Problem 14 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Construct the dual problem associated with the primal problem. Solve the primal problem. $$ \begin{aligned} \text { Minimize } & C=10 x &+y \\ \text { subject to } & 4 x+y & \geq 16 \\ x+2 y & \geq 12 \\ x & \geq 2 \\ x & \geq 0, y & \geq 0 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the primal problem is \( x = 2 \) and \( y = 4 \), resulting in a minimum value of \( C = 24 \). The corresponding dual problem is to maximize \( D = 16p_1 + 12p_2 + 2p_3 \) subject to: \( \begin{aligned} 4p_1 + p_2 + p_3 \leq 10 \\ p_1 + 2p_2 \leq 1 \\ p_1, p_2, p_3 \geq 0 \end{aligned} \).

Step 1: Converting primal problem to canonical form

The primal problem given is in the inequality form. In order to construct the canonical form, we need to convert these inequalities into equalities using non-negative slack variables. The given primal problem is: \( \begin{aligned} \text{Minimize } & C=10 x + y \\\ \text{subject to } & 4 x+y & \geq 16 \\\ x+2 y & \geq 12 \\\ x & \geq 2 \\\ x & \geq 0, \, y \geq 0 \end{aligned} \) To convert this into canonical form, introduce slack variables, \( s_1, s_2, s_3 \geq 0 \), such that: \( \begin{aligned} 4 x + y + s_1 &= 16 \\ x + 2 y + s_2 &= 12 \\ x - s_3 &= 2 \\ x,y,s_1,s_2,s_3 &\geq 0 \end{aligned} \) Now the problem is in canonical form and we can construct the dual problem.

Step 2: Construct the dual problem

For the resulting canonical primal problem, the dual problem will be as follows: Maximize \( D = 16p_1 + 12p_2 + 2p_3 \), subject to: \( \begin{aligned} 4p_1 + p_2 + p_3 \leq 10 \\ p_1 + 2p_2 \leq 1 \\ p_1, p_2, p_3 \geq 0 \end{aligned} \)

Step 3: Solving the primal problem

Using the graphical method, we will first represent each constraint as an equation by replacing the inequality signs with equal signs: 1. \( 4x + y = 16 \) 2. \( x + 2y = 12 \) 3. \( x = 2 \) Now we find the feasible region satisfying the constraints. After plotting the constraints on a graph, we find the feasible region with vertices (2,6), (2,4), and (4,4). Now we evaluate the objective function at these vertices to find the minimum value: 1. \( C(2,6) = 10(2) + 6 = 26 \) 2. \( C(2,4) = 10(2) + 4 = 24 \) 3. \( C(4,4) = 10(4) + 4 = 44 \) Since the objective is to minimize \( C \), we can see that the point (2,4) provides the smallest value for \( C \), which is 24. Therefore, the solution to the primal problem is \( x = 2 \) and \( y = 4 \), resulting in a minimum value of \( C = 24 \).

Key Concepts

Primal Problem

A primal problem in linear programming is the original problem that needs to be solved. Typically, it involves finding the maximum or minimum value of a linear objective function subject to a set of linear inequalities or equalities, known as constraints. In this exercise, the primal problem is about minimizing the cost function, given as \( C = 10x + y \). The constraints for the primal problem are:

• \( 4x + y \geq 16 \)
• \( x + 2y \geq 12 \)
• \( x \geq 2 \)
• \( x \geq 0, y \geq 0 \)

These constraints define the feasible region within which the solution must lie. Solving the primal problem involves finding the values for \( x \) and \( y \) that minimize \( C \), while satisfying all these constraints. The primal problem is often solved using methods like the graphical method, the simplex method, or computational optimization software.

Dual Problem

The dual problem corresponds to the primal problem and is constructed by converting the constraints and objective function according to specific rules. The significance of the dual problem is shown through the strong duality theorem, which states that if the primal has an optimal solution, so does the dual, and their objective values are equal. In this exercise, the dual problem is:
Maximize \( D = 16p_1 + 12p_2 + 2p_3 \), subject to:

• \( 4p_1 + p_2 + p_3 \leq 10 \)
• \( p_1 + 2p_2 \leq 1 \)
• \( p_1, p_2, p_3 \geq 0 \)

The dual problem flips the original objectives, shifting from minimization to maximization, and changes the constraints to inequality constraints in the opposite direction. Variables in the primal become coefficients for the objective function in the dual, and vice versa. Solving the dual can provide bounds to the primal problem or even an alternate way to achieve the solution.

Canonical Form

In linear programming, placing a linear program in canonical form makes it easier to analyze and solve. The canonical form requires all constraints to be equalities rather than inequalities, facilitated by adding slack variables. The canonical form for the given primal problem uses these slack variables:

• \( 4x + y + s_1 = 16 \)
• \( x + 2y + s_2 = 12 \)
• \( x - s_3 = 2 \)

The introduction of slack variables \( s_1, s_2, s_3 \geq 0 \) transforms the inequality constraints into equality constraints, while ensuring non-negativity. This form is especially critical when applying methods like the Simplex algorithm, which requires a start from the feasible region's corner (or vertex) and proceeds iterating towards optimization.

Slack Variables

Slack variables are added to linear programming problems to convert inequality constraints into equality ones, easing the process of finding a solution. For instance, consider the constraint \( 4x + y \geq 16 \). To turn this into an equality, we introduce a new variable \( s_1 \), leading to the equation \( 4x + y + s_1 = 16 \), where \( s_1 \geq 0 \).
Slack variables absorb the difference between the left and right sides of the inequality, indicating how much a constraint is not tight. They are always non-negative when solving a minimization problem. In our problem:

• \( s_1 \) corresponds to \( 4x + y \geq 16 \)
• \( s_2 \) for \( x + 2y \geq 12 \)
• \( s_3 \) for \( x \geq 2 \)

Through slack variables, each inequality constraint becomes an equality, making it possible to employ algebraic methods like the Simplex Method and to analyze the solutions more effectively by breaking them down into compositions of linear equations.