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Database Design 2009


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  • 1. Database Design 1 What is a Database?  A collection of data that is organised in a predictable structured way  Any organised collection of data in one place can be considered a database  Examples  filing cabinet  library  floppy disk 2 ©Chisholm Institute
  • 2. What is Data?  The heart of the DBMS.  Two kinds  Collection of information that is stored in the database.  A Metadata, information about the database. Also known as a data dictionary. 3 Relational Data Model  A relational database is perceived as a collection of tables.  Each table consists of a series of rows & columns.  Tables (or relations) are related to each other by sharing a common characteristic. (EG a customer or product (E m p table)  A table yields complete physical data independence. 4 ©Chisholm Institute
  • 3. Features of the relational data model  Logical and Physical separated  Simple to understand Easy to use understand. use.  Powerful nonprocedural (what, not how) language to access data.  Uniform access to all data.  Rigorous database design principles principles.  Access paths by matching data values, not by following fixed links. 5 Terminology  Relation  A 2-dimensional table of values with these properties:  No duplicate rows  Rows can be in any order y  Columns are uniquely named by Attributes  Each cell contains only one value Employee Job Manager Jack Secretary Jill Jill Executive Bozo Bozo Director Lulu Clerk Jill The special value is NULL which implies that there is no corresponding value for that cell. This may mean the value does not apply or that it is unavailable. Entire rows of NULLs are not allowed. 6 ©Chisholm Institute
  • 4. Terminology Tuple  Commonly referred to as a row in a relation. C l f d i l i Eg: Jack Clerk Jill Attribute • A name given to a column in a relation Each column must have a relation. unique attribute. This are often referred to as the fields. Employee Job Manager 7 Terminology: Domain  A pool of atomic values from which cells a given column take their values. Each attribute has a domain.  Attributes may share domains m y m Tom Mary Attribute Domain Bozo Kali........ Employee Person Name Typist Manager Job Job Name Clerk........ Manager Person Name Here again we use the same domain as above in employee. An attribute value (a value in a column labelled by the attribute) must be from the corresponding domain or may be NULL ( ). 8 ©Chisholm Institute
  • 5. Terminology:Relation Schema A Relational Schema is a named set of attributes. This refers to the structure only of a relation. It is derived from the traditional set notation displayed below EMPLOYEE = { Employee, Job, Manager } This is usually written in the modified version for database purposes: EMPLOYEE( Employee, Job, Manager ) referring to the Table EMPLOYEE Employee Job Manager 9 Terminology:Integrity Constraint and Domain Constraint An Integrity Constraint is a condition that prescribes what values are allowable in a relation. This permits the restriction of the type of value that can be placed in a particular cell. Eg. only numbers for telephone numbers The Domain Constraint is a condition on the allowable values for an attribute. e.g. Salary < $60,000 Employee Job Manager Salary Jack Secretary Jill 25,000 This restricts the EMPLOYEE salary to be under Jill Executive Bozo 40,000 a set value. Bozo Director 50,000 Lulu Clerk Jill 30,000 10 ©Chisholm Institute
  • 6. Terminology:Key Constraint  A condition that no value of an attribute or set of attributes be repeated in a relation. e.g. Employee(the attribute) has only unique values in EMPLOYEE (the relation) relation).  The following relation violates this constraint: EMPLOYEE Employee Job Manager Salary Jack appears twice. Jack Secretary Bozo 25,000 This means that This violates the Jack Secretary Jill 25,000 Key Constraint Jill Executive Bozo 40,000 Bozo Director 50,000 Lulu Clerk Jill 30,000 11 Terminology:Key Constraint An attribute (or set of attributes) to which a key constraint applies is called a key ( or candidate key). Every relation schema must have a key. EMPLOYEE Another possible key key. Employee Job Manager Salary The combination of Job and manager is Jack Secretary Bozo 25,000 also unique Key Kim Secretary Jill 25,000 Jill Executive Bozo 40,000 Bozo Director Bozo 50,000 Lulu Clerk Jill 30,000 Simple Key Composite Key: If a key constraint applies to a set of attributes, it is called a composite or Concatenated Key. Otherwise it is a simple key. 12 ©Chisholm Institute
  • 7. Terminology:Key Constraint A key cannot have a NULL ( ) value. For example, If we change the table so that the Employee Bozo does not have a manager then Job+Manager cannot be a key. Employee Job Manager Salary Jack Secretary Bozo 25,000 Kim K Secretary Jill J ll 25,000 25 000 Jill Executive Bozo 40,000 Bozo Director 50,000 Lulu Clerk Jill 30,000 13 Terminology:Key Constraint  A primary key is a special preassigned key that can always be used to uniquely identify tuples. We have to choose a Primary Key for every Relation. We must consider all of the Candidate Keys and choose between them them.  Employee is a primary key for EMPLOYEE is usually written as: EMPLOYEE( Employee, Job, Manager, Salary ) Employee Job Manager Salary Here we have chosen Jack Secretary Bozo 25,000 the Simple Key Employee Over the concatenated Kim Secretary Jill 25,000 option of both Jill Executive Bozo 40,000 Job and Manager Bozo Director Bozo 50,000 Lulu Clerk Jill 30,000 14 ©Chisholm Institute
  • 8. A Database is more than multiple tables you must be able to “relate” them Cus-code Cus-Name Area-Code Phone Agent-Code 10010 Ramus 615 844-2573 502 10011 Dunne 713 894-1238 501 10012 Smith 615 894-2205 502 10013 Olowaski 615 894-2180 502 10014 Orlando 615 222-1672 501 10015 O’Brian 713 442-3381 503 10016 Brown 615 297-1226 502 10017 Williams 615 290-2556 503 10018 Farris 713 382-7185 501 10019 Smith 615 297-3809 503 The link is through the Agent-Code Agent-Code Agent-Name Agent-AreaCode Agent-Phone 501 Alby 713 226-1249 502 Hahn 615 882-1244 503 Okon 615 123-5589 15 Terminology: Relational Database A Relational Database is just a set of Relations. For example EMPLOYEE Employee Job Manager Salary Jack Secretary Bozo 25,000 Kim Secretary Jill 25,000 Jill Executive Bozo 40,000 Bozo Director 50,000 Lulu Clerk Jill 30,000 JOB Job Salary y Secretary 25,000 , Which Attribute do you think Secretary 25,000 relates these two tables Executive 40,000 together? Director 50,000 Clerk 30,000 16 ©Chisholm Institute
  • 9. Terminology:Relational Database Schema A Relational Database Schema a set of Relation Schemas, together with a set of Integrity Constraints. For example the Relations that you have been looking at with the headings EMPLOYEE Employee Job Manager Salary JOB Job Salary are usually written as EMPLOYEE(Employee, Job, Manager) JOB(Job, Salary) Notice how the Primary Keys are underlined 17 Terminology :Referential Integrity Constraint This constraint says that – All the values in one column should also appear in another column. Look at the table below. Every entry in the Job column of the Employee table must appear in the Job column of the Job table EMPLOYEE FK PK JOB Employee Job Manager Job Salary Jack Secretary Bozo Secretary 25,000 Kim Secretary Jill Secretary S t 25,000 25 000 Jill Executive Bozo Executive 40,000 Bozo Director Director 50,000 Lulu Clerk Jill Clerk 30,000 PK FK 18 ©Chisholm Institute
