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Data Management

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DC10 PART-I DATABASE MANAGEMENT SYSTEMS TYPICAL QUESTIONS & ANSWERS OBJECTIVE TYPE QUESTIONS Each question carries 2 marks. Choose the correct or best alternative in the following: Q. 1 In the relational modes, cardinality is termed as: (A) Number of tuples. (B) Number of attributes. (C) Number of tables. (D) Number of constraints. Ans: A Q.

2 Relational calculus is a (A) Procedural language. (C) Data definition language. Ans: B Q. 3 The view of total database content is (A) Conceptual view. (C) External view. Ans: A Q. 4 Cartesian product in relational algebra is (A) a Unary operator. (B) a Binary operator. C) a Ternary operator. (D) not defined. Ans: B Cartesian product in relational algebra is a binary operator. (It requires two operands. e. g. , P X Q) Q. 5 DML is provided for (A) Description of logical structure of database. (B) Addition of new structures in the database system. (C) Manipulation & processing of database.

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(D) Definition of physical structure of database system. Ans: C DML is provided for manipulation & processing of database. (Data stored in the database is processed or manipulated using data manipulation language commands as its name) (B) Internal view. (D) Physical View. (B) Non- Procedural language. D) High level language.

1 DC10 Q. 6 ‘AS’ clause is used in SQL for (A) Selection operation. (C) Join operation. DATABASE MANAGEMENT SYSTEMS (B) Rename operation. (D) Projection operation. Ans: B ‘AS’ clause is used in SQL for rename operation. (e. g. , SELECT ENO AS EMPLOYEE_NO FROM EMP) Q. 7 ODBC stands for (A) Object Database Connectivity. (B) Oral Database Connectivity. (C) Oracle Database Connectivity. (D) Open Database Connectivity. Ans: D Q. 8 Architecture of the database can be viewed as (A) two levels. (B) four levels. (C) three levels. (D) one level. Ans: C Q. 9 In a relational model, relations are termed as (A) Tuples. B) Attributes (C) Tables. (D) Rows. Ans: Q. 10 The database schema is written in (A) HLL (C) DDL Ans: C Q. 11 In the architecture of a database system external level is the (A) physical level. (B) logical level. (C) conceptual level (D) view level. Ans: D Q. 12 An entity set that does not have sufficient attributes to form a primary key is a (A) strong entity set. (B) weak entity set. (C) simple entity set. (D) primary entity set. Ans: B Q. 13 In a Hierarchical model records are organized as (A) Graph. (B) List. (C) Links. (D) Tree. 2 (B) DML (D) DCL DC10 Ans: D Q. 14 DATABASE MANAGEMENT SYSTEMS

In an E-R diagram attributes are represented by (A) rectangle. (B) square. (C) ellipse. (D) triangle. Ans: C Q. 15 In case of entity integrity, the primary key may be (A) not Null (B) Null (C) both Null & not Null. (D) any value. Ans: A Q. 16 In tuple relational calculus P1 > P2 is equivalent to (B) P1 ? P2 (A) ¬P1 ? P2 (D) P1 ? ¬P2 (C) P1 ? P2 Ans: A In tuple relational calculus P1 (The logical implication expression A P2 is equivalent to ¬P1 ? P2. B, meaning if A then B,is equivalent to ¬A ? B) Q. 17 The language used in application programs to request data from the DBMS is referred to as the (A) DML (B) DDL (C) VDL (D) SDL Ans: A

Q. 18 A logical schema (A) is the entire database. (B) is a standard way of organizing information into accessible parts. (C) describes how data is actually stored on disk. (D) both (A) and (C) Ans: A Q. 19 Related fields in a database are grouped to form a (A) data file. (B) data record. (C) menu. (D) bank. Ans: B Related data fields in a database are grouped to form a data record. (A record is a collection of related fields) Q. 20 The database environment has all of the following components except: (A) users. (B) separate files. (C) database. (D) database administrator. Ans: A 3 DC10 Q. 21

DATABASE MANAGEMENT SYSTEMS The language which has recently become the defacto standard for interfacing application programs with relational database system is (A) Oracle. (B) SQL. (C) DBase. (D) 4GL. Ans: B Q. 22 The way a particular application views the data from the database that the application uses is a (A) module. (B) relational model. (C) schema. (D) sub schema. Ans: D Q. 23 In an E-R diagram an entity set is represent by a (A) rectangle. (B) ellipse. (C) diamond box. (D) circle. Ans: A Q. 24 A report generator is used to (A) update files. (C) data entry. Ans: B (B) print files on paper. (D) delete files. Q. 25

The property / properties of a database is / are : (A) It is an integrated collection of logically related records. (B) It consolidates separate files into a common pool of data records. (C) Data stored in a database is independent of the application programs using it. (D) All of the above. Ans: D Q. 26 The DBMS language component which can be embedded in a program is (A) The data definition language (DDL). (B) The data manipulation language (DML). (C) The database administrator (DBA). (D) A query language. Ans: B Q. 27 A relational database developer refers to a record as (A) a criteria. (B) a relation. (C) a tuple. (D) an attribute. Ans: C

Q. 28 The relational model feature is that there 4 DC10 (A) (B) (C) (D) DATABASE MANAGEMENT SYSTEMS is no need for primary key data. is much more data independence than some other database models. are explicit relationships among records. are tables with many dimensions. Ans: B Q. 29 Conceptual design (A) is a documentation technique. (B) needs data volume and processing frequencies to determine the size of the database. (C) involves modelling independent of the DBMS. (D) is designing the relational model. Ans:C Q. 30 The method in which records are physically stored in a specified order according to a key field in each record is (A) hash. B) direct. (C) sequential. (D) all of the above. Ans: A A method in which records are physically stored in a specified order according to a key field in each record is hash. (In hash method, a hash function is performed on the key value to determine the unique physical address of the record to store or retrieve) Q. 31 A subschema expresses (A) the logical view. (C) the external view. (B) the physical view. (D) all of the above. Ans: C A subschema expresses the external view. (External schemas are called also called as subschemas) Q. 32 Count function in SQL returns the number of (A) values. (B) distinct values. C) groups. (D) columns. Ans: A Count function in SQL returns the number of values. (Count function counts all the not null values in the specific column. If we want to count only distinct values than the DISTINCT keyword is also to be used) Q. 33 Which one of the following statements is false? (A) The data dictionary is normally maintained by the database administrator. (B) Data elements in the database can be modified by changing the data dictionary. (C) The data dictionary contains the name and description of each data element. (D) The data dictionary is a tool used exclusively by the database administrator.

Ans: B 5 DC10 Q. 34 DATABASE MANAGEMENT SYSTEMS An advantage of the database management approach is (A) data is dependent on programs. (B) data redundancy increases. (C) data is integrated and can be accessed by multiple programs. (D) none of the above. Ans: C Q. 35 A DBMS query language is designed to (A) support end users who use English-like commands. (B) support in the development of complex applications software. (C) specify the structure of a database. (D) all of the above. Ans: D Q. 36 Transaction processing is associated with everything below except (A) producing detail, summary, or exception reports. B) recording a business activity. (C) confirming an action or triggering a response. (D) maintaining data. Ans: C Q. 37 It is possible to define a schema completely using (A) VDL and DDL. (B) DDL and DML. (C) SDL and DDL. (D) VDL and DML. Ans: B Q. 38 The method of access which uses key transformation is known as (A) direct. (B) hash. (C) random. (D) sequential. Ans: B Q. 39 Data independence means (A) data is defined separately and not included in programs. (B) programs are not dependent on the physical attributes of data. (C) programs are not dependent on the logical attributes of data. (D) both (B) and (C).

Ans: D both (B) and (C) Q. 40 The statement in SQL which allows to change the definition of a table is (B) Update. (A) Alter. (C) Create. (D) select. Ans: A 6 DC10 Q. 41 DATABASE MANAGEMENT SYSTEMS E-R model uses this symbol to represent weak entity set ? (A) Dotted rectangle. (B) Diamond (C) Doubly outlined rectangle (D) None of these Ans: C Q. 42 SET concept is used in : (A) Network Model (B) Hierarchical Model (C) Relational Model (D) None of these Ans: A Q. 43 Relational Algebra is (A) Data Definition Language . (B) Meta Language (C) Procedural query Language (D) None of the above Ans: C Q. 44

