QUESTION Find the optimal solution to the Linear Programming (LP) Problem below, using Simplex Algorithm method: Maximize $$ \begin{array}{l}\pi=12 x_{1}+16 x_{2} \ \text { subject to: } \ \qquad \begin{array}{l}10 x_{1}+20 x_{2} \leq 120 \ 8 x_{1}+8 x_{2} \leq 80\end{array} \ \quad \text { where } x_{1}, x_{2}0\end{array} $$

Question

QUESTION Find the optimal solution to the Linear Programming (LP) Problem below, using Simplex Algorithm method: Maximize $$ \begin{array}{l}\pi=12 x_{1}+16 x_{2} \ \text { subject to: } \ \qquad \begin{array}{l}10 x_{1}+20 x_{2} \leq 120 \ 8 x_{1}+8 x_{2} \leq 80\end{array} \ \quad \text { where } x_{1}, x_{2}>0\end{array} $$

UpStudy AI · Accepted Answer

We wish to solve

Maximize π = 12x₁ + 16x₂

subject to

10x₁ + 20x₂ ≤ 120  
  8x₁ + 8x₂ ≤ 80  
  x₁, x₂ ≥ 0

using the simplex method. One common approach is to convert the inequalities into equalities by adding slack variables.

Step 1. Introduce Slack Variables

Let s₁ and s₂ be the slack variables for the two constraints, respectively; then

10x₁ + 20x₂ + s₁ = 120  
  8x₁ + 8x₂ + s₂ = 80

with s₁, s₂ ≥ 0.

Step 2. Set Up the Initial Simplex Tableau

We have decision variables x₁ and x₂ and slack variables s₁ and s₂. Our tableau (with the basic variables initially being s₁ and s₂) is:

x₁   x₂   s₁   s₂  RHS  
Row 1:  10   20   1   0   120  
Row 2:  8   8   0   1   80  
Obj:  –12 –16   0   0   0

Note: In the objective row we subtract the coefficients from the pivot form.

Step 3. Perform Simplex Iterations

a) Identify the entering variable.  
We look at the objective row; the most negative coefficient is –16 corresponding to x₂, so x₂ enters the basis.

b) Determine the leaving variable.  
Compute the ratios (RHS divided by the coefficient of x₂ in each row):

Row 1: 120 / 20 = 6  
  Row 2: 80 / 8 = 10

The smallest ratio is 6; therefore, s₁ leaves the basis and x₂ enters.

c) Pivot on the element in Row 1, Column x₂ (which is 20).  
Divide Row 1 by 20 to make the pivot element 1:

New Row 1:  
   x₁: 10/20 = 0.5  
   x₂: 20/20 = 1  
   s₁: 1/20 = 0.05  
   s₂: 0  
   RHS: 120/20 = 6

d) Eliminate x₂ from the other rows.

For Row 2, subtract 8 times the new Row 1 from the original Row 2:

Original Row 2: [8 8 0 1 80]  
  New Row 2 = Row 2 – 8×(New Row 1):

x₁: 8 – 8×0.5 = 8 – 4 = 4  
   x₂: 8 – 8×1 = 8 – 8 = 0  
   s₁: 0 – 8×0.05 = 0 – 0.4 = –0.4  
   s₂: 1 – 8×0 = 1  
   RHS: 80 – 8×6 = 80 – 48 = 32

For the Objective row, add 16 times the new Row 1 to it (because the coefficient for x₂ is –16):

Original Obj: [–12 –16 0 0 0]  
  New Obj = Obj + 16×(New Row 1):

x₁: –12 + 16×0.5 = –12 + 8 = –4  
   x₂: –16 + 16×1 = –16 + 16 = 0  
   s₁: 0 + 16×0.05 = 0 + 0.8 = 0.8  
   s₂: 0 + 16×0 = 0  
   RHS: 0 + 16×6 = 96

The updated tableau is now:

x₁   x₂   s₁   s₂  RHS  
Row 1 (x₂ basis): 0.5   1   0.05   0   6  
Row 2 (s₂ basis): 4   0  –0.4   1   32  
Obj:     –4   0   0.8   0   96

e) Check optimality.  
Since all the coefficients in the objective row for the decision variables (x₁ and x₂) are now nonnegative (–4 is negative only for x₁; however, we must check whether we can improve the objective further), we see that the coefficient for x₁ is –4, suggesting that entering x₁ can increase the objective value.

