Линейное программирование. Практикум. Учебное пособие для бакалавриата

Answers for questions and EXERCISES

a)Build the feasible region.

b)If the feasible region is not empty then set aside the gradient vector from the origin.

c)Construct a straight line (the level line) perpendicular to the gradient and move this line in the positive direction of the gradient vector.

d)The last intersection of the level line with the feasible region is the point of maximum.

7)

a)Build the feasible region.

b)If the feasible region is not empty then set aside the gradient vector from the origin.

c)Construct a straight line (the level line) perpendicular to the gradient and move this line in the positive direction of the gradient vector.

d)The first intersection of the level line with the feasible region is the point of maximum.

8) The last intersection of the level line with the feasible region is a side.

9) The first intersection of the level line with the feasible region is a side.

10) There is no last intersection of the level line with the feasible region when moving in the positive direction of the gradient.

Exercises

1)xmin = (3;2), fmin =12 .

2)xmax = (4;6), fmax =16 .

3)xmin = (1;2), fmin =7 .

4)xmax = (3;5), fmax =21 .

5)fmin = −∞.

6)xmin = (2 +4t;3 +t); t [0;1] , fmin = −20 .

7)xmax = (2 +2t;3 −2t); t [0;1] , fmax = −4 .

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Answers for questions and EXERCISES

6)

a)Choose j-th pivot column from the condition: the evaluation γj < 0 and at least one element aij > 0 ;

b)Choose k-th pivot row from the condition:

c)Recalculate the elements of the simplex table by using the Jordan-Gauss algorithm.

d)Repeat step 1.

7)

a)Choose j-th pivot column from the condition: the evaluation γj > 0 and at least one element aij > 0 ;

b)Choose k-th pivot row from the condition:

c)Recalculate the elements of the simplex table by using the Jordan-Gauss algorithm.

d)Repeat step 1.

8) All evaluations of free variables are non-negative

(γj ≥ 0) .

9) All evaluations of free variables are non-positive

(γj ≤ 0) .

10)All evaluations of free variables are non-negative, but there is at least one evaluation equal to zero.

11)All evaluations of free variables are non-positive, but there is at least one evaluation equal to zero.

12) All evaluations of free variables are positive

(γj > 0) .

13) All evaluations of free variables are negative

(γj < 0) .

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Answers for questions and EXERCISES

14)There is a negative evaluation and, in the column, corresponding to this evaluation all the elements are negative.

15)There is a positive evaluation and, in the column, corresponding to this evaluation all the elements are negative.

Exercises

1)xmax = (0;0;2;0;9), fmax =14 .

2)xmax = (0, 3t, 0, 3 −3t, 0,5 −3t); t [0,1]. fmax = 3 .

3)xmin = (4;1;0;0;5), fmin =2 .

4)fmax = +∞.

5)xmax = (6;9;2;0;0), fmax =14 .

Section 5

Questions

1)The method of artificial variable is used if a linear programming problem is written in the canonicals form and the system of main constraints is not reduced to the unit basis (it means that the basic and free variables are not determined and there is no basic solution).

2)We add an artificial variable in equation where there is no basic variable.

F = −y1 − −ym

where variables y1, ,ym are the artificial variables. 4) The artificial objective function is equal to zero

F = 0 .

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Answers for questions and EXERCISES

5) The artificial objective function is less than zero

F < 0 .

Exercises

1)fmax = +∞.

2)xmax = (2 +4t, 3 +t, 0, 7 −t, 7t); t [0,1]. fmax =20 .

3)xmax = (1;0;0;4), fmax = 3 .

4)The problem is infeasible.

5)The problem is infeasible.

6)xmax = (2;3;0;15;0), fmax = 3 .

Section 6

Questions

1)

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Answers for questions and EXERCISES

3)a) The non-trivial constraint of the type of equality in the primal problem corresponds to the condition of the absence of constraints in the primal problem and vice versa. b) The row of inequalities of the maximum problem goes into the column of inequalities of the minimum problem with the changing of sign on the opposite, and the column of inequalities goes into the row of inequalities with

preserving of the sign.

c) And conversely: the row of inequalities of the minimum problem goes into the column of inequalities of the maximum problem with the preserving of sign and the column of inequalities goes into the row of inequalities with changing of the sign on the opposite.

4)fmax = gmin .

5)The equilibrium theorem. The optimal solutions

Exercises

1) Construct the dual problem for the following linear programming problelems

g = 8y1 +8y2 −3 → min

≥−2,

a)−4y1 +5y2 ≥ 3,−y1 −5y2 ≥2,

y ≥ 0, y ≥ 0.

1 22y1 −7y2

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Answers for questions and EXERCISES

g = 3y1 +5y2 +2y3 → max

+y3

b)2y1 −8y2 −7y3 ≤2,−5y1 +6y2 +y3 ≤ −1,

y ≥ 0, y ≥ 0,y ≥ 0.

1 2 3y1 +2y2 ≤1,

g = y1 +5y2 −3 → max

c)y1 +3y2 ≤1,−y1 −2y2 ≤2,y1 +y2 ≤1.

g = y1 +5y2 +5 → max

d) 2y1 +4y2 ≥1,−y1 +3y2 ≥2,

3y1 +7y2 ≥ −1.

2) gmax = −34; ymax = (−6, 0, 0) .

Section 7

Questions

mn

1)z(X) = ∑∑cijxij → min

i=1 j=1

subject to

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Answers for questions and EXERCISES

2) A transportation problem is said to be closed if the total supply equals total demand

3)

a)finding an initial basic plan;

b)testing it for optimality;

c)improving the basic plan if the test fails.

Exercises

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