Assignment 2problem 1 An Auto Company Manufactures Cars And Trucks ✓ Solved

Assignment 2 Problem 1. An auto company manufactures cars and trucks. Each vehicle must be processed in the paint shop and in the body assembly shop. If the paint shop were painting only trucks, then 40 trucks per day could be painted. If the paint shop were painting only cars, then 60 cars per day could be painted.

If the body shop were only producing trucks, then it could process 50 per day. If the body shop were only producing cars, then it could process 50 per day. Each truck contributes 0 to profit, and each car contributes 0 to profit. (a) Formulate an LPP whose solution will determine a daily production schedule that maximizes the company profit. (b) Solve the LPP (find all optimal solutions of the LPP) obtained in (a) by using an appropriate part of the EPT (Extreme Point Theorem). Explain. (c) If in addition, the market demand requires at least 30 cars to be produced daily, what must be the daily production schedule to maximize the company profit (goal programming). Problem 2.

Given the 3-variable LPP: max z = 5x1 + 3x2 − 3x3 s.t. 2x1 + 3x3 ≤ 6 x2 ≤ 5 x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 . (a) Show that the FR of the given LPP is bounded. (b) Sketch the FR of the LPP and determine all extreme (corner) points of the FR. (c) Find an optimal solution of the LPP by using the EPT. Point out which part of the EPT you have applied. (d) How many optimal solutions does the given LPP have? Explain. Problem 3.

Solve the LPP: max z = 3x1 + 2x2 s.t. 3x1 + x2 ≤ 12 x1 + 3x2 ≤ 12 2x2 ≤ 7 x1 ≥ 0, x2 urs . Hint: In the given LPP, x2 urs means that x2 is unrestricted by sign, i.e., x2 can take positive and negative values; no sign restriction on x2. 1 Problem 4. Which of the Cases 1-4 (Unique Solution; Alternative Solutions; Infea- sible LPP, Unbounded LPP) apply to each of the following LPP: (a) max z = x1 + x2 (b) max z = 4x1 + x2 s.t. s.t. x1 + x2 ≤ 4 8x1 + 2x2 ≤ 16 x1 −x2 ≥ 5 5x1 + 2x2 ≤ 12 x1 ≥ 0, x2 ≥ 0 x1 ≥ 0, x2 ≥ 0 (c) max z = −x1 + 3x2 (d) max z = 3x1 + x2 s.t. s.t. x1 −x2 ≤ 4 2x1 + x2 ≤ 6 x1 + 2x2 ≥ 4 x1 + 3x2 ≤ 9 x1 ≥ 0, x2 ≥ 0 x1 ≥ 0, x2 ≥ 0 Problem 5.

Consider the following optimization problem: max z = 40x1 + 10x2 s.t. x1 + 2x2 ≤ 5 |x1 −x2| ≤ 2 x1 ≥ 0, x2 ≥ 0,. (a) By restating the absolute value constraint as a combination of two linear con- straints, show that the given optimization problem is in fact a Linear Programming Problem. (b) Solve graphically the given LPP either by using iso-cost lines or by using the EPT (Extreme Point Theorem), Part 1, if applicable. 2

Paper for above instructions

Assignment Solution

Problem 1: Auto Company Manufacturing Cars and Trucks

(a) Formulation of Linear Programming Problem (LPP)

