Sol to Univ Dorm

PART III

SOLUTIONS TO REVIEW QUESTIONS

AND EXERCISES

Database Systems: Instructor's Guide - Part III

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Solutions to Review Questions and Exercises

Part One Background 4

Chapter 1 Introduction to Databases............................................................................................................................................. 4

Chapter 2 Database Environment ................................................................................................................................................. 6

Chapter 3 The Relational Model................................................................................................................................................... 8

Chapter 4 Database Planning, Design, and Administration ....................................................................................................... 12

Part Two Methodology 14

Chapter 5 Entity-Relationship Modeling.................................................................................................................................... 14

Chapter 6 Normalization............................................................................................................................................................. 17

Chapter 7 Methodology - Conceptual Database Design ............................................................................................................ 22

Chapter 8 Methodology - Logical Database Design for Relational Model................................................................................ 23

Chapter 9 Methodology - Physical Database Design for Relational DBMSs........................................................................... 25

Chapter 10 Conceptual Database Design Methodology - Worked Example ............................................................................. 27

Chapter 11 Logical Database Design Methodology - Worked Example ................................................................................... 29

Chapter 12 Physical Database Design Methodology – Worked Example ................................................................................. 32

Part Three Database Languages 33

Chapter 13 SQL .......................................................................................................................................................................... 33

Chapter 14 Advanced SQL......................................................................................................................................................... 39

Chapter 15 QBE.......................................................................................................................................................................... 46

Part Four Selected Database Issues 47

Chapter 16 Security..................................................................................................................................................................... 47

Chapter 17 Transaction Management......................................................................................................................................... 49

Chapter 18 Query Processing ..................................................................................................................................................... 54

Part Five Current Trends 62

Chapter 19 Distributed DBMSs - Concepts and Design ............................................................................................................ 62

Chapter 20 Distributed DBMSs - Advanced Concepts .............................................................................................................. 67

Chapter 21 Introduction to Object DBMSs ................................................................................................................................ 73

Chapter 22 Object-Oriented DBMSs.......................................................................................................................................... 74


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Chapter 23 Object-Relational DBMSs ....................................................................................................................................... 84

Part Six Future Trends 89

Chapter 24 Web Technology and DBMSs ................................................................................................................................. 89

Chapter 25 Data Warehousing.................................................................................................................................................... 91

Chapter 26 OLAP and Data Mining........................................................................................................................................... 93


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Part One BackgroundChapter 1 Introduction to Databases

Review Questions

1.1 List four examples of database systems other than those listed in Section 1.1.

Some examples could be:

• A system that maintains component part details for a car manufacturer;• An advertising company keeping details of all clients and adverts placed with them;• A training company keeping course information and participants' details;• An organization maintaining all sales order information.

1.2 Discuss each of the following terms:

Data For end users, this constitutes all the different values connected with the variousobjects/entities that are of concern to them. (See also Section 1.3.3)

Database (See Section 1.3.1)Database Management System (See Section 1.3.2)Data Independence This is essentially the separation of underlying file structures from the

programs that operate on them, also called program-dataindependence. (See also Sections 1.2.2 and 1.3.1)

Security The protection of the database from unauthorized users, which may involvepasswords and access restrictions. (See also Section 1.6)

Integrity The maintenance of the validity and consistency of the database by use ofparticular constraints that are applied to the data. (See also Section 1.6)

Views These present only a subset of the database that is of particular interest to auser. Views can be customized, for example, field names may change, and theyalso provide a level of security preventing users from seeing certain data. (Seealso Section 1.3.2)

1.3 Describe the approach taken to the handling of data in the early file-based systems. Discuss thedisadvantages of this approach.

Focus was on applications for which programs would be written, and all the data required wouldbe stored in a file or files owned by the programs. (See also Section 1.2).

Clearly, each program was responsible for only its own data, which could be repeated in otherprogram’s data files. Different programs could be written in different languages, and would not beable to access another program’s files. This would be true even for those programs written in thesame language, because a program needs to know the file structure before it can access it. (Seealso Section 1.2.2).

1.4 Describe the main characteristics of the database approach and contrast it with the file-basedapproach.

Focus is now on the data first, and then the applications. The structure of the data is now keptseparate from the programs that operate on the data. This is held in the system catalog or datadictionary. Programs can now share data, which is no longer fragmented. There is also a reductionin redundancy, and achievement of program-data independence. (See also Section 1.3)

1.5 Describe the five components of the DBMS environment and discuss how they relate to eachother.

See Section 1.3.3.

1.6 Discuss the roles of the following personnel in the database environment:


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Data Administrator See Section 1.4.1Database Administrator See Section 1.4.1Logical Database Designer See Section 1.4.2Physical Database Designer See Section 1.4.2Application Programmer See Section 1.4.3End Users See Section 1.4.4

1.7 Discuss the advantages and disadvantages of database processing.

See Section 1.6

Exercises

1.8 Interview some users of database systems. Which DBMS facilities do they find most useful andwhy? Which DBMS facilities do they find least useful and why? What do these users perceive tobe the advantages and disadvantages of the DBMS?

Select a variety of users for a particular DBMS. If the users are using different DBMSs, group theanswers for the different systems, which will give an overall picture of specific systems.

1.9 Write a small program that allows entry and display of renter details including a renter number,name, address, telephone number, preferred number of rooms and maximum rent. The detailsshould be stored in a file. Enter a few records and display the details. Now repeat this process butrather than writing a special program, use any DBMS that you have access to. What can youconclude from these two approaches?

The program can be written in any appropriate programming language, such as Pascal,FORTRAN, C. It should adhere to basic software engineering principles including being well-structured, modular, and suitably commented. It is important to appreciate the process involvedeven in developing a small program such as this. The DBMS facilities to structure, store, andretrieve data are used to the same effect. The differences in the approaches, such as the effortinvolved, potential for extension, ability to share the data should be noted.

1.10 Study the DreamHome case study presented in Section 1.7. In what ways would a DBMS help thisorganization? What data can you identify that needs to be represented in the database? Whatrelationships exist between the data? What queries do you think are required?

It may be useful to review the file-based approach and the database approach here before tacklingthe first part of the exercise. Careful reading and thinking about how people might use theapplications should help in carrying out the rest of the exercise.

1.11 Study the Wellmeadows Hospital case study presented in Appendix A. In what ways would aDBMS help this organization? What data can you identify that needs to be represented in thedatabase? What relationships exist between the data?

The approach used for Exercise 1.10 should be used for this exercise also.


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Chapter 2 Database Environment

Review Questions

2.1 Discuss the concept of data independence and explain its importance in a database environment.

See Section 2.1.5

2.2 To address the issue of data independence, the ANSI-SPARC 3-level architecture was proposed.Compare and contrast the 3 levels of this model.

See Section 2.1

2.3 What is a data model? Discuss the main types of data models.

An integrated collection of concepts for describing data, relationships between data andconstraints on the data in an organization. (See also Section 2.3).

Object-based data models such as the entity-relationship model (see Section 2.3.1). Record-baseddata models such as the relational data model, network data model, and hierarchical data model(see Section 2.3.2).

2.4 Discuss the function and importance of conceptual modeling.

See Section 2.3.4.

2.5 Describe the types of facilities you would expect to be provided in a multi-user DatabaseManagement System.

Data Storage, Retrieval and Update Authorization ServicesA User-Accessible Catalog Support for Data CommunicationTransaction Support Integrity ServicesConcurrency Control Services Services to Promote Data IndependenceRecovery Services Utility Services

See also Section 2.4

2.6 Of the facilities in 2.4, which ones do you think would not be needed in a standalone PCDatabase Management System? Provide justification for your answer.

Concurrency Control Services - only single user.Authorization Services - only single user, but may be needed if different individuals are to use theDBMS at different times.Utility Services - limited in scope.Support for Data Communication - only standalone system.

2.7 Describe the main components in a DBMS and suggest which components are responsible foreach facility identified in question 2.5.

Query Processor, DML Preprocessor, Data Storage, Retrieval andQuery Optimizer, Data Manager UpdateDictionary Manager A User-Accessible CatalogData Manager Transaction SupportScheduler Concurrency Control ServicesUtilities Recovery ServicesAuthorization Control Authorization ServicesUtilities Support for Data CommunicationIntegrity Checker Integrity ServicesDatabase Manager, DDL Compiler, File Manager

Services to Promote Data Independence


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Utility Services

See also Sections 2.5 and 2.4.

2.8 What is meant by the term "client-server architecture" and what are the advantages of thisapproach? Compare the client-server architecture with two other architectures.

The client is a process that requires some resource, and the server provides the resource. Neitherneed reside on the same machine. Advantages include:

• Better performance• Likely reduction in hardware costs• Reduction in communication costs• Better consistency

See also Section 2.6.

2.9 Discuss the function and importance of the data dictionary.

See Section 2.7

Exercises

2.10 Analyze the DBMSs that you are currently using. Determine each system’s compliance with thefunctions that we would expect to be provided by a DBMS. What types of languages does eachsystem provide? What type of architecture does each DBMS use? Check the accessibility andextensibility of the data dictionary. Is it possible to export the data dictionary to another system?

To do this you will need to obtain appropriate information about each system. There should bemanuals available or possibly someone in charge of each system who could supply information.

2.11 Write a program that stores names and telephone numbers in a database. Write another programthat stores names and addresses in a database. Modify the programs to use external, conceptual,and internal schemas. What are the advantages and disadvantages of this modification?

The programs can be written in any suitable language and should be well structured andappropriately commented. Two distinct files result. The structures can be combined into onecontaining name, address, and tel_no, which can be the representation of both the internal andconceptual schemas. The conceptual schema should be created separately with a routine to mapthe conceptual to the internal schema. The two external schemas also must be created separatelywith routines to map the data between the external and the conceptual schema. The two programsshould then use the appropriate external schema and routines.

2.12 Write a program that stores names and dates of birth in a database. Extend the program so that itstores the format of the data in the database; in other words, create a data dictionary. Provide aninterface that makes this data dictionary accessible to external users.

Again, the program can be written in any suitable language. It should then be modified to add thedata format to the original file. This should not be difficult, if the original program is wellstructured. The interface for other users operates on the data dictionary and is separate from theoriginal program. A menu-based interface is adequate.

2.13 How would you modify this program to conform to a client-server architecture? What would bethe advantages and disadvantages of this modification?

The server should hold the data dictionary and the programs that operate on it. The user interfaceshould be separate, on the client, and call the data dictionary programs.


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Chapter 3 The Relational Model

Review Questions

3.1 Discuss each of the following concepts in the context of the relational data model:

(a) Relation(b) Attribute(c) Tuple(d) Intension and Extension(e) Degree and Cardinality.

Each term defined in Section 3.2.1.

3.2 Discuss the differences between the candidate keys and the primary key of a relation. Explainwhat is meant by a foreign key. How do foreign keys of relations relate to candidate keys?

The primary key is the candidate key that is selected to identify tuples uniquely within a relation.A foreign key is an attribute or set of attributes within one relation that matches the candidate keyof some (possibly the same) relation.

3.3 Define the two principal integrity rules for the relational model. Discuss why it is desirable toenforce these rules.

Two rules are Entity Integrity (Section 3.3.2) and Referential Integrity (Section 3.3.3).

3.4 Define the five basic relational algebra operations. Define the remaining three relational algebraoperations in terms of the five basic operations.

Five basic operations are:

• Selection and Projection (Unary)• Cartesian Product, Union and Set Difference (Binary).

There is also the Join, Intersection and Division operations:

• Can rewrite θ-Join in terms of the basic selection and Cartesian product operations:

R F S = σF(R × S)

• Can express the intersection operator in terms of the set difference operation:

R ∩ S = R - (R - S)

• Can express the division operator in terms of the basic operations:

T1 = ΠC(R)T2 = ΠC( (S x T1) - R)T = T1 - T2

3.5 What is a view? Discuss the difference between a view and a base relation. Explain what happenswhen a user accesses a database through a view.

View is the dynamic result of one or more relational operations operating on the base relations toproduce another relation. Base relation exists as a set of data in the database. A view does notcontain any data, rather a view is defined as a query on one or more base relations and a query onthe view is translated into a query on the associated base relations.


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Exercises

The following tables form part of a database held in a relational DBMS:-

Hotel (Hotel_No, Name, Address)Room (Room_No, Hotel_No, Type, Price)Booking (Hotel_No, Guest_No, Date_From, Date_To, Room_No)Guest (Guest_No, Name, Address)

where Hotel contains hotel details and Hotel_No is the primary key Room contains room details for each hotel and (Hotel_No, Room_No) forms the primary key Booking contains details of the bookings and the primary key comprises (Hotel_No, Guest_No,

and Date_From)and Guest contains guest details and Guest_No is the primary key.

3.6 Generate the relational algebra for the following queries:

(a) List all hotels.

HOTEL

(b) List all single rooms with a price below £20 per night.

σtype='S' AND price < 20(ROOM)

(c) List the names and addresses of all guests.

Πname, address(GUEST)

(d) List the price and type of all rooms at the Grosvenor Hotel.

Πprice, type(ROOM hotel_no (σname='Grosvenor Hotel'(HOTEL)))

(e) List all guests currently staying at the Grosvenor Hotel.

GUEST guest_no (σdate_from <= '01-01-99' AND date_to >= '01-01-99' (BOOKING hotel_no (σname='Grosvenor Hotel'(HOTEL))))

(substitute '01-01-99' for today’s date).

(f) List the details of all rooms at the Grosvenor Hotel, including the name of the gueststaying in the room, if the room is occupied.

(ROOM hotel_no (σname='Grosvenor Hotel'(HOTEL)) // Outer JoinΠguest.name, hotel.hotel_no, room.room_no(

(GUEST guest_no (σdate_from <= '01-01-99' AND date_to >= '01-01-99' (BOOKING hotel_no (σname='Grosvenor Hotel'(HOTEL))))


(g) List the guest details (Guest_No, Name and Address) of all guests staying at theGrosvenor Hotel.

Πguest_no, name, address(GUEST guest_no (σdate_from <= '01-01-99' AND date_to >= '01-01-99' (BOOKING hotel_no (σname='Grosvenor Hotel'(HOTEL)))))



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3.7 Using relational algebra, create a view of all rooms in the Grosvenor Hotel, excluding pricedetails. What would be the advantages of this view?

Πroom_no, hotel_no, type(ROOM hotel_no (σname='Grosvenor Hotel'(HOTEL)))

Security - hides the price details from people who should not see it.Reduced complexity - a query against this view is simpler than a query against the two underlyingbase relations.

3.8 Produce the equivalent tuple and domain relational calculus statements for the above queries.

Tuple Relational Calculus

(a) RANGE OF H IS HOTEL{H}

(b) RANGE OF R IS ROOM{R | R.Type = 'S' AND R.Price < 20}

(c) RANGE OF G IS GUEST{G.Name, G.Address}

(d) RANGE OF H IS HOTELRANGE OF R IS ROOM{R.Price, R.Type | ∃H (R.Hotel_No = H.Hotel_No AND H.Name = 'Grosvenor Hotel')}

(e) RANGE OF H IS HOTELRANGE OF G IS GUESTRANGE OF B IS BOOKING{G | �B ((B.Date_From <= '01-01-99' AND

B.Date_To >= '01-01-99') AND(B.Guest_No = G.Guest_No) AND∃H (B.Hotel_No = H.Hotel_No ANDH.Name = 'Grosvenor Hotel'))}

(f) Need to use Union of a relation containing all Rooms that are occupied with a relationextended by a NULL name for all unoccupied rooms.

(g) RANGE OF H IS HOTELRANGE OF G IS GUESTRANGE OF B IS BOOKING{G.Guest_No, G.Name, G.Address | �B((B.Date_From <= '01-01-99' AND

B.Date_To >= '01-01-99') AND(B.Guest_No = G.Guest_No) AND∃H (B.Hotel_No = H.Hotel_No ANDH.Name = 'Grosvenor Hotel'))}

Domain Relational Calculus

(a) {Hotel_No, Name, Address | ∃Hotel_No, Name, Address (HOTEL(Hotel_No, Name, Address)}

(b) {Room_No, Hotel_No, Type, Price | ∃Room_No, Hotel_No, Type, Price(ROOM(Room_No, Hotel_No, Type, Price) ANDType = 'S' AND Price < 20)}

(c) {Name, Address | ∃Name, Address (GUEST(Name, Address)}

(d) {Price, Type | ∃Room_No, Type, Price, Hotel_No


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(ROOM(Hotel_No, Type, Price) AND∃Name (HOTEL(Hotel_No, Name) ANDName = 'Grosvenor Hotel')}

(e) {Guest_No, Guest.Name, Guest.Address | ∃Guest_No, Name, Address,(GUEST(Guest_No, Name, Address) AND∃Date_From, Date_To, Hotel_No(BOOKING(Hotel_No, Guest_No, Date_From, Date_To) ANDDate_From <= '01-01-99' AND Date_To >= '01-01-99') AND∃Name (HOTEL(Hotel_No, Name) ANDName = 'Grosvenor Hotel')))}

(f) Rest similar to above.

3.9 Explain how the entity and referential integrity rules apply to these relations.

For each relation, the primary key must not contain any nulls.

Room is related to Hotel through the attribute Hotel_No. Therefore, the Hotel_No in Roomshould either be null or contain the number of an existing hotel in the Hotel relation. In this casestudy, it would probably be unacceptable to have a Hotel_No in Room with a null value.

Booking is related to Hotel through the attribute Hotel_No. Therefore, the Hotel_No in Bookingshould either be null or contain the number of an existing hotel in the Hotel relation. However,because Hotel_No is also part of the primary key, a null value for this attribute would beunacceptable. Similarly for Guest_No. Booking is also related to Room through the attributeRoom_No.

3.10 Analyze the RDBMSs that you are currently using. Determine the support the system provides forprimary keys, alternate keys, foreign keys, relational integrity and views. What types of relationallanguages does the system provide? For each of the languages provided, what are the equivalentoperations for the eight relational algebra operations?

This is a small student project, the result of which is dependent on the system analyzed.


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Chapter 4 Database Planning, Design, and Administration

Review Questions

4.1 Discuss the relationship between the information system lifecycle and the database applicationlifecycle.

See Sections 4.1 and 4.2.

4.2 Describe the purpose of each of the stage of the database application lifecycle.

See Section 4.2.

4.3 Identify some of the techniques available to help document the users' requirements specification.

See Section 4.2.3.

4.4 Describe the main aims of the conceptual and logical database design phases.

The aims of conceptual database design are described in Section 4.3.2 and the aims of logicaldatabase design are described in Section 4.3.3.

4.5 Explain why it is necessary to select the target database management system before commencingwith the physical database design phase. Describe the main aims of physical database design.

Physical database design is tailored to a specific DBMS, therefore it is essential the DBMS isdetermined before the physical design phase can begin. The aims of physical database design aredescribed in Section 4.3.4

4.6 Describe the prototype approach and identify the potential advantages of using this approach.

See Section 4.2.7.

