Abstract: Today I am going to present the topic of LPP means (Linear programing problems) web also read it LPP. Many definition of linear programing problems are available in modern days but we define very easy and conceptual definition. In which different properties and steps like Its Objectives, uses, importance and the some of the Methods used to solve the linear programing problems. In which we solve some Examples Linear programing problems according to Gauss’s Jordan elimination method (Reduced echelon form).
Keywords: Gaussian elimination, Gauss-Jordan elimination, Variables, Matrix, A, B, C, linear system, Borjan, MATLAB etc.

Defination of Linear programing Problems
Linear Programming. Linear programming is often used in business to find maximum profit or minimum cost. The first step in solving linear programming problems is to set up a function that represents cost, profit, or some other quantity to be maximized or minimized subject to the constraints of the problem.

Translating Real World Problem of Linear Programming

Abstract:-We have discussed about the Linear Programming in Mathematics. We define the Linear programming in different way. Its Objectives, uses, importance and the some of the Methods to solve the linear system. We have taken an example of linear system from the real world and solve it by Gauss’s Jordan elimination method (Reduced echelon form).
Keywords:- Gaussian elimination, Gauss-Jordan elimination, Variables, Matrix, A, B, C, linear system, Borjan, MATLAB etc


INTRODUCTION

 Linear algebra is the branch of mathematics deals with algebraic equations, spaces (vector and scalar), linear mappings between such spaces etc. Combined with the theory of calculus, linear algebra ensures to have methodologies to compute the solutions of system of equations (algebraic and differential).

Application of Linear Algebra

1)       DEFINITIONS:
A branch of mathematics that is concerned with mathematical structures closed under the operations of addition and scalar multiplication and that includes the theory of systems of linear equations, matrices, determinants, vector spaces, and linear transformations.

2)          Objective:-

In linear programming, the objective is always to maximize or to minimize some linear function of these decision variables.

  3)    Uses

1.      The most widely used application of linear algebra is definitely optimization, and the most widely used kind of optimization is linear programming. You can optimize budgets, your diet, and your route to work using linear programming, and this only scratches the surface of the applications. Here’s a series (still in progress) on the mathematics behind linear programming. The primary technique for solving them, called the simplex algorithm, is essentially a beefed up Gaussian elimination.


2.      The field of signal analysis gives one massively useful tools for encoding, analyzing, and manipulating “signals” that can be audio, images, video, or things like x-rays and light refracting through a crystal. The simplest way to understand the Fourier transform is as a linear map that performs a change of basis. Fourier analysis has even been used to make art. Here’s the first post in a series deriving Fourier analysis from scratch, although much of it can be abbreviated, skipped, or skimmed if you have a strong understanding of linear algebra. A discrete cousin of Fourier analysis has been part of many theoretical techniques in computer science as well.

3.      Quantum computing
4.      Greedy algorithms
5.      Community detection
6.      Facial recognition
7.      Graphics
8.      Signal analysis
9.      Error correcting codes
10.  Linear programming
11.  Ranking in search engines etc,



3)         Importance of Linear Programming

Many real-world problems can be approximated by linear models. These are the most Successful Applictaions
1.  Manufacturing
2.  Marketing
3.  Finance (investment)
4.  Advertising
5.  Agriculture

Methods To solve a linear system problem:

I.     Gaussian elimination

Gaussian elimination is an efficient method for solving any linear system using systematic elimination of variables.


II.Gauss-Jordan elimination

To perform row reduction on a matrix, one uses a sequence of elementary row operations to modify the matrix until the lower left-hand corner of the matrix is filled with zeros, as much as possible. There are three types of elementary row operations:
·         Swapping two rows,
·         Multiplying a row by a nonzero number,
·         Adding a multiple of one row to another row.
Using these operations, a matrix can always be transformed into an upper triangular matrix, and in fact one that is in row echelon form. Once all of the leading coefficients (the leftmost nonzero entry in each row) are 1, and every column containing a leading coefficient has zeros elsewhere, the matrix is said to be in reduced row echelon form. This final form is unique; in other words, it is independent of the sequence of row operations used. For example, in the following sequence of row operations (where multiple elementary operations might be done at each step), the third and fourth matrices are the ones in row echelon form, and the final matrix is the unique reduced row echelon form.

Using row operations to convert a matrix into reduced row echelon form is sometimes called Gauss–Jordan elimination.



