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

1)       DEFINITIONS:
A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources.
As its name implies, the linear programming model consists of linear objectives and linear constraints, which means that the variables in a model have a proportionate relationship.
Ø LP is an in which the objective is a line optimal decision-making tool ear function and the constraints on the decision problem are linear equalities and inequalities. It is a very popular decision support tool: in a survey of Fortune 500 firms, 85% of the responding firms said that they had used LP.

2)         Objective:

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

3)         Uses

·       Linear programming is a widely used mathematical modeling technique to determine the optimum allocation of scarce resources among competing demands. Resource typically include raw materials, manpower, machinery, time, money and space.
·       The technique is very powerful and found especially useful because of its application to many different types of real business problems in areas like finance, production, sales and distribution, personnel, marketing and many more areas of management.

4)     Importance of Linear Programming

Many real-world problems can be approximated by linear models.
There are well-known successful applications in:
a.  Manufacturing
b.  Marketing
c.   Finance (investment)
d.  Advertising
e.  Agriculture

5)         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

Gauss-Jordan elimination is a variation of Gaussian elimination. The difference is that the elementary row operations in a Gauss-Jordan elimination continue until we reach the reduced echelon form of the matrix.
This means a few extra row operations, but easier calculations in the final step since back substitution is now longer.

III.                    Inverse Method.

Problem

                A Borjan shoes store produces three type of article (A, B, & C). In these articles, each article has three quality of product, 1) Standard Quality, 2) Normal Quality, & 3) Low Quality. The products of Article A Respective to the above quality are 1, 4, & 2 costs Rs.99. And products of Article B Respective to the above quality are 1, 2, & 4 costs Rs.111. And products of Article C Respective to the above quality are 2, 4, & 1 costs Rs.96.
          Find the costs of each quality by using the Gauss-Jordan elimination method of Linear System (Linear Programming).

Solution

                Let A, B, C are three articles, having x1, x2, x3. And the values of these are A= 1, 4, 2 , B= 1, 2, 4 , C= 2, 4, 1. Their Costs are 99, 111, 96 respectively.
Can be written as:
                x1+ 4x2+ 2x3= 99
                x1+ 2x2+ 4x3= 111
                2x1+ 4x2+ 1x3= 96
We can write the problem in Matrix form:    
             1     4     2                               x1                                                           99
A=         1     2     4                   X=       x2                                           B=         111
  2     4     1                               x3                                                           96


           1     4     2    99
Ab=    1     2     4   111
           2     4     1    96 


Performing Rows operation to solve:


            1   4   2   99
R=       1   2   4   111                        by R3 – 2R2
            0   0   -7   -126

            1   4   2   99
R=       0   -2   2   12             by R2-R1
            0   0   -7   -126


            1   4   2   99
R=       0   1   -1   -6              by R2/-2
            0   0   -7   -126


            1   4   2   99
R=       0   1   -1   -6              by R3/-7
            0   0   1   18


            1   4   2   99
R=       0   1   0   12               by R2+R3
            0   0   1   18


            1   4   0   63
R=       0   1   0   12               by R1-2R3
            0   0   1   18


            1   0   0   15
R=       0   1   0   12               by R1-4R2
            0   0   1   18

Ans:
                   x1= 15
                   x2= 12
                   x3=18

Rank:

          Number of non-zero rows are called Rank.
So, rank of above solution is 3.
          R= 3

MATLAB Solution

>> Ab= [1 4 2 99;1 2 4 111;2 4 1 96]
Ab =
     1     4     2    99
     1     2     4   111
     2     4     1    96

>> w=rref(Ab)
w =
     1     0     0    15
     0     1     0    12
     0     0     1    18

Ans:            x1= 15         x2= 12        x3=18
>> z=rank(w)
z =
     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= 15
                   x2= 12
                   x3=18                           With Rank of 3

References

·       Vassilis Kostoglou

·       LINEAR PROGRAMMING: FOUNDATIONS AND EXTENSIONS Second Edition
(Robert J. Vanderbei)
·       Intro to OR (F.Hillier & J. Lieberman)
·       Linear Systems and Gaussian Elimination
(Eivind Eriksen)