1 Linear programming

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Operations Research

1 Linear programming

1.1 Modeling

Exercise 1. The clothing manufacturer should specify how many jackets and coats should be produced so that the pro…t from their sale is maximal. One type of fabric is used for production. The manufacturer has 150 m² of this fabric. According to orders, at least 20 jackets and at most 10 coats should be produced. For the production of one jacket and one coat 2; 5 m² and 4 m² of fabric are needed, respectively. When selling one jacket, the producer makes a pro…t of Euro 50 , one coat - Euro 60 . Write this task as the minimum linear programming problem in the matrix-vector form.

Model 1. Let us denote:

u¹ - number of jackets, u² - number of coats.

Restrictions imposed on the variables u¹, u² can be written as follows:

u¹ 20; u² 10;

2; 5u¹+ 4u² 150:

The cost function, which should be maximized, takes the form 50u¹+ 60u²

Therefore, taking into account the natural restrictions of the non-negativity of variables u¹, u² (standard nonnegativity constraints), we can write the examined problem in the form of the following minimum linear programming problem

8>

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:

h( 50; 60); (u¹; u²)i ! min :

u2 U = fu = (u¹; u²)2 R²; u 0;

2 66 64

1 0

0 1

2; 5 4 3 77 75

2 4 u¹

u² 3 5

2 66 64

20 10 150

3 77 75g :

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Exercise 2 (carpenter problem). The carpenter should specify how many tables, chairs, desks and cabinets should be produced so that the pro…t from their sale is maximal.

Two types of boards are used for production. The manufacturer has 1500 m of type I boards and 1000 m of type II boards, and has the capital of 860 working hours to carry out the planned production. According to the contracts, carpenter should produce at least 40 tables, at least 130 chairs, at least 30 desks and not more than 10 cabinets. For the production of each table, chair, desk and cabinet carpenter needs 5, 1, 9, 12 m of boards of type I and 2, 3, 4, 1 m of boards of type II, respectively. To make a table carpenter needs 3 hours of work, a chair -2 hours, a desk - 5 hours, a cabinet - 10 hours.

The producer makes a pro…t of Euro 50, Euro 20, Euro 60 and Euro 40 from the sale of one table, chair, desk and cabinet, respectively.

Model 2.

Let us denote:

u¹ - number of tables, u² - number of chairs u³ - number of desks, u⁴ - number of cabinets.

Total pro…t can be written in the following way 50u¹+ 20u²+ 60u³+ 40u⁴ Of course, we assume non-negativity of all variables:

u₁ 0; u₂ 0; u₃ 0; u₄ 0:

Moreover,

u1 40; u2 130; u3 30; u4 10:

The amount of the boards of type I used during the production can be written as 5u₁+ u₂ + 9u₃+ 12u₄

and it should be less or equal to 1500. So, the corresponding contraint is the following 5u₁+ u₂+ 9u₃+ 12u₄ 1500:

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Similarly, the amount of the boards of type II used during the production can be written as

2u₁+ 3u₂+ 4u₃+ 11u₄ and it should be less or equal to 1000:

2u₁+ 3u₂+ 4u₃+ 11u₄ 1000:

Time used to produce u1 tables, u2 chairs, u3 desks and u4 cabinets can be written as 3u₁+ 2u₂+ 5u₃+ 10u₄

and it should be less or equal to 860:

3u₁+ 2u₂+ 5u₃+ 10u₄ 860:

Finally, our problem can be written in the following form: maximize the value 50u¹+ 20u²+ 60u³+ 40u⁴

subject to:

u₁ 0; u₂ 0; u₃ 0; u₄ 0

u₁ 40; u₂ 130; u₃ 30; u₄ 10 5u₁+ u₂+ 9u₃+ 12u₄ 1500

2u₁+ 3u₂+ 4u₃+ 11u₄ 1000 3u₁+ 2u₂+ 5u₃+ 10u₄ 860

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The minimum matrix-vector form of the above proble is the following

h( 50; 20; 60; 40); (u¹; u2; u3; u4)i ! min :

u2 U = fu 2 Rⁿ; u₁ 0; u₂ 0; u₃ 0; u₄ 0;

2 66 66 66 66 66 66 66 64

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

5 1 9 12

2 3 4 1

3 2 5 10

3 77 77 77 77 77 77 77 75

u 2 66 66 66 66 66 66 66 64

40 130

30 10 1500 1000 860

3 77 77 77 77 77 77 77 75

g:

Exercise 3 (diet problem). The dietitian should prepare the composition of the morning porridge so that it contains the necessary daily requirement of the body for speci…c nutrients and is the cheapest. The dietitian has two types of ‡akes: I and II.

