Chapter 5 Logical Database Design and the Relational Model
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Chapter 5 79 Chapter 5 Logical Database Design and the Relational Model Chapter Objectives The purpose of this chapter is to describe in depth the major steps in logical database design, with more emphasis on the relational model. Logical database design is the process of transforming the conceptual data model (described in Chapters 3 and 4) into a logical data model. First, we provide a concise description of the relational data model, including the properties of relations. Next, we describe and illustrate the various types of integrity constraints associated with the relational model. This section introduces SQL table definitions, and the concept of well-structured relations. We then provide a detailed description of the process of transforming EER diagrams into relations. Next, we define normalization and describe the steps in normalizing relations. The chapter concludes with a discussion of merging relations, and techniques for dealing with typical issues that arise during this process. Specific student learning objectives are included in the beginning of the chapter. From an instructor's point of view, the objectives of this chapter are to: 1. Show students the position of logical database design within the overall database development process. This is a key chapter in the textbook, since students will begin to see how their databases will be implemented. 2. Provide students with a solid understanding of the relational data model, including the properties of relations, integrity constraints, and well-structured relations. 3. Discuss the principles and detailed steps involved in mapping EER diagrams to relations. Computer-assisted techniques are often used to speed up this process, but students should still understand the principles involved. 4. Provide students with a firm grasp on the principles of functional dependencies, determinants, and related concepts of normalization. 5. Emphasize why normalization is important to stable database design with the relational, and then present a concise description of the various normal forms and the normalization process. 6. Discuss some of the anomalies that arise when merging relations, and how to apply the principles we have learned to address these anomalies. Classroom Ideas 1. 2. 3. Motivate the need for logical database design. We sometimes start by showing the students the conceptual data model (E-R diagrams) for Pine Valley Furniture Company (Figure 3-22). Emphasize that this E-R diagram must be transformed through logical database design before it can be implemented. Review the position of logical database design in the overall database development process (see Figure 2-5). You might want to discuss who in the organization is usually responsible for this step, and what CASE tools might be appropriate. Discuss the relational data model, using Figures 5-1 through 5-4 as examples. 80 4. 5. 6. 7. 8. 9. 10. 11. Modern Database Management, Sixth Edition Introduce the important integrity constraints in the relational model using Figures 5-4 and 5-5 and Table 5-1. Emphasize that these constraints will be enforced by the DBMS, but must first be specified by the designer. Introduce SQL table definitions (Figure 5-6). Show how these definitions specify the referential integrity constraints that are diagrammed in Figure 5-5. Illustrate how anomalies can occur when relations are not well structured, using Figures 5-2b and 5-7. Emphasize the fact that much real-world data (including relational data) are not well structured. Discuss the process of transforming EER diagrams to relations (Figures 5-8 through 5-22). We suggest you reinforce these concepts by asking your students (in teams of two) to perform Exercise 6a in class immediately following the discussion. Preview the steps in normalization using Figure 5-23. You