304470-ch14 pV1-V59 11/7/06 10:01 AM Page 2 · 07-11-2006 · 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 2. 14.1 Voting Systems V3 Thus, you would expect about 44,347 Florida votes

Voting and ApportionmentOne of the most precious rights in our democracy is the right to vote. We have elections to select the president of the United States, senatorsand representatives, members of the United Nations General Assembly,baseball players to be inducted into the Baseball Hall of Fame, and even“best” performers to receive Oscar and Grammy awards. There are manyways of making the final decision in these elections, some simple, somemore complex.

Electing senators and governors is simple: Have some primary elec-tions and then a final election. The candidate with the most votes in thefinal election wins. Elections for president, as attested by the controver-sial 2000 presidential election, are complicated by our Electoral Collegesystem. Under this system, each state is allocated a number of electorsselected by their political parties and equal to the number of its U.S.senators (always two), plus the number of its U.S. representatives (whichmay change each decade according to the size of each state’s populationas determined in the census). These state electors cast their electoralvotes (one for president and one for vice president) and send them to thepresident of the Senate who, on the following January 6, opens and readsthem before both houses of Congress. The candidate for president withthe most electoral votes, provided that it is an absolute majority (one overhalf the total), is declared president. Similarly, the vice presidential can-didate with the absolute majority of electoral votes is declared vice pres-ident. At noon on January 20, the duly elected president and vice presi-dent are finally sworn into office.

In this chapter we will look at several voting methods, the “fairness”of these methods, how votes are apportioned or divided among voters orstates, and the fairness of these apportionments.

V1

C H A P T E R

14The concept of apportionment or fairdivision plays a vital role in theoperation of corporations, politics, andeducational institutions. For example,colleges and universities deal with largeissues of apportionment such as theallocation of funds. In Section 14.3, youwill encounter many different types ofapportionment problems.

14.1 Voting Systems*

14.2 Voting Objections

14.3 Apportionment Methods

14.4 ApportionmentObjections

*Portions of this section were developed byProfessor William Webb of WashingtonState University and funded by a NationalScience Foundation grant (DUE-9950436)awarded to Professor V. S. Manoranjan.

David Butow/Corbis SABA

For links to Internet sites related to Chapter 14, please accesscollege.hmco.com/PIC/bello9eand click on the Online Study Center icon.

Online Study Center

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 1

V2 14 Voting and Apportionment

Voting Systems

Monsieur Butterfly and Pizza Too

When you vote in a presidential election, you are not directly voting for the pres-ident! You are actually voting for electors, individuals who cast the electoralvotes on behalf of their party and states. They are the ones who elect the presi-dent. Originally, electors were free to cast their votes as they pleased, but manyof today’s electors are “bound” or “committed” by state law (25 states have suchlaws) to vote for the candidate who received the most popular votes in their state.In a typical U.S. election, voters vote for their first choices by using a ballot. Aso-called butterfly ballot used in Palm Beach County, Florida, during the 2000presidential election is shown. There was some confusion about votes cast forPat Buchanan (second hole) or Al Gore (third hole).

About 460,000 votes were cast in Palm Beach County, and of those, 3400were for Buchanan. Assuming that the remaining precincts in Florida wouldyield the same proportion of votes for Buchanan, how many of the approxi-mately 6 million votes cast in Florida would you project for Buchanan? Thinkabout it before you answer!

The proportion of votes for Buchanan in Palm Beach was

If the same proportion applies to Florida,

or equivalently

F � 44,348

4600F � 34 � 6,000,000

F6,000,000

�34

4600

3400460,000

�34

4600

GE

TT

I N G S T A R T ED

Marie-JeanAntoineNicolas deCaritat,Marquis deCondorcet,was bornSeptember17, 1743, in Ribemont,France. Condorcetdistinguished himself as awriter, administrator, andpolitician. His mostimportant work was theEssay on the Application ofAnalysis to the Probabilityof Majority Decisions(1785), in which he tried tocombine mathematics andphilosophy to apply tosocial phenomena. One ofthe major developments inthis work is known as theCondorcet paradox, a topiccovered in this chapter.

Looking AheadIn this chapter we shall studydifferent voting systems, the“flaws” or objections that can beraised about such systems, themethods used to fairly apportionresources among differentgroups, and the objections tothese apportionment methods.

A butterfly ballot used in Palm Beach County, Florida, during the 2000 presidential election.

HUMAN SIDE OF MATH

(1743–1794)

The

Gra

nger

Col

lect

ion

Robert Duyos/South Florida Sun-Sentinel

14.1

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 2

14.1 Voting Systems V3

Thus, you would expect about 44,347 Florida votes for Buchanan. (He actuallygot about 17,000 votes in Florida.) Moreover, the number of registered voters forBuchanan’s Reform Party in Palm Beach County was a mere 304 voters! Whatmight be some of the reasons for this discrepancy? �

There are two fundamentally different types of voting methods: preferential andnonpreferential. As the name suggests, a preferential voting system asks a voterto state a preference by ranking alternatives. This is usually done using a pref-erence table or preference schedule.

For example, suppose the Math Club wants to order some pizzas for theend-of-year party. The Pizza House offers a special: three different one-toppingpizzas—one jumbo, one large, and one medium—for only $20. The question is,Which topping to order on which pizza? The club members decide that themost popular topping should go on the jumbo pizza, the second-choice toppingon the large pizza, and the third choice on the medium pizza; the toppingchoices are pepperoni, sausage, mushrooms, or anchovies. Each club membercould fill out a preference ballot, and the results for the ballots might be sum-marized as in Table 14.1.

If you were only considering each person’s first choice, and Joan, Richard, andSuzanne were the only voters, sausage would win 2 votes to 1. We say that sausagereceived a majority (2 out of 3) of the first-place votes. A candidate with a major-ity of the votes is the one with more than half, or 50%, of the vote. Looking at thetable, you might argue that pepperoni is a better choice because each voter has itlisted as first or second choice. If all the Math Club members were voting, listingall the ballots would take a lot of space because with only four toppings, therewould be 4! � 24 different ballots to consider. If you had five toppings, therewould be 5! � 120 ballots. You can summarize the results of an election byshowing how often a particular outcome was selected with a preference table. Doit in steps.

1. Replace the word sausage with the letter S, pepperoni with the letter P, andso on.

2. If several people have exactly the same list of preferences, list them together.Suppose five people all vote S, P, M, A. This fact is shown by using a table ofvotes like Table 14.2. The number 5 at the top indicates that five people hadthe exact results S, P, M, A on their ballots. Note that the first choice S appearsat the top, the second choice P is next, and so on.

3. Assume that all the club members chose one of 3 or 4 different rankings. Thevoting methods we will study will work the same way no matter how manyof the 4! � 24 possible rankings were chosen.

TA B L E 14 .1

Choice Joan Richard Suzanne

First Sausage Sausage Pepperoni

Second Pepperoni Pepperoni Mushrooms

Third Mushrooms Anchovies Sausage

Fourth Anchovies Mushrooms Anchovies

5

S

P

M

A

TA B L E 14 . 2

PhotoDisc/Getty Images

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 3


Now we are ready to analyze the results of elections using different votingsystems: plurality, plurality with runoff, plurality with elimination, Borda count,and pairwise comparison.

A. The Plurality MethodAs you can see from the results of the 2000 presidential election in Table 14.3,if there are three or more candidates it is possible that no candidate receives amajority (more than 50%) of the votes. In this case, one method of selecting thewinner is to select the candidate with the most votes. This method is called theplurality method. In a U.S. presidential election, the candidate with the mostpopular votes does not necessarily win!

Now let us go back to our pizza ballots.

EXAMPLE 1 � Using the Plurality Method

The Math Club conducted an election, and the results were as shown in Table 14.4.

(a) Did any of the rankings get the majority of the votes?

(b) Which topping is the plurality winner?

(c) Which topping comes in second?

(d) Which topping comes in last?

SolutionSince plurality counts only first-place votes, we can see that A got 7 votes (seecolumn 1, with 7 at the top), S got 5 votes (column 2), and P got 4 � 2 � 6 votes.Mushroom was never at the top, so it got no votes.

(a) None of the rankings got a majority of the votes. Since there are 7 � 5 � 4 � 2 � 18 voters, more than 18/2 � 9 votes are needed for a majority.

(b) A (anchovies) is the plurality winner with 7 votes.

TA B L E 14 . 3 Results as of 6:00 P.M., EST, 11/17/2000

Candidates Votes Vote (%)

Gore 49,921,267 49

Bush 49,658,276 48

Nader 2,756,008 3

Buchanan 447,927 0

No winner declared

RF

G

R

D

Plurality Method

Each voter votes for one candidate. The candidate with the most first-placevotes is the winner.

TA B L E 14 . 4

7 5 4 2

A S P P

S P S M

M M M S

P A A A

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 4


(c) P (pepperoni) comes in second with 6 votes.

(d) M (mushrooms) comes in last with no votes. �

As we saw in Example 1, a plurality is not necessarily a majority. There maybe a situation with a large number of alternative choices where the winner mightnot get even 10% of the votes! Many political elections have only two candidates(or at least only two with a chance of winning). With only two choices, a plural-ity is necessarily a majority. However, there are also numerous instances withmany candidates, including primary elections, electing members to the BaseballHall of Fame, ranking football teams, and so on.

In Example 1, the jumbo pizza ended up with anchovies (A) as the toppingbut you may have noticed that anchovies was the last choice of 5 � 4 � 2 � 11voters. Since many people who don’t like anchovies really hate anchovies, itcould well be the case that these 11 people—a clear majority—might not evenwant any of the jumbo pizza. Although this means more pizza for the 7 peoplewho like anchovies, it doesn’t seem like the fairest way to choose. How can weovercome these difficulties? One way is to begin by eliminating all but the toptwo candidates and then make a head-to-head comparison between these two.Now the winner will have a majority! This variation of the plurality method iscalled plurality with runoff.

EXAMPLE 2 � Using the Plurality with Runoff Method

As you recall from Example 1, the election results were A, 7 votes; P, 6 votes; S,5 votes; and M, 0 votes. Find the winner using the plurality with runoff method.

SolutionSince the top two vote getters were A and P, all others are eliminated, and we run an election between A and P. Look at Table 14.5 and mentally (or you canactually do it with a pencil) cross out all the S and M entries. Now look at the first column. There are 7 people who prefer A to P. The second, third, andfourth columns have 5 � 4 � 2 � 11 people who prefer P to A. (Note that in these columns we are only concerned with the fact that P is preferred over A, not the particular value of the preference.) Eleven is the majority of the7 � 5 � 4 � 2 � 18 people voting in the election. Thus, A has 7 votes againstP’s 11, and P is the new winner using the plurality with runoff method. Thejumbo pizza will now have pepperoni! �

So far we have looked at the methods of plurality and plurality with runoff,two of the most widely used methods for political elections in many countries.Although they can be used to obtain a complete ranking of many alternatives,they are really designed to choose an overall winner. A major problem with bothmethods is that candidates who do not get either the most or second most first-place votes are immediately eliminated. Do we really want to place so muchemphasis on first-place votes?

Plurality with Runoff Method

Each voter votes for one candidate. If a candidate receives a majority ofvotes, that candidate is the winner. If no candidate receives a majority,eliminate all but the two top candidates and hold a runoff election. Thecandidate that receives a majority in the runoff election is the winner.

TA B L E 14 . 5

7 5 4 2

A S P P

S P S M

M M M S

P A A A

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 5


A fairly natural way to correct this emphasis on first-place votes is to usesome kind of system that assigns a point value to each of the rankings and thencounts points instead of votes. This kind of method is widely used in rankingsports teams such as in football polls, as well as in scoring track meets or select-ing winners in music or television award shows. Historically, this method goes back to the eighteenth century and is named for Jean-Charles Borda(1733–1799), a French mathematician and nautical astronomer.

B. The Borda Count Method

EXAMPLE 3 � Using the Borda Count Method

Find the winner of the election in Example 1 using the Borda count method.

SolutionAward 0, 1, 2, and 3 points to last, next to last, and so on. Counting thepoints for anchovies (A) in column 1 of Table 14.6, you get 7 first-placevotes, worth 3 points each, a total of 3 � 7 � 21 points. Sausage (S) gets2 � 7 in column 1, 3 � 5 in column 2, 2 � 4 in column 3, and 1 � 2 incolumn 4 for a total of 14 � 15 � 8 � 2 � 39 points. Pepperoni (P) gets10 points in column 2, 12 in 3 and 6 in 4 for a total of 28 points. Finally,mushrooms (M) get 7 points in column 1, 5 in 2, 4 in 3, and 4 in 4 for atotal of 20 points. Thus, using the Borda count method, the rankings are S(winner), P, A, and M with 39, 28, 21, and 20 points, respectively. �

C. The Plurality with Elimination MethodThis method is a variation of the plurality method and may involve a series ofelections.

TA B L E 14 . 6

Points 7 5 4 2

3 A S P P

2 S P S M

1 M M M S

0 P A A A

Plurality with Elimination (The Hare Method)

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

The Borda Count Method

Voters rank candidates from most to least favorable. Each last-place voteis awarded no point; each next-to-last-place vote is awarded one point,each third-from-last-place vote is awarded two points, and so on.* Thecandidate who receives the most points is the winner.

EXAMPLE 4 � Using the Plurality with Elimination Method

Consider the familiar pizza voting results. Which topping wins the election usingthe plurality with elimination method?

*Sometimes the last-place vote is awarded one point, next-to-last vote two points, and so on.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 6


SolutionFirst, let us count the number of first place votes in Table 14.7 to see if there is amajority.

A has 7 votes (first column).S has 5 votes (second column).P has 4 � 2 � 6 votes (third and fourth columns).M has no votes.

Since there are 7 � 5 � 4 � 2 � 18 voters, we need 10 votes for a majority. Noneof the toppings has a majority of the votes, but M received the fewest first-placevotes, so M is eliminated, and all selections in each column below M move upone place, as shown in Table 14.8.

Now A still has 7 votes, S has 5, and P has 4 � 2 � 6. Since S has the fewestvotes, S is eliminated, and we are down to just P and A, as shown in Table 14.9.

Now P is the clear majority winner with 5 � 6 � 11 votes. Thus, P is the win-ner of the election when we use the plurality with elimination method. �

If we look at Examples 1–4, we can see that A is the winner using the plu-rality method, P is the winner using the plurality with runoff method, S is thewinner using the Borda count method, and P is the winner using the pluralitywith elimination method. If a voting method is to indicate a group’s preference,the method used should not change the winner. This situation points out theimportance of deciding on the voting system to be used before the election takesplace. Of course, elections with only two candidates are easy because the winnerwill get at least half the votes—not only a plurality but also a majority. Thedifficulty arises when we have three or more candidates. If this is the case, wecan compare candidates the easiest way we know: two at a time. This is the basisof the next voting method.