Breakfast should contain at least 1 mg of vitamin B1, at least 12 mg of iron and have an energy value of 360 kcal. 100 g ‡akes of the …rst type contains 1; 2 mg of vitamin B1, 12 mg of iron and has an energy value of 358 kcal, while 100 g ‡akes of the second type contains 1; 5 mg of vitamin B1, 10 mg of iron and has an energy value of 390 kcal. In addition, 100 g of ‡akes of the …rst type cost PLN 0; 32 and 100 g of ‡akes of the second type - PLN 0; 36.

Model.

Let us denote:

u¹ - the amount of ‡akes I, u² - the amount of ‡akes II

Total costs can be written in the following way 0; 32u¹+ 0; 36u² It is natural to assume that

u1 0; u2 0:

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The amount of vitamin B1 in the prepared porridge can be written as 1; 2u₁+ 1; 5u₂

and it should be grater or equal to 1 mg.

The amount of iron can be written as

12u₁+ 10u₂ and it should be grater or equal to 12 mg.

Energy value of the porridge can be written as 358u₁+ 390u₂ and it should be equal to 360 kcal.

Finally, our problem can be written in the following form: minimize the value 0; 32u¹+ 0; 36u²

subject to:

u₁ 0; u₂ 0 1; 2u₁+ 1; 5u₂ 1

12u₁+ 10u₂ 12 358u₁+ 390u₂ = 360

The minimum matrix-vector form of the above problem is the following

h(0:32; 0:36); (u¹; u2)i ! min : u2 U = fu 2 Rⁿ; u₁ 0; u₂ 0;

2

4 1; 2 1; 5

12 10

3 5 u

2 4 1

12 3 5 ; h

358 390 i

u = 360g:

Exercise 4Write the following problem in the form of a linear programming problem.

The paint manufacturer must specify how many liters of white, green, blue and red paint

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should be produced to make the maximal pro…t from sales. Three raw materials are used in production: A, B and C. The producer has 230 liters of raw material A, 200 liters of raw material B and 170 liters - raw material C. Moreover, he has 160 working hours capital.

The accepted orders show that at least 125 liters of white paint should be produced, at least 135 liters - green paint, at most 205 liters - blue paint and not less than 175 liters - red paint. The amounts of raw materials needed to produce 1 liter of each paint are presented in the following table (in liters)

white green blue red A 0; 30 0; 60 0; 35 0; 15 B 0; 25 0; 20 0; 45 0; 55 C 0; 45 0; 20 0; 20 0; 30

In addition, it takes 15 minutes to produce 1 liter of each paint. Pro…t from the sale of 1 liter of white paint is 7 PLN, green - 6 PLN, blue - 7 PLN, red - 5 PLN.

Model.

Let us denote:

u¹ - the amount of white paint, u² - the amount of green paint, u³ - the amount of blue paint, u⁴ - the amount of red paint

Total pro…t that should be maximized can be written in the following way 7u₁+ 6u₂+ 7u₃+ 5u₄

Moreover,

u₁ 125; u₂ 135; u₃ 205; u₄ 175:

The amount of raw material A

0; 3u₁+ 0; 6u₂ + 0; 35u₃+ 0; 15u₄ used in production has to be less or equal to 230 liters.

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The amount of raw material B

0; 25u₁+ 0; 2u₂+ 0; 45u₃+ 0; 55u₄ has to be less or equal to 200 liters.

The amount of raw material C

0; 45u1 + 0; 2u2+ 0; 2u3+ 0; 3u4

has to be less or equal to 170 liters.

The amount of working hours

0; 25u₁+ 0; 25u₂+ 0; 25u₃+ 0; 25u₄ has to be less or equal to 160 hours.

Taking into account the non-negativity of variables u1, u2, u3, u4, our problem can be written in the following form: minimize

7u₁ + 6u₂+ 7u₃ + 5u₄ ! max subject to:

u₁ 0; u₂ 0; u₃ 0; u₄ 0 0; 3u₁ + 0; 6u₂+ 0; 35u₃+ 0; 15u₄ 230

0; 25u₁ + 0; 2u₂+ 0; 45u₃+ 0; 55u₄ 200 0; 45u₁+ 0; 2u₂+ 0; 2u₃+ 0; 3u₄ 170 0; 25u₁+ 0; 25u₂+ 0; 25u₃+ 0; 25u₄ 160

The minimum matrix-vector form of the above proble is the following h( 7; 6; 7; 5); (u¹; u₂; u₃; u₄)i ! min :

u2 U = fu 2 Rⁿ; u1 0; u2 0; u3 0; u4 0;

2 66 66 66 4

0; 3 0; 6 0; 35 0; 15 0; 25 0; 2 0; 45 0; 55 0; 45 0; 2 0; 2 0; 3 0; 25 0; 25 0; 25 0; 25

3 77 77 77 5

u 2 66 66 66 4

230 200 170 160 3 77 77 77 5

g:

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1.2 Equivalence of problems

Exercise 5. Write "the carpenter" problem in the canonical form.