  • 10. Referential Integrity Constraint Why does the following relational database violate the referential integrity constraints? EMPLOYEE FK PK JOB Employee Job Manager Job Salary Jack Secretary Bozo Director 50,000 Kim Secretary Jill Clerk 30,000 Bozo Director Lulu Clerk Jill PK FK In other words, Why can’t Employee(Job) be a Foreign Key to Job(Job), or Employee(Manager) be a Foreignfor the answers Click here Key to Employee(Employee)? 19 Why Use Relational Databases  Their major advantage is they minimise the need t store the same data i a number of d to t th d t in b f places  This is referred to as data redundancy 20 ©Chisholm Institute
  • 11. Example of Data Redundancy (1) 21 Example of Data Redundancy (2)  The names and addresses of all students are being b i maintained i th i t i d in three places l  If Owen Money moves house, his address needs to be updated in three separate places  Consider what might happen if he forgot to mg pp f f g let library administration know 22 ©Chisholm Institute
  • 12. Example of Data Redundancy (3) 23 Example of Data Redundancy (4)  Data redundancy results in:  wastage of storage space by recording duplicate f d d l information  difficulty in updating information  inaccurate inaccurate, out-of-date data being maintained out of date 24 ©Chisholm Institute
  • 13. Other Advantages of Relational Databases  Flexibility  relationships l h (links) are not implicitly defined by (l k ) l l d f d the data  Data structures are easily modified  Data can be added, deleted, modified or queried easily 25 Summary of Some Common Relational Terms  Entity - an object (person, place or thing) that we wish to store data about  Relationship - an association between two entities  Relation - a table of data  Tuple - a row of data in a table  Attribute - a column of data in a table  Primary Key - an attribute (or group of attributes) that uniquely identify individual records in a table  Foreign Key - an attribute appearing within a table that is a primary key in another table 26 ©Chisholm Institute
  • 14. Network Diagrams 27 Terminology: Network Diagram Referential Integrity constraints can easily be represented by arrows FK PK. The arrow points from the Foreign Key to the matching Primary Key g y y EMPLOYEE(Employee, Job, Manager) JOB(Job, Salary) A relational database schema with referential integrity constraints can also be represented by a network diagram. A Referential Integrity Constraint is notated as an arrow labeled by the foreign key. You must always write the label of the Foreign Key on the arrow. Sometimes the same attribute h s diff s tt ib t has different titl s i diff t titles in different t bl s t tables. EMPLOYEE Job JOB Manager Network Diagram Notice here, the label is Manager and not Employee. 28 ©Chisholm Institute
  • 15. Personnel Database: Consider the following Tables PRIOR_JOB EXPERTISE E_NUMBER PRIOR_TITLE E_NUMBER SKILL ASSIGNMENT SKILL 1001 Junior consultant 1001 Stock market E_NUMBER P_NUMBER AREA 1001 Research analyst 1001 Investments 1002 Junior consultant 1002 Stock market 1001 26713 Stock Market 1002 Research analyst 1003 Stock market 1002 26713 Taxation 1003 Junior consultant 1003 Investments 1003 23760 Investments 1004 Summer intern 1004 Taxation 1003 26511 Management 1005 Management 1004 26511 PROJECT 1004 28765 1005 23760 NAME P_NUMBER MANAGER ACTUAL_COST EXPECTED_COST New billing system 23760 Yates 1000 10000 Common stock issue 28765 Baker 3000 4000 Resolve bad debts 26713 Kanter 2000 1500 New office lease 26511 Yates 5000 5000 Revise documentation 34054 Kanter 100 3000 Entertain new client 87108 Yates 5000 2000 New TV commercial 85005 Baker 10000 8000 EMPLOYEE TITLE NAME E_NUMBER DEPARTMENT E_NUMBER CURRENT_TITLE Kanter 1111 Finance 1001 Senior consultant Yates 1112 Accounting 1002 Senior consultant Adams 1001 Finance 1003 Senior consultant Baker 1002 Finance 1004 Junior consultant Clarke 1003 Accounting 1005 Junior consultant Dexter 1004 Finance 29 Early 1005 Accounting Personnel Database Schema What are the connecting Foreign Keys to Primary Keys? Not FK, we will look at this later PROJECT (NAME, P_NUMBER, MANAGER, ACTUAL_COST, EXPECTED_COST )  ASSIGNMENT (E_NUMBER, P_NUMBER) SKILL (AREA)  PRIOR_JOB (E_NUMBER, PRIOR_TITLE)  EXPERTISE (E_NUMBER, SKILL)  TITLE (E NUMBER CURRENT TITLE ) (E_NUMBER, EMPLOYEE (NAME, E_NUMBER, DEPARTMENT) 30 ©Chisholm Institute
  • 16. Personnel Database Network Diagram SKILL EMPLOYEE PROJECT Once you have produced your Schema and identified the Primary and Foreign Keys you can create the Network Diagram.The Network Diagram shows each of the tables with their links. Each of the Tables (Relations) are represented in a rectangle as shown. They are then connected by arrows that show the FKs pointing to the PKs, The arrow head points towards the PK, while the FK name written is the same as the attribute of the table that has the th t bl th t h th FK i it in it. EXPERTISE PRIOR_JOB TITLE ASSIGNMENT 31 Personnel Database Network Diagram SKILL EMPLOYEE PROJECT EXPERTISE PRIOR_JOB TITLE ASSIGNMENT 32 ©Chisholm Institute
  • 17. Summary: Questions  What is a Relational Database?  What is a relation?  What are Constraints?  What is a Schema?  What is a Network Diagram and why is it used? 33 Summary: Answers  A relational database is based on the relational data model. It is one or more Relations(Tables) that are Related to each other  A relation is a table composed of rows (tuples) and columns, satisfying 5 properties • No duplicate rows • Rows can be in any order • Columns are uniquely named by Attributes • Each cell contains only one value • No null rows.  Constraints are central to the correct modeling of business information. Here we have seen them limit the set up of your tables: Referential Constraint  The Network Diagram is used to navigate complex database structures. It is a compact way to show the relationships between Relations (Tables) 34 ©Chisholm Institute
  • 18. Activities  Consider the following relational database schemas. h Suppliers(suppId, name, street, city,state) Part(partId,partName,weight,length,composition) Products(prodId, prodName,department) Supplies(partId,suppId) Uses(partId,prodId)  Make M k reasonable assumptions about the meaning of attribute and s n bl ss mpti ns b t th m nin f tt ib t nd relations, identify the primary and foreign keys and draw a network diagram showing the relations and foreign keys. 35 Answer Supplier Part Product P d Supplies Uses 36 ©Chisholm Institute
  • 19.  Show the foreign keys on the network diagrams Orders Ordnum ordDate custNumb 12489 2/9/91 124 Customer custNumb custName Address Balance credLim Slsnumber 124 Adams 48 oak st 418.68 500 3 SalesRep Slsnumber Name address totCom commRate 3 Mary 12 Way 2150 .05 Part Part Desc onHand IT wehsNumb unitPrice AX12 Iron 1.4 HW 3 17.95 37 OrLine ordNum Part ordNum quotePrice 38 ©Chisholm Institute
  • 20. Answer SalesRep Part SlsNumber Part Customer OrLine CustNumb orLine Orders 39 Activities  What problems many arise from this table?  What h data redundancies are there? d d d h  What changes would you make? (hint make another table.  What if I wanted to search by surname? 40 ©Chisholm Institute
  • 21. Activities  What is wrong with this table? 41 Functional Dependence FDD 42 ©Chisholm Institute
  • 22. Functional Dependency Diagrams A FUNCTIONAL DEPENDENCY DIAGRAM is a way of representing the structure of information needed to support a business or organization It can easily be converted into a design for a relational database to support the operations of the business. 43 Data Analysis and Database Design Using Functional Dependency Diagrams 1. The 1 Th steps of D f Data Analysis i FDD are l i in 1.1 Look for Data Elements 1.2 Look for Functional Dependencies 1.3 Represent Functional Dependencies in a diagram 1.4 Eli i 1 4 Eliminate R d d Redundant FFunctional i l Dependencies 2. Data Design, after we have our final version of the FDD 2.1 Apply the Synthesis Algorithm 44 ©Chisholm Institute
