Faculty Of Computer Studies
M359 - Form A
Relational databases: theory and practice
Midterm Examination Marking Guide
Spring – 2011-2012
Date ______
Number of Exam Pages:
(10) Time Allowed:
(
2 ) Hours
Instructions:
1. This exam consists of three parts:
Part I: 5 obligatory Multiple Choice
Questions (MCQ), each worth 2
marks for a total of 10 marks
Part II: 6 short essay questions of
which you can choose any 5. Each
question is worth 8 marks for a total
of 40 marks.
Part III: 6 problems of which you
can choose any 4. Each problem
is worth 12.5 marks for a total of 50
marks.
2. Write all your answers in the
answer booklet and not in this
questions booklet.
3. Use of calculators or any other
electronic devices is not permitted in
this exam.
Good Luck
:‫ هذا االمتحان مكون من ثالثة اجزاء‬.1
‫ اسئلة اجبارية من نوع‬5 :‫الجزء االول‬
‫االختيار المتعدد مخصص لكل سؤال درجتان‬
‫ درجات‬10 ‫باجمالى‬
‫ اسئلة اختبارات قصيرة يختار‬6 :‫الجزء الثانى‬
8 ‫ منها و مخصص لكل سؤال‬5 ‫الطالب اى‬
‫ درجة‬40 ‫درجات بإجمالى‬
4 ‫ مسائل يختار الطالب اى‬6 :‫الجزء الثالث‬
‫ درجة‬12,5 ‫منها و مخصص لكل سؤال‬
‫ درجة‬50 ‫بإجمالى‬
‫ برجاء كتابة االجابات فى ورقة االجابة و‬.2
.‫ليس فى ورقة االسئلة‬
‫ ال يسمح باستخدام االلة الحاسبة او اى من‬.3
.‫االدوات االلكترونية فى هذا االمتحان‬
‫حظ سعيد‬
Part I: answer ALL of the following 5 Multiple Choice Questions (MCQ), each worth 2
marks for a total of 10 marks.
1. Referring to the three schema architecture, which schema describes how data is placed
on the hard disk?
a. External schema
b. Storage schema
c. Logical schema
d. ER schema
2. What is the property of modern databases that allows the logical schema to be
changed without having to modify the code of existing application programs?
a. Physical data independence
b. Robustness
c. Logical data independence
d. None of the above
3. What is the end product of the analysis phase of the database development life cycle?
a. Relational database design
b. Complete database implementation
c. Conceptual Data Model (CDM)
d. Requirements Specifications
4. Which of the following will be considered a valid relational table?
a. a table that has duplicate tuples
b. a table whose rows have multiple values in each field
c. a table with multiple data types appearing in a single column
d. a table with only a single field
5. Which of the following database properties is undesirable?
a. Uncontrolled redundancy
b. Lossy joins
c. Lost dependencies
d. All of the above
Part II: Answer ANY 5 of the following 6 short essay questions. Each question is worth
8 marks for a total of 40 marks.
1. Explain 4 different means by which constrains are represented in the Conceptual
Data Model (CDM).
a.
b.
c.
d.
By specifying participation conditions
By specifying the degree of relationship (1:1, 1:n, or n:m)
By specifying entity unique identifiers
By specifying additional constraints
2. Briefly explain the different options usually available when specifying the system
action in response an attempted deletion of a referenced tuple that would result in
violating of the referential integrity constraint.
Restrict: Disallow the deletion.
Cascade: propagate the deletion by deleting all referring tuples.
Default: perform the deletion and apply a predefined default value in place of the
deleted value in the referring tuples.
3. Briefly explain the main drawbacks of the client-multi-server approach of distributed
database management
a. The user must manage the connections to remote servers
b. The user must be aware of where the data is located, thus losing the desirable
property of location transparency
4. A Mother-child relation is naturally 1:m. Briefly explain what is wrong with the
following database design for this relationship?
