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ECONOMIC PROBLEMS INTO LINEAR PROGRAMMING PROBLEMS
AND SOLVING THE SIMPLEX METHOD
S.M.Kamoldinov
Tashkent State University of Economics
kamoldinovs03@gmail.com
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As is known, problems related to the theory and application of
quantitative methods and models are solved by bringing them to the problem of linear
programming. The condition of certainty is understood as a situation in which all
parameters and conditions of system control are certain, that is, there is no random
effect. In such problems, the method of linear optimization is used, which aims to
create an optimal production plan, determine the optimal volume of trade, purchase
or transportation, optimal financial planning, and similar goals. Planning is one of the
main functions of management.
Annotation. This article deals with bringing economic problems to the
problem of linear programming and solving them graphically. Also, at first, linear
programming will be touched upon, its solution methods and mathematical
interpretation of the given problem: a special emphasis will be placed on solving it
through a linear function. As an example, the issue of optimal production planning
for the “Olmos” furniture factory is given.
Key words: Multi-argument linear function, argument, linear constraints,
extremum, graphical method, mathematical modeling, objective function.
If the number of variables in the mathematical model of a linear
programming problem is more than two (with some exceptions), the problem cannot
be solved graphically. The simplex method is used to solve such problems.
The simplex method is a method of successively moving from one basic
solution (one end of the solution polygon) to another until the objective function of a
linear programming problem takes the optimal (maximum or minimum) value. This
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method is a universal method that allows you to solve any linear programming
problem, unlike the graphical method, which is designed to solve problems with only
two variables. is considered .
The simplex method was proposed in 1947 by the American
mathematician R. Danzig, and since then it has been widely used in industrial
production to solve linear programming problems involving thousands of variables
and constraints. Before describing the simplex method, let us recall some concepts of
a system of linear equations.
To us
n
variable
m
Given a system of equations:
11 1
12 2
1
1
21 1
22 2
2
2
1 1
2 2
...
...
......................
...
n n
n n
m
m
mn n
m
a x
a x
a x
b
a x
a x
a x
b
a x
a
x
a x
b
(1)
In linear programming problems
(
)
ij
A
a
(
1, 2, ...,
;
1, 2, ..., )
i
m j
n
matrix color
r
m
is,
m
n
The situation is interesting.
If (1) the system
m
If the determinant of the matrix formed from the
coefficients before the variables is different from zero, then the basis for such
variables is are called variables .
remaining
n m
variables are either independent or non-independent. are
called variables .
If (1) the system
1
2
( ,
, ...,
)
n
x x
x
solutions
0
j
x
(
1, 2, ..., )
j
n
If the
condition is satisfied, such solutions are called feasible solutions, otherwise they are
called impossible solutions .
A solution to a system in which the non-basic variables are zero is called
a basic solution .
Analysis and results
For convenience in calculations, it is advisable to present the simplex
method in tabular form.
Simplex table.
The following algorithm is used to solve linear
programming problems using the simplex method.
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Step 1. Construct an initial simplex tableau;
Step 2. Check the solution for optimality. End the process when an
optimal solution is found;
Step 3. Find the state that leads to optimality;
Step 4. Switch to a new solution and return to step 2.
The general form of a simplex table is given in Table 1 (
m
– number of
conditions,
n
– number of variables)
Objective function
…
…
B
as
is
changers
C
oef
fi
ci
en
ts
of
the
obj
ect
ive
funct
ion
incl
uded
in
the
basi
s
B
as
is
sol
ut
ion
val
ues
Coefficients of the terms of the problem
j
F
j
j
C
F
line definition
j
j
C
F
A string defining the optimality criterion
Table 1. Simplex table view
• The first row of the table records all (main and additional) variables;
• Table
B
The first column, separated by the letter , lists the basic
variables.
• The second row of the table, starting from cell 3, lists the coefficients
of the objective function.
•
b
C
The coefficients of the variables included in the basis are placed in
the column (except for the last two rows).
• The coefficients of the conditions are given in the rows dedicated to the
basic variables.