  • 23. Starting points for drawing functional dependency diagrams To start the process of constructing our FDD we do the following:  We must Understand the data  We Examine forms, reports,data entry and output screens etc…  We Examine sample data  We consider Enterprise (business) rules  We examine narrative descriptions and conduct interviews.  We apply our Experiences/Practice and that of others 45 Enterprise Rules What are Enterprise / Business Rules? An enterprise rule (in the context of data analysis) is a statement made by the enterprise (organisation, company, officer in charge etc.) which constrains data in some way. ff h ) h h d Functional dependencies are the most important type of constraint on data and are often expressed in the form of enterprise rules. e.g No two employees may have the same employee number. An order is made by only one customer An employee can belong to only one department at a time. 46 ©Chisholm Institute
  • 24. Drawing FDDs - Data Elements We often refer to Data Elements during the FDD process  A data element is a elementary piece of recorded information  Every data element has a unique name.  A data element is either a Label, e.g PersonName, Address, g BulidingCode, or Measurement, e.g. Height, Age, Date  A data element must take values that can be written down. 47 Functional Dependency Diagrams Using the Method of Decomposition Given the Sample Data Tables Problem ONF Eliminate Repeating Groups OR, here is the same Attribute process using the FDD Universal & Functional Relation Dependencies approach 1NF Functional Eliminate Dependency Part Key Diagram Now we have the Dependencies Database Design 2NF Relation Method of 3NF Eliminate Non Key Synthesis Relation Dependencies 48 ©Chisholm Institute
  • 25. Data Element Examples Here are some examples PersonName h values Jeff, Jill, G Enid  P N has l ff ll Gio, E d  Address has values 1 John St, 25 Rocky Road  Height has values 171cm, 195cm  Age has values 21,52,93,2  Date has values 20th May 1947, 2nd March 1997  JobName has values Manager Secretary Clerk Manager, Secretary,  Manager might not be a data element, but ManagerName could be. It could be a value of another data element e.g. JobName 49 Drawing FDDs Data Elements Start drawing the Functional Dependency Diagram by representing the Data Elements. A Data Element is represented by its name placed in a box: Data El D t Element t Every data element must have a unique name in the functional dependency diagram. A data element cannot be composed of other data elements i.e. it cannot be broken down into smaller components m mp A Data Element is also known as an ATTRIBUTE, because it generally describes a property of some thing which we will later call an ENTITY 50 ©Chisholm Institute
  • 26. Drawing FDDs –Using Elements  A functional Dependency is a relationship between Attributes.  It is shown as an arrow e.g A B  It means that for every value of A, there is only one value for B  It reads “A determines B”.  A is called a determinant attribute attribute.  B is called the dependent attribute. 51 Data Element Examples Here are some examples of finding the Data Elements on a typical form Surname . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . On a form gives rise to the element Surname CREDIT CARD Bankcard Mastercard Visa Other On a form gives rise to the element CreditCardType 52 ©Chisholm Institute
  • 27. Functional Dependency Examples Students and their family names “Each student (identified by student number) has only one ( f y ) y family name” Students FamilyName 1 Smith 2 Jones 3 Smith 4 Andrews Considering the rules stated above we should be able to draw a FDD for this. What are the elements of interest? 53 FDDs Answer Students FamilyName 1 Smith 2 Jones 3 Smith 4 Andrews Data elements of interest are Student# and FamilyName. Students determine FamilyName (or FamilyName depends on Students) Students FamilyName Each student has exactly one family name, but the name could be the name of many students. So FamilyName does not determine Student# e.g. “Smith is the name of students 1 and 3 54 ©Chisholm Institute
  • 28. FDDs Examples Employees and the departments they work for. Department Name p Accounting g Department Name p Sales Employee Number 11 Employee Number 45 2 27 31 Enterprise Rule: “Each employee works on only one department” In this example the tables are representing some interesting data of th b sin ss W see that Empl f the business. We s th t Employees with the ID numbers 11,2 s ith th n mb s 11 2 and 31 all work in the Accounting Dept and that Employees with the ID numbers 45 and 27 work in the Sales Dept. Do you think that you could draw an FDD to represent this? Have a go and then check your answers 55 FDD Answers Employees and the departments they work for. Department Name Accounting Department Name Sales Employee Number E l N b 11 Employee Number 45 2 27 31 Data elements of interest are Employee# and DeptName” Employee# DeptName Employee# p y DeptName p 11 Acc So we could make this following Table 2 Acc 45 Sales 31 Acc 27 Acc 56 ©Chisholm Institute
  • 29. FDDs Examples The quantity of parts held in a warehouse and their suppliers “Parts are uniquely identified by part numbers” “Suppliers are uniquely identified by Supplier Names” “A part is supplied by only one supplier” A supplier “A part is held in only one quantity” Parts Suppliers Name QOH 1 Wang Electronics 23 2 Cumberland Enterprises 80 3 Wang Electronics 4 4 Roscoe Pty Ltd Pty. 58 Part# determines SupplierName & Part# determines QOH Parts SupplierName Parts QOH Should QOH be a determinant? No, common sense tells us that is not a reliable 57 choice. We could have had repeating values FDDs Examples Students and their subjects enrolled. “Each student is given a unique student number” “A subject is uniquely identified by its name” “A student may choose several subjects” A subjects Student SubjectName Data element of interest are 1 History Student# and SubjectName 1 Geography Student 1 Mathematics 1 History 2 English E li h SubjectName 2 English There us no functional dependency here. 3 Mathematics Student# does not determine 3 English SubjectName, 4 French nor does SubjectName determine Student# 4 Geography 58 ©Chisholm Institute
  • 30. FDDs Examples Results obtained by each student for each subject. “Each student is given a unique student number” b ” “A subject is uniquely identified by its name” “A student may choose several subjects” “A student is allocated a result for each subject” “Each student has only one name.” Data elements are Student#, StudentName, SubjectName and Grade 59 FDDs Examples Results obtained by each student for each subject. Student Subject Student Grade Name Name 1 Smith History A 1 Smith Geography B 1 Smith Mathematics A 2 Jones History C 2 Jones English C 3 Smith English A 3 Smith Mathematics A 4 Andrews English D 4 Andrews French C 4 Andrews Geography C Try and construct an FDD for this table considering 60 the given Business Rules and the Data Elements ©Chisholm Institute
  • 31. FDDs Examples Results obtained by each student for each subject. We can see that there is only one and only one student name for each student number, even th h t d t b though th h there might be more than one i ht b th student with the same name. So…. Student # StudentName But the subject grade for any student cannot be determined by the subject name or the student# by itself. A student can have many grades depending on the subject. How can we cater for subject this? 61 FDDs Answer Results obtained by each student for each subject. We need to combine the two Elements to say that there is one and only one grade for a student doing a particular subject. Here then is the complete diagram StudentName Student SubjectName Grade This is called the Composite Determinant 62 ©Chisholm Institute