D. The Pairwise Comparison Method

*Sometimes the last-place vote is awarded one point, next-to-last two points, and so on.

TA B L E 14 . 7

7 5 4 2

A S P P

S P S M

M M M S

P A A A

TA B L E 14 . 8

7 5 4 2

A S P P

S P S S

P A A A

TA B L E 14 . 9

7 5 6

A P P

P A A

Pairwise Comparison Method

Voters rank candidates from most to least favorable. Each candidate is thencompared with each of the other candidates. If candidate A is preferred tocandidate B, then A receives one point. If candidate B is preferred to can-didate A, then B receives one point. If there is a tie, each candidate receivesone-half point. The candidate who receives the most overall points is thewinner.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 7


For example, suppose we have three candidates: Alice, Bob, and Carol. Wehave to compare Alice versus Bob, Alice versus Carol, and Bob versus Carol. Wecould hold three separate elections, but it is possible to use the information in thepreference tables we have used before. As the number of candidates grows, sodo the number of head-to-head comparisons that need to be made. For n candi-dates, there are

such comparisons. Thus, for n � 10 candidates, we would need (10 � 9)/2 � 45head-to-head comparisons. Let us use our preference tables to calculate the winnerof all the possible head-to-head comparisons. The one clear-cut case is when onecandidate beats all the others. This case even has a special name: A candidate whobeats all the others in head-to-head comparisons is the Condorcet winner (namedafter the Marquis de Condorcet mentioned in the Human Side of Mathematics atthe beginning of the chapter, who, like Borda, was an eighteenth-century French-man). As you might suspect, a big problem with using Condorcet winners is thatoften there is no such winner, as we shall see in the following example.

EXAMPLE 5 � Using the Pairwise Comparison Method

The results of an election involving three candidates, A, B, and C, are shown inTable 14.10. Who wins the election using the pairwise comparison method?

SolutionTo determine the winner using the pairwise comparison method, we have tocompare A and B, A and C, and B and C.

Suppose the election is between just A and B (leave C out).

A: 2 votes from column 1 and 4 from column 3, a total of 6 votesB: 3 votes from column 2, a total of 3 votes

Thus, A beats B 6 votes to 3, and A is awarded one point.Now, let us compare A and C (leave B out).

A: 2 votes from column 1, a total of 2 votesC: 3 votes from column 2 and 4 votes from 3, a total of 7 votes

Thus, C beats A 7 votes to 2, and C is awarded one point.Finally, let us compare B and C (leave A out).

B: 2 votes from column 1 and 3 from column 2, a total of 5 votesC: 4 votes from column 3, a total of 4 votes

Thus, B beats C 5 votes to 4, and B is awarded one point.What a dilemma! All the candidates have one point. There is no Condorcet

winner in this election. �

As we mentioned at the beginning of the section, there are two fundamen-tally different types of voting methods: preferential (those using a preferencetable) and nonpreferential. We will now discuss a nonpreferential voting method:approval voting.

n(n � 1)2

� C(n, 2)

TA B L E 14 .10

2 3 4

A B C

B C A

C A B

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 8


E. Approval VotingApproval voting uses a different kind of preference table. The good news is thatthe table is much simpler in one respect: Each voter does not have to rank all thecandidates first, second, third, and so on. Instead, each voter simply approves(A) or disapproves (D) each candidate. Thus, if you are a voter, you can vote forone candidate, two candidates, three candidates, and so on. Voting for two ormore candidates doesn’t dilute your vote; each candidate that you approve ofgets one full vote. When the votes are counted, the candidate with the mostapproval votes wins.

EXAMPLE 6 � Using Approval Voting

In Table 14.11, each row corresponds to a different candidate (W, X, Y, and Z),and each column corresponds to a different voter. An A means “approve” and aD means “disapprove.” Which of the candidates wins using approval voting?

SolutionWe examine each of the rows and count only the A’s.

Row W has 3 A’s.Row X has 5 A’s.Row Y has 4 A’s.Row Z has 4 A’s.

This means that candidate X (row 2) wins with 5 votes. Y and Z are tied with 4votes each, and W is in last place with only 3 votes. �

Like all voting methods, approval voting has its deficiencies, but it has anumber of good features, too. It is simpler than the Borda count or plurality withelimination method, although not as simple as the plurality method. However,unlike the plurality method, it doesn’t rely only on first-place votes. It works wellwhen voters can easily divide the candidates into “good” and “bad” categories.Approval voting is also good in situations where more than one winner isallowed. This occurs, for example, in electing players to the Baseball Hall ofFame. To be elected, an eligible player has to be named on 75% of the ballots.The voters are members of the Baseball Writers’ Association of America. Theyadd one extra requirement: No one can vote for more than ten players.

TA B L E 14 .11

Candidate Voter 1 Voter 2 Voter 3 Voter 4 Voter 5 Voter 6 Voter 7 Voter 8

W A D A A D D D D

X A A D D A D A A

Y D D A D D A A A

Z D D A D A A D A

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 9


A The Plurality Method

1. Why did they have a vote recount in Floridaduring the 2000 presidential election? BecauseFlorida law requires a recount when the winningmargin in votes is less than 0.5% of the totalnumber of votes cast.

results are summarized in the preference schedulebelow.

E X E R C I S E S 1 4 . 1

Candidate Votes

Bush 2,911,872

Gore 2,910,942

Nader 97,419

Buchanan 17,472RF

G

D

R

a. What is the total number of votes shown in thetable?

b. What is the difference between the number ofvotes obtained by Bush and by Gore?

c. What percent difference (to three decimalplaces) is that?

d. Does the difference require a recount?

2. Who is the winner in Florida under the pluralitymethod?

3. Four candidates, A, B, C, and D, are running forclass president and receive the number of votesshown in the table.

6 7 3 4

D C A B

C B D A

B D B C

A A C D

a. How many votes were cast in the election?b. How many first-place votes are needed for a

majority?c. Did any candidate receive a majority of first-

place votes?d. Who is the winner using the plurality method?

4. Five hundred registered voters cast their prefer-ence ballots for four candidates, P, T, R, and S. The

Number of Voters

Place 130 120 100 150

First P T T S

Second R R R R

Third S S P P

Fourth T P S T

a. How many first-place votes are needed for amajority?

b. Did any candidate receive a majority of first-place votes?

c. Who is the winner by the plurality method?

5. Refer to the preference schedule in problem 4.a. Which two candidates have the most first-

place votes?b. Which candidate is the winner using the plu-

rality with runoff method?

6. The preference schedule below shows the rank-ings for four brands of auto tires, A, B, C, and D.

Number of Voters

Place 13 12 10

First A C D

Second B B A

Third D D B

Fourth C A C

a. How many votes were cast in the election?b. How many first-place votes are needed for a

majority?c. Did any brand receive a majority of first-place

votes?d. Who is the winner by the plurality method?

7. Refer to the preference schedule in problem 6.a. Which two brands have the most first-place

votes?b. Which brand is the winner using the plurality

with runoff method?

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 10


In problems 8–10, use the table below. A survey wasconducted at Tampa International Airport to find thefavorite vacation destination in Florida. The rankingfor four destinations, Busch Gardens (B), DisneyWorld (D), Epcot (E), and Sea World (S), are shown inthe table.

11. Which source wins using the plurality method?

12. Which source wins using the plurality with runoffmethod?

13. Which source wins using the Borda countmethod?

14. Which source wins using the plurality with elimi-nation method?

15. Which source wins using the pairwise comparisonmethod?

E Approval Voting

16. The results of a hypothetical election usingapproval voting are summarized in the tablebelow. An X indicates that the voter approves ofthe candidate; a blank indicates no approval. Whois the winner using approval voting?

17. In problem 16, who is the winner using approvalvoting if Collins drops out of the race?

18. Have you seen the new color choices for iMaccomputers? The iMac Club is sponsoring a week-end event, and each participant will vote for his orher favorite iMac color using approval voting. The

B The Borda Count Method

8. Find the winner and runner-up using the Bordacount method.

C The Plurality with Elimination Method

9. Find the winner using the plurality with elimina-tion method.

D The Pairwise Comparison Method

10. Find the winner using the pairwise comparisonmethod.

In problems 11–15, use the following information: Agroup of patients suffering from a severe cold wereinformed that they needed at least 60 mg of vitamin Cdaily. The possible sources of vitamin C were 1 orange(O), 2 green peppers (G), 1 cup of cooked broccoli (B),or cup of fresh orange juice (J). The rankings for thegroup are given in the table (above, right).

12

Number of Voters

Place 20 15 10

First D E S

Second B B B

Third E D D

Fourth S S E

Number of Voters

Place 5 11 8 6

First B O G J

Second J J J G

Third G G O O

Fourth O B B B

Table for Problem 16

Voters

Candidates Richard Sally Thomas Uma Vera Walter Yvette Zoe

Adams X X X X X X

Barnes X X X

Collins X X X X

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 11


possible colors are strawberry, lime, grape, tan-gerine, and blueberry. Here is a summary of theresults.

12 participants voted for strawberry.7 participants voted for strawberry andblueberry.

20 participants voted for grape and tangerine.18 participants voted for lime, grape, and

tangerine.23 participants voted for blueberry and lime.25 participants voted for tangerine.

Use approval voting to determine the club’sfavorite iMac color.

19. A college class has decided to take a vote todetermine which coffee flavors are to be served in the cafeteria. The choices are latte, cappuccino,mocha, and Americano. The winning coffee flavorwill be determined using approval voting on thebasis of the following responses:

12 students voted for latte and cappuccino.5 students voted for cappuccino, mocha, and

Americano.10 students voted for mocha and cappuccino.13 students voted for Americano and cappuccino.

The flavor with the most votes wins.a. How many total votes did latte receive?b. How many total votes did cappuccino receive?c. How many total votes did mocha receive?d. How many total votes did Americano receive?e. Which coffee is selected by the class using

approval voting?

20. The Math Club uses approval voting to choose afaculty adviser for the upcoming year on the basisof the following responses:

Anne and Fran voted for Mr. Albertson.Peter, Alex, and Jennifer voted for Ms. Baker

and Ms. Carr.William, Sam, Allison, and Betty voted for

Mr. Albertson, Ms. Baker, and Mr. Davis.Joe, Katie, and Paul voted for Ms. Carr and

Mr. Davis.Jonathan voted for Mr. Davis.

a. How many total votes did Mr. Albertsonreceive?

b. How many total votes did Ms. Baker receive?c. How many total votes did Ms. Carr receive?d. How many total votes did Mr. Davis receive?e. Which teacher is selected as faculty adviser

using approval voting?

In problems 21–30, use the following information: OnSeptember 23, 1993, 88 members of the InternationalOlympics Committee (IOC) met in Monte Carlo tochoose a site for the 2000 Summer Olympics. Fivecities made bids: Beijing, (BC), Berlin (BG), Istan-bul (I), Manchester (M), and Sydney (S). In the tablebelow is a summary of the site preferences of thecommittee members.

21. Does any city have a majority of the first-placevotes? If so, which city?

22. Which city has the most first-place votes? Howmany does it have?

23. Which city is selected if the committee decides touse the plurality method?

Table for Problems 21–30

Number of Votes

Choice 3 2 32 3 3 1 8 30 6

First I I BC M BG I M S BG

Second BC BC I BC BC S S M S

Third M BG BG BG I BC BG BG M

Fourth BG M M S S M I I BC

Fifth S S S I M BG BC BC I

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 12


24. Which city is selected if the committee decides touse the plurality with elimination method?

25. Suppose the committee decides to give 5 points toeach city for every first-place selection it gets, 4 points for every second-place selection, 3 pointsfor every third-place selection, 2 points for everyfourth-place selection, and 1 point for every fifth-place selection. If the winning city will be the citywith the most points, which city will be selected?

26. Which city is selected if the committee decides touse the “regular” 4-3-2-1-0 Borda count method?

27. Which city is selected if the committee decides touse the pairwise comparison method?

28. Rank the cities from first to last using the “regu-lar” Borda count method. (Remember, you foundthe Borda count winner in problem 26.)

29. Which city is selected if the committee decides touse approval voting? (Assume that each voterapproves only his or her first two choices.)

30. Rank the cities from first to last using approvalvoting.

The following information will be used in problems31–36. In July 2005, members of the InternationalOlympic Committee (IOC) met in Singapore to choosethe site for the 2012 Summer Olympics. Five citiesmade bids: London (L), Paris (P), Madrid (MA),Moscow (MO), and New York (NY). The results of theelection, which used the plurality with eliminationmethod, are shown below. (IOC members from coun-tries with candidate cities were ineligible to vote whiletheir nation’s city was still in the running.)

33. Did any city receive a majority of votes in thesecond round?

34. Why were four rounds of voting held?

35. If the plurality with runoff method were usedinstead, which cities would have faced off in therunoff election?

36. If the plurality method were used, which citywould have won the election?

Using Your Knowledge

Ace Cola has decided to begin a multimillion-dollar adcampaign to increase its lagging sales. The ads are tobe based on consumers’ preferring the taste of AceCola to its major competitors, Best Cola, Coala Cola,and Dkimjgo Cola.

Even koalas love

Ace

with

Ace

Changethepace ...

Source: http://news.bbc.co.uk/sport1/hi/other_sports/olympics_2012/4656529.stm.

31. Which city won the election?

32. Did any city receive a majority of votes in the firstround?

L P MA MO NY

First round 22 21 20 15 19Second round 27 25 32 16Third round 39 33 31Fourth round 54 50

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 13


An independent testing agency conducted a carefullycontrolled taste test on 50 randomly selected coladrinkers. Their results are summarized in the tablebelow. (In the table, A represents Ace, B representsBest, and so on.)

37. Use the plurality method to find the preferred cola.

38. Use the plurality with runoff method to find thepreferred cola.

39. Use the Borda count method to find the preferredcola.

40. As an expert in the mathematics of voting, you areapproached by Ace and offered a $25,000 consult-ing fee if you can show that Ace is really the num-ber one cola. Find a point assignment for theBorda count method in which Ace comes in first.(Hint: A gets a lot of second-place votes, so wewant to make second place worth proportionallymore. Remember, first place must still be worthmore than second, so make the gap between sec-ond and third place larger.)

Research Questions

1. In the 2000 presidential election, there were more than two candidates. Inhow many other presidential elections have there been more than twocandidates?

2. The 2000 presidential race was one of the closest in history. In what otheryears was the difference between the winner and runner-up less than 50electoral votes? Who were the winners and runners-up of these elections?

