Management of Information Systems Session 3

Transcription

1 Management of Information Systems Session 3 Prof. Dr. Frank G. Goethals Information Systems lifecycle The role of a manager: 1. Define requirements! System Conception Line Leadership IS Leadership Line Leadership Design & Programming & Testing Implementation Operation IS Line Leadership IS Information systems lifecycle 1

2 Objectives w.r.t. this class 1. It should be clear to you as a future manager that you have the task to reveal your requirements to people of the IT department in a rigorous way! 2. Feel the importance of an Information System (TopConsulting Game) 3. You should be able to build models: 3.1 create a correct BPMN diagram, 3.2 understand a routing scheme, 3.3 build a UML class diagram, to make clear to the IT department what you are actually doing (/where you want changes) create a relational database specification (normalisation) 3.5 get data out of the database using SQL Creating a UML class diagram 2

3 Data modelling Developing a conceptual data model = modelling the information requirements of groups of users in an organisation, reflecting the reality, without paying attention to implementation details and limitations that are caused by the computerized databasesystem that will be used in the end. Companies typically use UML class diagrams. Such a model is then translated later into a relational Database. UML class diagram (TopConsulting) Customer ClientName: string {id} ExcellentCustomer: boolean 1..1 Gives is given by 1..* Request Request-ID: integer {id} WeekReceived: integer NrOfStudents: Integer Is requested in 0..* Concerns 1..1 Training ClassName: string 3

4 UML class diagram (Loyola) Student StudNumber: integer FirstName: string LastName: string Birthyear: integer 0..* Follows Followed by 1..* Course CourseID: integer CourseName: string ContactHours: integer Teaches 0..* Taught by Professor 0..1 ProfID: integer ProfFirstName: string ProfLastName: string Building blocks of a UML class diagram (Supplier delivers products) Associations Classes provides Attributes Supplier number Supplier name Supplier Phone Price Quantity in Stock and

5 UML class diagrams provides Supplier number Supplier name Supplier Phone Price Quantity in Stock We need to find a way to store data concerning all suppliers and all products; not just one supplier or one product. Therefore, we make abstraction of details where different suppliers differ and and where different products differ. UML class diagram (Suppliers) Supplier SupNumber: integer{id} SupName: string SupPhone: integer 0..* Provides Provided by 1..* Product Price: integer ProdName: string {id} QuantityInStock: integer 5

6 UML diagrams: building blocks Classname Attributes Supplier Attribute types SupNumber: integer{id} SupName: string SupPhone: integer Signals primary key Every supplier thus has a different SupNumber UML diagrams: building blocks Asssociation: 2 directions Supplier SupNumber: integer{id} SupName: string SupPhone: integer 0..* Provides Provided by 1..* Product Price: integer ProdName: string {id} QuantityInStock: integer Cardinalities (Minimal cardinality: 0 or 1; Maximal cardinality: 1 or *) * * 6

7 Exercise 1: Music festivals You want to create a database with which music bands play on which summer festivals. On each festival there are one or more music bands. On the other hand some music bands don t play on any festival, while others play on one or more festivals. Each festival has got a unique name. You also want to keep track of the date of the festival and of the (one) contact person that is responsible for the festival organisation. You also store the year of foundation of the festival. Of the music bands, you can assume they have a unique name. You also want to store the date of foundation of the band and the official website URL of the band. Exercise 2: Pet shop You sell pets in your shop. Each animal has a name, a unique number, a gender, a breed and a birthdate. You want to be able to store for each animal what are its children. An animal can have zero, one or more children and for each animal you may have data stored on zero, one or both of its parents. You also want to store for which animal you have made publicity in some paper. For each paper in your database, you store the paper s (unique) name and the price for an advertisement. For each animal, publicity is made in zero, one or more papers. The other way around, a paper in your database has made publicity for at least one of your animals. You want to keep track of how many times you have made publicity for a specific animal in a specific paper. 7

