Slide 15-1 Copyright © 2005 Pearson Education, Inc. SEVENTH EDITION and EXPANDED SEVENTH EDITION

Slide 15-1 Copyright © 2005 Pearson Education, Inc.

SEVENTH EDITION and EXPANDED SEVENTH EDITION

Copyright © 2005 Pearson Education, Inc.

Chapter 15

Voting and Apportionment


15.1

Voting Methods


Example: Voting

Voting for Math Club President: Four students are running for president of the Math Club: Jerry, Thomas, Annette and Becky. The club members were asked to rank all candidates. The resulting preference table for this election is shown on the next slide.a) How many students voted in the election?b) How many students selected the candidates in this order: A, J, B, T?c) How many students selected A as their first choice?


Example: Voting continued

a) How many students voted in the election?Add the row labeled Number of Votes14 + 12 + 9 + 4 + 1 = 40Therefore, 40 students voted in the election.

TTTTBFourth

A

B

J

4

JBJAThird

BJAJSecond

AABTFirst

191214Number of Votes


Example: Voting continued

b) How many students selected the candidates in this order: A, J, B, T?

3rd column of numbers, 9 people

c) How many students selected A as their first choice?

9 + 1 = 10


Plurality Method

This is the most commonly used and easiest method to use when there are more than two candidates.

Each voter votes for one candidate. The candidate receiving the most votes is declared the winner.


Example: Plurality Method

Who is elected math club president using the plurality method? We will assume that each member would vote for the person he or

she listed in first place.

TTTTBFourth

A

B

J

4

JBJAThird

BJAJSecond

AABTFirst


From the table:Thomas received 14 votes Becky received 12 votes

Annette received 9 + 1 = 10 votes Jerry received 4 votes


Example: Plurality Method continued

Thomas would be elected since he received the most votes. Note that Thomas received 14/40, or 35%, of the first-place votes, which is less than a majority.


Borda Count Method

Voters rank candidates from the most favorable to the least favorable. Each last-place vote is awarded one point, each next-to-last-place vote is awarded two points, each third-from-last-place vote is awarded three points, and so forth. The candidate receiving the most points is the winner of the election.


Example: Borda Count

Use the Borda count method to determine the winner of the election for math club president.

Since there are four candidates, a first-place vote is worth 4 points, a second-place vote is worth 3 points, a third-place vote is worth 2 points, and a fourth-place vote is worth 1 point.


Example: Borda Count continued

Thomas 14 first place votes 0 second place 0 third place 26 fourth place 14(4) + 0 + 0 + 26(1) = 82

Annette 10 first place votes 12 second place 18 third place 0 fourth place 10(4) + 12(3) + 18(2) + 0 = 112

TTTTBFourth

A

B

J

4

JBJAThird

BJAJSecond

AABTFirst




Betty 12 first place votes 5 second place 9 third place 14 fourth place 12(4) + 5(3) + 9(2) + 14 = 95

Jerry 4 first place votes 23 second place 13 third place 0 fourth place 4(4) + 23(3) + 13(2) + 0 = 111

TTTTBFourth

A

B

J

4

JBJAThird

BJAJSecond

AABTFirst




Thomas-82 Annette-112 Betty-95 Jerry-111 Annette, with 112 points, receives the most

points and is declared the winner.


Plurality with Elimination

Each voter votes for one candidate. If a candidate receives a majority of votes, that candidate is declared the winner. If no candidate receives a majority, eliminate the candidate with the fewest votes and hold another election. (If there is a tie for the fewest votes, eliminate all candidates tied for the fewest votes.) Repeat this process until a candidate receives a majority.


Example: Plurality with Elimination

Use the plurality with elimination method to determine the winner of the election for president of the math club.

Count the number of first place votes Annette 10 Betty 12 Thomas 14 Jerry 4 TTTTBFourth

A

B

J

4

JBJAThird

BJAJSecond

AABTFirst



Example: Plurality with Elimination continued Since 40 votes were cast, a candidate must

have 20 first place votes to receive a majority. Jerry had the fewest number of first place votes so he is eliminated.

Redo the table. Thomas 14 Annette 10 Betty 16

T

A

B

4

TTTBThird

BBAASecond

AABTFirst



Example: Plurality with Elimination continued Still, no candidate received a majority. Annette

has the fewest number of first-place vote so she is eliminated.

