# Hw7s - ENGRI 1101 Engineering Applications of OR Fall 2008...

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Unformatted text preview: ENGRI 1101 Engineering Applications of OR Fall 2008 Homework 7 Homework 7 Solutions 1. (a) To make it easier to talk about this problem, suppose — as we did in class — that the rows of the matrix correspond to jobs, and the columns correspond to workers. We know that we have to assign exactly one job to person 1. So if we subtract 15 from every entry in the first column of the matrix, the cost of every overall assignment will go down by 15, since no matter which job we decide to assign to person 1, it will be 15 cheaper to do this than before. But this means that if it was optimal to assign job j to person 1 before, then it will still be optimal to do this now, since the cost of any feasible solution went down by 15. So the cheapest assignment on the original cost matrix is still the cheapest assignment with respect to the new cost matrix. (b) The first two steps of the Hungarian algorithm are straightforward: 5 7 11 14 6 2 9 3 4 7 12 6 8 5 9 5 →- 5 from row 1- 2 from row 2- 4 from row 3- 5 from row 4 → 0 2 6 9 4 0 7 1 0 3 8 2 3 0 4 0 → - 4 from column 3 → 0 2 2 9 4 0 3 1 0 3 4 2 3 0 0 0 We cannot find a 0-assignment yet, so we look for a 0-cover with the smallest number of rows/columns, which is given by column 1, column 2 and row 4 1 . The minimum uncovered entry is 1, so we subtract this from each row not in the 0-cover (i.e., rows 1, 2 and 3), and we add it to each column in the 0-cover (i.e., columns 1 and 2): 0 2 2 9 4 0 3 1 0 3 4 2 3 0 0 0 →- 1 from row 1- 1 from row 2- 1 from row 3 → - 1 1 1 8 3- 1 2 0- 1 2 3 1 3 0 0 0 → +1 to column 1 +1 to column 2 → 0 2 1 8 4 0 2 0 0 3 3 1...
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Hw7s - ENGRI 1101 Engineering Applications of OR Fall 2008...

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