UPSC MainsMANAGEMENT-PAPER-II201212 Marks150 Words
Q20.

Linear Programming: Door & Window Production

MNP Manufacturing produces ornate, decorative wood frame doors and windows. Each item produced goes through three manufacturing processes : cutting, sanding and finishing. Each door produced requires 1 hour in cutting, 30 minutes in sanding and 30 minutes in finishing. Each window requires 30 minutes in cutting, 45 minutes in sanding and 1 hour in finishing. In the coming week, MNP has 40 hours of cutting capacity, 40 hours of sanding capacity and 60 hours of finishing capacity available. Assume that all doors produced can be sold for a profit of ₹500 each and all windows can be sold for a profit of ₹400 each. Formulate a linear programming model to solve the problem. Use the graphical method to find the optimal solution.

How to Approach

This question requires applying linear programming techniques to a practical manufacturing problem. The approach involves first formulating the problem mathematically by defining decision variables, objective function, and constraints. Then, the graphical method should be used to identify the feasible region and determine the optimal solution that maximizes profit. The answer should clearly demonstrate the steps involved in both formulation and solution. Focus on accurate mathematical representation and interpretation of the graphical solution.

Model Answer

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Introduction

Linear programming is a powerful mathematical technique used for optimizing resource allocation in situations where objectives are linear and constraints are also linear. It finds extensive applications in operations management, particularly in production planning, inventory control, and transportation. This problem presents a classic scenario where MNP Manufacturing aims to maximize its profit by deciding the optimal number of doors and windows to produce, given limited resources – cutting, sanding, and finishing capacities. Formulating and solving this problem using linear programming will provide MNP with a data-driven decision-making tool.

Formulation of the Linear Programming Model

Let:

  • x = Number of doors produced
  • y = Number of windows produced

Objective Function: Maximize Profit (Z)

Z = 500x + 400y

Constraints:

  • Cutting Capacity: 1x + 0.5y ≤ 40
  • Sanding Capacity: 0.5x + 0.75y ≤ 40
  • Finishing Capacity: 0.5x + 1y ≤ 60
  • Non-negativity: x ≥ 0, y ≥ 0

Graphical Solution

To solve this graphically, we first convert the inequalities into equations and plot them on a graph. The feasible region is the area that satisfies all the constraints simultaneously.

Step 1: Plotting the Constraints

  • 1x + 0.5y = 40: When x=0, y=80; When y=0, x=40.
  • 0.5x + 0.75y = 40: When x=0, y=53.33; When y=0, x=80.
  • 0.5x + 1y = 60: When x=0, y=60; When y=0, x=120.

Step 2: Identifying the Feasible Region

The feasible region is the area bounded by the axes (x ≥ 0, y ≥ 0) and the constraint lines. We need to determine which side of each line satisfies the inequality.

Step 3: Finding the Corner Points

The optimal solution will lie at one of the corner points of the feasible region. The corner points are:

  • (0, 0)
  • (40, 0)
  • (0, 60)
  • Intersection of 1x + 0.5y = 40 and 0.5x + 0.75y = 40
  • Intersection of 0.5x + 0.75y = 40 and 0.5x + 1y = 60

Let's calculate the intersection points:

  • Intersection of 1x + 0.5y = 40 and 0.5x + 0.75y = 40: Multiplying the first equation by 0.5, we get 0.5x + 0.25y = 20. Subtracting this from the second equation, we get 0.5y = 20, so y = 40. Substituting y = 40 into the first equation, we get x + 0.5(40) = 40, so x = 20. Therefore, the intersection point is (20, 40).
  • Intersection of 0.5x + 0.75y = 40 and 0.5x + 1y = 60: Subtracting the first equation from the second, we get 0.25y = 20, so y = 80. Substituting y = 80 into the second equation, we get 0.5x + 80 = 60, so 0.5x = -20, which gives x = -40. This point is not in the feasible region as x must be non-negative.

Step 4: Evaluating the Objective Function at Corner Points

Corner Point (x, y) Z = 500x + 400y
(0, 0) Z = 0
(40, 0) Z = 500(40) + 400(0) = 20000
(0, 60) Z = 500(0) + 400(60) = 24000
(20, 40) Z = 500(20) + 400(40) = 10000 + 16000 = 26000

The maximum profit (Z) is ₹26,000, which occurs when x = 20 and y = 40.

Conclusion

Therefore, MNP Manufacturing should produce 20 doors and 40 windows to maximize its profit, given the available cutting, sanding, and finishing capacities. This solution utilizes all available resources efficiently. The graphical method provides a clear visual representation of the feasible region and helps in identifying the optimal production plan. Further analysis could involve sensitivity analysis to understand how changes in resource availability or profit margins would affect the optimal solution.

Answer Length

This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.

Additional Resources

Key Definitions

Linear Programming
A mathematical method for achieving the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships.
Objective Function
A mathematical expression that represents the goal or objective to be maximized or minimized in a linear programming problem.

Key Statistics

The global linear programming market was valued at USD 11.2 billion in 2023 and is expected to grow at a CAGR of 13.5% from 2024 to 2030.

Source: Grand View Research, 2024

Manufacturing contributed approximately 17% to India's GDP in 2023.

Source: National Statistical Office, Ministry of Statistics and Programme Implementation, 2023 (Knowledge Cutoff: Dec 2023)

Examples

Airline Crew Scheduling

Airlines use linear programming to determine the optimal assignment of flight crews to minimize costs while adhering to regulations regarding rest periods and crew qualifications.

Frequently Asked Questions

What if the constraints are not linear?

If the constraints are non-linear, other optimization techniques like non-linear programming or integer programming are required to find the optimal solution.

Topics Covered

OperationsManagementMathematicsLinear ProgrammingOptimizationResource Allocation