1. **State the problem:** Minimize $$z = 2x_1 + 3x_2$$ subject to $$2x_1 + x_2 \geq 4$$ $$x_1 - x_2 \geq -1$$ $$x_1, x_2 \geq 0$$ 2. **Convert inequalities to equalities for Big M method:** For constraints with $$\geq$$, subtract surplus variables and add artificial variables: $$2x_1 + x_2 - s_1 + a_1 = 4$$ $$x_1 - x_2 - s_2 + a_2 = -1$$ where $$s_1, s_2 \geq 0$$ are surplus variables and $$a_1, a_2 \geq 0$$ are artificial variables. 3. **Set up the Big M objective function:** Add a large penalty $$M$$ for artificial variables to force them out: $$Z = 2x_1 + 3x_2 + M a_1 + M a_2$$ 4. **Initial tableau variables:** Basic variables: $$a_1, a_2$$ Non-basic variables: $$x_1, x_2, s_1, s_2$$ 5. **Initial tableau:** \[ \begin{array}{c|cccccc|c} & x_1 & x_2 & s_1 & s_2 & a_1 & a_2 & RHS \\ \hline a_1 & 2 & 1 & -1 & 0 & 1 & 0 & 4 \\ a_2 & 1 & -1 & 0 & -1 & 0 & 1 & -1 \\ \hline Z & -2 & -3 & 0 & 0 & -M & -M & 0 \\ \end{array} \] Note: Objective row coefficients are negative of objective function coefficients. 6. **Apply pivot operations (only allowed are):** - Swap rows (Ri ↔ Rj) - Multiply a row by nonzero scalar (kRi) - Replace a row by itself plus scalar multiple of another (Ri → kRj + Ri) 7. **Goal:** Remove artificial variables $$a_1, a_2$$ from basis by driving their coefficients in Z-row to zero and then optimize. 8. **Step 1: Remove $$a_1$$ from Z-row** Add $$M \times$$ row 1 to Z-row: New Z-row = Z + M * (row 1): $$(-2 + 2M) x_1 + (-3 + M) x_2 + (-M) s_1 + 0 s_2 + 0 a_1 + (-M) a_2 = 4M$$ 9. **Step 2: Remove $$a_2$$ from Z-row** Add $$M \times$$ row 2 to Z-row: New Z-row = previous Z-row + M * (row 2): $$(-2 + 3M) x_1 + (-3) x_2 + (-M) s_1 + (-M) s_2 + 0 a_1 + 0 a_2 = 3M$$ 10. **Choose entering variable:** Most negative coefficient in Z-row is for $$x_2$$ with coefficient $$-3$$. 11. **Determine leaving variable:** Calculate ratios for positive coefficients in $$x_2$$ column: Row 1: $$\frac{4}{1} = 4$$ Row 2: $$\frac{-1}{-1}$$ negative denominator, ignore. So, row 1 leaves. 12. **Pivot on row 1, column $$x_2$$:** Make pivot element 1 by dividing row 1 by 1. Row 1: $$[2,1,-1,0,1,0,4] \to [2,1,-1,0,1,0,4]$$ (already 1 in $$x_2$$ column) 13. **Eliminate $$x_2$$ from other rows:** Row 2: new row 2 = row 2 + row 1 (since coefficient is -1): $$[1,-1,0,-1,0,1,-1] + [2,1,-1,0,1,0,4] = [3,0,-1,-1,1,1,3]$$ Z-row: new Z-row = Z-row + 3 * row 1 (since coefficient is -3): $$(-2 + 3M, -3, -M, -M, 0, 0, 3M) + 3 * (2,1,-1,0,1,0,4) = (-2 + 3M + 6, 0, -M - 3, -M, 3, 0, 3M + 12)$$ Simplify: $$Z = (4 + 3M) x_1 + 0 x_2 + (-M - 3) s_1 + (-M) s_2 + 3 a_1 + 0 a_2 = 3M + 12$$ 14. **Next entering variable:** Most negative coefficient in Z-row is $$-M - 3$$ for $$s_1$$ (assuming large $$M$$, negative). 15. **Determine leaving variable:** Check ratios for positive $$s_1$$ coefficients: Row 1: $$s_1 = -1$$ negative, ignore. Row 2: $$s_1 = -1$$ negative, ignore. No positive coefficients for $$s_1$$, so no leaving variable. 16. **Next entering variable:** Check $$s_2$$ coefficient $$-M$$ negative, no positive coefficients in constraints. 17. **Next entering variable:** Check $$a_1$$ coefficient $$3$$ positive, so no improvement. 18. **Next entering variable:** Check $$x_1$$ coefficient $$4 + 3M$$ positive, no improvement. 19. **Since no further pivots possible to reduce Z, and artificial variables remain in basis, problem is infeasible or requires further analysis.** 20. **Summary:** The Big M method tableau shows artificial variables cannot be removed with given pivot operations and constraints, indicating no feasible solution under current setup. **Final answer:** No feasible solution found using Big M method tableau with given pivot operations.