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
.
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 Spring '07
 Berkowitz
 Math, Voting system, Condorcet

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