will want to use this figure again to summarize normalization at the end of your discussion. Discuss the concepts of functional dependencies, determinants, and candidate keys. Start with your own examples on the board, then have your students give additional examples. Summarize using Figure 5-23. Discuss first through third normal forms, using Figures 5-24 through 5-26. Additional normal forms (BCNF and 4NF) are presented in Appendix B, if time permits. We strongly suggest for you to ask your students to work in small teams on one or more chapter-end exercises (Exercises 3 and 4 work well for this purpose). Answers to Review Questions 1. 2. Define each of the following terms: a. Determinant. The attribute on the left-hand side of the arrow in a functional dependency. b. Functional dependency. A constraint between two attributes or two sets of attributes. c. Transitive dependency. A functional dependency between two (or more) nonkey attributes. d. Recursive foreign key. A foreign key in a relation that references the primary key values of that same relation. e. Normalization. The process of decomposing relations with anomalies to produce smaller, well-structured relations. f. Composite key. A primary key that consists of more than one attribute. g. Relation. A named, two-dimensional table of data. h. Normal form. A state of a relation that results from applying simple rules regarding functional dependencies (or relationships between attributes) to that relation. i. Partial functional dependency. A functional dependency in which one or more nonkey attributes (such as Name) are functionally dependent on part (but not all) of the primary key. j. Enterprise Key. A primary key whose value is unique across all relations. f well-structured relation Chapter 5 3. 4. 5. 81 e anomaly a functional dependency j determinant g composite key d 1NF h 2NF i 3NF c recursive foreign key k relation b transitive dependency Contrast the following terms: a. Normal form; normalization. Normal form is a state of a particular relation regarding functional dependencies, while normalization is the process of decomposing relations with anomalies to produce smaller, wellstructured relations. b. Candidate key; primary key. A primary key is an attribute (or combination of attributes) that uniquely identifies a row in a relation. When a relation has more than one such attribute (or combination of attributes), each is called a candidate key. The primary key is then the one chosen by users to uniquely identify the rows in the relation. c. Functional dependency; transitive dependency. A functional dependency is a constraint between any two attributes (or two sets of attributes), while a transitive dependency is a functional dependency between two or more non-key attributes. d. Composite key; recursive foreign key. A composite key is a primary key that consists of more than one attribute, while a recursive foreign key is a foreign key in a relation that references the primary key values of that same relation. e. Determinant; candidate key. A determinant is on the left-hand side of the arrow in a functional dependency, while a candidate key uniquely identifies a row in a relation. f. Foreign key; primary key. A primary key uniquely identifies each row in a relation while a foreign key is a primary key in another table. Six important properties of relations are: a. Each relation in a database has a unique name. b. An entry at the intersection of each row and column is atomic (or single valued). c. Each row is unique. d. Each attribute within a table has a unique name. e. The sequence of columns is insignificant. f. The sequence of rows is insignificant. Describe two properties that must be satisfied by candidate keys: a. Unique identification: for every row, the value of the key must uniquely identify that row. b. Nonredundancy: no attribute in the key can be deleted without destroying the property of unique identification. 