Solution. Let us recall that "carpenter" problem has been written in the form

h( 50; 20; 60; 40); (u¹; u₂; u₃; u₄)i ! min :

u2 U = fu 2 Rⁿ; u₁ 0; u₂ 0; u₃ 0; u₄ 0;

2 66 66 66 66 66 66 66 64

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

5 1 9 12

2 3 4 1

3 2 5 10

3 77 77 77 77 77 77 77 75

u 2 66 66 66 66 66 66 66 64

40 130

30 10 1500 1000 860

3 77 77 77 77 77 77 77 75

g:

So, equivalent canonical form of this problem is the following

h( 50; 20; 60; 40; 0; 0; 0; 0; 0; 0; 0); (z¹; :::; z₄; z₅; :::; z₁₁)i ! min :

z2 Z = fz 2 R⁴⁺⁷; z 0;

2 66 66 66 66 66 66 66 64

1 0 0 0 1 0 ::: 0

0 1 0 0 0 1 ::: 0

0 0 1 0

0 0 0 1

5 1 9 12

2 3 4 1

3 2 5 10 0 0 ::: 1

3 77 77 77 77 77 77 77 75

z = 2 66 66 66 66 66 66 66 64

40 130

30 10 1500 1000 860

3 77 77 77 77 77 77 77 75

g:

Exercise 5a. Write the following general problem in the canonical form.

J (u) = u1 3u3 2u4+ 15u5 ! min :

u2 U = fu = (u¹; u₂; u₃; u₄; u₅)2 R⁵; u₁ 0; u₄ 0; u₅ 0; u₁+ 2u₃ 21;

u₂+ u₄+ 3u₅ 10; u₂ u₃+ 7u₅ = 2; 2u₁ 7u₄ = 9g:

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Solution. First, let us write our problem in matrix-vector form:

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:

J (u) =h(1; 0; 3; 2; 15); (u¹; :::; u₅)i ! min : u2 U = fu 2 R⁵; u₁ 0; u₁ 0; u₁ 0;

2

4 1 0 2 0 0 0 1 0 1 3

3 5 u

2 4 21

10 3 5 ;

2

4 0 1 1 0 7

2 0 0 7 0

3 5 u =

2 4 2

9 3 5g

:

In our case n = 5, I = f1; 4; 5g; J = f2; 3g; m = 2, s = 4; c = (1; 0; 3; 2; 15);

A = 2

4 1 0 2 0 0 0 1 0 1 3

3 5 ; A =

2

4 0 1 1 0 7

2 0 0 7 0

3 5 ;

b = 2 4 21

10 3 5 ; b =

2 4 2

9 3 5 :

Thus, p = 2 + 3 + 2 + 2 = 9 and the appropriate canonical problem has the form 8>

>>

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>>

:

J (z) =h(0; 0; 1; 2; 15; 0; 3; 0; 3); (v¹; v₂; u₁; u₄; u₅; w₂; w₃; w₂; w₃)i ! min : z 2 Z = fz = (v¹; v₂; u₁; u₄; u₅; w₂; w₃; w₂; w₃)2 R⁹; z 0;

2 66 66 66 4

1 0 1 0 0 0 2 0 2

0 1 0 1 3 1 0 1 0

0 0 0 0 7 1 1 1 1

0 0 2 7 0 0 0 0 0

3 77 77 77 5

z = 2 66 66 66 4

21 10 2 9

3 77 77 77 5 g

:

Exercise 6. Write the following general problem in the canonical form.

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>:

J (u) = u₁+ 2u₂+ 3u₃ ! min : u2 U = fu = (u¹; u₂; u₃)2 R³; u₁ 0;

10u₁+ 20u₂ + 30u₃ 11;

100u₁+ 200u₂+ 300u₃ 12 1000u₁+ 2000u₂+ 3000u₃ = 0g:

:

Solution. First, let us write our problem in matrix-vector form:

8>

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>>

:

J (u) =h(1; 2; 3); (u¹; :::; u₃)i ! min : u2 U = fu 2 R³; u₁ 0;

2

4 10 20 30 100 200 300

3 5 u

2 4 11

12 3 5 ; h

1000 2000 3000 i

u = [0]g :

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In our case n = 3, I = f1g; J = f2; 3g; m = 2, s = 3; c = (1; 2; 3);

A = 2

4 10 20 30 100 200 300

3

5 ; A = h

1000 2000 3000 i

;

b = 2 4 11

12 3

5 ; b = [0] :

Thus, p = 2 + 1 + 2 + 2 = 7 and the appropriate canonical problem has the form 8>

>>

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>>

>:

J (z) =h(0; 0; 1; 2; 3; 2; 3); (v¹; v₂; u₁; w₂; w₃; w₂; w₃)i ! min : z 2 Z = fz = (v¹; v₂; u₁; w₂; w₃; w₂; w₃)2 R⁷; z 0;

2 66 64

1 0 10 20 30 20 30

0 1 100 200 300 200 300 0 0 0; 1 0; 2 0; 3 0; 2 0; 3

3 77 75z =

2 66 64

11 12 0

3 77 75g

:

Exercise 7. Write the "diet" problem in the canonical form.