3. Name five advantages of approval voting.

References

http://home.capecod.net/~pbaum/vote2.htmhttp://web.archive.org/collections/e2k.htmlwww.infoplease.com/spot/closerace1.htmlwww.multied.com/elections/www.washingtonpost.com/wp-srv/onpolitics/elections/2000/results/

whitehouse/www.archives.gov/federal_register/electoral_college/popular_vote_

2000.htmlwww.archives.gov/federal_register/electoral_college/votes_

2000.htmlwww.sa.ua.edu/ctl/math103/

10 13 8 7 12

A B C C D

B A D B A

C D B A B

D C A D C

304470_ch14_pV1-V59 11/7/06 2:08 PM Page 14

14.2 Voting Objections V15

Voting Objections

GE

TT

I N G S T A R T ED

14.2

Disaster 2000

In the preceding section we studied five preferential voting systems: plurality,plurality with runoff, Borda count, plurality with elimination, and pairwise com-parison. We also studied a nonpreferential voting system: approval voting. Aswe pointed out, all these systems have advantages and disadvantages and some-times can produce different winners. Let us look at an actual example—the 2000presidential election (Table 14.12).

As you can see from the results, Gore had more votes and Bush had wonmore states, but neither had won the electoral vote (EV) because it took 270votes to win, and the 25 Florida electoral votes had not been decided as ofNovember 17. If Gore got the 25 Florida votes, he would win. On the other hand,if Bush got them, he would be president. Of course, by now you know the rest ofthe story! �

Is this fair? If we rely on the fact that Gore had the most votes, it would be fairto say that Gore was the winner. However, when we discussed the pluralitymethod, we defined a majority as more than 50%. Should Gore win then? Heshould certainly beat Nader and Buchanan! But Bush also clearly beats Naderand Buchanan. Who is the winner then? Of course, you know the actual answer,but to make the discussion more precise, we will introduce four criteria thatmathematicians and political scientists have agreed on as their fairness criteriafor a voting system: the majority criterion, the head-to-head (Condorcet) crite-rion, the monotonicity criterion, and the irrelevant alternatives criterion.

TA B L E 14 .12 Results as of 5:58 P.M., EST, 11/17/2000

Candidates Votes Vote (%) States Won EV

Gore 49,921,267 49 19 255

Bush 49,658,276 48 29 246

Nader 2,756,008 3 0 0

Buchanan 447,927 0 0 0

No winner declared Exit polls

RF

G

R

D

Web It Exercises

Have you heard of a cartogram? A cartogram is a map that relates regions basedon their populations rather than their geographic sizes. After the 2004 presiden-tial election, which pitted President George Bush against Senator John Kerry,much of the discussion centered on the “red states” versus the “blue states.” Goto www-personal.umich.edu/~mejn/election to see how a cartogram depicts thered state–blue state phenomenon. Which type of map represents the 2004 elec-tion results most accurately? Write a short essay about your conclusions.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 15


A. The Majority CriterionIt seems fair that if a candidate is the first choice of a majority of voters, then thatcandidate should be declared the winner. If this is not the case, then that votingmethod violates the majority criterion. Under this criterion (total number ofvotes), Gore should have been the winner. But as we know, Bush won theelection.

Majority Criterion

If a candidate receives a majority of first-place votes, then that candidateshould be the winner.

EXAMPLE 1 � Using the Majority Criterion

La Cubanita Restaurant is conducting a survey to find out which is the most pop-ular omelet among the western (W), bacon (B), and ham (H) omelets. The resultsof the survey are shown in Table 14.13.

CUBAN TOAST $ .99CHEESE TOAST $1.45WESTERN OMELETTE $2.95CHEESE OMELETTE $2.50HAM OMELETTE $2.80PLAIN OMELETTE $2.25BACON OMELETTE $2.80CAFE CON LECHE SM.$1.40 LG.$1.70HOT CHOCOLATE SM.$1.40 LG.$1.70

*SERVED ON CUBAN BREAD* CHEESE 45¢

BREAKFAST 7AM TO 11AM

TA B L E 14 .13

Number of Votes

Place 60 25 15

First W B B

Second B H W

Third H W H

(a) Which omelet is the winner using the Borda count method?

(b) Does the winner have a majority of votes?

SolutionUsing the Borda count method, W has 2(60) � 1(15) � 135 points

B has 1(60) � 2(25) � 2(15) � 140 pointsH has 1(25) � 25 points

(a) Using the Borda count method, the winner is B, the bacon omelette, with 140points.

(b) No. A majority of the people, 60 out of 100, chose the western omelette. �

Note that although a majority of the people (60 out of 100) preferred thewestern omelet, under the Borda count method, the bacon omelet wins. Thus, inthis example, the Borda count method violates the majority criterion; that is, acandidate with a majority of first-place votes can lose the election!

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 16


EXAMPLE 2 � Using the Majority Criterion

An election to select their favorite airline, A, B, or C, is conducted among 32 stu-dents. The results are shown in Table 14.14. Which airline should be selectedunder the specified method, and does the method satisfy the majority criterion?

(a) The plurality method

(b) The Borda count method

(c) The plurality with elimination method

(d) The pairwise comparison method

Solution

(a) Using the plurality method, B is the winner with 18 out of 32 votes. Note thatB received a majority of the votes, so the method of plurality does not vio-late the majority criterion. In general, a candidate who holds a majority offirst-place votes also holds a plurality of first-place votes.

TA B L E 14 .14

Number of Votes

Place 8 6 18

First A A B

Second B C A

Third C B C

The plurality method never violates the majority criterion.

Note that the converse is not true: If you have a plurality of the votes, you donot necessarily have a majority of the votes.

(b) Under the Borda count method we assign 0, 1, and 2 points to the third, sec-ond, and first places, respectively. The points for each airline are as follows:

A: 2(8) � 2(6) � 1(18) � 46 pointsB: 1(8) � 2(18) � 44 pointsC: 1(6) � 6 points

Thus, A is the winner under the Borda count method.

Since airline B is the one holding the majority of first-place votes (18 out of32), the Borda count method violates the majority criterion. Of course, theBorda count method does not always violate the majority criterion; it just hasthe potential to do so.

(c) Since B has the majority of the votes (18 out of 32), B is the winner underplurality with elimination, so the majority criterion is not violated. In gen-eral, a candidate who holds a majority of first-place votes wins the electionwithout having to hold a second election.

The Borda count method has the potential for violating the majoritycriterion.

The plurality with elimination method never violates the majoritycriterion.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 17


Head-to-Head (Condorcet) Criterion

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

(d) Using the pairwise comparison involves the following cases and outcomes:

A versus B (eliminate C)A: 8 � 6 � 14 B: 18 B wins 18 to 14. B is awarded 1 point.

A versus C (eliminate B)A: 8 � 6 � 18 � 32 C: 0 A wins 32 to 0. A is awarded 1 point.

B versus C (eliminate A)B: 8 � 18 � 26 C: 6 B wins 26 to 6. B is awarded 1 point.

Since B has 2 points, B wins the election under the pairwise comparisonmethod. In general, if a candidate holds a majority of first-place votes, thiscandidate always wins every pairwise (head-to-head) comparison.

Even though the Borda count method is the only method studied that violatesthe majority criterion, it does take into account the voters’ preferences byhaving all candidates ranked. �

B. The Head-to-Head (Condorcet) CriterionSuppose four candidates, A, B, C, and D, are running for chair of the mathemat-ics department. There are 20 voting members in the department, and the studentnewspaper performed a postelection survey of each of the 20 members in thedepartment. Among other things, the survey asked the voters whom they pre-ferred in a two-way race between candidate C (the one endorsed by students) andeach of the other candidates. Here are the results.

11 voters preferred candidate C over candidate A.

11 voters preferred candidate C over candidate B.

17 voters preferred candidate C over candidate D.

So, in head-to-head competition, candidate C won against each of the othercandidates. Wouldn’t it seem unfair if candidate C was not declared the winnerof the election? When the actual votes were tabulated, candidate A got 9 first-place votes, candidate B got no first-place votes, candidate C got 8 first-placevotes, and candidate D got 3 first-place votes. If candidate C is not declared thewinner, this would be a violation of the Condorcet criterion because C certainlywins when compared with every other candidate.

The pairwise comparison method never violates the majority criterion.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 18


EXAMPLE 3 � Using the Head-to-Head Criterion

Which sandwich is the most popular? La Cubanita restaurant conducted a surveyamong its customers to select the favorite sandwich from Cuban (C), pork (P),turkey (T), and vegetarian (V). The number of votes for each is shown in Table14.15. Which sandwich should be selected under the specified method, and doesthe method satisfy the head-to-head criterion?

(a) Head-to-head (b) Plurality

(c) Borda count (d) Plurality with elimination

(e) Pairwise comparison

CUBAN $3.49

$3.99$3.75

$4.19MEDIA NOCHE $3.29

VEGETARIAN $3.50TUNA $3.90

PORK $3.90STEAK $3.90BREADED

$3.90CHICKEN $3.50B.L.T.

SPECIAL

CLUBTURKEY $3.95HAM & CHEESE $3.35

*ADD LETTUCE & TOMATO* 30¢

SANDWICHES

TA B L E 14 .15

Number of Voters

Place 30 50 58 60 90

First V V T P C

Second T T V V P

Third P C P T T

Fourth C P C C V

Solution

(a) We need a total of six head-to-head comparisons. A further look seems toindicate that P is the winner. Let us see why.

P beats C in columns 1, 3, and 4 for 30 � 58 � 60 � 148 points, whereas C beats P in columns 2 and 5 for 50 � 90 � 140 points. Thus, P beats C.

Comparing P and T, we see that P beats T in columns 4 and 5, obtaining60 � 90 � 150 points, and T beats P in columns 1, 2, and 3, obtaining 30 � 50 � 58 � 138 points. Thus, P beats T 150 to 138.

Comparing P and V, we see that P beats V in columns 4 and 5, and Vbeats P in columns 1, 2 and 3, so the score is the same as in the precedingcomparison: P beats V 150 to 138.

Thus, P is the favored candidate when compared head-to-head withevery other candidate.

(b) Using the plurality method, C wins with 90 votes.

The plurality method has the potential for violating the head-to-headcriterion.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 19


The Borda count method has the potential for violating the head-to-headcriterion.

(d) Using plurality with elimination, T is eliminated in the first round, P in thesecond round, and C in the third round. (Check this!) Thus, V is the winner198 to 90 over C.

The plurality with elimination method has the potential for violating thehead-to-head criterion.

(e) As you recall, in the pairwise comparison method each candidate is rankedand compared with each of the other candidates. Each time, the preferredcandidate gets 1 point. Let us look at the comparisons.

C and PC: 50 � 90 � 140 P: 30 � 58 � 60 � 148 P wins and gets

1 point.C and T

C: 90 T: 30 � 50 � 58 � 60 � 198 T wins and gets 1 point.

C and VC: 90 V: 30 � 50 � 58 � 60 � 198 V wins and gets

1 point.P and T

P: 60 � 90 � 150 T: 30 � 50 � 58 � 138 P wins and gets 1 point.

P and VP: 60 � 90 � 150 V: 30 � 50 � 58 � 138 P wins and gets

1 point.T and V

T: 58 � 90 � 148 V: 30 � 50 � 60 � 140 T wins and gets 1 point.

Thus, using the pairwise comparison method, P is the winner with 3 points.

(c) Using the Borda count method, we assign 0, 1, 2, and 3 points to the fourth-, third-, second-, and first-place winners. The total points are

C: 1(50) � 3(90) � 320 points

P: 1(30) � 1(58) � 3(60) � 2(90) � 448 points

T: 2(30) � 2(50) � 3(58) � 1(60) � 1(90) � 484 points

V: 3(30) � 3(50) � 2(58) � 2(60) � 476 points

Using the Borda count method, T wins with 484 points.

The pairwise comparison method never violates the head-to-headcriterion.

�

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 20


This example shows that the plurality, Borda count, and plurality with elim-ination methods may potentially violate the head-to-head criterion. Next, weshall introduce a third criterion called the monotonicity criterion that can beused to evaluate the fairness of an election and explore the possibility that theplurality with elimination method may have some further flaws.

C. Monotonicity CriterionWhen the outcome of a first election is not binding—for example, when a strawpoll or survey is taken before the election—voters may change their preferencesbefore the actual election. If a leading candidate gains votes at the expense ofanother candidate, the chances of winning for the leading candidate shouldincrease because of the additional votes. However, this is not always the case!This strange result is a violation of the monotonicity criterion.

The Monotonicity Criterion

If a candidate is the winner of a first nonbinding election and then gainsadditional support without losing any of the original support, then the can-didate should be the winner of the second election.

We shall now discuss an example in which the winner of the first election(straw vote) gains additional votes before the actual election and still loses.

EXAMPLE 4 � Using the Monotonicity Criterion

Months before the actual vote to select the Heisman Trophy winner, it wasclaimed that one of the contenders was too old to win the trophy. A straw vote(first election) was conducted among 105 sportswriters, and the results were asshown in Table 14.16. After several weeks of heated discussion, five writersdecided to change their ballots and award their first-choice votes to JH. Theresults of the new election are shown in Table 14.17.

TA B L E 14 .16 First Election

Number of Voters

Place 42 30 23 10

First JH CW DB DB

Second DB JH CW JH

Third CW DB JH CW

TA B L E 14 .17 New Election

Number of Voters

Place 47 30 23 5

First JH CW DB DB

Second DB JH CW JH

Third CW DB JH CW

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 21


Using the plurality with elimination method,

(a) who is the winner of the first election?

(b) who is the winner of the new election?

(c) is the monotonicity criterion violated?

Solution

(a) Using the plurality with elimination method, the first election results in theelimination of CW and then a win by JH over DB with a majority vote of 72to 33. Thus, JH is the winner.

(b) In the new election, using the plurality with elimination method, DB, with23 � 5 � 28 points, is eliminated, and CW gets a majority of 30 � 23 � 53votes over JH’s 47 � 5 � 52 votes. This time CW is the winner.

(c) Although 5 voters changed from DB to JH in the new election, adding 5votes to JH’s total, JH’s win in the first election was not repeated in thesecond election.

The plurality with elimination method has the potential for violating themonotonicity criterion.

�

By the way, a similar situation actually occurred in 2000 when selecting theHeisman winner among Chris Weinke (CW), Josh Heupel (JH), and Drew Brees(DB). The election, however, was actually conducted using the Borda countmethod, awarding each candidate 1, 2, and 3 points for third, second, and firstplace, respectively. Even though Weinke had reached the ripe old age of 28, hewon by collecting 1628 points.

