Monash University

FIT9132

SAMPLE EXAM PAPER
SOLUTIONS


Page ​1​ of ​15


Q1. Relational Model (6 + 1 + 3 = 10 marks)

A company wishes to record the following attributes about their employees: employee ID,
department number, name, home address, education qualifications and skills which the
employee has.

A small sample of data is show below:

Employee
ID
Department
Number
Employee
Name
Home
Address
Qualification Skill
101 21 Given name:
Joe
Family name:
Bloggs
Street: 12
Wide Rd
Town: Mytown
Postcode:
1234
Bachelor of
Commerce
MBA
Project
Management
Hadoop
R
102 13 Given name:
Wendy
Family name:
Xiu
Street: 55
Narrow St
Town: Mytown
Postcode:
1234
Bachelor of
Computer
Science
Master of IT
Doctor of
Philosophy
SQL
PL/SQL
103 13 Given name:
Sarah
Family name:
Green
Street: 25
High St Rd
Town: Mytown
Postcode:
1234
Certificate IV in
Business
Administration
SQL
Java
Phyton


(a) Use this data​ to explain the difference between a simple attribute, a composite
attribute and a multivalued attribute. Your answer must include examples drawn from
this data.
[6 marks]


Simple - an attribute which cannot be subdivided eg. employeeid, department
number


Composite - an attribute which can be subdivided into additional attributes eg.
employee name, home address


Multivalued - an attribute which has many potential values eg. qualification, skill
Page ​2​ of ​15

(b) The following relations represent a publications database.

Authors write papers which are published in an edition of a journal.

Each edition of a journal is assigned a journal id and appoints an editor.

A given paper may be authored by several authors, in such cases each author is assigned
a position representing their contribution to the paper:

author (​author_id​, first_name, last_name)

author_paper (​author_id​, ​paper_id​, author_position)

paper (​paper_id​, paper_title, journal_id)

journal (​journal_id​, journal_title, month, year, editor)

● Primary keys are underlined
● editor in journal references author(author_id) – this is an author acting as the
journal editor
Write the relational algebra for the following queries (your answer must show an
understanding of query efficiency):
(i) Show all the journal titles.
[1 mark]
π ​journal_title​ (JOURNAL)

(ii) Show the paper title, journal title and month and year of publication for all papers
published before 2012
​[3 marks]
π ​paper_title, journal_title, month, year
(
(π ​journal_id, journal_title, month, year​ (σ ​year < 2012 ​(JOURNAL))
⨝
(π ​journal_id, paper_title​(PAPER))
)
OR
ANSWER1 = π ​journal_id, journal_title, month, year​ (σ ​year < 2012 ​(JOURNAL))
ANSWER2 = π ​journal_id, paper_title​(PAPER)
ANSWER3 = ANSWER1 ​⨝ ANSWER2
ANSWER4 = π ​paper_title, journal_title, month, year​ (ANSWER3)
Here ANSWER1 could be done in two steps, a select and then a project.
Page ​3​ of ​15

Q2 Database Design (20 marks)

