DBMS(Data Base Management System) Objective types Question

DBMS(Data Base Management System) 

.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. (B) Non- Procedural language.
(C) Data definition language. (D) High level language.
Ans: B
Q.3 The view of total database content is
(A) Conceptual view. (B) Internal view.
(C) External view. (D) Physical 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.
(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)
Q.6 ‘AS’ clause is used in SQL for
(A) Selection operation. (B) Rename operation.
(C) Join 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:c
Q.10 The database schema is written in
(A) HLL (B) DML
(C) DDL (D) DCL
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.
Ans: D
Q.14 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
(A) ¬P1 Ú P2 (B) P1 Ú P2
(C) P1 Ù P2 (D) P1 Ù¬P2
Ans: A In tuple relational calculus P1 P2 is equivalent to ¬P1 Ú P2.
(The logical implication expression A B, meaning if A then B,is equivalent to ¬A Ú 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
Q.21 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. (B) print files on paper.
(C) data entry. (D) delete files.
Ans: B
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
(A) is no need for primary key data.
(B) is much more data independence than some other database models.
(C) are explicit relationships among records.
(D) 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. (B) the physical view.
(C) the external 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
Q.34 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
(A) Alter. (B) Update.
(C) Create. (D) select.
Ans: A

Operating system | Types of OS | Functions and Services of OS

Operating System

An operating system is a program that acts as an interface between the user of a computer and the computer hardware. The purpose of operating system is to provide an environment in which a user can execute programs in a convenient and efficient manner.

Operating system is a first program that gets loaded into the memory through a process called booting.

Its primary objective is to improve the performance and efficiency of a computer system and make it user friendly.

Its secondary objective is to manage the resources of computer. 

Functions of operating system

• Process Management 
• Main Memory Management
• Secondary storage Management
• File Management 
• I\O Management
• Protection and Security
• Networking
• Command Interpretation

Operating System Services

• Program Execution
• I|O operations
• File system Manipulation
• Communication
• Error Direction
• Resources Allocation
• Accounting
• Protection

Classification of operating system

• Single user operating system
• Multi-user operating system
• Batch operating system  
• Multiprogramming operating system 
• Multitasking operating system
• Multiprocessing operating system
• Timesharing OS
• Real time OS
• Distribued system

Single user operating system

In Single user operating system , a user can access the computer at a particular time. The computer , that are based on this operating system,  have only single processor and execute only a single program at all the time.

Single user operating system is of two types

• ·         Single user, single –tasking operating system
• ·         Single user, multi- tasking operating system

Single user, single –tasking operating system

This operating system allows a single user to execute one program at a particular time

Example MS- DOS and Palm OS

Single user, multi- tasking operating system

 This operating system allows a single user to execute multiple programs at a same time.

Multi-user operating system

In this multiple number of users can access different resources of a computer at a same time. The access is provided using a network that consist of various personal computers attached to a mainframe computer system.

Example UNIX, Window 2000

Batch operating system  

it is one of oldest method of running the programs. The users did not interact directly with computer system. The users prepare their job that consisted of the programs, data and some control information . This job was usually in form of punched cards. Users submit their job to a computer operator, when a batch of programs have been collected, the operator loads this batch of programs into the computer at one time where they  are executed one after the after

The batch operating system was called a monitor that resides in the main memory is known as resident monitor

Multiprogramming operating system 

It is used for multi – user systems i.e the systems that are used by many users at the same time.

It refers to the concurrent execution of multiple jobs that may be of same user or different users.

Requirements of multiprogramming os

• ·         Large Memory
• ·         Memory Protection
• ·         Job status Preservation
• ·         Proper job Mix
• ·         CPU Scheduling

Multitasking operating system

It is used for single user systems I,e  the systems that are used by only one user at a time. It refers to the concurrent execution of multiple jobs of a same user.

