Линейное программирование. Практикум. Учебное пособие для бакалавриата

5. Method of artificial variables

a12 =1

is the pivot (x3 ↔ x2 ) .

After recalculation we have the following table.

In this table we have only one the negative evaluation

γ1 = −4 < 0 .

So, the first column is a pivot column. In this column we find the minimal ratio. We obtain

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5. Method of artificial variables

In this table all evaluations are non-negative, so the maximum is attained.

Fmax = 0

and

y1 = 0 .

So, the initial linear programming problem is feasible, and we have found its basic solution. We can begin the second phase. We get.

In this table all evaluations are non-negative, so the maximum is attained.

X1max* = (4, 5, 0, 0, 16) ,

fmax = f (X1max* )= −27 .

But the evaluation

γ3 = 0 .

So, we the alternative optimum. To find another optimal solution we need to select the third column as the pivot column. In this column we have only one positive number

a33 =2 .

Thus, a33 is a pivot (x5 ↔ x3 ) ..

After recalculation we get the table.

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5. Method of artificial variables

So,

X2max* = (10, 3, 8, 0, 10) .

And the optimal solution of the initial problem is

fmax = −27 .

Example. Find the solution of the linear programming problem f = x1 +2x2 +9 → max

x1 +x2 −x3 =2,x1 +4x2 +x4 =1,−x1 +x2 −x5 = 3,

xi ≥ 0.

Solution.

We have the basis variable x4 in the second equation. Therefore, we need to add an artificial variable y1 and y2 in the first and third equations. We obtain

F = −y1 −y2 → maxx1 +x2 −x3 +y1 =2,

x1 +4x2 +x4 =1,

−x1 +x2 −x5 +y2 = 3,

f −x1 −2x2 = 9xi ≥ 0,yj ≥ 0.

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5. Method of artificial variables

We find from the main constraints

y1 =2 −x1 −x2 +x3,

y2 = 3 +x1 −x2 +x5.

Using these expressions, we have

F= −y1 −y2 = −(2 −x1 −x2 +x3 )−(3 +x1 −x2 +x5 )= ,

=−5 +2x2 −x3 −x5.

Or

F −2x2 +x3 +x5 = −5 .

And get the following simplex table.

We have only one negative estimation

γ2 = −2 < 0 .

So, the second column is a pivot column. From this column we obtain

Thus,

a22 = 4

is the pivot (x5 ↔ x2 ) .

After recalculation we have.

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5. Method of artificial variables

In the table all evaluations are non-negative, so

Fmax = −9 /2 < 0 .

It shows us that the initial linear programming problem is infeasible.

Questions for self-control

1)For which linear programming problems do we use the method of artificial variables?

2)How we use the artificial variables?

3)What is the artificial objective function?

4)How to determine using the method of artificial

is infeasible?

Exercises for independent work

Find the solution of the linear programming problems by using the method of artificial variables.

f = −x1 −x2 −4x3 +5x4 → max

1) 4x1 +x2 = 4,

12x1 +3x2 −x3 +x4 =12,xi ≥ 0.

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5. Method of artificial variables

f =2x1 −x2 −2x3 +x4 +12 → max

+4x2 +x3 =10,

2)7x1 +x3 +4x4 = 42,x1 +3x2 −x5 =14,x ≥ 0.

i−x1

f = x1 −3x2 −5x3 −x4 +6 → max

3)x1 +x2 =1,

2x1 +11x2 +12x3 +3x4 =14,

xi ≥ 0.

f = x3 +x4 +x5 → max2x1 +3x2 −x3 =23,

4)x1 −2x2 −x4 =1,2x1 −x2 +x5 =1,x ≥ 0.

i

f = x2 +x3 +x5 −14 → max

+4x2 −x4 = 9,

5)−3x1 +x2 +3x4 = 3,x1 +5x2 +x3 +2x4 = 4,x ≥ 0.

i2x1

f = −6x1 +x3 +3 → max

−x3 = 3,

6)5x2 +x3 +x4 = 30,3x1 −x3 −x5 = 6,x ≥ 0.

i3x1 −x2

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6.DUALITY

6.1.Symmetric Dual problems

Example. Consider the planning problem

f = CT X → max

AX ≤ B,

X ≥ 0.

where A = (aij )isthetechnologicalmatrix, C = (c1,c2, ,cn )T is the price vector, B = (b1,b2, ,bm )T –is the vector of supply of the resources, X = (x1,x2, ,xn )T is the production plan vector. The alternative of the production is the selling

pT B = BT p to be minimizing. Thus, we have the linear

programming problem

g = BT p → min

AT p ≥ C,p ≥ 0.

Definition. The linear programming problems

are called the symmetric dual problems.

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6. DUALITY

Example. Construct the dual problem for the linear programming problem

f = −10x1 +6x2 +3x3 +4 → max

8x1 −7x2 −2x3 ≤1,−9x1 +5x2 +2x3 ≤ 4,

x1 ≥ 0,x2 ≥ 0,x3 ≥ 0,

Solution.

6.2. Asymmetric Dual problems

Definition. The linear programming problems

are called the asymmetric dual problems

Remark. This formulation is consistent with the symmetric dual problems if the equation AX = B will be written as the system of inequalities

AX ≤ B,

AX ≥ B,

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6. DUALITY

or

AX ≤ B,

−AX ≤ −B.

Example. Construct the dual problem for the linear programming problem

z = 4x1 −5x2 +8x3 +1 → max

x1 −2x2 +7x3 = 3,−4x1 −7x2 +6x3 = −8,

x1 ≥ 0,x2 ≥ 0,x3 ≥ 0.

Solution.

f = 3y1 −8y2 +1 → min

y1 −4y2 ≥ 4,−2y1 −7y2 ≥ −5,

7y1 +6y2 ≥ 8.

6.3. General formulation of the dual problems

Let the primal linear programming is defined by the augmented matrix

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6. DUALITY

Then the dual problem has the matrix

Algorithm (general case)

1)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.

2)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.

3)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.

Example. Construct the dual problem for the linear programming problem

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