Solution. Let recall the matrix-vector form of "diet" problem:

h(0:32; 0:36); (u¹; u₂)i ! min :

u2 U = fu 2 R²; u₁ 0; u₂ 0;

2

4 1:2 1:5

12 10

3 5 u

2

4 1

12 3 5 ; h

358 390 i

u = [360]g:

In this case, n = 2, I = f1; 2g; J = ?; m = 2, s = 3; c = (1; 2; 3);

A = 2

4 1:2 1:5

12 10

3

5 ; A = h

358 390 i

;

b = 2 4 1

12 3

5 ; b = [360] :

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So, p = 2 + 2 = 4 and he matrix-vector form of the "diet" problem is as follows:

8>

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>:

J (z) =h(0; 0; 0:32; 0:36); (v¹; v₂; u₁; u₂)i ! min : z 2 Z = fz = (v¹; v₂; u₁; u₂)2 R⁴; z 0;

2 66 64

1 0 1:2 1:5

0 1 12 10

0 0 358 390 3 77 75z =

2 66 64

1 12 360

3 77 75g

:

Exercise 8. Write the following general problem in the canonical form.

8>

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>>

>:

J (u) = 3u¹+ 5u²+ 7u³+ 9u⁴ ! min : u2 U = fu = (u¹; u²; u³; u⁴)2 R⁴; u² 0; u⁴ 0;

11u¹+ 12u²+ 13u³+ 14u⁴ 1;

21u¹+ 23u³ 1;

32u² 8;

41u¹+ 42u²+ 43u³+ 44u⁴ = 2;

51u¹+ 54u⁴ = 2g:

Solution. Let us write the above problem in matrix-vector form:

h(3; 5; 7; 9); (u¹; :::; u₄)i ! min :

u2 U = fu 2 R⁴; u₂ 0; u₄ 0;

2 66 64

11 12 13 14

21 0 23 0

0 32 0 0

3 77 75u

2 66 64

1 1 8 3 77 75; 2

441 42 43 44

51 0 0 54

3 5 u =

2 4 2

2 3 5g:

In this case, n = 4, I = f2; 4g; J = f1; 3g; m = 3, s = 5; c = (3; 5; 7; 9);

A = 2 66 64

11 12 13 14

21 0 23 0

0 32 0 0

3 77 75; A=

2

441 42 43 44

51 0 0 54

3 5 ;

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b = 2 66 64

1 1 8 3 77 75; b =

2 4 2

2 3 5 :

So, p = 3 + 2 + 2 + 2 = 9 and he matrix-vector form of the "diet" problem is as follows:

8>

>>

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>>

>:

J (z) =h(0; 0; 5; 9; 3; 7; 3; 7); (v¹; v₂; u₂; u₄; w₁; w₃; w₁; w₃)i ! min : z 2 Z = fz = (v¹; v₂; u₂; u₄; w₁; w₃; w₁; w₃)2 R⁹; z 0;

2 66 66 66 66 64

1 0 0 12 14 11 13 11 13

0 1 0 0 0 21 23 21 23

0 0 1 32 0 0 0 0 0

0 0 0 42 44 41 43 41 43

0 0 0 0 54 51 0 51 0

3 77 77 77 77 75

z = 2 66 66 66 66 64

1 1 8 2

2 3 77 77 77 77 75 g

:

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1.3 Geometrical method of solving of 2-dimensional linear pro- gramming problems

Exercise 9. Solve geometrically the following problem:

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>:

J (u) = u¹+ u² ! min :

u2 U = fu 2 R²; u 0;

2 66 66 66 4

2 1

1

2 1

1 1

1 0

3 77 77 77 5

u 2 66 66 66 4

2

1 2

2 3

3 77 77 77 5

:

Solution.

The unique solution to the problem is the point u = (1; 0).

8>

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>>

>:

J (u) = 2u¹+ u² ! min :

u2 U = fu 2 R²; u 0;

2 66 66 66 4

1 1

1 2

2 1

3 77 77 77 5

u 2 66 66 66 4

1 1 0 5

3 77 77 77 5

:

Solution.

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The problem possesses in…nite many solutions. The set of solutions is the following U =f (5

2; 0) + (1 )(10 3 ;5

3); 2 [0; 1]g:

It is the side of the polygon U , with end-points (⁵₂; 0) and (¹⁰₃;⁵₃).

