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Feasible Solutions: Definition and Properties Infeasible Solutions: Definition Optimal Feasible Solution: Definition
Vertices (Corner Points) of the Feasible Region


Solutions of an LPP: Feasible and Optimal



Feasible Solutions: Definition and Properties

Definition of Feasible Solution

A Feasible Solution to a Linear Programming Problem (LPP) is a set of values for the decision variables, say $x_1, x_2, \ldots, x_n$, that satisfies all the constraints of the problem simultaneously. This includes both the structural constraints (inequalities or equations representing resource limitations, requirements, etc.) and the non-negativity restrictions (if applicable, typically $x_i \geq 0$ for all $i$).

In the context of a two-variable LPP (with decision variables $x$ and $y$), a feasible solution corresponds to any point $(x, y)$ in the coordinate plane that lies within or on the boundary of the feasible region. The feasible region itself is the set of all such feasible solutions.


Properties of Feasible Solutions

Graph of a feasible region with several points marked inside and on the boundary, labeled as 'Feasible Solutions'.

The diagram above illustrates a typical feasible region (the shaded area). Any point located within this shaded area or along its boundary lines is considered a feasible solution because its coordinates $(x, y)$ satisfy all the linear inequalities that define the region.


Example of a Feasible Solution

Example 1. Consider the following LPP:

Maximize $Z = 2x + 3y$

Subject to the constraints:

$x + y \leq 5$

$x \leq 4$

$y \geq 1$

$x \geq 0, y \geq 0$

Determine if the point $(2, 2)$ is a feasible solution for this LPP.

Answer:

To check if the point $(2, 2)$ is a feasible solution, we must substitute $x=2$ and $y=2$ into each of the given constraints and verify if all constraints are satisfied.

The constraints are:

$x + y \leq 5$

... (1)

$x \leq 4$

... (2)

$y \geq 1$

... (3)

$x \geq 0, y \geq 0$

... (4)

Substitute $x=2$ and $y=2$ into each constraint:

  • For constraint (1): $2 + 2 = 4$. Since $4 \leq 5$, constraint (1) is satisfied.
  • For constraint (2): $x = 2$. Since $2 \leq 4$, constraint (2) is satisfied.
  • For constraint (3): $y = 2$. Since $2 \geq 1$, constraint (3) is satisfied.
  • For constraint (4): $x = 2, y = 2$. Since $2 \geq 0$ and $2 \geq 0$, constraint (4) is satisfied.

Since the point $(2, 2)$ satisfies all the constraints, it is a feasible solution for the given LPP.


Infeasible Solutions: Definition

Definition of Infeasible Solution

An Infeasible Solution to a Linear Programming Problem (LPP) is a set of values for the decision variables ($x_1, x_2, \ldots, x_n$) that violates at least one of the constraints or non-negativity restrictions specified in the problem formulation.

Graphically, for a two-variable LPP, an infeasible solution corresponds to any point $(x, y)$ in the coordinate plane that lies outside the feasible region. Such a point does not satisfy the conditions imposed by one or more constraints, making it an invalid or impossible solution in the context of the problem.

Graph showing a feasible region and several points marked outside it, labeled as 'Infeasible Solutions'.

These points represent combinations of decision variable values that cannot be achieved under the given circumstances or limitations, such as exceeding available resources or failing to meet minimum production quotas.


Distinction: Infeasible Solution vs. Infeasible LPP

It is crucial to differentiate between an "infeasible solution" and an "infeasible LPP":


Example of an Infeasible Solution

Example 1. Consider the same LPP as before:

Maximize $Z = 2x + 3y$

Subject to the constraints:

$x + y \leq 5$

$x \leq 4$

$y \geq 1$

$x \geq 0, y \geq 0$

Determine if the point $(6, 1)$ is a feasible solution for this LPP.

Answer:

To check if the point $(6, 1)$ is a feasible solution, we substitute $x=6$ and $y=1$ into each of the given constraints.

The constraints are:

$x + y \leq 5$

... (1)

$x \leq 4$

... (2)

$y \geq 1$

... (3)

$x \geq 0, y \geq 0$

... (4)

Substitute $x=6$ and $y=1$ into each constraint:

  • For constraint (1): $6 + 1 = 7$. Since $7$ is not $\leq 5$, constraint (1) is violated.
  • For constraint (2): $x = 6$. Since $6$ is not $\leq 4$, constraint (2) is violated.
  • For constraint (3): $y = 1$. Since $1 \geq 1$, constraint (3) is satisfied.
  • For constraint (4): $x = 6, y = 1$. Since $6 \geq 0$ and $1 \geq 0$, constraint (4) is satisfied.

Since the point $(6, 1)$ violates constraints (1) and (2), it is an infeasible solution for the given LPP.


Example of an Infeasible LPP

Example 2. Consider the following LPP:

Minimize $Z = x + y$

Subject to the constraints:

$x + y \leq 5$

$x + y \geq 7$

$x \geq 0, y \geq 0$

Is this LPP feasible?

Answer:

We need to find if there exists any point $(x, y)$ that satisfies all the given constraints simultaneously.

