Nguyễn Phương Anh _ Optimization Problem In 10Th Grade Math Program.pdf

HANOI PEDAGOGICAL UNIVERSITY 2 DEPARTMENT OF MATHEMATICS NGUYEN PHUONG ANH OPTIMIZATION PROBLEM IN 10TH GRADE MATH PROGRAM DEPARTMENT OF MATHEMATICS GRADUATION THESIS Major Applied mathematics Hanoi –[.]

HANOI PEDAGOGICAL UNIVERSITY 2 DEPARTMENT OF MATHEMATICS

NGUYEN PHUONG ANH

OPTIMIZATION PROBLEM IN 10TH GRADE MATH PROGRAM

DEPARTMENT OF MATHEMATICS

GRADUATION THESISMajor: Applied mathematics

Hanoi – 2023

MINISTRY OF EDUCATION AND TRAINING

HANOI PEDAGOGICAL UNIVERSITY 2

Supervisor

PhD Ngoc Chi Le

Hanoi – 2023

I hereby declare that this thesis is my own research work, not copied

by anyone else, which I have researched, read, translated, synthesized, andperformed by myself Theoretical content in the thesis I have used somereferences as presented in the references section The data and results inthe thesis are honest and have not been published in any other works

Ha Noi, April 2023

Student

Nguyen Phuong Anh

The topic "Optimal problem in 10th grade math program" is thecontent that I researched and made my graduation thesis after studying atthe Faculty of Mathematics, Hanoi Pedagogical University 2 During theresearch process In researching and completing my thesis, I have received

a lot of attention and help from teachers, colleagues, family, and friends.For the success of this thesis, I would like to express my sincere thanks to:

The Department of Mathematics, Hanoi Pedagogical University 2 ated a very good learning and training environment, providing me withuseful knowledge and skills to help me apply and facilitate my thesis

cre-Tutor PhD Ngoc Chi Le is a passionate teacher who has dedicatedlyguided and helped me throughout the process of researching and imple-menting the topic You have exchanged ideas and suggestions for me tocomplete this research thesis well

I would also like to thank the management board and the teaching staff

of Hanoi Pedagogical University 2 for giving me the opportunity to work

at the school to gain practical knowledge and experience for useful mation useful for the thesis

infor-Finally, I would like to thank my family and friends for always aging and creating the best conditions for me to strive to complete theresearch well

encour-Thank you sincerely!

Ha Noi, April 2023

Student

Nguyen Phuong Anh

1.2 Linear programming problem 10

2 Methods for solving linear programming

2.1 Simplex method 14

2.2 Graphical Method 23

3 Optimization problem in 10th grade math program 28

3.1 How to determine the experimental domain 28

3.2 Method of solving the real problem is reduced to the system

of first-order inequalities with two unknowns 29

3.3 Some practical examples 30

1 Rationale of the thesis

In all areas of life, we are always interested in the problem of findingthe best way to achieve the desired goal under certain constraints Opti-mal methods are a powerful tool to help decision makers come up with thebest quantitative and qualitative solutions In recent years, optimizationmethods have been widely and effectively applied in economics, engineer-ing, information technology and other sciences

One of the first classes of optimization problems that has been fullystudied in both theory and algorithm is the linear programming problem.Linear programming since its inception (in the late 30s - 40s of the twen-tieth century) has occupied an important position in optimization Thelinear model is a very common model in practice, and at the same time,the linear fork dependence is a simple dependency and is easy to studymathematically

Because optimization problem is a branch of mathematics with manyapplications in life and economics, in some economic or pedagogical (un-dergraduate) disciplines there is a subject on this problem