4.7 Outline the procedure for selecting a DBMS.

See Section 4.6.

4.8 Define the purpose and tasks associated with data administration and database administration.

See Section 4.7.

Exercises

4.9 Produce a corporate data model for the Wellmeadows Hospital case study described in Appendix A.


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4.10 Assume that you are responsible for selecting a new DBMS product for a group of users in yourorganization. To undertake this exercise, you must first establish a set of requirements for thegroup and then identify a set of features that a DBMS product must provide to fulfill therequirements. Describe the process of evaluating and selecting the best DBMS product.

The student should follow the approach to DBMS selection described in Section 4.6 andproduce a report that identifies a suitable DBMS product that meets the requirements of theorganization. The selection should be fully justified.

4.11 Assume that you are responsible for selecting a DBMS product for the Wellmeadows Hospitalcase study. Describe the process of evaluating and selecting the best DBMS product.

The student should follow the approach to DBMS selection described in Section 4.6 andproduce a report that identifies a suitable DBMS product that meets the requirements of theWellmeadows Hospital case study. The selection should be fully justified and any assumptionsmade about the case study should be highlighted..

4.12 Investigate whether data administration and database administration exists as distinct

functional areas within your organization. If identified, describe the organization,responsibilities, and tasks associated with each functional area.

The student should investigate the organization and identify whether data administration and databaseadministration exists as distinct functional areas. However, the student should be careful to note that thesefunctions may be named differently, merged as a single function or included as part of a larger IT/ISfunction. If identified, the student should compile a report documenting the organization, responsibilities,and tasks.

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Part Two MethodologyChapter 5 Entity-Relationship Modeling

Review Questions

5.1 Describe the purpose of high-level data models in database design.

The main purpose for developing a high-level data model is to support a user’s perception of dataand to conceal the more technical aspects associated with database design. Furthermore, aconceptual data model is independent of the particular DBMS and hardware platform that is usedto implement the database.

5.2 Describe the basic concepts of the Entity-Relationship (ER) model. Present the diagrammaticrepresentation of these concepts.

The basic concepts of the Entity-Relationship model include entity types, relationship types, andattributes.

See Section 5.1

5.3 Describe the constraints that may be placed on participating entities in a relationship.

There are two main types of restrictions on relationships called cardinality and participationconstraints.

See Section 5.2

5.4 Describe the problems that may occur when creating an ER model.

Problems may occur due to the misinterpretation of the meaning of certain relationships and theseproblems are referred to as connection traps. There are two main types of connection traps calledfan traps and chasm traps.

See Section 5.3

5.6 Describe the main concepts associated with the Enhanced Entity-Relationship model. Present thediagrammatic representation of these concepts.

The main concepts of the EER model are specialization/generalization, aggregation andcategorization. However, in the accompanying textbook, we focus only on specialization/generalization and categorization, which is associated with the related concepts of entity typesdescribed as superclasses or subclasses and the process of attribute inheritance.

See Section 5.4

Exercises

The University Accommodation Office Case Study

The Director of the University Accommodation Office requires you to design a database to assist withthe administration of the office. The requirements collection and analysis phase of the database designprocess based on the Director’s view has provided the following requirements specification for theAccommodation Office database.

(1) The data stored on each full-time student includes the matriculation number, name (first and lastname), home address (street, city/town, postcode), date of birth, sex, category of student (forexample, first year undergraduate (1UG), postgraduate (PG)), nationality, smoker (yes or no),special needs, any additional comments, current status (placed/waiting), and what course thestudent is studying on. The student information stored relates to those currently renting a room


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and those on the waiting list. Students may rent a room in a university owned hall of residence orstudent flat. When a student joins the University he or she is assigned to a member of staff whoacts as his or her Advisor of Studies. The Advisor of Studies is responsible for monitoring thestudent’s welfare and academic progress. The data held on a student’s Advisor includes their fullname, position, name of department, internal telephone number, and room number.

(2) Each hall of residence has a name, address, telephone number, and a hall manager whosupervises the operation of the hall. The halls provide only single rooms, which have a roomnumber, place number, and monthly rent rate. The place number uniquely identifies each room inall the halls controlled by the Accommodation Office and is used when renting a room to astudent.

(3) The Accommodation Office also offers student flats. These flats are fully furnished and providesingle room accommodation for groups of 3, 4, or 5 students. The information held on studentflats includes a flat number, address, and the number of single bedrooms available in each flat.The flat number uniquely identifies each flat. Each bedroom in a flat has a monthly rent rate, aroom number, and a place number. The place number uniquely identifies each room available inall student flats and is used when renting a room to a student.

(4) A student may rent a room in a hall or student flat for various periods of time. New leaseagreements are negotiated at the start of each academic year with a minimum rental period ofone semester (15 weeks) and a maximum rental period of one year, which includes Semesters 1, 2,and the Summer Semester. Each individual lease agreement between a student and theAccommodation Office is uniquely identified using a lease number. The data stored on each leaseincludes the lease number, duration of the lease (given as semesters), name, and matriculationnumber of the student, place number, room number, address details of the hall or student flat, thedate the student wishes to enter the room, and the date the student wishes to leave the room (ifknown).

(5) Student flats are inspected by staff on a regular basis to ensure that the accommodation is wellmaintained. The information recorded for each inspection is the name of the member of staff whocarried out the inspection, the date of inspection, an indication of whether the property was foundto be in a satisfactory condition (yes or no), and any additional comments.

(6) Some information is also held on members of staff on the Accommodation Office and includes thestaff number, name (first and last name), home address (street, city/town, postcode), date of birth,sex, position (for example, Hall Manager, Administrative Assistant, Cleaner), and location (forexample, Accommodation Office or Hall).

(7) The Accommodation Office also stores a limited amount of information on the courses run by theUniversity including the course number, course title (including year), course leader’s name,internal telephone number, and room number, and department name. Each student is associatedwith a single course.

(8) Whenever possible, information on a student’s next-of-kin is stored which includes the name,relationship, address (street, city/town, postcode), and contact telephone number.

5.6 Create an Enhanced Entity–Relationship (EER) model to represent the data requirements of theUniversity Accommodation Office case study. Develop the model using the following the steps:

(a) Identify entity types.(b) Identify relationship types and determine the cardinality and participation constraints of the

relationships.(c) Identify attributes and associate attributes with entity or relationship types.(d) Determine candidate and primary key attributes.(e) Specialize / generalize entity types (where appropriate).(f) Categorize entity types (where appropriate).(g) Draw the EER diagram.

State any assumptions you made when creating the EER model.

An example of a possible ER model for the University Accommodation Office case study is shownbelow.


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The student should identify any ambiguities in the specification for the University AccommodationOffice and state clearly his or her assumptions. For example, we may assume that we only holdinformation on a single next-of-kin per student and that some students do not provide this information.

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Chapter 6 Normalization

Review Questions

6.1 Describe the purpose of normalizing data.

When we design a database for a relational system, the main objective in developing a logical datamodel is to create an accurate representation of the data, its relationships and constraints. Toachieve this objective, we must identify a suitable set of relations. A technique that we can use tohelp identify such relations is called normalization. Normalization is a technique for producing aset of relations with desirable properties, given the data requirements of an enterprise.Normalization supports database designers by presenting a series of tests, which can be applied toindividual relations so that a relational schema can be normalized to a specific form to prevent thepossible occurrence of update anomalies.

See also Sections 6.1 and 6.4.

6.2 Describe the problems that are associated with redundant data.

A major aim of relational database design is to group attributes into relations so as to minimizeinformation redundancy and thereby reduce the file storage space required by the base relations.Another serious difficulty using relations that have redundant information is the problem ofupdate anomalies. These can be classified as insertion, deletion, or modification anomalies.

See Section 6.2

6.3 Describe the concept of functional dependency.

Functional dependency describes the relationship between attributes in a relation. For example, ifA and B are attributes of relation R, B is functionally dependent on A (denoted A → B), if eachvalue of A in R is associated with exactly one value of B in R.

Functional dependency is a property of the meaning or semantics of the attributes in a relation.The semantics indicate how the attributes relate to one another and specify the functionaldependencies between attributes. When a functional dependency is present, the dependency isspecified as a constraint between the attributes.


6.4 How is the concept of functional dependency associated with the process of normalization?

Normalization is a formal technique for analyzing relations based on their primary key (orcandidate keys in the case of BCNF) and functional dependencies. Normalization is oftenperformed as a series of tests on a relation to determine whether it satisfies or violates therequirements of a given normal form. Three normal forms were initially proposed, which arecalled first (1NF), second (2NF) and third (3NF) normal form. Subsequently, a stronger definitionof third normal form was introduced and is referred to as Boyce-Codd normal form (BCNF). Allof these normal forms are based on the functional dependencies among the attributes of a relation.

6.5 Provide a definition for first, second, third and Boyce-Codd normal forms.

First Normal Form (1NF) is a relation in which the intersection of each row and column containsone and only one value.

Second Normal Form (2NF) is a relation that is in first normal form and every non-primary-keyattribute is fully functionally dependent on the primary key.

Third Normal Form (3NF) is a relation that is in first and second normal form in which no non-primary-key attribute is transitively dependent on the primary key.


18

Boyce-Codd Normal Form (BCNF) is a relation in which every determinant is a candidate key.

See also Sections 6.5 to 6.8.

6.6 Describe the purpose of fourth (4NF) and fifth (5NF) normal form.

For 4NF see Section 6.11 and for 5NF see Section 6.12.

Exercises

The table shown in Figure 6.25 lists dentist/patient appointment data. A patient is given an appointment ata specific time and date with a dentist located at a particular surgery. On each day of patientappointments, a dentist is allocated to a specific surgery for that day.

StaffNo DentistName PatNo PatName AppointmentDate Time

SurgeryNo

S1011 Tony Smith P100 Gillian White 12/9/95 10.00 S15

S1011 Tony Smith P105 Jill Bell 12/9/95 12.00 S15

S1024 Helen Pearson P108 Ian MacKay 12/9/95 10.00 S10

S1024 Helen Pearson P108 Ian MacKay 14/9/95 14.00 S10

S1032 Robin Plevin P105 Jill Bell 14/9/95 16.30 S15

S1032 Robin Plevin P110 John Walker 15/9/95 18.00 S13

Figure 6.25: Lists Dentist/Patient Appointment Data.

6.7 The table shown in Figure 6.25 is susceptible to update anomalies. Provide examples of insertion,deletion, and update anomalies.

The student should provide examples of insertion, deletion and update anomalies using the datashown in the table. An example of a deletion anomaly is if we delete the details of the dentistcalled 'Helen Pearson', we also lose the appointment details of the patient called 'Ian MacKay'.

6.7 Describe and illustrate the process of normalizing the table shown in Figure 6.25 to Boyce-Coddnormal forms. State any assumptions you make about the data shown in this table.

The student should state any assumptions made about the data shown in the table. For example,we may assume that a patient is registered at only one surgery. Also, a patient may have morethan one appointment on a given day.


19

StaffNo ADate ATime StaffNo DName

PK

3NF / BCNF

fd1 fd2

StaffNo ADate

fd4

FK

FKPK

PatNo

SurgeryNo

fd5

FK

PK

StaffNo ADate ATime DName PatNo PName

PK

fd2

1NF

SurgeryNo

fd4

fd2 and fd4 violates 2NF

2NF

StaffNo ADate

Fd3’ violates 3NF

SurgeryNo StaffNo DName

fd1

fd3

StaffNo ADate ATime PatNo PName

fd5

PatNo PName

PK

Fd3’


20

An agency called Instant Cover supplies part-time/temporary staff to hotels within Strathclyde Region. Thetable shown in Figure 6.26 lists the time spent by agency staff working at various hotels. The NationalInsurance Number (NIN) is unique for every member of staff.

NIN ContractNo Hours EName H_No H_Loc

1135 C1024 16 Smith J H25 East Kilbride

1057 C1024 24 Hocine D H25 East Kilbride

1068 C1025 28 White T H4 Glasgow

1135 C1025 15 Smith J H4 Glasgow

Figure 6.26: Instant Cover’s Contracts.

6.8 The table shown in Figure 6.26 is susceptible to update anomalies. Provide examples of insertion,deletion, and update anomalies.

The student should provide examples of insertion, deletion and update anomalies using the datashown in the table. An example of an update anomaly is if we wish to change the name of theemployee called 'Smith J', we may only change the entry in the first row and not the last with theresult that the database becomes inconsistent.

6.9 Describe and illustrate the process of normalizing the table shown in Figure 6.26 to Boyce-CoddNormal Form. State any assumptions you make about the data shown in this table.

The student should state any assumptions made about the data shown in the table. For example,we may assume that a hotel may be associated with one or more contracts.


21

NIN Contract_No Hours EName H_No H_Loc

NIN Contract_No Hours

NIN EName

NIN Contract_No Hours Contract_No H_No

H_No H_ Loc

PK

PK

FK

PK

PK

fd1

fd2

fd3

fd4

fd2 and fd3 violates 2NF

2NF

3NF / BCNF

fd1 fd3

fd4

1NF

Contract_No H_No H_Loc

fd4 violates 3NF

FK

FK

NIN EName

PK

fd2


22

Chapter 7 Methodology - Conceptual Database Design

Review Questions

7.1 Describe the purpose of a design methodology.

A structured approach that uses procedures, techniques, tools, and documentation aids, to supportand facilitate the process of design. A design methodology consists of phases that contain steps,which guide the designer in the choice of techniques that are appropriate at each stage of theproject and also helps to plan, manage, control and evaluate database development projects.Furthermore, it is a structured approach for analyzing and modeling a set of requirements for adatabase in a standardized and organized manner.

See Section 7.1.

7.2 Describe the main phases involved database design.

See Section 7.1.2

7.3 Identify important factors in the success of logical database design.

See Section 7.1.3

7.4 Discuss the important role played by users in the process of database design.

The users' involvement throughout the database design phase is critical to providing the 'correct'system. In particular, the user should clarify any ambiguities in the specification that describes therequired system and also review continually the development of the database design. The processof developing the database design is repeated until the user is prepared to 'sign-off' the design asbeing a 'true' representation of the part of the enterprise that is being modeling.

See Sections 7.1.3 and 7.2.

7.5 Describe the main objective of conceptual database design.See Section 7.3 Step 1.

7.6 Describe what a user view represents and the approaches that may be used to identify userviews.

User views can be identified using various methods. First, by examining the data flow diagramsthat should have been produced previously to identify functional areas and possibly individualfunctions. Alternatively, by interviewing users, examining procedures, reports, forms, and/orobserving the enterprise in operation.

See also Section 7.3

7.7 Identify the main tasks associated with conceptual database design.See Section 7.3

7.8 Discuss the purpose of specialization/generalization of entity types, and discuss why this is an

optional step in conceptual database design.

See Section 7.3 Step 1.6. 7.9 Identify and describe the purpose of the documentation generated during conceptual database

design.

See Section 7.3 and in particular note the documentation generated at the end of each step.


23

Chapter 8 Methodology - Logical Database Design for Relational Model

Review Questions

8.1 Identify the three main phases of database design and discuss the purpose of logical databasedesign.

The three main phases of database design are conceptual, logical, and physical. For purpose oflogical database design see Section 8.1.

8.2 Describe the steps involved in refining a conceptual data model into a logical data model.

See Section 8.1 Step 2.

8.3 Describe the rules for deriving relations that represent strong entity types, weak entity types,one-to-one binary relationship types, one-to-many relationship types, multi-valued attributes,and superclass/subclass relationships.

See Section 8.1 Step 2.2.

8.4 Discuss how the technique of normalization can be used to validate the logical data model andthe relations derived from the model.

The logical data model can be validated using the technique of nomalization and against the transactionsthat the model is required to support. Normalization is used to improve the model so that it satisfiesvarious constraints that avoid unnecessary duplication of data. Normalization ensures that the resultantmodel is a closer model of the enterprise that it serves, it is consistent, and has minimal redundancy andmaximum stability.

See also Section 8.1 Step 2.3.

8.5 Discuss two approaches that can be used to validate that the logical data model is capable ofsupporting the transactions required by the user’s view.

Two possible approaches to ensure that the logical data model supports the requiredtransactions include: checking that all the information (entities, relationships and theirattributes) required by each transaction is provided by the model in documenting a descriptionof each transaction’s requirements, and diagrammatically representing the pathway taken byeach transaction directly on the ER diagram.


8.6 Describe the purpose of integrity constraints and identify the five main types of constraints.

The main types of constraints include: There are five types of integrity constraints: required data,attribute domain constraints, entity integrity, referential integrity, and enterprise constraints.


8.7 Describe the alternative strategies that can be applied if there exists a child occurrencereferencing a parent occurrence that we wish to delete.

There are several strategies to consider when there exists a child occurrence referencing theparent occurrence that we are attempting to delete: NO ACTION, CASCADE, SET NULL,SET DEFAULT, and NO CHECK.


8.8 Identify the tasks typically associated with merging local logical data models into a globallogical model.


24

See Section 8.1 Step 3 and in particular Step 3.1.


25

Chapter 9 Methodology - Physical Database Design for Relational DBMSs

Review Questions

9.1 Explain the difference between logical and physical database design. Why might these tasks becarried out by different people?

Logical database design is concerned with building the data model for the organization that iscompletely independent of any DBMS. Physical database design, however, is concerned withactually defining the data model using the DDL of a particular DBMS. Consequently, logicaldatabase design focuses on what the data model represents, whereas physical database designfocuses on how the data model is to be implemented. Different skills are required to undertakethese design phases, which are often found in different people.

See Section 9.1.

9.2 Describe the inputs and outputs of physical database design.

The inputs are the global logical data model and the data dictionary. The outputs are the baserelations, integrity rules, file organization specified, secondary indexes determined, user viewsand access rules.

See Section 9.3.

9.3 Describe the purpose of the main steps in the physical design methodology presented in thischapter.

Step 4 Produces a relational database schema from the global logical data model. Thisincludes integrity rules.

Step 5 Determines the file organizations for the base relations. This takes account ofthe nature of the transactions to be carried out, which also determine wheresecondary indexes will be of use. As a result of analyzing the transactions, thedesign may be altered by incorporating controlled redundancy into it.

Step 6 Designs the security measures for the database implementation. This includesdesigning the user views and the access rules on the relations and views.

Step 7 Monitors the database application systems and improves performance bymaking amendments to the design as appropriate.


9.4 "One of the main objectives of physical database design is to store data in an efficient way." Howmight we measure efficiency in this context?

This can be measured by the number of transactions that can be processed by the system in agiven time frame, or by the length of time it takes to complete one transaction, or by the amountof disk storage taken up by the database files.

See also Section 9.3 Step 5.

9.5 Under what circumstances would we want to denormalize a logical data model? Use examples toillustrate your answer.

Generally, if overall performance needs to be improved, controlled redundancy can be introduced.