8>

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:

J (u) = u¹ u² ! min : u2 U = fu = (u¹; u²)2 R²; u 0;

1

2u¹+ u² 2;

1

3u¹ u² = 1g

:

The problem has no solution.

In a factory products A and B are produced. In the production process three machines are used: M1, M2, M3. The machine M1 can work for 2400 minutes, M2 - 4000 minutes, M3 - 2700 minutes. The following table shows the operating time of each machine, that is needed to produce one unit of each product

A B

M1 3 6

M2 8 4

M3 9 3

The pro…t from the sale of a unit of product A is 90 euro, and of product B - 60 euro.

The production should be planed to maximize the total pro…t.

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1.4 Extreme points

Exercise 13. Find all extreme points of the set

U =fu = (u¹; u²; u³; u⁴)2 R⁴; u 0;

u¹+ u²+ 3u³+ u⁴ = 3; u¹ u²+ u³+ 2u⁴ = 1g:

Rozwi ¾azanie. It is easy to see that

rank 2

4 1 1 3 1

1 1 1 2

3 5 = 2:

1⁰ Let j1 = 1, j2 = 2. The columns A1 = 2 4 1

1 3

5, A₂ = 2 4 1

1 3

5 are linearly independent. Moreover, the solution of the problem

A₁v¹+ A₂v² = b;

i.e. 8

<

:

v¹+ v² = 3 v¹ v² = 1

is the pair v¹ = 2 0, v² = 1 0. So, the point v = (2; 1; 0; 0) is nonsingular extreme point of the set U with the basis A1, A2.

1 3 5, A₃ =

2 4 3

1 3

5 are linearly independent.

Moreover, the solution of the problem

A₁v¹+ A₃v³ = b;

i.e. 8

<

:

v¹+ 3v³ = 3 v¹+ v³ = 1

is the pair v¹ = 0 0, v³ = 1 0. So, the point v = (0; 0; 1; 0) is singular extreme point of the set U with the basis A1, A3.

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1 3 5, A₄ =

2 4 1

2 3

A₁v¹+ A₄v⁴ = b;

i.e. 8

<

:

v¹+ v⁴ = 3 v¹+ 2v⁴ = 1

is the pair v¹ = 5 0, v⁴ = 2 < 0. Thus, the columns A1, A4 are not the basis for any extreme point.

1 3

5, A₃ = 2 4 3

1 3

A₂v²+ A₃v³ = b;

i.e. 8

<

:

v²+ 3v³ = 3 v²+ v³ = 1

is the pair v² = 0 0, v³ = 1 0. So, the point v = (0; 0; 1; 0) is singular extreme point of the set U with the basis A2, A3.

1 3

5, A₄ = 2 4 1

2 3

A₂v²+ A₄v⁴ = b;

i.e. 8

<

:

v²+ v⁴ = 3 v²+ 2v⁴ = 1

is the pair v² = ⁵₃ 0, v⁴ = ₃⁴ 0. So, the point v = (0;⁵₃; 0;⁴₃) is nonsingular extreme point of the set U with the basis A2, A4.

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1 3 5, A₄ =

2 4 1

2 3

A₃v³+ A₄v⁴ = b;

i.e. 8

<

:

3v³+ v⁴ = 3 v³+ 2v⁴ = 1

is the pair v³ = 1 0, v⁴ = 0 0. So, the point v = (0; 0; 1; 0) is the singular extreme point of the set U with the basis A3, A4.

U =fu = (u¹; u²; u³)2 R³; u 0;

u¹+ 2u²+ 3u³ = 4; u¹+ 5u³ = 0g:

U = fu = (u¹; u²; u³)2 R³; u 0;

u¹+ u²+ 2u³ = 10; u¹+ 3u³ = 9; u¹+ 2u²+ 7u³ = 29g:

rank 2 66 64

1 1 2

1 0 3

1 2 7

3 77 75= 2:

1 1 1

3 77

75, A2 = 2 66 64

1 0 2

3 77

75 are linearly indepen-

dent (one can check it using the de…nition of linear independence of vectors). Moreover, the solution of the system

A₁v¹+ A₂v² = b;

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i.e. 8

>>

><

>>

>:

v¹+ v² = 10 v¹ = 9 v¹+ 2v² = 29

is the pair v¹ = 9 < 0, v² = 19 0. Thus, the columns A1, A2 are not the basis for any extreme point of the set U .

1 1 1

3 77

75, A3 = 2 66 64

2 3 7

3 77

75 are linearly indepen-

dent (one can check it using the de…nition of linear independence of vectors). Moreover, the solution of the system

A₁v¹+ A₃v³ = b;

i.e. 8

>>

><

>>

>:

v¹+ 2v³ = 10 v¹+ 3v³ = 9 v¹+ 7v³ = 29

is the pair v¹ = ¹²₅ > 0, v³ = ¹⁹₅ > 0. So, the point v = (¹²₅ ; 0;¹⁹₅ )is nonsingular extreme point of the set U with the basis A1, A3.

