A Linear Programming Problem Has Three Variables X1, X2, And

11 A Linear Programming Problem Has Three Variables X1 X2 And X3

A linear programming problem involves three variables, x₁, x₂, and x₃, with three constraints. The first constraint is given as 3x₁ + 2x₂ – 5x₃ ≤ 150, but the other two constraints and the objective function are not specified. It is known that the optimal solution has an objective function value of 1025 with x₁ = 0, x₂ = 75, and x₃ = 25. The objective function for the dual problem is described as min 150y₁ + 250y₂ + y₃.

The first dual constraint is 3y₁ + y₂ + 2y₃ ≥ 3, and the second is 2y₁ + 3y₂ – 3y₃ ≥ 12. The third constraint is unknown. Using the principles of strong duality and the complementary slackness theorems, this problem asks to determine the optimal solution (y₁, y₂, y₃) to the dual problem. This task can be accomplished without deriving the missing constraints, as the provided information suffices.

Paper For Above instruction

The application of duality theory in linear programming offers profound insights into solving optimization problems efficiently by exploring the relationships between primal and dual formulations. In the scenario described, understanding and leveraging the principles of strong duality and complementary slackness are essential in determining the optimal dual variables, given partial information about the primal problem's solution. This paper explores the theoretical foundations, necessary calculations, and logical deductions required to identify the dual solution in the absence of full constraint data, emphasizing the elegance and power of duality in linear programming.

Introduction

Linear programming (LP) provides a mathematical framework to optimize a linear objective function subject to linear constraints. Its applications span diverse disciplines including economics, engineering, and operations research. Among its core properties, the duality principle stands out as a critical concept that links primal and dual problems, granting powerful analytical tools to analyze and solve LPs. In particular, the strong duality theorem states that if both the primal and dual problems are feasible, their optimal objectives are equal. Complementary slackness conditions further connect the solutions of primal and dual problems, allowing inference of dual variables from primal solutions and vice versa.

The Structure of the Given Problem

In this specific problem, the primal LP involves three decision variables and three constraints, with the given optimal primal solution being (x₁=0, x₂=75, x₃=25) and an optimal objective value of 1025. The dual problem's objective aims to minimize 150y₁ + 250y₂ + y₃. The dual constraints are partially specified: the first two are known, while the third remains unknown.

Given the optimal primal solution and the dual problem's structure, the task involves using duality and complementary slackness to deduce the values of (y₁, y₂, y₃). The process hinges on the relationships stipulated by the theorems, which establish that for each constraint, either the constraint is tight and the corresponding dual variable is non-zero or the constraint is loose and the dual variable is zero.

Applying Strong Duality

The strong duality theorem asserts that at optimality, the objective values of primal and dual problems are equal. Thus, for our primal problem with an optimal value of 1025, the dual problem's optimal value also equals 1025. Formally, this means:

Objective value (primal) = Objective value (dual) = 1025.

In the dual problem, the objective function is: Minimize 150y₁ + 250y₂ + y₃, and at optimality, it equals 1025. Therefore:

150y₁ + 250y₂ + y₃ = 1025.

Using Complementary Slackness

The theorem states that for each primal constraint, either the constraint is active (binding), or the associated dual variable is zero; similarly, for each dual constraint, either it is tight, or the dual variable is zero. To apply this, we analyze the primal variables:

• x₁=0
• x₂=75
• x₃=25

Given that x₁=0, the primal constraint associated with x₁ (which is based on the first constraint 3x₁ + 2x₂ – 5x₃ ≤ 150) is likely not tight since increasing x₁ would improve the objective (or violate constraints). Conversely, the other variables suggest certain constraints are binding, which informs the values of dual variables y₁, y₂, and y₃.

Specifically, the primal's complementary slackness conditions imply:

• For the primal constraint related to the first constraint: 3y₁ + y₂ + 2y₃ = 3 (since the corresponding x₁ is zero and possibly inactive)
• For the second constraint: 2y₁ + 3y₂ – 3y₃ = 12

The third constraint remains unknown; however, the given information about optimality helps deduce the values of y₁, y₂, and y₃.

Solving for Dual Variables

Using the two known dual constraints:

1. 3y₁ + y₂ + 2y₃ = 3
2. 2y₁ + 3y₂ – 3y₃ = 12

Express y₂ from the first equation:

y₂ = 3 – 3y₁ – 2y₃

Substitute into the second equation:

2y₁ + 3(3 – 3y₁ – 2y₃) – 3y₃ = 12

Expand and simplify:

2y₁ + 9 – 9y₁ – 6y₃ – 3y₃ = 12

(2y₁ – 9y₁) + 9 – (6y₃ + 3y₃) = 12

-7y₁ + 9 – 9y₃ = 12

Rearranged as:

-7y₁ – 9y₃ = 12 – 9 = 3

Express y₁ in terms of y₃: y₁ = (–3 – 9y₃) / 7

Now, substitute y₁ back into y₂:

y₂ = 3 – 3 * [(–3 – 9y₃) / 7] – 2y₃

Calculating the numerator:

y₂ = 3 + (9 + 27y₃) / 7 – 2y₃

Express all terms over 7:

y₂ = (21/7) + (9 + 27y₃) / 7 – (14y₃)/7

Combine numerator:

y₂ = [21 + 9 + 27y₃ – 14y₃] / 7 = (30 + 13y₃) / 7

Given that the dual objective function is 150y₁ + 250y₂ + y₃ = 1025, substitute these expressions:

150 [(–3 – 9y₃) / 7] + 250 [(30 + 13y₃) / 7] + y₃ = 1025

Multiply through by 7 to clear denominator:

150 (–3 – 9y₃) + 250 (30 + 13y₃) + 7y₃ = 1025 * 7 = 7175

Compute each term:

-450 – 1350y₃ + 7500 + 3250y₃ + 7y₃ = 7175

Combine constants: (-450 + 7500) = 7050

Combine y₃ terms: (–1350 + 3250 + 7) y₃ = (1907) y₃

Equation becomes:

7050 + 1907 y₃ = 7175

Solve for y₃:

1907 y₃ = 7175 – 7050 = 125

y₃ = 125 / 1907 ≈ 0.0655

Now, back-substitute to find y₁ and y₂:

y₁ = (–3 – 9 * 0.0655) / 7 ≈ (–3 – 0.5895) / 7 ≈ (–3.5895) / 7 ≈ –0.5127

y₂ = (30 + 13 * 0.0655) / 7 ≈ (30 + 0.852) / 7 ≈ 30.852 / 7 ≈ 4.407

The approximate dual optimal solution is:

• y₁ ≈ –0.5127
• y₂ ≈ 4.407
• y₃ ≈ 0.0655

Note that exact solutions depend on precise calculations, but these approximate values satisfy the necessary optimality conditions consistent with the given information.

Conclusion

Applying the principles of strong duality and complementary slackness facilitated solving for the dual variables using known primal solutions and constraints. The derived values of y₁, y₂, and y₃ satisfy the dual problem's structure and the optimality conditions. This synergy underscores the utility of duality theory in efficiently solving linear programming problems, especially when some data are missing but the primal's optimal solution and spot constraints are known.

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