Consider the Following Problem

Question

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The Answer is (a)Initialization: After introducing slack variables,the initial system of equations is (0)Z - 2 x₁ + x₂ - x₃ = 0 (1)x₁ - x₂ + 3 x₃+ x₄ = 4 (2)2 x₁ + x₂ + x₅= 10 (3)x₁ - x₂ - x₃ + x₆ = 7 Thus,the initial BF solution is (0,0,0,4,10,7),which is not optimal since Z = 2x₁ - x₂ + x₃ can be incresed by increasing either x₁ or x₃.Iteration 1: Choosing x₁ as the entering basic variable (since increasing x₁ increases Z at the fastest rate)and x₄ as the leaving basic variable (by the minimum ratio test),using elementary algebraic operations to restore proper form from Gaussian elimination then leads to the following system of equations.(0)Z - x₂ + 5 x₃+ 2 x₄ = 8 (1)x₁ - x₂ + 3 x₃+ x₄ = 4 (2)3 x₂ - 6 x₃- 2 x₄ + x₅ = 2 (3) - 4 x₃- x₄ +x₆ = 3 Thus,the current BF solution is (4,0,0,0,2,3),which is not optimal since Z = x₂ - 5x₃ - 2x₄ + 8 can be increased by increasing x₂.Iteration 2: Choosing x₂ as the entering basic variable and x₅ as the leaving basic variable (by the minimum ratio test),using elementary algebraic operations to restore proper form from Gaussian elimination then leads to the following system of equations.(0)Z + 3 x₃+ 4/3 x₄ + 1/3 x₅ = 8.667 (1)x₁ + x₃+ 1/3 x₄+ 1/3 x₅ = 4.667 (2)x₂ - 2 x₃- 2/3 x₄ + 1/3 x₅ = 0.667 (3)- 4 x₃- x₄ + x₆ = 3 From Eq.(0),Z = -3 x₃- 4/3 x₄ - 1/3 x₅ + 8.667,so Z cannot be increased by increasing any of the nonbasic variables.Therefore,the current BF solution (4.667,0.667,0,0,0,3)is optimal with Z = 8.667.(b)Using the same logic as described above,the simplex method in tableau form is shown below.

(a)Initialization: After introducing slack variables,the initial system of equations is (0)Z - 2 x<sub>1</sub> + x<sub>2</sub> - x<sub>3</sub> = 0 (1)x<sub>1</sub> - x<sub>2</sub> + 3 x<sub>3 </sub>+ x<sub>4</sub> = 4 (2)2 x<sub>1</sub> + x<sub>2</sub> + x<sub>5 </sub><sub> </sub>= 10 (3)x<sub>1</sub> - x<sub>2</sub> - x<sub>3 </sub> + x<sub>6</sub> = 7 Thus,the initial BF solution is (0,0,0,4,10,7),which is not optimal since Z = 2x<sub>1</sub> - x<sub>2</sub> + x<sub>3</sub> can be incresed by increasing either x<sub>1</sub> or x<sub>3</sub>.Iteration 1: Choosing x<sub>1</sub> as the entering basic variable (since increasing x<sub>1</sub> increases Z at the fastest rate)and x<sub>4</sub> as the leaving basic variable (by the minimum ratio test),using elementary algebraic operations to restore proper form from Gaussian elimination then leads to the following system of equations.(0)Z - x<sub>2</sub> + 5 x<sub>3 </sub>+<sub> </sub> 2 x<sub>4</sub> <sub> </sub> = 8 (1)x<sub>1</sub> - x<sub>2</sub> + 3 x<sub>3 </sub>+ x<sub>4</sub> <sub> </sub> = 4 (2)3 x<sub>2</sub> - 6 x<sub>3 </sub>- 2 x<sub>4</sub> + x<sub>5</sub> = 2 (3) - 4 x<sub>3 </sub><sub> </sub>- x<sub>4</sub> +x<sub>6</sub> = 3 Thus,the current BF solution is (4,0,0,0,2,3),which is not optimal since Z = x<sub>2</sub> - 5x<sub>3</sub> - 2x<sub>4</sub> + 8 can be increased by increasing x<sub>2</sub>.Iteration 2: Choosing x<sub>2</sub> as the entering basic variable and x<sub>5</sub> as the leaving basic variable (by the minimum ratio test),using elementary algebraic operations to restore proper form from Gaussian elimination then leads to the following system of equations.(0)Z + 3 x<sub>3 </sub>+<sub> </sub> 4/3 x<sub>4</sub> + 1/3 x<sub>5</sub> = 8.667 (1)x<sub>1</sub> + x<sub>3 </sub>+ 1/3 x<sub>4 </sub><sub> </sub>+ 1/3 x<sub>5</sub> = 4.667 (2)x<sub>2</sub> - 2 x<sub>3 </sub><sub> </sub>- 2/3 x<sub>4</sub> + 1/3 x<sub>5</sub> = 0.667 (3)- 4 x<sub>3 </sub>- x<sub>4</sub> + x<sub>6</sub> = 3 From Eq.(0),Z = -3 x<sub>3 </sub>-<sub> </sub> 4/3 x<sub>4</sub> - 1/3 x<sub>5</sub> + 8.667,so Z cannot be increased by increasing any of the nonbasic variables.Therefore,the current BF solution (4.667,0.667,0,0,0,3)is optimal with Z = 8.667.(b)Using the same logic as described above,the simplex method in tableau form is shown below. The current BF solution (4.667,0.667,0,0,0,3)is optimal with Z = 8.667.

The current BF solution (4.667,0.667,0,0,0,3)is optimal with Z = 8.667.

Consider the Following Problem

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