D. The Irrelevant Alternatives CriterionThe fourth and last criterion we will study involves the removal (or introduction)of a candidate who has no chance of winning the election. For example, let usassume that we have an election among candidates A, B, and C with the resultsshown in Table 14.18.

It is easy to see that B is the winner using plurality, plurality with runoff, orpairwise comparison. But let us use the Borda count method. When we add upthe points, we find

A: 23 points B: 22 points C: 3 points

The race is close, but A wins out using the Borda count method.It seems that in deciding between A and B, what people think of C shouldn’t

matter; after all, C is completely out of the running. But look at the 3 voters rep-resented in column 2. Suppose that after thinking it over a little more, they alldecide that candidate C is even worse than they thought before and should bedropped to the bottom of their ballots or even dropped out of the election alto-gether! Note that the relative positions of A and B have not changed. With C atthe bottom, the results are shown in Table 14.19 on page V23.

TA B L E 14 .18

Numberof Voters

Place 4 3 9

First A A B

Second B C A

Third C B C

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 22


Our Borda point count now becomes

A: 23 points B: 25 points C: 0 points

B is now the winner using the Borda count method.In other words, because some voters changed their minds about C, or because

C dropped out of the race, the rankings of A and B were reversed. But therankings of A and B should depend on how voters view A and B and not on whatthey think of some other alternative. We call C an irrelevant alternative inranking A and B.

TA B L E 14 .19

Numberof Voters

Place 4 3 9

First A A B

Second B B A

Third C C C

The Irrelevant Alternatives Criterion

If a candidate is the winner of an election, and in a second election one ormore of the losing candidates is removed, then the winner of the firstelection should be the winner of the second election.

EXAMPLE 5 � Using the Four Fairness Criteria

A group of 50 students ranked professors A, B, and C as shown in Table 14.20.If the plurality method is used to select the top professor, does the method satisfythe four fairness criteria we have studied?

All the methods we have studied have the potential to violate the irrelevantalternatives criterion.

TA B L E 14 . 2 0

Number of Voters

Place 28 12 10

First C B A

Second B C B

Third A A C

Using the plurality method, C is the winner with 28 votes, which is a major-ity (56%) of the 50 votes cast. Thus, the majority criterion is satisfied. If we usethe pairwise comparison method, we see that C beats A 40 to 10, C beats B 28 to22, and B beats A 40 to 10, so C wins two points and the head-to-head criterionis satisfied.

The monotonicity criterion is satisfied if we assume that a second electionis held in which C picks up additional votes. C will certainly win the second elec-tion by plurality. Finally, if A or B drops out, C still wins by the plurality method,satisfying the irrelevant alternatives criterion. Thus, this particular electionsatisfies all four fairness criteria we have studied. Of course, each of the votingmethods can be made to violate at least one of the fairness criteria. �

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 23


Can we find a method that will satisfy all four criteria all the time? Thisquestion led to a long and futile search. In 1950, Kenneth Arrow, a U.S. econo-mist, made a very surprising discovery. He found that no voting method couldever satisfy these four conditions all the time. This idea it not restricted to the voting methods we know of now, but any voting method anybody mightthink of in the future as well. This fact is known as Arrow’s ImpossibilityTheorem. This discovery was a major factor in Arrow’s winning the NobelPrize in economics.

THEOREM 14.1 Arrow’s Impossibility Theorem

There is no possible voting method that will always simultaneously satisfyeach of the four fairness criteria:

1. The majority criterion

2. The head-to-head criterion

3. The monotonicity criterion

4. The irrelevant alternatives criterion

In simple terms, Arrow’s discovery means that we can never find a votingmethod that does everything we want.

Before you attempt the exercises, we summarize in Table 14.21 the four fair-ness criteria and indicate in Table 14.22 on page V25 which of the voting meth-ods we have studied satisfies a particular criterion.

TA B L E 14 . 21

Majority criterion If a candidate receives a majority of first-place votes, then that candidate should be the winner.

Head-to-head If a candidate is favored when compared head-to-head(Condorcet) with every other candidate, then that candidate criterion should be the winner.

Monotonicity criterion If a candidate is the winner of a first election and then gains additional support without losing any of theoriginal support, then that candidate should be thewinner of the second election.

Irrelevant alternatives If a candidate is the winner of an election and in a criterion second election one or more of the losing candidates

is removed, then the winner of the first electionshould be the winner of the second election.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 24


A The Majority Criterion

1. Who makes the best Cuban sandwich in Tampa?According to a panel of Tampa Tribune judgeswho rated each sandwich anywhere from 1 (low)to 5 (high), the best three Cuban sandwiches areproduced at Wrights Gourmet (W), Puccetti’sMarket (P), and La Segunda Central Bakery (S).Here is a table simulating the points in the voting.

a. Who is the winner using the plurality method?b. Who is the winner using the Borda count

method?c. Does the Borda count method violate the

majority criterion?d. Who is the winner using the pairwise compar-

ison method?

Best CubanSandwich

2. Of course, you cannot rely solely on professionaljudges, so the Tribune had readers vote for theirfavorite Cuban sandwich, and the outcome wasdifferent! According to the people, the three bestCuban sandwiches are produced at La Septima(L), West Gate Bakery (G), and the Cuban Sand-wich Shop (C). The approximate number of votesis shown in the table below.

E X E R C I S E S 1 4 . 2

TA B L E 14 . 2 2

PluralityBorda with Pairwise

Fairness Plurality Count Elimination ComparisonCriterion Method Method Method Method

Majority Always May not Always Alwayscriterion satisfies satisfy satisfies satisfies

Head-to-head May not May not May not Alwayscriterion satisfy satisfy satisfy satisfies

Monotonicity Always Always May not Alwayscriterion satisfies satisfies satisfy satisfies

Irrelevant May not May not May not May notalternatives satisfy satisfy satisfy satisfycriterion

Numberof Points

Place 25 5 20

First W S P

Second P W S

Third S P W

Number of Voters

Place 600 300 200

First L G C

Second G C L

Third C L G

a. Who is the winner using the plurality method?b. Who is the winner using the Borda count

method?c. Does the Borda count method violate the

majority criterion?d. Who is the winner using the pairwise compar-

ison method?

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 25


3. Do you drink coffee? Which kind do you prefer?Starbucks coffee offers latte (L), cappuccino (C),mocha (M), and Americano (A). The preferencesof 70 students surveyed at the University of SouthFlorida are shown in the table below.

a. In a head-to-head comparison, is there a com-mercial preferred to all others?

b. Is the head-to-head criterion satisfied if theplurality method is used to find the preferredcommercial? Explain your answer.

C The Monotonicity Criterion

6. A company is planning to relocate to one of thelarger counties in the United States: Los Angeles(L), Cook (C), or Harris (H). The Committee of100 is to use the plurality with elimination methodto select the county, and their preferences areshown in the table below. After careful delibera-tion, the ten voters who voted C, L, H changedtheir vote to L, C, H. Is the monotonicity criterionsatisfied? Explain your answer.

Number of Voters

Place 10 20 30 10

First C C L L

Second A M C C

Third M A M A

Fourth L L A M

a. Which flavor is the winner using the Bordacount method?

b. Is the majority criterion satisfied? Explainyour answer.

B The Head-to-Head (Condorcet) Criterion

4. The Performing Arts Center Board is consideringshowing three different plays this season: Cats(C), A Chorus Line (L), and Les Miserables (M).The ten members of the board rank the playsaccording to the preference table below.

Number of Voters

Place 2 5 3

First L C M

Second C L C

Third M M L

Number of Voters

Place 50 25 15 10

First S H B U

Second U S H B

Third H U S S

Fourth B B U H

Number of Voters

Place 25 30 10 35

First C H C L

Second H L L C

Third L C H H

a. In a head-to-head comparison, is there a playthat is preferred to all others?

b. Is the head-to-head criterion satisfied if theplurality method is used to determine the mostpopular play?

5. Did you watch the Super Bowl this year? Whichcommercials do you remember? Four of the all-time most memorable are Staples (S), 7 Up (U),Honda (H), and Budweiser (B). One hundredviewers were asked to watch and vote on thecommercial they preferred. The results are in thepreference table (above, right).

D The Irrelevant Alternatives Criterion

7. Which of the following has the highest cost ofliving: Washington, D.C. (D), Alaska (A), orHawaii (H)? The table below shows the responsesof 50 people who were asked that question. IfHawaii (H) is eliminated, is the irrelevant alterna-tives criterion satisfied? Explain your answer.

Number of Voters

Place 20 18 12

First D A H

Second A H D

Third H D A

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 26


8. Suppose that in problem 7 the responses of the 50people are as shown in the table below. If Alaska(A) is eliminated, is the irrelevant alternativescriterion satisfied? Explain your answer.

a. Who wins using pairwise comparisons?b. Does any candidate beat every other candidate

one on one, that is, in a head-to-head compari-son? If so, which one?

c. Who wins using the plurality method?d. Which fairness criteria, if any, are violated?

Explain.e. Suppose candidate C drops out, but the winner

is still chosen using the plurality method. Is thewinner the same as in part (c)? If not, whichcandidate does win?

f. Which fairness criteria, if any, are violated?Explain.

g. Who wins using the plurality with eliminationmethod? (Assume candidate C is now back in.)

h. Now suppose candidate A drops out, but thewinner is still chosen using the plurality withelimination method. Is the winner the same asin part (g)? If not, which candidate does win?

i. Which fairness criteria, if any, are violated?Explain.

11. The preference schedule below gives the results ofan election among four candidates, A, B, C, and D.

Number of Voters

Place 20 16 14

First D A H

Second A H D

Third H D A

9. The preference table below gives the results of anelection among three candidates, A, B, and C.

Number of Voters

Place 27 24 2

First A B C

Second C C B

Third B A A

Number of Voters

Place 20 19 5

First A B C

Second B C B

Third C A A

a. Who wins using the plurality method?b. Does any candidate get a majority of the first-

place votes? If so, which one?c. Who wins using the pairwise comparison

method?d. Does any candidate beat every other candidate

one-on-one, that is, in a head-to-head compar-ison? If so, which one?

e. Who wins using the Borda count method?f. Which fairness criteria, if any, are violated?

Explain.g. Suppose candidate B drops out, but the winner

is still chosen using the Borda count method. Isthe winner the same as in part (e)? If not, whichcandidate does win?

h. Which fairness criteria, if any, are violated?Explain.

10. The following preference table gives the results ofan election among three candidates, A, B, and C.

Number of Voters

Place 14 4 10 1 8

First A B C C D

Second B D B D C

Third C C D B B

Fourth D A A A A

a. Who wins using the plurality with eliminationmethod?

b. Who wins using the pairwise comparisonmethod?

c. Does any candidate beat every other candidateone on one, that is, in a head-to-head compari-son? If so, which one?

d. Which fairness criteria, if any, are violated?Explain.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 27


12. The preference table below gives the results of anelection among three candidates, A, B, and C.

a. Who wins using pairwise comparisons?b. Suppose candidate C drops out, but the winner

is still chosen using the pairwise comparisonmethod? Is the winner the same as in part (a)?If not, which candidate does win?

c. Which fairness criteria, if any, are violated?Explain.

15. When using the pairwise comparison method, howmany comparisons need to be made if there area. three candidates? b. four candidates?c. five candidates? d. n candidates?

16. When using the pairwise comparison method,how many comparisons must a candidate (say A)win to guarantee winning the election if there area. three candidates? b. four candidates?c. five candidates? d. n candidates?

In Other Words

17. Explain the majority criterion in your own words.

18. Explain why the plurality method always satisfiesthe majority criterion.

19. Explain why the pairwise comparison methodalways satisfies the majority criterion.

20. Explain the monotonicity criterion in your ownwords.

21. Explain why the plurality with elimination methodalways satisfies the monotonicity criterion.

22. Explain the Condorcet criterion in your own words.

23. Explain why the pairwise comparison methodalways satisfies the head-to-head (Condorcet)criterion.

24. Explain the irrelevant alternatives criterion inyour own words.

25. Which of the five election techniques that we havestudied is most likely to end in a tie? Explain.

Number of Voters

Place 7 8 10 4

First A B C A

Second B C A C

Third C A B B


b. Suppose that the Florida Supreme Court inval-idates the results of the election, and everyonemust revote. As it happens, everyone votesexactly as before except for the 4 voters in thelast column of the table. These 4 voters, whooriginally voted A, C, B, decide to switch theorder of their votes for A and C so that theirnew preference ballots are C, A, B. Who winsthis new election using the plurality with elim-ination method?


13. The preference table below gives the results of anelection among three candidates, A, B, and C

Number of Voters

Place 20 19 5

First A B C

Second B C B

Third C A A


b. Suppose candidate A drops out, but the winneris still chosen using the plurality with elimina-tion method. Is the winner the same as in part(a)? If not, which candidate does win?


14. The following preference table (above, right)gives the results of an election among five candi-dates, A, B, C, D, and E.

Number of Voters

Place 3 3 1 1 3 3 1 1

First B A C C B A B E

Second A C B E A D A D

Third C D A B D E E A

Fourth D E D D E C C C

Fifth E B E A C B D B

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 28

14.3 Apportionment Methods V29

Research Questions

1. There are several election procedures we have not discussed. Find out aboutBlack’s election procedure and then write a short paragraph about it.

2. Write a short paragraph about the work and discoveries of Kenneth Arrow.

3. Can you reduce warfare by adopting advanced methods of voting? Write ashort paragraph on this topic. For information, go tohttp://www.solutionscreative.com/voteadv.html.

4. Which election method is best? Write a research paper on the subject. Youcan get some background at http://electionmethods.org/.

5. How do Oscar nominees get chosen? Write a short essay on the subject. Goto http://electionmethods.org/.

6. You can’t have an effect on an election if you don’t vote.a. What was the voter turnout for the 2000 presidential election?b. What was the voter turnout for the 2004 presidential election?c. Which U.S. presidential election had the greatest turnout? What percent?d. Which state had the greatest turnout during the 2000 presidential election?

What percent?e. Which state had the greatest turnout during the 2004 presidential election?

What percent? To find out, go to www.fairvote.org/.

7. Which country had the best and worst average voter turnout in the 1990s?Go to www.fairvote.org/ to find out.

Apportionment Methods14.3

In 1787, at the Constitutional Convention in Philadelphia, delegates of the 13original states created a system of government with three branches: executive,legislative, and judicial. One of the most important issues was the representa-tion of the states in the legislative branch. Smaller states wanted equal represen-tation. In response, a Senate in which two senators represent each state wascreated. Larger states preferred proportional representation. Thus, a House ofRepresentatives, in which each state receives a number of representatives pro-portional to its population, was created. Unfortunately, the founding fathers didnot decide on the exact number of representatives for each state. In fact, Article1, Section 2, of the Constitution states:

Representatives shall be apportioned among the several states . . .according to their respective numbers. The number of representativesshall not exceed one for every thirty thousand, but each state shall haveat least one representative.

