14.2 14.1 The Apportionment Problem

Winnie has convinced two friends, Louise and Tim, to buy lottery tickets. Winnie buys 18 tickets, Louise buys 4, and Tim buys 1. They are very lucky and win the top prize: a necklace with 100 diamonds! They decide to divide the diamonds in proportion to the number of tickets each bought.

Table 14.1 shows how the diamonds should be shared—if it made sense to cut one of the diamonds into three pieces! They round each share to the nearest whole number: When the fractional part is less than 0.500, they round down, and if any fractional part is greater than or equal to 0.500, they round up. Winnie’s share, 78.26 diamonds, is rounded down to 78; Louise’s share, 17.39 diamonds, is rounded down to 17; and Tim’s 4.35 diamonds are rounded down to 4. Because all three shares were rounded down, there is one diamond left. Winnie declares that since it was her idea to participate in the lottery, that diamond is hers.

Table 14.1: TABLE 14.1 Sharing the Diamonds
Tickets Shares Diamonds
Winnie 18
Louise 4
Tim 1
Total 23 1 100

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Apportionment Problem DEFINITION

Suppose we have a set of numbers whose sum is a whole number. The apportionment problem is to round each number so that the rounded numbers still add up to the original sum.

Winnie, Louise, and Tim have an apportionment problem because their shares of the diamonds are not whole numbers but the sum of their shares is a whole number. Simple rounding leaves one diamond undistributed. Winnie imposed an arbitrary solution to this apportionment problem: That diamond is hers!

Self Check 1

An administrative assistant decides to keep track of his activities for a 24-hour period. He worked 10 hours (600 minutes), commuted to and from work for 2 hours 20 minutes (140 minutes), and the remaining 11 hours 40 minutes (700 minutes) were spent on personal activities. What percent of his time was devoted to each activity? Express your answer in whole percentages. How would you round the numbers so that they add up to 100%?

  • Convert all times to minutes, and note that 24 hours is 1440 minutes. Thus, the administrative assistant worked of the day. He commuted of the day, and spent of the day on personal activities. Rounding these percentages to their nearest whole numbers we get

    To make the sum 100%, one of the percentages must be rounded down. You have no guidance yet on how to do this, so there is no wrong way to choose which one.

Apportionment Method DEFINITION

An apportionment method is a procedure for solving apportionment problems without making arbitrary choices.

Because we will be discussing the apportionment of seats in the U.S. House of Representatives based on the the census of 1790, it is useful to see the relevant text from Article I, Section 2, of the Constitution.

Representatives … shall be apportioned among the several States which may be included within this Union, according to their respective Numbers, which shall be determined by adding to the whole Number of free Persons, including those bound to Service for a Term of Years, and excluding Indians not taxed, three fifths of all other Persons. The actual Enumeration shall be made within three Years after the first Meeting of the Congress of the United States, and within every subsequent Term of ten Years, in such Manner as they shall by Law direct. The Number of Representatives shall not exceed one for every thirty Thousand, but each State shall have at Least one Representative.

The italicized text includes the notorious “three-fifths rule” that, for the purpose of apportionment, a state’s population included three-fifths of the number of slaves residing in the state. In 1868, this wording was replaced by the part of the Fourteenth Amendment that was quoted at the beginning of this chapter. In this text, we use the term apportionment population to mean a state’s population for apportionment purposes. Until 1868, this would be the “Numbers” specified in the Constitution as quoted above. Currently, each state’s apportionment population includes those from the state who are stationed abroad for military or other United States government service.

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Howard Chandler Christy’s Scene at the Signing of the Constitution of the United States.

The Constitution leaves to Congress the task of determining the apportionment method, subject to the following guidelines:

There was a vigorous discussion of apportionment methods in President Washington’s cabinet. Alexander Hamilton, the Secretary of the Treasury, proposed an apportionment method that will be the subject of Section 14.2. The Secretary of State, Thomas Jefferson—Hamilton’s rival—proposed another method that was eventually adopted. We will see Jefferson’s method in Section 14.3.

