Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 13 - Voting and Apportionment - 13.1 Voting Methods - Exercise Set 13.1 - Page 851: 36

Answer

(a) Candidate A is declared the new department chair using the pairwise comparison method. (b) After Candidate E withdraws, we can see the new preference table below. After Candidate E withdraws, Candidate A is declared the new department chair using the pairwise comparison method.
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Work Step by Step

(a) With the pairwise comparison method, each candidate is compared with every other candidate. For each pair of candidates, if one candidate is ranked higher than the other candidate on a majority of ballots, then the higher-ranked candidate receives 1 point. If the two candidates tie, then they each receive 0.5 points. After all the comparisons have been made, the candidate which receives the most points is declared the winner. We can compare Candidate A and Candidate C. Candidate A: 5 + 3 + 3 = 11 Candidate C: 5 + 3 + 2 = 10 Since Candidate A is ranked higher than Candidate C on more ballots, Candidate A receives 1 point. We can compare Candidate A and Candidate B. Candidate A: 5 + 5 + 3 = 13 Candidate B: 3 + 3 + 2 = 8 Since Candidate A is ranked higher than Candidate B on more ballots, Candidate A receives 1 point. We can compare Candidate B and Candidate C. Candidate B: 5 + 3 + 3 = 11 Candidate C: 5 + 3 + 2 = 10 Since Candidate B is ranked higher than Candidate C on more ballots, Candidate B receives 1 point. We can compare Candidate A and Candidate D. Candidate A: 5 + 3 + 3 = 11 Candidate D: 5 + 3 + 2 = 10 Since Candidate A is ranked higher than Candidate D on more ballots, Candidate A receives 1 point. We can compare Candidate D and Candidate C. Candidate D: 3 + 3 + 2 = 8 Candidate C: 5 + 5 + 3 = 13 Since Candidate C is ranked higher than Candidate D on more ballots, Candidate C receives 1 point. We can compare Candidate B and Candidate D. Candidate B: 5 + 3 = 8 Candidate D: 5 + 3 + 3 + 2 = 13 Since Candidate D is ranked higher than Candidate B on more ballots, Candidate D receives 1 point. We can compare Candidate A and Candidate E. Candidate A: 5 + 3 + 2 = 10 Candidate E: 5 + 3 + 3 = 11 Since Candidate E is ranked higher than Candidate A on more ballots, Candidate E receives 1 point. We can compare Candidate B and Candidate E. Candidate B: 5 + 3 + 3 + 3 + 2 = 16 Candidate E: 5 Since Candidate B is ranked higher than Candidate E on more ballots, Candidate B receives 1 point. We can compare Candidate D and Candidate E. Candidate D: 5 + 3 + 3 + 2 = 13 Candidate E: 5 + 3 = 8 Since Candidate D is ranked higher than Candidate E on more ballots, Candidate D receives 1 point. We can compare Candidate E and Candidate C. Candidate E: 3 Candidate C: 5 + 5 + 3 + 3 + 2 = 18 Since Candidate C is ranked higher than Candidate E on more ballots, Candidate C receives 1 point. After all the pairwise comparisons have been made, we can add up the total number of points for each candidate. Candidate A: 1 + 1 + 1 = 3 points Candidate B: 1 + 1 = 2 points Candidate C: 1 + 1 = 2 points Candidate D: 1 + 1 = 2 points Candidate E: 1 point Since Candidate A received the most points, Candidate A is declared the winner. Candidate A is declared the new department chair using the pairwise comparison method. (b) Suppose Professor E withdraws before the votes are counted. Then each candidate who was below Candidate E on a ballot will move up one spot on that ballot. We can see the new preference table below. To find the new department chair, we can use the pairwise comparison method with the remaining four candidates. We can compare Candidate A and Candidate C. Candidate A: 5 + 3 + 3 = 11 Candidate C: 5 + 3 + 2 = 10 Since Candidate A is ranked higher than Candidate C on more ballots, Candidate A receives 1 point. We can compare Candidate A and Candidate B. Candidate A: 5 + 5 + 3 = 13 Candidate B: 3 + 3 + 2 = 8 Since Candidate A is ranked higher than Candidate B on more ballots, Candidate A receives 1 point. We can compare Candidate B and Candidate C. Candidate B: 5 + 3 + 3 = 11 Candidate C: 5 + 3 + 2 = 10 Since Candidate B is ranked higher than Candidate C on more ballots, Candidate B receives 1 point. We can compare Candidate A and Candidate D. Candidate A: 5 + 3 + 3 = 11 Candidate D: 5 + 3 + 2 = 10 Since Candidate A is ranked higher than Candidate D on more ballots, Candidate A receives 1 point. We can compare Candidate D and Candidate C. Candidate D: 3 + 3 + 2 = 8 Candidate C: 5 + 5 + 3 = 13 Since Candidate C is ranked higher than Candidate D on more ballots, Candidate C receives 1 point. We can compare Candidate B and Candidate D. Candidate B: 5 + 3 = 8 Candidate D: 5 + 3 + 3 + 2 = 13 Since Candidate D is ranked higher than Candidate B on more ballots, Candidate D receives 1 point. After all the pairwise comparisons have been made, we can add up the total number of points for each candidate. Candidate A: 1 + 1 + 1 = 3 points Candidate B: 1 point Candidate C: 1 point Candidate D: 1 point Since Candidate A received the most points, Candidate A is declared the winner. Candidate A is declared the new department chair using the pairwise comparison method.
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