82 6. 7. 8. 9. 10. Modern Database Management, Sixth Edition Three types of anomalies in tables: a. Insertion anomaly: a new row cannot be inserted unless all primary key values are supplied. b. Deletion anomaly: deleting a row results in losing important information not stored elsewhere. c. Modification anomaly: a simple update must be applied to multiple rows. Fill in the blanks. a. second b. first c. third A well-structured relation is a relation that contains a minimum amount of redundancy and allows users to insert, modify, and delete the rows in a table without errors or inconsistency. Well-structured relations are important because they promote database integrity. Describe how the following components of an E-R diagram are transformed to relations: a. Regular entity type: each entity type is transformed to a simple relation. Each simple attribute of the entity type becomes an attribute of the relation. b. Relationship (1:M): a relation is created for each of the two entity types participating in the relationship. The primary key attribute of the entity on the one-side of the relationship becomes a foreign key in the relation on the many-side of the relationship. c. Relationship (M:N): a new relation is created to represent this relationship. The primary key for each of the participating entity types is included in this new relation. d. Relationship (supertype/subtype): a separate relation is created for the supertype and each of its subtypes. The primary key of the supertype is assigned to each subtype, as well as attributes that are unique to the subtype. e. Multivalued attribute: a new relation is created to replace the multivalued attribute. The primary key of this new relation consists of two attributes: the primary key of the original relation, plus the multivalued attribute itself. f. Weak entity: a new relation is created corresponding to the weak entity. The primary key of this relation consists of the primary key of the owner relation, plus the partial identifier of the weak entity type. g. Composite attribute: the simple component attributes of the composite attribute are included in the new relation. Four typical problems in merging relations: a. Synonyms: two (or more) attributes have different names but the same meaning. Solution: convince users to standardize on a single name. b. Homonyms: a single attribute has more than one meaning. Solution: create new attribute names that capture the separate meanings. c. Transitive dependency: merging relations produces transitive dependencies. Chapter 5 11. 12. 13. 14. 15. 16. 17. 18. 83 Solution: create 3 NF relations by removing the transitive dependency. d. Supertype/ subtype: may be implied by content of existing relations. Solution: create new relations that explicitly recognize this relationship. Three conditions that imply a relation is in second normal form: a. The primary key consists of a simple attribute. b. No nonkey attributes exist in the relation. c. Every nonkey attribute is functionally dependent on the full set of primary key attributes. Integrity constraints enforced in SQL CREATE TABLE commands: a. Entity integrity: enforced by NOT NULL clause. b. Referential integrity: enforced by FOREIGN KEY REFERENCES statement. Relationships between entities are represented by foreign key values in one relation that match primary key values in another relation. A 1:M unary relationship is represented by a recursive foreign key whose values reference the primary key values of the same relation. An M:N ternary relationship is represented by a new associative relation whose primary key consists of the primary key attributes of the participating entity types. All of the non-key attributes of a relation are functionally dependent on the primary key of that relation. A foreign key must not be null when the minimum cardinality is one. Primary keys can be transformed into Enterprise keys