103s11chap1Bsolutions

103s11chap1Bsolutions - Math 103, Spring 2011, Solutions to...

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Math 103, Spring 2011, Solutions to Chapter 1 homework 33 total points Problem 34: 6 points (a) Round 1: D has fewest first place votes, namely 0, so D is eliminated first. Round 2: B has only 5 first place votes while A and C each have 6, so B is eliminated. Round 3: Only A and C remain, and C now has 5+6=11 first place votes while A still has only 6. A is eliminated and C is the winner under plurality with elimination. (b) Now the preference schedule is # of voters: 4 5 6 2 1st choice A B C C 2nd choice D C A A 3rd choice B A D D 4th choice C D B B which can be consolidated to # of voters: 4 5 8 1st choice A B C 2nd choice D C A 3rd choice B A D 4th choice C D B First round: D is eliminated, resulting in # of voters: 4 5 8 1st choice A B C 2nd choice B C A 3rd choice C A B Second round: A is eliminated, resulting in # of voters: 4 5 8 1st choice B B C 2nd choice C C B
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Third round: C is eliminated, and B is the winner in the official vote. (c) The results of (a) and (b) imply that the plurality with elimination method violates the monotonicity criterion . Changing some ballots in favor of the original winner, and making no other changes, has caused
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This note was uploaded on 01/20/2012 for the course MATH 103 taught by Professor Berkowitz during the Spring '07 term at Rutgers.

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103s11chap1Bsolutions - Math 103, Spring 2011, Solutions to...

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