•
0
P
The column contains the values of the basic variables.
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• Last
j
j
C
F
The line is aimed at determining the optimality criterion.
•
j
F
using the latest information
j
j
C
F
is a row, the last cell of which
contains the current value of the objective function.
This
1
2
1
2
1
2
1
2
2
10
2
14
0,
0
2
3
max
x
x
x
x
x
x
F
x
x
We determine the solution of the problem using the simplex method. To bring
the problem into canonical form, we use the following addition
1
s
,
2
s
We introduce
variables:
1
2
1
1
2
2
1
2
1
2
1
2
1
2
2
10
2
14
0,
0,
0,
0
2
3
0
0
max
x
x
s
x
x
s
x
x
s
s
F
x
x
s
s
(2)
There are a total of 4 equations in the system of equations, i.e. 2 basic
ones
1
x
,
2
x
and 2 additional ones
1
s
,
2
s
There are variables. The vector of coefficients
of the objective function is
C
, and the matrix of coefficients of the constraint
conditions is
A
and the right-hand side vectors of the conditions
B
are defined as
follows:
1
2
3
4
( ;
;
;
)
(2; 3; 0; 0)
C
c c c c
11
12
13
21
22
23
1
2 1
,
2
1 1
a
a
a
A
a
a
a
1
2
10
14
b
B
b
Step 1. Construct an initial simplex tableau
Let's construct a simplex table for our example above. The objective
function
1
2
1
2
2
3
0
0
z
x
x
s
s
Write it in the form of a table, taking into account the system (2)
We fill in the following ( Table 2 ).
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B
b
C
0
P
1
x
2
x
1
s
2
s
2
3
0
0
1
s
0
10
1
2
1
0
1
s
0
14
2
1
0
1
j
F
0
0
0
0
0
j
j
C
F
2
3
0
0
Table 2. Elementary simplex table
2, 3, and 4 of the initial simplex tableau consist directly of the objective
function and the system coefficients (
A
matrix,
C
and
B
pay attention to vectors).
j
F
The row elements are found as follows. The vector consisting of the coefficients of
the objective function in the basis is scalar multiplied by the vectors in the condition
column. That is,
0
0
b
C
a vector
1
1
2
A
is scalar multiplied by a vector, etc. In
this way
j
F
all elements of the row are found.
j
j
C
F
The row elements are the
coefficients of the objective function, respectively
j
F
is obtained by subtracting the
elements of the row. Since the variables not included in the basis are equal to zero
1
0
x
,
2
0.
x
the value of the basis variables is taken from the last column:
1
10,
s
2
14.
s
j
F
The number in the last cell of the row is the value of the objective function
at the initial step.
0.
F
In the first step, the values of the last row match the coefficients of the
objective function. This completes the first step.
Step Two. Check the result for optimality
Optimality of the obtained result
j
j
C
F
is determined by the non-negativity
of all numbers in the row. If
j
j
C
F
all elements in the row are zero or negative If the
result obtained is optimal and the process is completed. If there is at least one positive
element among these elements, then optimality has not been achieved and the solution
can be improved.
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In our example, the result is not optimal because the last row contains
two positive numbers. This completes the verification of the optimality condition.
Step Three. Finding the optimality-directing state
We determine the maximum element from the last row of the initial table,
which is equal to 3. The column containing the largest positive element in the last row
of the simplex table is called the decisive column (Table 3).
B
b
C
0
P
1
x
2
x
1
s
2
s
0
/
ij
P
a
2
3
0
0
1
s
0
10
1
2
1
0
10/2=5
1
s
0
14
2
1
0
1
14/1=14
j
F
0
0
0
0
0
j
j
C
F
2
3
0
0
Table 3. Determination of the decisive element
In the table given in Table 3, the decisive column is indicated by an arrow. In
order to find the decisive row, we introduce an additional column and divide the
elements of the column by the elements of the decisive column. We take the smaller
of the resulting numbers:
min{5, 14} 5.