to eliminate key ripple effects as a database evolves. Answers to Problems and Exercises 1. Transforming E-R diagrams to relations: a. EMPLOYEE Employee_ID Employee_Name Address EMPLOYEE Employee_ID Skill b. FLIGHT Flight_No Date No_of_Passengers Date_Employed 84 Modern Database Management, Sixth Edition c. EMPLOYEE Employee_ID Employee_Name COMPLETION Employee_ID Course_ID Date_Completed COURSE Course_ID Course_Title d. EMPLOYEE Employee_ID Employee_Name CERTIFICATE Certificate_No Employee_ID COURSE Course_ID Course_Title e. COURSE Course_ID Course_Title Course_ID Prereq_ID Course_ID Date_Completed Chapter 5 85 f. MOVIE Movie_Name VIDEO TAPE Copy_No Movie_Name g. PRODUCT Product_ID Product_ID 2. Effective_Date Price Transforming EER diagrams to relations: a. VEHICLE Vehicle_ID Price Make Model Engine_Displacement CAR C_Vehicle_ID No_of_Passengers TRUCK T_Vehicle_ID Car_Type Capacity 86 Modern Database Management, Sixth Edition b. RESPONSIBLE PHYSICIAN Physician_ID PATIENT Patient_ID Admit_Date Physician_ID OUTPATIENT O_Patient_ID Checkback_Date RESIDENT PATIENT R_Patient_ID Date_Discharged BED Bed_ID BedID Chapter 5 87 c. PART Part_No Description Location Manufactured? MANUFACTURED PART M_Part_No PURCHASED PART P_Part_No SUPPLY LINE P_Part_No Supplier_ID Unit_Price SUPPLIER Supplier_ID Supplier_Name PURCHASED? Quantity_on_Hand 88 Modern Database Management, Sixth Edition d. PERSON SSN Name Address Date_of_Birth Sex EMPLOYEE E_SSN Salary Date_Hired FACULTY F_SSN STAFF Rank Position S_SSN ALUMNUS A_SSN DEGREES A_SSN Degree Year Date STUDENT ST_SSN Major_Department GRADUATE GS_SSN UNDERGRADUATE Test_Score US_SSN Class_Standing Chapter 5 89 e. STUDENT Student_ID Student_Name REGISTRATION Student_ID Course_ID Section_No Semester SECTION Course_ID Section_No Semester COURSE Course_ID Course_Name QUALIFICATION Course_ID Faculty_ID Date_Qualified ASSIGNMENT Faculty_ID Course_ID Section_No FACULTY Faculty_ID 3. Faculty_Name The normal form for the relations are: a. 3NF b. 3NF c. 2NF CLASS (Course_No, Section_No, Room) ROOM (Room, Capacity) Semester 90 Modern Database Management, Sixth Edition d. 4. 5. 1NF COURSE (Course_No, Course_Name) CLASS (Course_No, Section_No, Room) ROOM (Room, Capacity) 3NF relations for Millennium College are: OBJECT(OID,Object_Type) INSTRUCTOR (OID,Instructor_Name, Instructor_Location) COURSE (OID.Course_No, Course_Title, Instuctor_Name) STUDENT (OID,Section_No, Student_Name, Major) OUTCOME (OID,Student_No, Course_No, Grade) Transforming an E-R diagram to relations (parts a and b) CUSTOMER Customer_ID Customer_Name Customer_Address CARD ACCOUNT Account_ID Expiration_Date Card_Type Customer_ID DEBIT CARD D_Account_ID Bank_No C_Account_ID Current_Balance CHARGES Merchant_ID C_Account_ID Date MERCHANT MERCHANT_ID Merchant_Address Amount Chapter 5 91 part c: using an enterprise key OBJECT OID Object_Type CUSTOMER OID Customer_ID Customer_Name Customer_Address Account_ID Expiration_Date Card_Type Customer_OID D_Account_OID Bank_No C_Account_OID Current_Balance Merch_OID C_Account_ID Date Amount Merch_ID Merch_Address CARD_ACCOUNT OID DEBIT_CARD OID CREDIT_CARD OID CHARGES OID MERCHANT OID 92 6. Modern Database Management, Sixth Edition Transforming Table 5-2 to relations: a. PART SUPPLIER Part_No 1234 1234 5678 5678 5678 b. c. Description Logic Chip Logic Chip Memory Chip Memory Chip Memory Chip Vendor_Nam Fast Chips Smart Chips Fast Chips Quality Chips Smart Chips Part_No Vendor_Name Part_No, Vendor_Name Address Cupertino Phoenix Cupertino Austin Phoenix Unit_Cost 10.00 8.00 3.00 2.00 5.00 Description Address Unit_Cost Insert anomaly: we cannot insert a new vendor unless we also include a part number. Delete anomaly: if we delete part information, we also lose information about a vendor who supplies that part. Modification anomaly: if a vendor address changes, we have to modify all records (or rows) for that vendor. d. Part_No e. f. Description Vendor_Name 1NF PART SUPPLIER Part_No Description Part_No Vendor_Name Vendor_Name Address Unit_Cost Addres Unit_Cost Chapter 5 7. 