Historically, at least four apportionment methods have been implemented.

1792–1841: The Jefferson method1842–1851, 1901–1941: The Webster method1852–1900: The Hamilton (Vinton) method1941–present: The Hill-Huntington method

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 29


We shall study the Hamilton, Jefferson, and Webster methods and omit the Hill-Huntington method because of its complexity. Instead, we will examine a simi-lar method known as the John Quincy Adams method. The various methods willbe presented in order of mathematical complexity rather than chronologicalorder. Keep in mind that apportionment methods are not limited to governingbodies. Budget allocations, Super Bowl tickets, faculty and student senate seats,and many other items have to be fairly distributed, or apportioned.

Funding the Community College System

The community college system in the State of Florida consists of 28 colleges.In 1999, the general budget allocation for the system amounted to more than 1 billion dollars—$1,102,817,849 to be exact. How can we fairly distribute themoney among the 28 colleges? Here are some possibilities.

1. Divide the $1,102,817,849 equally among the 28 colleges. Each college gets

Is this fair? Consider this: Miami-Dade had 98,924 students, whereas theFlorida Keys had 4068. Under this equal allocation method, each will get thesame amount, despite the disparity in their student populations.

2. We can also base funding on the number of students attending. In a recentyear, the total student population in the Florida community college systemwas 737,864, so funding for each college would be proportional to the num-ber of students attending that college. The amount for each college would be

Is this method fair? Keep in mind that a college with, say, 1000 students eachtaking a 3-hour course will need to fund 3 � 1000 � 3000 student semesterhours, whereas a college with 1000 full-time equivalent students (FTEs) willhave to fund 40 � 1000 � 40,000 student semester hours. Note: One FTE isequivalent to 40 student semester hours.

3. Perhaps a fairer distribution would be to allocate the money on the basis ofthe number of FTEs in each college. The formula for the allocation to eachcollege would then be

Can you think of any other way of fairly apportioning the $1,102,817,849 to the28 colleges? �

A. Apportionment ProblemsIn Getting Started, we considered several formulas for apportioning money. Tomake these formulas more standard and the resulting allocation quotas more pre-cise, we define the standard divisor (SD) and the standard quota (SQ) as follows:

Number of FTEs in the collegeTotal number of FTEs

� $1,102,817,849

Number of students in the college737,864

� 1,102,817,849

1,102,817,84928

� $39,386,351.75

GE

TT

I N G S T A R T ED

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 30


EXAMPLE 1 � Finding the Standard Divisor

The five largest community colleges in Florida received about $324 million intotal general education and general fund revenues. The number of FTEs (to thenearest 1000) in each of the five colleges is shown in Table 14.23.

Formula for Standard Divisor Formula for Standard Quota

SQ �population in the group

SDSD �

total population in the grouptotal number to be apportioned

TA B L E 14 . 2 3

Miami Jacksonville Broward Valencia Daytona Total

30,000 17,000 13,000 12,000 9000 81,000

Find the standard divisor SD and the standard quota SQ for each college.

Solution

The total population in the group is the total number of FTEs, that is, 81,000. Thetotal number to be apportioned is $324,000,000. Thus,

means that 1 FTE gets $4000

The standard quota SQ for each college is given in Table 14.24.

ba 14000

SD �81,000

324,000,000�

14000

� 0.00025

SD �total population in the group

total number to be apportioned

TA B L E 14 . 2 4

Miami SQ � � � $120,000,000

Jacksonville SQ � � � $68,000,000

Broward SQ � � � $52,000,000

Valencia SQ � � � $48,000,000

Daytona SQ � � � $36,000,0009,000

1>4000population in the group

SD

12,0001>4000

population in the groupSD

13,0001>4000


17,0001>4000


30,0001>4000


You may have noticed that since each FTE gets $4000, each college’s allocationwill be ($4000 � number of FTEs). �

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 31


B. The Alexander Hamilton Method of ApportionmentOne of the earliest apportionment methods was the Hamilton method. Proposedto President George Washington in 1791, the method was promptly vetoed by thepresident—the first presidential veto in U.S. history! First, we give the proce-dure for apportioning a number of items into various groups using the Hamiltonmethod and then discuss the presidential objections to the method.

Hamilton’s Method

1. Find

2. Find

3. Round SQ down to the nearest integer (lower quota).Each state should get at least that many seats but must get at least oneseat.

4. Apportion additional seats one at a time to the states with the largestfractional part of the standard quotas.

SQ �state population

SD

SD �total population

total seats to be apportioned

EXAMPLE 2 � Using the Hamilton Method

Table 14.25 shows the population of the 15 states in the Union according to the1790 census. Use the Hamilton method to apportion the 105 seats in the Houseof Representatives.

TA B L E 14 . 2 5

State Population SQ Rounded Down Seats

Virginia 630,560 18.31 18 18Massachusetts 475,327 13.80 13 14Pennsylvania 432,879 12.57 12 13North Carolina 353,523 10.27 10 10New York 331,589 9.63 9 10Maryland 278,514 8.09 8 8Connecticut 236,841 6.88 6 7South Carolina 206,236 5.99 5 6New Jersey 179,570 5.21 5 5New Hampshire 141,822 4.12 4 4Vermont 85,533 2.48 2 2Georgia 70,835 2.06 2 2Kentucky 68,705 2.00 2 2Rhode Island 68,446 1.99 1 2Delaware 55,540 1.61 1 2

Total 3,615,920 98 105

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 32


SolutionWe use the four steps.

1. Since the total population is 3,615,920 and we have to apportion 105 seats,

2. We first find SQs for Virginia, Massachusetts, and Delaware.

For Virginia,

For Massachusetts,

For Delaware,

The SQ for all states, to two decimal places, is shown in column 3 of Table 14.25.

3. The SQs rounded down to the nearest integer are in column 4 of Table 14.25.Accordingly, Virginia, Massachusetts, and Delaware will get 18, 13, and 1seat, respectively. Note that the total seats in column 4 add up to 98. Whatabout the 105 � 98 � 7 seats that are left over? See step 4!

4. The additional seats are apportioned, one at a time, to the states with thelargest fractional parts (South Carolina, Rhode Island, Connecticut, Massa-chusetts, New York, Delaware, and Pennsylvania). The actual number ofseats apportioned is in column 5 of Table 14.25. �

Note that the Hamilton method assigns 2 seats to Delaware, a state with apopulation of 55,540. However, it was stipulated in the Constitution that eachseat in the House would represent a population of at least 30,000. Logically, 2seats would have to represent 60,000 people, but Delaware only had a popula-tion of 55,540! Partially on the basis of this flaw, President Washington vetoedthe use of the Hamilton plan to apportion the first House of Representatives.Instead, the Jefferson method, which assigned an extra seat to Virginia, Jeffer-son’s home state, was used.

We shall use the Jefferson method to apportion the 105 seats in the originalHouse of Representatives after we give one more example using the Hamiltonmethod.

EXAMPLE 3 � Using the Hamilton Method

The second floor of Brandon Hospital houses five intensive care units: Medical(M), Surgical (S), Cardiac (C), Transitional (T), and Progressive (P). The maxi-mum number of patients that each unit can house is shown in Table 14.26. Thetotal for all units is 90. The hospital has bought 50 recliners to be distributedamong the five units. Use the Hamilton method to apportion the 50 recliners onthe basis of the number of patients in each unit.

SolutionThe standard divisor SD � 90/50 � 1.8, and the standard quota SQ is the num-ber of patients in each unit divided by 1.8, as shown in column 3. Next, we

SQ �55,540

34,437.33� 1.61

SQ � 475,32734,437.33

� 13.80

SQ � 630,56034,437.33

� 18.31

SD �3,615,920

105� 34,437.33

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 33


round down each SQ and enter the result in column 4. Note that the sum of allthe rounded-down numbers in column 4 is 47. We apportion the 3 remainingrecliners, one by one, to the units with the highest fractional parts: 0.89 (P), 0.67(C), and 0.67 (S). Column 5 shows the actual number of recliners apportionedto each unit, with the bold numbers reflecting the extra recliner. Note that thesum of the numbers in column 5 is 50, the total number of recliners.

TA B L E 14 . 2 6

Unit Patients SQ Rounded Down Actual Number

Medical 15 15/1.8 � 8.330 8 8

Surgical 30 30/1.8 � 16.67 16 16 � 1 � 17

Cardiac 12 12/1.8 � 6.670 6 6 � 1 � 7

Transitional 8 8/1.8 � 4.440 4 4

Progressive 25 25/1.8 � 13.89 13 13 � 1 � 14

Totals 90 50 47 50

Patient Rooms 231-276

Patient Rooms 211-220

M.I.C.U., S.I.C.U. & C.C.U.

P.C.U. & T.C.U.

North Elevators

I.C.U. & C.C.U. Waiting

Now you know how to apportion recliners as well as seats! �

C. The Thomas Jefferson Method of ApportionmentIn Examples 2 and 3 some of the groups (Delaware and the Surgical Care Unit,for example) received additional items when applying step 4 in Hamilton’smethod. Can we modify the standard quota SQ to overcome the possibleinequity? Jefferson’s method attempts to do this by using a modified divisorMD that is slightly lower than the standard divisor SD to obtain a modifiedquota MQ that is slightly higher than the standard quota SQ. Does this soundconfusing? Just remember that if you have the fraction

and you make the denominator SD slightly lower, the new modified quota MQwill be slightly higher. Here are the steps to apportion items using Jefferson’smethod.

SQ �population in the group

SD

Jefferson’s Method

1. Find a modified divisor MD such that when each modified quota MQis rounded down to the nearest integer, the sum of the resulting integersequals the number of items to be apportioned.

2. The apportionment for each group corresponds to the rounded downMQs found in step 1.

As we shall see, the challenge is to find that “magical” MD!

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 34


EXAMPLE 4 � Using the Jefferson Method

Use the Jefferson method to apportion the 50 recliners of Example 3.

SolutionTable 14.27 shows the first four columns in Example 3. The standard divisor inExample 3 was 1.8, so let us make the modified divisor MD slightly lower—say,1.7—to obtain the modified quotas shown in column 5 (rounded to two decimalplaces). We show the rounded down numbers corresponding to the actual num-ber of recliners apportioned in the last column.

TA B L E 14 . 2 7Rounded Modified Rounded

Unit Patients SQ Down Quota Down

Medical 15 � 8.33 8 � 8.82 8Surgical 30 � 16.67 16 � 17.65 17Cardiac 12 � 6.67 6 � 7.06 7Transitional 8 � 4.44 4 � 4.71 4Progressive 25 � 13.89 13 � 14.71 14

Total 90 50 47 50

251.7

251.8

81.7

81.8

121.7

121.8

301.7

301.8

151.7

151.8

Note that the sum of the rounded-down numbers in the last column adds up to 50as required. �

EXAMPLE 5 � Using the Jefferson Method

Use the Jefferson method to apportion the 1794 House of Representatives shownin Table 14.28. Note that the total U.S. population was 3,615,920, 105 seats wereto be apportioned, and

MQ �total population

MD

TA B L E 14 . 2 8State Population MQ Rounded Down

Virginia 630,560 19.11 19Massachusetts 475,327 14.40 14Pennsylvania 432,879 13.12 13North Carolina 353,523 10.71 10New York 331,589 10.05 10Maryland 278,514 8.44 8Connecticut 236,841 7.18 7South Carolina 206,236 6.25 6New Jersey 179,570 5.44 5New Hampshire 141,822 4.30 4Vermont 85,533 2.59 2Georgia 70,835 2.15 2Kentucky 68,705 2.08 2Rhode Island 68,446 2.07 2Delaware 55,540 1.68 1

Total 3,615,920 109.57 105

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 35


SolutionThe modified divisor MD � 33,000 was mercifully supplied by Congress. Wecalculate some modified quotas and show the rest in the table.

For Virginia, � 19.11

For Massachusetts, � 14.40

For Delaware, � 1.68

The rounded-down quotas corresponding to the number of seats under theJefferson method are shown in the last column of Table 14.28. �

D. The Daniel Webster Method of ApportionmentThe feasibility of the Jefferson method hinges on finding the “magic” modifieddivisor MD and was attacked on constitutional grounds. However, Jeffersonpointed out that the Constitution only required that apportionment be based onpopulation, and the modified divisor MD produced quotas that indeed reflectedthe population and, moreover, did so equally, since all states used the same divi-sor. The Jefferson method was used without incident until after the 1820 census,when a major flaw (to be discussed in the next section) was uncovered. When thesame flaw appeared again following the 1830 census, the method was replacedby one proposed by Daniel Webster in 1832. By that time, the country had grownfrom 15 states with 3,615,920 people to 24 states with 12,860,702 people. Theappeal of Webster’s method was its mathematical simplicity. Modified quotas arenot rounded down to the nearest integer. Instead, they are rounded to the nearestinteger using the mathematical rules we have studied: round up for fractions of

or more and down for fractions that are less than . The bad news is that youstill have to find that modified “magic” divisor MD. The good news is that theMD is usually supplied for us.

12

12

55,54033,000

475,32733,000

630,56033,000

Webster’s Method

1. Find MD, the modified divisor.

2. Find MQ, the modified quota for each group.

3. Round MQ in the usual manner for each group (up for 0.5 or more,down for less than 0.5).

4. The apportionment for each group corresponds to the values obtainedin step 3, and the sum of the apportionments for all groups must equalthe total number of items to be apportioned.

MQ �total population

MD

304470_ch14_pV1-V59 11/7/06 2:08 PM Page 36


EXAMPLE 6 � Using the Webster Method

Use the Webster method to apportion the five states shown with their respectivepopulations in the 1830 census in Table 14.29. The modified divisor MD selectedby Webster was 49,800, and the total population was 12,860,702. (Source:www.census.gov/population/censusdata/table-16.pdf.)

SolutionThe modified quota MQ is found by dividing the state population by 49,800.

For New York, MQ � 1,918,608/49,800 � 38.53, or 39New York gets 39 seats.

For Pennsylvania, MQ � 1,348,233/49,800 � 27.07, or 27Pennsylvania gets 27 seats.

For Kentucky, MQ � 687,917/49,800 � 13.81, or 14Kentucky gets 14 seats.

For Vermont, MQ � 280,652/49,800 � 5.64, or 6Vermont gets 6 seats.

For Louisiana, MQ � 215,739/49,800 � 4.33, or 4Louisiana gets 4 seats.