  • 32. FDDs Examples Customer Orders Order Part# CustomerName Address 454 12 David Smith 1 John St, Hawthorn 454 23 David Smith 1 John St, Hawthorn 455 32 Emily Jones 45 Grattan St, Parkville 455 49 Emily Jones 45 Grattan St, Parkville 455 54 Emily Jones 45 Grattan St, Parkville 456 12 Mary Ho 44 Park St, Hawthorn St 456 54 Mary Ho 44 Park St, Hawthorn Validating functional dependencies Using simple data and populating the table, check there is only one value of the dependent. 63 FDDs Examples “Orders is uniquely identified by its names” “Customers are uniquely identified by their names” “A customer has only one address” “An order belongs to only one customer” “A part may be ordered only once one each order” Order Parts Ordered CustomerName Address 454 23, 12 David Smith 1 John St, Hawthorn 455 54, 49, 32 Emily Jones 45 Grattan St, Parkville 456 54, 12 Mary Ho 44 Park St, Hawthorn Order CustomerName Address Part# 64 ©Chisholm Institute
  • 33. FDDs Examples Employees and their tax files numbers “Each employee has a unique employee number” “Each employee has a unique tax file number ” Employee TaxFile# Employee# determines taxfile# 1 1024-5321 Employee# Taxfile# 2 3456-3294 3 8246-7106 Taxfile# determines Employee# 4 8861 6750 8861-6750 Taxfile# Employee# 5 1234-4765 Taxfile# Employee# Alternative keys 65  Obtain Tutorial 1 from your tutor. 66 ©Chisholm Institute
  • 34. Functional Dependency Diagrams Database Design Let’s look at the process of converting the FDD into a schema. We have a 12 step process to do so, that has an iterative component to it (loop). The 12 steps are outlined in the next series of slides. 67 Functional Dependency Diagram Preparation 1. Represent each d t element as a box. 1 R t h data l t b 2. Represent each functional dependency by an arrow. 3. Eliminate augmented dependencies. 4. Eliminate transitive dependencies. 5. Eliminate pseudo-transitive dependencies. By this t B thi stage, intersecting attributes should have b i t ti tt ib t h ld h been eliminated. 68 ©Chisholm Institute
  • 35. Deriving 3NF Schema: Synthesis Algorithm 6. Pick any (unmarked) arrow in the diagram. 7. Follow it back to its source, and write down the name of the source. S S 8. Follow all arrows from the source data item, and write down the names of their destinations. A S B S, A, B, C C S is now the key of a 3NF relation (S , A, B, C). 69 Synthesis Algorithm: Deriving 3NF Schema 9. Mark all the arrows just processed. A S B C 10. If there are any unmarked arrows in the diagram, go back to step 6. 11. Finally, determine the Universal Key. Any attribute which is not determined by any other attribute (ie. has no arrow going into it) is part of the Universal Key. U1 U2 U3 12. If the universal key is not already contained in any of the above relations, make it into a relation. The universal key is the key of the new relation. 70 ©Chisholm Institute
  • 36. A Fully Worked Example  We will now work from a given set of forms to produce an FDD then use the 12 steps to produce the Schema. The forms that p p f follow show the time spent by a particular employee on a particular project. They contain details of the employee along with details of the project. In addition they also state the hours that the employee has spent on any one project to date. This is important to the FDD. Notice also that the employee can have many previous titles and have a number of skills. This also has to be dealt with in the FDD and then later after we have used the synthesis technique to create the Schema. Have h d th nth i t hni t t th S h m H a good look at the forms on the next 2 slides and try to develop the FDD yourself. 71 Personnel Database Forms 1 EMPLOYEE ______________________________________________________________________________________________________________ NAME E_NUMBER DEPARTMENT LOCATION CURRENT TITLE PRIOR_TITLES SKILLS_ SKILLS ______________________________________________________________________________________________________________ Adams 1001 Finance 9th Floor Senior consultant Junior consultant Stock market Research analyst Investments ______________________________________________________________________________________________________________ PROJECTS ______________________________________________________________________________________________________________ NAME TIME_SPENT P_NUMBER MANAGER ACTUAL_COST EXPECTED_COST ______________________________________________________________________________________________________________ Resolve bad debts 35 26713 Kanter 2000 1500 ______________________________________________________________________________________________________________ We say that this table is in “zero normal form” (0NF) This is because the cells have multiple values, eg. Prior titles and Skills. The next slide shows forms that demonstrate that an employee can work on many projects. 72 ©Chisholm Institute
  • 37. Personnel Database Forms 2 EMPLOYEE __________________________________________________________________________________________________________ NAME E_NUMBER DEPARTMENT LOCATION CURRENT TITLE PRIOR_TITLES SKILLS __________________________________________________________________________________________________________ Baker 1002 Finance 9th Floor Senior consultant Junior consultant Stock market Research analyst _____________________________________________________________________________________________________________________ _ PROJECTS __________________________________________________________________________________________________________ NAME TIME_SPENT P_NUMBER MANAGER_NUM ACTUAL_COST EXPECTED_COST __________________________________________________________________________________________________________ Res bad debts 18 26713 Kanter 2000 1500 __________________________________________________________________________________________________________ ________________________________________________________________________________________________________________ EMPLOYEE _________________________________________________________________________________________________________ NAME E_NUMBER DEPARTMENT LOCATION CURRENT TITLE PRIOR_TITLES SKILLS _________________________________________________________________________________________________________ Clarke 1003 Accounting 8th Floor Senior consultant Junior consultant Stock market Investments _________________________________________________________________________________________________________ PROJECTS _________________________________________________________________________________________________________ NAME TIME_SPENT P_NUMBER MANAGER_NUM ACTUAL_COST EXPECTED_COST _________________________________________________________________________________________________________ New billing system 26 23760 Yates 1000 10000 New office lease 10 26511 Yates 5000 5000 ___________________________________________________________________________________________________________________________ 73 Personnel Database FD Diagram From the forms given we can produce the following FDD EXPECTED_COST _ PROJECT_NAME ACTUAL_COST TIME_SPENT MANAGER_NUM P_NUMBER EMPLOYEE_NAME PRIOR_TITLE E_NUMBER CURRENT_TITLE SKILL DEPARTMENT_NAME LOCATION 74 ©Chisholm Institute
  • 38. Personnel Database FD Diagram -Synthesis Let us just consider the section of the FDD that looks at the project number as the determinant EXPECTED_COST PROJECT_NAME ACTUAL_COST MANAGER_NUM P_NUMBER By using the synthesis method we can choose an arrow, trace it back to the source, and gather together all of the attributes that the source points to. Try this and see if you can create the schema for this table. 75 Personnel Database FD Diagram - Synthesis Again, if we choose another arrow that has not been chosen before and follow it back to the determinant we find DEPARTMENT_NAME DEPARTMENT NAME is a determinant. Gathering all of the d t min nt G th in ll f th attributes that it points to we only have the location attribute. Hence this is a simple table consisting of DEPARTMENT_NAME as the Primary key and LOCATION as the only other attribute. DEPARTMENT_NAME LOCATION So the table DEPT(DEPARTMENT_NAME, LOCATION) is created 76 ©Chisholm Institute