For high school students, just consider the simple form of a tion problem problem presented in the 10th grade Algebra program.With the current organization of the national high school exam in the form

optimiza-of multiple choice, in my opinion, optimization problem is an importantproblem and it is very likely that it will appear in the national high school

exam because this is a form of math that comes from the essential needs

of life

2 Aims of the thesis

The thesis presents about the optimization problem in the 10th grademath program, specifically the real problem that is reduced to the system

of first order inequalities with two unknowns

3 Object and scope of thesis

3.1 Object: The optimization problem in the 10th grade math program.3.2 Scope: We will present about linear programming, optimization prob-lems and solving methods optimization problem in 10th grade mathprogramming

4 Structure of the thesis

In addition to the introduction, conclusion and list of references, thestructure of the thesis consists of 3 chapters:

Chapter 1: Preparatory knowledge

Present the optimization problem and linear programming problems.Chapter 2: Methods for solving linear programming problemsPresenting simplex methods and Graphical method to solve linearprogramming problems

Chapter 3: The optimization problem in the 10th grade mathprogram

Presenting solving methods the real problem that is reduced to thesystem of first order inequalities with two unknowns

Chapter 1

Preparatory knowledge

1.1 Optimization problem

General optimization problem

Minimum or maximum of the function f (x)

• f (x): objective function with n variables

• gi(x) , i = 1, m: Constraints Each inequality is a constraint

• Set D=



x ∈ X ⊂ Rn

gi(x) ≤, =, ≥ bi; i = 1, m



is a bound main

do-• x = (x1, x2, , xn) ∈ D: Acceptable option or acceptable solution

• Option x∗ ∈ D make the target function maximum (minimum) iscalled the optimal solution or optimal solution Namely: f (x∗) ≥

f (x) , ∀x ∈ D for the maximum problem or f (x∗) ≤ f (x) , ∀x ∈ D

for the minimum problem Then f∗ = f (x∗) is called the optimalvalue of the problem

Optimization problems are classified into:

Based on the acceptable domain X:

Figure 1.1: The acceptable domain X

bring the problem to be solved to the problem of linear programming:

The optimal solution to this problem is y0 = 0, 04; y1 = 0, 08; y2 =

0, 16with an optimal target value of 0.2

The optimal solution of the original math table is: x1 = 2, x2 = 4with anoptimal target value of fmin = 0, 2

Optimal production programming problem

A company wants to produce 2 types of products A and B using terials I, II, III The cost of production for one unit of output is given inthe following table:

(Unit of

raw materials) The interest for 1 unit of class A and B products is 4 and 5(Monetary Units), respectively It is necessary to plan production so thatthe company earns the most profit provided that the material is limited

as above

The symbols x1 and x2 represent the quantity of class A and B products

to produce, respectively The mathematical model of the above problem

takes the form of a linear programming problem:

• cj: Interest on a unit of product type j;

• aij: Raw material cost rate type i



i = 1, m

for a unit of producttype j;

• bi: Material reserve of type i;

1.2 Linear programming problem

The problem of programming can be stated in the form of:

f (x) = c1x1 + c2x2 + + cnxn → min / max

Provided that

a11x1 + a12x2 + + a1nxn ≤, =, ≥ b1

Where A = aij
m×n called the constraints, b ∈ Rmor b = (b1, , bm)T

called right and point vectors x ∈ Rm or x = (x1, , xn)T are called ables that need to be optimal

vari-Linear function x 7→ ⟨c, x⟩ is objective function or cost function

• The problem given by the constraint (1.1) programming is called thestandard form It is called canonical form when the constraint (1.1)

is replaced by an inequality Ax ≤ b

• Since linear equations can be converted into linear inequalities andvice versa, any linear programming problem can be returned to theform above

• Point x ∈ Rn satisfying all the constraints of the problem is called anacceptable point or an option The set of all options, denoted by X,

is called the bound set or acceptable set of the problem, i.e., the X

set of solutions of the system (1.1)

• Experiment x ∈ X¯ is called the optimization of the problem ⟨c, ¯x⟩ ≥

⟨c, x⟩ with ∀x ∈ X

• A sub-matrix B level k × k consisting of columns of A called a base

if it is inverse Suppose B is a base By using a permutation we canassume that B consists of the first k columns of A and the remainingcolumns k × (n − k)- matrix N, called the non-base part of A

