MTH 108
Exam 1 on Chapters 1 and 3
Fall 2003
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1. The Board of Directors of the XY Z Corporation is holding an election to choose a new
Chairman of the Board. The candidates are Allen, Beckman, Cole, Dent, and Emery.
The preference schedule for the election is given in the following table.
Number of Voters
First choice
Second choice
Third choice
Fourth choice
Fifth choice
10
C
A
D
B
E
8
B
A
D
E
C
5
D
A
B
C
E
4
A
D
B
E
C
3
E
C
A
D
B
(a) Find the winner of the election under the Plurality method.
Sol: A : 4, B : 8, C : 10, D : 5, E : 3 C is the winner.
(b) Is there a condorcet candidate? If so, who?
Sol: A vs B : A wins, A vs C : A wins, A vs D : A wins, A vs E : A wins.
So, A is the condorcet candidate.
(c) Do your answers to (a) and (b) demonstrate whether or not the Plurality Method
violates the Condorcet Criterion?
Sol: Yes, it does. It says the Plurality Method violates the Condorcet Criterion.
because the Condorcet Candidate is A, but A is not the winner under the plurality
method of tabulating the votes.
2. The members of the Tasmania State University soccer team are having an election
to choose the captain of the team from among the four seniors- Anderson, Bergman,
Chow, and Delgado. The preference schedule for the election is given in the following
table.
Number of Voters 4 1 9 8 5
First choice
B B D A C
Second choice
A A C D D
Third choice
D D B B B
Fourth choice
C C A C A
Find the winner of the election using the Method of Pairwise Comparisons.
Sol: A vs B : B wins
A vs C : C wins
A vs D : D wins
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B vs C : C wins
B vs D : D wins
C vs D : D wins
Points: A : 0, B : 1, C : 2, D : 3. So D is the winner.
3. A math class is asked by the instructor to vote among four possible dates for the final
exam - A (Dec. 15), B (Dec. 20), C (Dec. 21), and D (Dec. 23). The following is the
class preference schedule.
Number of Voters
First choice
Second choice
Third choice
Fourth choice
10
C
D
B
A
3
A
C
D
B
8
A
D
B
C
5
B
D
C
A
2
B
C
D
A
Find the winner of the election using the Plurality With Elimination Method.
Sol: There are 28 voters, so 15 are needed for a majority.
Round 1: A : 11, B : 7, C : 10, D : 0. Eliminate D.
Round 2: A : 11, B : 7, C : 10. Eliminate B.
Round 3: A : 11, C : 17. C has the majority and so C is the winner.
4. For this question, refer to the information given in problem number 3. Find the winner
of the election using the Borda Count Method.
Sol: A : 10 · 1 + 3 · 4 + 8 · 4 + 5 · 1 + 2 · 1 = 61. B : 10 · 2 + 3 · 1 + 8 · 2 + 5 · 4 + 2 · 4 = 67.
C : 10 · 4 + 3 · 3 + 8 · 1 + 5 · 2 + 2 · 3 = 73. D : 10 · 3 + 3 · 2 + 8 · 3 + 5 · 3 + 2 · 2 = 79.
D is the winner.
5. Suppose there is an election between candidates A, B, C, D, E. The method of pairwise
comparisons is used and the results of the pairwise comparisons are as follows.
A vs B A wins
A vs C A wins
A vs D A wins
A vs E A wins
B vs C C wins
B vs D B wins
B vs E E wins
C vs D C wins
C vs E E wins
D vs E E wins
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Rank the candidates using the extended pairwise comparisons method.
Sol: A : 4, B : 1, C : 2, D : 0, E : 3. First: A, Second: E, Third: C, Forth: B, Fifth:
D.
6. Candy and Cookie want to divide an chocolate-vanilla cake worth $16 using the dividerchooser method. Candy values chocolate three times as much as vanilla. How much in
terms of dollars is the chocolate side worth to Candy and how much is the vanilla side
worth?
Sol: The chocolate side is worth 3 out of 4 parts. And so is worth 16 ·
The vanilla side is worth 1 out of 4 parts. And so is worth 16 ·
1
4
3
4
= 12 dollars.
= 4 dollars.
7. Susan and Veronica want to divide an orange-pineapple cake worth $10 using the
divider-chooser method. Susan puts a value of $7.50 on the orange side and $2.50 on
the pineapple side while Veronica puts a value of $4 on the orange side and $6 on the
pineapple side. Susan is the divider of the cake.
(a) From the choices given, pick one cut that is a valid way for Susan to cut the cake
so that each piece is worth a fair share to Susan.
Sol: The first picture shows the cut on the left, LeftSide, with an interior angle
120o of the orange side. This piece is 23 of the orange side and so is worth 23 ·7.50 = 5
dollars. The other piece must also be worth 5 dollars. This IS a valid way for
Susan to cut the cake.
The second picture shows the cut on the left, LeftSide, with an angle of 90o of the
orange side and an angle of 135o of the pineapple side. That is 21 of the orange
and 34 of the pineapple. Thus LeftSide is worth 12 · 7.50 = 3.75 plus 34 · 2.50 = 1.88
dollars. So LeftSide is worth 5.63 which is more than half. So this is not a valid
way for Susan to cut the cake.
(b) Using the cut you chose is part (a), show the value of each piece in Veronica’s
eyes.?
Sol: Looking only at the First Cut, we see that LeftSide is only worth 23 · 4 = 2.67
dollars to Veronica. So she would select the piece on the right RightSide.
8. Three players (one divider and two choosers ) are going to divide a plot of land fairly
using the lone divider method. Suppose the choosers value the sections as follows.
s1
s2
s3
Chooser 1 25% 50% 25 % What is a fair division of the land? Say which piece
Chooser 2 40 % 20 % 40 %
is given to the divider and each of the choosers.
Sol: The divider could get S3 , Chooser 1 could get S2 , and Chooser 2 could get S1 .
9. Ed and Elizabeth are getting a divorce and spitting the car, the house, and the truck
using the method of sealed bids. Ed bids $30,000 on the car, $70,000 on the house and
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$20,000 on the truck. Elizabeth bids $33,000 on the car, $60,000 on the house, and
$22,000 on the truck. Describe the final outcome of this fair-division problem.
car
house
truck
total
fair share
Ed
Eliz.
30,000 33,000
70,000 60,000
20,000 22,000
120,000 115,000
60,000 57,500
In the initial allocation, the car and the truck go to Elizabeth. They are worth $55,000
to Elizabeth and she thinks a fair share would be $57,500, so she will also get $2,500
from the estate.
The house goes to Ed. He will have to pay the estate $10,000.
The estate has 10, 000 − 2500 = 7500 so they split and each gets an additional $ 3750.
10. Four players (A, B, C, and D) agree to divide the 15 items shown below by lining
them up in order and using the method of markers. The players’ bids are as indicated.
Describe the initial allocation of the items to each player and name the pieces that are
left over.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
A : A1 is between 2 and 3. A2 is between 5 and 6. A3 is between 11 and 12.
B : B1 is between 4 and 5. B2 is between 7 and 8. B3 is between 12 and 13.
C : C1 is between 3 and 4. C2 is between 7 and 8. C3 is between 10 and 11.
D : D1 is between 3 and 4. D2 is between 6 and 7. D3 is between 10 and 11.
Sol: A gets 1 and 2.
D gets 4, 5, 6.
C gets 8, 9, 10.
B gets 13, 14, 15.
Leftovers: 3, 7, 11, 12.
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