120 lines
3.4 KiB
Markdown
120 lines
3.4 KiB
Markdown
## Lecture 1 (13:00) - Data Modelling
|
||
|
||
1. Using the Crow’s foot notation, draw an Entity-Relationship diagram for the following specification. You diagram must show both the functionalities and membership classes of relationships.
|
||
|
||
Entities:
|
||
Student
|
||
Programme
|
||
Module
|
||
|
||
Student >-|---|-- Programme (M:1)(m:m)
|
||
Programme --|---|-< Module (1:M)(m:m)
|
||
|
||
1. Transform the conceptual data model shown below into a logical (relational) model.
|
||
- 
|
||
- `Barrack`
|
||
Attributes:
|
||
BarrackID (Primary Key)
|
||
…
|
||
|
||
- `Regiment`
|
||
Attributes:
|
||
RegimentID (Primary Key)
|
||
…
|
||
BarrackID (Foreign Key)
|
||
|
||
- `Soldier`
|
||
Attributes:
|
||
SoldierID (Primary Key)
|
||
…
|
||
ReigmentID (Foreign Key)
|
||
|
||
1. Convert Table 17.1 to Second Normal Form
|
||
- 
|
||
- `Order`
|
||
Attributes:
|
||
OrderNo (Primary Key)
|
||
PartNo (Foreign Key)
|
||
Quantity
|
||
Cost
|
||
|
||
- `Part`
|
||
Attributes:
|
||
PartNo (Primary Key)
|
||
Description
|
||
Price
|
||
|
||
1. Write a definition of Third Normal Form
|
||
Third normal form determines a table that is in second normal form, with no transitive dependencies. All fields in the table rely on the primary key.
|
||
|
||
## Tutorial 2 (15:00) - Mobile Phones and Energy Generation
|
||
|
||
1. A mobile phone company requires you to implement a database with the following two relations:
|
||
|
||
customer(cust-number, name, age, nationality)
|
||
handset(handsetID, size, cust-number*)
|
||
|
||
Where the attribute cust-number is nine characters long, the attribute handsetID is six characters and both of the attributes age and size are integers.
|
||
|
||
The database must comply with the following constraints.
|
||
|
||
- A customer must have a name.
|
||
- Unless otherwise specified, the nationality of a customer is assumed to be British.
|
||
|
||
Write SQL commands which create the two relations. (You are not required to populate the relations with data.)
|
||
|
||
```sql
|
||
CREATE TABLE customer(
|
||
cust-number CHAR(9) PRIMARY KEY
|
||
name VARCHAR(20) CONSTRAINT name_not_null NOT NULL
|
||
age INTEGER
|
||
nationality VARCHAR(20) DEFAULT "British"
|
||
);
|
||
```
|
||
|
||
```sql
|
||
CREATE TABLE handset(
|
||
handsetID VARCHAR(6) PRIMARY KEY
|
||
size INTEGER
|
||
cust-number FOREIGN KEY REFERENCES customer(cust-number)
|
||
);
|
||
```
|
||
|
||
1. The United Kingdom is facing an energy crisis. Suppose that the government has set up a database of fuels and countries that can supply them. A sample of the data in the database is shown in Tables 18.1, 18.2 and 18.3.
|
||
|
||
- i. Define a SQL subquery which lists the names of countries which supply carbon neutral fuels. Do not use joins.
|
||
|
||
```sql
|
||
SELECT name FROM country
|
||
WHERE countryID = (
|
||
SELECT countryID FROM supplies
|
||
WHERE fuelID = (
|
||
SELECT fuelID FROM fuel
|
||
WHERE carbon-neutral = "Yes"
|
||
)
|
||
);
|
||
```
|
||
|
||
- ii. Define a SQL join query which also lists names of countries which supply carbon neutral fuels. Do not use subqueries.
|
||
|
||
```sql
|
||
SELECT name FROM country
|
||
INNER JOIN supplies
|
||
INNER JOIN fuel
|
||
ON country.countryID = supplies.countryID
|
||
ON supplies.fuelID = fuel.fuelID
|
||
WHERE fuel.carbon-neutral = "Yes";
|
||
```
|
||
|
||
```sql
|
||
SELECT DISTINCT name
|
||
FROM country, supplies, fuel
|
||
WHERE country.countryID = supplies.countryID
|
||
AND supplies.fuelID = fuel.fuelID
|
||
AND carbon-neutral = "Yes";
|
||
```
|
||
|
||
- iii. If the amount of data stored in the tables was huge, would it be better to use your answer to question (i) or your answer to question (ii). Briefly justify your answer.
|
||
|
||
The answer to ii would be better for large datasets; they are faster since less queries are being used, hence less CPU time and memory access.
|