93 Transforming Table 5-3 to relations: a. Student _ID Student _Name b. c. Campus _Address Major Course _Title Course_I Instructor _Name 1NF STUDENT Student_ID Student_Name Campus_Address REGISTRATION Student_ID COURSE_ID Grade COURSE COURSE_ID Course_Title Instructor_Name INSTRUCTOR Instructor_Name Instructor_Location Major Instructor _Location Grade 94 Modern Database Management, Sixth Edition d. STUDENT Student_ID Student_Name Campus_Address Major REGISTRATION Student_ID Course_ID Grade COURSE Course_ID Course_Title Instructor_Name INSTRUCTOR Instructor_Name Instructor_Location 8. attribute version: EMPLOYEE Employee_ID Employee_Name SKILL Employee_ID Skill_Code Skill_Title Skill_Type Chapter 5 95 relationship version: EMPLOYEE Employee_ID Employee_Name POSSESSES Employee_ID Skill_Code SKILL Skill_Code Skill_Title Skill_Type The attribute version of the 3NF relations is similar to Figure 5-10. However, we have a much clearer definition of a primary key in this version. One main advantage of the relationship version is that we do not have to store skill_title and skill_type many times. If a skill title was changed or types were reclassified, this would make things much easier since update anomalies are eliminated. Suggestions for Field Exercises 1. For this exercise, we suggest you interview at least two organizations: a manufacturing company and a service sector organization (you may choose to combine this exercise with Field Exercise 2 in Chapter 4). First, determine what methodology (if any) each uses for conceptual design: E-R diagrams, object diagrams, etc. Then determine how these models are transformed to logical data models (relational schema, object-oriented designs, etc.). To what extent are these activities automated through the use of CASE tools? If the target data model is relational, determine the role of normalization: who is responsible for normalization, to what level is it performed, and how are users involved (if at all) in these activities? 2. We suggest you first perform this exercise as an in-class exercise with student participation in the process. Bring a copy of your own document to class, and ask the students to volunteer a document as well. This provides students with valuable “hands-on” experience in the bottom-up design process. 3. For this exercise you may choose to assign a sample relational schema (such as Figure 6-13b or 6-19b) as a basis for comparing the CASE tools. 4. This exercise is really a continuation of Exercise 2 above, now possibly applied to a more complex document. Use a report (or other document) that has detail lines and requires the use of normalization skills. 96 Modern Database Management, Sixth Edition Project Case Project Questions 1. 2. 3. 4. Mountain View Community Hospital will continue to use relational technology for several reasons: a. The present IS staff is trained and experienced in using this technology. b. The present relational systems are stable and support existing operations quite well. c. Conversion to newer technology would be costly and would entail a number of risks. Yes, Mountain View Community Hospital should use normalization in designing its relational database. Normalization helps avoid anomalies that impair data quality. Entity integrity and referential integrity are important: a. Entity integrity helps assure that two real-world entities (such as patient or tests) are not confused. b. Referential integrity helps assure that one real-world entity (such as a test result) is not lost or disassociated from its owner entity (such as patient). All users of data in the organization should be consulted during the normalization process to ensure that the meaning and usage of data have been understood correctly. Project Exercises (See the next page) Chapter 5 1. 