The Constitution does not specify the number of seats in the House of Representatives. It assigns to Congress the primary responsibility for all matters regarding apportionment, subject to the above guidelines. To accomplish the reapportionment of the House of Representatives, which is done following the decennial censuses, an apportionment bill must be passed by the House and the Senate, and signed into law by the president. Like any bill passed by Congress, the president has the option to veto it.

Congress interpreted the phrase, “one [seat] per thirty Thousand,” to mean that the number of seats in the House cannot be more than the total apportionment population of the United States, divided by 30,000. According to the census of 1790, the total apportionment population was 3,615,920, which, divided by 30,000, yields a quotient between 120 and 121. Thus, in the first apportionment bill, passed in 1792, Congress set the House size to be 120. The apportionment that resulted is shown in Table 14.2.

Table 14.2: TABLE 14.2 The Congressional Apportionment that George Washington Vetoed
State Population District Population Apportionment
Virginia 630,560 30,027 21
Massachusetts 475,327 29,708 16
Pennsylvania 432,879 30,920 14
North Carolina 353,523 ? 12
New York 331,589 ? 11
Maryland 278,514 30,946 9
Connecticut 236,841 29,605 8
South Carolina 206,236 29,462 7
New Jersey 179,570 29,928 6
New Hampshire 141,822 28,364 5
Vermont 85,533 28,511 3
Georgia 70,835 35,418 2
Kentucky 68,705 34,352 2
Rhode Island 68,446 34,223 2
Delaware 55,540 27,770 2
Totals 3,615,920 120

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This bill was the subject of the first presidential veto in U.S. history. (President Washington only vetoed two bills during his time in office.) The reason for the veto can be explained in terms of district population.

District Population DEFINITION

The population of a state, divided by the number of seats apportioned to it, is the state’s district population.

President Washington’s interpretation of “one [seat] per thirty Thousand” was that no state was allowed to have a district population less than 30,000—and 8 of the 15 states had district populations that violated that limit. His veto message, probably written by Jefferson, also contains a statement about fractional parts being unrelated to the state’s populations. This indicates that the apportionment was determined by the Hamilton method, which we will see in Section 14.2.

Self Check 2

Fill in the blank entries in Table 14.2, and determine which state’s district population was too small: New York’s or North Carolina’s.

  • North Carolina was apportioned 12 seats, so its district population is its population of 353,523 divided by 12, or 29,460, which is below the limit. New York’s population, 331,589, divided by its apportionment of 11 seats, is equal to 30,144, so New York’s district population does satisfy the requirement.

Now let’s see how to set up an apportionment problem. Although many apportionment problems do not involve the House of Representatives, our terminology refers to states, populations, and a house size. The first step in solving an apportionment problem is to identify the states, populations, and house size.

EXAMPLE 1 Sharing the Diamonds

In the problem of rounding shares of the diamonds so that their sum is 100, the house size is 100. The participants, Winnie, Louise, and Tim, correspond to the states, and the numbers of tickets each bought correspond to the populations of the states.

Standard Divisor DEFINITION

The standard divisor is the total population divided by the house size. If is the total population, is the house size, and is the standard divisor, then

The standard divisor is the average district population. It is not the average of the district populations for each state, but the average district population for the nation as a whole.

The second step in addressing an apportionment problem is to determine the standard divisor. Continuing with our example of sharing the diamonds, the house size is 100 so the standard divisor is the total population divided by 100. The total population was the number of tickets bought, 23, and the standard divisor was .

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Self Check 3

A total of 123 pieces of hard candy are to be divided among three families, according to the number of children in each family. The Browns have five children, the Joneses have three children, and the Robinsons have four children. Identify the states, house size, and populations for this apportionment problem. What is the standard divisor?

  • The states are the families (Browns, Joneses, and Robinsons); the house size is the number of candies to be apportioned (123); the populations are Browns (5), Joneses (3), and Robinsons (4), for a total of 12. The standard divisor is the total population divided by the house size, .