• Let x be a vector with a component xB and xN, where xB is a kdimensional vector and xN is a (n − k)-dimensional vector satisfying:

-BxB = b,

xN = 0

• If xB is a positive vector, then x is a solution of (1.1) and is called anacceptable base solution (attached to base B) If apart xB without azero component, it is called non-degenerate; If it’s the opposite, it’scalled degenerate

• Let B be an acceptable basis, we call it an optimal basic if the sponding base solution is the optimal solution of the problem (LP)

corre-We will decompose a vector of value c into the base vector cB andnon-base vectors cN Vector: ¯cN = cN− B−1N
TcB called a reducedcost vector

Theorem 1.1.1 (Theorem 3.1.1 [1] pp 48-49)

Let X be other empty The following four conditions are equivalent.(i) (LP) has an optimal solution

(ii) (LP) has the optimum solution reached at the vertex

(iii) The cost function is non-positive for each asymptotic of X

(iv) The cost function is blocked on X.

Description of proof

Each of the constraints given above is converted to the form:

a11x1 + a12x2 + + a1nxn = b1

Chapter 2

Methods for solving linear programming problems

2.1 Simplex method

The basis of the simplex method

Consider the problem of canonical linear programming:

• If the problem has a solution, then there is an extreme solution

• If the problem has an optimal solution, then there is also an optimalpolar solution

• The number of extreme solutions is finite

Thus we can find an optimal solution (or a solution to the problem) inthe set of extreme solutions This set is finite So Dantzig proposed thefollowing algorithm (called Simplex method):

Starting from an extreme solution x0 Check if x0 is the optimal solution

If x0 is not the optimal option, then try to improve it so that anotherextreme option x1 is better than x0 in the sense f x1
< f x0
Thisprocess repeats several times Since the number of boundary polar options

is finite, after a finite number of iterative steps we must find the optimalsolution.

Conversely, find l such that: xˆs := ¯b1

¯

a 1s = minn¯bi

¯

a is : ¯ais > 0o

Step 5 Make a new acceptable basis Bk+1word Bkby crossing out the al

column and replacing it as The vertex corresponds to xk+1 received from

xk by setting variables xs = ˆxs > 0and variables xl = 0

Step 6 Inverse matrix calculation Bk+1−1 of the new basis Bk+1 and return

to Step 2

Find an acceptable facility

We consider the constraint of the problem (LP)

Ax = b

x ≥ 0

By multiplying both sides of the equality by (−1) when the right side is

a vector b with only non-negative coordinates Furthermore, when we seethat the other set of constraints is empty, we can remove some equations

so that the remaining constraint has no superfluous equations and we havethe same set of solutions From this, we assume there are two conditionsset as the following limitation: (1) the vector b is positive, and (2) there is

no superfluous equality

Since choosing an acceptable basis to start the simplex algorithm is biguous, we include a vector of dummy variable y = (y1, , ym)T andconsider the linear problem

am-Find the maximum:

The problem (LP) has acceptable solution if and only if the problem

y1 + + ym has a minimum value of 0 with y = 0

Find the inverted matrix

We see the base matrixBk andBk+1 uniquely different at one column,that is, they are adjacent to each other This helps us to calculate theinverse matrix of Bk+1 from the inverse matrix of Bk Symbol D is m × m

- matrix, called a matrix for change of basic, is a unit matrix minus thecolumn l by vectors

ls 00 −¯ams

Clause 2.1.2 With matrix D above, we have Bk+1−1 = DBk−1 In lar, if the first basis is a unit matrix, then Bk−1 = Dk D1 where D is thetransformation matrix

particu-Simplex table

To solve the problem (LP):