The final results are shown in Table 14.29.

TA B L E 14 . 2 9

State Population MQ � Apportionment

New York 1,918,608 38.53 39

Pennsylvania 1,348,233 27.07 27

Kentucky 687,917 13.81 14

Vermont 280,652 5.64 6

Louisiana 215,739 4.33 4

Population49,800

E. The John Quincy Adams Method of ApportionmentAs we have mentioned, by 1830, politicians were once again struggling over theapportionment method to be used. The debate was so intense that former presi-dent John Quincy Adams, at the time a representative from Massachusetts, wrotein his memoirs:

I passed an entirely sleepless night again. The iniquity of the Apportion-ment bill, and the disreputable means by which so partial and unjust adistribution of the representation had been effected, agitated me so that Icould not close my eyes.

Mr. Adams was referring to a proposal by James K. Polk of Tennessee, whichused Jefferson’s method of apportionment with an increased divisor of 47,700.This increase favored the representation of some states but hurt the representa-tion of some of the New England states. As a consequence, Adams proposed anew apportionment method that was similar to Jefferson’s but rounded upinstead of down and, unfortunately, still used a “magic” divisor. Here is the pro-cedure for Adams’s apportionment method.

�

304470_ch14_pV1-V59 11/7/06 2:08 PM Page 37


EXAMPLE 7 � Using the Adams Method

Use Adams’s method to apportion the recliners of Example 3.

SolutionWe have to find the modified “magic” divisor. Recall that in Example 4 the standarddivisor 1.8 was slightly lowered to 1.7 to obtain the desired modified quotas. If themodified quotas are rounded up, the sum will be 9 � 18 � 8 � 5 � 15 � 55. Toreduce this number, let us increase the divisor to 1.9 and round up the quota asshown in Table 14.30.

Adams’s Method

1. Find a modified divisor MD such that when each group’s modifiedquota MQ is rounded up to the nearest integer, the sum of the resultingintegers equals the number of items to be apportioned.

2. The apportionment for each group corresponds to the rounded up MQsfound in step 1.

TA B L E 14 . 3 0

Unit Patients Modified Quota Rounded Up

Medical 15 � 7.89 8Surgical 30 � 15.79 16Cardiac 12 � 6.32 7Transitional 8 � 4.21 5Progressive 25 � 13.16 14

Totals 90 50

251.9

81.9

121.9

301.9

151.9

Our modified “magic” divisor 1.9 did the trick; the rounded-up values, whichcorrespond to the number of recliners each unit will get, add up to 50. �

Before you attempt the exercises, Table 14.31 gives a summary of theapportionment methods we have studied and their important features.

TA B L E 14 . 31

Method Divisor Round the Quota Apportionment

Hamilton’sSD �

Jefferson’s MD is less than SD. Down to the nearest Apportion to each group its modified lowerinteger quota.

Webster’s MD is less than, greater than, To the nearest Apportion to each group its modified or equal to SD. integer rounded quota.

Adams’s MD is greater than SD. Up to the nearest Apportion to each group its modified upperinteger quota.

Distribute left-over items to the groups with the largest fractional part until all itemsare distributed.

Down to the nearest integer

total populationseats to be apportioned

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 38


As you can see from Table 14.31, Hamilton’s method rounded the standardquotas down to the nearest integer, Jefferson’s method rounded the modified quo-tas down to the nearest integer, Webster’s method rounded the modified quotasto the nearest integer, and Adams’s method rounded the modified quotas up to thenearest integer.

A Apportionment Problems

In problems 1–7, use the following table:

How much money will each of these two univer-sities receive if the money is allocated accordingto the number of students in each institution?Answer to the nearest dollar.

3. The University of South Florida has 18,176 FTEs,whereas the University of West Florida has 4556.How much money will each of these two univer-sities receive if the money is allocated accordingto the number of FTEs in each institution? Answerto the nearest dollar.

B The Alexander Hamilton Method ofApportionment

4. Suppose that the state decides to apportion 200 newteaching positions on the basis of the number ofstudents in each university.a. Use the Hamilton method to find the standard

divisor SD.b. Use the Hamilton method to find the standard

quota SQ for Florida Atlantic and the Univer-sity of Central Florida. Answer to three deci-mal places.

5. Suppose the state decides to apportion 200 newteaching positions on the basis of the number ofFTEs in each university.a. Use the Hamilton method to find the standard

divisor SD.b. Use the Hamilton method to find the standard

quotient SQ for Florida Atlantic and the Uni-versity of Central Florida. Answer to three dec-imal places.

6. Use the Hamilton method to apportion the 200new teaching positions to each of the ten universi-ties on the basis of the number of students.

7. Use the Hamilton method to apportion the 200new teaching positions to each of the 10 universi-ties on the basis of the number of FTEs. Is thenumber of positions for each university the sameas that obtained in problem 6?

E X E R C I S E S 1 4 . 3

University Headcount FTE

University of Florida 41,652 29,646Florida State University 30,389 21,195Florida A&M University 11,324 8064University of South

Florida 31,555 18,176Florida Atlantic University 19,153 10,725University of West Florida 7790 4556University of Central

Florida 30,009 18,312Florida International

University 30,096 17,434University of North Florida 11,360 6697Florida Gulf Coast

University 2893 1558

Subtotal E & G 216,221 136,363

1. The university system in the State of Florida con-sists of the ten universities listed in the tableabove. In a recent year, the general budget alloca-tion for the system amounted to more than 2 bil-lion dollars: $2,001,102,854.a. If the state decides to apportion the money

equally among the ten universities, how muchwill each university get? Answer to the nearestdollar.

b. How much money will each student be allo-cated if the money is apportioned equallyamong all the students? Answer to the nearestdollar.

c. How much money will each FTE (full-timeequivalent) be allocated if the money is appor-tioned equally among all FTEs? Answer to thenearest dollar.

2. The University of Florida has 41,652 students,whereas Florida Gulf Coast University has 2893.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 39


8. Let us go back to the five intensive care units ofExample 3. Suppose the hospital buys 75 newintravenous (IV) pumps to be distributed amongthe five units on the basis of the number of patientsin each unit. Use the Hamilton method to fill in theblanks in the table below and apportion the 75 IVunits.

a. Find each county’s standard quota.b. Using Hamilton’s method, find each county’s

apportionment.

C The Thomas Jefferson Method ofApportionment

11. What leisure activities do you participate in? Inthe table below are five activities and the approx-imate number of participants (in millions) in each.

Rounded Actual Unit Patients SQ Down Number

Medical 15

Surgical 30

Cardiac 12

Transitional 8

Progressive 25

90

9. According to the Centers for Disease Control andPrevention, the states reporting the highest annualnumber of AIDS cases in a recent year are asshown in the table below. In that same year,federal spending on AIDS research amounted to$9,988 million.

California 5637

Florida 5683

New York 7655

Texas 3715

New Jersey 2061

Hillsborough 940,484

Manatee 243,531

Orange 817,206

Pasco 330,704

Polk 457,347

Suppose the federal government wishes to allo-cate an additional $100 million for AIDS researchon the basis of the number of cases in each of thesestates.a. Find each state’s standard quota.b. Find each state’s apportionment using Hamil-

ton’s method

10. Do you know where the 2012 Summer Olympicswill be held? One of the possible venues wasFlorida.

The populations of five counties where someof the events were to be held are as shown in thefollowing table. Suppose $500 million is allocatedto these five counties on the basis of their respectivepopulations.

Exercise 150

Sports 90

Charity work 85

Home repair 130

Computer hobbies 80

Salvation Army $1230

YMCA $ 630

Fidelity Investments $ 570

American Cancer Society $ 560

American Red Cross $ 540

Sources: National Endowment forthe Arts; Statistical Abstract of theUnited States.

Suppose you wish to allocate $100 million to pro-mote leisure activities on the basis of the numberof participants.a. Find the modified quota for each activity using

the divisor 5.25.b. Find how much money should be apportioned

to each activity using Jefferson’s method.

12. In the table below, the five U.S. charities receivingthe highest donations (in millions) in a recent yearare shown.

Source: The Chronicle of Philanthropy.

304470_ch14_pV1-V59 11/7/06 2:08 PM Page 40


Suppose you are a philanthropist willing to donate$150 million to these five charities on the basis ofreceived donations.a. Find the modified quota for each charity using

the divisor 23.3.b. Find how much money should be apportioned

to each charity using Jefferson’s method.

D The Daniel Webster Method ofApportionment

13. How much do you spend on your pet annually? Inthe table below are the average annual costs (to thenearest $10) spent per household for several typesof pets.

Suppose the Immigration and NaturalizationService is planning on granting 700,000 visas nextyear.a. Find the modified quota using the divisor

0.928.b. How many visas should be allocated to each

continent using Webster’s method?

15. The acreage of five county parks in HillsboroughCounty is shown in the table below. Suppose thecounty wishes to distribute 75 new park rangersamong these five parks.

Texas 21

California 20

Florida 12

Ohio 10

Colorado 7

Lake Park 600

E. G. Simmons 470

Lettuce Lake 240

Lithia Springs 160

Eureka Springs 30

Europe 90,000

Asia 220,000

North America 255,000

South America 45,000

Africa 40,000

Dogs $190

Cats $110

Birds $ 10

Horses $230

Source: U.S. Pet Ownershipand Demographic Sourcebook.

For every $500 spent on each of these four typesof pets,a. find the modified quota using the divisor 1.08.b. how much money should be apportioned to

each pet category using Webster’s method?

14. What continents do immigrants to the UnitedStates come from? The table below shows thenumber of immigrants from each continent admit-ted to the United States in a recent year.

Source: U.S. Immigration andNaturalization Service.

a. Find the modified quota for each park using thedivisor 20.5.

b. Find the number of rangers that should be allo-cated to each park using Adams’s method.

16. Has your telephone area code been changedlately? With the popularity of cell phones increas-ing, more area codes are needed. The table belowshows the number of existing area codes in fivestates. Unfortunately, there are a limited numberof area codes that can be allocated to states. Sup-pose we wish to allocate 25 new area codes to thefive states listed on the basis of the number of areacodes they already have.

a. Find the modified quota for each state usingthe divisor 3.08.

b. Find the number of area codes allocated toeach state using Adams’s method.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 41


Suppose the state decides to allocate $100 million tothese five counties.

17. a. Find the standard divisor.b. Find each county’s standard quota.c. Find each county’s apportionment using Ham-

ilton’s method.

18. a. Find each county’s modified quota using thedivisor 916.05.

b. Find each county’s apportionment using Jef-ferson’s method.

19. a. Find each county’s modified quota using thedivisor 948.60.

b. Find each county’s apportionment usingAdams’s method.

20. Find each county’s apportionment using Web-ster’s method with the standard quota.

In problems 21–24, use the top-rated AIDS treatmenthospitals and their scores given in the table below.

22. a. Find each hospital’s modified quota using thedivisor 1.6745.

b. Find each hospital’s apportionment usingJefferson’s method.

23. a. Find each hospital’s modified quota using thedivisor 1.7238.

b. Find each hospital’s apportionment usingAdams’s method.

24. Find each hospital’s apportionment using Web-ster’s method.

Using Your Knowledge

Students have suggested that many of the apportion-ment problems can be done using ratio and proportion.Suppose you have four sports, A, B, C, and D, and theStudent Government Association wishes to distribute$200,000 among the four on the basis of their averageattendance of 2000, 4000, 6000, and 8000 spectators,respectively.

25. a. How much will each sport get if the $200,000is distributed proportionately to its attendance?

b. How much will each sport get if the $200,000is apportioned using Hamilton’s, Jefferson’s,Adams’s, and Webster’s methods?

c. Are the answers the same in parts (a) and (b)?

In Other Words

Describe in your own words how to calculate

26. a standard divisor. 27. a standard quota.

28. Look at Table 14.31 on page V38. Name themethod we have studied that rounds the modifiedquotaa. up. b. down. c. in the usual manner.

County Dade Broward Hillsborough Orange Pinellas

Headcount 33,000 21,000 17,000 11,000 11,000

Hospital Points

San Francisco General 100

Johns Hopkins Hospital 72

Massachusetts General 62

Univ. of Calif. at San Francisco 56

Memorial Sloan-Kettering 50

Source: U.S. News and World Report.

Suppose the best 200 AIDS specialists are to beassigned to these 5 hospitals on the basis of the numberof points in the survey.

21. a. Find the standard divisor.b. Find each hospital’s standard quota.c. Find each hospital’s apportionment using Ham-

ilton’s method.

In problems 17–20, use the headcount enrollment bycounty in the Florida State University System (to thenearest thousand) given in the following table:

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 42

14.4 Apportionment Objections V43

Research Questions

1. The 1824 election was marred by the so-called Corrupt Bargain. Describethe events that marred the election. What was claimed to be corrupt in theelection?

2. The 2000 presidential election was disputed in court but was by no meansthe first disputed election in history. What was the first disputed presidentialelection in U.S. history? What was the dispute about?

3. Write a short paragraph describing the present method of apportionment forthe House of Representatives. Make sure you mention the names andtechniques involved with this method.

4. As we have discovered from this section, most of the apportionment methodsused for the U.S. House of Representatives rely on the “magical” modifieddivisor MD. It is also apparent that a precise formula for computing this MDis not clearly prescribed (see Table 14.31 on page V38). See if you canuncover the mystery of how actual MDs have been obtained throughouthistory.

5. In December 2005, Iraq held an election to select members of itsparliament. This was only the second proportional election ever held in Iraq(a proportional election is an electoral system in which all parties arerepresented in proportion to their voting strength). How were the membersof parliament selected? Go to www.fairvote.org/blog/?p�20 and find out.Write a short essay on your findings.

Apportionment Objections14.4

As we saw in Section 14.2, there were several objections, or flaws, associatedwith the voting methods we studied. Similarly, there are several objections, orflaws, associated with Hamilton’s apportionment method. These objections areparadoxical and depend mainly on three factors: the number of items to beapportioned (objection: the Alabama paradox), the population in the group(objection: the population paradox), and the addition of one or more newgroups that require apportionment (objection: the new-states paradox).

The main objection to Jefferson’s, Webster’s, and Adams’s methods is thateach has the potential of violating the quota rule.

The Quota Rule

The apportionment for every group under consideration should alwaysequal either the group’s upper quota (rounding up) or its lower quota(rounding down)

Even though Hamilton’s method will always satisfy the quota rule andJefferson’s, Webster’s, and Adams’s methods will not, the method has equallyserious flaws. The three main objections to Hamilton’s method are the Alabamaparadox, the population paradox, and the new-states paradox.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 43


The Alabama Paradox

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

EXAMPLE 1 � Finding the Standard Divisor and Quota

In the 1880 Census, the population of the United States was 50,189,209, and thepopulation of Alabama was 1,262,505. Find the standard divisor and the stan-dard quota when

(a) 299 seats are to be apportioned.