Types of Multitasking operating system

• ·         Cooperative Multitasking
• ·         Pre-emptive Multitasking

Multiprocessing operating system

This concept is implemented in systems with multiple processors. Multiprocessing refers to the simultaneous execution of two or more processes by system with multiple processor.

In multiprocessing , several programs segments of one or more programs are executed simultaneously

Multiprocessing systems are cateforized into two different types

• ·         Symmetric multiprocessing
• ·         Asymmetric multiprocessing

·        Timesharing OS

The time sharing system executes one job up to certain period of time and then switches to another job and so on. Concepts like job scheduling, memory swapping are used for handling timesharing systems.It seems to execute many jobs at the same time. It is impossible to predict response in advance in timesharing system

Real time OS

This OS has well defined fixed time constraints, process must be done within the defined constraints or system will fail. It execute one jon dedicatedly. Real time OS give users the ability to prioritize tasks so that the most critical task can always take control of the process when needed

Real time OS are categorized into two different types

• ·         Soft real-time OS
• ·         Hard real-time OS

·          Distributed system

A distributed system is a collection of processors located in geographical dispered physical location. In this the workload is distributed or divided between two or more computers that are linked together by a communication network. That is the different processors communicate using communication links, such as telephone lines and buses.

(OR) Operation Research | Assignment Problem | MAX case[HAM method] | MIN case | Unbalanced assignment problem

Assignment problem

HAM (HUNGARIAN METHOD ) OR MIN CASE

The Hungarian mathematician D. Konig developed simpler & more efficient method of solving assignment problem which is known as Hungarian techniques or method.

Steps to follow

1) Prepare cost matrix, if it is not square matrix it  means problem is unbalanced, so to convert it into balanced matrix add a dummy row with zero.

2) Starting with this smallest the first row, locate the smallest cost element in each row of the cost table. Now subtract this smallest element from each element in that row. As a result there shall be at least one zero in each row of this new table.

3) In the reduced cost table obtained in step 2, consider each column and locate the smallest element in it .subtract the smallest element in each column from every element of that column.

4) Make the assignment for the reduced matrix obtained from step no 2 & 3 in the  following way.

·        Examine the rows successively, until a row with exactly one zero is found make an assignment to this single zero by putting square around it and cross out all other zeros appearing in the corresponding column as they will not be used to male any other assignment in that column. Proceed in this manner until all rows have been examined.

·        Examine the columns successively, until a column with exactly one zero is found make an assignment to this single zero by putting square around it and cross out all other zeros appearing in the corresponding column as they will not be used to male any other assignment in that row. proceed in this manner until all columns have been examined.

·        Repeat step 4(i) and 4(ii) until all zeros in the rows and columns are either marked or crossed out. If the no. of assignment made are equal to no. of rows / columns. Then it is an optimum solution and there is exactly one assignment in each row and in each column.

5) Draw the minimum number of horizontal and vertical lines necessary to cover all the zeros in the reduced matrix obtained from step 4 in the following way

· Mark all rows that do not have any assignment

· Mark all columns that have zero in marked rows [step 5(a)]

·  Market all rows that have assignment in marked columns. [step 5(b)]

·   Repeat step 5(a) and 5(b) until no more rows and columns can be marked

· Draw straight lines through all unmarked rows and marked columns

6) If the number of lines drawn are equal to 'n' that is equal to number of rows or columns then it is an optimum solution, otherwise go to step 7

7) Select the smallest element among all the uncovered elements. Subtract this smallest element from all the uncovered elements and add it to the element which lies at the intersection of two lines. Then we obtain another reduced matrix for fresh assignment

8) Go to step 4 an repeat the procedure until the number of assignment become equal to the number of rows and columns. In such a case we shall observe that every row and column has an assignment. Then the current solution is the optimum solution