Quota DEFINITION

In an apportionment problem, the quota for a state is the exact share that would be allocated to the state if a whole number were not required.

Because the standard divisor is equal to , we can also say that

This formula emphasizes that a state’s quota is its fair share of the seats in the House, but it is less efficient for calculation.

The third step in solving an apportionment problem is to calculate the quota for each state. In the diamond-sharing example, the number of tickets is interpreted as the population of a state. The quota for each participant is calculated by dividing its population by the standard divisor. If there are many states, this may entail a bit of work. With a calculator, it is helpful to store the standard divisor in memory to avoid having to enter it repeatedly. If you would like to use a spreadsheet to compute an apportionment, there are pointers in the introduction to the Writing Projects (page 619).

Self Check 4

Find the quota for each family in the problem of apportioning the hard candy in Self Check 3.

  • The quotas are: Browns, ; Joneses, ; Robinsons, .

The quotas in the diamond-sharing scenario, for Winnie, for Louise, and for Tim, are shown in the right column of Table 14.1.

All apportionment methods start with these three initial steps leading to the determination of the quotas. The next step, rounding the quotas to obtain whole numbers whose sum is the house size, is where the various methods differ.

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EXAMPLE 2 The High School Mathematics Teacher

A high school has one mathematics teacher who teaches all geometry, precalculus, and calculus classes. She has time to teach a total of five sections, and 100 students are enrolled as follows: 52 for geometry, 33 for precalculus, and 15 for calculus. How many sections of each course should be scheduled?

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This is an apportionment problem because the number of sections is specified (5), and the number of sections allotted to each course must be a whole number. The three courses correspond to the states; the number of students enrolled in each course corresponds to each state’s population; and the total number of sections to be taught, 5, is the house size. Thus, the populations of geometry, precalculus, and calculus are 52, 33, and 15, respectively. The total population is 100, so the standard divisor is . Table 14.3 displays the calculations of the quotas.

Table 14.3: TABLE 14.3 Calculation of the Quotas for High School Mathematics Courses
Course Population Quota Rounded
Geometry 52 3
Precalculus 33 2
Calculus 15 1
Totals 100                   5.00 6

The sum of the quotas is the house size (in this case, 5). it is tempting to round each quota to the nearest whole number, as in the right column of the table, but this makes 6 sections in all—too many! The purpose of an apportionment method is to find an equitable way to round a set of numbers such as these quotas without increasing or decreasing the original sum.

EXAMPLE 3 California’s Quota

The Census Bureau recorded the apportionment population of the United States—as of April 1, 2010—to be 309,183,463. There are 435 seats in the House of Representatives; therefore, the standard divisor is

California’s quota was determined by dividing its apportionment population, 37,341,989, by this standard divisor. Thus,

This quota is slightly more than the quota that was computed with the 2000 census data, 52.447 seats. However, California’s apportionment, which must be a whole number, was unchanged at 53 seats.

Self Check 5

The 2010 apportionment populations of Rhode Island and North Carolina were 1,055,247 and 9,565,781, respectively. Find their quotas. Given that the apportionments were 2 seats in the House for Rhode Island and 13 for North Carolina, find the district populations for these states.

  • Rhode Island’s quota was , and North Carolina’s quota was . The district populations are for Rhode Island, and for North Carolina.

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Ideally, each state’s apportionment should be close to its quota. It is unrealistic to expect that any state will be apportioned its exact quota because each apportionment is required to be a whole number and the quota is unlikely to be a whole number. In choosing an apportionment method, we must decide what we mean by the phrase “each state’s apportionment should be close to its quota.”

Apportionment always involves rounding, and there are many ways to round. “Rounding down” means discarding the fractional part of a number to obtain a whole number that we will denote as . Thus, , , and . “Rounding up” gives the next whole number, denoted as . Thus, , but

No apportionment method is perfect for all occasions. Our goal is to understand how to choose a method that is appropriate for a particular apportionment problem.