Линейное программирование. Ч.1. Учебное пособие для бакалавриата

4. Simplex method

Solution.

To use simplex method, we have to express the objective function in terms of basic variables. We get from the main constraints the expression for the basic variables

x3 =1+x1 −x2,x4 =1−x1 +x2,x5 =2−x1 −x2.

Substituting these expressions into the objective function we find

f =2x1 −x2 +3(1+x1 −x2) −2(1−x1 +x2) +(2−x1 −x2).

Or

f −6x1 +7x2 =3 .

As a result, we get the following simplex table.

We have the basic solution that is the starting corner point of the feasible region

X1 =(0, 0, 1, 1, 2).

And

f (X1 )=3 .

We have only one the negative evaluation

γ1 = −6 <0 .

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

min 1,2 =1 .

1 1

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4. Simplex method

Or

f −2x1 −3x2 =0 .

So, we find the following simplex table.

We have the basic solution

X1 =(0, 0, 3, 1). And the following values

f (X1 )=0 .

We have only one the negative evaluation

γ1 = −3 <0 .

Thus, we choose the first column as pivot column. In this column there is only one positive number a21 =3 and we use it as is a pivot.

After recalculation we have

In this table we have only one the negative evaluation

γ1 = −8 <0 .

But all numbers in the first column are negative. It means that the problem is unbounded.

Thus,

fmax = +∞.

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4. Simplex method

Exercises for independent work

Find the solution of the linear programming problems by using the simplex method.

5)2x1 +3x2 +x4 =18,2x1 −x2 +x5 =10,x ≥0.

i 1 2 3

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5. METHOD OF ARTIFICIAL VARIABLES

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). The method is also known as the two-phase simplex method.

Remark. All free terms in the system of equations of the main constraints must be non-negative.

b1 ≥0,

bm ≥0.

We have the following linear programming problem

f = c1x1 + +cxn +c0 →maxa11x1 + +a1nxn = b1,,

........................................

am1x1 + +amnxn = bmxi ≥0.

xi ≥0, yj ≥0.

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5. Method of artificial variables

The function F is called the artificial objective function. The variables y1, ,ym are called the artificial variables (or balance variables). They are the basic variables in new system of the main constraints. So, this problem has the basic solution

X1 =(0, ,0,b1,b2, ,bm )≥0

Thus, we can use the standard simplex method for the linear programming problem (5).

Remark. We add an artificial variable only in equation where there is no basic variable.

When we solve the problem (7), the following alternatives are possible.

1)Fmax <0 then the initial linear programming problem is infeasible.

2)Fmax =0 (it means that all artificial variables are equal to zero) then the initial linear programming problem is feasible, and we have found its basic solution. In this case, we can already apply the standard simplex method to the initial problem. That is, we can begin the second phase of solving the initial problem.

Example. Find the solution of the linear programming

problem

f = −x1 −3x2 −8 →max

−x1 +x2 +x3 =1,x1 +3x2 −x4 =19,3x1 +x2 +x5 =33,

x ≥0.

Solution.

We have the basis variables x3 and x5 in the first and third equations of the system of main constraints. Therefore, we need to add an artificial variable y1 only in the second equation. We get

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5. Method of artificial variables

From here we find that

y1 =191 −x1 −3x2 +x4 .

Substituting this expression into the objective function we get

F = −y1 = −19 +x1 +3x2 −x4 .

Or

F −x1 −3x2 +x4 = −19 .

As a result, we get the following simplex table.

We have the negative evaluation

γ2 = −3 <0 .

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

Thus,

a12 =1

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5. Method of artificial variables

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

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

Fmax =0

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5. Method of artificial variables

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

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