1 0 2

3 77 75, A3 =

2 66 64

2 3 7

3 77

75are linearly independent

(one can check it using the de…nition of linear independence of vectors). Moreover, the solution of the system

A₂v²+ A₃v³ = b;

i.e. 8

>>

><

>>

>:

v²+ 2v³ = 10 3v³ = 9 2v²+ 7v³ = 29

is the pair v² = 4 > 0, v³ = 3 > 0. So, the point v = (0; 4; 3) is nonsingular extreme point of the set U with the basis A2, A3.

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U =fu = (u¹; u²; u³; u⁴)2 R⁴; u 0;

u¹+ u⁴ = 0; 2u² + u⁴ = 3; 3u³ = 0g:

rank 2 66 64

1 0 0 1 0 2 0 1 0 0 3 0

3 77 75= 3:

1⁰ Let j1 = 1, j2 = 2, j3 = 3. The columns A1 = 2 66 64

1 0 0

3 77 75, A2 =

2 66 64

0 2 0

3 77 75, A3 =

2 66 64

0 0 3

3 77 75 are

linearly independent (one can check it using the notion of the determinant of the matrix).

Moreover, the solution of the system

A₁v¹+ A₂v²+ A₃v³ = b;

i.e. 8

>>

><

>>

>:

v¹ = 0 2v² = 3 3v³ = 0

are three numbers v¹ = 0 0, v² = ⁹₆ 0, v³ = 0 0:So, the point v = (0;⁹₆; 0; 0)is the singular extreme point of the set U with the basis A1, A2, A3.

1 0 0

3 77 75, A2 =

2 66 64

0 2 0

3 77 75, A4 =

2 66 64

1 1 0

3 77 75 are

linearly dependent (one can check it using the notion of the determinant of the matrix).

So, they are not the basis for any extreme point of the set U .

1 0 0

3 77 75, A3 =

2 66 64

0 0 3

3 77 75, A4 =

2 66 64

1 1 0

3 77 75 are

linearly independent (one can check this fact using the notion of the determinant of the

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matrix). Moreover, the solution of the system

A₁v¹+ A₃v³ + A₄v⁴+ = b;

i.e. 8

>>

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>>

>:

v¹+ v⁴ = 0 v⁴ = 3 3v³ = 0

are three numbers v¹ = 3 < 0, v³ = 0 0, v⁴ = 3 0: So, the columns A1, A3, A4 are not the basis for any extreme point of the set U .

0 2 0

3 77 75, A3 =

2 66 64

0 0 3

3 77 75, A4 =

2 66 64

1 1 0

3 77 75 are

linearly independent (one can check this fact using the notion of the determinant of the matrix). Moreover, the solution of the system

A₂v²+ A₃v³+ A₄v⁴ = b;

i.e. 8

>>

><

>>

>:

v⁴ = 0 2v²+ v⁴ = 3

3v³ = 0

are three numbers v² = ³₂ > 0, v³ = 0 0, v³ = 0 0: Thus, the point v = (0;³₂; 0; 0) is the singular extreme point of the set U with the basis A2, A3, A4.

U =fu = (u¹; u²; u³; u⁴)2 R⁴; u 0;

2u¹ 3u²+ 4u³+ u⁴ = 3;

u¹+ u² u³ = 10g

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1.5 Simplex method

Exercise 18. Create the simplex table for the problem 8>

>>

><

>>

:

J (u) = u¹ u²+ 2u⁴ ! min : u2 U = fu = (u¹; u²; u³; u⁴)2 R⁴; u 0;

2u¹ 3u²+ 4u³+ u⁴ = 3;

u¹+ u² u³ = 10g and the extreme point

v = (33 5 ;17

5 ; 0; 0):

Exercise 19. Solve the problem 8>

>>

<

>>

>:

J (u) = u¹+ 2u²+ 3u³+ 4u⁴ ! min : U = fu 2 R⁴; u 0 ,

2

4 1 1 3 1

1 1 1 2

3 5 u =

2 4 3

1 3 5g

using the simplex method with the initial extreme point v = (2; 1; 0; 0).