Key to represent relationship between tables is called (A) Primary key (B) Secondary Key (C) Foreign Key (D) None of these Ans: C Q. 45 _______ produces the relation that has attributes of R1 and R2 (A) Cartesian product (B) Difference (C) Intersection (D) Product Ans: A Q. 46 The file organization that provides very fast access to any arbitrary record of a file is (A) Ordered file (B) Unordered file (C) Hashed file (D) B-tree Ans: C Q. 47 DBMS helps achieve (A) Data independence (C) Neither (A) nor (B) Ans: D (B) Centralized control of data (D) both (A) and (B) Q. 48 Which of the following are the properties of entities? A) Groups (B) Table 7 DC10 (C) Attributes Ans: Q. 49 C DATABASE MANAGEMENT SYSTEMS (D) Switchboards In a relation (A) Ordering of rows is immaterial (B) No two rows are identical (C) (A) and (B) both are true (D) None of these. Ans: C Q. 50 Which of the following is correct: (A) a SQL query automatically eliminates duplicates. (B) SQL permits attribute names to be repeated in the same relation. (C) a SQL query will not work if there are no indexes on the relations (D) None of these Ans: D Q. 51 It is better to use files than a DBMS when there are (A) Stringent real-time requirements. B) Multiple users wish to access the data. (C) Complex relationships among data. (D) All of the above. Ans: B Q. 52 The conceptual model is (A) dependent on hardware. (B) dependent on software. (C) dependent on both hardware and software (D) independent of both hardware and software. Ans: D . Q. 53 What is a relationship called when it is maintained between two entities? (A) Unary (B) Binary (C) Ternary (D) Quaternary Ans: B Q. 54 Which of the following operation is used if we are interested in only certain columns of a table? (A) PROJECTION (B) SELECTION (D) JOIN (C) UNION Ans: A 8 DC10 Q. 55

DATABASE MANAGEMENT SYSTEMS Which of the following is a valid SQL type? (A) CHARACTER (B) NUMERIC (C) FLOAT (D) All of the above Ans: D Q. 56 The RDBMS terminology for a row is (A) tuple. (B) relation. (C) attribute. (D) degree. Ans: A Q. 57 Which of the following operations need the participating relations to be union compatible? (A) UNION (B) INTERSECTION (C) DIFFERENCE (D) All of the above Ans: D Q. 58 The full form of DDL is (A Dynamic Data Language (C) Data Definition Language Ans: C (B) Detailed Data Language (D) Data Derivation Language Q. 59 Which of the following is an advantage of view? A) Data security (B) Derived columns (C) Hiding of complex queries (D) All of the above Ans: D Q. 60 Which of the following is a legal expression in SQL? (A) SELECT NULL FROM EMPLOYEE; (B) SELECT NAME FROM EMPLOYEE; (C) SELECT NAME FROM EMPLOYEE WHERE SALARY = NULL; (D) None of the above Ans: B Q. 61 The users who use easy-to-use menu are called (B) Naive users. (A) Sophisticated end users. (C) Stand-alone users. (D) Casual end users. Ans: B Q. 62 Which database level is closest to the users? (B) Internal (A) External (C) Physical (D) Conceptual Ans: A 9 DC10 Q. 63

DATABASE MANAGEMENT SYSTEMS Which are the two ways in which entities can participate in a relationship? (A) Passive and active (B) Total and partial (C) Simple and Complex (D) All of the above Ans: B Q. 64 The result of the UNION operation between R1 and R2 is a relation that includes (A) all the tuples of R1 (B) all the tuples of R2 (C) all the tuples of R1 and R2 (D) all the tuples of R1 and R2 which have common columns Ans: D Q. 65 Which of the following is a comparison operator in SQL? (A) = (B) LIKE (C) BETWEEN (D) All of the above Ans: D Q. 66 A set of possible data values is called (A) attribute. (B) degree. C) tuple. (D) domain. Ans: D Q. 67 Which of the operations constitute a basic set of operations for manipulating relational data? (A) Predicate calculus (B) Relational calculus (C) Relational algebra (D) None of the above Ans:C Q. 68 Which of the following is another name for weak entity? (A) Child (B) Owner (C) Dominant (D) All of the above Ans: A Q. 69 Which of the following database object does not physically exist? (A) base table (B) index (C) view (D) none of the above Ans: C Q. 70 NULL is (A) the same as 0 for integer (B) the same as blank for character (C) the same as 0 for integer and blank for character 10

DC10 (D) not a value Ans: D Q. 71 DATABASE MANAGEMENT SYSTEMS Which of the following is record based logical model? (A) Network Model (B) Object oriented model (C) E-R Model (D) None of these Ans: A Q. 72 A data dictionary is a special file that contains: (A) The name of all fields in all files. (B) The width of all fields in all files. (C) The data type of all fields in all files. (D) All of the above. Ans: D Q. 73 A file manipulation command that extracts some of the records from a file is called (A) SELECT (B) PROJECT (C) JOIN (D) PRODUCT Ans: A Q. 74

The physical location of a record is determined by a mathematical formula that transforms a file key into a record location is : (A) B-Tree File (B) Hashed File (C) Indexed File (D) Sequential file. Ans: B Q. 75 Using Relational Algebra the query that finds customers, who have a balance of over 1000 is (A) ? Customer_name( ? balance >1000(Deposit)) (B) ? Customer_name( ? balance >1000(Deposit)) (C) ? Customer_name( ? balance >1000(Borrow)) (D) ? Customer_name( ? balance >1000(Borrow)) Ans: A Q. 76 A primary key is combined with a foreign key creates (A) Parent-Child relation ship between the tables that connect them. B) Many to many relationship between the tables that connect them. (C) Network model between the tables that connect them. (D) None of the above. Ans: A Q. 77 In E-R Diagram derived attribute are represented by 11 DC10 (A) Ellipse (C) Rectangle Ans B Q. 78 Cross Product is a: (A) Unary Operator (C) Binary Operator Ans: C Q. 79 DATABASE MANAGEMENT SYSTEMS (B) Dashed ellipse (D) Triangle (B) Ternary Operator (D) Not an operator An instance of relational schema R (A, B, C) has distinct values of A including NULL values. Which one of the following is true? A) A is a candidate key (B) A is not a candidate key (C) A is a primary Key (D) Both (A) and (C) Ans: B Q. 80 Consider the join of a relation R with relation S. If R has m tuples and S has n tuples, then the maximum size of join is: (A) mn (B) m+n (C) (m+n)/2 (D) 2(m+n) Ans: A Q. 81 The natural join is equal to : (A) Cartesian Product (B) Combination of Union and Cartesian product (C) Combination of selection and Cartesian product (D) Combination of projection and Cartesian product Ans: D Q. 82 Which one of the following is not true for a view: (A) View is derived from other tables.

(B) View is a virtual table. C) A view definition is permanently stored as part of the database. (D) View never contains derived columns. Ans: C Q. 83 A primary key if combined with a foreign key creates (A) Parent-Child relationship between the tables that connect them. (B) Many to many relationship between the tables that connect them. (C) Network model between the tables that connect them. (D) None of the above. Ans: A 12 DC10 Q. 84 DATABASE MANAGEMENT SYSTEMS In E-R Diagram relationship type is represented by (A) Ellipse (B) Dashed ellipse (C) Rectangle (D) Diamond Ans: D Q. 85 Hierarchical model is also called (A) Tree structure (C) Normalize Structure Ans: A B) Plex Structure (D) Table Structure Q. 86 To delete a particular column in a relation the command used is: (A) UPDATE (B) DROP (C) ALTER (D) DELETE Ans: C Q. 87 The ______ operator is used to compare a value to a list of literals values that have been specified. (A) BETWEEN (B) ANY (C) IN (D) ALL Ans: A Q. 88 A logical schema A) is the entire database B) is a standard way of organizing information into a accessible part C) describe how data is actually stored on disk D) none of these Ans: D Q. 89 A B-tree of order m has maximum of _____________ children (A) m (B) m+1 (C) m-1 (D) m/2 Ans: A Q. 0 _____________ function divides one numeric expression by another and returns the remainder. (A) POWER (B) MOD (C) ROUND (D) REMAINDER Ans: B Q. 91 A data manipulation command the combines the records from one or more tables is called (A) SELECT (B) PROJECT (C) JOIN (D) PRODUCT 13 DC10 Ans: C Q. 92 DATABASE MANAGEMENT SYSTEMS In E-R diagram generalization is represented by (A) Ellipse (B) Dashed ellipse (C) Rectangle (D) Triangle Ans: D Q. 93 _________ is a virtual table that draws its data from the result of an SQL SELECT statement. (A) View (B) Synonym (C) Sequence (D) Transaction Ans: A Q. 94

The method of access which uses key transformation is known as (A) Direct (B) Hash (C) Random (D) Sequential Ans: B Q. 95 A table joined with itself is called (A) Join (C) Outer Join Ans: B (B) Self Join (D) Equi Join Q. 96 _________ data type can store unstructured data (A) RAW (B) CHAR (C) NUMERIC (D) VARCHAR Ans: A Q. 97 Which two files are used during operation of the DBMS (A) Query languages and utilities (B) DML and query language (C) Data dictionary and transaction log (D) Data dictionary and query language Ans: C 14 DC10 PART-II DATABASE MANAGEMENT SYSTEMS DESCRIPTIVES Q. 1 What is a database?

Describe the advantages and disadvantages of using of DBMS. (7) Ans: Database – A database is a collection of related data and/or information stored so that it is available to many users for different purposes. Advantages Of DBMS 1. Centralized Management and Control – One of the main advantages of using a database system is that the organization can exert, via the DBA, centralized management and control over the data. 2. Reduction of Redundancies and Inconsistencies – Centralized control avoids unnecessary duplication of data and effectively reduces the total amount of data storage required.

