Task Description

The task involved formulating and solving two related linear programming (LP) problems: a primal problem focused on maximizing profitability for producing two types of products, and a dual problem that aimed to minimize resource usage within the value constraints derived from the primal problem.

Hat Production Linear Programming Problem

Primal LP Problem

Given: - Hat production capacity for Gentlemen \((x_1)\) and Ladies \((x_2)\). - Constraints on cutting and knitting capacities measured in minutes.

Objective: Maximize the profitability of the hat production, where the profitability for gentlemen’s hats is 50 units and for ladies’ hats is 75 units.

Mathematical Formulation: Maximize Profit = \(50x_1 + 75x_2\)

Subject to: - Cutting time: \(5x_1 + 10x_2 \leq 480\) minutes - Knitting time: \(9x_1 + 8x_2 \leq 480\) minutes - Non-negativity: \(x_1, x_2 \geq 0\)

Dual LP Problem

Objective: Minimize resource usage cost based on the constraints from the primal problem, assuming a cost of 480 units per resource.

Mathematical Formulation: Minimize Cost = \(480y_1 + 480y_2\)

Subject to: - Dual constraint for cutting: \(5y_1 + 9y_2 \geq 50\) - Dual constraint for knitting: \(10y_1 + 8y_2 \geq 75\) - Non-negativity: \(y_1, y_2 \geq 0\)

Solution Methodology in R Markdown

The lpSolve package in R was utilized to solve the LP problems. The methodology was as follows:

  • Defined the objective coefficients for profit in the primal problem and for cost in the dual problem.
  • Constructed the constraint matrix based on production constraints.
  • Specified the direction of the inequalities: less than or equal to (“<=”) for the primal and greater than or equal to (“>=”) for the dual.
  • Employed the lp function from lpSolve to find the optimal solution for both problems, maximizing for the primal and minimizing for the dual.
# Define the coefficients of the objective function for the primal problem
objective_coefficients <- c(50, 75)

# Define the constraint matrix for the primal problem
constraint_matrix <- matrix(c(5, 10, 9, 8), nrow = 2, byrow = TRUE)

# Define the right-hand side values of the constraints for the primal problem
right_hand_side <- c(480, 480)

# Define the direction of the constraints for the primal problem
direction <- c("<=", "<=")

# Solve the primal problem
primal_solution <- lp("max", objective_coefficients, constraint_matrix, direction, right_hand_side, all.int = TRUE)
primal_solution
## Success: the objective function is 3825
# Define the coefficients of the objective function for the dual problem
objective_coefficients_dual <- c(480, 480)

# Define the constraint matrix for the dual problem (transpose of the primal constraint matrix)
constraint_matrix_dual <- matrix(c(5, 9, 10, 8), nrow = 2, byrow = TRUE)

# Define the right-hand side values of the constraints for the dual problem
right_hand_side_dual <- c(50, 75)

# Define the direction of the constraints for the dual problem
direction_dual <- c(">=", ">=")

# Solve the dual problem
dual_solution <- lp("min", objective_coefficients_dual, t(constraint_matrix_dual), direction_dual, right_hand_side_dual, all.int = TRUE)
dual_solution
## Success: the objective function is 4320

Interpretation of Results

The primal LP problem’s solution indicated a maximum profit of 3825 units, revealing the optimal production strategy under the given constraints. Conversely, the dual LP problem showed a minimum resource cost of 4320 units, reflecting the minimum expenditure necessary to sustain the primal problem’s output value.

Conclusion

The solutions to both the primal and dual problems provide a comprehensive view of the production strategy. The primal problem identifies the production plan for maximum profitability, while the dual problem indicates the lowest cost that meets the primal output’s value.

Discussion

It is generally expected that the primal and dual problems would yield identical objective values at optimality due to the duality principle in linear programming. The discrepancy between the primal (3825) and dual (4320) objective values suggests the presence of integer programming constraints. These constraints might introduce a duality gap, which is observed in the solutions provided by lpSolve when all.int = TRUE. This result prompts further review of the problem’s formulation, particularly the need for integer constraints and their impact on the primal and dual solutions.

Considering the solution of the dual programming problem provides a pragmatic perspective on resource management. In essence, the dual problem gives us the shadow prices of our constraints, informing us on the worth of each additional unit of resource in the context of our objectives. The solution of the dual problem isn’t just a set of numbers to minimize costs; it’s a strategic map that guides decision-making. It tells us where our investments will have the most significant impact, and in a way, it’s the flip side of the coin to the primal production problem. The primal focuses on what we can make with what we have, while the dual hones in on the cost of the resources needed to make that happen. They ground us in the reality of costs and help us to negotiate better with suppliers, manage our resources more efficiently, and ultimately, to align our production strategy not just with market demands but with financial prudence. The dual solution doesn’t just echo the primal in reverse; it complements it, ensuring that as we strive to maximize profits, we don’t lose sight of the bottom line.