Student, Sport_Name varchar NOT NULL, PRIMARY KEY (StudentId, Sport_Name));

COMP302 Appendix 31

APPENDIX D

EER Diagram

Division Department

Employee

Project

Works_On

Operates

Employs Manages

Runs

Owns

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