Examples:If queries on staff always required the branch address, this attribute could be posted intostaff. The effect on updating would be minimal if branch data was relatively static, and itremoves a join from the query.


26

If the number of staff for each branch was often required with branch details, a derivedattribute could be placed in branch. This would remove the need to access andrepeatedly count the relevant records in staff. When staff joined or left a branch, thisattribute would required updating.


27

Chapter 10 Conceptual Database Design Methodology - Worked Example

Exercises

The Wellmeadows Hospital case study

10.1 Identify user views for the Medical Director and Charge Nurse in the Wellmeadows Hospital casestudy, described in Appendix A.

See Appendix A

10.2 List the users' requirements specification for each of these views.

The student should examine the case study in detail and identify and compile a user specificationfor the Medical Director’s and the Charge Nurse’s views. As an additional task the student mayalso compile a specification for the Personnel Officer’s view.

Of course to complete this exercise will require that the student makes some ascertions about theprecise requirements of each user view. Any assumption should be documented along with eachview.

Examples of specification for the Medical Director’s and Charge Nurse’s views are shown below.

Medical Director

The Director is responsible for the overall management of the hospital and must maintain controlover the use of resources (including staff, beds, and supplies) in the provision of cost-effectivetreatment for all patients.

1. The hospital is composed of many wards. Each ward is managed by a Charge Nurse.The information to be held on each ward includes the ward name, number (e.g. W1),phone number, location (e.g. Block E), number of beds and the name of the ChargeNurse.

2. Each ward is allocated staff including (e.g. Charge nurse, senior and junior nurses,doctors, consultants, auxiliaries).

3. The hospital maintains a central stock of surgical (e.g. syringe, bandages) and non-surgical (e.g. plastic bags, aprons). The details of surgical and non-surgical suppliesincludes item number and name, item description, quantity in stock, re-order level, costper unit. The supplies used by each ward is monitored.

The hospital also maintains a stock of pharmaceutical supplies (e.g. antibiotics, painkillers). The details of pharmaceutical supplies includes drug number and name,description, dosage, quantity in stock, re-order level, cost per unit. The pharmaceuticalsupplies used by each ward is monitored.

4. The details of the suppliers of the surgical, non-surgical and pharmaceutical items arestored. The information stored includes the supplier name and number, address, phoneand fax number.

5. Patients are normally referred to the hospital for treatment by their local doctor. Thedetails of local doctors are stored including the their name, clinic number, address andphone number.

6. The details of patients referred to the hospital includes the patient number, name (firstand last name), address, phone number, date of birth, martial status, next-of-kin details(name, relationship, address and phone number).


28

7. When a patient is referred by their doctor to attend the hospital, the patient is given anappointment and is examined by a consultant. The details of the appointment are storedincluding the consultant’s name and number, appointment number, date, time andexamination room (e.g. Room E112).

As a result of the examination, the patient is either recommended to attend the outpatientclinic or placed on a waiting list until a bed can be found in a particular ward.

8. The details of outpatients are stored. The information stored includes the patient detailsas stated earlier (see 6) and the date and time of the appointment at the outpatient clinic.

9. The details of patients currently placed in a ward and those on the waiting list for a placeon a ward are stored. The information stored includes the patient details as stated earlier(see 6) and the date placed on waiting list, ward required, expected duration of stay, dateplaced in the ward and date left the ward.

Charge Nurse

The Charge Nurse has overall responsibility for the management of a single ward. The ChargeNurse is allocated a budget to run the ward and must ensure that all resources (staff, beds andsupplies) are used effectively in the care of patients.

The Charge Nurse and other senior medical staff are responsible for the allocation of beds topatients on the waiting list.

1. The information to be held on each ward includes the details of staff allocated to eachward including the staff number, name, address, phone number, position, number ofhours worked per week and shift (e.g. early, late).

2. The information stored on each patient on the waiting list includes the patient number,name (first and last name), address, phone number, date of birth, martial status, next-of-kin details, date placed on waiting list, required ward, date placed in ward, expectedduration of stay, date left ward.

3. When a patient enters the ward they are allocated a bed with a unique bed number. Eachpatient is prescribed medication and the details of this medication includes the patientnumber, drug number and name, units per day, start and finish date. The medication(pharmaceutical supplies) given to each patient is monitored.

4. Staff are allocated to work in wards, as required. The Charge Nurse of each ward isresponsible for creating a staff rota which ensures that the correct complement of staffare on duty for each shift (early, late, night). Nursing staff may be specifically allocatedto patients who required specialist care.

5. When required the Charge Nurse may obtain surgical, non-surgical and pharmaceuticalsupplies from the central stock of supplies held by the hospital. The information to bestored includes the requisition number, staff name and number, ward number, itemnumber (or drug number), quantity required, date ordered and date received.

10.3 Create local conceptual data models for each of the user views. State any assumptionsnecessary to support your design.

The student should create local conceptual data models using the requirements specification forthe Medical Director and the Charge Nurse. This should include an ER model representing eachuser view and the supporting documentation that describes the models. Throughout the process ofdesign, the student should clearly state any assumptions necessary to support his or her design.


29

Chapter 11 Logical Database Design Methodology - Worked Example

Exercises

The Wellmeadows Hospital case study

11.1 Create and validate the local logical data models for each of the user views of the WellmeadowsHospital case study identified in Exercise 10.1.

The student should refine the local conceptual models to create local logical data models based onthe Medical Director and Charge Nurse views. The logical models representing each user viewshould also be validated. The student should produce an ER model for each user view and thesupporting documentation that describes each model. Throughout the process of design, thestudent should clearly state any assumptions necessary to support his or her design.

11.2 Merge the local data models to create a global logical data model of the Wellmeadows Hospitalcase study. State any assumptions necessary to support your design.

Once the local data models have been validated, the student should demonstrate the viewintegration approach to create a global logical data model. The student should produce an ERmodel of the global data model, representing both user views and the supporting documentationthat describes the model. Throughout the process of design, the student should clearly state anyassumptions necessary to support his or her design.

11.3 Create or update the supporting documentation for the global logical data model of theWellmeadows Hospital case study.

An example of an ER model and the relational schema of the global data model of the WellmeadowsHospital case study is shown below. Note that this answer also includes the Personnel Officer’s view.


30

Ward (WardNo, WName, Location, TotalBeds, TelExtn, CNStaffNo)Primary Key WardNoAlternate Key TelExtnForeign Key CNStaffNo is NOT NULL references Staff(StaffNo) on delete SET DEFAULTon update CASCADE

Staff (StaffNo, FName, LName, Address, TelNo, DOB (Date_of_Birth), Sex, (NationalInsurance Number) NIN, Position, Salary, SScale, WeekHrs, ContType, TypePay)

Primary Key StaffNoAlternate Key NIN

Qualification (QDate, QType, Institution, StaffNo)Primary Key QType, StaffNoForeign Key StaffNo is NOT NULL references Staff(StaffNo) on delete CASCADE onupdate CASCADE

Work_Experience (SDate, FDate, Position, OrgName, StaffNo)Primary Key OrgName, StaffNoForeign Key StaffNo is NOT NULL references Staff(StaffNo) on delete CASCADE onupdate CASCADE

Staff_Rota (Shift, WeekNo, StaffNo, WardNo)Primary Key StaffNo, WeekNoForeign Key StaffNo is NOT NULL references Staff(StaffNo) on delete CASCADE onupdate CASCADE)Foreign Key WardNo is NOT NULL references Ward(WardNo) on delete CASCADE onupdate CASCADE)

Patient (PatNo, FName, LName, Address, TelNo, DOB, Sex, MStatus, DateReg, DocName,ClinicNo, NName, NRelationship, NAddress, NTelNo)Primary Key PatNo

Doctor Clinic

NOK

Supply OPClinic Appointmt InPatient

PatientItem Drug Medication

Staff Ward Bed

M1

1 1

1 1 1

1

Supplier Requisition

M

MMM

MM

M

MMMMM

M

1M

MM

M

M

1

M

M

1

1

111 1

N

1

M

M N

Attends

Refers

RelatedToTakesForms

InRequires

Has

WorksIn

WorksIn

IsGoesTo

Examines

HasMakes

ForAssignedTo

Receives

Provides

1

Qualificatn WorkExp

HasHas

11

M M

d


31

Foreign Key DocName, ClinicNo is NOT NULL references Doctor(DocName, ClinicNo) ondelete NO ACTION on update CASCADE

Doctor (DocName, ClinicNo, Address, TelNo)Primary Key DocName, ClinicNo

Appointment (AppNo, PatNo, ConsStaffNo, ADate, ATime, RoomNo)Primary Key AppNoForeign Key PatNo is NOT NULL references Patient(PatNo) on delete NO ACTION onupdate CASCADE)Foreign Key ConStaffNo is NOT NULL references Staff(StaffNo) on delete NO ACTION onupdate CASCADE)

OutPatient_Appointment (OutPatDate, OutPatTime, PatNo)Primary Key OutPatDate, PatNoForeign Key PatNo is NOT NULL references Patient(PatNo) on delete CASCADE on updateCASCADE)

InPatient_Allocation (ListDate, WardReq, Duration, PlacedDate, ExLeaveDate, ActLeaveDate,PatNo, BedNo)Primary Key PatNo, ListDateForeign Key PatNo is NOT NULL references Patient(PatNo) on delete CASCADE on updateCASCADE)Foreign Key WardReq is NOT NULL references Ward(WardNo) on delete NO ACTION onupdate CASCADE

Medication (PatNo, DrugNo, UnitsDay, AMethod, SDate, FDate)Primary Key PatNo, DrugNo, SDateForeign Key PatNo is NOT NULL references Patient(PatNo) on delete CASCADE on updateCASCADE)Foreign Key DrugNo is NOT NULL references Pharmaceutical(DrugNo) on delete NOACTION on update CASCADE)

Pharmaceutical (DrugNo, DName, Description, Dosage, MAdmin, QStock, RLevel, UnitCost ,SupplierNo)Primary Key DrugNoForeign Key SupplierNo is NOT NULL references Supplier(SupplierNo) on delete NOACTION on update CASCADE

Non-Surgical/Surgical (ItemNo, IName, IDescription, QStock, RLevel, UnitCost, SupplierNo)Primary Key ItemNoForeign Key SupplierNo is NOT NULL references Supplier(SupplierNo) on delete NOACTION on update CASCADE

Requisition (ReqNo, CNStaffNo, WardNo, ItemDrugNo, QuantReq, DateOrder, DateReceive)Primary Key ReqNoForeign Key CNStaffNo is NOT NULL references Staff(StaffNo) on delete NO ACTION onupdate CASCADEForeign Key WardNo is NOT NULL references Ward(WardNo) on delete NO ACTION onupdate CASCADEForeign Key ItemDrugNo is NOT NULL references Non-Surgical/Surgical(ItemNo) andPharmaceutical(DrugNo) on delete NO ACTION on update CASCADE

Supplier (SupplierNo, SName, SAddress, TelNo, FaxNo)Primary Key SupplierNoAlternative Key TelNoAlternative Key FaxNo


32

Chapter 12 Physical Database Design Methodology – Worked Example

Exercises

12.1 Create a physical database design for the logical design of the DreamHome case study(described in Chapter 11) based on the DBMS that you have access to.

The assumption made here is that any DBMS that is being used is a relational DBMS. It isimportant to know the facilities that are provided by the DBMS, and understand how to makeuse of them for physical database design. Assumptions made may made, for example, on theperformance of transactions. The student should produce the required documentation for thetarget DBMS.

12.2 Implement the DreamHome database using the physical design created in Section 12.1.

The student should implement the physical database design created in Exercise 12.1 using thetarget DBMS.

12.3 Investigate whether your DBMS can accommodate the two new requirements for theDreamHome case study given in Step 7 of this chapter.

Again the student will have to investigate the functionality of the target DBMS to assesswhether the new requirements can be made. If yes, the student should implement the newrequirements and also suggest further enhancements.

12.4 Create a physical database design for the Wellmeadows Hospital case study (described inAppendix A) based on the DBMS that you have access to.

The student should create a physical database design for the Wellmeadows Hospital case studybased on the logical design created in Exercise 11.3.

12.5 Implement the Wellmeadows Hospital database using the physical design created in 12.4.

The student should implement the physical database design created in Exercise 12.4 using thetarget DBMS.


33

Part Three Database LanguagesChapter 13 SQL

Review Questions

13.1 What are the two major components of SQL and what function do they serve?

A data definition language (DDL) for defining the database structure.A data manipulation language (DML) for retrieving and updating data.

13.2 What are the advantages and disadvantages of SQL?

Advantages• Satisfies ideals for database language• (Relatively) Easy to learn• Portability• SQL standard exists• Both interactive and embedded access• Can be used by specialist and non-specialist.

Disadvantages• Impedance mismatch - mixing programming paradigms with embedded access• Lack of orthogonality - many different ways to express some queries• Language is becoming enormous (SQL-92 is 6 times larger than predecessor)• Handling of nulls in aggregate functions• Result tables are not strictly relational - can contain duplicate tuples, imposes an ordering on

both columns and rows.

13.3 Explain the function of each of the clauses in the SELECT statement. What restrictions areimposed on these clauses?

FROM Specifies the table or tables to be used.WHERE Filters the rows subject to some condition.GROUP BY Forms groups of rows with the same column value.HAVING Filters the groups subject to some condition.SELECTSpecifies which columns are to appear in the output.ORDER BY Specifies the order of the output.

If the SELECT list includes an aggregate function and no GROUP BY clause is being used togroup data together, then no item in the SELECT list can include any reference to a column unlessthat column is the argument to an aggregate function.

When GROUP BY is used, each item in the SELECT list must be single-valued per group.Further, the SELECT clause may only contain:

• Column names.• Aggregate functions.• Constants.• An expression involving combinations of the above.

All column names in the SELECT list must appear in the GROUP BY clause unless the name isused only in an aggregate function.

13.4 What restrictions apply to the use of the aggregate functions within the SELECT statement? Howdo nulls affect the aggregate functions?

An aggregate function can be used only in the SELECT list and in the HAVING clause.


34

Apart from COUNT(*), each function eliminates nulls first and operates only on the remainingnon-null values. COUNT(*) counts all the rows of a table, regardless of whether nulls or duplicatevalues occur.

13.5 Explain how the GROUP BY clause works. What is the difference between the WHERE andHAVING clauses?

SQL first applies the WHERE clause. Then it conceptually arranges the table based on thegrouping column(s). Next, applies the HAVING clause and finally orders the result according tothe ORDER BY clause.

WHERE filters rows subject to some condition; HAVING filters groups subject to somecondition.

13.6 What is the difference between a subquery and a join? Under what circumstances would you notbe able to use a subquery?

With a subquery, the columns specified in the SELECT list are restricted to one table. Thus,cannot use a subquery if the SELECT list contains columns from more than one table.

Exercises

The following tables form part of a database held in a relational DBMS:

Hotel (Hotel_No, Name, Address)Room (Room_No, Hotel_No, Type, Price)Booking (Hotel_No, Guest_No, Date_From, Date_To, Room_No)Guest (Guest_No, Name, Address)

where Hotel contains hotel details and Hotel_No is the primary key Room contains room details for each hotel and (Hotel_No, Room_No) forms the primary key Booking contains details of the bookings and the primary key comprises (Hotel_No, Guest_No

and Date_From)and Guest contains guest details and Guest_No is the primary key.

Simple Queries

13.7 List full details of all hotels.

SELECT * FROM hotel;

13.8 List full details of all hotels in London.

SELECT * FROM hotel WHERE address LIKE '%London%';

Strictly speaking, this would also find rows with an address like: '10 London Avenue, New York'.

13.9 List the names and addresses of all guests in London, alphabetically ordered by name.

SELECT name, address FROM guest WHERE address LIKE '%London%'ORDER BY name;

13.10 List all double or family rooms with a price below £40.00 per night, in ascending order of price.

SELECT * FROM room WHERE price < 40 AND type IN ('D', 'F')ORDER BY price;

(Note, ASC is the default setting).

13.11 List the bookings for which no date_to has been specified.


35

SELECT * FROM booking WHERE date_to IS NULL;

Aggregate Functions

13.12 How many hotels are there?

SELECT COUNT(*) FROM hotel;

13.13 What is the average price of a room?

SELECT AVG(price) FROM room;

13.14 What is the total revenue per night from all double rooms?

SELECT SUM(price) FROM room WHERE type = 'D';

13.15 How many different guests have made bookings for August?

SELECT COUNT(DISTINCT guest_no) FROM bookingWHERE (date_from <= DATE'1999-08-01' AND date_to >= DATE'1999-08-01') OR

(date_from >= DATE'1999-08-01' AND date_from <= DATE'1999-08-31');

Subqueries and Joins

13.16 List the price and type of all rooms at the Grosvenor Hotel.

SELECT price, type FROM roomWHERE hotel_no =

(SELECT hotel_no FROM hotelWHERE name = 'Grosvenor Hotel');

13.17 List all guests currently staying at the Grosvenor Hotel.

SELECT * FROM guestWHERE guest_no =

(SELECT guest_no FROM bookingWHERE date_from <= CURRENT_DATE AND

date_to >= CURRENT_DATE ANDhotel_no =

(SELECT hotel_no FROM hotelWHERE name = 'Grosvenor Hotel'));

13.18 List the details of all rooms at the Grosvenor Hotel, including the name of the guest staying in theroom, if the room is occupied.

SELECT r.* FROM room r LEFT JOIN(SELECT g.name, h.hotel_no, b.room_no FROM Guest g, Booking b, Hotel hWHERE g.guest_no = b.guest_no AND b.hotel_no = h.hotel_no AND

h.name= 'Grosvenor Hotel' ANDb.date_from <= CURRENT_DATE ANDb.date_to >= CURRENT_DATE) AS XXX

ON r.hotel_no = XXX.hotel_no AND r.room_no = XXX.room_no;

13.19 What is the total income from bookings for the Grosvenor Hotel today?

SELECT SUM(price) FROM booking b, room r, hotel hWHERE (b.date_from <= CURRENT_DATE AND

b.date_to >= CURRENT_DATE) ANDr.hotel_no = h.hotel_no and r.room_no = b.room_no;


36

13.20 List the rooms that are currently unoccupied at the Grosvenor Hotel.

SELECT * FROM room rWHERE room_no NOT IN

(SELECT room_no FROM booking b, hotel hWHERE (date_from <= CURRENT_DATE AND

date_to >= CURRENT_DATE) ANDb.hotel_no = h.hotel_no AND name = 'Grosvenor Hotel');

13.21 What is the lost income from unoccupied rooms at the Grosvenor Hotel?

SELECT SUM(price) FROM room rWHERE room_no NOT IN


date_to >= CURRENT_DATE) ANDb.hotel_no = h.hotel_no AND name = 'Grosvenor Hotel');

Grouping

13.22 List the number of rooms in each hotel.

SELECT hotel_no, COUNT(room_no) AS count FROM roomGROUP BY hotel_no;

13.23 List the number of rooms in each hotel in London.

SELECT hotel_no, COUNT(room_no) AS count FROM room r, hotel hWHERE r.hotel_no = h.hotel_no AND address LIKE '%London%'GROUP BY hotel_no;

13.24 What is the average number of bookings for each hotel in August?

SELECT AVG(X)FROM ( SELECT hotel_no, COUNT(hotel_no) AS X

FROM booking bWHERE (b.date_from <= DATE'1999-08-01' AND

b.date_to >= DATE'1999-08-01') OR (b.date_from >= DATE'1999-08-01' AND

b.date_from <= DATE'1999-08-31')GROUP BY hotel_no);

Yes - this is legal in SQL-92!

