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the current solution optimal extract the solutiona

The current solution optimal extract the solutionafter completing the iterations

2. Find the optimal solution using simplex method.

Min Z=10x1 +5x2

Problem Statement

Minimize: Z = 10x1 + 5x2

Subject to: 2x1 + 5x2 ≥ 150 3x1 + x2 ≥ 120 x1, x2 ≥ 0

  1. 2x1 + 5x2 − s1 + a1 = 150
  2. 3x1 + x2 − s2 + a2 = 120
  • s1, s2: Surplus variables (to account for the excess over the minimum requirement).
  • a1, a2: Artificial variables (introduced to handle the constraints).

The objective function is to minimize Z = 10x1 + 5x2. To handle minimization, we use the Big-M Method, where we assign a large penalty M to the artificial variables in the objective function:

Z = 10x1 + 5x2 + Ma1 + Ma2

Basis x1 x2 s1 s2 a1 a2 RHS
a1 2 5 -1 0 1 0 150
a2 3 1 0 -1 0 1 120
Z -10 -5 0 0 -M -M 0

Iteration 1:

    • For a1: 150/2 = 75
    • For a2: 120/3 = 40
    • The smallest ratio is 40, so a2 leaves the basis.
  1. Pivot on the element in row a2 and column x1:

    • Divide row a2 by 3 to make the pivot element 1: $\text{New~row~}x_{1}:\left\lbrack 1,\frac{1}{3},0, - \frac{1}{3},0,\frac{1}{3},40 \right\rbrack$
    • Update the other rows to eliminate x1:
      • For row a1: New row a1 = Old row a1 − 2 × New row x1 $\text{New~row~}a_{1}:\left\lbrack 0,\frac{13}{3}, - 1,\frac{2}{3},1, - \frac{2}{3},70 \right\rbrack$
      • For row Z: New row Z = Old row Z + 10 × New row x1 $\text{New~row~}Z:\left\lbrack 0, - \frac{5}{3},0, - \frac{10}{3}, - M,M - \frac{10}{3},400 \right\rbrack$

Updated Table:

Basis x1 x2 s1 s2 a1 a2 RHS
a1 0 $$\frac{13}{3}$$ -1 $$\frac{2}{3}$$ 1 $$- \frac{2}{3}$$ 70
x1 1 $$\frac{1}{3}$$ 0 $$- \frac{1}{3}$$ 0 $$\frac{1}{3}$$ 40
Z 0 $$- \frac{5}{3}$$ 0 $$- \frac{10}{3}$$ -M $$M - \frac{10}{3}$$ 400

Iteration 2:

  1. Identify the entering variable: The most negative coefficient in the Z-row is $- \frac{5}{3}$ (for x2), so x2 enters the basis.

    • For a1: $70/\frac{13}{3} = \frac{210}{13} \approx 16.15$
    • For x1: $40/\frac{1}{3} = 120$
    • The smallest ratio is $\frac{210}{13}$, so a1 leaves the basis.
    • Divide row a1 by $\frac{13}{3}$ to make the pivot element 1: $\text{New~row~}x_{2}:\left\lbrack 0,1, - \frac{3}{13},\frac{2}{13},\frac{3}{13}, - \frac{2}{13},\frac{210}{13} \right\rbrack$
    • Update the other rows to eliminate x2:
      • For row x1: $\text{New~row~}x_{1} = \text{Old~row~}x_{1} - \frac{1}{3} \times \text{New~row~}x_{2}$ $\text{New~row~}x_{1}:\left\lbrack 1,0,\frac{1}{13}, - \frac{5}{13}, - \frac{1}{13},\frac{5}{13},\frac{350}{13} \right\rbrack$
      • For row Z: $\text{New~row~}Z = \text{Old~row~}Z + \frac{5}{3} \times \text{New~row~}x_{2}$ $\text{New~row~}Z:\left\lbrack 0,0, - \frac{5}{13}, - \frac{20}{13}, - M + \frac{5}{13},M - \frac{20}{13},\frac{2650}{13} \right\rbrack$

Updated Table:

Basis x1 x2 s1 s2 a1 a2 RHS
x2 0 1 $$- \frac{3}{13}$$ $$\frac{2}{13}$$ $$\frac{3}{13}$$ $$- \frac{2}{13}$$ $$\frac{210}{13}$$
x1 1 0 $$\frac{1}{13}$$ $$- \frac{5}{13}$$ $$- \frac{1}{13}$$ $$\frac{5}{13}$$ $$\frac{450}{13}$$
Z 0 0 $$- \frac{5}{13}$$ $$- \frac{20}{13}$$ $$- M + \frac{5}{13}$$ $$M - \frac{20}{13}$$ $$\frac{2650}{13}$$

Step 5: Extract the Solution

After completing the iterations, we obtain:

  • $x_{1} = \frac{550}{13}34.61$
  • $x_{2} = \frac{210}{13} \approx 16.15$
  • $Z = \frac{2650}{13} \approx 203.85$
  • x1 = 34.61
  • x2 = 16.15
  • Minimum cost Z = 426.85

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