Let \( x_1 \) be the number of cars produced daily, and \( x_2 \) be the number of trucks produced daily. The profit contributions from cars and trucks, respectively, are represented as follows:
- Profit from cars: \( 200x_1 \)
- Profit from trucks: \( 300x_2 \)
Thus, the objective function to maximize profit can be formulated as:
\[
\text{Maximize } z = 200x_1 + 300x_2
\]
Next, we must derive the constraints based on the given data.
Paint Shop Constraints:
- The paint shop can paint 60 cars per day if focused solely on cars or 40 trucks if focused solely on trucks:
\[
\frac{x_1}{60} + \frac{x_2}{40} \leq 1 \quad \text{(total capacity constraint for the paint shop)}
\]
This inequality can be multiplied through by 120 (the least common multiple of 60 and 40) to remove denominators:
\[
2x_1 + 3x_2 \leq 120
\]
Body Assembly Shop Constraints:
- The body shop can process 50 cars or trucks per day:
\[
x_1 \leq 50 \quad \text{and} \quad x_2 \leq 50
\]
The constraints can be summarized as follows:
1. \( 2x_1 + 3x_2 \leq 120 \)
2. \( x_1 \leq 50 \)
3. \( x_2 \leq 50 \)
Additionally, non-negativity constraints apply:
\[
x_1 \geq 0 \quad \text{and} \quad x_2 \geq 0
\]
The complete LPP is thus:
Objective:
Maximize \( z = 200x_1 + 300x_2 \)
Subject to:
1. \( 2x_1 + 3x_2 \leq 120 \)
2. \( x_1 \leq 50 \)
3. \( x_2 \leq 50 \)
4. \( x_1, x_2 \geq 0 \)

(b) Solving the LPP using the Extreme Point Theorem (EPT)

To find the feasible region, we can sketch the constraints on a graph.
1. Constraint \( 2x_1 + 3x_2 = 120 \) intersects the axes at points \( (60, 0) \) and \( (0, 40) \).
2. Constraint \( x_1 = 50 \) intersects the x-axis at \( (50, 0) \).
3. Constraint \( x_2 = 50 \) intersects the y-axis at \( (0, 50) \).
Finding feasible corner points:
- Intersection of \( 2x_1 + 3x_2 = 120 \) and \( x_1 = 50 \):
Substituting \( x_1 = 50 \):
\[
2(50) + 3x_2 = 120 \implies 100 + 3x_2 = 120 \implies x_2 = \frac{20}{3} \approx 6.67
\]
- Intersection of \( 2x_1 + 3x_2 = 120 \) and \( x_2 = 50 \):
Substituting \( x_2 = 50 \):
\[
2x_1 + 3(50) = 120 \implies 2x_1 + 150 = 120 \implies 2x_1 = -30 \quad (\text{Not feasible, as } x_1 \geq 0)
\]
Thus, the corner points are:
1. \( (0, 0) \)
2. \( (50, \frac{20}{3}) \)
3. \( (0, 40) \)
4. \( (50, 0) \)
Calculating the profit at each corner point:
- At \( (0, 0): z = 0 \)
- At \( (50, \frac{20}{3}): z \approx 200(50) + 300 \times \frac{20}{3} = 10000 + 2000 = 12000 \)
- At \( (0, 40): z = 300 \times 40 = 12000 \)
- At \( (50, 0): z = 200 \times 50 = 10000 \)
Hence, the optimal solutions for maximum profit:
1. \( (50, \frac{20}{3}) \)
2. \( (0, 40) \)
By observing the values, it's clear we have a unique maximum profit of \( 12000 \).

(c) Adding Market Demand Constraint and Finding Optimal Production Schedule

If market demand requires at least 30 cars produced daily (\( x_1 \geq 30 \)), we need to update the constraints:
1. \( 2(30) + 3x_2 \leq 120 \implies 60 + 3x_2 \leq 120 \implies x_2 \leq 20 \)
2. \( x_1 \leq 50 \)
3. \( x_2 \leq 50 \)
Now, feasible bounds for \( x_2 \) under the new demand constraint (and simple inequality adjustments):
- \( x_2 \) can be at most 20.
With this new constraint:
Given \( x_1 = 30 \) and \( x_2 = 20 \):
Profit becomes \( z = 200 \times 30 + 300 \times 20 = 6000 + 6000 = 12000 \).
Thus, the new optimal production schedule under market demand conditions is:
- Produce 30 cars and 20 trucks per day for a maximum profit of \( 12000 \).

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