13.25 What is the most commonly booked room type for each hotel in London?

SELECT MAX(X)FROM ( SELECT type, COUNT(type) AS X

FROM booking b, hotel h, room rWHERE r.room_no = b.room_no AND b.hotel_no = h.hotel_no AND

h.address LIKE '%London%'GROUP BY type);

13.26 What is the lost income from unoccupied rooms at each hotel today?

SELECT hotel_no, SUM(price) FROM room rWHERE room_no NOT IN



37

date_to >= CURRENT_DATE) ANDb.hotel_no = h.hotel_no)

GROUP BY hotel_no;

Creating and Populating Tables

13.27 Using the CREATE TABLE statement, create the Hotel, Room, Booking and Guest tables.

CREATE TABLE hotel( hotel_no CHAR(4) NOT NULL,name VARCHAR(20) NOT NULL,address VARCHAR(50) NOT NULL);

CREATE TABLE room( room_no VARCHAR(4) NOT NULL,hotel_no CHAR(4) NOT NULL,type CHAR(1) NOT NULL,price DECIMAL(5,2) NOT NULL);

CREATE TABLE booking(hotel_no CHAR(4) NOT NULL,guest_no CHAR(4) NOT NULL,date_from DATETIME NOT NULL,date_to DATETIME NULL,room_no CHAR(4) NOT NULL);

CREATE TABLE guest( guest_no CHAR(4) NOT NULL,name VARCHAR(20) NOT NULL,address VARCHAR(50) NOT NULL);

13.28 Insert records into each of these tables.

INSERT INTO hotelVALUES ('H111', 'Grosvenor Hotel', 'London');

INSERT INTO roomVALUES ('1', 'H111', 'S', 72.00);INSERT INTO guestVALUES ('G111', 'John Smith', 'London');INSERT INTO bookingVALUES ('H111', 'G111', DATE'1999-01-01', DATE'1999-01-02', '1');

13.29 Update the price of all rooms by 5%.

UPDATE room SET price = price*1.05;

13.30 Create a separate table with the same structure as the Booking table to hold archive records.Using the INSERT statement, copy the records from the Booking table to the archive tablerelating to bookings before 1st January 1990. Delete all bookings before 1st January 1990 fromthe Booking table.

CREATE TABLE booking_old( hotel_no CHAR(4) NOT NULL,guest_no CHAR(4) NOT NULL,date_from DATETIME NOT NULL,date_to DATETIME NULL,room_no VARCHAR(4) NOT NULL);

INSERT INTO booking1(SELECT * FROM bookingWHERE date_to < DATE'1990-01-01');

DELETE FROM bookingWHERE date_to < DATE'1990-01-01';


38

General

13.31 Investigate the SQL dialect on any DBMS that you are currently using. Determine the complianceof the DBMS with the ISO standard. Investigate the functionality of any extensions the DBMSsupports. Are there any functions not supported?

This is a small student project, the result of which is dependent on the dialect of SQL being used.

13.32 Show that a query using the HAVING clause has an equivalent formulation without a HAVINGclause.

Hint: Allow the students to show that the restricted groups could have been restricted earlier witha WHERE clause.

13.33 Show that SQL is relationally complete.

Hint: Allow the students to show that each of the relational algebra operations can be expressed inSQL.


39

Chapter 14 Advanced SQL

Review Questions

14.1 Discuss the advantages and disadvantages of views.

See Section 14.1.7.

14.2 Describe how the process of view resolution works.

Described in Section 14.1.3.

14.3 What restrictions are necessary to ensure that a view is updatable?

ISO standard specifies the views that must be updatable in a system that conforms to the standard.Definition given in SQL-92 is that a view is updatable if and only if:

• DISTINCT is not specified; that is, duplicate rows must not be eliminated from the queryresults.

• Every element in the SELECT list of the defining query is a column name (rather than aconstant, expression, or aggregate function) and no column appears more than once.

• The FROM clause specifies only one table; that is, the view must have a single source tablefor which the user has the required privileges. If the source table is itself a view, then thatview must satisfy these conditions. This, therefore, excludes any views based on a join, union(UNION), intersection (INTERSECT), or difference (EXCEPT).

• The WHERE clause does not include any nested SELECTs that reference the table in theFROM clause.

• There is no GROUP BY or HAVING clause in the defining query.

In addition, every row that is added through the view must not violate the integrity constraints ofthe base table (Section 14.1.5).

14.4 Discuss the functionality and importance of the Integrity Enhancement Feature (IEF).

Required data: NOT NULL of CREATE/ALTER TABLE.Domain constraint:CHECK clause of CREATE/ALTER TABLE andCREATE DOMAIN.Entity integrity: PRIMARY KEY (and UNIQUE) clauseof CREATE/ALTER TABLE.

Referential integrity: FOREIGN KEY clause of CREATE/ALTER TABLE.Enterprise constraints: CHECK and UNIQUE clauses ofCREATE/ALTER TABLE and (CREATE) ASSERTION.

14.5 Discuss how the Access Control mechanism of SQL works.

Each user has an authorization identifier (allocated by DBA).Each object has an owner. Initially, only owner has access to an object but the owner can passprivileges to carry out certain actions on to other users via the GRANT statement and take awaygiven privileges using REVOKE.

14.6 Discuss the difference between interactive SQL, static embedded SQL, and dynamic embeddedSQL.

Interactive SQL: SQL statements usually input interactively from a terminal.Static and dynamic embedded SQL refers to the embedding of SQL statements in a high-levelprogramming language such as C. Embedded SQL statements are changed by a preprocessorprovided with the DBMS vendor into functions calls.

The basic difference between static and dynamic embedded SQL is that static SQL does not allowhost variables to be used in place of table names or column names.


40

Exercises

Answer the following questions using the relational schema from the Exercises of Chapter 13.

14.7 Create a view containing the hotel name and the names of the guests staying at the hotel.

CREATE VIEW hotel_data(hotel_name, guest_name)AS SELECT h.name, g.name FROM hotel h, guest g, booking b WHERE h.hotel_no = b.hotel_no AND g.guest_no = b.guest_no AND

b.date_from <= CURRENT_DATE ANDb.date_to >= CURRENT_DATE;

14.8 Create a view containing the account for each guest at the Grosvenor Hotel.

CREATE VIEW booking_out_todayAS SELECT g.guest_no, g.name, g.address, r.price*(b.date_to - b.date_from)

FROM guest g, booking b, hotel h, room rWHERE g.guest_no = b.guest_no AND r.room_no = b.room_no AND

b.hotel_no = h.hotel_no AND h.name = 'Grosvenor Hotel' ANDb.date_to = CURRENT_DATE;

14.9 Give the users Manager and Deputy full access to these views, with the privilege to pass theaccess on to other users.

GRANT ALL PRIVILEGES ON hotel_dataTO manager, deputy WITH GRANT OPTION;

GRANT ALL PRIVILEGES ON booking_out_todayTO manager, deputy WITH GRANT OPTION;

14.10 Give the user Accounts SELECT access to these views. Now revoke the access from this user.

GRANT SELECT ON hotel_data TO accounts;GRANT SELECT ON booking_out_today TO accounts;

REVOKE SELECT ON hotel_data FROM accounts;REVOKE SELECT ON booking_out_today FROM accounts;

14.11 Create the Hotel table using the Integrity Enhancement Features of SQL.

CREATE DOMAIN HOTEL_NUMBER AS CHAR(4);

CREATE TABLE hotel(hotel_no HOTEL_NUMBER NOT NULL,name VARCHAR(20) NOT NULL,address VARCHAR(50) NOT NULL,PRIMARY KEY (hotel_no));

14.12 Now create the Room, Booking, and Guest tables using the Integrity Enhancement Features ofSQL with the following constraints:

(a) Type must be one of Single, Double, or Family.(b) Price must be between £10 and £100.(c) Rid must be between 1 and 100.(d) Date_From and Date_To must be greater than today’s date.(e) The same room cannot be double booked.(f) The same guest cannot have overlapping bookings.


41

CREATE DOMAIN ROOM_TYPE AS CHAR(1)CHECK(VALUE IN ('S', 'F', 'D'));

CREATE DOMAIN HOTEL_NUMBERS AS HOTEL_NUMBERCHECK(VALUE IN (SELECT hotel_no FROM hotel));

CREATE DOMAIN ROOM_PRICE AS DECIMAL(5,2)CHECK(VALUE BETWEEN 10 AND 100);

CREATE DOMAIN ROOM_NUMBER AS VARCHAR(4)CHECK(VALUE BETWEEN '1' AND '100');

CREATE TABLE room(room_no ROOM_NUMBER NOT NULL,hotel_no HOTEL_NUMBERS NOT NULL,type ROOM_TYPE NOT NULL DEFAULT 'S'price ROOM_PRICE NOT NULL,PRIMARY KEY (room_no, hotel_no),FOREIGN KEY (hotel_no) REFERENCES hotel

ON DELETE CASCADE ON UPDATE CASCADE);

CREATE DOMAIN GUEST_NUMBER AS CHAR(4);

CREATE TABLE guest(guest_no GUEST_NUMBER NOT NULL,name VARCHAR(20) NOT NULL,address VARCHAR(50) NOT NULL);

CREATE DOMAIN GUEST_NUMBERS AS GUEST_NUMBERCHECK(VALUE IN (SELECT guest_no FROM guest));

CREATE DOMAIN BOOKING_DATE AS DATETIMECHECK(VALUE > CURRENT_DATE);

CREATE TABLE booking(hotel_no HOTEL_NUMBERS NOT NULL,guest_no GUEST_NUMBERS NOT NULL,date_from BOOKING_DATE NOT NULL,date_to BOOKING_DATE NULL,room_no ROOM_NUMBER NOT NULL,PRIMARY KEY (hotel_no, guest_no, date_from),FOREIGN KEY (hotel_no) REFERENCES hotel

ON DELETE CASCADE ON UPDATE CASCADE,FOREIGN KEY (guest_no) REFERENCES guest

ON DELETE NO ACTION ON UPDATE CASCADE,FOREIGN KEY (hotel_no, room_no) REFERENCES room

ON DELETE NO ACTION ON UPDATE CASCADE,CONSTRAINT room_bookedCHECK (NOT EXISTS (SELECT *

FROM booking bWHERE b.date_to > booking.date_from ANDb.date_from < booking.date_to ANDAND b.room_no = booking.room_no ANDb.hotel_no = booking.hotel_no)),

CONSTRAINT guest_bookedCHECK (NOT EXISTS (SELECT *

FROM booking bWHERE b.date_to > booking.date_from ANDb.date_from < booking.date_to ANDAND b.guest_no = booking.guest_no)));


42

14.13 Investigate the embedded SQL functionality of any DBMS that you use. Determine the complianceof the DBMS with the ISO standard. Investigate the functionality of any extensions the DBMSsupports. Are there any functions not supported?


14.14 Write a small program that prompts the user for guest details and inserts the record into the guesttable.

#include <stdio.h>#include <stdlib.h>EXEC SQL INCLUDE sqlca;

main(){EXEC SQL BEGIN DECLARE SECTION;

char guest_no[5]; /* input guest number */char name[21]; /* input guest name */char address[51]; /* input address */

EXEC SQL END DECLARE SECTION;

/* Connect to database */EXEC SQL CONNECT 'hoteldb';if (sqlca.sqlcode < 0) exit (-1);

/* Prompt for data */printf("Enter guest number: "); scanf("%s", guest_no);printf("Enter guest name: "); scanf("%s", name);printf("Enter address: "); scanf("%s", address);

EXEC SQL INSERT INTO guestVALUES :guest_no, :name, :address;

/* Check success */if (sqlca.sqlcode >= 0)

printf("Insert successful\n");else

printf("Insert unsuccessful\n");

/* Finally, disconnect from the database */EXEC SQL DISCONNECT;}


43

14.15 Write a small program that prompts the user for booking details, checks that the specified hotel,guest and room exists and inserts the record into the booking table.



char hotel_no[5]; /* input hotel number */char guest_no[5]; /* input guest number */char date_from[26]; /* input from date */char date_to[26]; /* input to date */char room_no[5]; /* input room number */char hname[21]; /* return hotel name */char gname[21]; /* return guest name */char type[2]; /* return room type */



/* Prompt for data */printf("Enter hotel number: "); scanf("%s", hotel_no);

EXEC SQL SELECT name INTO :hname FROM hotel WHERE hotel_no = :hotel_no;if (sqlca.sqlcode < 0) {

printf("Hotel %s does not exist ... exiting\n", hotel_no);exit(-1);

}printf("Enter guest number: "); scanf("%s", guest_no);

EXEC SQL SELECT name INTO :gname FROM guest WHERE guest_no = :guest_no;if (sqlca.sqlcode < 0) {

printf("Guest %s does not exist ... exiting\n", guest_no);exit(-1);

}printf("Enter room number: "); scanf("%s", room_no);

EXEC SQL SELECT type INTO :type FROM room WHERE room_no = :room_no;if (sqlca.sqlcode < 0) {

printf("Room %s does not exist ... exiting\n", room_no);exit(-1);

}printf("Enter from date: "); scanf("%s", date_from);printf("Enter to date: "); scanf("%s", date_to);

EXEC SQL INSERT INTO bookingVALUES :hotel_no, :guest_no, :date_from, :date_to, :room_no;

/* Check success */if (sqlca.sqlcode >= 0)

printf("Insert successful\n");else

printf("Insert unsuccessful\n");/* Finally, disconnect from the database */EXEC SQL DISCONNECT;}

14.16 Write a program that increases the price of every room by 5%.



44

main(){/* Connect to database */EXEC SQL CONNECT 'hoteldb';if (sqlca.sqlcode < 0) exit(-1);

/* Display message for user and update the table */printf("Updating ROOM table\n");EXEC SQL UPDATE room SET price = price*0.05;if (sqlca.sqlcode >= 0) /* Check success */

printf("Update successful\n");else

printf("Update unsuccessful\n");

/* Commit the transaction */EXEC SQL COMMIT;

/* Finally, disconnect from the database */EXEC SQL DISCONNECT;}

14.17 See overleaf.

14.18 Write a program that allows the user to insert data into any user-specified table.

Several ways to write this program. For example:

a) Could use static SQL and after user enters table name, have a switch that branches to theappropriate piece of code to prompt for the relevant data and perform an INSERTstatement. This solution is not very elegant.

b) Use the program given in Example 14.17 of the book and type in the relevant INSERTstatement.


45

14.17 Write a program that calculates the account for every guest checking out of the Grosvenor Hoteltoday.



char guest_no[5]; /* guest number */char gname[21]; /* guest name */char address[51]; /* guest address */double balance; /* guest address */



/* Establish SQL error handling */EXEC SQL WHENEVER SQLERROR GOTO error;EXEC SQL WHENEVER NOT FOUND GOTO done;

/* Declare cursor for selection */EXEC SQL DECLARE booking_out_cursor CURSOR FOR

SELECT g.guest_no, g.name, g.address, r.price*(b.date_to - b.date_from)FROM guest g, booking b, hotel h, room rWHERE g.guest_no = b.guest_no AND r.room_no = b.room_no AND

b.hotel_no = h.hotel_no AND h.name = 'Grosvenor Hotel' ANDb.date_to = CURRENT_DATE

ORDER by g,name;

/* Open the cursor to start of selection */EXEC SQL OPEN booking_out_cursor;

printf("Guest Number\t Guest Name\t Guest Address\t Balance\n\n");

/* Loop to fetch each row of the result table */for ( ; ; ) {/* Fetch next row of the result table */EXEC SQL FETCH booking_out_cursor

INTO :guest_no, :gname, :address, :balance;

/* Display data */printf("%s\t%s\t%s\t%f\n", guest_no, gname, address, balance);

}/* Error condition - print out error */error:

printf("SQL error %d\n");done:/* Close the cursor before completing */EXEC SQL WHENEVER SQLERROR continue;EXEC SQL CLOSE booking_out_cursor;EXEC SQL DISCONNECT;}


46

Chapter 15 QBE

Exercises

The student should attempt the following exercises using the QBE facility or similar of theavailable DBMS.

15.1 Create the sample tables of the DreamHome case study, shown in Figure 3.3 and carry out theexercises demonstrated in this chapter, using (where possible) the QBE facility of your DBMS.

15.2 Create the following additional select QBE queries for the sample tables of the DreamHomecase study, using (where possible) the QBE facility of your DBMS.(a) Retrieve the branch number, address, and telephone number for all branch offices.(b) Retrieve the staff number, position, and salary for all members of staff working at

Branch Office B3.(c) Retrieve the details of all flats in Glasgow.(d) Retrieve the details of all female members of staff who are older than 25 years old.(e) Retrieve the full name and telephone of all customers who have viewed flats in

Glasgow.(f) Retrieve the total number of properties, according to property type.(g) Retrieve the total number of staff working at each branch office, ordered by branch

number.

15.3 Create the following additional advanced QBE queries for the sample tables of theDreamHome case study, using (where possible) the QBE facility of your DBMS.(a) Create a parameter query that prompts for a property number and then displays the

details of that property.(b) Create a parameter query that prompts for the first and last names of a member of

staff and then displays the details of the property that the member of staff isresponsible for.

(c) Add several more records into the Property_for_Rent tables to reflect the fact thatproperty owners ‘Carol Farrel’ and ‘Tony Shaw’ now own many properties inseveral cities. Create a select query to displays for each owner, the number ofproperties he or she owns in each city. Now, convert the select query into a crosstabquery and assess whether the display is more or less useful when comparing thenumber of properties owned by each owner in each city.

(d) Introduce an error into your Staff table by entering an additional record for themember of staff called ‘David Ford’, assigned a new staff number. Use the FindDuplicates query to identify this error.

(e) Use the Find Unmatched query to identify those members of staff who are notassigned to manage property.

(f) Create an autolookup query that fills in the details of an owner, when a new propertyrecord is entered into the Property_for_Rent table and the owner of the propertyalready exists in the database.

15.4 Use action queries to carry out the following tasks on the sample tables of the DreamHomecases study, using (where possible) the QBE facility of your DBMS.(a) Create a cut-down version of the Property_for_Rent table called

PROPERTY_GLASGOW, which has the Pno, Street, Pcode and Type fields of theoriginal table and contains only the details of properties in Glasgow.