Removing redundancy eliminates inconsistencies. 3. Data Sharing – A database allows the sharing of data under its control by any number of application programs or users. 4. Data Integrity – Data integrity means that the data contained in the database is both accurate and consistent. Centralized control can also ensure that adequate checks are incorporated in the DBMS to provide data integrity. 5. Data Security – Data is of vital importance to an organization and may be confidential. Such confidential data must not be accessed by unauthorized persons.

The DBA who has the ultimate responsibility for the data in the DBMS can ensure that proper access procedures are followed. Different levels of security could be implemented for various types of data and operations. 6. Data Independence – Data independence is the capacity to change the schema at one level of a database system without having to change the schema at the next level. It is usually considered from two points of view: physical data independence and logical data independence. Physical data independence is the capacity to change the internal schema without having to change conceptual schema.

Logical data independence is the capacity to change the conceptual schema without having to change external schemas or application programs. 7. Providing Storage Structures for Efficient Query Processing – Database systems provide capabilities for efficiently executing queries and updates. Auxiliary files called indexes are used for this purpose. 8. Backup and Recovery – These facilities are provided to recover databases from hardware and/or software failures. Some other advantages are: Reduced Application Development Time Flexibility Availability of up-to-date Information Disadvantages Of DBMS 1.

Cost of Software/Hardware and Migration – A significant disadvantage of the DBMS system is cost. 15 DC10 2. DATABASE MANAGEMENT SYSTEMS Reduced Response and Throughput – The processing overhead introduced by the DBMS to implement security, integrity, and sharing of the data causes a degradation of the response and throughput times. Problem with Centralization – Centralization also means that the data is accessible from a single source namely the database. This increases the potential of security breaches and disruption of the operation of the organization because of downtimes and failures. 7) 3. Q. 2 Explain five duties of Database Administrator. Ans: 1. DBA administers the three levels of the database and, in consultation with the overall user community, sets up the definition of the global view or conceptual level of the database. 2. Mappings between the internal and the conceptual levels, as well as between the conceptual and external levels, are also defined by the DBA. 3. DBA ensures that appropriate measures are in place to maintain the integrity of the database and that the database is not accessible to unauthorized users. 4.

DBA is responsible for granting permission to the users of the database and stores the profile of each user in the database. 5. DBA is responsible for defining procedures to recover the database from failures with minimal loss of data. Q. 3 Explain the terms primary key, candidate key and foreign key. Give an example for each. (7) Ans: Primary Key – Primary key is one of the candidate keys that uniquely identifies each row in the relation. Candidate Key – A candidate key of an entity set is a minimal superkey, that uniquely identifies each row in the relation.

Foreign Key – Let there are two relations (tables) R and S. Any candidate key of the relation R which is referred in the relation S is called the foreign key in the relation S and referenced key in the relation R. The relation R is also called as parent table and relation S is also called as child table. For example: STUDENT Enrl No 11 15 6 33 Roll No 17 16 6 75 Name Ankit Vats Vivek Rajput Vanita Bhavya City Delhi Meerut Punjab Delhi Mobile 9891663808 9891468487 9810618396 16 DC10 GRADE DATABASE MANAGEMENT SYSTEMS

Grade Roll No Course 6 C A 17 VB C 75 VB A 6 DBMS B 16 C B Roll No is the primary key in the relation STUDENT and Roll No + Course is the primary key of the relation GRADE. Enrl No and Roll No are the candidate keys of the relation STUDENT. Roll No in the relation GRADE is a foreign key whose values must be one of those of the relation STUDENT. Q. 4 Differentiate between logical database design and physical database design. Show how this separation leads to data independence. (7) Ans: Basis Task Logical Database Design Maps or transforms the conceptual schema (or an ER schema) from the high-level data model into a relational database schema.

Physical Database Design The specifications for the stored database in terms of physical storage structures, record placement, and indexes are designed. Choice of The mapping can proceed in two The following criteria are often used criteria stages: to guide the choice of physical database design options: System-independent mapping Response Time but data model-dependent Tailoring the schemas to a Space Utilization specific DBMS Transaction Throughput Result DDL statements in the language of An initial determination of storage the chosen DBMS that specify the structures and the access paths for conceptual and external level the database files.

This corresponds schemas of the database system. But to defining the internal schema in if the DDL statements include some terms of Data Storage Definition physical design parameters, a Language. complete DDL specification must wait until after the physical database design phase is completed. The database design is divided into several phases. The logical database design and physical database design are two of them. This separation is generally based on the concept of three-level architecture of DBMS, which provides the data independence.

Therefore, we can say that this separation leads to data independence because the output of the logical database design is the conceptual and external level schemas of the database 17 DC10 DATABASE MANAGEMENT SYSTEMS system which is independent from the output of the physical database design that is internal schema. Q. 5 Consider the following relation schemes: (2? 7=14) ? Project (Project#, Project_name, chief_architect) Employee (Emp#, Empname) Assigned_To (Project#, Emp#) Give expression in Tuple calculus and Domain calculus for each of the queries below: (i) Get the employee numbers of employees who work on all projects. ii) Get the employee numbers of employees who do not work on the COMP123 project. Ans: (i) Tuple Calculus: ? u (u ? ASSIGNED_TO ? {t[Emp#] | t ? ASSIGNED_TO ? ?p (p ? PROJECT p[Project#] = u[Project#] ? t[Emp#] = u[Emp#]))} Domain Calculus: {e | ? p ( ? ASSIGNED_TO ? ?p1 ( ? PROJECT ? ASSIGNED_TO))} (ii) Tuple Calculus: {t[Emp#] | t ? ASSIGNED_TO ? ¬? u (u ? ASSIGNED_TO ? u[Project#] = ‘COMP123’? t[Emp#] = u[Emp#])} Domain Calculus: {e | ? p ( ? ASSIGNED_TO ? ?p1, e1 ( ? ASSIGNED_TO ? p1 ? ‘COMP123’ ? e1 ? e))} Q. 6 What is ODBC? How does Oracle act as ODBC and give examples of front end uses with ODBC. 7) Ans: ODBC – Open DataBase Connectivity (ODBC) enable the integration of SQL with a general-purpose programming language. ODBC expose database capabilities in a standardized way to the application programmer through an application programming interface (API). Using ODBC, an application can access not just one DBMS but several different ones simultaneously. ODBC achieve portability at the level of the executable by introducing an extra level of indirection. All direct interaction with a specific DBMS happens through a DBMSspecific driver. A driver is a software program that translates the ODBC calls into DBMSspecific calls.

Drivers are loaded dynamically on demand since the DBMSs the application is going to access are known only at run-time. Available drivers are registered with a driver manager. The Oracle database driver translates the SQL commands from the application into equivalent commands that the Oracle DBMS understands and takes the result from the DBMS and translate into equivalent form for the application. Example: Let there be a DSN named EMPLOYEE through, which we want to access the Oracle database in Visual Basic. Dim CN As New ADODB. Connection Dim RS As New ADODB. Recordset CN. Open “DSN=employee”, “scott”, “tiger” 18 DC10 RS.

Open “Select * From Emp”, CN Q. 7 DATABASE MANAGEMENT SYSTEMS Define the five basic operators of relational algebra with an example each. (7) Ans: Five basic operators of relational algebra are: 1. Union (? ) – Selects tuples that are in either P or Q or in both of them. The ? duplicate tuples are eliminated. R=P? Q 2. Minus (–) – Removes common tuples from the first relation. R=P–Q 3. Cartesian Product or Cross Product (? ) – The cartesian product of two ? relations is the concatenation of tuples belonging to the two relations and consisting of all possible combination of the tuples. R=P? Q For Example: P: ID 101 103 104 R=P?

Q ID 100 101 103 104 R=P? Q P. ID 101 101 103 103 104 104 P. Name Jones Jones Smith Smith Lalonde Lalonde 19 Q. ID 100 104 100 104 100 104 Q. Name John Lalonde John Lalonde John Lalonde Name John Jones Smith Lalonde Name Jones Smith Lalonde R=P–Q ID 101 103 Name Jones Smith Q: ID 100 104 Name John Lalonde DC10 DATABASE MANAGEMENT SYSTEMS 4. Projection (? ) – The projection of a relation is defined as a projection of all its tuples ? over some set of attributes, i. e. , it yields a vertical subset of the relation. It is used to either reduce the number of attributes (degree) in the resultant relation or to reorder attributes.

The projection of a relation T on the attribute A is denoted by ? A(T). 5. Selection (? ) – Selects only some of the tuples, those satisfy given criteria, from the ? relation. It yields a horizontal subset of a given relation, i. e. , the action is defined over a complete set of attribute names but only a subset of the tuples are included in the result. R = ? B(P) For Example: EMPLOYEE: Id Name Name 101 Jones Jones 103 104 106 Smith Lalonde Byron Smith Lalonde Byron Projection of relation EMPLOYEE over attribute Name EMPLOYEE: Result of Selection Id Name Id Name 101 Jones 104 Lalonde 103 104 106 Smith Lalonde Byron 106 Byron

Result of Selection over EMPLOYEE for ID > 103 Q. 8 Explain entity integrity and referential integrity rules in relational model. Show how these are realized in SQL. (7) Ans: Entity Integrity Rule – No primary key value can be null. Referential Integrity Rule – In referential integrity, it is ensured that a value that appears in one relation for a given set of attributes also appears for a certain set of attributes in another relation. In SQL, entity integrity and referential integrity rules are implemented as constraints on the relation called as primary key constraint and reference key constraint respectively.

