Mat540 Homework Week 6: Linear Programming Problems And Solu
Mat540 Homework Week 6: Linear Programming Problems and Solutions
This assignment involves formulating and solving multiple linear programming problems related to production, resource allocation, and cost minimization. The tasks include developing LP models for given scenarios and solving these models using graphical analysis. The problems cover manufacturing, resource allocation, and optimization under constraints, emphasizing understanding how to translate real-world problems into LP models and interpret solutions.
Paper For Above instruction
Introduction
Linear programming (LP) is a mathematical method used for optimizing a linear objective function subject to linear equality and inequality constraints. It has widespread applications across industries for resource allocation, production planning, and cost minimization. This paper addresses multiple LP problems, providing formulation, solution approaches through graphical analysis, and insights into resource utilization and decision-making processes.
Problem 1: Producing Two Products on Assembly Lines
Problem Description
A company produces two products processed on two assembly lines. Assembly line 1 has 100 hours available; line 2 has 42 hours. Product 1 requires 10 hours on line 1, and on line 2, it needs 7 hours; Product 2 needs 10 hours on line 1 and 3 hours on line 2. The profit per unit is $6 for product 1 and $4 for product 2. The goal is to maximize profit through optimal production quantities.
Formulation
Let x₁ = number of units of product 1, x₂ = number of units of product 2.
- Objective Function:
Maximize Z = 6x₁ + 4x₂
- Constraints:
Based on available hours:
- Assembly line 1: 10x₁ + 10x₂ ≤ 100
- Assembly line 2: 7x₁ + 3x₂ ≤ 42
- Non-negativity:
- x₁, x₂ ≥ 0
Solution via Graphical Analysis
Graphical analysis involves plotting the constraints on a coordinate plane and identifying the feasible region where all constraints are satisfied. The corner points of this feasible region represent potential optimal solutions. Calculating the objective function Z at each vertex determines the maximum profit. The feasible region is bounded by the lines:
- 10x₁ + 10x₂ = 100 (line 1)
- 7x₁ + 3x₂ = 42 (line 2)
Points of intersection and axes intercepts are evaluated for optimality, leading to the production plan that maximizes profit within resource constraints.
Problem 2: Furniture Production—Chairs and Tables
Problem Description
Pinewood Furniture produces chairs and tables from labor and wood. The available resources per day are 80 hours of labor and 36 board-feet of wood. Each chair requires 8 hours of labor and 2 board-feet, while each table requires 10 hours and 6 board-feet. Demand limits chairs to 6 units per day. Profits are $400 per chair and $100 per table. The goal is to maximize profit.
Formulation
Let x₁ = number of chairs, x₂ = number of tables.
- Objective Function:
Maximize Z = 400x₁ + 100x₂
- Constraints:
Resources:
- Labor: 8x₁ + 10x₂ ≤ 80
- Wood: 2x₁ + 6x₂ ≤ 36
Demand Constraint:
- x₁ ≤ 6
Non-negativity:
- x₁, x₂ ≥ 0
Solution via Graphical Analysis
Graphing these constraints reveals the feasible region. The optimal solution is at a vertex where profit is maximized, considering the binding constraints such as resource limits and demand caps. The intersection points are evaluated to identify the production quantities that maximize profit while respecting resource and demand constraints.
Problem 3: Resource Utilization after Optimization
Question
Given the optimal production quantities of chairs and tables from Problem 2, calculate the unused amounts of labor and wood. This involves substituting the optimal values into resource constraint equations and computing the difference between available and utilized resources.
Problem 4: Drug Composition Optimization
Problem Description
Elixer Drug Company combines two ingredients to produce a drug with specific antibiotic content. Each ingredient contributes units of three antibiotics in different proportions, with cost per gram known. Constraints include minimum antibiotic requirements.
Formulation
Let x₁ = grams of ingredient 1, x₂ = grams of ingredient 2.
- Objective Function:
Minimize C = 80x₁ + 50x₂
- Constraints:
- Antibiotic 1: 3x₁ + x₂ ≥ 6
- Antibiotic 2: 2x₁ + 6x₂ ≥ 4
- Antibiotic 3: 2x₁ + 6x₂ ≥ 12
- Non-negativity:
- x₁, x₂ ≥ 0
Solution via Graphical Analysis
The feasible region, determined by the intersection of the constraints, allows identifying the least-cost combination of ingredients by evaluating the objective function at the vertices of this region. The solution provides the optimal grams of each ingredient to meet antibiotic requirements at minimum cost.
Problem 5: Coat and Slacks Production
Problem Description
A clothier produces coats and slacks utilizing wool and labor. The constraints involve available resources, and profits are known. The goal is to maximize profit by choosing optimal production quantities.
Formulation
Let x₁ = coats, x₂ = slacks.
- Objective Function:
Maximize Z = 50x₁ + 40x₂
- Constraints:
- Wool: 3x₁ + 5x₂ ≤ 150
- Labor: 10x₁ + 4x₂ ≤ 200
- Demand constraint for slacks:
- x₂ ≤ 6
- Non-negativity:
- x₁, x₂ ≥ 0
Solution via Graphical Analysis
By plotting the resource constraints, the feasible region is determined. Evaluating at the vertices, considering demand limits, reveals the production quantities that maximize profit. This solution balances resource availability with market demand.
Problem 6: Linear Programming Graph Optimization
Objective Function and Constraints
Maximize Z = 5x₁ + 8x₂ subject to:
- 3x₁ + 5x₂ ≤ 50
- 2x₁ + 4x₂ ≤ 40
- x₁ ≤ 8
- x₂ ≤ 10
- x₁, x₂ ≥ 0
The graphical solution involves plotting each constraint, identifying the feasible region, and evaluating the objective function at the vertices to find the maximum Z.
Conclusion
The problems detailed above exemplify essential concepts in linear programming, such as model formulation, graphical solutions, and resource optimization. These methods enable decision-makers to allocate resources efficiently, maximize profits, or minimize costs under complex constraints. The graphical approach, though limited to two variables, offers visual insight into feasible solutions and optimal points, reinforcing theoretical understanding with practical application.
References
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