(b) Remove all records of property viewings that do not have an entry in the Commentfield.

(c) Update the salary of all members of staff, except Managers, by 12.5%. (d) Create a table called NewRenter, which contains the details of potential renters of

property. Append this information into the original Renter table.

15.5 Using the sample tables of the DreamHome case study, create equivalent QBE queries for theSQL examples given in Chapter 15.


47

Part Four Selected Database IssuesChapter 16 Security

Review Questions

16.1 Explain the purpose and scope of database security.

The purpose is clearly concerned with the protection of the data. However, the scope is wider thanthat of the DBMS alone, and takes into account the database environment. Consequently, it alsoconsiders the hardware, software, and users.


16.2 List six different types of threat that could affect a database system, and for each, describe thecontrols that you would use to counteract each of them.

For threats see Section 16.1.1 and for computer-based countermeasures see Section 16.2 and fornon-computer based controls see Section 16.3.

16.3 Explain the following:

(a) authorization See Section 16.2.1(b) backup See Section 16.2.3 and 16.2.6(c) encryption See Section 16.2.5(d) contingency plan See Section 16.3.1(e) personnel controls See Section 16.3.2(f) Escrow agreement See Section 16.3.4(g) privacy See Section 16.8(h) data protection See Section 16.8

16.4 Outline the stages of risk analysis and provide a brief explanation of each stage.

See Section 16.7.

Exercises

16.5 Identify an important computer-based application in your computing environment and determine:

(a) the type of data the application uses and produces.(b) the integrity checks that are required.(c) how the system performs the integrity checks.

The select application may be used by a single user, group of users, department, or the entireorganization. Investigating the application may require any or all of the following to beundertaken:

• users to be interviewed• forms to be obtained or perused• analysts, designers and developers to be interviewed• use of the system.

16.6 Select a DBMS used within your computing environment and evaluate how well it supports all theintegrity controls described in the chapter.

The DBMS may be either mainframe or PC-based. A good method of evaluating a system is toinvestigate how the integrity constraints for an application are implemented. You may find thatbasic constraints can be easily handled, but more complex ones, such as conditional businessrules, require much more effort.

16.7 Carry out an investigation of your computing environment and:


48

(a) List all potential threats and breaches of security you can identify.(b) List all examples you find of countermeasures against potential threats.(c) Determine whether or not any risk assessment is undertaken and if not, how security

breaches are dealt with.

It is better to focus on a particular area of concern here, rather than attempt to cover an entiredepartment’s or organization’s operation. The student will need to know how procedures arecarried out and how the application functions. If all exercises are being undertaken, it may beuseful to build on the work carried out for 14.6.

16.8 Determine the nature and extent of the personal data used by your computing environment and:

(a) Identify who has responsibility for setting guidelines for its collection and usage.(b) Determine which staff handle personal data and whether they receive any special

training regarding the handling of this type of data. Is the training regularly updated?(c) Give details of the arrangements for access to personal data.

This is an investigation into the policies, procedures and practices carried out within the student’senvironment. The student should not require access to any data, only the type of data collectedand stored. Again, it may be easier to focus on a particular application area, such as a department.

16.9 Discover what legislation exists that applies to data of a personal nature. Does it apply to alldata, or just that held on computer?

A useful starting point is to determine whether there is someone who is responsible for managingthe legislation, such as a Data Protection Registrar, and obtain information from him/her. Thefocus here could simply be national or federal legislation, although the student may wish to checkout state or provincial legislation where it exists.

16.10 Consider the DreamHome case study introduced in Section 1.7. List the potential threats thatshould be considered in this environment and propose countermeasures to overcome them.

This should be tackled in a similar manner to Exercise 16.7 in determining the potential threatsand any countermeasures.

16.11 Consider the Wellmeadows Hospital case study introduced in Appendix A. List the potentialthreats that should be considered in this environment and propose countermeasures to overcomethem.

This should be tackled in a similar manner to Exercise 16.7 in determining the potential threatsand any countermeasures.


49

Chapter 17 Transaction Management

Review Questions

17.1 Explain what is meant by a transaction. Why are transactions important units of operation in aDBMS?

Transaction: An action or series of actions, carried out by a single user or application program,which accesses or changes the contents of the database (Section 17.1).

A logical unit of work that transforms the database from one consistent state to another. Also theunit of concurrency and recovery control.

17.2 The consistency and reliability aspects of transactions are due to the "ACIDity" properties oftransactions. Discuss each of these properties and how they relate to the concurrency control andrecovery mechanisms. Give examples to illustrate your answer.

ACID properties discussed in Section 17.1.1.

17.3 Discuss, with examples, the types of problems that can occur in a multiuser environment whenconcurrent access to the database is allowed.

Lost update problem, the uncommitted dependency problem and the inconsistent analysisproblem (Examples 17.1 - 17.3).

17.4 Give full details of a mechanism for concurrency control that can be used to ensure the types ofproblems discussed in 17.3 cannot occur. Show how the mechanism prevents the problemsillustrated from occurring. Discuss how the concurrency control mechanism interacts with thetransaction mechanism.

Should discuss 2PL, timestamping, or an optimistic technique. Solutions to above problems for2PL given in Examples 17.6 - 17.8.

Transaction is the unit of concurrency control.

17.5 Discuss the types of problems that can occur with locking-based mechanisms for concurrencycontrol and the actions that can be taken by a DBMS to prevent them.

Deadlock/livelock (see end of Section 17.2.3 and Section 17.2.4.)

17.6 Explain the concepts of serial, non-serial, and serializable schedules. State the rules forequivalence of schedules.

See Section 17.2.2.

17.7 Discuss the difference between conflict serializability and view serializability.

See Section 17.2.2.

17.8 Discuss the types of failure that may occur in a database environment. Explain why it would beunreasonable for a multiuser DBMS not to provide a recovery mechanism.

Failures given in Section 17.3.1. Answer to second part is based on the amount of work thatpotentially could be lost following a failure.

17.9 Discuss how the log file (or journal) is a fundamental feature in any recovery mechanism. Explainwhat is meant by forward and backward recovery and describe how the log file is used in forwardand backward recovery. What is the significance of the write-ahead log protocol? How docheckpoints affect the recovery protocol?


50

Log file contains before and after-images of updates to the database. Before images can be used toundo changes to the database; after-images can be used to redo changes.

Log file also contains a checkpoint record, which can speed up the time for recovery following afailure.

17.10 Discuss the following advanced transaction models:(a) nested transactions,(b) sagas,(c) multi-level transactions,(d) dynamically restructuring transactions.

See Section 17.4.

Exercises

17.11 Analyze the DBMSs that you are currently using. What concurrency protocol does the DBMSuse? What type of recovery mechanism is used? What support is provided for the advancedtransaction models discussed in Section 17.4?


17.12 (a) Explain what is meant by the constrained write rule, and explain how to test whether aschedule is serializable under the constrained write rule. Using the above method,determine whether the following schedule is serializable:

S = [R1(Z), R2(Y), W2(Y), R3(Y), R1(X), W1(X), W1(Z), W3(Y), R2(X), R1(Y), W1(Y),W2(X), R3(W), W3(W)]

where Ri(Z)/Wi(Z) indicates a read/write by transaction i on data item Z.

Constrained write rule: transaction updates a data item based on its old value, which is first readby the transaction. A precedence graph can be produced to test for serializability.

Cycle in precedence graph, which implies that schedule is not serializable.

(b) Would it be sensible to produce a concurrency control algorithm based onserializability? Give justification for your answer. How is serializability used instandard concurrency control algorithms?

No - interleaving of operations from concurrent transactions is typically determined by operatingsystem scheduler. Hence, it is practically impossible to determine how the operations will beinterleaved beforehand to ensure serializability. If transactions are executed and then you test forserializability, you would have to cancel the effect of a schedule if it turns out not to beserializable. This would be impractical!

17.13 Produce a wait-for-graph for the following transaction scenario and determine whether deadlockexists.

T1 T2

T3


51

Transaction Data items locked by transaction Data items transaction is waiting for

T1 X2 X1, X3

T2 X3, X10 X7, X8

T3 X8 X4, X5

T4 X7 X1

T5 X1, X5 X3

T6 X4, X9 X6

T7 X6 X5

Cycles in graph implies that deadlock exists.

17.14 Write an algorithm for shared and exclusive locking. How does granularity affect this algorithm?

read_lock(X):

B: if LOCK (X) = "unlocked"then begin

LOCK (X) = "read-locked";no_of_reads(X) = 1end

else if LOCK(X) = "read-locked"then no of_reads(X) = no_of _reads(X) + 1else begin

wait (until LOCK (X) = "unlocked" andthe lock manager wakes up the transaction);

goto Bend

write_lock (X):

B: if LOCK (X) = "unlocked"then LOCK (X) = "write-locked"else begin

wait (until LOCK(X) = "unlocked" andthe lock manager wakes up the transaction);

goto Bend;

T2 T3

T1

T4 T5

T6

T7


52

unlock_item (X):

if LOCK (X) = "write-locked"then begin

LOCK (X) = "unlocked";wakeup one of the waiting transactions, if anyend

else if LOCK(X) = "read-locked"then begin

no of_reads(X) = no_of_reads(X) - 1;if no of reads(X) = 0then begin

LOCK (X) = "unlocked";wakeup one of the waiting transactions, if anyend;

end;

Algorithm 1 Locking and unlocking operations for two-mode (read-write orshared-exclusive) locks

17.15 Write an algorithm that checks whether the concurrently executing transactions are in deadlock.

Boolean function deadlock_detection

Input: A table called Wait_for_Table containingTransaction_id; Data_Item_Locked; Data_Item_Waiting_For

Output: Boolean flag indicating whether system is deadlocked.begin

Deadlock = FALSE;Transaction_stack = NULL;for next transaction in Wait_for_Table while not Deadlockbegin

push next transaction_id into Transaction_stack;for next Data_Item_Waiting_For of transaction on top of stack and

not Deadlock and not Transaction_stack = NULLbegin

D_next = next Data_Item_Waiting_For;find Tran_id of transaction which has locked D_next;if Tran_id is in stackthen

Deadlock = TRUE;else

push Tran_id to Transaction_stack;endpop Transaction_stack;

endreturn deadlock;

end

17.16 Explain why stable storage cannot really be implemented. How would you simulate stablestorage?

Information residing on stable storage is never lost. Theoretically, this cannot be guaranteed. Toimplement an approximation of stable storage, we need to replicate information in severalnonvolatile storage media with independent failure modes and update the information in acontrolled manner. Although a large number of copies reduces the probability of a failure, it isusually reasonable to simulate stable storage with only two copies.

Block transfer can result in: successful completion, partial failure (destination block has incorrectinformation) and total failure (destination blocks not written to).


53

If a data transfer occurs, the system must detect it and recover the block to a consistent state. Todo so, the system maintains two physical blocks for each logical database block, written asfollows:

1. Write information to first physical block.2. When first is successfully complete, write same information to second physical block.3. Output is complete, only after the second write successfully completes.

17.17 Would it be realistic for a DBMS to dynamically maintain a wait-for-graph rather than create iteach time the deadlock detection algorithm runs? Explain your answer.

Yes, could do this by maintaining the WFG in memory and only update directed edges thatchange.


54

Chapter 18 Query Processing

Review Questions

18.1 What are the objectives of query processing?

The aims of query processing are to transform a query written in a high-level language,typically SQL, into a correct and efficient execution strategy expressed in a low-levellanguage (implementing relational algebra), and to execute the strategy to retrieve the requireddata (see Section 18.1).

18.2 How does query processing in relational systems differ from the processing of low-level querylanguages for network and hierarchical systems?

In first generation network and hierarchical database systems, the low-level procedural querylanguage is generally embedded in a high-level programming language such as COBOL, and itis the programmer’s responsibility to select the most appropriate execution strategy. Incontrast, with declarative languages such as SQL, the user specifies what data is requiredrather than how it is to be retrieved. This relieves the user of the responsibility of determining,or even knowing, what constitutes a good execution strategy and makes the language moreuniversally usable (see start of chapter).

18.4 What are the typical stages of query decomposition?

The typical stages of query decomposition are analysis, normalization, semantic analysis,simplification, and query restructuring (see Section 18.2).

18.5 What is the difference between conjunctive and disjunctive normal form?

Conjunctive normal form: A sequence of conjuncts that are connected with the ∧(AND) operator. Each conjunct contains one or more terms connected by the ∨ (OR) operator.A conjunctive selection contains only those tuples that satisfy all conjuncts.

Disjunctive normal form: A sequence of disjuncts that are connected with the ∨(OR) operator. Each disjunct contains one or more terms connected by the ∧ (AND) operator.A disjunctive selection contains those tuples formed by the union of all tuples that satisfy thedisjuncts.

See Section 18.2 under normalization.

18.6 How would you check the semantic correctness of a query?

See Section 18.3 under semantic analysis.

18.7 State the transformation rules that apply to:(a) Selection operations. See Section 18.3.1: Rules 1, 2, 4, 6, 9.(b) Projection operations. See Section 18.3.1: Rules 3, 4, 7, 10.(c) Theta-join operations. See Section 18.3.1: Rules 5, 6, 7, 11.

18.8 State the heuristics that we should apply to improve the processing of a query.

See Section 18.3.2.

18.9 What type of statistics should a DBMS hold to be able to derive estimates of relational algebraoperations?

See Section 18.4.1.

18.10 Under what circumstances would the system have to resort to a linear search whenimplementing a selection operation?


55

(1) Unordered file, with no indexes.(2) Composite predicate where one of the terms contains an OR condition and the term

requires a linear search.(3) Composite predicate where no attribute can be used for efficient retrieval.

18.11 What are the main strategies for implementing the join operation?

Main strategies for implementing join are: block nested loop join, indexed nested loop join,sort-merge join, hash join. See Section 18.4.3.

18.12 What is the difference between materialization and pipelining?

Materialization - output of one operation is stored in a temporary relation for processing bynext operation.Pipelining - pipeline results of one operation to another operation without creating a temporaryrelation to hold the intermediate result.

See Section 18.5.

18.13 Discuss the difference between linear and non-linear relational algebra trees. Give examplesto illustrate your answer.

Linear tree - relation on one side of each operator is always a base relation. See Section 18.5under Linear Trees.

18.14 What are the advantages and disadvantages of left-deep trees?

The inner relation of a join needs to be examined for each tuple of the outer relation, so innerrelations must always be materialized. With a left-deep tree, the inner relation is always a baserelation, and so already materialized.

Exercises

18.15 Calculate the cost of the three strategies cited in Example 18.1 if the Staff relation has 10,000tuples, Branch has 500 tuples, there are 500 Managers (one for each Branch), and there are10 London branches.

Costs: (1) 10,010,500(2) 30,500(3) 11,520.

18.16 Using the Hotel schema given in the Exercises of Chapter 13, determine whether the followingqueries are semantically correct:

(a) SELECT r.type, r.priceFROM room r, hotel hWHERE r.hotel_number = h.hotel_number AND

h.hotel_name = ‘Grosvenor Hotel’ ANDr.type > 100;

Not semantically correct: hotel_number and hotel_name not in schema; type is character stringand so cannot be compared with an integer value (100).

(b) SELECT g.guest_no, g.nameFROM hotel h, booking b, guest gWHERE h.hotel_no = b.hotel_no AND h.hotel_name = ‘Grosvenor Hotel’;

Not semantically correct: hotel_name not in schema; Guest table not connected to remainderof query.


56

(c) SELECT r.room_no, h.hotel_noFROM hotel h, booking b, room rWHERE h.hotel_no = b.hotel_no AND h.hotel_no = ‘H21’ ANDb.room_no = r.room_no AND type = ‘S’ AND b.hotel_no = ‘H22’;

Not semantically correct: hotel_no cannot be both H21 in Hotel and H22 in Booking.

18.17 Again, using the Hotel schema given in the Exercises of Chapter 13, draw a relational algebratree for each of the following queries and use the heuristic rules given in Section 18.3.2 totransform the queries into a more efficient form:

(a) SELECT r.rno, r.type, r.priceFROM room r, booking b, hotel hWHERE r.room_no = b.room_no AND b.hotel_no = h.hotel_no AND

h.hotel_name = ‘Grosvenor Hotel’ AND r.price > 100;

(b) SELECT g.guest_no, g.nameFROM room r, hotel h, booking b, guest gWHERE h.hotel_no = b.hotel_no AND g.guest_no = b.guest_no AND

h.hotel_no = r.hotel_no AND h.hotel_name = ‘Grosvenor Hotel’ ANDdate_from >= ‘1-Jan-98’ AND date_to <= ‘31-Dec-98’;

Discuss each step and state any transformation rules used in the process.

See Figures 18.1 and 18.2 overleaf.

18.18 Using the Hotel schema, assume the following:

There is a hash index with no overflow on the primary key attributes, Room_No/Hotel_No inRoom.There is a clustering index on the foreign key attribute Hotel_No in Room.There is a B+-tree index on the Price attribute in Room.A secondary index on the attribute Type in Room.

ntuples(Room) = 10000 bfactor(Room) = 200ntuples(Hotel) = 50 bfactor(Hotel) = 40ntuples(Booking) = 100000 bfactor(Booking) = 60ndistincthotel_no(Room) = 50ndistincttype(Room) = 10ndistinctprice(Room) = 500minprice(Room) = 200 maxprice(Room) = 50nlevelshotel_no(I) = 2nlevelstype(I) = 2nlevelsprice(I) = 2 nlfblocksprice(I) = 50

(a) Calculate the cardinality and minimum cost for each of the following selectionoperations:

S1: σroom_no=1 ∧ hotel_no=1(Room)Use Hash: cost = 1; Linear search cost is: [10000/200]/2 = 25

S2: σtype='D'(Room)Use equality condition on clustering index:SCtype(Room) = [10000/10] = 1000Cost: 2+ [1000/200] = 5Linear search cost is: 50

S3: σhotel_no=2(Room)Use equality on clustering index:SChotel_no(Room) = [10000/50] = 200Cost: 2+ [200/200] = 3


57

Linear search cost is: 50

S4: σprice>100(Room)Use inequality on a secondary B+-tree:Cost: 2 + [50/2 + 10000/2] = 5027Linear search cost is: 50

S5: σtype='S' ∧ hotel_no=3(Room)If use secondary index searches on each of the two components, costs are 5 and 3, respectively(from above). Optimizer would then choose search via clustering index on hotel_no, and checkthe remaining predicate in memory.Linear search cost is: 50

S6: σtype='S' ∨ price < 100(Room)Use the secondary indexes on Type and Price, then take the union of the two sets.Linear search cost is: 50

(b) Calculate the cardinality and minimum cost for each of the following join operations:J1: Hotel hotel_no Room

Assume nbuffer = 100Block Nested Loop 102, buffer has only 1 block for Hotel and Room.