These constraints can be specified with relation at the time of creation of the relations or after the creation of the relations by altering the definition of the relations. For example: CREATE TABLE DEPT (DEPTNO DNAME (EMPNO NUMBER PRIMARY KEY, VARCHAR2(15)); NUMBER PRIMARY KEY, 20 CREATE TABLE EMP DC10 ENAME JOB DEPTNO Q. 9 DATABASE MANAGEMENT SYSTEMS VARCHAR2(15), VARCHAR2(10), NUMBER REFERENCES DEPT(DEPTNO)); What are the advantages of embedded query language? Give an example of a embedded SQL query. (7) Ans: Embedded query language – SQL can be implemented in two ways.

It can be used interactively or embedded in a host language or by using API. The use of SQL commands within a host language (e. g. , C, Java, etc. ) program is called embedded query language or Embedded SQL. Although similar capabilities are supported for a variety of host languages, the syntax sometimes varies. Some of the advantages of embedded SQL are: SQL statements can be used wherever a statement in the host language is allowed. It combines the strengths of two programming environments, the procedural features of host languages and non-procedural features of SQL.

SQL statements can refer to variables (must be prefixed by a colon in SQL statements) defined in the host program. Special program variables (called null indicators) are used to assign and retrieve the NULL values to and from the database. The facilities available through the interactive query language are also automatically available to the host programs. Embedded SQL along with host languages can be used to accomplish very complex and complicated data access and manipulation tasks. Example: The following Embedded SQL statement in C inserts a row, whose column values are based on the values of the host language variables contained in it.

EXEC SQL INSERT INTO Sailors VALUES (:c_sname, :c_sid, :c_rating, :c_age); Q. 10 Consider the following relations: (3. 5 x 2=7) S (S#, SNAME, STATUS, CITY) SP (S#, P#, QTY) P (P#, PNAME, COLOR, WEIGHT, CITY) Give an expression in SQL for each of queries below: (i) Get supplier names for supplier who supply at least one red part (ii) Get supplier names for supplier who do not supply part P2. Ans:(i) SELECT SNAME FROM S WHERE S# IN (SELECT S# FROM SP WHERE P# IN (SELECT P# FROM P WHERE COLOR = RED’)) (ii) SELECT SNAME FROM S WHERE S# NOT IN (SELECT S# FROM SP WHERE P# = ‘P2’) Q. 11 Define a view and a trigger.

Construct a view for the above relations which has the information about suppliers and the parts they supply. The view contains the S#, SNAME, P# , PNAME renamed as SNO, NAME, PNO, PNAME. (7) 21 DC10 DATABASE MANAGEMENT SYSTEMS Ans: View – A view is a virtual table which is based on the one or more physical tables and/or views. In other words, a view is a named table that is represented, not by its own physically separate stored data, but by its definition in terms of other named tables (base tables or views). Trigger – A trigger is a procedure that is automatically invoked by the DBMS in the response to specified changes to the database.

Triggers may be used to supplement declarative referential integrity, to enforce complex business rules or to audit changes to data. Command: CREATE VIEW SUP_PART (SNO, NAME, PNO, PNAME) AS SELECT S. S#, SNAME, P. P#, PNAME FROM S, SP, P WHERE S. S# = SP. S# AND P. P# = SP. P# Q. 12 Differentiate between the following: (i) (iv) Theta Join. (ii) Equi Join. (iii) Natural Join Outer Join. (10) Ans:(i) Theta Join – The theta join operation is an extension to the natural-join operation that allows us to combine selection and a Cartesian product into a single operation. Consider relations r(R) and s(S), and let ? e a predicate on attributes in the schema R ? S. The theta join operation r ? s is defined as follows: r ? s = ?? (r x s) (ii) Equi Join – It produces all the combinations of tuples from two relations that satisfy a join condition with only equality comparison (=). (iii) Natural Join – Same as equi-join except that the join attributes (having same names) are not included in the resulting relation. Only one sets of domain compatible attributes involved in the natural join are present. (iv) Outer Join – If there are any values in one table that do not have corresponding value(s) in the other, in an equi-join that will not be selected.

Such rows can be forcefully selected by using the outer join. The corresponding columns for that row will have NULLs. There are actually three forms of the outer-join operation: left outer join ( X), right outer join (X ) and full outer join ( X ). Q. 13 What are temporary tables? When are they useful? Justify with an example. (4) Ans: Temporary tables exists solely for a particular session, or whose data persists for the duration of the transaction. The temporary tables are generally used to support specialized rollups or specific application processing requirements.

Unlike a permanent table, a space is not allocated to a temporary table when it is created. Space will be dynamically allocated for the table as rows are inserted. The CREATE GLOBAL TEMPORARY TABLE command is used to create a temporary table in Oracle. CREATE GLOBAL TEMPORARY TABLE ( 22 DC10 DATABASE MANAGEMENT SYSTEMS ) ON COMMIT {PRESERVE|DELETE} ROWS; Q. 14 Draw and explain the three level architecture of the database system. (7) Ans: A DBMS provides three levels of data is said to follow three-level architecture. The goal of the three-schema architecture is to separate the user applications and the physical database.

The view at each of these levels is described by a schema. The processes of transforming requests and results between levels are called mappings. In this architecture, schemas can be defined at the following three levels: External Level or Subschema – It is the highest level of database abstraction where only those portions of the database of concern to a user or application program are included. Any number of user views (some of which may be identical) may exist for a given global or conceptual view. Each external view is described by means of a schema called an external schema or subschema.

Conceptual Level or Conceptual Schema – At this level of database abstraction all the database entities and the relationships among them are included. One conceptual view represents the entire database. This conceptual view is defined by the conceptual schema. There is only one conceptual schema per database. The description of data at this level is in a format independent of its physical representation. It also includes features that specify the checks to retain data consistency and integrity. Internal Level or Physical Schema – It is closest to the physical storage method used.

It indicates how the data will be stored and describes the data structures and access methods to be used by the database. The internal view is expressed by the internal schema. 23 DC10 Q. 15 DATABASE MANAGEMENT SYSTEMS Explain (a) Heap file (b) Sorted file. Also discuss their advantages and disadvantages. Ans: Heap File is an unordered set of records, stored on a set of pages. This class provides basic support for inserting, selecting, updating, and deleting records. Temporary heap files are used for external sorting and in other relational operators. A sequential scan of a heap file (via the Scan class) is the most basic access method.

Sorted file The sort utility shall perform one of the following functions: 1. Sort lines of all the named files together and write the result to the specified output. 2. Merge lines of all the named (presorted) files together and write the result to the specified output. 3. Check that a single input file is correctly presorted. Comparisons shall be based on one or more sort keys extracted from each line of input (or, if no sort keys are specified, the entire line up to, but not including, the terminating ), and shall be performed using the collating sequence of the current locale.

Q. 16 Describe a method for direct search? Explain how data is stored in a file so that direct searching can be performed. Ans: For a file of unordered fixed length records using unspanned blocks and contiguous allocation, it is straight forward to access any record by its position in the file. If the file records are numbered 0,1,2,—,r-1 and the records in each block are numbered 0,1,—bfr-1; where bfr is the blocking factor, then ith record of the file is located in block [(i/bfr)] and is the (I mod bfr)th record in that block.

Such a file is often called a relative or direct file because records can easily be accessed directly by their relative positions. Accessing a record based on a search condition; however, it facilitates the construction of access paths on the file, such as the indexes. Q17 Explain the integrity constraints: Not Null, Unique, Primary Key with an example each. Is the combination ‘Not Null, Primary Key’ a valid combination. Justify. (7) Ans: Not Null – Should contain valid values and cannot be NULL. Unique – An attribute or a combination of two or more attributes must have a unique value in each row.

The unique key can have NULL values. 24 DC10 DATABASE MANAGEMENT SYSTEMS Primary Key – It is same as unique key but cannot have NULL values. A table can have at most one primary key in it. For example: STUDENT Roll No Name 17 Ankit Vats 16 Vivek Rajput 6 Vanita 75 Bhavya Roll No is a primary key. City Delhi Meerut Punjab Delhi Mobile 9891663808 9891468487 NULL 9810618396 Name is defined with NOT NULL, means each student must have a name. Mobile is unique. ‘Not Null, Primary Key’ is a valid combination. Primary key constraint already includes ‘Not Null’ constraint in it but we can also add ‘Not Null’ constraint with it.

The use of ‘Not Null’ with ‘Primary Key’ will not have any effect. It is same as if we are using just ‘Primary Key’. Q. 18 Explain the followings : (i) Nested Queries. (ii) Cursors in SQL. (iii) RDBMS. (iv) View (v) Application Programming Interface (14) Ans: (i) Nested Queries – A SELECT query can have subquery(s) in it. When a SELECT query having another SELECT query in it, is called as nested query. Some operations cannot be performed with single SELECT command or with join operation. There are some operations which can be performed with the help of nested queries (also referred to as subqueries).

