Problem 16 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Construct the dual problem associated with the primal problem. Solve the primal problem. $$ \begin{array}{ll} \text { Minimize } & C=40 x+30 y+11 z \\ \text { subject to } & 2 x+y+z \geq 8 \\ & x+y-z \geq 6 \\ x & \geq 0, y \geq 0, z \geq 0 \end{array} $$

Step-by-Step Solution

Answer

In summary, we have constructed the dual problem associated with the given primal problem as follows: Maximize \(D(u_1, u_2) = 8u_1 + 6u_2\) subject to: (1) \(2u_1 + u_2 \leq 40\) (2) \(u_1 + u_2 \leq 30\) (3) \(u_1 - u_2 \leq 11\) (4) \(u_1, u_2 \geq 0\) However, due to the three-dimensional nature of the primal problem, the graphical method is not suitable for solving the primal problem and one can consider using the simplex method instead.

Step 1: Identify the Objective Function and Constraints

The given primal problem is: Minimize \(C = 40x + 30y + 11z\) subject to: (1) \(2x + y + z \geq 8\) (2) \(x + y - z \geq 6\) (3) \(x \geq 0\) (4) \(y \geq 0\) (5) \(z \geq 0\)

Step 2: Construct the Dual Problem

To construct the dual problem, we will start by introducing dual variables \(u_1\) and \(u_2\) for inequalities (1) and (2), respectively. Moreover, the primal problem is a minimization problem, so the dual problem will be a maximization problem. The dual problem will be: Maximize \(D(u_1, u_2) = 8u_1 + 6u_2\) subject to: (1) \(2u_1 + u_2 \leq 40\) (2) \(u_1 + u_2 \leq 30\) (3) \(u_1 - u_2 \leq 11\) (4) \(u_1, u_2 \geq 0\)

Step 3: Solve the Primal Problem using Graphical Method

To solve the primal problem using the graphical method, we need to plot the feasible region for the primal problem. As the primal problem is three-dimensional, it is difficult to solve the problem using the graphical method. In this case, one way to make the problem more accessible is by using the simplex method. Alternatively, if we are only interested in the construction of the dual problem, we have already done that in Step 2.

Key Concepts

Dual Problem

In linear programming, when you have a primal problem, you can construct a corresponding dual problem. This dual problem often helps in gaining insights into the original problem and can sometimes be easier to solve. If the primal is a minimization problem, its dual will be a maximization problem.
In the given exercise, since the primal problem involves minimizing the objective function, the dual problem is constructed to maximize a new function. Here's how it works:

• Each constraint in the primal problem becomes a variable in the dual problem. In the exercise, two constraints lead to dual variables: \( u_1 \) and \( u_2 \).
• The coefficients from the primal constraints become the right-hand side values of the dual constraints.
• The objective function's coefficients in the primal problem become the coefficients in the dual constraints.

The solution involves setting up the dual problem using these rules, transforming the primal's inequality expressions into maximization expressions that use \( u_1 \) and \( u_2 \).

Primal Problem

The primal problem in linear programming implies that we are dealing with the original optimization problem. In this case, it revolves around the idea of minimizing an objective function subject to several constraints.
In our exercise, the primal problem is given by:

• Objective: Minimize \( C = 40x + 30y + 11z \)
• Constraints: Three inequalities that involve variables \( x, y, z \).

All variables are non-negative, they cannot be less than zero (\( x, y, z \geq 0 \)). Primal problems can be solved using different methods, such as the graphical method or simplex method depending on the number of dimensions involved. The primal problem here is three-dimensional, making it challenging to solve graphically, and often necessitates advanced methods such as the simplex algorithm.

Objective Function

The objective function in linear programming indicates what needs to be optimized, which could mean either maximizing or minimizing its value. For our exercise, the aim is to minimize the objective function.
The exercise's objective function is given by:

• \( C = 40x + 30y + 11z \).

This function is expressed in terms of decision variables: \( x, y \), and \( z \). Each variable is usually associated with a coefficient that represents its "cost" (for minimization) or "profit" (for maximization).
For solving, you'd adjust the values of these variables, subject to the constraints, to find the smallest possible value of \( C \). This is crucial since optimizing the objective function often underlies the main goal of the problem.

Constraints

Constraints are crucial in linear programming problems as they define the feasible region. This is the area where all possible solutions must lie. In essence, they restrict the values that decision variables can take, forming the boundaries of the problem.
In the given exercise, the constraints are:

• \(2x + y + z \geq 8\)
• \(x + y - z \geq 6\)
• Non-negativity: \(x, y, z \geq 0\)

These inequalities limit the potential values of \( x, y, \) and \( z \) in the primal problem. In the context of creating the dual problem, these conditions are translated into different forms, which will guide the bounds for dual variables \( u_1 \) and \( u_2 \).
Handling constraints accurately is important because any violation can lead to infeasibility, where no solution satisfies all constraints simultaneously.