辅导案例-FIT9132

  • June 25, 2020

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

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