(b) 300 seats are to be apportioned.

Solution

(a) The standard divisor is

� 167,856.89

and the standard quota is

� 7.52

(b) The standard divisor is

� 167,297.36

and the standard quota is

� 7.55 �

Does Example 1 prove that Alabama would get 8 seats when 299 seats wereapportioned and only 7 when 300 seats were apportioned? The answer is no. Toprove this, we would have to find the standard quotas for all the states and then

1,262,505167,297.36

50,189,209300

1,262,505167,856.89

50,189,209299

GE

TT

I N G S T A R T ED

Sweet Home Alabama

The Alabama paradox first surfaced after the 1870 census. With 270 members in the House of Representatives, Rhode Island got 2 representatives, but when the house size was increased to 280, Rhode Island lost a seat. After the1880 census, C. W. Seaton (chief clerk of the U.S. Census Office) computedapportionments for all house sizes between 275 and 350 members. He thenwrote a letter to Congress pointing out that if the House of Representativeshad 299 seats, Alabama would get 8 seats, but if the House of Representativeshad 300 seats, Alabama would only get 7 seats! Again, this meant a loss of 1 seatfor Alabama, even though the total number of house seats could be increasedfrom 299 to 300. This objection or flaw has come to be known as the Alabamaparadox. (Source: www.sa.ua.edu/ctl/math103/.)

A. The Alabama Paradox

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 44


EXAMPLE 2 � The Alabama Paradox and the Hamilton Method

Consider a country with a population of 50,000 and three states, A, B, and C,with the populations shown in Table 14.32. Show using Hamilton’s apportion-ment method that the Alabama paradox occurs if the number of seats is increasedfrom 50 to 51.

SolutionWhen the number of seats to be apportioned is 50, the standard divisor is

� 1000. The standard quotas SQs and the rounded-down values RDs areshown in columns 3 and 4 of Table 14.32. The state with the largest fractionalpart (0.5) is C. Thus, C gets the extra seat.

50,00050

TA B L E 14 . 3 2

State Population SQ RD Extra Final

A 25,200 � 25.2 25 0 25

B 23,300 � 23.3 23 0 23

C 1500 � 1.5 1 1 2

Total 50,000 49 50

15001000

23,3001000

25,2001000

TA B L E 14 . 3 3


A 25,200 � 25.70 25 1 26

B 23,300 � 23.77 23 1 24

C 1500 � 1.53 1 0 1

Total 50,000 49 51

1500980.39

23,300980.39

25,200980.39

If we repeat the apportionment in Table 14.33 using 51 seats, the standarddivisor is � 980.39. Again, we show the standard quotas SQs and therounded-down values in columns 3 and 4. The states with the largest fractionalparts are B (0.77) and A (0.70), so each gets an extra seat.

50,00051

Here we go again! Even though we increased the number of seats from 50 to 51, state C, formerly with 2 seats, ended up losing 1 seat—the Alabama paradox! �

Note that this paradox can only occur when the number of objects to beapportioned increases. Thus, it seems reasonable to expect that if we hold thesize of the House of Representatives to 435, as it has been for many years, no

assign the leftover seats to the states with the largest fractional parts. Unfortu-nately, there were a total of 38 states in 1880, so the task would be monumentalindeed. To learn more you can go to www.ctl.ua.edu/math103/ and look under“Apportionment.” For now, let us try a simple example.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 45


The Population Paradox

The population paradox occurs when the population of group A isincreasing faster than the population of group B, yet A loses items to group B.

The population paradox was discovered around 1900, when it was shown that astate could lose seats in the House of Representatives as a result of an increasein its population. (Virginia was growing much faster than Maine—about 60%faster—but Virginia lost a seat in the house, whereas Maine gained a seat.)

EXAMPLE 3 � The Population Paradox and the Hamilton Method

Table 14.34 shows the number of students taking mathematics, English, andscience courses during the fall and spring semesters. If 100 full-time teachingpositions are to be apportioned among the three departments on the basis of theirrespective course enrollments,

(a) how many positions will each department get in the fall using Hamilton’smethod?

(b) how many positions will each department get in the spring using Hamilton’smethod?

(c) is the apportionment fair? Explain your reasoning

(d) is this apportionment an example of the population paradox?

SolutionSince there are 10,000 students and 100 positions, the standard divisor for the fallis 10,000/100 � 100. The standard quotas SQs and the number of positions foreach department during the fall are shown in columns 3 and 4 of Table 14.34. Thenumber of students for the spring is 10,030, so the standard divisor SQ is10,030/100 � 100.3. The new standard quotas New SQs and the number ofpositions per department for the spring are shown in columns 6 and 7.

TA B L E 14 . 3 4

Fall Semester Spring Semester

Subject Number SQ Positions Number New SQ Positions

Math 951 9.51 10 961 961/100.3 � 9.58 9

English 1949 19.49 19 1969 1969/100.3 � 19.63 20

Science 7100 71.00 71 7100 7100/100.3 � 70.79 71

Total 10,000 100 10,030 100

objections or paradoxes should surface. Unfortunately, this is not the case. If thepopulation of one or more states changes, one state could lose a seat to anotherstate, even if its population is growing at a faster rate than the state that loses theseat. This paradox is known as the population paradox.

B. The Population Paradox

304470_ch14_pV1-V59 11/7/06 2:08 PM Page 46


(a) In the fall, mathematics gets 10 positions, English 19, and science 71.

(b) In the spring, mathematics gets 9 positions, English 20, and science 71.

(c) No. The rate of growth for mathematics was � � 1.05%.

The rate of growth for English was � � 1.03%.

Thus, mathematics was growing at a faster rate (1.05%) than English(1.03%), but despite this, mathematics lost 1 position (from 10 to 9), whereasEnglish gained 1 position (from 19 to 20).

(d) Yes. In this instance the mathematics population was increasing faster thanthe English population, yet mathematics lost one position to English. �

C. The New-States ParadoxAs we have mentioned before, the objections to Hamilton’s apportionment methodoccur when the number of items to be apportioned changes (Alabama paradox) andwhen the population in the group changes (population paradox). We consider onemore paradox that occurs when we add one or more groups that require appor-tionment (new-states paradox). The new-states paradox means that adding a newstate with its fair share of seats can affect the number of seats due to other states.This has actually happened! The paradox was discovered in 1907 when Oklahomabecame a state. Before Oklahoma became a state, the House of Representativeshad 386 seats. Comparing Oklahoma’s population with that of other states, it wasclear that Oklahoma should have 5 seats, so the house size was increased by 5 to391 seats. The intent was to leave the number of seats unchanged for the otherstates. However, when the apportionment was mathematically recalculated, Mainegained a seat (from 3 to 4), and New York lost a seat (from 38 to 37).

201949

1969 � 19491949

10951

961 � 951951

The New-States Paradox

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

EXAMPLE 4 � The New-States Paradox and the Hamilton Method

Suppose that in Example 3 the art department enrolls 600 students in the fall andthat 5 additional positions are allocated, bringing the total number of students to10,600 and the total number of positions to 105.

(a) Use the Hamilton method to find the new fall apportionment for mathemat-ics, English, science, and art.

(b) How does the new fall apportionment compare with the original fall appor-tionment of Example 3?

(c) Does the new apportionment seem fair?

(d) Show that the new apportionment results in an occurrence of the new-statesparadox.

Solution

(a) To find the new fall semester apportionment, note that there are now 10,600students and 105 positions, so the new standard divisor is � 100.95.10,600

105

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 47


The standard quotas and the final number of positions are shown in Table 14.35.

TA B L E 14 . 3 5

New Fall Semester Original Fall Semester

Subject Number SQ Positions Number SQ Positions

Math 951 9.42 10 951 9.51 10

English 1949 19.31 19 1949 19.49 19

Science 7100 70.33 70 7100 71.00 71

Art 600 5.94 6

Total 10,600 105 10,000 100

(b) The right side of Table 14.35 shows the original fall semester apportion-ments. As you can see, the science department lost 1 position (from 71 to 70)in the new apportionment, even though it did not lose any students. Presum-ably, the art department should have gotten the 5 new positions, but it got 6instead.

(c) The new apportionments do not seem fair because science lost 1 position toart.

(d) The addition of one group (art) changed the apportionment of another group(science) and, by definition, is an occurrence of the new-states paradox. �

Now we have seen that Hamilton’s method satisfies the quota rule but canproduce the paradoxes we have studied in this section. Jefferson’s, Webster’s,and Adams’s methods can violate the quota rule because they are based on thephilosophy that quotas can be conveniently modified. On the other hand, they donot produce the paradoxes we have studied. Can we find a perfect apportionmentmethod that not only satisfies the quota rule but also avoids these paradoxes?Two mathematicians, Michael Balinski and H. Payton Young, proved in 1980 thatthere is no such method. Their result is called Balinski and Young’s Impossi-bility Theorem.

Balinski and Young’s Impossibility Theorem

There is no apportionment method that satisfies the quota rule and avoidsparadoxes.

Table 14.36 on page V49 shows which methods violate the quota rule or producesome of the paradoxes studied.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 48


So there we have it. Just as we could find no perfect voting method, there is alsono perfect apportionment method!

TA B L E 14 . 3 6

Characteristic Hamilton’s Jefferson’s Adams’s Webster’s

May violate the quota rule No Yes Yes Yes

May result in the Alabama paradox Yes No No No

May result in the population paradox Yes No No No

May result in the new-states paradox Yes No No No

A The Alabama Paradox

1. Three Greek letter societies, �, �, and �, have thenumbers of members as shown in the table below.If Hamilton’s method is used to apportion seats inthe Pan Hellenic Council, does the Alabama para-dox occur if the number of seats is increaseda. from 30 to 31? b. from 60 to 61?

3. Which are the four best theme parks in the UnitedStates? According to Inside Track, a publicationthat rates theme parks and attractions, the fourbest parks are Busch Gardens (B), King’s Island(K), Walt Disney World (W), and Six Flags MagicMountain (S). The number of votes received byeach of the parks in a survey of 1020 persons is asshown in the table below. If Hamilton’s method isused to apportion 71 free tickets to visit the parkson the basis of the number of votes obtained in thesurvey, does the Alabama paradox occur if thenumber of tickets is increased to 72? If so, whichpark loses a ticket?

E X E R C I S E S 1 4 . 4

Society Members

� 3220

� 5000

� 9780

Total 18,000

2. A country consists of four states, A, B, C, and D,with the populations shown in the table below. IfHamilton’s method is used to apportion the legis-lature, does the Alabama paradox occur if thenumber of seats is increaseda. from 104 to 105? b. from 114 to 115?

City Population

A 1800

B 3720

C 2330

D 2150

Total 10,000

Park Votes

B 405

K 306

W 204

S 105

Total 1020

4. The total number of minority students (in thou-sands) studying allopathic medicine in 1998 is as shown in the following table. If Hamilton’smethod is used to allocate scholarships on thebasis of enrollment, does the Alabama para-dox occur when the number of scholarships isincreased from 24 to 25?

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 49


a. Which group(s) lose 1 scholarship?b. Which group(s) gain 1 scholarship?c. Which group(s) stay the same?

B The Population Paradox

5. The number of Medicare enrollees (in thousands)for Delaware (D), Nebraska (N), and Kansas (K) ina recent year and the estimated projection for afuture year are as shown in the table below. Sup-pose the federal contribution is $11 million for eachyear and Hamilton’s method is used to apportionthe money on the basis of the number of enrollees.a. How much will each state get in the most

recent year? (Answer to the nearest million.)b. How much will each state get using the pro-

jected population? (Answer to the nearest mil-lion.)

c. Which state has the higher percent increase ofenrollees, Delaware or Kansas?

d. Does the population paradox occur whenHamilton’s method is used to allocate the $11 million? Explain.

b. How many seats will each of the counties getin 10 years?

c. Does the population paradox occur? Explain.

Race Students

White 44

Black 5

Hispanic 4

Asian 12

Total 65

Source: National Center forHealth Statistics.

State Number Projection

Delaware 110 122

Nebraska 250 300

Kansas 380 428

Total 740 850

Source: Health Care Financing Administration.

6. A state consists of three counties, A, B, and C,with the present populations and in 10 yearsshown in the following table. Suppose 100 seatsare to be apportioned on the basis of populationusing Hamilton’s method.a. How many seats will each of the counties get

now?

County Now 10 Years

A 89,000 97,000

B 12,500 14,500

C 22,500 24,700

Total 124,000 136,200

County Now 10 Years

A 89,000 97,000

B 125,000 145,000

C 225,000 247,000

Total 439,000 489,000

7. The populations of three counties, A, B, and C, atpresent and in 10 years are shown in the tablebelow. “Lucky” Fulano, the state governor, sug-gested a 13-seat legislature apportioned accordingto population using Hamilton’s method.a. How many seats will each of the counties get

now?b. How many seats will each of the counties get

in 10 years?c. Does the population paradox occur? Explain.d. Which of the three counties will be unhappy

with reapportionment and why?

8. A nation consists of five states, A, B, C, D, and E,with populations 300, 156, 346, 408, and 590,respectively. Suppose 50 seats are to be appor-tioned on the basis of population using Hamilton’smethod.a. How many seats will each state receive?b. If the populations of states C and E were to

increase by 16 and 2, respectively, how manyseats would each state receive then?

c. Does the population paradox occur? Explain.d. Which of the five states will be unhappy with

reapportionment and why?

C The New-States Paradox

9. A company has two divisions: production (P) andsales (S). The number of employees in each is

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 50


shown in the table below. There are 41 managersto be apportioned between the two divisions Pand S.a. Find each division apportionment using

Hamilton’s method.b. Suppose a new advertising division (A) with

114 employees and 8 new managers is to beadded. Does the new-states paradox occurusing Hamilton’s method? Explain.

11. A country has two states, A and B, with the popu-lation (in hundreds) shown in the table below and100 seats in the legislature.a. Find the apportionments for A and B using

Hamilton’s method.b. Suppose a third state C with a population of

263 (hundred) is added with 6 additional seats.Does the new-states paradox occur usingHamilton’s method? Explain.

Division Number

P 402

S 156

A 114

Total 672

10. A country has two states, A and B, with the popu-lations (in thousands) shown in the table belowand 30 seats in the legislature.a. Find the apportionments for A and B using

Hamilton’s method.b. Suppose a third state C with a population of 76

(thousand) is added with 6 additional seats.Does the new-states paradox occur usingHamilton’s method? Explain.