Monash Computing Students Society (MCSS) is one of the student clubs at Monash
University.
Students are welcome to join as a member. When a student joins MCSS, a member id is
assigned, and the students first name, last name, date of birth, email and phone number
will be recorded. This club has an annual membership fee. When a member has paid
the membership fee for the current year, the current year is recorded against the year of
membership as part of their membership details.
MCSS hosts several events throughout the year. The events are currently categorised
into ​Professional Events​, ​General Events​, and ​Social Events​. MCSS would like to be
able to add further categories as they develop new events, When an event is scheduled,
MCSS assigns an event id to the event. The event date and time, description, location,
allocated budget, the ticket price and the discount rate (eg 5%) for members. Some
events are organised as free events for members. In this situation, the discount rate is
recorded as 100% for members. For all events, only members can purchase the tickets.
However, members can buy additional tickets for their friends or family at full price. For
each of the sales, the receipt number, number of tickets sold, total amount paid and the
member id are recorded.
Some events attract some sponsorships. The sponsor may be an organisation or an
individual. The sponsors provide financial support to the event. Some events may have
several sponsors. The amount of financial support provided by each sponsor is recorded
for the event. Each sponsor is identified by a sponsor id. The name, contact email and
sponsor type are also recorded. A sponsor may support several events throughout the
year.
For some events such as career night, MCSS may also invite some guest speakers to
share their experience. The database records all guests’ information, the guests full
name, email and phone number are recorded. If a guest comes from an organisation or
an individual that provides a sponsorship to any of the MCSS events (does not have to
be at the event where the guest speaks), this fact will also be recorded. A guest may be
invited to several events.
Create ​a logical level diagram using Crow’s foot notations​ to represent the "Monash
Computing Students Society" data requirements described above. Clearly state any
assumptions you make when creating the model.
This model must be created using LucidChart (​you MUST NOT use SQL Developer
Data Modeller​). After you have built your model in LucidChart, export it as a png image
and add it to the answer paper via the MS Word Insert - Picture menu.
Please note the following points:
● Be sure to include all relations, attributes and relationships (unnecessary
relationships must not be included)
● Identify clearly the Primary Keys (P) and Foreign Keys (F), as part of your
design
● Surrogate keys must not be added
● In building your model you must conform to FIT9132 modelling requirements
● The following are​ NOT required​ on your diagram
● verbs/names on relationship lines
● indicators (*) to show if an attribute is required or not
● data types for the attributes
Page ​4​ of ​15


Monash Computing Students Society (MCSS) Logical Model




Page ​5​ of ​15

Q3. Normalisation (10 marks)


The Super Electronics Invoice shown below displays the details of an invoice for the client
Alice Paul.

Super Electronics
INVOICE

Client Number: C3178713 ​Invoice No.: ​132
Client Name: Alice Paul ​Invoice​ ​Date:​ 02/11/2018
Client Address:​ 43 High Street,
Caulfield, VIC 3162
Client Phone:​ 0411 245 718

ItemID Item Name Purchase
Price
Expected
Delivery Date
Quantity Cost
316772 Soniq S55UV16B 55" 499.00 2 weeks 1 499.00
452550 Microsoft Surface Pro 1198.00 1-3 weeks 1 1198.00
483041 Delonghi Digital Coffee 299.00 Same Day 2 598.00
SUB TOTAL: $ 2295.00
DELIVERY: $145.00
ORDER TOTAL: $2440.00


Represent this form in UNF. In creating your representation you should note that Super
Electronics wish to treat the client name and address as simple attributes.

Convert your UNF to first normal form (1NF) and then continue the normalisation to third
normal form (3NF). At each normal form show the appropriate dependencies for that normal
form, if there are none write "No Dependencies"

Do not add new attributes during the normalisation​. Clearly write the relations in each
step from the unormalised form (UNF) to the third normal form (3NF). Clearly, indicate
primary keys on all relations from 1NF onwards.

[​10 marks​]

Page ​6​ of ​15


UNF
INVOICE (invoice_nbr, inv_date, client_number, client_name, client_address, client_phone,
(item_id, item_name, item_purchase_price, item_delivery_time, qty_ordered, line_cost)
sub_total, delivery_fee, order_total)

ii) Remove repeating groups and identify the primary key for each relation
1NF
INVOICE (​invoice_nbr​, inv_date, client_number, client_name, client_address, client_phone,
sub_total, delivery_fee, order_total)
INVOICE_LINE (​invoice_nbr​, ​item_id​, item_name, item_purchase_price, item_delivery_time,
qty_ordered, line_cost)

Partial Dependencies:
item_id -> item_name

iii) Remove partial dependency and identify the primary key for each relation
2NF
INVOICE (​invoice_nbr​, inv_date, client_number, client_name, client_address, client_phone,
sub_total, delivery_fee, order_total)
INVOICE_LINE (​invoice_nbr​, ​item_id​, item_purchase_price, item_delivery_time, qty_ordered,
line_cost)
ITEM (​item_id​, item_name)