Mother
Living -Child
Mother-ID
Child-ID
Mother-Name
Child-ID Child-Name
1
1
1
2
3
1
2
3
4
5
Alia
Laila
Sarah
Kawthar
Mariam
1
2
3
4
5
Fahd
Sally
Ramy
Lamis
Wagdy
The problem is that the relation Mother is un-normalized (not in 2NF) because
Mother-Name will not be dependent on the whole primary key, rather it will be
dependent on only part of the primary key (Mother-ID). This will result in possible
insertion, deletion and update anomalies.
5. Explain three important advantages of SQL routines.
SQL routines provide the following advantages:
a. Security: only execute privileges can be given to users, without any other
explicit privileges. This enhances security.
b. Efficiency: SQL routines can be stored on the server side, thus resulting in
lower network traffic
c. The separation of concerns: developers will spate the internal details of how a
certain procedure is implemented from the way it is used.
6. Briefly explain what is meant by a lost dependency (also called a hidden
dependency), and how this problem can be corrected.
A lost dependency is a situation that arises due to the incompatibility of normalization
and the preservation of all functional dependencies in the ER diagram. It results in
not being able to directly represent some functional dependencies in the ER diagram,
thus losing part of the problem specifications. This problem can be corrected by
specifying constraints in the additional constraints section to restore the lost
dependencies.
Part III: Answer ANY 4 of the following 6 problems. Each problem is worth 12.5
marks for a total of 50 marks.
1. Develop a Conceptual Data Model (CDM) for a student advising database, including the ER diagram, entity types, additional constraints and assumptions (limitations are not
needed). Each student has a unique ID, a name, a telephone and an address, and must
be assigned to exactly one tutor as an advisor. Not every tutor is an advisor, but a tutor
may advise many students. A student has taken zero or more courses in the past, for
which we need to record the year and semester, in which the course was taken, and the
grade achieved. Courses have unique IDs and we need to keep information about the
course ID, name and description.
E-R Diagram:
Tutor
Student
StudentCourse
Course
Entity Types:
Tutor(TID, TName)
Student (SID, SName, Tel, Address)
Student-Course (TID, SID, Year, Semester, Grade)
Course (CID, CName, Description)
Additional Constraints:
c.1 Student-Course is a weak entity type dependent on Student. Matching entities must
have matching values of SID.
c.2 Student-Course is a weak entity type dependent on Course. Matching entities must
have matching values of CID.
Assumptions: (Students may make different assumptions, but their assumptions must
match their diagrams):
a.1 A course may be stored in the database which has never been offered before.
2. Give an E-R diagram to represent the following occurrence diagram, assuming that the
diagram is representative of the application constraints.
E1
Role A
Role B
Role A
Role A
Role B
Role B
A:
Role B
Role A
E1
3. Convert the following conceptual model into a relational representations
R
C
Entity types:
C(c1, c2)
D(d1, d2)
D
relation C
c1: c1domain
c2: c2domain
primary key: c1
relation R
c1: c1domain
d1d1domain
primary key a1, b1
foreign key c1 references C
foreign key d1 references D
constraint (project C over c1) difference
(project R over c1)) is empty
constraint (project D over d1)
difference (project R over d1)) is empty
relation D
d1: d1domain
d2: d2domain
primary key d2
4. for the following Employee database, write each of the following queries in relational
algebra:
Worker (ID, WName, HourlyRate, CID)
Employee (EID, EName, Salary, CID)
Company (CID, CName)
Where ID is a unique ID over the set of all workers and all employees.
a. Get the names of all employees with a salary > 2000
project (select employee where Salary > 2000) over EName
b. Get the names of all employees, along with the name of the company in which they work
project (Employee join Company) over EName, CName
c. Get the names of all employees who are also employed as workers.