Find the maximum

f (x) = cTx

Provided that

Ax ≥ b,

x ≥ 0

We assume that b is the positive vector and the matrix A is written as

(BN ), where B is the acceptable basis For shortening, the vector c isplaced at the top of the table

The original simplex table had the form, symbol T

c

= c

c

0

By multiplying the right side of table T by the expansion matrix S

• The value of the target function at this base solution is equal to

⟨c, x⟩ = (cB)TB−1b, as opposed to the given value in the last column

in the upper-right corner

• Shortened price vector ¯cN is calculated above in the middle of thetable If all these vector coordinates are negative, then the currentbase vertex is optimal

• If some coordinates of the shortened price vector are positive, selectthe index s with ¯csthe largest Get variable xs will be the basis Onevariable xl with a satisfying l-index ¯bl

Simplex table T∗ obtained by multiplying to the right of matrix T by trix S

ma-1 0 1

¯ls

0 0

0 1

¯ls

0

.

.

ls

.

0 0

¯ls

1

We find the matrix of transformations D with the lower part of thetable above The rotating element of the simplex algorithm is the ¯ls.Example 2.1.1:

Consider the problem

Find the maximum

5x1 + 4x2 + 6x3

Provided that

x1 − x2 + x3 ≤ 203x1 + 2x2 + 4x3 ≤ 423x1 + 2x2 ≤ 30

x1, x2, x3 ≥ 0

The problem is equivalent to the standard form

Find the maximum

5x1 + 4x2 + 6x3

Provided that

x1 − x2 + x3 + x4 = 20

3x1 + 2x2 + 4x3 + x5 = 423x1 + 2x2 + x6 = 30



=



424



We see that ˆt get at i = 42

4 corresponding to the base variable x3 fore, put x3 into the premises and put x5 out of the premises A rotating

There-element is an There-element ¯23 = 4 with rotating rows l = 2 and rotating umn s = 3 Transformation matrix D1 and the expansion matrix S1 be

optimal The non-base variable with a positive shortened is x1, x2 ened vector ¯c1 = 12, ¯c2 = 1) We again put the non-base variable x2 intothe base variable by the largest shortened price vector ¯c2 = 1 We have

= {15, 21} = 15

We see that ˆt get at i = 15 corresponding to the base variable x6

There-fore, put x2into the premises and put x6out of the2 premises A rotatingelement is an element ¯32 = 2 with rotating rows l = 3 and rotating col-umn s = 2 Matrix S2 be

and by Theorem 1.1.2, the optimal current basis I get it right

away x1 = 0, x2 = 15, x3 = 3 The optimal value is 78 (as opposed to thenumber in the upper right corner of the table) End of the algorithm

2.2 Graphical Method

We will first discuss the steps of the algorithm:

Step 1: Formulate the LP (Linear programming) problem We have alreadyunderstood the mathematical formulation of an LP problem in a previoussection Note that this is the most crucial step as all the subsequent stepsdepend on our analysis here

Step 2: Construct a graph and plot the constraint lines

The graph must be constructed in n dimensions, where n is the number ofdecision variables This should give you an idea about the complexity ofthis step if the number of decision variables increases

One must know that one cannot imagine more than 3-dimensions anyway!The constraint lines can be constructed by joining the horizontal and ver-tical intercepts found from each constraint equation

Step 3: Determine the valid side of each constraint line

This is used to determine the domain of the available space, which canresult in a feasible solution How to check? A simple method is to put thecoordinates of the origin (0, 0) in the problem and determine whether theobjective function takes on a physical solution or not If yes, then the side

of the constraint lines on which the origin lies is the valid side Otherwise

it lies on the opposite one

Step 4: Identify the feasible solution region

The feasible solution region on the graph is the one which is satisfied by allthe constraints It could be viewed as the intersection of the valid regions

of each constraint line as well Choosing any point in this area would result

in a valid solution for our objective function

Step 5: Plot the objective function on the graph