BCA 06 (Introduction to Data Base Management System) TNOU BCA First year Assignment 2012

BCA 06 (Introduction to Data Base Management System) TNOU BCA First year Assignment 2012

PART A

Short Answer Questions:

1.       What are the draw backs of commercial database?

Answer:-
Existing commercial DBMS, both small and large has proven inadequate for these applications. The traditional database notion of storing data in two-dimensional tables, or in flat files, breaks down quickly in the face of complex data structures and data types used in today’s applications. Research in model and process complex data has gone in two directions:
a. Extending the functionality of RDBMS

b. Developing and implementing OODBMS that is based on object oriented programming paradigm.
The main elements of RDBMS are based on Ted Codd’s 13 rules for a relational system, the concept of relational integrity, and normalization. The three fundamentals of a relational database are that all information must be held in the form of a table, where all data are described using data values. The second fundamental is that each value found in the table columns does not repeat. The final fundamental is the use of Standard Query Language (SQL).
The language important to OODBMS is data definition and manipulation language (DDML). The use of this language allows persistent data to be created, updated, deleted, or retrieved. An OODBMS needs a computational versus a relational language because it can be used to avoid impedance mismatch. DDML allows users to define a database, including creating, altering, and dropping tables and establishing constraints. DDMLs are used to maintain and query a database, including updating, inserting, modifying, and querying data.

2.      (i) List factors which motivate the move to acquire the DBMS approach?

Answer:

A Database Management System (DBMS) approach to storing data is a smart move for a business to make. There are many advantages to organizing your data in a database. These range from reducing data redundancy to implementing data access controls. Not only can a company benefit internally from using a DBMS, its clients can be confident in the security of their personal information.

ii)What is the Difference between B-Tree and B+Tree ?

Difference between B and B+ tree :

B tree indices are similar to B+ tree indices. The primary distinction between the two approaches is that a B-tree eliminates the redundant storage of search key values. Search keys are not repeated in B tree indices.

Simple B+ tree

Simple B-Tree

Given below the major difference between B tree and B+ tree structure.

1.      In a B tree search keys and data stored in internal or leaf nodes. But in B+-tree data store only leaf nodes.

2.      Searching of any data in a B+ tree is very easy because all data are found in leaf nodes otherwise in a B tree data cannot found in leaf node.

3.      In B tree data may found leaf or non leaf node. Deletion of non leaf node is very complicated. Otherwise in a B+ tree data must found leaf node. So deletion is easy in leaf node.  

4.      Insertion of a B tree is more complicated than B+ tree. 

5.      B +tree store redundant search key but b-tree has no redundant value.

6.      In B+ tree leaf node data are ordered in a sequential linked list but in B tree the leaf node cannot stored using linked list.

Many database system implementers prefer the structural simplicity of a b+ tree.

3.      I) What are the Basic relational Algebra Operations?

Answer:

Relational Algebra:

A query language is a language in which user requests information from the database. it can be categorized as either procedural or nonprocedural. In a procedural language the user instructs the system to do a sequence of operations on database to compute the desired result. In nonprocedural language the user describes the desired information without giving a specific procedure for obtaining that information. Relational algebra is a collection of operations used to manipulate relations (tables). These operations enable the users to specify the retrieval requests which results in a new relation built from one or more relations.

Relational Algebra is a Procedural language, which specifies, the operations to be performed on the existing relations to derive result relations. It is a procedural language which means that user has to specify what is required and what is the sequence of steps performed on the database to obtain the required output. Whenever the operations are performed on the existing relations to produce new relations then the original relations(s) are not effected i.e. they remain the same, and the resultant relation obtained can act as an input to some other operation, so relational algebra operations can be composed together into a relational algebra expression. Composing relational algebra operation into relational expression is similar to composing arithmetic operations (+, -, *) into arithmetic expressions. R1+R2 is a relational expression where R1 and R2 are relations.

Relational Algebra is a formal and non-user friendly language. It illustrates the basic operations required for any Data Manipulation languages but it is very less commonly used in the commercial languages because it lacks the syntactic details, although it acts as a fundamental technique for extracting data from the database.

3( ii)Write Short notes on Client Server Approach ?

Answer:

Client-server is just one approach to managing network applications The primary alternative, peer-to-peer networking, models all devices as having equivalent capability rather than specialized client or server roles. Compared to client-server, peer to peer networks offer some advantages such as more flexibility in growing the system to handle large number of clients. Client-server networks generally offer advantages in keeping data secure.

The client-server model was originally developed to allow more users to share access to database applications. Compared to the mainframe approach, client-server offers improved scalability because connections can be made as needed rather than being fixed. The client-server model also supports modular applications that can make the job of creating software easier. In so-called "two-tier" and "three-tier" types of client-server systems, software applications are separated into modular pieces, and each piece is installed on clients or servers specialized for that subsystem.

               

                                                                Part B

Answer All the Questions:

1.       What is Normalization ? Explain the Normalized Relations.

Answer:

The normalization is the process of organizing the fields and tables of a relational database to minimize redundancy and dependency. Normalization usually involves dividing large tables into smaller (and less redundant) tables and defining relationships between them. The objective is to isolate data so that additions, deletions, and modifications of a field can be made in just one table and then propagated through the rest of the database via the defined relationships.

Normalization of a data is a process of analyzing the given relation schema based on their FDs and primary keys to achieve the desirable properties of

• Minimizing redundancy
• Minimizing the insertion, deletion and update anomalies.

First Normal Form:

First Normal form imposes a very basic requirement on relations; unlike the other normal forms, it does not require additional information such as functional dependencies.
         
A domain is atomic if elements of the domain are considered to be indivisible units. We say that a relation schema R is in first normal form (1NF) if the domains of all attributes of R are atomic.

A set of names is an example of a nonatomic value. For example, if the schema of a relation employee included an attribute children whose domain elements are sets of names, the schema would not be in first normal form.

Composite attributes, such as an attribute address with component attributes street and city, also have nonatomic domains.

Second Normal Form:

2NF is based on the concept of full functional dependency.

A functional dependency X -> Y is a full functional dependency if removal of any attribute A from X means that the dependency does not hold any more.
         
A relational schema R is in 2NF every non prime attribute A in R is fully functionally dependent on the PK of R.

The relation schema can be 2NF normalized into a number of 2NF relations in which non-prime attributes are associated only with the part of the PK on which they are fully functionally dependent.

Third Normal Form:

BCNF requires that all nontrivial dependencies be of the form α -> ß, where α is a superkey. 3NF relaxes this constraint slightly by allowing nontrivial functional dependencies whose left side is not a superkey.

A relation schema R is in third normal form (3NF) with respect to a set F of functional dependencies if, for all functional dependencies in F+ of the form α -> ß.

• α -> ß is a trivial functional dependency.
• α is a superkey of R.
• Each attribute A in ß – α is contained in a candidate key for R.