Solution. It is easy to see that

r = rankA = 2

and, in consequence, the basic coordinates of the point v are v1, v2. According to our notations (see lecture) u = (u¹; u²), v = (2; 1), c = (1; 2), B =

2 4 1 1

1 1

3

5. Thus

B ¹ = 1 2

2

4 1 1

1 1

3 5

T

= 2 4

1 2

1 2 1 2

1 2

3 5 ;

and, consequently, 2 4 ^1;3

2;3

3

5 = B ¹A₃ = 2 4 2

1 3 5 ; 2

4 ^1;4

2;4

3

5 = B ¹A4 = 2 4

3 2 1 2

3 5

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as well as

3 = c; B ¹A₃ c₃ = 1;

4 = c; B ¹A₄ c₄ = 7 2: So, the simplex table for point v = (2; 1; 0; 0) is of the form

u¹ u² u³ u⁴ u¹ 1 0 2 ³₂ 2 u² 0 1 1 ¹₂ 1

0 0 1 ⁷₂ 4

:

Clearly, for this table there is the case 3⁰. We have

3 > 0

I₃ =fi 2 f1; 2g; i;3 > 0g = f1; 2g:

So,

min

i2I³

vⁱ

i;3

= minf2 2;1

1g = 2 2

Thus, k = 3, s = 1. The basis of the next extreme point is the system of columns A₂; A₃:

Usin the theorem charizing the extreme points we …nd the next extrem point w:

2 4 1

1 3 5 w²+

2 4 3

1 3 5 w³ =

2 4 3

1 3 5 ;

so, w = (0; 0; 1; 0). Moreover, B = 2

4 1 3

1 1 3

5 and, in consequence,

B ¹ = 1 4

2

4 1 1

3 1 3 5

T

= 2 4

1 4

3 4 1 4

1 4

3 5 :

It implies that 2

4 ^2;1

3;1

3

5 = B ¹A₁ = 2 4

1 2 1 2

3 5 ;

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2 4 ^2;4

3;4

3

5 = B ¹A₄ = 2 4

5 4 3 4

3 5

and

1 = c; B ¹A₁ c₁ = 1 2;

4 = c; B ¹A₄ c₄ = 17 4 : The simplex table for point w = (0; 0; 1; 0) takes the form

u¹ u² u³ u⁴ u² ¹₂ 1 0 ⁵₄ 0 u³ ¹₂ 0 1 ³₄ 1

1

2 0 0 ¹⁷₄ 3

:

We see that for this table there is the case 1⁰. It means that the point w = (0; 0; 1; 0) is a solution of our problem.

>>

><

>>

:

J (u) = u₁ u₂+ u₄ ! min :

u2 U = fu = (u¹; u₂; u₃; u₄)2 R⁴; u 0;

2u¹ 2u²+ 4u³+ u⁴ = 2;

u¹+ u²+ u⁴ = 0g

;

using the simplex method with the initial extreme point = (0; 0;¹₂; 0) and its basis formed by columns

2 4 2

1 3 5,

2 4 4

0 3 5.

>>

><

>>

:

J (u) = u¹+ 2u²+ 3u³+ 4u⁴ ! min : u2 U = fu = (u¹; u²; u³; u⁴)2 R⁴; u 0;

u¹+ u²+ 3u³+ u⁴ = 3;

u¹ u² + u³+ 2u⁴ = 1g

;

using the simplex method with the initial extreme point v = (0;5

3; 0;4 3).

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Exercise 22. Create the simplex table for the problem 8>

>>

><

>>

:

J (u) = u¹+ 3u² 5u³+ u⁴ 4u⁵ ! min : u2 U = fu 2 R⁵; u 0;

u¹+ u² 4u³ + u⁴ 3u⁵ = 3;

u¹ 4u³+ 2u⁴ 5u⁵ = 6g:

and the extreme point v = (0; 0; 0; 3; 0) with basic coordinates v¹, v⁴.

1.6 Choice of a "starting" extreme point

Exercise 23. Using an auxiliary problem, verify that the set U =fu = (u¹; u²; u³; u⁴)2 R⁴; u 0;

u¹ + u²+ 3u³+ u⁴ = 3;

u¹ u²+ u³+ 2u⁴ = 1g is non empty and …nd an extreme point of it.

Rozwi ¾azanie. Let us consider the following auxiliary problem 8>

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>>

>:

J (z) = u⁵+ u⁶ ! min :

Z =fz = (u¹; :::; u⁶)2 R⁶; z 0 , 2

4 1 1 3 1 1 0

1 1 1 2 0 1

3 5 z =

2 4 3

1 3 5g

In this case b = 2 4 1

0 3

5 0and the point z0 = (0; b) = (0; 0; 0; 0; 3; 1) is an extreme point

of the set Z with the basis C5 = 2 4 1

0 3 5, C₆ =

2 4 0

1 3 5.

