For each part, determine whether the first point in the consequence \(X^{(1)}\) dominates the second point in the consequence space \(X^{(2)}\): (i) (min, max): \(X^{(1)} = (1.2, 3... For each part, determine whether the first point in the consequence \(X^{(1)}\) dominates the second point in the consequence space \(X^{(2)}\): (i) (min, max): \(X^{(1)} = (1.2, 3.5), X^{(2)} = (1.5, 3.0)\) (ii) (min, max, min): \(X^{(1)} = (10.5, 1.5, -10.0), X^{(2)} = (5.0, 0.5, -12)\) (iii) (max, max, min): \(X^{(1)} = (5, 5, 3), X^{(2)} = (2, 5, 4)\) Read more

Steps to Solve

1. Analyze part (i): (min, max); $X^{(1)} = (1.2, 3.5)$, $X^{(2)} = (1.5, 3.0)$

   • First dimension should be minimized: We compare the first dimension of $X^{(1)}$ and $X^{(2)}$, which are 1.2 and 1.5 respectively. Since 1.2 < 1.5, $X^{(1)}$ is better in the first dimension.
   • Second dimension should be maximized: We compare the second dimension of $X^{(1)}$ and $X^{(2)}$, which are 3.5 and 3.0 respectively. Since 3.5 > 3.0, $X^{(1)}$ is better in the second dimension.
   • Since $X^{(1)}$ is better than $X^{(2)}$ in both dimensions, $X^{(1)}$ dominates $X^{(2)}$.

2. Analyze part (ii): (min, max, min); $X^{(1)} = (10.5, 1.5, -10.0)$, $X^{(2)} = (5.0, 0.5, -12)$

   • First dimension should be minimized: We compare the first dimension of $X^{(1)}$ and $X^{(2)}$, which are 10.5 and 5.0 respectively. Since 10.5 > 5.0, $X^{(1)}$ is worse in the first dimension.
   • Second dimension should be maximized: We compare the second dimension of $X^{(1)}$ and $X^{(2)}$, which are 1.5 and 0.5 respectively. Since 1.5 > 0.5, $X^{(1)}$ is better in the second dimension.
   • Third dimension should be minimized: We compare the third dimension of $X^{(1)}$ and $X^{(2)}$, which are -10.0 and -12.0 respectively. Since -10.0 > -12.0, $X^{(1)}$ is worse in the third dimension.
   • $X^{(1)}$ is better in the second dimension, but worse in the first and third dimensions. Thus, $X^{(1)}$ does not dominate $X^{(2)}$.

3. Analyze part (iii): (max, max, min); $X^{(1)} = (5, 5, 3)$, $X^{(2)} = (2, 5, 4)$

   • First dimension should be maximized: We compare the first dimension of $X^{(1)}$ and $X^{(2)}$, which are 5 and 2 respectively. Since 5 > 2, $X^{(1)}$ is better in the first dimension.
   • Second dimension should be maximized: We compare the second dimension of $X^{(1)}$ and $X^{(2)}$, which are both 5. Thus, neither is better.
   • Third dimension should be minimized: We compare the third dimension of $X^{(1)}$ and $X^{(2)}$, which are 3 and 4 respectively. Since 3 < 4, $X^{(1)}$ is better in the third dimension.
   • Since $X^{(1)}$ is better or equal in all dimensions, and strictly better in at least one dimension, $X^{(1)}$ dominates $X^{(2)}$.