  • 39. Personnel Database FD Diagram - Synthesis EMPLOYEE_NAME EMPLOYEE NAME E_NUMBER CURRENT_TITLE Likewise for the section of the FDD based around the E_NUMBER, creating the following table for the Employees details. DE DEPARTMENT_N ME MEN NAME EMPLOYEE (EMPLOYEE_NAME, E_NUMBER, DEPARTMENT, CURRENT TITLE ) 77 Personnel Database FD Diagram - Synthesis Here we have a slightly more complicated one. The Time spent on the project is dependent on both the Project number and the Employee name, name as it is the time spent by a particular employee on a particular project. This is demonstrated by the boxing of both the above attributes together pointing to the TIME_SPENT P_NUMBER TIME_SPENT E_NUMBER Try to create the Assignment table for this part of the FDD.When you think you have it have a look at ours and see if you are right. 78 ©Chisholm Institute
  • 40. Personnel Database FD Diagram - Synthesis P_NUMBER TIME_SPENT E_NUMBER The main difference here is that when choosing the arrow to follow back to the determinant we find that we have 2. This is OK, we just have to make sure that in the table both of them are the primary Key. We have a Composite Primary Key consisting P_NUMBER and E_NUMBER. When we then gather up all of the attributes that they point to together we get TIME_SPENT. Hence the table is written as ASSIGNMENT (E_NUMBER, P_NUMBER, TIME_SPENT) See the composite primary key 79 Personnel Database FD Diagram - Universal Key Now, the last part of the synthesis is often forgotten. We must collect up all of the attributes that do not have arrows pointing into them and place them in the one table called the Universal Key. Every attribute collected then becomes part of the composite Primary Key. In this case we have the following attributes inside the box below. Notice how Skill is there, as it sits by itself. Nothing is its determinant. P_NUMBER PRIOR_TITLE PRIOR TITLE SKILL E_NUMBER UK (E_NUMBER, P_NUMBER, PRIOR_TITLE, SKILL) 80 ©Chisholm Institute
  • 41. Foreign Keys  In the Synthesis Algorithm, a foreign key will arise from any attribute that is: A. both a determinant and part of another determinant, OR B. both a determinant and a dependent. TIME_SPENT ASSIGNMENT (E_NUMBER, P_NUMBER, TIME_SPENT) A. P_NUMBER E_NUMBER EMPLOYEE (E_NUMBER, DEPARTMENT_NAME) B. DEPARTMENT_NAME LOCATION DEPT(DEPARTMENT_NAME, LOCATION) 81 ISA = Is A In the case of the manager we say that the manager number is contained within the employee number  Every MANAGER value is a E_NUMBER value. MANAGER_NUM ISA E_NUMBER MANAGER_NUM EMPLOYEE PROJECT  Gives rise to a new Foreign Key 82 ©Chisholm Institute
  • 43. A Fully Worked Example We now have to take care of the multi-valued areas such as skills and prior titles. Our FDD synthesis takes care of everything up to that. titles that It converts the FDD to what we call “Third normal Form”. We know that an individual can have many skills and many Prior Titles. They can also work on many Projects. Knowing the Employee number will not tell us one and only one value of the Skills that they have. We show this on the extended FDD with a double arrow notation.The notation for such a relationship is shown here where E_NUMBER is a determinant for many values of skill. Consequently the resulting representation shown on the next slide can be constructed, giving rise p , g g to the splitting of the UK to form three more relations E_NUMBER SKILL 85 Personnel Database Multivalued Dependency-Decomposition MultiValued Dependency ASSIGN (E_NUMBER, (E NUMBER P_NUMBER, P_NUMBER) PRIOR_TITLE Employees are associated with MVDs Projects, Titles and Skills E_NUMBER independently. There is no direct relationship between SKILL Projects, Titles and Skills. PRIOR_JOB (E_NUMBER, PRIOR_TITLE) EXPERTISE (E_NUMBER, SKILL) Hence we have the three new relations ASSIGN, PRIOR_JOB and EXPERTISE 86 ©Chisholm Institute
  • 45. Final Personnel Database Network Diagram DEPT DEPARTMENT_NAME MANAGER_NUM EMPLOYEE PROJECT E_NUMBER E_NUMBER E_NUMBER P_NUMBER EXPERTISE PRIOR_JOB ASSIGNMENT 89 Personnel Database FD Diagram - Synthesis EXPECTED_COST PROJECT_N ME OJE NAME ACTUAL_COST MANAGER P_NUMBER Choosing any of the arrows and following it back leads you to the project number (P N b ) Thi is then the P i j t b (P_Number). This i th th Primary K Key. If you then th gather all of the attributes that P_Number points to and place them in the brackets you get the table Project with P_Number as the primary Key. PROJECT (PROJECT_NAME,P_NUMBER, MANAGER, ACTUAL_COST, EXPECTED_COST ) 90 ©Chisholm Institute
  • 46. Role Splitting In Functional Dependency Diagrams  In a Functional Dependency Diagram any group of attributes can be related in only one way way.  For example, a pair of attributes can be related by an FD or not.  Sometimes data can be related in more one way.  For example, a department can have an employee as its head or as a member.  The member relationship is represented in the FDD: E_NUMBER DEPARTMENT_NAME  But the head relationship is represented in the FDD: DEPARTMENT_NAME E_NUMBER 91 Role Splitting In Functional Dependency Diagrams  We c n ch s t W can choose to split the E NUMBER attribute into E NUMBER and th E_NUMBER tt ibut int E_NUMBER nd HOD.  But the foreign key constraint that a Head of Department is an Employee is lost on the FDD. E_NUMBER DEPARTMENT_NAME FDD Synthesis HOD ISA NetworkD DEPARTMENT_NAME EMPLOYEE DEPT HOD 92 ©Chisholm Institute
  • 47. Role Splitting In FDDs  Alternatively, we can choose to split the DEPARTMENT_NAME attribute into EMPLOYING_DEPT and HEADED_DEPT.  But h f B the foreign key constraint that an Employing k h E l Department must be a Headed Department is again lost on the FDD. E_NUMBER EMPLOYING_DEPT FDD Synthesis S nth sis HEADED_DEPT ISA NetworkD EMPLOYING_DEPT EMPLOYEE DEPT E_NUMBER 93 Role Splitting Example Consider this example. We have the Employee p p y with many Skills, Prior Titles, as before but we also have equipment that belongs to a particular employee, such as a computer and a fax. An employee can have many different pieces of equipment. It is worthwhile recognizing them on the diagram and then decomposing them into smaller relations as part of the schema ll l f h h 94 ©Chisholm Institute
  • 48. Suppose each item of equipment (identified by SERIAL#) belongs to an employee. SERIAL# DESCRIPTION PRIOR_TITL E MVDs EMPLOYEE_NAME SKILL E_NUMBER CURRENT_TITLE UK ISA HOD DEPARTMENT_NAME LOCATION •MVDs not necessarily embodied in the UK. •Better to decompose on MVDs first. •MVDs partition attributes into independent sets. 95  Obtain Tutorial 2 from your tutor. 96 ©Chisholm Institute
  • 49. ENTITY RELATIONSHIP ANALYSIS In this area of the course we concentrate an another modelling technique called Entity Relationship Modelling (ERM or ER). The first stage of this process will look at the following: ER Data Model and Notation Strong E titi St Entities Discovering Entities, Attributes Identifying Entities Discovering Relationships 97 Critique of FD Analysis We originally concentrated on the modelling technique called Functional Dependency Diagrams. They have limitations as follows:  Disadvantages of FDD Does not represents real world objects, but only data; Cannot represent MVDs or specialization; Cannot represent multiple relationships without artificial splitting of attributes; Entities fragmented during analysis; 98 ©Chisholm Institute
