Consider the Following Convex Programming Problem

Question

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The Answer is The KKT conditions are 1 a) $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ 2 a) $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ 1 b) $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ 2 b) $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ 3) $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ 4) $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ 5)x₁

\ge

0,x₂

\ge

0,6)u

\ge

0.(a)Consider (x₁,x₂)= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x₁,x₂)= (1,1).So,(x₁,x₂)= (1,1)is not optimal.(b)There are eight different cases where either x₁ = 0 or x₁ > 0,either x₂ = 0 or x₂ > 0,and either u = 0 or u > 0.Consider the case where x₁ > 0,x₂ > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x₁ - 3 $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ - u = 0 8 - 2x₂ - u = 0 x₁ + x₂ - 2 = 0 x₁ ≥ 0,x₂ ≥ 0,u ≥ 0 The third equation gives x₂ = 2 - x₁,so the second equation then reduces to 4 + 2x₁ = u,which then reduces the first equation to 10 - 4x₁ - 3 $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ - (4 + 2x₁)= 0.Solving this quadratic equation gives x₁ = $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ - 1,so $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ and $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that $The KKT conditions are 1 a) 2 a) 1 b) 2 b) 3) 4) 5)x1 \ge 0,x2 \ge 0,6)u \ge 0.(a)Consider (x1 ,x2 )= (1,1).From 2a),we have u = 3; from 2b),we have u = 6.Thus,the KKT conditions cannot be met at (x1 ,x2 )= (1,1).So,(x1 ,x2 )= (1,1)is not optimal.(b)There are eight different cases where either x1 = 0 or x1 > 0,either x2 = 0 or x2 > 0,and either u = 0 or u > 0.Consider the case where x1 > 0,x2 > 0,and u > 0,so that the relevant conditions reduce to the following.10 - 4x1 - 3 - u = 0 8 - 2x2 - u = 0 x1 + x2 - 2 = 0 x1 ≥ 0,x2 ≥ 0,u ≥ 0 The third equation gives x2 = 2 - x1,so the second equation then reduces to 4 + 2x1 = u,which then reduces the first equation to 10 - 4x1 - 3 - (4 + 2x1)= 0.Solving this quadratic equation gives x1 = - 1,so and satisfies all the KKT conditions.Since we are dealing with a convex programming problem,it follows that is optimal.$ is optimal.

Consider the Following Convex Programming Problem

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