MAX CASE

In some cases, the element of assignment problem may represent revenue or profit instead of cost, so that the objective will be to maximize total revenue or profit. The problem of MAX can also be solved by using HAM method that we can use for solving the MIN problem.in this the MAX problem will be converted into MIN problem by selecting the largest element among all the profit matrix and then subtract it from all the elements in the matrix including itself. The apply HAM(Discussed above)

Profit = sale revenue – production cost

UNBALANCED ASSIGNMENT PROBLEM

If the cost matrix of assignment problem is not a square matrix i.e. number of rows not equal to number of columns , then the assignment problem is called Unbalanced assignment problem. In such case add a dummy row or column depending on need with zero cost to convert it in to a square matrix then apply HAM

Android | Intents

INTENTS

Intents are used to start and stop activities and services, or to broadcast message, or to request an action to be performed on a particular piece of data.

Intent is a parameter used in start activity() method. When the start activity() method called with intent parameter, the android system match the intent action with activity on android system. That activity is then launch.

An intent is represented android. Content . intent class. It is in the content package because intents can be used to access content provides.

“an intent is an abstract description of an operation to be performed” 

an intent object is basically a passive data object. There are 3 types od intent objects that can be used inside the android OS to communication with activities, services and broadcast receiver. In fact, there is one intent type for each intent object based message can contain up to 7 kinds of information parts

1. Name: The name of the class the intent are targeting bb using package name and class name
2. Action:  A predefined type of action that is to be performed
3. Data: The actual data such as phone number to call
4. Category: Android has predefined intents that are the parts of OS that are divided into various types of categories for easy access.
5. Type: It specify type of data using a MIME format
6. Flag: This allow on\off flags to be sent with the intents
7. Extras: It all extra information included in intent

Passing Information with Intents

Intents can be used to passed data between activities. You can use an intent to include addition of data called extras, within the intent. For this put Extra() method will be used

 Example : 

Intentin = newIntent (getApplicationContext (), HelpActivity.class);

In.putExtra(com.androidbook.chippy.LEVEL, 23);

startActivity(in)

when help activity class launches, get Intent() method used to retrieve the intent. Then extra information can be extracts using appropriate method.

Intent nn = getIntent();

Int helpLevel= nn.getIntExtra(com.androidbook.chippyLEVEL,23)

" parent activity that launch sub activity use startActivityForResult() the result is passed as a parameter to the onActivityResult()"

Method with an Intent parameter.

• Use Intent to launch other application
• Launching the build in web browser and supply a URL address
• Launching the web browser and supply a search string
• Launching the build in Dialer Application
• Launching the build in  Application
• Launching the ringtone picker

There are two broadcast categories of intents

Explicit intents use component portion of intent object and generally use these when working with application you have developed

Implicit intents they don’t specify components within intent object

OR(Operation Research) | Transportation problem | its Methds

Transportation problem

The Transportation problems are one of the types of the L.P.P in which objective is to transport various quantities of a single homogeneous commodity, to different destination  in such a way that the total transportation cost is minimum.

Terminology used in Transportation problem

Feasible solution  feasible solution is a set of non negative values (x>=0) which satisfies the row and column sum conditions is called a feasible solution

Basic feasible solution If the total number of positive allocations is exactly equal to m+n-1 i.e. one less than the sum of number of rows(m) and columns (n) that solution is called a basic feasible solution.

Optimal solution A feasible solution is said to be optimal of it minimizes the total transportation problem

Balanced Transportation problem A Transportation problem. In which the total supply from all the sources equal the total demand in all the destinations.

Unbalanced Transportation problem A Transportation problem In which the total supply from all the sources are not equal the total demand in all the destinations.

Degenerate Basic feasible solution A basic feasible solution that contains less than (m+n-1) non negative allocations.

Non degenerate Basic feasible solution If the total number of positive allocation is exactly equal to ( m+n-1) and these allocation are in independent positions that BFS solution is called non degenerate BFS.