8 UML class diagrams Unary associations Associations with an attribute?! 15 Exercise 4: Car owners You want to keep track of who owns or owned a specific car. A car owner has a unique ID, a name and an address. A registered car owner usually owns one or more cars (including second hand cars), but might exceptionally have no car for a short time span. You want to store what owner sold a car to what other owner (without storing what cars were sold exactly). Every owner bought cars from 1 or more other owners, but an owner did not necessarily sell a car. A car always belongs to exactly one car owner at a specific moment. Every car has a unique car registration number and is of a specific brand. You want to keep track of the year in which a car was bought by some owner. (You may assume a car owner only owns a specific car once: he does not sell it and buy it back later.) Cars are produced in exactly one plant. Each plant has a plant number and is located in a city. Also explicitly model that one Plant can produce cars for different brands. 8

9 3.4 Defining the tables in your database: normalization Companies store data in tables All tables together form a relational database Data for the Entity Type Student Field = Attribute Field (Attribute) value Record = Entity Last Name Studentnr DoB Major Smith /11/84 IS Mclaughlin /1/85 FIN Smith /1/81 MKT McGuyver /3/88 IS 9

10 - Must be unique Primary key - Thus: in your table with data: in the column of the primary key, every row should have a different value. A value that appears on one row in that column cannot appear on another row in that column. What if we just enter all information in one table? Imagine you have a book store Problems? 1. Redundancy (customer name, phone, book name, author, price) 2. Adding a potential customer who has not ordered yet Adding a book recently received from supplier (i.e., it has not been ordered by any customer) 3. Deleting an order (such as order # 4) 4. Modifying an attribute (Changing price of book named selfish gene ) Cause? Data is not structured properly (i.e., un-normalized) 10

11 Companies store data in tables All tables together form a relational database Logical Structure of the database: Tables are connected using Foreign Keys Primary Key Primary Key Supplier(SupplierNumber, SupplierName, SupplierPhone) Foreign Key 1 supplier can make many products Each product can be made by only one supplier Product(ProductID, ProductPrice, ProductQuantity, SupplierNumber ) 11

12 Primary Key Logical Structure of the database: Tables are connected using Foreign Keys Primary Key Foreign Key Foreign Key 1 customer can order many books 1 book can be ordered by many customers Primary Key Textual Representation of Tables (Relations) Customer (CustomerID, Name, Phone) Book (ISBN, Author, BookName, Price) Order (OrderNumber, CustomerID, ISBN, Date) Relational Databases Every table is called a relation. Supplier(SupplierNumber, SupplierName, SupplierPhone) Product(ProductID, ProductPrice, ProductQuantity, SupplierNumber) 12

13 Normalisation of relations Normalisation is a process that is applied to a set of relations in order to distribute the attribute types across several (new) relations so that the data model does not contain contradictions or redundancies and so that the use of the data does not cause anomalies. 1st normalisation step 2nd normalisation step 3rd normalisation step 0NF 1NF 2NF 3NF First normal form - 1NF A relation is in first normal form if all attribute types are atomic, single-valued attribute types. Single-valued: the attribute can only have one value e.g. a customer can only have one name Atomic: the attribute type cannot be split up meaningfully in smaller pieces e.g. a customer s first name CANNOT be split up meaningfully in smaller pieces <-> his address CAN be split up meaningfully. 13

14 First normal form - 1NF R 01 (Dnr, Dname, Dlocation) Assumption: a department can have several locations (and in one location there can be several departments) Solution: for an attribute type that is multi-valued: remove the attribute type Dlocation from the relation and create a new relation where Dlocation is stored together with the primary key from the relation where Dlocation was removed (Dnr) R 11 (Dnr, Dname) R 12 (Dnr, Dlocation) First normal form - 1NF R 01 (Enr, Ename, Project (Pnr, Pname, Pduration, Worked)) R 11 (Enr, Ename) R 12 (Enr, Pnr, Pname, Pduration, Worked) 14

