9.2 Linear Programming---Geometric Method (Linear Programming---Geometric Method)

This article is the reading notes of "Linear algebra and its applications"

Linear Programming Problem

Generally speaking, a linear programming problem involves a system of linear inequalities in variables $x_1,...,x_n$ and a linear functional $f$ from ${\mathbb{R}}^n$ into $\mathbb{R}$. The system typically has many free variables, and the problem is to find a solution $\mathbf{x}$ that maximizes or minimizes $f(\mathbf{x})$.

EXAMPLE 1
The Shady-Lane grass seed company blends two types of seed mixtures, EverGreen and QuickGreen. Each bag of EverGreen contains 3 pounds of fescue seed, 1 pound of rye seed, and 1 pound of bluegrass. Each bag of QuickGreen contains 2 pounds of fescue, 2 pounds of rye, and 1 pound of bluegrass. The company has 1200 pounds of fescue seed, 800 pounds of rye seed, and 450 pounds of bluegrass available to put into its mixtures. The company makes a profit of $2 on each bag of EverGreen and $3 on each bag of QuickGreen that it produces. Set up the mathematical problem that determines the number of bags of each mixture that Shady-Lane should make in order to maximize its profit.
SOLUTION
Let $x_1$ be the number of bags of EverGreen and $x_2$ the number of bags of QuickGreen that are produced. The problem is summarized mathematically as

The example shows how a linear programming problem involves finding the maximum (or minimum) of a linear function, called the objective function, subject to certain linear constraints. In many situations, the constraints take the form of linear inequalities and the variables are restricted to nonnegative values.

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Here is a precise statement of the so-called canonical(最简洁的) form of a linear programming problem.

To minimize a function $h(\mathbf{x})$, replace it with the problem of maximizing the function $-h(\mathbf{x})$. A constraint inequality of the sort

can be replaced by

An equality constraint

can be replaced by two inequalities

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With an arbitrary canonical linear programming problem, two things can go wrong.

• If the constraint inequalities are inconsistent, then $\mathcal{F}$ is the empty set.
• If the objective function takes on arbitrarily large values in $\mathcal{F}$, then the desired maximum does not exist.

In the former case, the problem is said to be infeasible; in the latter case, the problem is called unbounded.

Fortunately, these are the only two things that can go wrong.

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See Section 8.5: The implicit definition of polytopes and THEOREM 16

Geometric Method

Theorem 6 describes when an optimal solution exists, and it suggests a possible technique for finding one. That is, evaluate the objective function at each of the extreme points of $\mathcal{F}$ and select the point that gives the largest value.

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This works well in simple cases such as the next two examples. The geometric approach is limited to two or three dimensions, but it provides an important visualization of the nature of the solution set and how the objective function interacts with the feasible set to identify extreme points.

EXAMPLE 5

SOLUTION
Figure 1 shows the shaded feasible set. (For simplicity, points in this section are displayed as ordered pairs or triples.)

There are five extreme points, corresponding to the five vertices of the feasible set. The table below shows the value of the objective function at each extreme point. Evidently, the maximum is $96$ at $x_1=30$ and $x_2=12$.

Another geometric technique that can be used when the problem involves two variables is to graph several level lines for the objective function. These are parallel lines, and the objective function has a constant value on each line. (See Fig. 2.) The values of the objective function $f(x_1,x_2)$ increase as $(x_1,x_2)$ moves from left to right. The level line farthest to the right that still intersects the feasible set is the line through the vertex $(30,12)$. Thus, the point $(30,12)$ yields the maximum value of $f(x_1,x_2)$ over the feasible set.

EXAMPLE 6

SOLUTION
Each of the five inequalities above determines a “half-space” in ${\mathbb{R}}^3$—a plane together with all points on one side of the plane. The feasible set of this linear programming problem is the intersection of these half-spaces, which is a convex set in the first octant(卦限) of $R^3$.

When the first inequality is changed to an equality, the graph is a plane that intercepts each coordinate axis 50 units from the origin and determines the equilateral(等边的) triangular region shown in Fig. 3.

Since $(0,0,0)$ satisfies the inequality, so do all the other points “below” the plane. In a similar fashion, the second (in)equality determines a triangular region on a plane (shown in Fig. 4) that passes somewhat closer to the origin. The two planes intersect in a line that contains segment $EB$.

The quadrilateral surface $BCDE$ forms a boundary of the feasible set, because it is below the equilateral triangular region. Beyond $EB$, however, the two planes change position relative to the origin, so the planar region $ABE$ forms another bounding surface for the feasible set. The vertices of the feasible set are the points $A,B,C,D,E$, and $0$ (the origin). See Fig. 5.

To find the coordinates of $B$ , solve the system

Obtain $x_2=30$, and find that $B$ is $(20,30,0)$. For $And$ , solve

Obtain $x_3=10$, and find that $E=(40,0,10)$.

Now that the feasible set and its extreme points are clearly seen, the next step is to examine the objective function $f(x_1,x_2,x_3)=2x_1+3x_2+4x_3$. The sets on which $f$ is constant are planes, all having $(2,3,4)$ as a normal vector to the plane. This normal vector has a direction different from the normal vectors $(1,1,1)$ and $(1,2,4)$ to the two faces $BCDE$ and $ABE$ . So the level sets of $f$ are not parallel to any of the bounding surfaces of the feasible set. Figure 6 shows just the feasible set and a level set on which $f$ has the value $120$. This plane passes through $C$ ,$E$, and the point $(30,20,0)$ on the edge of the feasible set between $A$ and $B$, which shows that the vertex $B$ is “above” this level plane. In fact, $f(20,30,0)=130$. Thus the unique solution of the linear programming problem is at $B=(20,30,0)$.