Methods for initial basic feasible solution (IBFS)

North West Corner Method ( NWCM)

Lowest Cost Entry Method (LCM )

Vogel’s Approximation Method  ( VAM)

North West Corner Method ( NWCM)

The idea is to obtain IBFS i.e.  a set of allocations that satisfies the totals of row and column. The methods simply consist of making allocations to each row in turn. Apportioning as much as possible to its first cell and proceeding in this manner to its following cells until the row is exhausted. The algorithm involved under NWCM consists of the following steps

• The method starts at the north west corner cell of the table i.e. upper left corner of the requirement table or top most left corner of the table i.e. cell (1, 1) and allocate the maximum possible amount and adjust the associated amount of supply and demand by subtracting the allocated amount.
• Move to the cell (1’2) towards right hand side, if there is still any quantity available left. Otherwise move towards down to cell ( 2,1 )  and allocate the maximum possible amount. Move to the cell (2,2), if supply equals demand at cell (1,1)
• Repeat steps 1 and 2 moving down towards the lower right corner of the transportation cost table until the rim requirements are satisfied.

Lowest Cost Entry Method (LCM )

This method give better starting solution them NWCM  as it reduces the computation cost and amount of time necessary to get optimal solution.

Steps to be followed

• Select the cell with the lowest transportation cost  in the whole matrix arbitrarily.
• Allocate an many units as possible to the cell determined instep 1 and eliminate that row/ column in which either supply is exhausted or demand is satisfied.
• Re- select the cell with the lowest transportation cost from the remaining cost matrix or reduced table and repeated the step 2 , till the rim requirement is satisfied.

Vogel’s Approximation Method  ( VAM)

VAM is also known as UNIT COS TPENALTY METHOD  . In this method, transportation problem can be solved in lesser time and solution to it may be considered as the appropriate solution as compared to the methods i.e. it is an algorithm used to find out relatively efficient IFS by considering the ‘PENALTY COST’  of not using the cheapest available route. This method is better than earlier, methods and is usually used to get a better starting point to the solution of the problem

Steps to be followed

• From the transportation cost table we determine the penalty for each row and each column. These penalties are calculated for each row(column) by subtracting the least cost element in that row (column) from the second lowest cost element in the same row (column).
• After this we identify the row or column with the largest penalty among all the rows and columns. In case the penalties corresponding to two or more rows or columns are equal then we use any arbitrary tie breaking choice.
• In the largest penalty row(column) identify the least cost cell if it happens to be in ith row and jth column then we allocate maximum possible number of units keeping the row and column constraints in mind. Let the number of units allocated be X_ij then X_ij is min{s_i, d_j} in the cell (i,j).Remove the row (or column) from further consideration for making allocation for which supply (or demand) is exhausted in making allocations at a stage.
• Again compute row and column penalties for the reduced transportation table, go to step 2 . Repeat the procedure until all the requirements are satisfied

Optimality test

After finding IFS by any initial basic feasible solution method , the next important step is to see, can we decrease the total cost further by making changes in the occupied cells ? This is known as testing optimality

Two Methods for testing optimality

Stepping stone Method

Stepping stone method It is an interactive technique from moving an initial feasible solution to an optimal feasible solution.

Steps to be followed

• Write the transportation table incorporating the original supply, demand and costs of transportation
• Determine an initial basic feasible solution by using any one of the methods discussed earlier.(prefer VAM ,as it provide either optimum or the nearest the optimum solution)
• Make sure that the number of allocations are  m+n-1 allocations should be in independent positions.
• Evaluate cost change of shipping goods via each unoccupied cell, which can be done as follows:   

  I. Select an unoccupied cell to be evaluated    
  II.  Starting from this unoccupied cell, trace a closed loop using the most direct route through at least three allocated cells used in the solution and then come back to the starting unoccupied cell. In the process of moving from one occupied cell to another,(i) move only vertically or horizontally but never diagonally, (ii) both empty or occupied cells may be skipped over, (iii) an even number of at least four cells must participate in a closed path. The calls at the turning points are known as stepping stones.   