52, all of Hotel fits into bufferIndexed Nested Loop 152, using primary key indexSort-Merge 302 unsorted

52 sortedHash 156 if hash index fits in memory

J2: Hotel hotel_no BookingBlock Nested Loop 33336, buffer has only 1 block for Hotel andBooking.

16669, all of Hotel fits into bufferIndexed Nested Loop 152, using primary key indexSort-Merge 250007 unsorted


J3: Room room_no BookingBlock Nested Loop 833400, buffer has only 1 block for Room andBooking.

16717, all of Room fits into bufferIndexed Nested Loop 30050, using clustering indexSort-Merge 250305 unsorted


J4: Room hotel_no HotelBlock Nested Loop 150, buffer has only 1 block for Room and Hotel.

52, all of Room fits into bufferIndexed Nested Loop 30050, using clustering indexSort-Merge 302 unsorted


J5: Booking hotel_no HotelBlock Nested Loop 50001, buffer has only 1 block for Hotel andBooking.

17008, if (nbuffer-2) blocks for Booking16669, if all of Booking fits into buffer

Indexed Nested Loop 1666716667, using linear searchSort-Merge 250007 unsorted


J6: Booking room_no Room


58

Block Nested Loop 850017, buffer has only 1 block for Booking andRoom.

25171, if (nbuffer-2) blocks for Booking16717, all of Booking fits into buffer

Indexed Nested Loop 1666716667, using clustering indexSort-Merge 250305 unsorted


(c) Calculate the cardinality and minimum cost for each of the following projectionoperations:P1: Πhotel_no(Hotel)

Duplicate elimination using sorting: 4Duplicate elimination using hashing: 4 (cardinality of result stays same

because Hotel_No is key attribute)

P2: Πhotel_no(Room)

Duplicate elimination using sorting: 350Duplicate elimination using hashing: 51 (cardinality of result estimated as

SChotel_no(Room) = 200, occupying 1block)

P3: Πprice(Room)


SCprice(Room) = 20, occupying 1block)

P4: Πtype(Room)


SChotel_no(Room) = 1000, occupying 5blocks)

P5: Πhotel_no, price(Room)

Duplicate elimination using sorting: 350Duplicate elimination using hashing: 51 (cannot be any more than cost of

P2, P3)

18.19 Modify the block nested loop join and the indexed nested loop join algorithms presented inSection 18.4.3 to read (nbuffer - 2) blocks of the outer relation R at a time, rather than oneblock at a time.

Hint: Add an extra outer loop to each algorithm to read (nbuffer-2) blocks of R in each time ratherthan process each block of R individually.


59

Π r.room_no, r.type, r.price

BR

σr.room_no=b.room_no ∧ b.hotel_no=h.hotel_no ∧

h.hotel_name=’Grosvenor Hotel’ ∧ r.price >

×

×

H


×

σb.hotel_no=h.hotel_no

B

R

× H

σr.price>100

σr.room_no=b.room σh.hotel_name='Grosvenor Hotel'


B

R

Hσr.price>100

σ h.hotel_name='Grosvenor Hotel'

b.hotel_no=h.hotel_no

r.room_no=b.room_no


BR H

σr.price>100 σh.hotel_name='Grosvenor Hotel'

b.hotel_no=h.hotel_no

r.room_no=b.room

Π b.room_no,b.hotel_no

Π h.hotel_no


60

Π g.guest_no,g.name

B×

σ b.hotel_no=h.hotel_no ∧ r.room_no=b.room_no ∧ h.hotel_no=r.hotel_no ∧

h.hotel_name=’Grosvenor Hotel’ ∧ date_from>= '1-Jan-98' ∧ date_to <='31-Dec-98'

×

×

G


×

σg.guest_no=b.guest_no

B

×

G

σh.hotel_no=r.hotel_no

σh.hotel_no=b.hotel_no

σh.hotel_name='Grosvenor Hotel'


B

H

B

σ h.hotel_name='Grosvenor Hotel'

g.guest_no=b.gues_no

h.hotel_no=b.hotel_no

HR

×

H

R

σ date_from>= '1-Jan-98' ∧ date_to <='31-Dec-98'

σ date_from>= '1-Jan-98' ∧ date_to <='31-Dec-98'h.hotel_no=r.hotel_no

R


61


R

B

G

σh.hotel_name='Grosvenor Hotel'

g.guest_no=b.guest_no

hotel_no=r.hotel_no

Π r.hotel_no


h.hotel_no=b.hotel_no

σ date_from>= '1-Jan-98' ∧ date_to <='31-Dec-98'

Π b.hotel_noΠ h.hotel_no

H


62

Part Five Current TrendsChapter 19 Distributed DBMSs - Concepts and Design

Review Questions

19.1 Explain what is meant by a DDBMS and discuss the motivation in providing such a system.

See Section 19.1.1; motivation given at start of Section 19.1.

19.2 Compare and contrast a DDBMS with distributed processing. Under what circumstances wouldyou choose a DDBMS over distributed processing?

Distributed processing defined at end of Section 19.1.1. Would choose a DDBMS, for example, ifeach site needed control over its own data, sites had their own existing DBMSs, communicationcosts would be significantly reduced, and so on.

19.3 Compare and contrast a DDBMS with a parallel DBMS. Under what circumstances would youchoose a DDBMS over a parallel DBMS?

Parallel DBMS defined at end of Section 19.1.1.

19.4 Discuss the advantages and disadvantages of a DDBMS.

See Section 19.1.2.

19.5 One problem area with DDBMSs is that of distributed database design. Discuss the issues thathave to be addressed with distributed database design. Discuss how these issues apply to theglobal system catalog.

The question that is being addressed is how the database and the applications that run against itshould be placed across the sites. Two basic alternatives: partitioned or replicated. In partitionedscheme database is divided into a number of disjoint partitions each of which is placed at adifferent site. Replicated designs can be fully or partially replicated.

Two fundamental design issues are fragmentation and distribution. Mostly involves mathematicalprogramming to minimize combined cost of storing the database, processing transactions againstit, and communication. Problem is NP-hard; therefore proposed solutions are based on heuristics.

The global system catalog (GSC) is only relevant if we talk about a distributed DBMS or multi-DBMS that uses a global conceptual schema. Problems are similar to above. Briefly, a GSC maybe either global to entire database or local; it may be maintained centrally at one site, or in adistributed fashion over a number of sites; finally, replication - there may be a single copy of thedirectory or multiple copies. These three dimensions are orthogonal to one another.

19.6 What are the strategic objectives for the definition and allocation of fragments?

See start of Section 19.4.

19.7 Define and contrast alternative schemes for fragmenting a global relation. State how you wouldcheck for correctness to ensure that the database does not undergo semantic change duringfragmentation.

Alternative schemes are: primary horizontal, vertical, mixed, and derived horizontalfragmentation (see Section 19.4).Correctness rules are: completeness, reconstruction, and disjointness (see Section 19.4 again).

19.8 What layers of transparency should be provided with a DDBMS? Give justification for youranswer.

See Section 19.5.


63

19.9 A DDBMS must ensure that no two sites create a database object with the same name. Onesolution to this problem is to create a central name server. What are the disadvantages with thisapproach? Propose an alternative approach that overcomes these disadvantages.

See Section 19.5.1 - Naming Transparency.

Problems with the central name server, which has the responsibility for ensuring uniqueness of allnames in the system, are:

• loss of some local autonomy• performance problems, if the central site becomes a bottleneck• low availability, if the central site fails, the remaining sites cannot create any new database

objects.

An alternative solution, is to prefix an object with the identifier of the site that created it. Forexample, a relation BRANCH created at site S1 might be named S1.BRANCH. Similarly, wewould need to be able to identify each fragment and each of its copies. Thus, copy 2 of fragment 3of the branch relation created at site S1 might be referred to as S1.BRANCH.F3.C2. However, thisresults in loss of distribution transparency.

An approach which resolves the problems with both these solutions uses aliases for each databaseobject. Thus, S1.BRANCH.F3.C2 might be known as local_branch by the user at site S1. TheDDBMS has the task of mapping aliases to the appropriate database object.

Exercises

A multinational engineering company has decided to distribute its project management information at theregional level in mainland Britain. The current centralised relational schema is as follows:-

Employee (NIN, First_Name, Last_Name, Address, Birth_Date, Sex, Salary, Tax_Code,Dept_No)

Department (Dept_No, Dept_Name, Manager_NIN, Business_Area_No, Region_No)Projects (Proj_No, Proj_Name, Contract_Price, Project_Manager_NIN, Dept_No)Works_On (NIN, Proj_No, Hours_Worked)Business (Business_Area_no, Business_Area_Name)Region (Region_No, Region_Name)

where Employee contains employee details and the national insurance number NIN is the key. Department contains department details and Dept_No is the key. Manager_NIN identifies

the employee who is the manager of the department. There is only one managerfor each department.

Projects contains details of the projects in the company and the key is Proj_No. Theproject manager is identified by the Project_Manager_NIN, and thedepartment responsible for the project by Dept_No.

Works_on contains details of the hours worked by employees on each project and (NIN,Proj_No) forms the key.

Business contains names of the business areas and the key is Business_Area_No.and Region contains names of the regions and the key is Region_No.

Departments are grouped regionally as follows:

Region 1: Scotland; Region 2: Wales; Region 3: England

Information is required by business area which covers: Software Engineering, Mechanical Engineeringand Electrical Engineering. There is no Software Engineering in Wales and all Electrical Engineeringdepartments are in England. Projects are staffed by local department offices.

As well as distributing the data regionally, there is an additional requirement to access the employee dataeither by personal information (by Personnel) or by work related information (by Payroll).


64

19.10 Draw an Entity-Relationship (ER) diagram to represent this system.

For simplicity use crow’s foot notation:

19.11 Produce a distributed database design for the above system and include:

(a) a suitable fragmentation schema for the system;(b) in the case of primary horizontal fragmentation, a minimal set of predicates;(c) the reconstruction of global relations from fragments.

State any assumptions necessary to support your design.

Possible solution as follows:

Don't fragment Business/Region - replicate relations at all sites - only contain a small number ofrecords.

DepartmentUse primary horizontal fragmentation for Department with minterm predicates :

D1 Region = 'Scotland' and Business_area = 'SE'D2 Region = 'Scotland' and Business_area = 'ME'D3 Region = 'Wales' and Business_area = 'ME'D4 Region = 'England' and Business_area = 'SE'D5 Region = 'England' and Business_area = 'ME'D6 Region = 'England' and Business_area = 'EL'

Reconstruction: D1 ∪ D2 ∪ D3 ∪ D4 ∪ D5 ∪ D6

EmployeeUse vertical fragmentation for Employee:

E1: Πnin,first_name,last_name,address,birth_date,sex,dept_noEmployeeE2: Πnin,salary,tax_codeEmployee

Then used derived fragmentation on fragment E1:

Eii: E1 dept_no Di 1 ≤ i ≤ 6

Reconstruction: (E11 ∪ E12 ∪ E13 ∪ E14 ∪ E15 ∪ E16 ) nin E2

ProjectsUse derived fragmentation for Projects:

Pi: Projects dept_no Di 1 ≤ i ≤ 6

Employee Works_On

Project

Business

Department

Region


65

Reconstruction: (P1 ∪ P2 ∪ P3 ∪ P4 ∪ P5 ∪ P6 )

Works_onUse derived fragmentation for Works_on:

Wi: Works_on ninE1i 1 ≤ i ≤ 6

19.12 Repeat Exercise 19.11 for the DreamHome case study presented in Section 1.7.

Possible solution as follows:

Don't fragment Branch - replicate relations at all sites - only contain a small number of records.

Property_for_RentUse primary horizontal fragmentation for Property_for_Rent with minterm predicates (forexample):

P1j asking_price ≤ 39999 AND bno = jP2j 40000 ≤ asking_price ≤ 69999 AND bno = jP3j asking_price ≥ 70000 AND bno = j

1 ≤ j ≤ maximum number of branches

Reconstruction: ∪ (P1i ∪ P2i ∪ P3i)i=1

StaffAssume salaries paid by head office (branch 1 say), so use vertical fragmentation first:

S1: Πsno, fname, lname, address, tel_no, bnoStaffS2: Πsno, position, sex, dob, salary, ninStaff

Then use horizontal fragmentation on fragment S1:

S1i: σbno= i S1 1 ≤ i ≤ j

Reconstruction: (S11 ∪ S12 ∪ S13 … ∪ S1j ) nin S2

RenterUse horizontal fragmentation:

Ri: σbno= i Renter 1 ≤ i ≤ j

Reconstruction: (R1 ∪ R2 ∪ R3 … ∪ Rj )

Viewing, OwnerUse derived fragmentation for Viewing and Owner:

Vik: Viewing SJ Pik 1 ≤ i ≤ 3, 1 ≤ k ≤ jOik: Owner SJ Pik 1 ≤ i ≤ 3, 1 ≤ k ≤ j

Reconstruction: as for Property_for_Rent

19.13 In Section 19.5.1 when discussing naming transparency, we proposed the use of aliases touniquely identify each replica of each fragment. Provide an outline design for the implementationof this approach to naming transparency.

FUNCTION map(name){IF name appears in the replica table


66

THENresult = name of replica of name;

IF name appears in the fragment tableTHEN {

result = expression to construct fragment;FOR each iname IN result {replace iname in result with map(iname);}

}RETURN result;}

IF name appears in the alias tableTHENexpression = map(name);ELSEexpression = name;


67

Chapter 20 Distributed DBMSs - Advanced Concepts

Review Questions

20.1 In a distributed environment, locking-based algorithms can be classified as centralized, primarycopy, or distributed. Compare and contrast these algorithms.

See Section 20.2.3.

20.2 One of the most well-known methods for distributed deadlock detection was developed byObermarck. Explain how Obermarck’s method works and how deadlock is detected and resolved.

See Section 20.3 under Distributed Deadlock Detection.

20.3 Outline two alternative two-phase commit topologies to the centralized topology.

Alternative topologies: linear and distributed 2PC. See end of Section 20.4.3.

20.4 Explain the term non-blocking protocol and explain why two-phase commit protocol is not a non-blocking protocol.

A non-blocking protocol should cater for both site and communication failures to ensure that thefailure of one site will not affect processing at another site. In other words, operational sitesshould not be left blocked.

In the event that a participant has voted COMMIT but has not received global decision and isunable to communicate with any other site that knows the decision, that site is blocked. Although2PC has a cooperative termination protocol that reduces the likelihood of blocking, blocking isstill possible and the blocked process will just have to keep on trying to unblock as failures arerepaired.

20.5 Discuss how the three-phase commit protocol is a non-blocking protocol in the absence ofcomplete site failure.

See Section 20.4.4.The basic idea of 3PC is to remove the uncertainty period for participants who have voted commitand are waiting for the global abort or global commit from the coordinator. 3PC introduces a thirdphase, called pre-commit, between voting and global decision.

20.6 Compare and contrast the different ownership models for replication. Give examples toillustrate your answer.

The main types of ownership are master/slave, workflow, and update-anywhere, sometimesreferred to as peer-to-peer or symmetric replication. See Section 20.6.1.

20.7 Compare and contrast the database mechanisms for replication.

Discuss table snapshots versus database triggers - see Section 20.6.1 under both these headings.

Exercises

20.8 Give full details of the centralized two-phase commit protocol in a distributed environment.Outline the algorithms for both coordinator and participants.

Algorithm (a) 2PC coordinator algorithm(a) begin

STEP C1 VOTE INSTRUCTIONwrite 'begin global commit' message to logsend 'vote' message to all participantsdo until votes received from all participants


68

waiton timeout go to STEP C2b

end-do

STEP C2a GLOBAL COMMITif all votes are 'commit'then begin

write 'global commit' record to logsend 'global commit' to all participants

end

STEP C2b GLOBAL ABORTat least one participant has voted abort or coordinator has timed outelse begin

write 'global abort' record to logsend 'global abort' to all participants

endend-if

STEP C3 TERMINATIONdo until acknowledgement received from all participants

waitend-dowrite 'end global transaction record' to logfinish

end

Algorithm (b) 2PC participants algorithm

(b) begin

STEP P0 WAIT FOR VOTE INSTRUCTIONdo until 'vote' instruction received from coordinator

waitend-do

STEP P1 VOTEif vote = 'commit' then send 'commit' to coordinatorelse send 'abort' and go to STEP P2bdo until global vote received from coordinator

waitend-do

STEP P2a COMMITif global vote = 'commit'

then perform local commit processing

STEP P2b ABORTat least one participant has voted abortelse perform local abort processing

end-if

STEP P3 TERMINATIONsend acknowledgement to coordinatorfinish

end


69

Algorithm Cooperative termination protocol for 2PCbegin

do while P0 is blocked

STEP 1 HELP REQUESTED FROM PiP0 sends a message to Pi asking for help to un-blockif Pi knows the decision (Pi received global commit/abort or Pi unilaterally aborted)

then beginPi conveys decision to P0P0 unblocks and finishes

endend-if

STEP 2 HAS Pi VOTED?if Pi has not voted

then beginPi unilaterally abortsP0 told to abortP0 unblocks and finishes

endend-if

STEP 3 Pi CANNOT HELP; TRY Pi+1next Pi

end-do end

Algorithm 2PC participant restart following failure

begindo while Pr is blocked

STEP 1 ASCERTAIN STATUS OF Pr IMMEDIATELY PRIOR TO FAILURE if Pr voted'commit'

then go to STEP 2else begin

Pr voted 'abort' prior to failure or had not votedPr aborts unilaterallyPr recovers independently and finishesend

end-if

STEP 2 IS GLOBAL DECISION KNOWN?if Pr knows global decision

then beginPr takes action in accordance with global decisionPr recovers independently and finishesend

end-if

STEP 3 Pr CANNOT RECOVER INDEPENDENTLY AND ASKS FOR HELPPr asks for help from participant Pr+1 using the cooperative termination protocol

end-doend


70

20.9 Give full details of the three-phase commit protocol in a distributed environment. Outline thealgorithms for both coordinator and participants.

Algorithm (a) 3PC coordinator algorithm(a) begin

STEP C1 VOTE INSTRUCTIONwrite 'begin global commit' message to logsend 'vote' message to all participantsdo until votes received from all participants

waiton timeout go to STEP C2b

end-do

STEP C2a PRE-COMMITif all votes are 'commit'then begin

write 'pre-commit' message to logsend 'pre-commit' message to all participants

end

STEP C2b GLOBAL ABORTat least one participant has voted abort or coordinator has timed outelse begin

write 'global abort' record to logsend 'global abort' to all participants

go to STEP 4end

end-if

STEP C3 GLOBAL COMMITdo until all (pre-commit) acknowledgements received

waitend-dowrite 'global commit' record to logsend 'global commit' to all participantsend

STEP C4 TERMINATIONdo until acknowledgement received from all participants

waitend-dowrite 'end global transaction record' to logfinish

end

Algorithm (b) 3PC participants algorithm

(b) begin

STEP P0 WAIT FOR VOTE INSTRUCTIONdo until 'vote' instruction received from coordinator

waitend-do

STEP P1 VOTEif participant is prepared to committhen send 'commit' message to coordinatorelse send 'abort' message to coordinator and go to STEP P2bdo until global vote received from coordinator


71

waitend-do

STEP P2a PRE-COMMITif global instruction = 'pre-commit'

then go to STEP P3 (and wait for global commit)end-if

STEP P2b ABORTat least one participant has voted abortperform local abort processinggo to STEP P4

STEP P3 COMMITdo until 'global commit' received from coordintaor

waitend-doperform local commit processing

STEP P4 TERMINATIONsend acknowledgement to coordinatorfinish

end

20.10 Analyze the DBMSs that you are currently using and determine the support each provides forthe X/Open DTP model and for data replication.