  • 50. Conceptual Data Analysis By using the ER technique we have the following advantages:  Data Analysis from the User's Point of View  Models the Real World  Independent of T h l I d d t f Technology  Able to be validated in user terms 99 Entity Relationship Data Model Features The Th real value of using this type of modelling is that it l l f sin t p f m d llin th t considers the design in context to the environment where it comes from. We have these Entities that have there own identifying attributes, real things and real people. They can be observed in the environment. ERM has the following features:  Populations of Real World objects represented by Entities  Objects have Natural Identity  Entities have Attributes which have values  Entities related by Relationships  Constraints  Subtypes 100 ©Chisholm Institute
  • 51. Occurrences versus Entities 56 Jack Ackov 28 Jill Hill Let’s consider these two instances. Here we have both Jack and Jill, aged 56 and 23 respectively. By themselves they exist as people in their environment. In this case we consider them to be two customers. If we wish to model them and all of the possible customers that we have p Entity Occurrences we need to create an Entity Class for Entity Instances all possibilities. Objects 101 Occurrences versus Entities 56 Jack Ackov 28 Jill Hill Customer# CustName CUSTOMER Entity Occurrences Entity Classes Entity Instances Entity Types Objects Entity Sets These are the Tuples of p This will convert to the schema the table below below with Customer# being the Primary Key Customer# CustName 56 Jack Ackov CUSTOMER(Customer#, CustName) 28 Jill Hill 102 ©Chisholm Institute
  • 52. 56 Jack Ackov 28 Jill Hill Here we have Jack and Jill placing orders for particular items of stock. They appear to order different amounts of each. For instance Jack orders 3 bikes. Each item being ordered also has a Stock#, Price and 3 4 1 Description. These are 12 individual instances of the process so we need to be able to represent any possibility of thi i our ibilit f this in model. See how we do this on the next page. 156 Cup of Tea 234 Pussy Cat 103 23 50 Bike 1 25 56 Jack Ackov 28 Jill Hill Customer# CustName CUSTOMER 3 4 1 12 ORDERS Quantity ITEM Stock# Price Desc 23 50 Bike 156 1 Cup of Tea 234 25 Pussy Cat 104 ©Chisholm Institute
  • 53. Occurrences to Entities to Schemas Customer# CustName CUSTOMER(Customer#, CustName) 56 Jack Ackov 28 Jill Hill Customer# Stock# Quantity ORDERS(Customer#, Stock#, Quantity) 56 23 3 56 156 12 28 156 4 28 234 1 Stock# Price Desc ITEM(Stock#, Price, Desc) 23 50 Bike 156 1 Cup of Tea 234 25 Pussy Cat 105 ENTITIES  Entities are classes of objects about which we wish to store information information.  Examples are:  People: Employees, Customers, Students,..... STRONG  Places: Offices, Cities, Routes, Warehouses,...  Things: Equipment, Products, Vehicles, Parts,....  Organizations: Suppliers, Teams, Agencies, Depts,...  Concepts: Projects, Orders, Complaints, Accounts,......  Events: Meetings, Appointments. WEAK 106 ©Chisholm Institute
  • 54. STRONG ENTITIES  An entity is Existence Independent if an instance can exist in isolation.  For example, CUSTOMER is existence independent of ORDER, but ORDER is existence dependent on CUSTOMER. The ORDER is by a particular customer for a/many particular item(s)  An entity is identified if each instance can be uniquely distinguished by its attributes (or relationships).  For example, CUSTOMER is identified by Customer#, PERSON is id tifi d b N identified by Name+Address+DoB, ORDER is id tifi d b Add ss D B identified by Customer#+Date+Time. 107 STRONG ENTITIES  An entity is STRONG if it can be identified by its (own) immediate attributes. Otherwise it is weak.  For example, CUSTOMER and PERSON are strong entities, but ORDER is weak because it requires an attribute of another entity to identify it. ORDER would be strong if it had an Order#.  Existence independent entities are always strong. 108 ©Chisholm Institute
  • 55. The Method: How to Develop the ERM  Step1: Search for Strong Entities and Attributes  Step2. Attach attributes and identify strong entities.  Step3. Step3 Search for relationships. relationships  Step4. Determine constraints.  Step5. Attach remaining attributes to entities and relationships.  Step6. Expand multivalued attributes, and relationship attributes.  Represent attributed relationships and/or multivalued attributes in a Functional Dependency Diagram.  Step7. Identify weak entities.  Step8. Step8 Iterate steps 4,5,6,7,8 until no further expansion is possible 45678 possible.  Step9. Look for generalization and specialization; Analyze Cycles; Convert domain-sharing attributes to entities. 109 The 1 Search for Method strong entities 2 Narrative and attributes Identify Attributes & strong Forms F Entities entities titi 3 Strong entities Search for 7 relationships Identify 4&5 weak entities Determine Identified constraints and weak Relationships attach attributes entities Entity-Relationship Weak Entities 6 Diagram Expand attributed relationships and/or multivalued attributes 6’ Functional Represent attributed Dependency relationships and/or multivalued attributes Diagrams110 as Functional Dependencies ©Chisholm Institute
  • 56. Step1: Search for Strong Entities and Attributes  1 Entities  relevant nouns  many instances  have properties (attributes or relationships)  identifiable by properties  2 Strong Entities  independent existence  identifiable by own single-valued attributes •3 Attributes 3 –printable names, measurements –domain of values –no properties –dependent existence 111 A worked example finding strong Entities A customer is identified by a customer#. A customer has a name and an address. A customer may order quantities Here we have a scenario. of many items. An item may Try to firstly identify all of be ordered by many the strong entities followed customers. An item is and all of the attributes. identified by a stock#. An Can you also identify a weak item has a description and a entity? Are there any attributes that you have price. A stock item may have missed? many colours. Any item ordered by a customer on the same day is part of the same order Narrative 112 ©Chisholm Institute
  • 57. Worked Example Continued Let us take and place it around the nouns. These lead us to what we will consider to be A customer is identified by a the strong entities. If we then customer#. A customer has a place the around items name and an address. A that we think would be the customer may order quantities of attributes, we can see if if any of the identified Entities are many items. An item may be strong. You will notice that the ordered by many customers. An item has a description, price, item is identified by a stock#. colour and stock # and a An item has a description and a customer h a customer t has t price. A stock item may have number, name, and address. many colours. Any item ordered These a Existence Independent by a customer on the same day is Entities, and hence they must be part of the same order strong. 113 Narrative Worked Example Continued We have our Entities and the attributes displayed before us. Customer and Item are strong entities as they are Existence Independent. What about Order? Order O d cannot b t be identified completely by any of its own attributes. Conceptual Schema It is dependent on the attributes of the other 2 CUSTOMER ITEM entities to be identified. Address Customer# Date An order is made up of a Quantity Stock# Description customer ordering an Price Customer Name item. We need the Colour customer# and the item# ORDER to identify the order 114 ©Chisholm Institute
