Assignment 2 Problem 1: An Auto Company Manufactures Cars

Assignment 2problem 1 An Auto Company Manufactures Cars And Trucks

Formulate an LP to maximize profits given constraints related to processing capacities of the paint shop and body shop for cars and trucks, with profit contributions for each vehicle type, and considering additional demand constraints.

Use the Extreme Point Theorem to find all optimal solutions of the LP formulated in part (a) and explain the reasoning.

When market demand requires at least 30 cars daily, determine the daily production schedule that maximizes profit, applying goal programming.

Sample Paper For Above instruction

Introduction

The manufacturing of cars and trucks involves complex capacity constraints and profit maximization objectives. This paper presents a comprehensive approach to formulating and solving a linear programming problem (LPP) for an auto company's production schedule, considering capacity limitations in the paint and body shops, profit contributions, and additional demand constraints. The aim is to determine an optimal production plan that maximizes profit while satisfying market demand requirements.

Problem Formulation and Constraints

The first step involves formulating the linear programming model. Let x_c denote the number of cars produced daily, and x_t denote the number of trucks produced daily. The constraints are derived from processing capacities of the paint shop and body shop.

Paint shop constraints:

- Trucks: 40 trucks per day if painting only trucks, so 40 units max.

- Cars: 60 cars per day if painting only cars, so 60 units max.

Hence:

- x_t ≤ 40

- x_c ≤ 60

Body shop constraints:

- Trucks: 50 trucks per day if producing only trucks, so 50 units.

- Cars: 50 cars per day if producing only cars, so 50 units.

Thus:

- x_t ≤ 50

- x_c ≤ 50

However, since both shops process all vehicles, simultaneous constraints apply:

- x_t ≤ min(40, 50) = 40

- x_c ≤ min(60, 50) = 50

Profit contributions:

- Truck: $300

- Car: $200

Objective function:

maximize Z = 300x_t + 200x_c

Additional constraints:

- Non-negativity:

x_c ≥ 0, x_t ≥ 0

Thus, the LP model is:

Maximize Z = 300x_t + 200x_c

Subject to:

x_t ≤ 40

x_t ≤ 50

x_c ≤ 50

x_c ≤ 60

x_c ≥ 0

x_t ≥ 0

Simplifying constraints for clarity:

x_t ≤ 40 (most restrictive)

x_c ≤ 50 (most restrictive)

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Solution Using Extreme Point Theorem

The feasible region is bounded by the intersection of the constraints. Key corner points (extreme points) are identified at intersections:

- (x_t, x_c) = (0,0)

- (40,0)

- (0,50)

- (40,50)

Calculate profit at these points:

- (0, 0): Z=0

- (40, 0): Z=30040 + 2000= $12,000

- (0, 50): Z=3000 + 20050= $10,000

- (40, 50): Z=30040 + 20050= $12,000 + $10,000= $22,000

The maximum profit occurs at (x_t=40, x_c=50).

All other points within the feasible region yield lower profits, and thus, the optimal solution is to produce 40 trucks and 50 cars daily, maximizing profit at $22,000.

The EPT confirms that optimal solutions occur at corner points, and since the maximum profit is at (40, 50), this is the unique optimal solution under these constraints.

Impact of Market Demand Constraint (at least 30 cars)

When the market requires a minimum of 30 cars, the constraints are:

x_c ≥ 30

with x_c ≤ 50, x_t ≤ 40, so feasible points lie along the line x_c ≥ 30, bounded by x_t from 0 to 40.

To maximize profit, choose the maximum number of trucks (x_t=40) and produce the minimum required cars (x_c=30).

Calculate profit:

Z = 30040 + 20030= $12,000 + $6,000= $18,000

But if we produce more cars (say 50), profit increases:

Z = 30040 + 20050= $12,000 + $10,000= $22,000

Thus, the optimal schedule under demand constraint is:

- x_c=50 (maximum cars)

- x_t=40 (maximum trucks)

Profit in this case remains at $22,000, while satisfying the demand for at least 30 cars.

Conclusion

The LP formulation effectively models manufacturing constraints and profit maximization, with the Extreme Point Theorem facilitating solution identification. The company should produce 40 trucks and 50 cars daily to maximize profit, considering capacity constraints, with the option to produce more cars if market demand increases. When minimum demand constraints are added, the optimal production remains at maximum capacity for both vehicles within the given limits, ensuring maximum profitability.

References

• Hillier, F. S., & Lieberman, G. J. (2010). Introduction to Operations Research (9th ed.). McGraw-Hill.
• Winston, W. L. (2004). Operations Research: Applications and Algorithms (4th ed.). Thomson Brooks/Cole.