Линейное программирование. Практикум. Учебное пособие для бакалавриата

3. Geometric Method

Fig. 3

Similarly, in the case of a linear programming problem for a minimum, we have.

Algorithm of Geometrical method (minimum)

1)Build the feasible region.

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

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

4)The first level line touching of the feasible region is the point of minimum.

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3. Geometric Method

Example. Solve the linear programming problem.

f = 3x1 +x2 −10 → max(min)

7x1 +x2 ≤29,3x1 +2x2 ≥25,

4x1 −x2 ≤15,x1 ≥ 0, x2 ≥ 0.

Solution.

First let us find the feasible region. The boundaries are given by the following equations.

The vector n = (3;1)T is the gradient vector (the normal vector of the level line).

Fig. 4

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3. Geometric Method

The first level line touching of the feasible region is the point D (see the Figure 3). So D is the point of minimum. The point D is the intersection of the lines l2 and l3 . To find D, it is necessary to solve the system of linear equations.

3x1 +2x2 =25, .

4x1 −x2 =15.

We obtain

x1 = 4, D(4;1)

x2 =1.

The last level line touching of the feasible region is the point B (see the Figure 3). So, B is the point of maximum. The point B is the intersection of the lines l1 and l3 (B = l1 ∩l3 ). To find D, it is necessary to solve the system of linear

equations.

7x1 +x2 =29,

4x1 −x2 =15.

So

x1 = 4, B(5;5)

x2 =1.

Now substituting the coordinates of points D and B into the objective function, we find the maximum and minimum.

fmin = f(D) = 3x1 +x2 −10 = 3 4 +1−10 = 3 , fmax = f(B) = 3x1 +x2 −10 = 3 5 +5 −10 =10 .

Example. Solve the linear programming problem. f = −x1 +x2 +4 → min (max)

x1 −x2 +x3 = 3,−2x1 +x2 +x4 =2,

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0.

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3. Geometric Method

Solution. The problem is written in the canonicals form. To apply the geometric method, we need to transform the problem into a standard form. We have from the constraints

x3 = 3 −x1 +x2 ≥ 0,x4 =2 +2x1 −x2 ≥ 0,

By removing the basic variables x3, x4 from the system of main constraints, we get the linear programming problem written in standard form

f = −x1 +x2 +4 → min (max)

x1 −x2 ≤ 3,2x1 −x2 ≥ −2,x1 ≥ 0,x2 ≥ 0.

Let us find the feasible region. Its boundaries are given by the following equation

l1 :2x1 −x2 = −2, l2 : x1 −x2 = 3, l3 : x1 = 0,

l4 : x2 = 0.

The intersection of these lines belonging to the feasible region gives us the following corner points O(0;0); A(0;2);

B(3;0).

The gradient vector n = (−1;1)T (see the Figure 5) is perpendicular to the ray BD (n BD). So, the first intersection of the level line with the feasible region is the ray BD. Thus, all the points of the ray BD are the points of minimum.

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3. Geometric Method

Fig. 5

The vector m = (1;1)T is the direction vector of the ray

BD. So

Xmin = B +t m = (3,0)+t(1,1) = (3 +t,t), t ≥ 0.

Note that the initial problem has 4 variables. Let us find the basis variables:

x3 = 3 −x1 +x2 = 3 −(3 +t) +t = 0,x4 =2 +2x1 −x2 =2 +2(3 +t) −t = 8.

Thus, the point of minimum is

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3. Geometric Method

Exercises for independent work

Solve linear programming problems using the geometric method.

f= x1 +x2 → maxx1 −2x2 ≤2,

2)x1 −x2 ≥ −2,3x1 +x2 ≤18,x1 ≥ 0, x2 ≥ 0.

f= x1 +3x2 → min

x1 −x2 ≤ −1,

3)3x1 −x2 ≥ −3,2x1 +x2 ≥ 4,x1 ≥ 0, x2 ≥ 0.

f=2x1 +3x2 → max5x1 +3x2 ≤ 30,

4)x1 −3x2 ≥ 4,2x1 +3x2 ≥ −9,x1 ≥ 0, x2 ≥ 0.

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3. Geometric Method

f = x2 −x4 +1 → min (max)

x1 +2x2 +x3 = −6,

10)2x1 +x2 −x4 =25x1 +2x2 +x5 =5,

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0, x5 ≥ 0.

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