Three jobs are to be assigned to three machines. Cost for each job-machine combination appears in the table below. Perform the first two steps of the assignment method (subtract the smallest number in each row and subtract the smallest number in each column; then cover with straight lines) . At this point in the problem-solving process $$\begin{array}{llll}& \text{ Machine A }& \text{ Machine B }& \text{ Machine C }\\ \text{ Tob 1 }& 11& 14& 6\\ \text{ Tob 2 }& 8& 10& 11\\ \text{ Job 3 }& 9& 12& 7\end{array}$$

A) the row for Job 1 contains the values 5, 6, and 0.
B) calculations are complete, and the problem is ready for an optimal set of assignments.
C) the column for Machine B indicates that it should be retired.
D) Job 1 should be performed on Machine B.
E) Job 1 should be performed on Machine A.

EFirst, subtract the smallest number in each row from all elements of that row. For Job 1, subtract 6 (the smallest number) from all elements, resulting in 5, 8, and 0. For Job 2, subtract 8 (the smallest number) from all elements, resulting in 0, 2, and 3. For Job 3, subtract 7 (the smallest number) from all elements, resulting in 2, 5, and 0. Then, subtract the smallest number in each column from all elements of that column. The smallest numbers in the columns are now 0, 2, and 0, respectively. After subtraction, the table looks like this: $$\begin{array}{llll}& \text{Machine A}& \text{Machine B}& \text{Machine C}\\ \text{Job 1}& 5& 6& 0\\ \text{Job 2}& 0& 0& 3\\ \text{Job 3}& 2& 3& 0\end{array}$$ This confirms choice A, as the row for Job 1 contains the values 5, 6, and 0 after the first two steps.Choice E is incorrect based on the given table and the steps performed; the assignment method does not suggest Job 1 should be performed on Machine A at this stage. The method involves minimizing the cost, and the choice does not directly follow from the steps described. The correct interpretation of the table after the first two steps does not inherently suggest an optimal assignment without further analysis.