  • 58. Step2. Identify Strong Entities. We now attach the attributes that belong to each of the Strong Entities. Notice that there are some left that belong to neither Customer or Item. We will look at this later. Item later Conceptual Schema Customer# Stock# Price CUSTOMER ITEM Desc Address Colour CustName Qty Date Both Customer and Item have what we call a Natural Identity 115 Another Example of the Difference Between Weak and Strong Entities Here is another example of a common occurrence that H n th x mpl f mm n n th t demonstrates the difference between a strong entity and a weak entity  A strong entity is identified by its own attributes.  Bidders make purchases of goods at the auction. BIDDER and a GOOD have independent existence, hence are strong, but PURCHASE requires attributes of BIDDER and GOOD. The Purchase is the identified by d the Bibbers name and the Goods description. These are 2 attributes that belong to both the Bidder and the Good respectively. 116 ©Chisholm Institute
  • 59. Additional Rules for Entities For an Entity to exist we have the following additional rules:  There must be more than one instance of an entity.  The company provides superannuation for its workers. Here there is only one instance of COMPANY so it is not a valid entity. We do not model anything that only has one instance  Each instance of an entity must be potentially distinguishable by its properties.  Members send five dollars to the association. A dollar does not normally have distinguishing attributes. 117 Step3. Search for Relationships. We can now identify Relationships that have the following properties:  Relationships  Have associate entities  Are relevant must be worth recording  Can be"structural" verbs in the narrative persistent, rather than transient relationships  Can be "abstract" nouns in the narrative nonmaterial connections, eg. Enrolment  Can be verbalizable in the narrative eg. Student EnrolledIn Unit  Have 2 (binary)or more associated entities.(3-Ternary, up to n-ary for n associated entities) 118 ©Chisholm Institute
  • 60. Relationships:  A relationship must be relevant. It should indicate a structural, persistent (extending over time) p ( g ) association between entities. Students enrol in units selected from the handbook.  A relationship should not usually indicate a procedural event (one that occurs momentarily, th n s forgott n.). then is forgotten.). Students read about units selected from the handbook. 119 Relationships and the Worked Example. We can now deal with the order. The order is a relationship between the Customer and the Item. It is for a set Quantity on a given Date. Conceptual Schema Customer# Stock# Price CUSTOMER ORDERS ITEM Desc Address Colour CustName Qty Date 120 ©Chisholm Institute
  • 61. Entity Relationship Analysis 2 We will now concentrate on the following areas of good ERM  Cardinality and Participation Constraints  Expanding to Weak Entities  Identifying Weak Entities  Derived Attributes and Relationships  Ternary Relationships 121 These are Steps 4,5 & 6 from the Original Diagram Unidentified Strong entities Unattched Attributes weak entities 4 & 5 Determine Identified 7 Relationships constraints and weak Identify attach attributes entities weak entities 6 Expand attributed Entity-Relationship relationships, Weak Entities Diagram domain sharing & multivalued attributes 122 ©Chisholm Institute
  • 62. Step4. Determine constraints: Cardinality(How many participate To complete this we “fix a single instance at one end and ask how many (one or many) are involved at the other end”. Look t the l ti ship h L k at th relationship where the Customer Orders an th C st m O d s Item. Consider a single Customer. Can they order many items at the one time? Yes We have seen this. So we position a crows foot (<) at the point where the line touches the Entity Item. We then ask if an Item can be ordered by many Customers? Yes So agin we place a crows foot at the Customers end. ORDERS CUSTOMER ITEM From left to right-A Cust can order many Items From right to left- An Item can be ordered by many Cust 123 Step4. Determine constraints: Cardinality. Again to complete this task we “Fix a single instance at one end and ask how many (one or many) are involved at the l d h CUSTOMER other end”. All of the Customers live in a City. A Customer can only live in one City(unless they are politicians) In this case we must place a single straight line (|) at the intersection of the LIVES IN relationship line and the Entity City. However, a city can have many Customers. We show this by placing crows foot (>) at the end near the Customer CITY 124 ©Chisholm Institute
  • 63. Step4. The Resulting ER with the Cardinality Constraints in Place ORDERS CUSTOMER ITEM Many CUSTOMERs can ORDER an Many ITEMs ITEM. can be {Colour} ORDERed by LIVES INMany CUSTOMERs a An ITEM can LIVE IN a CUSTOMER. can have CITY. many Colours. A CUSTOMER can LIVE IN only one CITY CITY. 125 Step4.Determine constraints: Participation. Again, we “Fix a single instance at one end and ask if any must (might or must) be involved at the other end”. We ask “Does the Customer have to order an Item? Well Does Well, some would say that they do not they are not Customers! But we know that we must be able to recognise our Customers even though at present they do not have an order with us. So, in this case they do not have to place an order. This is then not mandatory, and we show it by placing the O beside the cardinality constraint. An Item does not have to be on an order as well, so it also gets the O notation. , g ORDERS CUSTOMER ITEM 126 ©Chisholm Institute
  • 64. Step4.Determine constraints: Participation. This is also the case for the Customer living in th City. D s th customer have to live in the Cit Does the st m h t li the City? In this case Yes, as we class all areas as being within a City. Hence we place the “|” symbol beside the cardinality CUSTOMER constraint next to the Entity City. The next one is difficult. Does a City have to have a Customer living in it. You might think No here, but are you prepared to record all of , y p p f the cities in the world just to make sure? LIVES IN Common sense tells us that we have to make this mandatory so we only keep a record of the cities where our Customers live. CITY 127 Step4. The Resulting ER with the Participation Constraints in Place ORDERS CUSTOMER ITEM An ITEM might be ordered by a CUSTOMER. A CUSTOMER might LIVES IN CITY must have A order a ITEM. a CUSTOMER LIVing IN it it. A CUSTOMER must LIVE IN a CITY CITY. 128 ©Chisholm Institute
  • 65. Step4. Determine constraints: Validation by Population. CUSTOMER ITEM ORDERS Cust# An important method of {Colour} evaluating the proposed model LIVES IN is to populate with instances Stock# that demonstrate that the constraints that you have identified will work. CITY CityName 129 Step4. Tables Created to Validate CUSTOMER ITEM ORDERS Cust# Stock# Cust# 12 77 {Colour} LIVES IN 23 77 Stock# CityName Cust# 12 88 Ayr y 12 99 Ayr 23 13 Tully 13 Stock# Colour 77 Pink CITY CityName 77 Blue 130 ©Chisholm Institute
  • 66. Step5. Attach remaining attributes to entities and relationships. In the previous lectures we looked at a worked problem with a Customer ordering an Item. Here we were able to identify Entities from the narration. Next N t we also listed the attributes which h l d us identify the Strong Entities. ls list d th tt ib t s hi h helped s id tif th St E titi s We noticed that there were some Attributes, Qty and Date, left that could not be attached to any of the strong entities. They, in fact, belong to the Relationship that was associated with the two Entities. Customer# Stock# ORDERS Price CUSTOMER ITEM Desc Address Colour CustName Qty Date 131 Step5. Attach remaining attributes to entities and relationships. The quantity attribute cannot be attached to the Customer, as the Customer will order different quantities of various items at any time. It cannot also be attached to the Item. It must therefore be attached to the relationship between them, being the d th order. This is also the situation for th ls th sit ti f the Date that the order was placed. 132 ©Chisholm Institute
