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Consider the following linear program: maximizesubjectto3x1+4x2+3x3+6x42x1+x2x3+x4x1+x2+x3+x4x2+2x3+x4x1,x2,x3,x412=8100. After transforming the problem into standard form and apply Simplex method, we obtain the final tableau as follow: a)
Consider the following linear program: maximizesubjectto3x1+4x2+3x3+6x42x1+x2x3+x4x1+x2+x3+x4x2+2x3+x4x1,x2,x3,x412=8100. After transforming the problem into standard form and apply Simplex method, we obtain the final tableau as follow: a) Derive the dual problem of the linear program (1) and calculate a dual solution based on complementarity conditions. Given that the optimal solution to the primal solution is unique, investigate whether the dual solution is unique. b) Do the optimal solution and the objective function value change if we - decrease the objective function coefficient for x3 to 0 ? - increase the objective function coefficient for x3 to 9 ? - decrease the objective function coefficient for x4 to 5 ? - increase the objective function coefficient for x1 to 7 ? e) Find the possible range for adjusting the coefficient 8 of the second constraint such that the current basis is kept optimal. Consider the following linear program: maximizesubjectto3x1+4x2+3x3+6x42x1+x2x3+x4x1+x2+x3+x4x2+2x3+x4x1,x2,x3,x412=8100. After transforming the problem into standard form and apply Simplex method, we obtain the final tableau as follow: a) Derive the dual problem of the linear program (1) and calculate a dual solution based on complementarity conditions. Given that the optimal solution to the primal solution is unique, investigate whether the dual solution is unique. b) Do the optimal solution and the objective function value change if we - decrease the objective function coefficient for x3 to 0 ? - increase the objective function coefficient for x3 to 9 ? - decrease the objective function coefficient for x4 to 5 ? - increase the objective function coefficient for x1 to 7 ? e) Find the possible range for adjusting the coefficient 8 of the second constraint such that the current basis is kept optimal
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