15 Second normal form - 2NF A relation is in 2NF if the relation is in 1NF and every nonprime attribute is completely functionally dependent upon every possible primary key of the relation. Non-prime attribute: <-> A prime attribute is an attribute that is a part of primary key of the relation. A primary key is a minimal collection of attribute types in a relation for which it can be said that with each value of the key exactly one entity is indicated. Second normal form - 2NF A relation is in 2NF if the relation is in 1NF and every nonprime attribute is completely functionally Unique dependent number upon every possible primary key of the relation. Non-unique name Number name Complete functional dependence: 1. A collection of attributes Y is functionally dependent upon a collection of attributes X if, at every moment, the values of the X-component of an entity allow to determine the values of the Y-component: X Y. 2. A functional dependence is a complete functional dependence if dropping an attribute from X would imply that the functional dependence is lost. {Enr, Pnr} Worked {Enr, Pnr} PName 15

16 Second normal form - 2NF A relation is in 2NF if the relation is in 1NF and every non-prime attribute is completely functionally dependent upon every possible primary key of the relation. R 11 (Enr, Ename) R 12 (Enr, Pnr, Pname, Pduration, Worked) Solution: the non-prime attributes that are partially functionally dependent have to be moved to another, new relation which also includes the primary key upon which the attributes are completely functionally dependent. R 21 (Enr, Ename) R 22 (Enr, Pnr, Worked) R 23 (Pnr, Pname, Pduration) Third normal form - 3NF A relation is in 3NF if the relation is in 2NF and there are no non-prime attributes transitively dependent upon a possible primary key. Transitively dependent: X Z and Z Y Y is said to be transitively dependent upon X if a collection of attributes Z exists for which it is correct to say that Z is functionally dependent upon X and Y is functionally dependent upon Z, while Z is not a possible primary key of the relation. 16

17 Third normal form - 3NF A relation is in 3NF if the relation is in 2NF and there are no non-prime attributes transitively dependent upon a possible primary key. R 21 (Enr, Ename, Dnr, Dname, Mngr) Solution : move the non-prime attributes that are transitively dependent to another, new relation which gets as a primary key the attribute Dnr via which the transitive dependence was determined R 31 (Enr, Ename, Dnr) R 32 (Dnr, Dname, Mngr) Normalization of relations Exercise 2a Normalise the following relation in 0NF: Student(Studentnr, StudentName, StreetName, HouseNumber, PostalCode, City, Coursenr, CourseName, Profnr, ProfName) Assume that - a student only has one address, - a student follows several courses - different courses may have the same name, - a course is being taught by one professor

18 Normalization of relations Exercise 2b What needs to be changed to this model if a course could be taught by several professors? 35 Normalization of relations Exercise 3 You want to create a MySQL Database for a PetShop. You sell pets in your shop. Each animal has a name, a unique number, a gender, a breed and a birthdate. An animal can have zero, one or more children and for each animal you may have data stored on zero, one or both of its parents. You also want to store for which animal you have made publicity in some paper. For each paper you store the paper s (unique) name and the price for an advertisement. For each animal, publicity is made in zero, one or more papers. The other way around, a paper in your database has made publicity for at least one of your animals. You want to keep track of how many times you have made publicity for a specific animal in a specific paper ( NumberAdvertisements ).You should assume that different animals can have the same name, but that you only use the same name for animals of the same gender. The price for an advertisement is always the same in a specific paper. Write down the relations that would be used in the normalisation process. That is, write down the relations in 1NF, 2NF and 3NF, given the following 0NF: Animal (Aname, Anr, Agender, Abreed, Abirthdate, Achild, Pname, Padvertisementprice, Numberadvertisements) 36 18

19 Normalization of relations Exercise 4a Develop a relational database model, to store the information on the sales invoices of your company. An example of a sales invoice is given in the figure on the next slide. You may assume the following: - every invoice has a unique number, - clients may have the same name, - different products can have the same description. To create the 0NF, you can just put all attribute types on which you want to store data in a single relation called Sale. Hint: the Total and the Total amount are not included in the model. They can easily be derived by linking other attribute types. 37 Exercise 4a Quantity 0NF: Sale(InvoiceNo, ClientNo, Date, ClientName, StreetName, HouseNo, PostalCode, City, ProductNo, Description, Quantity, price) 38 19