For example, we want to compute the second highest salary: SELECT MAX(SAL) FROM EMP WHERE SAL < (SELECT MAX(SAL) FROM EMP) Some operations can be performed both by Join and subqueries. The Join operation is costlier in terms of time and space. Therefore, the solution based on subqueries is preferred. (ii) Cursors in SQL – An object used to store the output of a query for row-by-row processing by the application programs. Cursors are constructs that enable the user to name a private memory area to hold a specific statement for access at a later time. Cursors are used to process multi-row result sets one row at a time.

Additionally, cursors keep track of which row is currently being accessed, which allows for interactive processing of the active set. (iii) RDBMS – RDBMS is a database management system (DBMS) that stores data in the form of relations. Relational databases are powerful because they require few assumptions about how data is related or how it will be extracted from the database. As a result, the same database can be viewed in many different ways. An important feature of 25 DC10 DATABASE MANAGEMENT SYSTEMS relational system is that a single database can be spread across several tables.

This differs from flat-file databases, in which each database is self-contained in a single table. (iv) View – A view is a relation (virtual rather than base) and can be used in query expressions, that is, queries can be written using the view as a relation. In other words, a view is a named table that is represented, not by its own physically separate stored data, but by its definition in terms of other named tables (base tables or views). The base relations on which a view is based are sometimes called the existing relations.

The definition of a view in a create view statement is stored in the system catalog. The syntax to create a view is:CREATE [OR REPLACE] VIEW [()] AS WITH {READ ONLY|CHECK OPTION [CONSTRAINT ]}; (v) Application Programming Interface – Commercial SQL implementations take one of the two basic techniques for including SQL in a programming language – embedded SQL and application program interface (API). In the application program interface approach, the program communicates with the RDBMS using a set of functions called the Application Program Interface (API).

The program passes the SQL statements to the RDBMS using API calls and uses API calls to retrieve the results. In this method, the precompiler is not required. Consider the following relational schema: (7) PERSON (SS#, NAME, ADDRESS) CAR (REGISTRATION_NUMBER, YEAR, MODEL) ACCIDENT (DATE, DRIVER, CAR_REG_NO) OWNS (SS#, LICENSE) Construct the following relational algebra queries: (i) Find the names of persons who are involved in an accident. (ii) Find the registration number of cars which were not involved in any accident. Ans: (i) ? NAME(PERSON) ? ?DRIVER(ACCIDENT) (ii) ?

REGISTRATION_NUMBER(CAR) – ? CAR_REG_NO(ACCIDENT) Q. 19 Q. 20 What is a key? Explain Candidate Key, Alternate Key and Foreign Key. (7) Ans: Key – A single attribute or a combination of two or more attributes of an entity set that is used to identify one or more instances (rows) of the set (table) is called as key. Candidate Key – A candidate key is a minimal superkey, which can be used to uniquely identify a tuple in the relation. Alternate Key – All the candidate keys except primary key are called as alternate keys. Foreign Key – Let there are two relations (tables) R and S.

Any candidate key of the relation R which is referred in the relation S is called the foreign key in the relation S and referenced key in the relation R. The relation R is also called as parent table and relation S is also called as child table. Q. 21 What is data independence? Explain the difference between physical and logical data independence. (7) Ans: Data independence is the capacity to change the schema at one level of a database system without having to change the schema at the next level. The three-schema architecture allows the feature of data independence.

Data independence occurs 26 DC10 DATABASE MANAGEMENT SYSTEMS because when the schema is changed at some level, the schema at the next level remains unchanged; only the mapping between the two levels is changed. Types of data independence are: Physical Data Independence – It is capacity to change the internal schema without having to change conceptual schema. Hence, the external schemas need not be changed as well. Changes to the internal schema may be needed because some physical files had to be reorganized to improve the performance of retrieval or update.

If the same data as before remains in the database, the conceptual schema needs not be changed. Logical Data Independence – It is the capacity to change the conceptual schema without having to change external schemas or application programs. The conceptual schema may be changed to expand the database (by adding a record type or data item), to change constraints, or to reduce the database (by removing a record type or data item). Only the view definition and the mappings need be changed in a DBMS that supports logical data independence.

Changes to constraints can be applied to the conceptual schema without affecting the external schemas or application programs. Q. 22 Write short notes on: (i) Weak and strong entity sets. (ii) Types of attributes. (iii) Oracle Instance. (iv) Mid square method of hashing. (4 x 4 = 16 ) Ans: (i) Weak and Strong entity sets: A strong entity set has a primary key. All tuples in the set are distinguishable by that key. A weak entity set has no primary key unless attributes of the strong entity set on which it depends are included.

Tuples in a weak entity set are partitioned according to their relationship with tuples in a strong entity set. Tuples within each partition are distinguishable by a discriminator, which is a set of attributes. A strong entity set has a primary key. All tuples in the set are distinguishable by that key. A weak entity set has no primary key unless attributes of the strong entity set on which it depends are included. Tuples in a weak entity set are partitioned according to their relationship with tuples in a strong entity set. Tuples within each partition are distinguishable by a discriminator, which is a set of attributes. ii) Types of attributes:An attribute’s type determines the kind of values that are allowed in the attribute. For example, the value version 1 is not valid for an attribute defined as an integer, but the value 1 is valid. Numeric types (such as integer or real) can also be limited to a predefined range by their attribute definition. Choice :An attribute with a list of predefined values. ID Reference: An attribute with a value that is a Unique ID value from another element. It is typically used for element-based cross-references.

ID References: An attribute with a value of one or more Unique ID values from another element. Integer: An attribute with a whole number value (no decimal parts). Examples of valid integers are 22, -22, and +322. An integer can be defined to fall within a range. Integers: An attribute with a value of one or more integers. Enter each number on a separate line in the Attribute Value text box. Real An attribute with a real number value, with or without a decimal part (the value can also be expressed in scientific notation). Examples of valid real numbers are 2, 22. 4, 0. 22, and 2. e-1. A real number can be defined to fall within a range. 27 DC10 DATABASE MANAGEMENT SYSTEMS Reals: An attribute with a value of one or more real numbers. Enter each number on a separate line in the Attribute Value text box. String: An attribute with a value of a series of characters (text). Strings: An attribute with a value of one or more strings. Enter each string on a separate line in the Attribute Value text box. Unique ID: An attribute with a value of a unique text string. An element can have only one ID attribute (which can be of type Unique ID or Unique IDs).

All ID values must be unique in the document or book. An element with a Unique ID attribute can be the source for an element-based cross-reference. Unique IDs: An attribute with a value of one or more unique text strings. Enter each string on a separate line in the Attribute Value text box. (iii) Oracle Instances: An instance is the (executed) Oracle software and the memory they use. It is the instance that manipulates the data stored in the database. It can be started independent of any database. It consists of: 1) A shared memory area that provides the communication between various processes. ) Upto five background processes which handled various tasks. Whenever an oracle instance starts, the file ‘INIT. ORA’ is executed. (iv) Mid square method of hashing: In midsquare hashing, the key is squared and the address selected from the middle of the squared number. Mid square method * Square K. * Strip predetermined digits from front and rear. * e. g. , use thousands and ten thousands places. Q. 23 Consider the following relational schemas: EMPLOYEE (EMPLOYEE_NAME, STREET, CITY) WORKS (EMPLOYEE_NAME, COMPANYNAME, SALARY) COMPANY (COMPANY_NAME, CITY) Specify the table definitions in SQL.

Ans: CREATE TABLE EMPLOYEE ( EMPLOYEE_NAME VARCHAR2(20) PRIMARY KEY, STREET VARCHAR2(20), CITY VARCHAR2(15)); CREATE TABLE COMPANY ( COMPANY_NAME VARCHAR2(50) PRIMARY KEY, CITY VARCHAR2(15)); CREATE TABLE WORKS ( EMPLOYEE_NAME VARCHAR2(20) REFERENCES EMPLOYEE(EMPLOYEE_NAME, COMPANYNAME VARCHAR2(50) REFERENCES COMPANY(COMPANY_NAME, SALARY NUMBER(6), CONSTRAINT WORKS_PK PRIMARY KEY(EMPLOYEE_NAME, COMPANY_NAME)); (5) Q. 24 Give an expression in SQL for each of queries below: 28 (9) DC10 DATABASE MANAGEMENT SYSTEMS (i) Find the names of all employees who work for first Bank Corporation. ii) Find the names and company names of all employees sorted in ascending order of company name and descending order of employee names of that company. (iii) Change the city of First Bank Corporation to ‘New Delhi’ Ans: SELECT EMPLOYEE_NAME FROM WORKS WHERE COMPANYNAME = ‘First Bank Corporation’; (ii) SELECT EMPLOYEE_NAME, COMPANYNAME FROM WORKS ORDER BY COMPANYNAME, EMPLOYEE_NAME DESC; (iii) UPDATE COMPANY SET CITY = ‘New Delhi’ WHERE COMPANY_NAME = ‘First Bank Corporation’; (i) Q. 25 Discuss the correspondence between the E-R model construct and the relation model construct.