State Population

A 268

B 104

C 76

Total 448

State Population

A 4470

B 520

C 263

Total 5253

Research Questions

1. When was the Alabama paradox discovered? Discuss the details of thediscovery.

2. When was the population paradox discovered? Which states were involved?Discuss the details of the discovery.

3. When was the new-states paradox discovered? Which states were involved?Discuss the details of the discovery.

4. The results of the 2000 Census led to changes in the makeup of the U.S.House of Representatives. Some states gained delegates, and some states lostdelegates. Which ones? Go to www.census.gov/population/www/censusdata/apportionment.html to find out.

In Other Words

12. Describe in your own words the Alabama paradox.

13. Describe in your own words the populationparadox.

14. Describe in your own words the new-statesparadox.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 51


Section Item Meaning

14.1 Plurality method Each voter votes for one candidate and the candidate with the mostfirst-place votes is the winner.

14.1 Plurality with Each voter votes for one candidate. If the candidate receives a runoff method majority of votes, that candidate is the winner. Otherwise, eliminate

all but the two top candidates and hold a runoff election. Thecandidate that receives a majority is the winner.

14.1 Borda count method Each voter ranks the candidates from most to least favorable, witheach last-place vote awarded no point; each next to last place isawarded one point, each third from last place is awarded two points,and so on. The candidate that receives the most points is the winner.

14.1 Plurality with Each voter votes for one candidate. If a candidate receives a majorityelimination method of votes, that candidate is the winner. If no candidate receives a

majority, eliminate the candidate with the fewest votes and holdanother election. (If there is a tie for fewest votes, eliminate allcandidates tied for fewest votes.) Repeat this process until a candidatereceives a majority.

14.1 Pairwise comparison Each voter ranks candidates from most to least favorable. Each method candidate is then compared with each of the other candidates. If A is

preferred to B, A gets 1 point. If B is preferred to A, B gets 1 point. Ifthere is a tie, each candidate receives point. The candidate with themost overall points is the winner.

14.1 Approval voting Voters approve or disapprove each candidate. The candidate with themost approval votes wins.

14.2 Majority criterion If a candidate receives a majority of first-place votes, then thatcandidate should be the winner.

14.2 Head-to-head criterion If a candidate is favored when compared head-to-head with everyother candidate, then that candidate should be the winner.

14.2 Monotonicity criterion If a candidate is the winner of a first election and then gains additionalsupport without losing any of the original support, then the candidateshould be the winner of the second election.

14.2 Irrelevant alternatives If a candidate is the winner of an election, and in a second election criterion one or more of the losing candidates are removed, then the winner of

the first election should be the winner of the second election.

14.2 Arrow’s Impossibility There is no voting method that will always simultaneously satisfy allTheorem of the four fairness criteria.

14.3 Standard divisor (SD) Total population in a group/total number to be apportioned

14.3 Standard quota (SQ) Population in the group/standard divisor

14.3 Hamilton’s method A method in which the standard quota is rounded down and additionalseats are apportioned to the states with the largest fractional part of thestandard quotas.

12

Chapter 14 Summary

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 52

Chapter 14 Summary V53

Section Item Meaning

14.3 Jefferson’s method A method using a modified divisor so that when each group’smodified quota is rounded down to the nearest integer, the sum of theintegers equals the number of items to be apportioned.

14.3 Webster’s method A method using a modified divisor so that when each group’smodified quota is rounded in the usual manner, the sum of the integersequals the number of items to be apportioned.

14.3 Adams’s method A method using a modified divisor so that when each group’smodified quota is rounded up to the nearest integer, the sum of theintegers equals the number of items to be apportioned.

14.4 Quota rule The apportionment for every group under consideration should alwaysequal either the group’s upper quota or its lower quota.

14.4 Alabama paradox Occurs when an increase in the total number of items to beapportioned results in a loss of items for a group.

14.4 Population paradox Occurs when the population of group A increases faster than thepopulation of group B, yet A loses items to group B.

14.4 New-states paradox Occurs when the addition of a new group changes the apportionmentof another group.

14.4 Balinski and Young’s There is no apportionment method that satisfies the quota rule and Impossibility Theorem avoids paradoxes.

Research Questions

1. What if you had an election and nobody came? The electoral commission inEngland is working on new election methods that would increase voterturnout. Research which methods are being used to do that and what thevoter turnout was in last year’s local elections and in the 2005 generalelection in England. Is the turnout greater or less than that in U.S. local andgeneral elections?

2. What election methods are used for events other than presidential elections?Find out! What are the procedures and voting methods used fora. the Grammy Awards?b. the Latin Grammy Awards?c. the Academy Awards?d. the Nobel Prize?e. the Heisman Trophy?

3. Write a research paper chronologically discussing all voting methods thathave been used to apportion the U.S. House of Representatives. Whatmethod is currently used and what are the current objections to this method?

4. What impact did the 2000 census have on apportionment? Write a researchpaper on this question, paying particular attention toa. which states gained two seats as a result.b. which states lost two seats as a result.

304470_ch14_pV1-V59 11/7/06 2:08 PM Page 53


References (for most of the preceding information)

http://news.bbc.co.uk/1/hi/uk_politics/2988307.stmhttp://heismanmemorialtrophy.com/www.oscars.org/index.htmlwww.infoplease.com/ipa/A0150533.htmlwww.almaz.com/nobel/www.improb.com/ig/ig-pastwinners.htmlwww.census.gov/prod/2001pubs/c2kbr01-7.pdf

1. The results of an election involving four candidates, A, B, C, and D, areshown in the table to the left.a. Did any of the candidates receive a majority?b. Which candidate is the plurality winner?c. Which candidate comes in second?d. Which candidate comes in last?

2. Using the plurality with runoff method, who is the winner of the election inproblem 1?

3. Using the Borda count method, who is the winner of the election inproblem 1?

4. Using the plurality with elimination method, who is the winner of theelection in problem 1?

5. The results of an election involving three candidates, A, B, and C, areshown in the table to the left. Using the pairwise comparison method, whois the winnera. between A and B?b. between A and C?c. between B and C?d. of the election?

6. The results of a hypothetical election using approval voting aresummarized in the table below. An X indicates that the voter approves ofthe candidate; a blank indicates no approval.

Chapter 14 Practice Test

10 8 7 5

A C D D

C D C B

B B B C

D A A A

5 3 9

A B C

B C A

C A B

VOTERS

Candidates Thomas Uma Vera Walter Yvette Zoe

Adams X X X X X

Barnes X X X

Collins X X

a. Who is the winner using approval voting?b. Who is the winner if Adams drops out of the race?

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 54

Chapter 14 Practice Test V55

7. An election to select their spring break destination, Aruba (A), Bahamas(B), or Cancun (C), is conducted among 64 students. The results are asshown in the table to the left. Which destination should be selected underthe specified method? Does the method satisfy the majority criterion?a. The plurality methodb. The Borda count methodc. The plurality with elimination methodd. The pairwise comparison method

8. A restaurant conducted a survey among its customers to select theirfavorite entree from chicken (C), pork (P), turkey (T), and vegetarian (V).The number of votes for each was as shown in the table below. Whichentree should be selected under the specified method, and does the methodsatisfy the head-to-head criterion?a. Head-to-head b. Pluralityc. Borda count d. Plurality with eliminatione. Pairwise comparison

Number of Voters

Place 36 16 12

First B A A

Second A B C

Third C C B

Number of Voters

Place 56 24 20

First C B A

Second B C B

Third A A C

Number of Voters

Place 15 25 29 30 45

First V V T P C

Second T T V V P

Third P C P T T

Fourth C P C C V

9. A group of 100 students ranked the three courses they liked best as shownin the table to the left. If the plurality method is used to select the topcourse, does the plurality method satisfya. the majority criterion? Explain.b. the head-to-head criterion? Explain.

10. As in problem 9, does the plurality method satisfya. the monotonicity criterion if we assume a second election is undertaken

and C gains additional support without losing any of the original sup-port? Explain.

b. the irrelevant alternatives criterion if we assume that either A or Bdrops out and a second election is undertaken? Explain.

11. Explain in your own words the meaning of Arrow’s ImpossibilityTheorem.

12. Five universities received $162 million in total general education andgeneral fund revenues. The number of FTEs (to the nearest 1000) in eachof the five universities is as shown in the table below.

A B C D E Total

15,000 8500 6500 6000 4500 40,500

Find the standard divisor SD and the standard quota SQ for each university.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 55


13. A college is composed of five departments: mathematics (M), English (E),languages (L), art (A), and chemistry (C), with the number of facultymembers shown in the table below. If 100 new positions are to beapportioned using Hamilton’s method and on the basis of the number offaculty in each department, what is the actual number of positionsapportioned to each of the departments?

Department Faculty MQ Actual Number

Mathematics 30

English 60

Language 24

Art 16

Chemistry 50

Totals 180 90

14. Use the Jefferson method to apportion 90 (instead of 100) positions inproblem 13. (Hint: The modified divisor must be slightly less than 2.)

15. In 1830, the population of Florida was 34,730. Use Webster’s method tofind the number of seats for Florida if the modified divisor was 49,800.

16. Use Adams’s method to apportion 90 faculty positions in problem 13.(Hint: The modified divisor must be slightly more than 2.)

17. Consider a country with a population of 100,000 and three states in thelegislature, A, B, and C, with the populations shown in the table below.Suppose 50 seats are apportioned using Hamilton’s method as shown inthe first table. Fill in the blanks in the second table and determine if theAlabama paradox occurs when the number of seats is increased from 50 to51 using Hamilton’s apportionment method. Explain your answer.


A 50,400 50,400/2000 � 25.2 25 0 25B 46,600 46,600/2000 � 23.3 23 0 23C 3000 3000/2000 � 1.5 1 1 2

Total 100,000 49 50


A 50,400B 46,600C 3000

Total 100,000 51

18. The table on page V57 shows the number of students taking mathematics,English, and science courses during the fall and spring semesters. Supposethat 100 full-time teaching positions were apportioned during the fallsemester using Hamilton’s method with the results shown.

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 56

Answers to Practice Test V57

a. How many positions will each department get in the spring usingHamilton’s method?

b. Is the apportionment fair? Explain your answer.c. Is this apportionment an example of the population paradox? Explain.

Fall Semester Spring Semester

Subject Number Fall SQ Positions Number Spring New SQ Positions

Math 476 9.5180 10 484 9.6338 9English 975 19.4961 19 990 19.7054 20Science 3550 70.9858 71 3550 70.6608 71

Total 5001 100 5024 100

19. A country has two states, A and B, with the population (in thousands)shown in the table to the left and 41 seats in the legislature.a. Find the apportionments for A and B using Hamilton’s method.b. Suppose a third state C with a population of 228 (thousands) is added

with 8 additional seats. Does the new-states paradox occur usingHamilton’s method? Explain.

20. Explain the meaning of Balinski and Young’s Impossibility Theorem inyour own words.

State Population

A 804B 312C 228

Total 1344

IF YOU ANSWER MISSED REVIEW

Question Section Example(s) Page(s)

Answers to Practice Test

1. a. No b. Dc. A d. B

2. D wins the runoff 20 to 10.3. C wins with 63 points.4. D5. a. A wins 14 to 3.

b. C wins 12 to 5.c. C wins 9 to 8.d. C is the winner.

6. a. Adams b. Barnes7. a. B. Yes

b. A. No. B has the majority but A wins underthe Borda count method.

c. B. Yes d. B. Yes8. a. P. Yes b. C. No

c. T. No d. V. Noe. P. Yes

9. a. Yes. C has the majority and wins under theplurality method.

b. Yes. C has the majority and wins head-to-head against all other candidates.

1 14.1 1 V4–V5

2 14.1 2 V53 14.1 3 V64 14.1 4 V6–V75 14.1 5 V8

6 14.1 6 V97 14.2 2 V17–V18

8 14.2 3 V19–V20

9 14.2 5 V23

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 57




10. a. Yes. C will still win the second electionunder the plurality method.

b. Yes. C will still win the second electionunder the plurality method when either A orB drops out.

11. There is no voting method that will alwayssimultaneously satisfy each of the four fairnesscriteria.

12. SD � 40,500/162,000,000 � 0.00025Standard Quotas (in millions)For A, 15,000/0.00025 � 60For B, 8500/0.00025 � 34For C, 6500/0.00025 � 26For D, 6000/0.00025 � 24For E, 4500/0.00025 � 18Note that the total is $162 million.

13.

14. Use 1.95 as the modified divisor, then rounddown.

15. 34,730/49,800 � 0.6974, or 116. Use 2.05 as the modified divisor, then round up.

10 14.2 5 V23

11 14.2 V24

12 14.3 1 V31

13 14.3 3 V32–V34

14 14.3 4 V35

14 14.3 6 V3716 14.3 7 V38

SQ RD Actual Number

30/1.8 � 16.67 16 16 � 1 � 1760/1.8 � 33.33 33 3324/1.8 � 13.33 13 1316/1.8 � 8.89 8 8 � 1 � 950/1.8 � 27.78 27 27 � 1 � 28

100 97 100

MQ Actual

30/1.95 � 15.38 1560/1.95 � 30.77 3024/1.95 � 12.31 1216/1.95 � 8.21 850/1.95 � 25.64 25

MQ Actual

� 14.63 15� 29.27 30� 11.71 12� 7.80 8� 24.39 2550

2.05

162.05

242.05

602.05

302.05

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 58

Answers to Practice Test V59



17.

The Alabama paradox occurs because eventhough the number of seats was increased from50 to 51, state C lost 1 seat, from 2 to 1.

18. a.

b. No.c. Yes. Mathematics lost 1 position and English

gained 1 even though mathematics wasgrowing at a faster rate, � 0.01681, thanEnglish, � 0.01538.

19. a. A gets 30 seats, and B 11.b. Yes. The addition of the new state C caused

A to lose 1 seat (from 30 to 29) and B to gain1 (from 11 to 12).

20. There is no apportionment method that satisfiesthe quota rule and avoids all paradoxes.

15975

8476

17 14.4 2 V45

18 14.4 3 V46–V47

19 14.4 4 V47–V48

20 14.4 V47–V48

SQ Extra Final

� 25.70 25 1 26� 23.77 23 1 24� 1.53 1 0 13000

1960.78

46,6001960.78

50,4001960.78

Position Number New SQ Positions

10 484 9.6338 919 990 19.7054 2071 3550 70.6608 71

304470-ch14_pV1-V59 11/7/06 10:01 AM Page 59

Documents

304470-ch14 pV1-V59 11/7/06 10:01 AM Page 2 · 07-11-2006 · 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 2. 14.1 Voting Systems V3 Thus, you would expect about 44,347 Florida votes