The constraints are:

$x + y \leq 5$

... (1)

$x + y \geq 7$

... (2)

$x \geq 0, y \geq 0$

... (3)

A feasible solution must satisfy both constraint (1) and constraint (2).

Constraint (1) requires that the sum of $x$ and $y$ must be less than or equal to 5.

Constraint (2) requires that the sum of $x$ and $y$ must be greater than or equal to 7.

It is impossible for a number ($x+y$) to be simultaneously less than or equal to 5 AND greater than or equal to 7. These two conditions are contradictory.

Therefore, there is no point $(x, y)$ that can satisfy both $x + y \leq 5$ and $x + y \geq 7$ at the same time, regardless of the non-negativity constraints.

The set of points satisfying all constraints is empty. Hence, the feasible region is empty, and the LPP is **infeasible**.

Graphical Interpretation:

Graphing the constraints $x+y \leq 5$ (region below the line $x+y=5$) and $x+y \geq 7$ (region above the line $x+y=7$) along with $x \geq 0, y \geq 0$ would show that the region satisfying $x+y \leq 5$ and the region satisfying $x+y \geq 7$ are disjoint (they do not overlap). There is no common area satisfying both, resulting in an empty feasible region.



Optimal Feasible Solution: Definition

Definition of Optimal Feasible Solution

An Optimal Feasible Solution (often simply called the Optimal Solution) to a Linear Programming Problem (LPP) is a specific feasible solution that yields the best possible value for the objective function. "Best possible" means the maximum value if the objective is to maximize, and the minimum value if the objective is to minimize.

Formally, a feasible solution $(x_1^*, x_2^*, \ldots, x_n^*)$ is an optimal feasible solution if:

  1. It is a feasible solution, meaning it satisfies all constraints and non-negativity restrictions of the LPP.
  2. The value of the objective function, $Z(x_1^*, x_2^*, \ldots, x_n^*)$, is the optimal value. This means:
    • If the objective is to maximize $Z$, then $Z(x_1^*, \ldots, x_n^*) \geq Z(x_1, \ldots, x_n)$ for all other feasible solutions $(x_1, \ldots, x_n)$.
    • If the objective is to minimize $Z$, then $Z(x_1^*, \ldots, x_n^*) \leq Z(x_1, \ldots, x_n)$ for all other feasible solutions $(x_1, \ldots, x_n)$.

In essence, among all the combinations of decision variable values that are allowed by the constraints (i.e., all feasible solutions), the optimal feasible solution is the one that achieves the specified goal (maximum profit, minimum cost, etc.).


Properties and Location of Optimal Solutions

The primary goal in solving an LPP is to identify this optimal feasible solution(s) and determine the corresponding optimal value of the objective function Z.


Example of an Optimal Feasible Solution

Example 1. Consider an LPP aiming to maximize $Z = 3x + 2y$ subject to certain constraints that define a feasible region. Suppose the feasible region has vertices at points A(0,0), B(4,0), C(2,3), and D(0,2). Evaluate the objective function at point C(2,3) and explain why it *could* be the optimal feasible solution.

Answer:

To evaluate the objective function $Z = 3x + 2y$ at the point C(2,3), we substitute $x=2$ and $y=3$ into the expression for $Z$:

$Z$ at C(2,3) $= 3(2) + 2(3)$

$= 6 + 6$

$= 12$

The value of the objective function at point C(2,3) is 12.

Point C(2,3) is given as a vertex of the feasible region. According to the Fundamental Theorem of Linear Programming, if an optimal solution exists for this maximization problem, it must occur at one of the vertices of the feasible region.

While we have only evaluated the objective function at one vertex (C), the process to find the *actual* optimal feasible solution involves evaluating $Z$ at all the vertices (A, B, C, and D in this case) and comparing the resulting values. The vertex that gives the highest value of $Z$ among all vertices is the optimal feasible solution, and that highest value is the optimal value.

Therefore, point C(2,3) is a candidate for the optimal feasible solution because it is a vertex of the feasible region. Whether it *is* the optimal solution depends on the values of $Z$ obtained at the other vertices (A, B, and D).

To confirm if C(2,3) is the optimal solution, we would evaluate Z at the other vertices:

  • At A(0,0): $Z = 3(0) + 2(0) = 0$
  • At B(4,0): $Z = 3(4) + 2(0) = 12$
  • At D(0,2): $Z = 3(0) + 2(2) = 4$

Comparing the values (0, 12, 12, 4), the maximum value of Z is 12, which occurs at both B(4,0) and C(2,3). Thus, both B(4,0) and C(2,3) are optimal feasible solutions, and the optimal value of Z is 12. This illustrates a case of multiple optimal solutions occurring at adjacent vertices.


Vertices (Corner Points) of the Feasible Region

Definition of Vertices (Corner Points)

A vertex or corner point of the feasible region in a graphical representation of a two-variable LPP is a point in the coordinate plane where two or more boundary lines of the feasible region intersect. These points are the "corners" of the polygonal feasible region (if bounded) or the corner points bordering an unbounded feasible region.