Show how each E-R model construct can be mapped to the relational model using the suitable example? Ans: An entity-relationship model (ERM): An entity-relationship model (ERM) is an abstract conceptual representation of structured data. Entity-relationship modeling is a relational schema database modeling method, used in software engineering to produce a type of conceptual data model (or semantic data model) of a system, often a relational database, and its requirements in a top-down fashion. Diagrams created using this process are called entity-relationship diagrams, or ER diagrams or ERDs for short.

ER-to-Relational Mapping Algorithm: 1) Step 1: Mapping of regular entity types: For each strong entity type E, create a relation T that includes all the simple attributes of a composite attribute. 2) Step2: Mapping of weak entity types: For each weak entity type W with owner entity type E, create relation R and include all simple attributes (or simple components of composite attributes) of W as attributed of R. In addition, include as foreign key attributes of R, the primary key attribute (s) of relation(s) that correspond to the owner(s) and the partial key of the weak entity type W, if any. ) Mapping of relationship types: form a relation R, for relationship with primary keys of participating relations A and B as foreign keys in R. In addition to this, any attributes of relationship become an attribute of R also. 4) Mapping of multivalued attributes: For each multilvalued attribute A, create a new relation R. This relation R will include an attribute corresponding to A, plus primary key attribute K-as a foreign key in R-of the relation that represents the entity type or relationship type that has A as an attribute. Q. 26 Explain the concepts of relational data model. isadvantages. Also discuss its advantages and (7) Ans: Relational Data Model – The relational model was first introduced by Prof. E. F. Codd of the IBM Research in 1970 and attracted immediate attention due to its simplicity and 29 DC10 DATABASE MANAGEMENT SYSTEMS mathematical foundation. The model uses the concept of a mathematical relation (like a table of values) as its basic building block, and has its theoretical basis in set theory and first-order predicate logic. The relational model represents the database as a collection of relations.

The relational model like all other models consists of three basic components: a set of domains and a set of relations operation on relations integrity rules Advantages • Ease of use – The revision of any information as tables consisting of rows and columns is quite natural and therefore even first time users find it attractive. • Flexibility – Different tables from which information has to be linked and extracted can be easily manipulated by operators such as project and join to give information in the form in which it is desired. Security – Security control and authorization can also be implemented more easily by moving sensitive attributes in a given table into a separate relation with its own authorization controls. If authorization requirement permits, a particular attribute could be joined back with others to enable full information retrieval. • Data Independence – Data independence is achieved more easily with normalization structure used in a relational database than in the more complicated tree or network structure. It also frees the users from details of storage structure and access methods. Data Manipulation Language – The possibility of responding to ad-hoc query by means of a language based on relational algebra and relational calculus is easy in the relational database approach. Provides simplicity in the data organization and the availability of reasonably simple to very powerful query languages. Disadvantages • Performance – If the number of tables between which relationships to be established are large and the tables themselves are voluminous, the performance in responding to queries is definitely degraded. Unsuitable for Hierarchies – While the relational database approach is a logically attractive, commercially feasible approach, but if the data is for example naturally organized in a hierarchical manner and stored as such, the hierarchical approach may give better results. Q. 27 Consider the following relational schema: (14) Doctor(DName,Reg_no) Patient(Pname, Disease) Assigned_To (Pname,Dname) Give expression in both Tuple calculus and Domain calculus for each of the queries: (i) Get the names of patients who are assigned to more than one doctor. ii) Get the names of doctors who are treating patients with ‘Polio’. Ans: (i)Tuple Calculus: {p[PName] | p ? PATIENT ? ?a1, a2 (a1 ? ASSIGNED_TO ? a2 ? ASSIGNED_TO ? p[PName] = a1[PName] ? a1[PName] = a2[PName] ? a1[DName] ? a2[DName])} Domain Calculus: 30 DC10 DATABASE MANAGEMENT SYSTEMS {p | ? p1, d1, p2, d2 ( ? PATIENT ? ? ASSIGNED_TO ? ? ASSIGNED_TO ? p1 = p2 ? d1 ? d2)} (ii)Tuple Calculus: {u[Dname] | u ? ASSIGNED_TO ? ?t (t ? PATIENT ? t[Disease] = ‘Polio’ ? t[PName] = u[PName])} Domain Calculus: {d | ? p1, p2, s2 ( ? ASSIGNED_TO ? ? PATIENT ? p1 = p2 ? 2 = ‘Polio’)} Q. 28 What are the features of embedded SQL? Explain. (7) Ans: Embedded SQL – SQL can be implemented in two ways. It can be used interactively or embedded in a host language or by using API. The use of SQL commands within a host language (e. g. , C, Java, etc. ) program is called embedded query language or Embedded SQL. Although similar capabilities are supported for a variety of host languages, the syntax sometimes varies. Some of the features of embedded SQL are: SQL statements can be used wherever a statement in the host language is allowed.

It combines the strengths of two programming environments, the procedural features of host languages and non-procedural features of SQL. SQL statements can refer to variables (must be prefixed by a colon in SQL statements) defined in the host program. Special program variables (called null indicators) are used to assign and retrieve the NULL values to and from the database. The facilities available through the interactive query language are also automatically available to the host programs.

Embedded SQL along with host languages can be used to accomplish very complex and complicated data access and manipulation tasks. Q. 29 What is the purpose of tables, private synonyms and public synonyms? If there are multiple objects of same name on an Oracle database, which order are they accessed in? Ans: The purpose of table is to store data. If we use the PUBLIC keyword (or no keyword at all), anyone who has access to the database can use our synonym. If the database is not ANSI-compliant, a user does not need to know the name of the owner of a public synonym.

Any synonym in a database that is not ANSI-compliant and was created in an Informix database server is a public synonym. In an ANSI-compliant database, all synonyms are private. If you use the PUBLIC or PRIVATE keywords, the database server issues a syntax error. If you use the PRIVATE keyword to declare a synonym in a database that is not ANSI-compliant, the unqualified synonym can be used by its owner. Other users must qualify the synonym with the name of the owner. Q. 30 Explain the followings: (i) Temporary Tables (ii) Integrity Constraints. Ans: 31 (14)

DC10 DATABASE MANAGEMENT SYSTEMS (i) Temporary Tables – Temporary tables exists solely for a particular session, or whose data persists for the duration of the transaction. The temporary tables are generally used to support specialized rollups or specific application processing requirements. Unlike a permanent table, a temporary table does not automatically allocate space when it is created. Space will be dynamically allocated for the table as rows are inserted. The CREATE GLOBAL TEMPORARY TABLE command is used to create a temporary table in Oracle.

CREATE GLOBAL TEMPORARY TABLE ( ) ON COMMIT {PRESERVE|DELETE} ROWS; (ii)Integrity Constraints – A database is only as good as the information stored in it, and a DBMS must therefore help prevent the entry of incorrect information. An integrity constraint is a condition specified on a database schema and restricts the data that can be stored in an instance of the database. If a database instance satisfies all the integrity constraints specified on the database schema, it is a legal instance. A DBMS enforces integrity constraints, in that it permits only legal instances to be stored in the database.

Integrity constraints are specified and enforced at different times: When the DBA or end user defines a database schema, he or she specifies the integrity constraints that must hold on any instance of this database. When a database application is run, the DBMS checks for violations and disallows changes to the data that violate the specified integrity constraints. Many kinds of integrity constraints can be specified in the relational model, such as, Not Null, Check, Unique, Primary Key, etc. Q. 31 Explain different types of failures that occur in Oracle database. (7)

Ans: Types of Failures – In Oracle database following types of failures can occurred: Statement Failure • Bad data type • Insufficient space Insufficient Privileges (e. g. , object privileges to a role) User Process Failure • User performed an abnormal disconnect • User’s session was abnormally terminated • User’s program raised an address exception User Error • User drops a table • User damages data by modification Instance Failure Media Failure • User drops a table • User damages data by modification Alert Logs • Records informational and error messages • All Instance startups and shutdowns are recorded in the log 32

DC10 • DATABASE MANAGEMENT SYSTEMS Every Create, Alter, or Drop operation on a rollback segment, tablespace, or database is record in the log Recovery Views DB Verify • Used to insure that a datafile is valid before a restore Q. 32 What is ODBC? What are the uses of ODBC? Under what circumstances we use this technology? (7) Ans: ODBC – Open DataBase Connectivity (ODBC) enable the integration of SQL with a general-purpose programming language. ODBC expose database capabilities in a standardized way to the application programmer through an application programming interface (API).

In contrast to Embedded SQL, ODBC allows a single executable to access different DBMSs without recompilation. Thus, while Embedded SQL is DBMSindependent only at the source code level, applications using ODBC are DBMSindependent at the source code level and at the level of the executable. All direct interaction with a specific DBMS happens through a DBMS-specific driver. A driver is a software program that translates the ODBC calls into DBMS-specific calls. Drivers are loaded dynamically on demand since the DBMSs the application is going to access are known only at run-time. Available drivers are registered with a driver manager.