  • 67. Step5. Attach remaining attributes to entities and relationships. Conceptual Schema Customer# Stock# Price CUSTOMER ORDERS ITEM Desc Address Add Qty Date {Colour} CustName 133 Step6.Expand multi-valued attributes, domain sharing attributes and binary relationship attributes. Once we have identified the Strong Entities, Relationships and attached all Attributes to either the Strong Entities or Relationships, we are required to expand the diagram as much as possible to permit us to complete the process. This requires us to move in 2 directions. We must first look at all of the binary relationships to see what the cardinality constraints are between them. If they are “many-to-many” they y y y y must be carefully considered and expanded where appropriate. We then must look at what we call Multi-valued Attributes and Domain Sharing Attributes. The process is shown on the following diagram. 134 ©Chisholm Institute
  • 68. Step6 Entity-Relationship Diagram Many-to-many M Multi-valued Multi valued Attributes Relationships with Attributes Domain Sharing Attributes Expand Expand Multi-valued and relationships domain sharing with attributes attributes Associative Entities Characteristic Entities Dependent Entities 135 Step6 In the worked example we have a Many-to-Many relationship with 2 attributes . When we have a Many-to-Many relationship with attached attributes we are required to create an Associative Entity that bridges the 2 Entities. Conceptual Schema Customer# Stock# Price CUSTOMER ORDERS ITEM Desc Address Qty Date {Colour} CustName 136 ©Chisholm Institute
  • 69. Step6 Between Customer and Item we create the Weak (Associative) Entity called Order. We have to redo the constraints. A customer can place many orders or none. An order can come from only one customer, and must be from a customer. An order is f many , f for y items and must be for at least one item, and an item can be on many orders but does not have to appear on an order. These have all been placed in the diagram shown below in their correct position. Associative Entity Stock# Customer# MAKES FOR Price CUSTOMER ORDER ITEM Desc Qty Address Date CustName 137 Step6 We have also noticed that an item can come in many colours. This is a multi-valued attribute. We can show this in our extended diagram by having a relationship between the Item and the Colour, where colour is the only attribute of the entity. In this case we are also saying that the entity colour of the item is optional (IE natural if requested) and that the only colours to be recorded are those that are used. Associative Entity Stock# Customer# MAKES FOR Price CUSTOMER ORDER ITEM Desc D Qty Address Date HAS CustName COLOUR Characteristic Entity 138 Colour ©Chisholm Institute
  • 70. Step6. Expand domain sharing attributes. Managers supervise Workers. All employees are residents of a City. Employees who live in different cities from their managers get a special allowance City allowance. City SUPERVISES MANAGER WORKER Allowance Characteristic Entity CITY OF OF CityName SUPERVISES MANAGER WORKER Allowance 139 Step7. Identify weak entities. Clarify the notion of instance. Weak entities are often ambiguous and difficult to agree on on. Attributes may be part of a key for a weak entity, but at least one (one-must) relationship for identification is required. So when we convert this into a table it will require one of the PKs from the strong entities as part of its own composite PK. Validation, not design. The purpose of identification is not to allocate a primary key, but to validate the concept. We have to be able to justify the concept of the relationship in the real world. Never invent keys. I know that it is tempting but you must reflect the business as it is. 140 ©Chisholm Institute
  • 71. Step7. Identify weak entities. Conceptual Schema C t l S h Customer# Stock# FOR Price CUSTOMER ORDER ITEM Desc Qty Address MAKES Date HAS CustName An ORDER is uniquely COLOUR identified by the Colour CUSTOMER and the Date. 141 Step7. Identify weak entities. Conceptual Schema C t l S h Customer# Stock# FOR Price CUSTOMER ORDER ITEM Desc Qty Address MAKES Date HAS CustName Here we still have the relationship COLOUR between Order and Item that is many to many with attributes. We must expand this. Colour 142 ©Chisholm Institute
  • 72. Step8. Iterate until no further expansion is possible. An intersection entity is We introduce the weak entity orderline that one that is identified by for one item. It is fully dependent on the on y y ts r at onsh ps. only by its relationships. attributes of Order and Item to be identified Conceptual Schema Stock# Date HAS FOR Price ORDER ORDERLINE ITEM Desc Customer# MADE BY Qty Qt HAS CUSTOMER An ORDERLINE COLOUR is identified by an Address ITEM on an Colour CustName ORDER. 143 Example 2  Ted’s Computer courses is a company that offers a number of computer courses to client ff b f t t li t companies. A client may request several courses at one time.  A course has a course code, a description and a list of resources required.  Every course has a number of suitable E h b f it bl instructors who are qualified to deliver it. An instructor has a name, address, and a telephone. 144 ©Chisholm Institute
  • 73. Example 2 cont.  A client company can request that a course begin on any date nominated. This course can be nominated offered repeatedly on many dates.  The cost of the course is negotiated for each offering.  Ted’s Company requires the details of all the attendees.  When a course is offered an instructor is assigned from the list of instructors for that course. 145 Example 2 cont.3  Each course is offered as a series of usually 4 four hour sessions. Each session has a time and f h i E h i h ti d a place, again negotiated with the client.  Develop and E-R Diagram for the database application.  The next slide show you the forms filled out when a course is offered. h i ff d 146 ©Chisholm Institute
  • 74. Course Specifications Form 147 Course Offering Form 148 ©Chisholm Institute
  • 75. Course Attendance Form 149 Solution 2 - 1  First get the major entities;  Client l  Course  Resources  Instructors  Attendees  Session 150 ©Chisholm Institute
  • 76. Solution 2 - 2  Lets look at the client. 151 Solution 2 - 3  Lets look at the Course. 152 ©Chisholm Institute
  • 77. Solution 2 - 4  Look at the relationship between the client and the th course. 153 Solution 2 - 5  The course entity has resources which is a multiple d lti l dependency value and i d lt with as d l d is dealt ith follows. 154 ©Chisholm Institute
  • 78. Solution 2 - 6  Next step is the Cardinality Constraints of the three entities th titi 155 Solution 2 - 7  Lets include the Instructor entity 156 ©Chisholm Institute
  • 79. Solution 2 - 8  For simplicity we are only going to add the attributes that are primary k tt ib t th t i keys.  A client can order a course on a set date. Where does the date belong? 157 Solution 2 - 9  Lets look at the attendees entity. The attendees attend a course requested b a tt d tt d t d by client. This is a binary relationship. 158 ©Chisholm Institute
  • 80. Solution 2 - 10  This binary relationship creates a weak entity called C ll d Course Off i Offering. 159 Solution 2 - 11  There is one more entity that needs to be added, S dd d Sessions. i 160 ©Chisholm Institute
  • 81. Solution 2 - 12  So far it looks like. 161 Solution 2 - 13  Now lets look at the many to many relationships that th t are still there. till th 162 ©Chisholm Institute
  • 82. Solution 2 - 14  Creating weak entities. Resources Qty ResourcesUsed CourseOffering CourseId Course Attendances teachingStaff Attendees name Consultant/Instructor 163 Solution 2 - 15  Now lets do the Network Diagram. 164 ©Chisholm Institute
  • 83. Solution 2 - 16  The database schema. 165 ©Chisholm Institute