20.11 You have been asked by the Managing Director of DreamHome to investigate the datadistribution requirements of the organization and to prepare a report on the potential use of adistributed DBMS. The report should compare the technology of the centralized DBMS withthat of the distributed DBMS, and should address the advantages and disadvantages ofimplementing a DDBMS within the organization, and any perceived problem areas. Thereport should also address the possibility of using a replication server to address thedistribution requirements. Finally, the report should contain a fully justified set ofrecommendations proposing an appropriate solution.

A well-presented report is expected. Justification must be given for any recommendationsmade.

20.12 Consider six transactions T1, T2, T3, T4, and T5 with:

T1 initiated at site S1 and spawning an agent at site S2,T2 initiated at site S3 and spawning an agent at site S1,T3 initiated at site S1 and spawning an agent at site S3,T4 initiated at site S2 and spawning an agent at site S3,T5 initiated at site S3.

The locking information for these transactions is shown in Table 1.


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Transaction Data items locked bytransaction

Data items transaction iswaiting for

Site involved inoperations

T1 x1 x8 S1

T1 x6 x2 S2

T2 x4 x1 S1

T2 x5 S3

T3 x2 x7 S1

T3 x3 S3

T4 x7 S2

T4 x8 x5 S3

T5 x3 x7 S3

Table 1

(a) Produce the local wait-for-graphs (WFGs) for each of the sites. What can you conclude from thelocal WFGs?

Conclusion: There is no local deadlock at any site.

(b)Using the above transactions, demonstrate how Obermarck's method for distributed deadlockdetection works. What can you conclude from the global WFG?

Cycle at site 1, so move WFG from Site 1 to site 3. The resulting WFG shows a cycle:

which implies system is in global deadlock and one of the transactions must be selected to beaborted and restarted.

T1 T2

T3

T1 T4 T2 T3

T4 T5

Site 1 Site 2 Site 3

T1 T2

T3

T1 T4 T2 T3

T4 T5

Site 1 Site 2 Site 3

TextText

Text

Sites 1 and 3

T1 T2 T3

T4 T5

Text


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Chapter 21 Introduction to Object DBMSs

Review Questions

21.1 Discuss the general characteristics of advanced database applications.

See Section 21.1.

21.2 Discuss why the weaknesses of relational DBMSs may make them unsuitable for advanceddatabase applications.

See Section 21.2.

21.3 Discuss each of the following concepts in the context of an object data model:

(a) abstraction, encapsulation, and information hiding; See Section 21.3.1(b) objects and attributes; See Section 21.3.2(b) object identity; See Section 21.3.3(c) methods and messages; See Section 21.3.4(d) classes, subclasses, superclasses, and inheritance; See Section 21.3.5/6(e) overloading; See Section 21.3.7(f) polymorphism and dynamic binding. See Section 21.3.8

Give examples using the DreamHome sample data shown in Figure 3.3. - Expect examplessimilar to the ones in the above referenced sections.

Exercises

21.4 Investigate one of the advanced database applications discussed in Section 21.1, or a similarone that handles complex, interrelated data. In particular, examine its functionality, and thedata types and operations it uses. Map the data types and operations to the object-orientedconcepts discussed in Section 21.3.

This is a small student project, the result of which is dependent on the application investigated.However, expect the student to cover not just standard concepts such as objects, attributes,classes, superclasses, but also concepts such as overloading, complex objects, dynamic binding.

21.5 Analyze the relational DBMSs that you are currently using. Discuss the object-orientedfeatures provided by the system. What additional functionality do these features provide?


21.6 For the DreamHome case study introduced in Section 1.7, suggest attributes and methods thatwould be appropriate for Branch, Staff, and Property_for_Rent classes.

The attributes should be similar to those documented in the textbook - see, for example, theappendices at the end of Chapter 8. The student would be expected to come up with standardget/set methods, such as Get_Staff_Salary, Put_Staff_Salary, and then methods for registeringnew properties for rent, new owners, new renters, appointments for viewings, and so on.


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Chapter 22 Object-Oriented DBMSs

Review Questions

22.1 Compare and contrast the different definitions of object-oriented data models.

See Section 22.1.

22.2 What is a persistent programming language and how does it differ from an OODBMS?

A language that provides its users with the ability to (transparently) preserve data acrosssuccessive executions of a program, and even allows such data to be used by differentprograms (See Section 22.1.1). Main difference between PPL and OODBMS is that theOODBMS tends to provide more DBMS-related services.

22.3 Discuss the difference between the two-level storage model used by conventional DBMSs andthe single-level storage model used by OODBMSs.

See Section 22.2.

22.4 How does this single-level storage model affect data access?

See Section 22.2.1.

22.5 Discuss the main strategies that can be used to create persistent objects.

Checkpointing, serialization, and explicit paging. Explicit paging includes reachability-based andallocation-based persistence (see Section 22.3.1).

22.6 What is pointer swizzling? Discuss the different approaches to pointer swizzling.

Action of converting OIDs to main memory pointers (See Section 22.3.3).

22.7 Discuss the types of transaction protocols that can be useful in design applications.

See Sections 22.4.1 and 19.4.

22.8 Discuss why version management may be a useful facility for some applications.

There are many applications that need access to the previous state of an object. For example, thedevelopment of a particular design is often an experimental and incremental process, the scope ofwhich changes with time. It is therefore necessary in databases that store designs to keep track ofthe evolution of design objects and the changes made to a design by various transactions.

22.9 Discuss why schema control may be a useful facility for some applications.

Engineering design is an incremental process and evolves with time. To support this process,applications require considerable flexibility in dynamically defining and modifying the databaseschema. For example, it should be possible to modify class definitions, the inheritance structure,and specifications of attributes and methods without requiring system shutdown.

22.10 Compare and contrast the different architectures for an OODBMS.

See Section 22.4.4.

Exercises

22.11 For the relational schema in the exercises at the end of Chapter 3, suggest a number of methodsthat would be applicable to the system. Produce an object-oriented schema for the system.


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Some methods might be:

create_new_hotel destroy_hotel change_hotel_namecreate_new_room destroy_room change_room_pricechange_room_typecreate_new_guest destroy_guest change_guest_namechange_guest_addresscreate_new_booking destroy_booking change_room_numberchange_dates

interface Hotel {(extent hotelskey Hotel_No)

attribute string Hotel_No;attribute string Name;attribute string Address;attribute Set<struct<string room_no, char type, float price>> rooms;

relationship List<Booking> has_booking inverse Booking::booking_for;

create_new_hotel();destroy_hotel() raises(no_such_hotel);create_new_room();destroy_room(in room_no:string) raises(no_such_room);change_room_price(in room_no:string, in float) raises(no_such_room);change_room_type(in room_no:string, in char) raises(no_such_room);

}

interface Guest {(extent guestskey Guest_No)

attribute string Guest_No;attribute string Name;attribute string Address;

relationship List<Booking> booked_for inverse Booking::booking_by;

create_new_guest();destroy_guest() raises(no_such_guest);

}

interface Booking {(extent bookingskey (Hotel_No, Guest_No, Date_From))

attribute string Hotel_No;attribute string Guest_No;attribute date Date_From;attribute date Date_To;attribute string Room_No;

relationship <Hotel> booking_for inverse Hotel::has_booking;relationship Guest booked_by inverse Guest::booking_for;

create_new_booking(in Hotel, in Guest) raises(no_such_hotel, no_such_guest, hotel_full)destroy_booking() raises(no_such_booking);change_room_number();change_dates();

}


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22.12 Using the schema produced above, show how the following queries would be written in OQL:

(a) List all hotels.

hotel

or to sort them:

sort h IN hotel by h.name

(b) List all single rooms with a price below £20.00 per night.

SELECT h.roomsFROM h IN hotelWHERE h.rooms.price < 20

(c) List the names and addresses of all guests.

SELECT STRUCT(name:g.name, address:g.address)FROM g IN guest

(d) List the price and type of all rooms at the Grosvenor Hotel.

type prices {attribute price : float; type: char;}prices (SELECT STRUCT(price:h.rooms.price, type:h.rooms.type)

FROM h IN hotelWHERE h.name = 'Grosvenor Hotel'

(e) List all guests currently staying at the Grosvenor Hotel.

SELECT gFROM g IN guests b IN g.booked_for h IN b.booking_forWHERE b.date_from <= '01-01-99' AND

b.date_to >= '01-01-99' ANDh.name = 'Grosvenor Hotel'

(substitute '01-01-99' for today's date).

(f) List the details of all rooms at the Grosvenor Hotel, including the name of the gueststaying in the room, if the room is occupied.

define occupied asSELECT STRUCT(h.rooms,

SELECT g.nameFROM g IN guest b IN g.booked_for h IN b.booking_forWHERE b.date_from <= '01-01-99' AND

b.date_to >= '01-01-99' ANDh.name = 'Grosvenor Hotel')

FROM h IN hotelUNIONSELECT STRUCT(h.rooms, nil)FROM h IN hotel, y IN occupiedWHERE h.rooms.room_no != y.room_no


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(g) List the guest details (guest_no, name and address) of all guests staying at theGrosvenor Hotel.

SELECT STRUCT(guest_no:g.guest_no, name:g.name, address:g.address)FROM g IN g IN guest

b IN g.booked_for h IN b.booking_for

WHERE b.date_from <= '01-01-99' ANDb.date_to >= '01-01-99' ANDh.name = 'Grosvenor Hotel'


22.13 Produce an object-oriented database design for the DreamHome case study presented in Section1.7. State any assumptions necessary to support your design.

Partial solution is as follows:

interface Branch {(extent brancheskey Branch_No)

attribute string Branch_No;attribute <struct<string Street, string Area, string City, string Postcode>> Address;attribute string Tel_No;attribute string Fax_No;attribute date Manager_Start_Date;attribute float Bonus_Payment;attribute string Car_Allowance;

relationship Set<Renter> registers inverse Renter::is_registered_with;relationship Staff has inverse Staff::works_in;relationship Manager is_managed_by inverse Manager::manages;relationship Set<Property_for_Rent> offers inverse Property_for_Rent is_offered_by;

destroy_branch(in Branch) raises(no_such_branch);}

interface Staff {(extent staffkey Staff_No, NIN)

attribute string Staff_No;attribute <struct<string FName, string LName>>;attribute string Address;attribute string Tel_No;attribute char Sex;attribute date DOB;attribute string Position;attribute float Salary;attribute date Date_Joined;attribute string NIN;attribute <struct<string NName, string Relationship, string Address, string Tel_No>> Next_of_Kin;

relationship Branch works_in inverse Branch::has;relationship Set<Property_for_Rent> oversees inverse Property_for_Rent is_overseen_by;relationship Set<Inspection> carryout inverse Inspection::is_carried_out_by;

destroy_staff(in Staff) raises(no_such_person);increase_salary(in Staff, in float) raises(no_such_person);oversee_property(in Staff, in Property_for_Rent) raises(no_such_person, no_such_property);move_branch(in Staff, in from::Branch, in to::Branch) raises(no_such_person, no_such_branch);


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}

interface Manager::Staff {(extent managers)

relationship Branch manages inverse Branch::is_managed_by;}

interface Secretary::Staff {(extent secretaries)

attribute integer Typing_Speed;

relationship Set<Staff> supports inverse Staff::is_supported_by;}

interface Supervisor::Staff {(extent supervisors)

relationship Set<Staff> supervises inverse Staff::is_supervised_by;}

interface Property_for_Rent {(extent rentalskey Property_No)

attribute string Property_No;attribute <struct<string Street, string Area, string City, string Postcode>> Address;attribute string Type;attribute integer Rooms;attribute float Rent;

relationship Staff is_overseen_by inverse Staff::oversees;relationship Branch is_offered_by inverse Branch::offers;relationship Set<Inspection> undergoes inverse Inspection::inspection_for;relationship Owner is_owned_by inverse Owner::owns;relationship Set<Viewing> takes inverse Viewing::is_taken_for;relationship Set<Lease_Agreement> associated_with inverse Lease_Agreement:: is_associated_with;

destroy_pfp(in Property_for_Rent) raises(no_such_property);rent_out(in Property_for_Rent) raises(no_such_property);

}

interface Renter {(extent renterskey Renter_No)

attribute string Renter_No;attribute <struct<string FName, string LName>>;attribute string Address;attribute string Tel_No;attribute string Pref_Type;attribute float Max_Rent;

relationship Branch is_registered_with inverse Branch::registers;relationship Set<Viewing> attends inverse Viewing::is_attended_by;relationship Set<Lease_Agreement> holds inverse Lease_Agreement::is_held_by;

rent_property(in Renter, in Property_for_Rent) raises(no_such_renter, no_such_property);register(in Renter, in Branch) raises(no_such_renter, no_such_branch);arrange_viewing(in Renter, in Property_for_Rent) raises(no_such_renter, no_such_property);destroy_renter(in Renter) raises(no_such_renter);

}


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22.14 Produce an object-oriented database design for the Wellmeadows Hospital student projectpresented in Appendix A. State any assumptions necessary to support your design.

The following is a sample schema, not all methods are shown:

interface Ward {(extent wardskey WardNo)

attribute string WardNo;attribute string WName;attribute string Location;attribute integer TotalBeds;attribute string TelExtn;

relationship List<Waiting_List> has_waiting_listinverse Waiting_List::is_waiting_for{order_by Waiting_List::ListDate, WardReq};

relationship List<Staff_Rota> has_staff_rotainverse Staff_Rota::on_shift_for{order_by Staff_Rota::WardNo, WeekNo};

relationship Set<Requisition> needs_req inverse Requisition::req_for;relationship Nurse is_managed_by inverse Nurse::manages;

create_new_ward();destroy_ward(in Ward) raises(no_such_ward);assign_charge_nurse(in Nurse) raises(no_such_nurse);

}

interface Staff {(extent staffkey StaffNo, NIN)

attribute string StaffNo;attribute <struct<string FName, string LName>>;attribute string Address;attribute string TelNo;attribute date DOB;attribute char Sex;attribute string NIN;attribute string Position;attribute float Salary;attribute integer SScale;attribute integer WeekHrs;attribute char ContType;attribute char TypePay;attribute Set<struct< date QDate, string QType, string Institution>> Qualification;attribute Set<struct< date SDate, date FDate, string Position, string OrgName>> Work_Experience;

relationship List<Staff_Rota> has_rotainverse Staff_Rota::shift_for{order_by Staff_Rota::StaffNo, WeekNo};

destroy_staff(in Staff) raises(no_such_person);increase_salary(in Staff, in float);

}

interface Nurse::Staff {(extent nurses)

relationship Ward manages inverse Ward::is_managed_by;relationship Set<Requisition> makes_req inverse Requisition::made_by;


80

create_new_nurse();make_requisition(in Nurse, in Pharmaceutical) raises(no_such_drug);

}

interface Consultant::Staff {(extent consultants)

relationship List<Appointment> seesinverse Appointment::is_seen_by{order_by Appointment::ConsStaffNo, Adate, ATime};

create_new_consultant();cancel_appointment(in Consultant, in Appointment);

}

interface Staff_Rota {(extent rotaskey (StaffNo, WeekNo))

attribute integer Shift;attribute integer WeekNo;attribute string StaffNo;attribute string WardNo;

relationship Ward on_shift_for inverse Ward::has_staff_rota;relationship Staff shift_for inverse Staff::has_rota;

create_new_rota();}

interface Patient {extent patients;key PatNo;

attribute string PatNo;attribute <struct<string FName, string LName>>;attribute string Address;attribute string TelNo;attribute date DOB;attribute char Sex;attribute char MStatus;attribute date DateReg;attribute <struct< string NName, string NRelationship, string NAddress, string NTelNo>> Next-Of-Kin;attribute Set<struct<date OutPatDate, time OutPatTime>> outpatient_appointment;

relationship Doctor has_gp inverse Doctor::gp_for;relationship List<Appointment> has_appt inverse Appointment::is_appt_for;relationship List<Medication> takes inverse Medication::is_taken_by;relationship List<Waiting_List> waits_for inverse Waiting_List::is_on;

create_new_patient();change_doctor(in from:Doctor, in to:Doctor) raises(unknown_doctor);cancel_appointment(in Appointment) raises(no_appointment);

}

interface Doctor {(extent gpskey (DocName, ClinicNo))

attribute string DocName;attribute string ClinicNo;


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attribute string Addressattribute string TelNo;

relationship List<Patient> gp_for inverse Patient::has_gp;

create_new_doctor();}

interface Appointment {(extent appointmentskey AppNo)

attribute string AppNo;attribute date ADate;attribute time ATime;attribute string RoomNo;

relationship Patient is_appt_for inverse Patient::has_appt;relationship Consultant is_seen_by inverse Consultant::sees;

}

interface Waiting_List {(extent waitingkey (PatNo, ListDate))

attribute date ListDate;attribute interval Duration;attribute date PlacedDate;attribute date ExLeaveDate;attribute date ActLeaveDate;attribute string BedNo;

relationship Ward is_waiting_for inverse Ward::has_waiting_list;relationship Patient is_on inverse Patient::waits_for;

add_to_list(in Patient, in Ward) raises(no_such_patient, no_such_ward);remove_from_list(in Patient, in Ward) raises(no_such_patient, no_such_ward);

}

interface Medication {(extent medicationskey (PatNo, DrugNo, SDate))

attribute string UnitsDay;attribute string AMethod;attribute date SDate;attribute date FDate;

relationship Patient is_taken_by inverse Patient::takes;relationship Pharmaceutical contains inverse Pharmaceutical::contained_in;

add_medication(in Patient, in Pharmaceutical) raises(no_such_patient, no_such_drug);change_medication(in Patient, in Pharmaceutical) raises(no_such_patient, no_such_drug);;

}

interface Pharmaceutical {(extent drugskey DrugNo)

attribute string DrugNo;attribute string DName;attribute string Description;attribute string Dosage;


82

attribute string MAdmin;attribute integer QStock;

relationship List<Medication> contained_in inverse Medication::contains;relationship List<Requisition> req_in inverse Requisition::reqs_for;relationship Supplier is_supplied_by inverse Supplier::supplies;

create_drug();remove_drug(in Pharmaceutical) raises(no_such_drug);

}

interface Non-Surgical/Surgical {(extent supplieskey ItemNo)

attribute string ItemNo;attribute string IName;attribute string IDescription;attribute integer QStock;attribute string RLevel;attribute float UnitCost;

relationship List<Requisition> req_in inverse Requisition::reqs_for;

create_item();remove_Item(in Non-Surgical/Surgical) raises(no_such_item);

}

interface Requisition {(extent requisitionskey ReqNo)

attribute string ReqNo;attribute integer QuantReq;attribute date DateOrder;attribute date DateReceive;relationship Ward req_for inverse Ward::needs_req;relationship Nurse made_by inverse Nurse::makes_req;relationship Pharmaceutical reqs_for inverse Pharmaceutical::req_in;relationship Non-Surgical/Surgical reqs_for inverse Non-Surgical/Surgical::req_in;deliver_req(in Requisition) raises(no_such_req);

}

interface Supplier {(extent supplierskey SupplierNo)

attribute string SupplierNo;attribute string SName;attribute string SAddress;attribute string TelNo;attribute string FaxNo;

relationship Set<Pharmaceutical> supplies inverse Pharmaceutical::is_supplied_by;}

22.15 You have been asked by the Managing Director of DreamHome to investigate and prepare areport on the applicability of an Object-Oriented DBMS for the organization. The reportshould compare the technology of the relational DBMS with that of the Object-OrientedDBMS, and should address the advantages and disadvantages of implementing an OODBMSwithin the organization, and any perceived problem areas. Finally, the report should contain afully justified set of conclusions on the applicability of the OODBMS for DreamHome.