The driver translates the SQL commands from the application into equivalent commands that the DBMS understands. An application that interacts with a data source through ODBC selects a data source, dynamically loads the corresponding driver, and establishes a connection with the data source. ODBC achieves portability at the level of the executable by introducing an extra level of indirection. In addition, using ODBC, an application can access not just one DBMS but several different ones simultaneously. Q. 33 List any two significant differences between a file processing system and a DBMS. 4) Ans: File Processing System vs. DBMS Data Independence – Data independence is the capacity to change the schema at one level of a database system without having to change the schema at the next level. In file processing systems the data and applications are generally interdependent, but DBMS provides the feature of data independence. Data Redundancy – Data redundancy means unnecessary duplication of data. In file processing systems there is redundancy of data, but in DBMS we can reduce data redundancy by means of normalization process without affecting the original data.

If we do so in file processing system, it becomes too complex. Q. 34 Differentiate between various levels of data abstraction. (5) Ans: Data Abstraction – Abstraction is the process to hide the irrelevant things from the users and represent the relevant things to the user. Database systems are often used by non-computer professionals so that the complexity must be hidden from database system users. This is done by defining levels of abstract as which the database may be viewed, there are logical view or external view, conceptual view and internal view or physical view. 3 DC10 DATABASE MANAGEMENT SYSTEMS o External View – This is the highest level of abstraction as seen by a user. It describes only the part of entire database, which is relevant to a particular user. o Conceptual View – This is the next higher level of abstraction which is the sum total of Database Management System user’s views. It describes what data are actually stored in the database. It contains information about entire database in terms of a small number of relatively simple structure. o Internal View – This is the lowest level of abstraction.

It describes how the data are physically stored Q. 35 What are the various symbols used to draw an E-R diagram? Explain with the help of an example how weak entity sets are represented in an E-R diagram. (6) Ans: Various symbols used to draw an E-R diagram Symbol Meaning ENTITY TYPE WEAK ENTITY TYPE RELATIONSHIP TYPE ATTRIBUTE KEY ATTRIBUTE MULTIVALUED ATTRIBUTE COMPOSITE ATTRIBUTE DERIVED ATTRIBUTE E1 1 R N E2 CARDINALITY RATIO 1:N Weak Entity Sets – An entity set that does not have a key attribute is called weak entity set.

A weak entity must participate in an identifying relationship type with an owner or identifying entity type Entities are identified by the combination of: A partial key of the weak entity type The primary key of the identifying entity type Example: Suppose that a DEPENDENT entity is identified by the dependent’s first name and birhtdate, and the specific EMPLOYEE that the dependent is related to. DEPENDENT is a weak entity type with EMPLOYEE as its identifying entity type via the identifying relationship type DEPENDENT_OF 34 DC10 FNAME DATABASE MANAGEMENT SYSTEMS BDATE ADDRESS EMPLOYEE DEPENDENTS_OF

DEPENDENT Name Birth Date Relationship Q. 36 Define the following terms: a) Primary key. b) DML c) Multivalued attribute d) (8) Relationship instance Ans: Primary Key – Primary key is one of the candidate keys. It should be chosen such that its attribute values are never, or very rarely, changed. b) Data Manipulation Language (DML) – A data manipulation language is a language that enables users to access or manipulate data as organized by the appropriate data model. c) Multivalued Attribute – Multivalued attribute may have more than one value for an entity. For example, PreviousDegrees of a STUDENT. ) Relationship Instance – A relationship is an association among two or more entities. An instance of relationship set is a set of relationships. Q. 37 Define a table in SQL called Client, which is used to store information about the clients. Define CLIENT_NO as the primary key whose first letter must start with ‘C’. Also ensure that the column ‘NAME’ should not allow NULL values. Column name Data type Size CLIENT_NO Varchar2 6 NAME Varchar2 20 ADDRESS1 Varchar2 30 ADDRESS2 Varchar2 30 CITY Varchar2 15 STATE Varchar2 15 PINCODE Number 6 BAL_DUE Number 10, 2 (7) 35 DC10

DATABASE MANAGEMENT SYSTEMS Ans: CREATE TABLE CLIENT ( CLIENT_NO VARCHAR2(6) PRIMARY KEY CHECK (CLIENT_NO LIKE ‘C%’), NAME VARCHAR2(20) NOT NULL, ADDRESS1 VARCHAR2(30), ADDRESS2 VARCHAR2(30), CITY VARCHAR2(15), STATE VARCHAR2(15), PINCODE NUMBER(6), BAL_DUE NUMBER(10,2)) Q. 38 An orchestra database consists of the following relations: (3. 5 x 2=7) CONDUCTS (conductor, composition) REQUIRES (composition, Instrument) PLAYS (Player, Instrument) Give the relational calculus queries for the following: (i) List the compositions and the players. (ii) List the compositions which require the ‘violin’ and the ‘congo’

Ans: (i) Tuple Calculus: {r[Composition] || p[Player] | r ? REQUIRES ? p ? PLAYS ? r[Instrument] = p[Instrument]} Domain Calculus: {c || p | ? i1, i2 ( ? REQUIRES ? ? PLAYS ? i1 = i2)} (ii) Tuple Calculus: {r[Composition] | r ? REQUIRES ? ?u (u ? REQUIRES ? r[Composition] = u[Composition] ? r[Instrument] = ‘violin’ ? u[Instrument] = ‘congo’)} Domain Calculus: {c | ? i1, c2, i2 ( ? REQUIRES ? ? REQUIRES ? c 1 = c ? i 1 = ‘violin’ ? i 2 = ‘congo’)} Q. 39 Perform the following with syntax and a suitable example (i) Create a table from existing table. ii) Insert data in your table from another table. Ans: (i) Create table as e. g, To create a new table ‘N_emp’ with employee names and their identification numbers only from employee table, statement is to create table N_emp as select empname, empid from employee (ii) insert into e. g, To insert tuples from employee into N_emp created above, use following statement Insert into N_emp select empname, empid from employee Q. 40 What is an INDEX as defined in ORACLE? Write the syntax of creating an INDEX. Create an index for the table Client, field CLIENT_NO of Q. (2+2+3) 36 DC10

DATABASE MANAGEMENT SYSTEMS Ans: Indexes in Oracle – Index is typically a listing of keywords accompanied by the location of information on a subject. In other words, An index can be viewed as an auxiliary table which contains two fields: the key and the location of the record of that key. Indexes are used to improve the performance of the search operation. Indexes are not strictly necessary to running Oracle, they do speed the process. Syntax of Creating an Index: CREATE [BITMAP] [UNIQUE] INDEX ON ( [, ] . . . ); Command: CREATE INDEX client_client_no ON client(client_no); Q. 41

Consider the following relational database: STUDENT (name, student#, class, major) COURSE (course name, course#, credit hours, department) SECTION (section identifier, course#, semester, year, instructor) GRADE_REPORT (student#, section identifier, grade) PREREQUISITE (course#, presequisite#) Specify the following queries in SQL on the above database schema. (3. 5 x 4=14) (i) Retrieve the names of all students majoring in ‘CS’ (Computer Science). (ii) Retrieve the names of all courses taught by Professor King in 1998 (iii) Delete the record for the student whose name is ‘Smith’ and whose student number is 17. iv) Insert a new course Ans: (i) SELECT NAME FROM STUDENT WHERE MAJOR = ‘CS’ (ii) SELECT COURSE_NAME FROM COURSE C, SECTION S WHERE C. COURSE# = S. COURSE# AND INSTRUCTOR = ‘KING’ AND YEAR = 1998 OR SELECT COURSE_NAME FROM COURSE WHERE COURSE# IN (SELECT COURSE# FROM SECTION WHERE INSTRUCTOR = ‘KING’ AND YEAR = 1998) (iii) DELETE FROM STUDENT WHERE NAME = ‘Smith’ AND STUDENT# = 17 (iv) INSERT INTO COURSE VALUES(‘Knowledge Engineering’, ‘CS4390’, 3, ‘CS’) Q. 42 Explain the concept of a data model. What data models are used in database management systems? 7) Ans: Data Model – Model is an abstraction process that hides irrelevant details while highlighting details relevant to the applications at hand. Similarly, a data model is a collection of concepts that can be used to describe structure of a database and provides the necessary means to achieve this abstraction. Structure of database means the data types, relationships, and constraints that should hold for the data. In general a data model consists of two elements: • A mathematical notation for expressing data and relationships. • Operations on the data that serve to express queries and other manipulations of the data. 7 DC10 Data Models used in DBMSs: DATABASE MANAGEMENT SYSTEMS Hierarchical Model – It was developed to model many types of hierarchical organizations that exist in the real world. It uses tree structures to represent relationship among records. In hierarchical model, no dependent record can occur without its parent record occurrence and no dependent record occurrence may be connected to more than one parent record occurrence. Network Model – It was formalised in the late 1960s by the Database Task Group of the Conference on Data System Language (DBTG/CODASYL).

It uses two different data structures to represent the database entities and relationships between the entities, namely record type and set type. In the network model, the relationships as well as the navigation through the database are predefined at database creation time. Relational Model – The relational model was first introduced by E. F. Codd of the IBM Research in 1970. The model uses the concept of a mathematical relation (like a table of values) as its basic building block, and has its theoretical basis in set theory and first-order predicate logic.

The relational model represents the database as a collection of relations. Object Oriented Model – This model is based on the object-oriented programming language paradigm. It includes the features of OOP like inheritance, object-identity, encapsulation, etc. It also supports a rich type system, including structured and collection types. Object Relational Model – This model combines the features of both relational model and object oriented model. It extends the traditional relational model with a variety of features such as structured and collection types.