New preference table Betty 26 Thomas 14 Betty is the winner. T

B

4

TTTBSecond

BBBTFirst



Pairwise Comparison Method

Voters rank the candidates. A series of comparisons in which each candidate is compared with each of the other candidates follows. If candidate A is preferred to candidate B, A receives one point. If candidate B is preferred to candidate A, B received 1 point. If the candidates tie, each receives ½ point. After making all comparisons among the candidates, the candidate receiving the most points is declared the winner.


Example: Pairwise Comparison

Use the pairwise comparison method to determine the winner of the election for math club president.

Comparisons needed:

( 1) 4(3)6

2 2

n nc


Example: Pairwise Comparison continued Thomas versus Jerry

T = 14 J = 12 + 9 + 4 + 1 = 26 Jerry = 1

Thomas versus Annette T = 14 A = 12 + 9 + 4 + 1 = 26 Annette = 1

Thomas versus Betty T = 14 B = 12 + 9 + 4 + 1 = 26 Betty = 1

TTTTBFourth

A

B

J

4

JBJAThird

BJAJSecond

AABTFirst



Example: Pairwise Comparison continued continued

Betty versus Annette B = 12 + 4 = 16 A = 14 + 9 + 1 = 24 Annette = 1

Betty versus Jerry B = 12 + 1 = 13 J = 14 + 9 + 4 = 27 Jerry = 1

Annette versus Jerry A = 12 + 9 + 1 = 22 J = 14 + 4 =

18 Annette = 1 Annette would win the

election since she received 3 points, the most points from the pairwise comparison method.

TTTTBFourth

A

B

J

4

JBJAThird

BJAJSecond

AABTFirst



15.2

Flaws of Voting


Majority Criterion

If a candidate receives a majority (more than 50%), of the first-place votes, that candidate should be declared the winner.


Head-to-Head Criterion

If a candidate is favored when compared head-to-head with every other candidate, that candidate should be declared the winner.


Monotonicity Criterion

A candidate who wins a first election and then gains additional support without losing any of the original support should also win a second election.


Irrelevant Alternative Criterion

If a candidate is declared the winner of an election and in a second election one or more of the other candidates is removed, the previous winner should still be declared the winner.


Summary of the Voting Methods and Whether They Satisfy the Fairness Criteria

May not satisfyMay not satisfy

May not satisfy

May not satisfy

Irrelevant alternatives

Always satisfies

May not satisfy

Always satisfies

Always satisfies

Monotonicity

Always satisfies

May not satisfy

May not satisfy

May not satisfy

Head-to-head

Always satisfies

Always satisfies

May not satisfy

Always satisfies

Majority

Pairwise comparison

Plurality with elimination

Borda countPlurality


15.3

Apportionment Methods


Apportionment

The goal of apportionment is to determine a method to allocate the total number of items to be apportioned in a fair manner.

Four Methods Hamilton’s method Jefferson’s method Webster’s method Adam’s method


Definitions

total populationStandard divisor =

number of items to be allocated

population for the particular groupStandard quota =

standard divisor


Example

A Graduate school wishes to apportion 15 graduate assistantships among the colleges of education, business and chemistry based on their undergraduate enrollments. Find the standard quotas for the schools.


Example continued

8020divisor = 534.67

15

14.999Standard quota

8020188029403200Population

TotalChemistryBusinessEducation

3200

534.675.985

2940

534.675.498

1880

534.673.516


Hamilton’s Method

1. Calculate each group’s standard quota. 2. Round each standard quota down to the

nearest integer (the lower quota). Initially, each group receives its lower quota.

3. Distribute any leftover items to the groups with the largest fractional parts until all items are distributed.


Example: Apportion the 15 graduate assistantships

15456Hamilton’s

13355Lower quota


8020188029403200Population


3200

534.675.985

2940

534.675.498

1880

534.673.516


The Quota Rule

An apportionment for every group under consideration should always be either the upper quota or the lower quota.


Jefferson’s Method

1. Determine a modified divisor, d, such that when each group’s modified quota is rounded down to the nearest integer, the total of the integers is the exact number of items to be apportioned. We will refer to the modified quotas that are rounded

down as modified lower quotas. 2. Apportion to each group its modified lower quota.


