Linear Programming: A Furniture Company
linear Programming1 A Furniture Company
A furniture company is producing tables and chairs for the coming week. The production process involves two stages: wood cutting and assembly. In the wood cutting process, 1 hour is required for each chair and 2 hours for each table. During assembly, 4 hours are needed for each chair and 3 hours for each table. The company has a maximum of 70 labor hours available for wood cutting and 180 hours for assembly in the upcoming week. Additionally, inventory constraints limit production to no more than 100 chairs and 30 tables. The profit margins are $5 per chair and $6 per table. The goal is to develop a linear programming model to maximize total profit, and then use Excel Solver to find the optimal production quantities.
Paper For Above instruction
The manufacturing of furniture involves multiple constraints related to resource availability, production capacity, and profit maximization. In this context, the primary goal for the furniture company is to determine optimal production levels of chairs and tables to maximize profit while adhering to resource limitations. This problem can be formulated as a linear programming (LP) model, which allows decision-makers to systematically analyze trade-offs and find the most profitable production mix under infrastructure constraints.
Formulation of the Linear Programming Model
Let us define the decision variables:
- \(x\): number of chairs produced
- \(y\): number of tables produced
Objective Function:
Maximize profit, subject to profit margins:
\[
\text{Maximize } Z = 5x + 6y
\]
Constraints:
1. Wood Cutting Labor Hours:
\[
1x + 2y \leq 70
\]
2. Assembly Labor Hours:
\[
4x + 3y \leq 180
\]
3. Production Quantity Limits:
\[
x \leq 100,\quad y \leq 30
\]
4. Non-negativity:
\[
x, y \geq 0
\]
This model captures the essential resource limitations, production constraints, and profit goal.
Standard Form Conversion
To solve this LP using the simplex method or in Excel Solver, it is conventional to convert the inequalities to equations by adding slack variables:
- Slack variable \( s_1 \) for wood cutting:
\[
1x + 2y + s_1 = 70,\quad s_1 \geq 0
\]
- Slack variable \( s_2 \) for assembly:
\[
4x + 3y + s_2 = 180,\quad s_2 \geq 0
\]
- Additional slack variables \( s_3 \) and \( s_4 \) for production constraints:
\[
x + s_3 = 100,\quad s_3 \geq 0
\]
\[
y + s_4 = 30,\quad s_4 \geq 0
\]
- Non-negativity constraints:
\[
x, y, s_1, s_2, s_3, s_4 \geq 0
\]
Finding the Feasible Region
The feasible region is bounded by the constraints. Each inequality boundary intersects with others to form vertices (extreme points). These points are critical to the LP solution because the optimal value occurs at one of them. To identify all extreme points, we set pairs of constraints equal and solve for \(x\) and \(y\):
1. Intersection of \(1x + 2y = 70\) and \(4x + 3y = 180\)
2. Intersection with the production limit lines \(x=0\) or \(y=0\)
3. Intersections with constraints \(x = 100\) and \(y=30\)
By solving these systems, we identify candidate points.
Extreme Points Calculation
1. Solve \(1x + 2y = 70\) and \(4x + 3y = 180\):
Multiply the first by 3:
\[
3x + 6y = 210
\]
Subtract from the second:
\[
(4x + 3y) - (3x + 6y) = 180 - 210 = -30
\]
\[
x - 3y = -30
\]
\[
x = 3y - 30
\]
Substitute into \(1x + 2y = 70\):
\[
(3y - 30) + 2y = 70
\]
\[
5y - 30 = 70
\]
\[
5y = 100
\]
\[
y = 20
\]
\[
x = 3(20) - 30 = 60 - 30 = 30
\]
So one vertex is at \((x, y) = (30, 20)\).
2. Check boundary points at production limits:
- At \(x=0\), from \(1(0)+2y \leq 70\):
\[
2 y \leq 70 \Rightarrow y \leq 35
\]
But production is limited to \( y \leq 30 \), so the maximum feasible \( y \) at \( x=0 \) is 30. Check if this satisfies the assembly constraint:
\[
4(0)+3(30) = 90 \leq 180 \quad \text{(OK)}
\]
Similarly, at \( y=0 \),
\[
1x \leq 70 \Rightarrow x \leq 70
\]
but production limit imposes \( x \leq 100 \), so feasible maximum at \( y=0 \):
\[
4x \leq 180 \Rightarrow x \leq 45
\]
and \( x \leq 70 \) from resource, so feasible at \( x=45, y=0 \).
Now, check the production capacity constraints:
- At \(x=100, y=0\):
Check the resource constraints:
\[
1(100)+2(0)=100 \leq 70 \quad \text{(Not feasible)}
\]
so \(x=100, y=0\) not feasible.
- At \(x=0, y=30\):
\[
1(0)+2(30)=60 \leq 70 \quad \text{(feasible)}
\]
Check assembly:
\[
4(0)+3(30)=90 \leq 180
\]
Feasible.
Furthermore, check at \(x=0, y=30\), the profit:
\[
Z = 5(0) + 6(30) = 180
\]
Similarly, at \(x=45, y=0\) (feasible):
\[
Z= 5(45)+6(0)= 225
\]
but the resource constraints need validation:
- At \(x=45, y=0\):
Wood:
\[
1(45)+2(0)=45 \leq 70
\]
Assembly:
\[
4(45)+3(0)=180 \leq 180
\]
Feasible and these give us the extreme point at \((45,0)\).
Optimal Solution
Evaluate profit at the identified extreme points:
- \((30, 20)\): \(Z=5(30)+6(20)=150+120=270\)
- \((45, 0)\): \(Z=225\)
- \((0, 30)\): \(Z=180\)
Comparing these, the maximum profit occurs at \((30, 20)\) with a profit of $270.
Slack/Surplus Variables at the Optimal Solution
At the optimal point \((x,y) = (30,20)\),
- Wood constraint:
\[
1(30) + 2(20)=30+40=70
\]
\[
s_1 = 70 - 70=0 \quad (\text{binding constraint})
\]
- Assembly constraint:
\[
4(30)+3(20)=120+60=180
\]
\[
s_2=180-180=0
\]
- Production constraints:
\[
x=30 \leq 100 \Rightarrow s_3=100-30=70
\]
\[
y=20 \leq 30 \Rightarrow s_4=30-20=10
\]
Slack variables indicate unused capacity: 70 units of chair production capacity unused and 10 units of table capacity unused.
Conclusion
The optimal production plan is to produce 30 chairs and 20 tables, achieving a maximum profit of $270 under the resource constraints. The constraints are binding for labor hours but not for production capacity limits, which allows some flexibility.
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