Problem 17 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Construct the dual problem associated with the primal problem. Solve the primal problem. $$ \begin{aligned} \text { Minimize } & C=6 x+8 y+4 z \\ \text { subject to } & x+2 y+2 z \geq 10 \\ & 2 x+y+z \geq 24 \\ & x+y+z \geq 16 \\ x & \geq 0, y \geq 0, z \geq 0 \end{aligned} $$

Step-by-Step Solution

Answer

The optimal solution to the primal problem is \(x = 0, y = 8, z = 0\), and the minimized cost is \(C = 64\). The associated dual problem is the maximization problem: Maximize \(D = 10p_1 + 24p_2 + 16p_3\) subject to: \(p_1 + 2p_2 + p_3 \leq 6\) \(2p_1 + p_2 + p_3 \leq 8\) \(2p_1 + p_2 + p_3 \leq 4\) with \(p_1, p_2, p_3 \geq 0\).

Step 1: Rewrite as canonical problem

We first rewrite the inequality constraints as equality constraints by introducing slack variables s1, s2, and s3. Minimize \(C = 6x + 8y + 4z\) subject to: \(x + 2y + 2z + s_1 = 10\) \(2x + y + z + s_2 = 24\) \(x + y + z + s_3 = 16\) with \(x, y, z, s_1, s_2, s_3 \geq 0\).

Step 2: Create the dual problem

Now we create the dual problem. The primal problem is a minimization problem, so the dual will be a maximization problem. Let \(p_1, p_2\), and \(p_3\) be the dual variables corresponding to the primal constraints. Then the dual problem will look like: Maximize \(D = 10p_1 + 24p_2 + 16p_3\) subject to: \(p_1 + 2p_2 + p_3 \leq 6\) \(2p_1 + p_2 + p_3 \leq 8\) \(2p_1 + p_2 + p_3 \leq 4\) with \(p_1, p_2, p_3 \geq 0\).

Step 3: Solve the primal problem

Now, let's solve the primal problem using simplex method. First, we arrange the simplex tableau: | s1 | s2 | s3 | x | y| z| C | RHS | |----|----|----|---|---|---|---|-----| | 1 | 0 | 0 | 1 | 2 | 2 | 0 | 10 | | 0 | 1 | 0 | 2 | 1 | 1 | 0 | 24 | | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 16 | | 0 | 0 | 0 |-6 |-8 |-4 | 1 | 0 | We can easily see that the last row has all negative values, so the tableau is optimal, and the current solution is the optimal solution to the primal problem. The primal optimal solution is: \(x = 0, y = 8, z = 0\) with \(C = 6x + 8y + 4z = 6(0) + 8(8) + 4(0) = 64\). Thus, the optimal solution to the primal problem is \(x = 0, y = 8, z = 0\), and the minimized cost is \(C = 64\).

Key Concepts

Primal Problem

The primal problem is the original problem in linear programming tasked with finding the minimum or maximum value of an objective function. In our exercise, we are required to minimize the cost function: \( C = 6x + 8y + 4z \).
The solution space is subject to constraints defined by inequalities:

• \( x + 2y + 2z \geq 10 \)
• \( 2x + y + z \geq 24 \)
• \( x + y + z \geq 16 \)
• \( x, y, z \geq 0 \)

To convert it into a canonical form suitable for solving, slack variables \(s_1, s_2,\) and \(s_3\) are introduced to turn these inequalities into equalities so that each constraint becomes a direct solution point in the space:

• \( x + 2y + 2z + s_1 = 10 \)
• \( 2x + y + z + s_2 = 24 \)
• \( x + y + z + s_3 = 16 \)

Slack variables  take on values that adjust each constraint to equality, helping in tracking resource surpluses. Each slack variable is also constrained to be non-negative \( (s_1, s_2, s_3 \geq 0) \) just like the main variables \( (x, y, z \geq 0) \).

Dual Problem

The dual problem corresponds to the primal problem but swaps the roles of the constraints and the objective. In linear programming, the primal and dual problems are closely related, and one's solution provides valuable insights into the other.
For our minimization primal problem, the dual becomes a maximization problem. Dual variables \( p_1, p_2, \) and \( p_3 \) are introduced as the coefficients of the constraints from the primal problem. Breaking the primal constraints establishes the dual's objective function and constraints:

• Maximize \( D = 10p_1 + 24p_2 + 16p_3 \)

The inequalities forming constraints in the dual are:

• \( p_1 + 2p_2 + p_3 \leq 6 \)
• \( 2p_1 + p_2 + p_3 \leq 8 \)
• \( 2p_1 + p_2 + p_3 \leq 4 \)

Reflecting reversed roles, the dual problem highlights resource value and limitations exposed in the primal solution space.

Simplex Method

The simplex method is a systematic procedure used to solve linear programming problems. Named for its exploration of feasible solutions in a piecewise-linear path, it processes through vertices  of the polytope defined by the constraints.
As applied in the current scenario, the primal problem is setup in a simplex tableau format:

• Top rows represent constraints converted to equality using slack variables \(s_1, s_2, s_3\).
• The bottom row corresponds to the cost function scaled to coefficients of objective function values.

The tableau is verified by checking whether the bottom row has only non-positive coefficients for decision variables.In this exercise, the tableau immediately indicated optimality. With non-positive coefficients across the bottom row, the current basic feasible solution is confirmed optimal:

• x = 0, y = 8, z = 0

The optimal vertex along the feasible region of the solution space has been found using simplex iterations.

Optimization

Optimization involves finding the best solution from all feasible solutions. In linear programming, optimization implies either maximizing or minimizing a given objective function within certain constraints.
There is a constant effort to effectively allocate limited resources represented typically through constraints. For this exercise, the optimized solution for the primal problem is achieved when:

• The cost function \( C = 6x + 8y + 4z \) reaches minimum value.
• The Minimal achieved with \( x = 0, y = 8, z = 0 \).
• Here, the total minimized cost is \( C = 64 \).

These solutions provide a mathematical sense of unequally distributing resources \((x, y, z)\) to meet the least cost possible under given constraints. Effective optimization means strategic adjustments to  reach an outcome that aligns with the defined goals  within specified limits.