III.                    Inverse Method.


We can calculate the Inverse of a Matrix by:
·         Step 1: calculating the Matrix of Minors,
·         Step 2: then turn that into the Matrix of Cofactors,
·         Step 3: then the Adjugate, and
·         Step 4: multiply that by 1/Determinant.
But it is best explained by working through an example!

Step 1: Matrix of Minors

The first step is to create a "Matrix of Minors". This step has the most calculations.
For each element of the matrix:
·         ignore the values on the current row and column
·         calculate the determinant of the remaining values
Put those determinants into a matrix (the "Matrix of Minors")

Determinant

For a 2×2 matrix (2 rows and 2 columns) the determinant is easy: ad-bc
Think of a cross:
·         Blue means positive (+ad),
·         Red means negative (-bc)


(It gets harder for a 3×3 matrix, etc)

The Calculations

Here are the first two, and last two, calculations of the "Matrix of Minors" (notice how I ignore the values in the current row and columns, and calculate the determinant using the remaining values):

And here is the calculation for the whole matrix:

Step 2: Matrix of Cofactors

This is easy! Just apply a "checkerboard" of minuses to the "Matrix of Minors". In other words, we need to change the sign of alternate cells, like this:

Step 3: Adjugate (also called Adjoint)

Now "Transpose" all elements of the previous matrix... in other words swap their positions over the diagonal (the diagonal stays the same):

Step 4: Multiply by 1/Determinant

Now find the determinant of the original matrix. This isn't too hard, because we already calculated the determinants of the smaller parts when we did "Matrix of Minors".

In practice we can just multiply each of the top row elements by the cofactor for the same location:
Elements of top row: 3, 0, 2
Cofactors for top row: 2, −2, 2
Determinant = 3×2 + 0×(−2) + 2×2 = 10
(Just for fun: try this for any other row or column, they should also get 10.)
And now multiply the Adjugate by 1/Determinant:

We take an Example 

Problems
A Khadi clouth store produces three type of article (A, B, & C). In these articles, each article has three quality of branded product 1 small, product 2 medium, product 3 Large, The products of Article A Respective to the above quality are 3, 2, & 4 costs Rs.18. And products of Article B Respective to the above quality are 2, 3, & 1 costs Rs.17. And products of Article C Respective to the above quality are 1, 1, & 5 costs Rs.11. Find the costs of each quality by using the Gauss-Jordan elimination method of Linear programing.

 Solution

                Let A, B, C are three articles, having x1, x2, x3. And the values of these are A= 3, 2, 4,  B= 2, 3, 1,  C= 1, 1, 5. Their Costs are 18, 17, 11 respectively.
Can be written as:
                3x1+ 2x2+ 4x3= 18
                2x1+ 3x2+ x3= 17
                x1+ x2+ 5x3= 11
We can write the problem in Matrix form:    
          3     2     4                               x1                                                         18
A=     2     3     1                   X=       x2                                           B=         17
           1     1     5                               x3                                                                    11

           3     2     4    18
Ab=    2     3     1   17
            1    1     5    11

Performing Rows operation to solve:

           1    1     5    11
Ab=    2     3     1   17                  R13
           3     2     4    18

           1    1      5     11
Ab=    0     1    -9    -5                 by R2 – 2R1                                                                                 0     -1  -11   -15                   R3 – 3R1                                 

 

           1     1       5      11
Ab=    0     1     -9      -5                 R3 + R2   
           0     0    -20     -20


           1    1       5       11
Ab=    0     1     -9      -5                 R3 /-20   
           0     0      1        1


           1    1       5      11
Ab=    0     1     0        4                R2 + 9R3                                       
           0     0     1        1


           1     1       0       6
Ab=    0     1     0        4                R1 + 5R3                                       
           0     0     1        1


           1     0      0       2
Ab=    0     1     0        4                R1 + R2                                       
           0     0     1        1


           1     0       0      2
Ab=    0     1     0        4               
           0     0     1        1

X= 2 ,   Y=4,    Z=1

According To MATLAB Solution
>> T=[3,2,4,18; 2,3,1,17; 1,1,5,11]T =

 

     3     2     4    18

     2     3     1    17

     1     1     5    11

 

>> Z=rref(T)

 

Z =

 

     1     0     0     2

     0     1     0     4

     0     0     1     1

X=2,          Y=4,                    Z= 1

>> W=rank(Z)

 

W =3

    Result:

                The problem which is taken is solved b Gauss’s Jordan Elimination method. The Solution is correctly solved by reduced echelon form. We have the cost of each quality is below:
x1= 2,   x2= 4,     x3=1                           With Rank of 3