So, x₁ is the next candidate to enter the basis.

f) Determine the leaving variable for x₁.  
Compute the ratios:

For Row 1: Coefficient of x₁ = 0.5, so ratio = 6 / 0.5 = 12  
For Row 2: Coefficient of x₁ = 4, so ratio = 32 / 4 = 8

The minimum ratio is 8, so s₂ (Row 2) leaves the basis and x₁ enters.

g) Pivot on the element in Row 2, Column x₁ (which is 4).  
Divide Row 2 by 4:

New Row 2:  
   x₁: 4/4 = 1  
   x₂: 0/4 = 0  
   s₁: –0.4/4 = –0.1  
   s₂: 1/4 = 0.25  
   RHS: 32/4 = 8

Then eliminate x₁ from the other rows.

For Row 1, eliminate x₁:
  New Row 1 = Row 1 – 0.5×(New Row 2):

x₁: 0.5 – 0.5×1 = 0  
   x₂: 1 – 0.5×0 = 1  
   s₁: 0.05 – 0.5×(–0.1) = 0.05 + 0.05 = 0.10  
   s₂: 0 – 0.5×0.25 = –0.125  
   RHS: 6 – 0.5×8 = 6 – 4 = 2

For the Objective row, eliminate x₁:
  New Obj = Obj + 4×(New Row 2):

x₁: –4 + 4×1 = 0  
   x₂: 0 + 4×0 = 0  
   s₁: 0.8 + 4×(–0.1) = 0.8 – 0.4 = 0.4  
   s₂: 0 + 4×0.25 = 1  
   RHS: 96 + 4×8 = 96 + 32 = 128

The new tableau is:

x₁   x₂   s₁   s₂  RHS  
Row 1 (x₂ basis): 0   1   0.10  –0.125  2  
Row 2 (x₁ basis): 1   0  –0.1   0.25  8  
Obj:     0   0   0.4   1   128

h) Check for optimality.  
Now the coefficients in the objective row for x₁ and x₂ are 0. Since no improvement is possible by bringing any nonbasic variable into the basis, we have reached an optimal solution.

Step 4. Read the Optimal Solution

The basic variables are:
  x₂ (Row 1) = 2  
  x₁ (Row 2) = 8

All other variables (s₁ and s₂) are nonbasic (their values can be computed if needed, but they are not part of the profit function).

The maximum value of the objective is 128 (given in the RHS of the objective row).

Thus, the optimal solution is:

x₁ = 8, x₂ = 2  with maximum π = 128

Alternate Verification by Corner-Point Calculation

An alternative approach is to consider the constraints in their original form:
  Constraint 1: 10x₁ + 20x₂ ≤ 120 ⟹ x₁ + 2x₂ ≤ 12  
  Constraint 2: 8x₁ + 8x₂ ≤ 80 ⟹ x₁ + x₂ ≤ 10

The feasible region is bounded by:
  Vertex A: (0,0)     π = 0
  Vertex B: Intersection of x₁ + 2x₂ = 12 with x₁ = 0 ⟹ (0,6)  π = 12×0 + 16×6 = 96
  Vertex C: Intersection of x₁ + x₂ = 10 with x₂ = 0 ⟹ (10,0)  π = 12×10 + 16×0 = 120
  Vertex D: Intersection of x₁ + 2x₂ = 12 and x₁ + x₂ = 10  
Solve: From x₁ + x₂ = 10, we have x₁ = 10 – x₂. Substitute into x₁ + 2x₂ = 12:  
  (10 – x₂) + 2x₂ = 12 ⟹ 10 + x₂ = 12 ⟹ x₂ = 2, thus x₁ = 8  
  π = 12×8 + 16×2 = 96 + 32 = 128

This confirms the simplex solution.

Final Answer:
The optimal solution is x₁ = 8 and x₂ = 2, and the maximum value of π is 128.
The optimal solution is $$ x_1 = 8 $$ and $$ x_2 = 2 $$, with the maximum value of $$ \pi = 128 $$.

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