20 Normalization of relations Exercise 4b What changes in your solution to exercise 4a if we assume that the Customers have a unique ClientName? What changes in your solution to exercise 4a if we assume that several Customers can live on the same address? What changes in your solution to exercise 4a if a customer can have several addresses? 39 Normalization of relations: Exercise 5 Create a relational database model for the ER model you created in the Personnel Administration exercise: Personnel-administration You need to store information on employees (their name, gender, address, birthdate and a unique employee number). You have to note what employee is controlling which other employees. More specifically each employee is controlling zero, one or more employees. On the other hand, each employee is directly controlled by zero or one employees. For each employee you want to keep track of the department (s)he is working for. Each employee works for exactly one department and each department has one or more employees. For each department you store the department name and location and a unique department number. A few employees manage an entire department (at most one department) and each department is managed by exactly one employee. Each employee can work on zero, one or more projects and each project can employ zero, one or more employees. For each project, you store its name and duration and a unique project number. You also want to store how many hours each employee has worked on a specific project. You also model explicitly that each department is executing zero, one or more projects and each project is assigned to exactly one department. Assume employees only have one address and departments only have one location. 0NF: Employee (Enr, Ename, Eaddress, Egender, Ebirthdate, LeaderEnr, Dnr, Dname, Dlocation, DMngr, Pnr, Pname, Pduration, HoursWorked, PDnr) MIS - Prof. Dr. F.G. Goethals 40 20

21 So now you can design a database What will you do with it? 3.5 Getting data out of a database using SQL Searching data Structured data vs. Unstructured data Search with SQL (Structured Query Language) Search with Google, 21

22 SQL (Structured Query Language) SELECT attribute types (= columns you want to see) FROM table names (= tables used in Select or Where statement) WHERE selection criteria (= rows you want to select); SELECT Book.Title, Book.ISBN, Book.Author FROM Book; MIS - Prof. Dr. F.G. Goethals SQL SELECT Book.Title, Book.ISBN FROM Book WHERE Book.Author="Paul McCartney"; 22

23 Combining data from different tables Book (Title, Author, Date, ISBN, Publisher) Author (Name, Birthdate, Country) SQL across different tables SELECT Book.Title, Book.ISBN, Book.Author, Author.Country FROM Book, Author WHERE Book.Author=Author.Name; MIS - Prof. Dr. F.G. Goethals 23

24 SQL across different tables SELECT Book.Title, Book.ISBN, Book.Author, Author.Country FROM Book, Author SQL SELECT Book.Title, Book.ISBN, Book.Author, Author.Country FROM Book, Author WHERE Author.Name = "Paul McCartney and Book.Author = Author.Name; 24

25 SQL Exercise1 Given the database used in the examples in the online movie: Book (Title, Author, Date, ISBN, Publisher) Author (Name, Birthdate, Country) Publisher (Name, StreetAddress, CityAndZipcode, Country) Retrieve the following data: 1. The name of all authors. 2. The country of the Publisher called Avon. 3. The title of the books written by Richard Bach and the birth date of that author. 4. The title and ISBN number of the books together with the author name and country of residence of the author as well as the publisher name and publisher country of the books of which the publisher lives in the U.S.A. 49 Given the following relational database: PurchaseOrder (ID, Date) Product (ID, Name, AvailableQuantity, OrderPoint, SupplierID) Supplier (ID, Name, Phone, City) ProductPurchase (ProductID, PurchaseOrderID, quantity, price) SQL Exercise 2 This database tracks the purchase orders that were placed with suppliers for specific products. Each product can only be bought with one supplier. Several products can be bought in one purchase order (an n:m relation resulting in a separate table: ProductPurchase where the link between products and purchase orders are made). The price in the ProductPurchase table is the unit price that was used for the specific product in this specific purchase order. Retrieve the following data: a) The productids of all products bought on 14/09/2010. b) The product names of all products bought on 14/09/2010 c) The names of the suppliers where products were ordered on 14/09/2010 together with the names of those products. d) The date of the purchase orders for the product with product name corn where the purchase (unit) price was below