Transitive Dependencies:
client_number -> client_name, client_address, client_phone

iv) Remove transitive dependency and identify the primary key for each relation
3NF
INVOICE (​invoice_nbr​, inv_date, client_number, sub_total, delivery_fee, order_total)
CLIENT (​client_number​, client_name, client_address, client_phone)
INVOICE_LINE (​invoice_nbr​, ​item_id​, item_purchase_price, item_delivery_time, qty_ordered,
line_cost)
ITEM (​item_id​, item_name)
Page ​7​ of ​15


Full Dependencies:
invoice_nbr -> inv_date, client_number, sub_total, delivery_fee, total_cost
client_number -> client_name, client_address, client_phone
invoice_nbr, item_id -> item_purchase_price, item_delivery_time, qty_ordered, line_cost
item_id -> item_name
Page ​8​ of ​15


Q4. SQL ( 6 + 10 + 10 + 4 + 4 + 6 + 10 = 50 marks)

A. SQL: 40 marks

The following relational model depicts an employee system:


The schema file to create these tables is listed in Appendix A.

Given this model and assuming the tables have been created and populated in an Oracle
database, provide the SQL statements for the following requirements.

When coding SQL you ​must ensure​ each clause you use, such as SELECT, FROM, WHERE,
GROUP BY, HAVING, ORDER BY, CREATE, ALTER etc ​starts on a new line​.



Page ​9​ of ​15

i. Display the course code, course name and duration for all those courses which are from the
course category "GEN" or "BLD", order the output with the course with the longest duration first.
Where two courses have the same duration, order their output by the course code.



[​6 marks​]
SELECT
crscode,
crsdesc,
crsduration
FROM
course
WHERE
crscategory = 'GEN'
OR crscategory = 'BLD'
ORDER BY
crsduration DESC,
crscode;

ii. For each department list the department name, the department location, the name of the
manager and the number of employees in that department. The name of the manager must be
output in a column called "MANAGERS NAME" and the number of employees must be output in a
column called "TOTAL EMPLOYEES". Order the output by the number of employees in the
department.


[​10 marks​]
SELECT
deptname,
deptlocation,
e1.empname AS "MANAGERS NAME",
COUNT(*) AS "TOTAL EMPLOYEES"
FROM
( department d
JOIN employee e1
ON d.empno = e1.empno )
JOIN employee e2
ON d.deptno = e2.deptno
GROUP BY
deptname,
deptlocation,
e1.empname
ORDER BY
COUNT(*);
Page ​10​ of ​15

iii. List for all employees, the employee number, name, birthdate and the number of ​different
courses they have registered for. Note that some employees may repeat a course, this repeat does
not count as a different course. Order the output by employee number. Sample output will have the
form (only partial shown):


[​10 marks​]

SELECT
e.empno,
empname,
TO_CHAR(empbdate, 'dd-Mon-yyyy') AS dob,
COUNT(DISTINCT r.crscode) AS crscount
FROM
employee e
LEFT JOIN registration r
ON e.empno = r.empno
GROUP BY
e.empno,
empname,
empbdate
ORDER BY
e.empno;

iv. Add a new department to the DEPARTMENT table, this department's number will be 10 higher
than the highest current department number and will be called EXAM and is located in BOSTON,
the department does not currently have a manager assigned. No sequences are available or may
be created.
[​4marks​]
INSERT INTO department VALUES (
(
SELECT
MAX(deptno)
FROM
department
) + 10,
'EXAM',
'BOSTON',
NULL
);

COMMIT;


Page ​11​ of ​15

v. The employee named KING who has a job as the only company DIRECTOR has been assigned
to manage the new EXAM department. Record this in the database.
[​4 marks​]