Let us apply to the auxiliary problem the simplex method. In this case r = 2, j1 = 5, j2 = 6, z = (u⁵; u⁶), v = (3; 1), c = (1; 1), B =

2 4 1 0

0 1 3 5. So,

B ¹ = 2 4 1 0

0 1 3 5 ;

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and 2 4 ^5;1

6;1

3

5 = B ¹C₁ = C₁ = 2 4 1

1 3 5 ; 2

4 ^5;2

6;2

3

5 = B ¹C₂ = C₂ = 2 4 1

1 3 5 ; 2

4 ^5;3

6;3

3

5 = B ¹C₃ = C₃ = 2 4 3

1 3 5 ; 2

4 ^5;4

6;4

3

5 = B ¹C₄ = C₄ = 2 4 1

2 3 5 :

as well as

1 = c; B ¹C₁ c₁ =h(1; 1); (1; 1)i 0 = 2;

2 = c; B ¹C₂ c₂ =h(1; 1); (1; 1)i 0 = 0;

3 = c; B ¹A₃ c₃ =h(1; 1); (3; 1)i 0 = 4;

4 = c; B ¹A₄ c₄ =h(1; 1); (1; 2)i 0 = 3:

The simplex table for point v = (0; 0; 0; 0; 3; 1) is of the form u¹ u² u³ u⁴ u⁵ u⁶

u⁵ 1 1 3 1 1 0 3

u⁶ 1 1 1 2 0 1 1

2 0 4 3 0 0 4

:

For the above simplex table there is case 3⁰. We have

1 > 0

I_v;1 =fjⁱ 2 f5; 6g; ji;1> 0g = f5; 6g;

and

jmini2Iv;1

v^jⁱ

ji;1

= minf3 1;1

1g = 1

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So, k = 1, js = 6. The basis of the next extreme point will be the system of columns C₁; C₅:

Using the theorem characterizing the extreme points, we …nd the next extreme point w:

2 4 1

1 3 5 w¹+

2 4 1

0 3 5 w⁵ =

2 4 3

1 3 5 ;

and w = (1; 0; 0; 0; 2; 0). Moreover, B = 2 4 1 1

1 0 3

B ¹ = 1 1

2

4 0 1

1 1 3 5

T

= 2 4 0 1

1 1

3 5 ; 2

4 ^1;2

5;2

3

5 = B ¹C₂ = 2

4 1

2 3 5 ; 2

4 ^1;3

5;3

3

5 = B ¹C₃ = 2 4 1

2 3 5 ; 2

4 ^1;4

5;4

3

5 = B ¹C4 = 2 4 2

1 3 5 ; 2

4 ^1;6

5;6

3

5 = B ¹C₆ = 2 4 1

1 3 5 :

and

2 =h(0; 1); ( 1; 2)i 0 = 2;

3 =h(0; 1); (1; 2)i 0 = 2;

4 =h(0; 1); (2; 1)i 0 = 1;

6 =h(0; 1); (1; 1)i 1 = 2:

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So, the simplex table for the point w = (1; 0; 0; 0; 2; 0) is the following u¹ u² u³ u⁴ u⁵ u⁶

u¹ 1 1 1 2 0 1 1

u⁵ 0 2 2 1 1 1 2

0 2 2 1 0 2 2

:

For the above simplex table there is case 3⁰. We have

2 > 0

I_v;2 =fjⁱ 2 f1; 5g; ji;2 > 0g = f5g;

and

min

ji2Iv;2

v^jⁱ

ji;1

= minf2 2g = 1

Thus, k = 2, js = 5. The basis of the next extreme point will be the system of columns C₁; C₂:

Using the theorem characterizing the extreme points, we …nd the next extreme point w:

2 4 1

1 3 5 w¹+

2 4 1

1 3 5 w² =

2 4 3

1 3 5 ;

and w = (2; 1; 0; 0; 0; 0). Moreover, B = 2 4 1 1

1 1

3

B ¹ = 1 2

2

4 1 1

1 1

3 5

T

= 2 4

1 2

1 2 1 2

1 2

3 5 ; 2

4 ^1;3

2;3

3

5 = B ¹C₃ = 2 4 2

1 3 5 ; 2

4 ^1;4

2;4

3

5 = B ¹C₄ = 2 4

3 2 1 2

3 5 ;

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2 4 ^1;5

2;5

3

5 = B ¹C₅ = 2 4

1 2 1 2

3 5 ; 2

4 ^1;6

2;6

3

5 = B ¹C₆ = 2 4

1 2 1 2

3 5 :

and

3 =h(0; 0); (2; 1)i 0 = 0;

4 = (0; 0); (3 2; 1

2) 0 = 0;

5 = (0; 0); (1 2;1

2) 1 = 1;

6 = (0; 0); (1 2; 1

2) 1 = 1:

The simplex table for the point w = (1; 0; 0; 0; 2; 0) is of the form u¹ u² u³ u⁴ u⁵ u⁶

u¹ 1 0 2 ³₂ ¹₂ ¹₂ 2 u² 0 1 1 ¹₂ ¹₂ ¹₂ 1

0 0 0 0 1 1 0

:

For the above simplex table there is case 1⁰. Since J1(2; 1; 0; 0; 0; 0) = 0, therefore U 6= ?.

Moreover, since the point z = (2; 1; 0; 0; 0; 0) is a solution of the auxiliary problem, therefore the point v1 = (2; 1; 0; 0) is an extreme point of the set U .