Abstract: - We have talked about the Linear Programming in Mathematics. We
characterize the Linear programming in various manner. Its Objectives, uses,
significance and the a portion of the Methods to understand the direct
framework. We have taken a case of direct framework from this present reality
and explain it by Gauss' Jordan disposal technique (Reduced echelon form).
Keywords: - Gaussian elimination, Gauss-Jordan
elimination, Variables, Matrix, A, B, C, linear system, Borjan, MATLAB etc.
Translating
Real World Problem of Linear Programming
Abstract: - We have talked about the Linear Programming in
Mathematics. We characterize the Linear programming in various manner. Its
Objectives, uses, significance and the a portion of the Methods to understand
the direct framework. We have taken a case of direct framework from this
present reality and explain it by Gauss' Jordan disposal technique .
Keywords: - Gaussian elimination,
Gauss-Jordan elimination, Variables, Matrix, A, B, C,
linear system, Borjan, MATLAB etc.
INTRODUCTION
1)
DEFINITIONS:
Linear programming is a mathematical method
that is used to determine the best possible outcome or solution from a given
set of parameters or list of requirements, which are represented in the form of
linear relationships. It is most often used in computer modeling or simulation
in order to find the best solution in allocating finite resources such as
money, energy, manpower, machine resources, time, space and many other
variables. In most cases, the "best outcome" needed from linear
programming is maximum profit or lowest cost.
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 used to
obtain optimal solutions for operations research.
·
Using linear programming allows
researchers to find the best, most economical solution to a problem within all
of its limitations, or constraints.
·
Many fields use linear programming techniques
to make their processes more efficient.
4)
Importance of Linear Programming
Many
real-world problems can be approximated by linear models. Importance of linear
programming. With linear programming we can easily solve
business problem. It is very benefited for increase the profit or decrease the
cost of business. Linear programming solve problem under
different limitions and conditions , so It is easy for manager to work under
limitations and conditions 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
In linear algebra, Gaussian elimination is an algorithm for
solving systems of linear equations. ... Using row operations to convert a
matrix into reduced row echelon form is sometimes called Gauss–Jordan
elimination.
III.
Inverse Method.
Problem
A Company produces three type of
bikes (A, B, & C). In these bikes, each article has three quality of
product, 1) Standard Quality, 2) Normal Quality, & 3) Low Quality. The products
of bikes A Respective to the above quality are 2, 4, & 1 costs Rs.15. And
products of bikes B Respective to the above quality are 2, 3, & 2 costs
Rs.21. And products of bike C Respective to the above quality are 3, 2, & 2
costs Rs.19.
Find the costs of each quality by
using the Gauss-Jordan elimination method of Linear System.
Solution
Let A, B, C are three articles,
having x1, x2, x3. And the values of these are A= 2,4,1, B= 2,3,2,
C= 3,2,2. Their Costs are 15, 21, 19 respectively.
Can be
written as:
2X1+ 4X2+ 1x3=
15
2x1+ 3x2+ 2x3=
21
3x1+ 2x2+ 2x3=
19
We can write
the problem in Matrix form:
A= 2 3 2 X= x2 B= 21
3 2
2 x3 19
Ab= 2
3 2 21
3
2 2 19
Performing Rows operation to solve:
R= 2
3 2 21 by R1 – 2R3
3
2 2 19
R= 0 3
4 29 by R2 – 2R1
0
2 5 31 by R3 – 3R1
R= 0
1 -1 -2 by R3 – R2
0
2 5 31
R= 0
1 -1 -2 by R3 – 2R2
0
0 7 35
R= 0
1 -1 -2 by R3 /7
0
0 1 5
R= 0
1 0 3 by R2 + R3
0
0 1 5
R= 0
1 0 3
0
0 1 5
X= 1,
Y= 3, Z= 5
MATLAB Solution
>> T=[4,2,1,15; 2,3,2,21; 3,2,2,19]
T =
4 2
1 15
2 3
2 21
3 2
2 19
>> W=rref(T)
W =
1 0
0 1
0 1
0 3
0 0
1 5
>> X=rank(W)
X = 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=
1
x2=
3
x3=5 With Rank of 3
ACKNOWLEDGEMENT:
Specially thanks to my respected Professor
Muhammad
Shahbaz Malik from
Dunyapur.
Also,
thanks to my fellows, they helped to make this Assignment.
References
· LINEAR
PROGRAMMING: FOUNDATIONS AND EXTENSIONS Second Edition
(Robert J.
Vanderbei)
· Intro
to OR (F.Hillier & J. Lieberman)
· Linear
Systems and Gaussian Elimination
(Eivind Eriksen)
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