UPDATE department
SET
empno = (
SELECT
empno
FROM
employee
WHERE
empname = 'KING'
AND empjob = 'DIRECTOR'
)
WHERE
deptname = 'EXAM';

COMMIT;

vi. The company has decided that they wish to record, for each department, the number of
employees currently working in the department. Modify the database structure to allow this data to
be recorded. Initially, following your modification, the number of employees in each department
should be set to 0 - this will be updated at a later stage, you do not need to code this later update.
[​6 marks​]

ALTER TABLE department ADD deptcount NUMBER(3, 0) DEFAULT 0 NOT NULL;

B. NOSQL: 10 marks

(i) Given this sample data:



and this select statement

SELECT
JSON_OBJECT( '_id' VALUE empno,
'name' VALUE JSON_OBJECT(
'initial' VALUE empinit,
'familyName' VALUE empname
),
'position' VALUE empjob,
'birthDate' VALUE to_char(empbdate,'dd-mm-yyyy'),
'courseInfo' VALUE JSON_ARRAYAGG(
JSON_OBJECT(
'code' VALUE crscode,
'date' VALUE to_char(offbegindate,'dd-mm-yyyy'),
'evaluation' VALUE regevaluation
Page ​12​ of ​15

)
)
FORMAT JSON )
|| ','

FROM PAYROLL.employee NATURAL JOIN payroll.registration
GROUP BY empno, empinit, empname, empjob,empbdate
ORDER BY empname;

Write the JSON formatted text for one of the employees listed in the table.

"_id": 7876,
"name": {
"initial": "AA",
"familyName": "ADAMS"
},
"position": "TRAINER",
"birthDate": "30-12-1983",
"courseInfo": [
{
"code": "SQL",
"date": "12-04-2016",
"evaluation": 2
},
{
"code": "PLS",
"date": "11-09-2017",
"evaluation": null
},
{
"code": "JAV",
"date": "13-12-2016",
"evaluation": 5
}
]
}
[​4 marks​]

(ii) Assume that the collection name is employees, write the MongoDB command to show all
employees who have a job as ‘MANAGER’

db.collection.find({
"position": "MANAGER"
})
[​2 marks​]

(iii) Write the MongoDB command to show all employees who have a surname of ‘JONES’ or
‘SCOTT’

db.employees.find({
$or:[{"name.familyName": "JONES"},{"name.familyName":"SCOTT"}]
})
[​4 marks​]
Page ​13​ of ​15

Q5. Transaction Management (5 + 5 = 10 marks)

a. ​Given two transactions:

T1 – R(X), W(X)

T2 – R(Y), W(Y), R(X), W (X)

Where R(X) means Read(X) and W(X) means Write(X).

i. If we wish to complete both of these transactions, explain the difference
between a ​serial​ and ​non-serial ​ ordering of these two transactions.
Provide an example of each as part of your answer.
ii. What transaction ACID property does a non-serial ordering of these two
transactions potentially violate.
[ ​4+ 1​ = ​5 marks​]


i.

Serial – all of one transaction followed by all of the other

T1 R(X), T1 W(X), T2 R(Y), T2 W(Y), T2 R(X), T2 W(X)

Non-Serial – interleaving of the transactions

T1 R(X), T2 R(Y), T2 W(Y), T1 W(X), T2 R(X), T2 W(X)


ii.

Isolation or Consistency



Page ​14​ of ​15

b. A ​write through ​ database has five transactions running as listed below (the time
is shown horizontally from left to right):



At time tc a checkpoint is taken, at time tf the database fails due to a power
outage.

Explain for each transaction what recovery operations will be needed when the
database is restarted and why.
[​5 marks​]

T1 – nothing required, committed before checkpoint

T2 – ROLL FORWARD, committed after checkpoint and before fail

T3 – ROLL BACK, never reached commit

T4 – ROLL FORWARD, started after checkpoint committed before fail

T5 - ROLL BACK, never reached commit
Page ​15​ of ​15