Modified divisor = 480

15366Jefferson

3.91676.1256.67Modified quota


8020188029403200Population


3200

534.675.985

2940

534.675.498

1880

534.673.516


Webster’s Method

1. Determine a modified divisor, d, such that when each group’s modified quota is rounded to the nearest integer, the total of the integers is the

exact number of items to be apportioned. We will refer to the modified quotas that are rounded down as

modified rounded quotas. 2. Apportion to each group its modified lower quota.


Adams’s Method

1. Determine a modified divisor, d, such that when each group’s modified quota is rounded up to the nearest integer, the total of the integers is the

exact number ot items to be apportioned. We will refer to the modified quotas that are rounded down as

modified upper quotas. 2. Apportion to each group its modified lower quota.


15.4

Flaws of the Apportionment Methods


Three Flaws of Hamilton’s Method

The three flaws of Hamilton’s method are: the Alabama paradox, the population paradox, and the new-state paradox. These flaws apply only to Hamilton’s method and do not

apply to Jefferson’s method, Webster’s method, or Adam’s method.

In 1980 the Balinski and Young’s Impossibility Theorem stated that there is no perfect apportionment method that satisfies the quota rule and avoids any paradoxes.


Alabama Paradox

The Alabama paradox occurs when an increase in the total number of items to be apportioned results in a loss of an item for the group.


Example: Demonstrating the Alabama Paradox A large company with braches in three cities, must distribute

30 cell phones to the three offices. The cell phones will be apportioned based on the number of employees in each office shown in the table below.

900489250161Employees

Total321Office


Example: Demonstrating the Alabama Paradox continued Apportion the cell phones using Hamilton’s

methods. Does the Alabama paradox occur using

Hamilton’s method if the number of new cell phones increased from 30 to 31. Explain.


Example: Demonstrating the Alabama Paradox continued Based on 30 cell phones, the table is as follows:


291685Lower Quota

301686Hamilton’s apportionment

16.38.335.37Standard Quota

Total321Office


Example: Demonstrating the Alabama Paradox continued Based on 31 cell phones, the table is as follows:


291685Lower Quota


16.848.615.55Standard Quota

Total321Office


Example: Demonstrating the Alabama Paradox continued When the number of cell phones increased from

30 to 31, office one actually lost a cell phone, while the other two offices actually gained a cell phone under Hamilton’s apportionment.


Population Paradox

The Population Paradox occurs when group A loses items to group B, although group A’s population grew at a faster rate than group B’s.


Example: Demonstrating Population Paradox A school district with five elementary schools has funds for 54 scholarships. The student

population for each school is shown in the table below.

5400106311339331538733Population in 2003

5450111111339331540733Population in 2005

D E TotalCBASchool


Example: Demonstrating Population Paradox continued Apportion the scholarships using Hamilton’s

method. If the school wishes to give the same number

of scholarships two years later, does a population paradox occur?


Solution

Based on the population in 2003, the table is as follows:

16

15

15.38

1538

B

9

9

9.33

933

C

11

11

11.33

1133

D

54

52

5400

Total

11

10

10.63

1063

E

733Population in 2003

7Lower Quota


7.33Standard Quota

ASchool


Solution continued

Based on the population in 2005, the table is as follows:

15

15

15.258

1540

B

9

9

9.24

933

C

11

11

11.23

1133

D

54

53

5450

Total

11

11

11.01

1111

E

733Population in 2005

7Lower Quota


7.262Standard Quota

ASchool


Solution continued

In the school district in 2005, school B actually gives one of its scholarships to school A, even though the population in school B actually grew by 2 students.


New State Paradox

The new-states paradox occurs when the addition of a new group changes the apportionment of another group.


Summary

Small states

Small states

Large statesLarge statesAppointment method favors

NoNoNoYesMay produce the new-states paradox

NoNoNoYesMay produce the population paradox

NoNoNoYesMay produce the Alabama paradox

YesYesYesNoMay violate the quota rule

WebsterAdamsJeffersonHamilton

Apportionment Method

Documents

Slide 15-1 Copyright © 2005 Pearson Education, Inc. SEVENTH EDITION and EXPANDED SEVENTH EDITION