MN405 Data and Information Management Assignment - Data Model Development and Implementation, Melbourne Institute of Technology, Australia
The purpose of this assignment is to develop data models and map Database System into a standard development environment to gain understanding of data model development. Then implement the data model using a commercially available database management system development tool.
On completion of this assignment students will be able to:
a. Understand the fundamental principles of the networking and data requirements of a network.
b. Identify organisational information requirements.
c. Model organisational information requirements using conceptual data modelling techniques.
d. Convert the conceptual data models into relational data models and verify their structural characteristics with normalisation techniques.
e. Implement and utilise a relational database using a database system.
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This assignment is to develop a data models, map and implement a Database System in a standard development environment to gain understanding of data model development and implementation.
Case Study on Mindful Body Fitness Company
1. Complete the information-level design for a database.
a. Identify and list entities described in all user requirements given above.
Identify all entities - The following entities are used in Mind full body fitness company
b. Add attributes to these entities and represent them as a collection of tables and attributes.
User Views as Tables
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c. Outline 3 business rules that describe the relationships between entities.
- A Coach can work on only one program.
- A Customer can register with a coach
- A Coach can maintain multiple customer report
- An Exercise can have multiple programs
- A Coach can have one login details
- Members can have one membership id
- A Coach can train multiple customers
d. Determine the functional dependences.
CustomerID -> FName
CustomerID -> LName
CustomerID -> Address
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e. Then normalise these tables. Make the normalization to 3NF. State for every step in the normalization, which functional dependency causes it.
First Normal Form (1NF)
The table should fulfil the first normal Form by using following rules
- Single atomic column values
- Column value should be same domain
- All columns in a table should have unique values
The above rules are satisfied in our fitness database. The fitness database of the tables is contained primary key to have unique values to maintain column values.
Second Normal Form (2NF)
The table should fulfil the Second Normal Form by using following rules
- Should fulfil the First Normal form
- Partial dependency not allowed
Already fitness database satisfied first form and all entities are has primary key and remaining columns are depends on primary key of the table.
Third Normal Form (3NF)
The table should fulfil the third Normal Form by using following rules
- Should fulfil the second Normal form
- Doesn't have Transitive dependency
Fitness database already satisfied second normal form and a non-prime attribute depends on other non-prime attributes rather than depending upon the prime attributes or primary key is called transitive dependency. In our database each entity does not have transitive dependency so it is fulfilled third normal from.
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2. Represent the structure of your database visually by using an entity-relationship (E-R) diagram.
3. Report Generation
a. Write an SQL query to generate one example of useful information that can be obtained from this database.
b. For additional 5 marks you will execute more queries (at least 2) on the table join from the database you created in Question 3 and include the screen shots of the outputs and all SQL statements.
Print Customer with payment Details
SELECT Customer.CustomerID, Customer.FirstName, Payment.PaymentID, Payment.PaymentAmount, Payment.PaymentDescrip, Payment.PaymentDate
FROM Customer INNER JOIN Payment ON Customer.CustomerID = Payment.PaymentCustomerID;
SELECT Member.MemberID, Member.MemberName, Member.MemberAddress, Member.MemberPhone, Member.MemberEmail
Coach & Customer Details
SELECT Coach.CoachID, Coach.FirstName, Customer.FirstName, Customer.CustomerID
FROM Coach INNER JOIN Customer ON Coach.CoachID = Customer.CoachID;
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