 III. After tracing the closed path, place (+) and (-) sign alternatively in the cells on each turn of the loop, beginning with a plus (+) sign in the empty cell.   
 IV.   Find the ‘net effect on the cost’ along the closed loop. For this purpose add the unit cost figures found in each cell containing a plus sign and then subtracting the unit costs in each cell having the minus sign.    
  V. Calculate the net change in cost for all the empty cells.

• Check the sign of each of the net cost changes. If all the net cost changes are greater than or equal to zero(>=0) ; it indicate that solution is optimum, if at least one net cost change is negative, it is possible to improve the current solution bt decreasing total transportation cost
• Choose the empty  cell having the most negative net cost change and determine the  maximum quantity that can be assigned to a cell marked with minus sign on the closed path.add this quantity to the empty cell and to all other cells marked with a (+) sign. Subtract this quantity from the cells with a minus(-) sign
• Go back to step 3 and repeat the procedure until an optimum solution is obtained

MODI (Modified Distribution Method)

This method allow us to compute improvement indices quickly for each unbalalnced square without drawing all of the closed paths. modi method provide a new means of finding the unused route with the largest negative improvement index. Once the largest index is identified we are required to trace only one closed path, just as with the stepping stone approach , this path helps to determine the maximum number of units that can be shipped via the best unused route.

1.From the given data construct a transportation table with the given cost of transportation and rim conditions.

2. Determine in initial basic feasible solution using a suitable method(i.e. NWCM, LCEM or VAM)

3.For the current basic feasible solution with m+n-1 occupied cells, calculate index numbers R_i =(i=1,2….m) and K_j=(j=1,2….n) for rows and columns respectively.

For calculating values of R_i  and K_{j ,} the following relationship for occupied cells is used,

C_ij = R_{i   +}  K_j             for all I,j

4.For occupied cells, the opportunity cost by using the formula

D_ij = C_ij - R_{i   +}  K_j             for all I,j

5. Now the opportunity cost of an unoccupied cell is determined by using the formula

Opportunity cost = actual cost – implied cost

D_ij = C_ij – (R_{i   +}  K_j )

6.Examine unoccupied cells evaluation for D_ij

• If D_ij >0 then the cost of transportation will increase, i.e. and optimal solution has been arrived at.
• If D_ij =0 then the cost of transportation will remain unchanged. But there exists an alternative solution.
• If D_ij <0 then an improved solution can be obtained by introducing cell (I,j) in the basis and go to step 7

7.Select an unoccupied cell with largest negative opportunity cost among all   unoccupied cells.

8. Construct a closed path for the unoccupied cell determined in step 7 and assign plus(+) and minus (-) sign alternatively beginning  with plus sign for the selected unoccupied cell in clockwise or other direction

9. Assign  as many units as possible to the unoccupied cell satisfying rim conditions.the smallest allocation in a cell with negative sign on the closed pathnindicated the number of units that can be transportation to the unoccupied cells. This quantity is added to all the occupied cells on the pathnmarked with minus sign

10.Go to step 4 and repeat procedure until all dij >+0 i.e. an optimal solution is reached. Calculate the association total  transportation cost

   

OR(Operations Research) | Linear programming | Assumptions | Applications

  Linear Programming

The word Linear Programming consist of two words

Linear This word is used to show the relationship between decision variables which are directly proportional. For example if we increase the production of product, it will proportionate increase the profit, this is called a linear relationship.

Programming It implies the planning of activities in such a way that the activities with available resources yield optimal results .

This Linear Programming states the  planning of decision variables which are directly proportional to achieve desired optimal results.

Or

Linear Programming involves the planning of activities to obtain an optimum result.

