Question

Consider the following standard maximum problem: Maximize \(\quad \mathrm{u}=4 \mathrm{x}+2 \mathrm{y}+\mathrm{z}\) subject to: \(\quad \mathrm{x}+\mathrm{y} \leq 1\) \(\mathrm{x}+\mathrm{z} \leq 1\) and \(\mathrm{x} \geq 0, \mathrm{y} \geq 0, \mathrm{z} \geq 0\) Identify the basic feasible points (extreme points) of the constraint set. Determine which ones, if any are degenerate.

Short answer

The basic feasible points (extreme points) of the constraint set are A(0,1,0), B(1,0,0), C(0,0,1), and E(1,0,0). All of these points are degenerate.

Step-by-step solution

Sketch the constraints and determine the feasible region

To determine the feasible region, we can sketch the constraints on a 3D coordinate system. Since x, y, and z are non-negative, we only need to consider the first octant. The two main constraints are: 1. \(x + y \leq 1\) 2. \(x + z \leq 1\) We can also write them in the following forms: 1. \(y \leq 1 - x\) 2. \(z \leq 1 - x\) Now, we have a feasible region determined in the first octant by the intersection of these constraints.

Find the basic feasible points (extreme points)

To find the extreme points, we need to determine the vertices formed by the intersection of these constraints. Let's consider the following points of intersection: 1. Intersection of \(x + y = 1\) and \(x = 0\): Point A (0, 1, 0) 2. Intersection of \(x + y = 1\) and \(y = 0\): Point B (1, 0, 0) 3. Intersection of \(x + z = 1\) and \(x = 0\): Point C (0, 0, 1) 4. Intersection of \(x + z = 1\) and \(z = 0\): Point D (1, 0, 0) 5. Intersection of \(x + y = 1\) and \(x + z = 1\): Point E (1, 0, 0) So, we have the following extreme points: A(0,1,0), B(1,0,0), C(0,0,1), and E(1,0,0).

Check for degenerate points

A degenerate point is a basic feasible point that has fewer than n linearly independent binding constraints, where n is the number of variables (in this case, 3). In our problem, every binding constraint has the equation of a plane in 3D space. At each of our basic feasible points, the intersection of all the planes is a line segment, which has two linearly independent binding constraints (not three). This indicates that all of our basic feasible points are degenerate. In conclusion, the extreme points of the constraint set are A(0,1,0), B(1,0,0), C(0,0,1), and E(1,0,0), and all of them are degenerate points.

Key concepts

Feasible Region

When discussing linear programming, the term "feasible region" refers to the set of all possible points that satisfy the given constraints of the problem. These constraints are usually linear inequalities, and the feasible region is essentially where they all overlap. In this particular problem, we are considering a 3-dimensional space with variables \(x, y,\) and \(z\), all of which are constrained to be non-negative. The inequalities \(x + y \leq 1\) and \(x + z \leq 1\) further restrict the space.
The feasible region can be visualized as a portion of the first octant of the 3D coordinate system. The first octant is where all coordinate values are zero or positive. By plotting the constraints \(x + y \leq 1\) and \(x + z \leq 1\) in this space, we find where the solutions overlap, forming a polyhedron or a shape-like figure bounded by planes.
The feasible region includes every point where these conditions are true. Identifying this region is crucial, as it contains all potential solutions to the problem that you'll evaluate to find the optimal one.

Extreme Points

In linear programming, extreme points (or vertices) of the feasible region are crucial because they are candidates for the optimal solution. They are basically corners or intersections within the feasible region. For this exercise, extreme points represent possible solutions formed where the constraints intersect in the first octant.
To locate the extreme points, we observe where the inequalities \(x + y = 1\) and \(x + z = 1\) intersect with each other and the axes. Points like A(0,1,0), B(1,0,0), and C(0,0,1) arise along these intersections where one or two variables become zero, leading the constraint equations to determine the exact location in space.
Finding these intersections involves setting one or more variables to zero and solving the resulting equations. This information pinpoints the extreme corners of our feasible region. Each extreme point is checked to identify which one maximizes or minimizes the function we are dealing with. A crucial aspect here is that, in linear programming, if there is a solution, it'll often lie at one of these extreme points.

Degenerate Solutions

A degenerate solution in the context of linear programming refers to a situation where an extreme point of the feasible region does not correspond to a unique linear combination. In other words, it's a vertex where the solution is less 'rich' than expected, meaning it meets fewer independent constraints than the number of variables.
This can happen in our exercise because each basic feasible point should theoretically satisfy three constraints - one for each variable \(x, y, z\). However, at some points, fewer constraints are actively defining the vertex, making these solutions degenerate.
Degenerate solutions complicate algorithms and solution finding, as they might indicate that an alternative set of extremities or directions could reach the same optimal value, potentially leading to multiple solutions. In the exercise, points like A, B, C, and E are considered degenerate because they don't require all three constraints to define them fully, meaning fewer active constraints make them the resulting intersection in the feasible region.