project (workers intersect Employees) over WName
5. For each of the following relations and functional dependencies, determine:
1. All candidate keys
2. The highest normal form to which the relation conforms
3. The reasons why it conforms to the normal form identified in 2. above and does
not comply with the next higher normal form (if applicable)
4. A normalization of the relation into the next higher normal form (if applicable)
Books (BookID, ISBN, BookTitle, Author)
FDs:
FD1: BookID  ISBN, BookTitle, Author
FD2: ISBN  BookID, BookTitle, Author
1. Candidate key 1: BookID
Candidate key 2: ISBN
2. BCNF
3. It is in BCNF because every non-trivial determinant is a candidate key.
4. Not applicable
6. For the same worker-employee-company database of question 4, write the following SQL
queries:
Worker (ID, WName, HourlyRate, CID)
Employee (EID, EName, Salary, CID)
Company (CID, CName)
Where ID is a unique ID over the set of all workers and all employees.
a) Write a SQL query to retrieve the names of all workers whose Hourly-Rate is above 20.
select WName
from Worker
where HourlyRate > 20
b) Write a SQL query to retrieve the names of all companies and the number of employees
employed by each company.
select CName, Count(*)
from Employee inner join Company on Employee.CID = Company.CID
group by CName
c) Write a SQL query to retrieve for each company, the name of the company and the
name(s) of the top paid employee(s) in that company.
select CName, EName
from Employee E1 inner join Company on E1.CID = company.CID
where Salary = (select Max(Salary)
from Employee E2
where E1.CID = E2.CID)
Reference Sheet
Properties of Functional Dependencies:
Property 1: combining functional dependencies
If A ↦ B and A ↦ C, then A ↦ B, C
Property 2: extending determinants
If A ↦ C and A is a subset of B, then B ↦ C
Property 3: transitivity
If A ↦ B and B ↦ C then A ↦ C
Property 4: augmentation
If A ↦ B, then A, C ↦ B, C
Normal Forms:
 1NF: A relation is in first normal form (1NF) if and only if it has no duplicate tuples
and in each tuple, each value of every attribute is a single value.
 2NF: A relation is in second normal form (2NF) if and only if every non-primary key
attribute is fully functionally dependent on the primary key.
 3NF: A relation is in third normal form (3NF) if and only if it is in 2NF and no
nonprimary key attribute is transitively dependent on the primary key.
Definition:
An attribute A is transitively dependent (TD) on a set of attributes X in a relation R if there
is a set of attributes Y such that all the following properties hold:
TD(i) X ↦ Y and Y ↦ A.
TD(ii) It is not true that Y ↦ X.
TD(iii) A is not an attribute of either X or Y.
We included TD(ii) to rule out the situation where Y is an alternate key.
 BCNF: A relation is in Boyce–Codd normal form (BCNF) if and only if each
irreducible determinant of a non-trivial FD is a candidate key.
Definitions:
A determinant A in A ↦ B is irreducible if there is no proper subset S of A such that S ↦
B
A trivial FD is one in which the right hand side is a subset of the left hand side
Definition:
Weak Entity type:
Weak entity types are those entity types that 1. Have a mandatory relationship with another entity
type where 2. The identifier of that entity type is the same as, or a subset of, the weak entity type,
and 3. That entity type is at the :1 end of this relationship.
General forms of relational algebra expressions:
select:
select <relation> where <selection condition>
project:
project <relation> over <attribute list>
combining select and project:
project (select <relation> where <selection condition>) over <attribute list>
alias:
<alias name> alias (<relational algebra expression>)
join:
<relation1> join <relation2>
rename:
<relation> rename (<old attribute1 name> as <new attribute1 name>, <old attribute2 name> as
<new attribute2 name>, ... )
join with renaming:
<relation> join (<relation> rename (<old attribute1 name> as <new attribute1 name>, <old
attribute2 name> as <new attribute2 name>, ... ))
divide:
divide <relation1> by <relation2>
union:
<relation1> union <relation2>
intersection:
<relation1> intersection <relation2>
difference:
<relation1> difference <relation2>
times:
<relation1> times <relation2>
general constraint:
constraint <condition>