“Linear Programming is the analysis of problems in which a linear function  of a number of variables is to be maximized (or minimized) when those variables are subject to a number of restraints in the form of linear inequalities”

Assumptions of Linear Programming

Linearity or proportionality A basic assumption of Linear Programming is that there exists proportionality in the objective function and the constraints. It implies that if a product gives a profit of Rs 25 the profit earned from the sale of 10 such products will be Rs 250.

Finite Choice Another assumption of a Linear Programming is that a limited number of choices are available to decision maker and the decision variable do not have negative values. This assumption is a realistic one, as it is not possible to produce or use negative quantities

Additivity It means that if m1 minutes are required to produce a product p1 on machine A and m2 minutes are required to produce a product p2, the total time required to make product p1 and p2 on machine A is m1+m2 minutes. This however, may not happen because of the change over time from product p1 to p2. Similarly the total profit is determined y the sum of profit is calculated by each of the products separately

Certainty A Linear Programming model also assumes that the various parameters, such as coefficients of the constraints, objective function coefficients and resources values are known certainty and they do not change with the passage of time. So, availability of materials, labour etc. the cost or profit per unit of the product, market demand of the product are assumed to be known certainly

Continuity One more assumption Linear programming is that the decision variables are continuous. It implies, combinations output with fractional values, in the context of production problems are possible.

Applications

Product Mix Linear programming helps in determining the quantity of different products to be manufactured knowing the marginal contribution of each product and amount of available resources used by each product

Transportation  problem Transportation n problems are faced by business organization which have number of availability centres (plants, warehouse etc.) with gives capacities which feed number of requirements cemtres (warehouse, markers etc.) with given requirements.

Blending problem This type of problem is faced by the chemical, food and petroleum industry etc. while deciding production of a product which can be made from a variety of available raw materials of different composition and prices.

Media selection LP has also been used in advertising firld as a decision aid in selecting the effective media. Media helps the marketing managers in allocation a fixed budget across various advertising media like newspapers, magazines, radio and television etc.

Diet problem it involves determination of combination of different nutrients such as proteins, vitamins, carbohydrates etc. for different foods to satisfy the minimum daily nutritional requirements at the minimum cost and minimizes the cost of raising live-stock.

Travelling salesman’s problem The problem of finding the shortest route for a salesman starting from a given city, visiting each of the specified cities, and returning to the original point of departure can be handled by the assignment technique of  linear programming.

Capital investment This type of problem arises because of the different ways in which a fixed amount of capital can be allocated to a number of activities. The total return depends upon the manner in which the allocation is made. This , the objective may  be to find that allocation which maximizes the total return

Fuzzy set | objective type question

Fuzzy Set

    1.       The height h(A) of a fuzzy set A is defined as h(A) =sup A(X) where X belongs to A. Then set A called normal when

• H(A)=0
• H(A)=1
• H(A)=0.0
• H(A)=2

    2.       Fuzzy logic is a form of

• 2-valued logic
• Crisp set logic
• Many valued logic
• Binary set logic

    3.       What is fuzzy approximate Theorem (FAT)

• A fuzzy system can model any continuous system
• The conversion of fuzzy logic to probability
• A continuous system can model any fuzzy system
• Fuzzy paths covering a series of fuzzy rules

    4.       What is the way of represent of uncertainty

• Fuzzy  logic
• Probability
•  Entropy
• All of the above

    5.       How fuzzy logic different from conventional control methods

• IF & THEN approach
• FOR approach
• WHILE approach
• DO approach

Data Mart|Maintenance of data mart

Maintenance of data mart

Data Mart provide complete information about a particular subject to the users. The users can find solution to all their queries related to that particular subject. Hence, the proper & adequate, sufficient maintenance of a data mart id necessary.

Periodic maintenance of a data mart means loading, refreshing that data in it. Refreshing may be done on daily, weekly, monthly, quarterly or yearly basis. Following are the points that can help in proper maintenance of data mart.