Q. 43 Briefly explain the differences between a stand alone query language, embedded query language and a data manipulation language. (7) Ans: Stand alone Query Language – The query language which can be used interactively is called stand alone query language. It does not need the support of a host language. Embedded Query Language – A query language (e. g. , SQL) can be implemented in two ways. It can be used interactively or embedded in a host language. The use of query language commands within a host language (e. . , C, Java, etc. ) program is called embedded query language. Although similar capabilities are supported for a variety of host languages, the syntax sometimes varies. Data Manipulation Language (DML) – A data manipulation language is a language that enables users to access or manipulate data as organized by the appropriate data model. Q. 44 Consider the following relations for a database that keeps track of business trips of salespersons in a sales office: SALESPERSON (SSN, Name, start_year, Dept_no)

TRIP (SSN, From_city, To_city, Departure_Date, Return_Date, Trip_ID) EXPENSE(TripID, Account#, Amount) Specify the following queries in relational algebra: (4×3 =12) (i) Give the details (all attributes of TRIP) for trips that exceeded $2000 in expenses. (ii) Print the SSN of salesman who took trips to ‘Honolulu’ 38 DC10 DATABASE MANAGEMENT SYSTEMS (iii) Print the trip expenses incurred by the salesman with SSN= ‘234-56-7890’. Note that the salesman may have gone on more than one trip. List them individually Ans: (i) ? TRIP. * (? mount > 2000 (TRIP EXPENSE)) (ii) ? SSN (? to_city = ‘Honolulu’ (TRIP)) (iii) ? EXPENSE. tripid, amount (? SSN = ‘234-56-7890’ (TRIP EXPENSE)) Q. 45 What is the difference between a key and a superkey? (2) Ans: Key – A key a single attribute or a combination of two or more attributes of an entity set that is used to identify one or more instances (rows) of the set (table). It is a minimal combination of attributes. Super Key – A super key is a set of one or more attributes that, taken collectively, allows us to identify uniquely a tuple in the relation. Q. 6 Why are cursors necessary in embedded SQL? (2) Ans: A cursor is an object used to store the output of a query for row-by-row processing by the application programs. SQL statements operate on a set of data and return a set of data. On other hand, host language programs operate on a row at a time. The cursors are used to navigate through a set of rows returned by an embedded SQL SELECT statement. A cursor can be compared to a pointer. Q. 47 Write a program in embedded SQL to retrieve the total trip expenses of the salesman named ‘John’ for the relations of Q. 4 (6) Ans: EXEC SQL BEGIN DECLARE SECTION; long total_expenses; EXEC SQL END DECLARE SECTION; EXEC SQL SELECT SUM(AMOUNT) INTO :total_expenses FROM EXPENSE WHERE TRIPID IN (SELECT TRIP_ID FROM TRIP WHERE SSN = (SELECT SSN FROM SALEPERSON WHERE NAME = ‘John’)); printf(“\nThe total trip expenses of the salesman John is: %ld”, total_expenses); Q. 48 What are views? Explain how views are different from tables. (6) Ans: A view in SQL terminology is a single table that is derived from other tables. These other tables could be base tables or previously defined views.

A view does not necessarily exist in physical form; it is considered a virtual table, in contrast to base tables, whose tuples are actually stored in the database. This limits the possible update operations that can be applied to views, but it does not provide any limitations on querying a view. A view represents a different perspective of a base relation(s). The definition of a view in a create view statement is stored in the system catalog. Any attribute in the view can be updated as long as the attribute is simple and not derived from a computation involving two or more 39 DC10 DATABASE MANAGEMENT SYSTEMS ase relation attribute. View that involve a join may or may not be updatable. Such views are not updatable if they do not include the primary keys of the base relations. Q. 49 What do you mean by integrity constraints? Explain the two constraints, check and foreign key in SQL with an example for each. Give the syntax. (8) Ans:Integrity Constraints –An integrity constraint is a condition specified on a database schema and restricts the data that can be stored in an instance of the database. If a database instance satisfies all the integrity constraints specified on the database schema, it is a legal instance.

A DBMS enforces integrity constraints, in that it permits only legal instances to be stored in the database. CHECK constraint – CHECK constraint specifies an expression that must always be true for every row in the table. It can’t refer to values in other rows. Syntax: ALTER TABLE ADD CONSTRAINT CHECK(); FOREIGN KEY constraint – A foreign key is a combination of columns with values based on the primary key values from another table. A foreign key constraint, also known as referential integrity constraint, specifies that the values of the foreign key correspond to actual values of the primary or unique key in other table.

One can refer to a primary or unique key in the same table also. Syntax: ALTER TABLE ADD CONSTRAINT FOREIGN KEY() REFERENCES () ON {DELETE | UPDATE} CASCADE; Q. 50 Define the following constraints for the table client of Q. 37 (i) BAL_DUE must be at least 1000. (ii) NAME is a unique key. (6) Ans: (i)ALTER TABLE CLIENT ADD CONSTRAINT CLIENT_BAL_DUE_C1 CHECK(BAL_DUE < 1000); (ii) ALTER TABLE CLIENT ADD CONSTRAINT CLIENT_NAME_U UNIQUE(NAME); Q. 51 What are the different types of database end users? Discuss the main activities of each. (7)

Ans: End-Users – End-users are the people whose jobs require access to the database for querying, updating, and generating reports; the database primarily exists for their use. The different types of end-users are: Casual end-users – occasionally access the database, need different information each time Naive or Parametric end-users – includes tellers, clerks, etc. , make up a sizable portion of database end-users, main job function revolves around constantly querying and updating the database 40 DC10 DATABASE MANAGEMENT SYSTEMS Sophisticated end-users – includes engineers, scientists, business analyst, etc. use for their complex requirements Stand-alone users – maintain personal databases by using ready-made program packages, provide easy-to-use menu-based or graphics-based interfaces Q. 52 Discuss the typical user friendly interfaces and the types of users who use each. (7) Ans: User-friendly interfaces provided by a DBMS may include the following: • Menu-Based Interfaces for Web Clients or Browsing – These interfaces present the user with lists of options, called menus, that lead the user through the formulation of a request. Pull-down menus are a very popular techniques in Web-based user interfaces.

They are also used in browsing interfaces, which allow a user to look through the contents of a database in an exploratory and unstructured manner. • Forms-Based Interfaces – A forms-based interface displays a form to each user. Forms are usually designed and programmed for naive users and interfaces to canned transactions. Many DBMSs have forms specification languages. • Graphical User Interfaces (GUIs) – A GUI typically displays a schema to the user in diagrammatic form. The user can then specify a query by manipulating the diagram.

In many cases, GUIs utilizes both menus and forms. Most GUIs use a pointing device to pick certain parts of the displayed schema diagram. • Natural Language Interfaces – These interfaces accept requests written in English or some other language and attempt to “understand” them. A natural language interface usually has its own “schema,” which is similar to the database conceptual schema, as well as a dictionary of important words. • Interfaces for Parametric Users – Parametric users, such as bank tellers, often have a small set of operations that they must perform repeatedly.

The interfaces for these users usually have a small set of abbreviated commands with the goal of minimizing the number of keystrokes required for each request. • Interfaces for the DBA – Most database systems contain privileged commands that can be used only by the DBA’s staff. These include commands for creating accounts, setting system parameters, granting account authorization, changing a schema, and a reorganizing the storage structures of a database. Q. 53 With the help of an example show how records can be deleted and updated in QBE. 5) (i) Increase Pay_Rate of employees with the skill of ‘cook’ by 10%. (ii) Delete employee record for EMP# 123459 Ans: (i) EMPLOYEE U. Emp# EX EX Name Skill Cook Pay_Rate PX PX * 1. 1 (ii) EMPLOYEE D. Q. 54 Emp# 123459 Name Skill Pay_Rate Describe cardinality ratios and participation constraints for relationship types. 41 (4) DC10 DATABASE MANAGEMENT SYSTEMS Ans: Cardinality Ratios – The cardinality ratios for a relationship type specifies the maximum number of relationship instances that an entity can participate in.

The possible cardinality ratios for relationship types are one-to-one (1:1), one-to-many or many-to-one (1:M or M:1), and many-to-many (M:N). Participation Constraints – The participation constraint specifies whether the existence of an entity depends on its being related to another entity via the relationship type. This constraint specifies the minimum number of relationship instances that each entity can participate in. It is sometimes called the minimum cardinality constraint. There are two types of participation constraints – total and partial. Q. 55

Information about a bank is about customers and their account. Customer has a name, address which consists of house number, area and city, and one or more phone numbers. Account has number, type and balance. We need to record customers who own an account. Account can be held individually or jointly. An account cannot exist without a customer. Arrive at an E-R diagram. Clearly indicate attributes, keys, the cardinality ratios and participation constraints. (10) Ans: P ho ne_ no cit y ar ea Ho us e_ no na me ad d r e ss name t yp e t yp e d eta il s CU ST OM E R ( I. M) O wn s a cco u nt

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Data Management. (2016, Oct 14). Retrieved from https://graduateway.com/data-management-2/

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