97 Relational schemas for Mountain View Community Hospital. a. Schema for E-R diagram (Exercise 2, Chapter 3): WARD Ward_No Ward_Name Employee_No ASSIGNED Ward_No Employee_No Hours EMPLOYEE Employee_No Employee_Name BED BED_NO Ward_No Room_No PATIENT_NO PATIENT Patient_No Patient_Name Physician_ID Physician_ID Treatment_No PERFORMS Patient_No TREATMENT PHYSICIAN Physician_ID Treatment_No Physician_Name CONSUMES Patient_No Item_No Date Quantity ITEM Item_No Results Description Unit_Cost TREATMENT_NAME 98 Modern Database Management, Sixth Edition b. Schema for EER diagram (Exercise 1, Chapter4): PERSON PERSON_ID Name Address City_State_Zip Birth_Date PATIENT PA_PERSON_ID Contact_Date PH_PERSON_ID PHYSICIAN PH_PERSON_ID Pager_No Specialty VOLUNTEER V_Person_ID Skill EMPLOYEE Date_Hired E_Person_ID NURSE N_Person_ID Certificate Name STAFF S_Person_ID Job_Class TECHNICIAN T_Person_ID Skill LAB ASSIGN T_Person_ID Name LABORATORY Name Location CARE CENTER Name Location Phone Chapter 5 To PATIENT 99 RESIDENT R_Person_ID Date_Admitted OUTPATIENT O_Person_ID (OTHER) BED Bed_No Room_No R_Person_ID VISIT O_Person_ID 2. 3. 4. Date Comments The functional dependencies are diagrammed in the above figures. All of the relations are 3NF. First, we will create enterprise keys for the E-R diagram: OBJECT OID Object_Type WARD OID Ward_No Ward_Name Employee_OID Ward_OID Employee_OID Hours Employee_No Employee_Name Bed_No Ward_OID ASSIGNED OID EMPLOYEE OID BED OID Room_No Patient_OID to Patient 100 Modern Database Management, Sixth Edition PATIENT OID Patient_No Patient_Name Physician_OID Patient_OID Physician_OID Treatment_OID Physician_ID Physician_Name Treatment_No Treatment_Name Patient_OID Item_OID Date Item_No Description Unit_Cost PERFORMS OID Results PHYSICIAN OID TREATMENT OID CONSUMES OID ITEM OID Quantity Chapter 5 101 EER Diagram: OBJECT OID Object_Type PERSON OID Person_ID Name Address Birth_Date City_State_Zip Phone PATIENT OID PA_PERSON_OID Contact_Date PH_OID PH_PERSON_OID Pager_No Speciality PHYSICIAN OID VOLUNTEER OID V_PERSON_OID Skill OID E_PERSON_OID Date_Hired OID N_PERSON_OID EMPLOYEE NURSE Certificate CARE_OID to CARE_CENTER STAFF OID S_PERSON_OID Job_Class T_PERSON_OID Skill TECHNICIAN OID 102 Modern Database Management, Sixth Edition LAB_ASSIGN OID to TECHNICIAN LAB_OID TECH_OID LABORATORY OID Location Name to NURSE CARE CENTER OID Location Name to PATIENT RESIDENT OID R_PERSON_OID Date_Admitted to PATIENT OUTPATIENT OID O_PERSON_OID (Other) OID Bed_No Room_No RES_OID OID OUTPATIENT_OID Date Comments BED VISIT Chapter 5 5. 103 Following are some sample CREATE TABLE commands. Please note, in order for these to work correctly in Oracle, they must be executed in the order specified in this solution. a. E-R diagram: CREATE TABLE OBJECT (OID VARCHAR2(5) Primary Key, Object_Type VARCHAR2(20)); CREATE TABLE EMPLOYEE (OID VARCHAR2(5) Primary Key, Employee_NO VARCHAR2(5), Employee_Name VARCHAR2(20), Foreign Key (OID) References Object(OID)); CREATE TABLE WARD (OID VARCHAR2(5) Primary Key, Ward_NO VARCHAR2(5), Ward_Name VARCHAR2(20), Employee_OID VARCHAR2(5) references EMPLOYEE(OID), Foreign Key (OID) References OBJECT(OID)); CREATE TABLE ASSIGNED (OID VARCHAR2(5) Primary Key, WARD_OID VARCHAR2(5) references WARD(OID), EMPLOYEE_OID VARCHAR2(5) references EMPLOYEE(OID), Hours NUMBER(4,2), Foreign Key (OID) References OBJECT(OID)); CREATE TABLE PHYSICIAN (OID VARCHAR2(5) Primary Key, Physician_ID VARCHAR2(5), Physician_Name VARCHAR2(20), Foreign Key (OID) References OBJECT(OID)); CREATE TABLE PATIENT (OID VARCHAR2(5) Primary Key, Patient_No VARCHAR2(5), Patient_Name VARCHAR2(20), PHYSICIAN_OID VARCHAR2(5) references PHYSICIAN(OID), Foreign Key (OID) References OBJECT(OID)); 104 Modern Database Management, Sixth Edition CREATE TABLE BED (OID VARCHAR2(5) Primary Key, Bed_No VARCHAR2(3), Ward_OID VARCHAR2(5) references WARD(OID), Room_No VARCHAR2(5), Patient_OID VARCHAR2(5) references