 1)Upgrading of data mart from data sources:  The data mart should be upgraded at regular intervals from operational data sources. So, that all the latest data related to the particular object can be retrieved easily & the users can efficiently solve their problems. Hence the data mart provides the adequate knowledge to the users & is maintained properly. The data can not only be taken from operational data source but it can also be taken from data warehouse.

  2)Backup Management: The backups are taken for only one reason:

Protection against data loses: Backups are useless if they can’t be used when they are required backup package have their own tools for managing backup & recovery.

Features:

Scheduling

Backup data tracking 

Data  Base awareness

 3) Metadata Management:  Mata data is data about data. It is like a card index describing how information is structured within the data warehoused. Metadata is used for a variety of purpose. The structure of meta data will different between each process because the focus is different. Hence the meta data should ne managed properly & efficiently so that the data mart produce accurate result. 

  4)  Hardware and Software requirements: Data mart  require additional hardware & software that are used to run different applications. The data mart database will require additional processing power & storage capacity to handle the user queries. Data mart should not be used as an alternative to aggregation. This will increase the costs hence the hardware & software should be maintained in such a way that overheads can be easily avoided.

  5) Network access: The process of loading data into each data mart will be affected by the available physical connections between both items of hardware. In many cases each data  mart will be in a different geographical location from the data warehouse.

Hence, such issue should be taken care of & should be handled in an efficient manner So, the data mart are not defected

 6)Time window constraints: The determination of how many data mart are feasible will vary, depending on a number of factors namely

• Time window available
• Network capacity
• Volume of data being transferred
• Mechanism being used to insert

We should calculate the approximate time necessary to complete the data mart load in order to determine . if the architecture is feasible

In time window constraints we have to:

1)calculate the data volume being sent into each data mart

2)Calculate elapsed time to transfer access the LAN & WAN into each mart

3)Calculate elapsed time to load the data into each mart

Data Warehouse| Data Mart & its Types

Data mart(DM)

Data mart usually organized as ID model as star schema made of a fact table & multiple dimensions table.

Definitions :

“smaller local data warehouse called data mart”

“ it is a subset of data warehouse& it support a particular region , business unit, or business function.”

“ it is a collection of subject area organized for decision support based on  the need of a given department”

Data mart is a simple form of data warehouse that :

·         Focus on simple subject

·         Controlled by single department

·         Smaller & less complexity as compare to data warehouse

Types of Data Mart

    1.       Dependent DM

    2.       Independent DM

    3.       Hybrid DM

    4.       Temporal DM

    5.       Non – Temporal  DM

   1)      Dependent data mart : These Data Mart draw data from a central data warehouse that has already been created . In dependent data mart the process of extraction, transformation is simplified because summarized clean data has been loaded into the data warehouse

Dependent data mart are usually build to achieve improved performance and availability of lower communication costs.  

2)      Independent data mart : These data mart draw data directly from operation or external sources of data. In independent data mart the ETL process has to go through number of stages i.e. the process is not simplified. The data is collected from different external sources. Independent data mar t are basically created to have a solution for a particular problem in a short period of time.

3)      Hybrid data mart : such type of data mart contain the features of dependent data mart and independent data mart . in such data mart’s the input is come from both data warehouse as well as ODS(operational data source) . this type of data mart are specially used to handle the integration of data from multiple source. The ETL(external source ) process in hybrid data marat also has to go through a number of steps because datmart is collected contain features of both data warehouse as well as external source

4) Temporal data mart : These are the data mart’s where: 

1. all the history is displayed i.e  each dimension reflects a number of facts and historical hierarchies

2. Dimensions reflects the hierarchies at the time when event took place

3. Are used for  depth trend analysis

4. Too slow for users who want to view the current position

5) Non Temporal data mart: These are the data mart’s where:

1.All the history is not retained

2.Dimensions reflects the current hierarchies

3.Are used for current analysis 

4.Helps in reducing the size of data & simplified the data in dimensions