Therefore, the third row of the table is the
decisive row, and this row is indicated by an arrow. The element located at the
intersection of the decisive row and the decisive columns is the decisive is called the
element . In our example, the decisive element is equal to 2 and is shown in red in the
table. This completes step 3.
Step 4. Switch to a new solution
The transition to a new solution begins with the exchange of basic variables.
The basic variable at the beginning of the solution row is exchanged with the variable
in the solution column, and the corresponding coefficients are also exchanged. The
elements in rows 3 and 4 of the simplex table are recalculated using the Gauss-Jordan
method using the solution element. The Gauss-Jordan method proceeds as follows:
1) The decisive row is divided by the decisive element. The remaining
elements of the decisive column are filled with zeros.
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2) The remaining rows are recalculated using the "rectangle" method. You are
familiar with this method from the topic of solving a system of linear equations using
the Gauss-Jordan method.
Let's mention the "rectangle" method.
( , )
ij
a i j
a
Let us define the element at
the intersection of
( , )
a s k
the -row and -column
j
of a table such as . Let
i
– be the
decisive element and
( , )
a i j
the element to be recalculated. The table
( , )
a s k
and
( , )
a i j
Using the cells containing the values, we can construct a right rectangle as
shown in Table 4 below.
( , )
a i j
…
( , )
a i k
…
…
( , )
a s j
…
( , )
a s k
Table 4. Rectangle method
( , )
a i j
new value of
*
( , )
a i j
is calculated using the following formula:
*
( , )
( , )
( , )
( , )
( , )
a s j
a i k
a i j
a i j
a s k
As a result of the recalculation, we arrive at the following table 5. Thus, a
second table was constructed, and a new simplex table was created.
To speed up calculations, it is advisable to use the following rules.
B
b
C
0
P
1
x
2
x
1
s
2
s
2
3
0
0
2
x
3
5
1/2
1
1/2
0
1
s
0
9
3/2
0
–1/2
1
j
F
15
3/2
3
3/2
0
j
j
C
F
1/2
0
–3/2
0
Table 5. Second simplex table
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• If the decisive row contains elements equal to 0, the value of the
corresponding column elements remains unchanged in the new table;
• If the decisive column contains elements equal to 0, the corresponding
row in the new table remains unchanged. We proceed to the second step and check
the optimality criterion again.
The last row of the new simplex table contains a positive element
1
2
Since
there is no optimal solution, we will construct a new table. Now the decisive column
1
x
is the one corresponding to the only positive element in the last row (Table 6).
B
b
C
0
P
1
x
2
x
1
s
2
s
0
/
ij
P
a
2
3
0
0
2
x
3
5
1/2
1
1/2
0
10
1
s
0
9
3/2
0
–1/2
1
6
j
F
15
3/2
3
3/2
0
j
j
C
F
1/2
0
–3/2
0
Table 6. Determination of the decisive element
Minimum value in the last column
min{10, 6}
6
Since the decisive line is the
fourth line. So,
1
x
The variable enters the basis,
2
s
and comes out of the basis. We
construct a new table according to the above rule (Table 7).
B
b
C
0
P
1
x
2
x
1
s
2
s
2
3
0
0
2
x
3
2
0
1
2/3
–1/3
1
x
2
6
1
0
–1/3
2/3
j
F
18
2
3
4/3
1/3
j
j
C
F
0
0
–4/3
–1/3
Table 7. The last simplex table
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If we look at the last row of the resulting table, all the elements in the row are
non-negative, which indicates that we have reached the optimal solution. Optimal
plan from the table
1
6,
x
2
2
x
,
1
0
s
and
2
0
s
and the optimal value of the
objective function
max
(6; 2)
2 6
3 2 18
F
F
In the last table, the optimal value
of the objective function is formed in the pink cell.
Conclusion
Solving a linear programming problem using the simplex method is a
universal method, in which there is no limit to the number of variables, as in the
graphical method. In addition, the algorithm for solving a linear programming
problem using the simplex method is available in many application programs. In
particular, linear programming problems can be solved using the simplex method
using MS Excel and POM QM for Windows .
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