Each constraint inequality in an LPP defines a half-plane. The feasible region is the intersection of all these half-planes (including those from non-negativity restrictions like $x \geq 0$ and $y \geq 0$). The boundary of each half-plane is a straight line. The vertices of the feasible region are precisely the points where these boundary lines intersect, provided the intersection point lies within or on the boundary of the feasible region defined by all other constraints.

Graph of a feasible region polygon with its vertices (corner points) clearly marked.

In the image above, the shaded area represents the feasible region. The points where the lines forming its boundary meet are the vertices.


Method for Finding Vertices

Vertices are determined by finding the points of intersection of the linear equations that form the boundaries of the feasible region. For a two-variable LPP, the steps typically involve:

  1. Convert each inequality constraint into a linear equation by replacing the inequality sign ($\leq, \geq, <, >$) with an equality sign ($=$).
  2. Identify pairs of these linear equations whose intersection points could potentially be vertices of the feasible region. These pairs usually correspond to lines that form adjacent boundaries of the feasible region.
  3. Solve the system of equations for each chosen pair to find the coordinates $(x, y)$ of the intersection point. Standard methods like substitution or elimination can be used.
  4. For each intersection point found, verify if it satisfies all the original constraints (including non-negativity). If a point satisfies all constraints, it is a vertex of the feasible region. Intersection points that lie outside the feasible region (violating one or more constraints) are not vertices of the feasible region, even though they are intersection points of the boundary lines.

Significance: The Corner Point Theorem

Vertices are of paramount importance in Linear Programming because of the **Fundamental Theorem of Linear Programming**, also known as the **Corner Point Theorem**. It states:

Theorem: If an optimal solution (maximum or minimum value of the objective function) exists for a Linear Programming Problem with a feasible region that is a convex polygon, then the optimal solution must occur at one (or more) of the vertices (corner points) of the feasible region.

This powerful theorem provides the basis for the widely used **Corner Point Method** (or Extreme Point Method) for solving LPPs, especially graphically. Instead of evaluating the objective function at every single point in the feasible region (which is impossible as there are infinitely many), the theorem tells us we only need to check the finite number of vertices. The optimal value must lie among the values calculated at these corners.

The steps derived from this theorem for solving a two-variable LPP graphically are:

  1. Graph the feasible region.
  2. Identify and find the coordinates of all the vertices of the feasible region.
  3. Evaluate the objective function $Z$ at each vertex.
  4. For a maximization problem, the vertex yielding the largest value of $Z$ is the optimal feasible solution. For a minimization problem, the vertex yielding the smallest value of $Z$ is the optimal feasible solution. The corresponding value of $Z$ is the optimal value.

Note: For unbounded feasible regions, an optimal solution may or may not exist. If it exists, it will still be at a corner point.


Example of Finding Vertices

Example 1. Find the vertices of the feasible region defined by the constraints:

$x + y \leq 4$

$x \geq 0$

$y \geq 0$

Answer:

First, we convert the inequalities into boundary line equations:

$x + y = 4$

... (1)

$x = 0$ (y-axis)

... (2)

$y = 0$ (x-axis)

... (3)

The feasible region is bounded by these lines in the first quadrant (due to $x \geq 0, y \geq 0$). The vertices are the intersection points of these lines that satisfy all the original inequalities.

We find the intersection points of these pairs of lines:

Intersection of (2) and (3):

Line (2) is $x=0$. Line (3) is $y=0$. The intersection is the origin (0,0).

Check if (0,0) is feasible:

  • $x + y \leq 4$: $0 + 0 = 0 \leq 4$ (Satisfied)
  • $x \geq 0$: $0 \geq 0$ (Satisfied)
  • $y \geq 0$: $0 \geq 0$ (Satisfied)

Since all constraints are satisfied, (0,0) is a vertex.

Intersection of (2) and (1):

Line (2) is $x=0$. Line (1) is $x+y=4$. Substitute $x=0$ into $x+y=4$:

$0 + y = 4$

$y = 4$

The intersection point is (0,4).

Check if (0,4) is feasible:

  • $x + y \leq 4$: $0 + 4 = 4 \leq 4$ (Satisfied)
  • $x \geq 0$: $0 \geq 0$ (Satisfied)
  • $y \geq 0$: $4 \geq 0$ (Satisfied)

Since all constraints are satisfied, (0,4) is a vertex.

Intersection of (3) and (1):

Line (3) is $y=0$. Line (1) is $x+y=4$. Substitute $y=0$ into $x+y=4$:

$x + 0 = 4$

$x = 4$

The intersection point is (4,0).

Check if (4,0) is feasible:

  • $x + y \leq 4$: $4 + 0 = 4 \leq 4$ (Satisfied)
  • $x \geq 0$: $4 \geq 0$ (Satisfied)
  • $y \geq 0$: $0 \geq 0$ (Satisfied)

Since all constraints are satisfied, (4,0) is a vertex.

Other potential intersection points (e.g., intersection of lines from constraints not adjacent in the feasible region, or intersection points outside the first quadrant) would be checked but found to violate at least one constraint, and thus would not be vertices of the feasible region.

The vertices of the feasible region are (0,0), (0,4), and (4,0).