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22.16 Using the rules for schema consistency given in Section 22.4.3, consider each of the followingmodifications and state what the effect of the change should be to the schema:

(a) Adding an attribute to a class(b) Deleting and attribute from a class(c) Making a class S a superclass of a class C(d) Removing a class S from the list of superclasses of a class C.

See Section 22.4.3.


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Chapter 23 Object-Relational DBMSs

Review Questions

23.1 What typical functionality would be provided by an ORDBMS?

Many different answers here – see, for example, Section 23.2.1.Expect standard DBMS functionality, plus object management capabilities (types, inheritance,etc), plus ability to extend query optimizer, and define new index types (see Section 23.5).

23.2 What are the advantages and disadvantages of extending the relational data model?

See Section 23.1 under Advantages and Disadvantages.

23.3 What are the main features of the forthcoming SQL standard?

See start of Section 23.4.

23.4 Discuss the extensions required to query processing and query optimization to fully supportthe ORDBMS?

See Section 23.5.

23.5 What are the security problems associated with the introduction of user-defined methods andsuggest some solutions to these problems?

See paragraph before start of Section 23.5.1.

Exercises

23.6 Analyze the relational DBMSs that you are currently using. Discuss the object-orientedfacilities provided by the system. What additional functionality do these facilities provide?


23.7 Consider the relational schema for the Hotel case study given in the Exercises of Chapter 11.Redesign this schema to take advantage of the new features of SQL3/SQL4. Add user-definedfunctions that you consider appropriate.

One possible solution as follows:

CREATE DOMAIN HOTEL_NUMBER AS CHAR(4);CREATE DOMAIN ROOM_TYPE AS CHAR(1)

CHECK(VALUE IN ('S', 'F', 'D'));CREATE DOMAIN ROOM_PRICE AS DECIMAL(5,2)

CHECK(VALUE BETWEEN 10 AND 100);CREATE ROOM_NUMBER AS VARCHAR(4)

CHECK(VALUE BETWEEN '1' AND '100');CREATE DOMAIN GUEST_NUMBER AS CHAR(4);CREATE DOMAIN BOOKING_DATE AS DATETIME

CHECK(VALUE > CURRENT_DATE);

CREATE TYPE hotel_type(hotel_no HOTEL_NUMBER NOT NULL,name VARCHAR(20) NOT NULL,address VARCHAR(50) NOT NULL);

CREATE TABLE hotel OF hotel_type(oid REF(hotel_type) VALUES ARE SYSTEM GENERATED,PRIMARY KEY (hotel_no));


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CREATE TYPE rooms_type(room_no ROOM_NUMBER NOT NULL,hotel_no REF(hotel_type) NOT NULL,type ROOM_TYPE NOT NULL DEFAULT 'S'price ROOM_PRICE NOT NULL);

CREATE TABLE room OF rooms_type(PRIMARY KEY (room_no, hotel_no),FOREIGN KEY (hotel_no) REFERENCES hotel

ON DELETE CASCADE ON UPDATE CASCADE);

CREATE TYPE guest_type(guest_no GUEST_NUMBER NOT NULL,name VARCHAR(20) NOT NULL,address VARCHAR(50) NOT NULL);

CREATE TABLE guest OF guest_type(oid REF(guest_type) VALUES ARE SYSTEM GENERATED,PRIMARY KEY (guest_no));

CREATE TYPE booking_type(hotel_no REF(hotel_type) NOT NULL,guest_no REF(guest_type) NOT NULL,date_from BOOKING_DATE NOT NULL,date_to BOOKING_DATE NULL,room_no ROOM_NUMBER NOT NULL);

CREATE TABLE booking OF booking_type(PRIMARY KEY (hotel_no, guest_no, date_from),FOREIGN KEY (hotel_no) REFERENCES hotel

ON DELETE CASCADE ON UPDATE CASCADE,FOREIGN KEY (guest_no) REFERENCES guest

ON DELETE NO ACTION ON UPDATE CASCADE,FOREIGN KEY (hotel_no, room_no) REFERENCES room

ON DELETE NO ACTION ON UPDATE CASCADE,CONSTRAINT room_bookedCHECK (NOT EXISTS (SELECT *

FROM booking bWHERE b.date_to > booking.date_from ANDb.date_from < booking.date_to ANDAND b.room_no = booking.room_no ANDb.hotel_no = booking.hotel_no)),

CONSTRAINT guest_bookedCHECK (NOT EXISTS (SELECT *

FROM booking bWHERE b.date_to > booking.date_from ANDb.date_from < booking.date_to ANDAND b.guest_no = booking.guest_no)));

23.8 Create SQL3/SQL4 statements for the queries given in Exercise 13.7 - 13.26.

Depends on the solution to the previous question. For example, using the above schema, thesolution to 13.16 would be:

SELECT price, typeFROM room rWHERE r–>hotel.name = 'Grosvenor Hotel';


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23.9 Create an insert trigger that sets up a mailshot table recording the names and addresses of allguests who have stayed at the hotel during the days before and after New Year for the past twoyears.

The aim of this question is to show the difficulty of creating a trigger that does not necessarilyneed to be tied to a modification to a table.

CREATE TRIGGER insert_mailshot_tableAFTER INSERT ON bookingBEGIN

INSERT INTO mailshot(SELECT g.guest_no, g.name, g.addressFROM guest g, booking b1, booking b2WHERE g.guest_no = b1.guest_no AND b1.guest_no = b2.guest_no AND((b1.date_from <= DATE'1998-12-31' AND b1.date_to>=DATE'1998-12-31') OR(b1.date_from >= DATE'1998-12-31' AND b1.date_from <= DATE'1999-01-02') )AND((b1.date_from <= DATE'1997-12-31' AND b1.date_to>=DATE'1997-12-31') OR(b1.date_from >= DATE'1997-12-31' AND b1.date_from <= DATE'1998-01-02') )ANDNOT EXISTS (SELECT * FROM mailshot m

WHERE m.guest_no = g.guest_no);END;

23.10 Repeat Exercise 23.7 for the multinational engineering case study in the Exercises of Chapter19.

Look for solution that uses the features of SQL3/4 such as types with inheritance, SETs, andobject references.

23.11 Create an object-relational schema for the DreamHome case study presented in Section 1.7.Add user-defined functions that you consider appropriate.

Look for solution that uses the features of SQL3/4 such as types with inheritance, SETs, andobject references.

23.12 Create an object-relational schema for the Wellmeadows case study presented in Appendix A.Add user-defined functions that you consider appropriate.

Partial solution (see Exercise for 11.3):

CREATE TYPE ward_type(WardNo VARCHAR(4) NOT NULL,WName VARCHAR(20) NOT NULL,Location VARCHAR(20) NOT NULL,TotalBeds INTEGER,TelExtn VARCHAR(4) NOT NULL,Consultant REF(consultant_type),FUNCTION assign_charge_nurse(W ward_type RESULT, N nurse_type)

RETURNS ward_type...RETURN;

END,);

CREATE TYPE qualification_type(QType VARCHAR(5) NOT NULL,QDate DATE,Institution VARCHAR(30),


87

);

CREATE TYPE work_experience_type(OrgName VARCHAR NOT NULL,SDate DATE,FDate DATE,Position VARCHAR(10)

);

CREATE TYPE name_type (FName VARCHAR(15) NOT NULL,LName VARCHAR(15) NOT NULL

);

CREATE TYPE staff_type( PRIVATE

date_of_birth DATE_CHECK(date_of_birth < CURRENT_DATE), PUBLIC

StaffNo VARCHAR(5) NOT NULL,Name name_type,Address VARCHAR(50) NOT NULL,TelNo VARCHAR(13) NOT NULL,Sex CHAR NOT NULL,NIN VARCHAR(9) NOT NULL,Position VARCHAR(10) NOT NULL,Salary DECIMAL(6,2) NOT NULL,SScale INTEGER NOT NULL,WeekHrs INTEGER NOT NULL,ContType CHAR NOT NULL,TypePay CHAR NOT NULL,qualification SET(qualification_type),work_exp SET(work_experience_type)

FUNCTION get_age (P person_type) RETURNS INTEGER

RETURN/* code to calculate age from date_of_birth */

END,

FUNCTION set_age (P person_type RESULT, DOB: DATE) RETURNS person_type

RETURN/* set date_of_birth */

END) NOT FINAL;

CREATE TYPE nurse_type UNDER staff_type (FUNCTION make_requisition(...);BEGIN

...END

);

CREATE TYPE consultant_type UNDER staff _type (FUNCTION cancel_appointment(...);BEGIN

...END

);

CREATE TABLE ward OF ward_type(oid REF(ward_type) VALUES ARE SYSTEM GENERATED,PRIMARY KEY(wardno));

CREATE TABLE nurse OF nurse_type(


88

oid REF(nurse_type) VALUES ARE SYSTEM GENERATED,PRIMARY KEY(staffno));

CREATE TABLE consultant OF consultant_type(oid REF(consultant_type) VALUES ARE SYSTEM GENERATED,PRIMARY KEY(staffno));

23.13 You have been asked by the Managing Director of DreamHome to investigate and prepare areport on the applicability of an Object-Relational DBMS for the organization. The reportshould compare the technology of the relational DBMS with that of the Object-RelationalDBMS, and should address the advantages and disadvantages of implementing an ORDBMSwithin the organization, and any perceived problem areas. The report should also consider theapplicability of an Object-Oriented DBMS, and a comparison of the two types of systems forDreamHome should be included. Finally, the report should contain a fully justified set ofconclusions on the applicability of the ORDBMS for DreamHome.



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Part Six Future TrendsChapter 24 Web Technology and DBMSs

Review Questions

24.1 Discuss each of the following terms:(a) Internet, intranet, and extranet. See Section 24.1(b) World-Wide Web. See Section 24.1.1(c) HyperText Transfer Protocol (HTTP). See Section 24.1.2(d) HyperText Markup Language (HTML). See Section 24.1.3(e) Uniform Resource Locators (URLs). See Section 24.1.4

24.2 Compare and contrast the two-tier client-server architecture for traditional DBMSs with thethree tier client-server architecture. Why is the latter architecture more appropriate for theWeb?

See Section 24.2.2 (additional information on client-server architecture can be found inSection 2.6.3).

24.3 Discuss the advantages and disadvantages of the Web as a database platform.

See Section 24.2.3.

24.4 Compare and contrast the Common Gateway Interface and server extensions, as approachesfor integrating databases onto the Web.

See Section 24.6.1.

24.5 Discuss, with examples, the security problems that can arise in a Web environment. Whatmechanisms are available to prevent these problems from occurring?

See Section 24.11.

Exercises

24.6 Examine the Web functionality provided by any DBMS that you currently use. Compare thefunctionality of your system with the approaches discussed in Section 24.3 to 24.10.


24.7 Examine the security features provided by the Web interface to your DBMS. Compare thesefeatures with the features discussed in Section 24.11.


24.8 Using an approach to Web-DBMS integration, create a series of forms that display the basetables of the DreamHome case study.

This depends on the type of system/layout preferred. Students should not use any automaticfacilities, such as those provided my Access, but should design the layout themselves.

24.9 Extend the implementation of Exercise 24.8 to allow the base tables to be updated from theWeb browser.

Again, depends on the technology used.

24.10 Repeat Exercises 24.8 and 24.9 for the Wellmeadows case study.

As above.


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24.11 Using any Web browser, look at some of the following Web sites and discover the wealth ofinformation held there:(a) W3C: http://www.w3c.org(b) Microsoft: http://www.microsoft.com(c) Oracle: http://www.oracle.com(d) Informix: http://www.informix.com(e) IBM: http://www.ibm.com(f) Sybase: http://www.sybase.com(g) Javasoft http://www.javasoft.com(h) Gemstone http://www.gemstone.com(i) Objectivity http://www.objectivity.com(j) ObjectStore http://www.idc.com(k)O2 Technology http://www.o2tech.com(l)Poet http://www.poet.com

24.12 You have been asked by the Managing Director of DreamHome to investigate and prepare areport on the feasibility of making the DreamHome database accessible from the Internet. Thereport should examine the technical issues, the technical solutions, address the advantagesand disadvantages of this proposal, and any perceived problem areas. The report shouldcontain a fully justified set of conclusions on the feasibility of this proposal for DreamHome.



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Chapter 25 Data Warehousing

Review Questions

25.1 Describe what is meant by the following terms, when describing the characteristics of the datain a data warehouse:(a) Subject-oriented(b) Integrated(c) Time-variant(d) Non-volatile

See Section 25.1.2.

25.2 Discuss how Online Transaction Processing (OLTP) systems differ from data warehousingsystems.

See Section 25.1.4.

25.3 Discuss the main benefits and problems associated with data warehousing.

For the main benefits of data warehousing see Section 25.1.3 and for the main problemsassociated with data warehousing see Section 25.1.5.

25.4 Present a diagrammatic representation of the typical architecture and main components of adata warehouse.

For a diagram of the typical architecture of a data warehouse see Figure 25.1.

25.5 Describe the characteristics and main functions of the following components of a datawarehouse.

(a) load manager See Section 25.2.2(b) warehouse manager See Section 25.2.3(c) query manager See Section 25.2.4(d) metadata See Section 25.2.8(e) end-user access tools. See Section 25.2.9

25.6 Discuss the activities associated with each of the five primary information flows or processeswithin the data warehouse:(a) Inflow See Section 25.3.1(b) Upflow See Section 25.3.2(c) Downflow See Section 25.3.3(d) Outflow See Section 25.3.4(e) Metaflow. See Section 25.3.5

25.7 What are the three main approaches taken by vendors to provide data extraction, cleansing,and transformation tools?

The three approaches are code generators, database data replication tools and dynamic transformationengines. See Section 25.4.1 for a description of each approach.

25.8 Describe the specialized requirements of a relational database management system (RDBMS)suitable for use in a data warehouse environment.

See Section 25.4.2

25.9 Discuss how parallel technologies can support the requirements of the data warehouse.

See last topic discussed in Section 25.4.2 under the heading Parallel database technologies.

25.10 Discuss the importance of managing metadata and how this relates to the integration of thedata warehouse.


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See Section 25.4.3.

25.11 Discuss the main tasks associated with the administration and management of a datawarehouse.See Section 25.4.4.

25.12 Discuss how data marts differ from data warehouses and discuss the main reasons forimplementing a data mart.

For a discussion on how data marts differ form data warehouses see introductory paragraphsof Section 25.5 and for reasons for implementing a data mart see Section 25.5.1.

25.12 Identify the main issues associated with the development and management of data marts.See Section 25.5.2.

25.13 Describe an approach for the development of the database component of a data warehousethat is capable of supporting decision-making using star, snowflake, and starflake schemas.See Section 25.6.

Exercises

25.15 Design a database suitable for decision-support for the DreamHome case study (see Section1.7). The design should be based on the query requirements for the Director of theorganization.

The student should follow the approach described in Section 25.6. The student must firstidentify and document the types of decision-support queries that the Director of theDreamHome may wish to ask. The student should then produce a design to meet theserequirements.

25.16 Design a database suitable for decision-support for the Wellmeadows case study (seeAppendix A). The design should be based on the query requirements for the Director of thehospital.

The student should follow the approach described in Section 25.6. The student must firstidentify and document the types of decision-support queries that the Director of theWellmeadows Hospital may wish to ask. The student should then produce a design to meetthese requirements.

25.17 Design a database suitable for decision-support for your organization. The design should bebased on the query requirements of the staff your organization.

The student should follow the approach described in Section 25.6. The student must firstidentify and document the types of decision-support queries that the senior staff of yourorganization may wish to ask. The student should then produce a design to meet theserequirements.

25.18 You are asked by the Managing Director of DreamHome to investigate and report on theapplicability of data warehousing for the organization. The report should compare datawarehouse technology with relational DBMSs and should identify the advantages anddisadvantages, and any problem areas associated with implementing a data warehouse. Thereport should reach a fully justified set of conclusions on the applicability of a datawarehouse for DreamHome.



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Chapter 26 OLAP and Data Mining

Review Questions

26.1 Discuss the major characteristics of Multi-dimensional OLAP (MOLAP) and how a datacubesupports multi-dimensional queries.

See Section 26.1.

26.2 Describe the architecture, characteristics, and issues associated with each of the followingcategories of OLAP tools:

(a) MOLAP,(b) ROLAP,(c) MQE.

See Section 26.1.3.

26.3 Discuss some of the possible extensions to SQL that support data analysis and decision-support applications.

See Section 26.1.4.

26.4 Discuss how data mining can realize the value of a data warehouse.

See introductory paragraphs of Section 26.2 and see Section 26.2.1. See also Section 26.2.8.

26.5 Describe the characteristics, associated techniques, and typical applications for each of thefollowing main data mining operations:

(a) Predictive modeling, See Section 26.2.3(b) Database segmentation, See Section 26.2.4(c) Link analysis, See Section 26.2.5(d) Deviation detection. See Section 26.2.6

See Section 26.2.2 for introduction to techniques.

Exercises

26.6 You are asked by the Managing Director of DreamHome to investigate and report on theapplicability of data mining for the organization. The report should describe the technologyand provide a comparison with traditional querying and reporting tools of relational DBMSs.The report should also identify the advantages and disadvantages, and any problem areasassociated with implementing data mining. The report should reach a fully justified set ofconclusions on the applicability of data mining for DreamHome


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