PATIENT(OID), Foreign Key (OID) References OBJECT(OID)); CREATE TABLE TREATMENT (OID VARCHAR2(5) Primary Key, Treatment_No NUMBER(5), Treatment_Name VARCHAR2(20), Foreign Key (OID) References OBJECT(OID)); CREATE TABLE PERFORMS (OID VARCHAR2(5) Primary Key, PATIENT_OID VARCHAR2(5) references PATIENT(OID), PHYSICIAN_OID VARCHAR2(5) references PHYSICIAN(OID), TREATMENT_OID VARCHAR2(5) references TREATMENT(OID), RESULTS VARCHAR2(20), Foreign Key (OID) References OBJECT(OID)); CREATE TABLE ITEM (OID VARCHAR2(5) Primary Key, Item_No VARCHAR2(5), Description VARCHAR2(20), Unit_Cost NUMBER(5,2), Foreign Key (OID) References OBJECT(OID)); CREATE TABLE CONSUMES (OID VARCHAR2(5) Primary Key, PATIENT_OID VARCHAR2(5) references PATIENT(OID), ITEM_OID VARCHAR2(5) references ITEM(OID), DATE DATE, QUANTITY NUMBER(5), Foreign Key (OID) References OBJECT(OID)); Chapter 5 b. EER diagram: CREATE TABLE OBJECT (OID VARCHAR2(5) Primary Key, Object_Type VARCHAR2(20)); CREATE TABLE PERSON (OID VARCHAR2(5) Primary Key, Person_ID VARCHAR2(5), Name VARCHAR2(20), Address VARCHAR2(30), Birth_Date Date, City VARCHAR2(20), State VARCHAR2(2), Zip VARCHAR2(10), Phone VARCHAR2(14), foreign key (OID) references OBJECT(OID)); CREATE TABLE PHYSICIAN (OID VARCHAR2(5) Primary Key, PH_PERSON_OID VARCHAR2(5) references PERSON(OID), Pager_No VARCHAR2(14), Speciality VARCHAR2(20), foreign key (OID) references OBJECT(OID)); CREATE TABLE PATIENT (OID VARCHAR2(5) Primary Key, PA_PERSON_OID VARCHAR2(5) references PERSON(OID), Contact_Date Date, PH_OID VARCHAR2(5) references Physician(OID), foreign key (OID) references OBJECT(OID)); CREATE TABLE VOLUNTEER (OID VARCHAR2(5) Primary Key, V_PERSON_OID VARCHAR2(5) references PERSON(OID), Skill VARCHAR2(20), foreign key (OID) references OBJECT(OID)); CREATE TABLE EMPLOYEE (OID VARCHAR2(5) Primary Key, E_PERSON_OID VARCHAR2(5) references PERSON(OID), Date_Hired Date, foreign key (OID) references OBJECT(OID)); 105 106 Modern Database Management, Sixth Edition CREATE TABLE CARE_CENTER (OID VARCHAR2(5) Primary Key, Name VARCHAR2(20), Location VARCHAR2(20), foreign key (OID) refences OBJECT(OID)); CREATE TABLE NURSE (OID VARCHAR2(5) Primary Key, N_PERSON_OID VARCHAR2(5) references EMPLOYEE(OID), Certificate VARCHAR2(2), CARE_OID VARCHAR2(5) references CARE_CENTER(OID), foreign key (OID) references OBJECT(OID)); CREATE TABLE STAFF (OID VARCHAR2(5) Primary Key, S_PERSON_OID VARCHAR2(5) references EMPLOYEE(OID), Job_Class Number(2), foreign key (OID) references OBJECT(OID)); CREATE TABLE TECHNICIAN (OID VARCHAR2(5) Primary Key, T_PERSON_OID VARCHAR2(5) references EMPLOYEE(OID), Skill VARCHAR2(10), foreign key (OID) references OBJECT(OID)); CREATE TABLE LABORATORY (OID VARCHAR2(5) Primary Key, Name VARCHAR2(20), Location VARCHAR2(20), foreign key (OID) references OBJECT(OID)); CREATE TABLE LAB_ASSIGN (OID VARCHAR2(5) Primary Key, TECH_OID VARCHAR2(5) references TECHNICIAN(OID), LAB_OID VARCHAR2(5) references LABORATORY(OID), foreign key (OID) references OBJECT(OID)); CREATE TABLE RESIDENT (OID VARCHAR2(5) Primary Key, R_PERSON_OID VARCHAR2(5) references PATIENT(OID), Date_Admitted Date, foreign key (OID) references OBJECT(OID)); Chapter 5 107 CREATE TABLE OUTPATIENT (OID VARCHAR2(5) Primary Key, O_PERSON_OID VARCHAR2(5) references PATIENT(OID), foreign key (OID) references OBJECT(OID)); CREATE TABLE BED (OID VARCHAR2(5) Primary Key, Bed_No VARCHAR2(5), Room_No VARCHAR2(5), RES_OID VARCHAR2(5) references RESIDENT(OID), foreign key (OID) references OBJECT(OID)); CREATE TABLE VISIT (OID VARCHAR2(5) Primary Key, O_PERSON_OID VARCHAR2(5) references OUTPATIENT(OID), Date Date, Comments VARCHAR2(50), foreign key (OID) references OBJECT(OID)); 6. You can use this exercise (or a selected subset) to illustrate the problems of merging relations described in the chapter. You can also use this exercise to anticipate the design of a data warehouse that consolidates user views (see Chapter 11).
