Chapter 1: Problem 59
Explain why the Borda count method satisfies the monotonicity criterion.
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Chapter 1: Problem 59
Explain why the Borda count method satisfies the monotonicity criterion.
These are the key concepts you need to understand to accurately answer the question.
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An election with six candidates \((A, B, C, D, E,\) and \(F)\) is decided using the method of pairwise comparisons. If \(A\) loses four pairwise comparisons, \(B\) and \(C\) both lose three, \(D\) loses one and ties one, and \(E\) loses two and ties one, (a) find how many pairwise comparisons \(F\) loses. (b) find the winner of the election.
The following simple variation of the conventional Borda count method is sometimes used: last place is worth 0 points, second to last is worth 1 point,..., first place is worth \(N-1\) points (where \(N\) is the number of candidates). Explain why this variation is equivalent to the conventional Borda count described in this chapter (i.e., it produces exactly the same winner and the same ranking of the candidates).
An election with five candidates \((A, B, C, D,\) and \(E)\) is decided using the method of pairwise comparisons. If \(B\) loses two pairwise comparisons, \(C\) loses one, \(D\) loses one and ties one, and \(E\) loses two and ties one, (a) find how many pairwise comparisons \(A\) loses. (b) find the winner of the election.
An election was held using the conventional Borda count method. There were four candidates \((A, B, C,\) and \(D)\) and 110 voters. When the points were tallied (using 4 points for first, 3 points for second, 2 points for third, and 1 point for fourth), \(A\) had 320 points, \(B\) had 290 points, and \(C\) had 180 points. Find how many points \(D\) had and give the ranking of the candidates.
Explain why the method of pairwise comparisons satisfies the monotonicity criterion.
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