Alexander Hamilton’s Method and Upper and Lower Quotas DEFINITIONS

With the Hamilton method, each state receives either its lower quota, —its quota rounded down—or its upper quota, —obtained by rounding the quota up. The states that receive their upper quotas are those whose quotas have the largest fractional parts.

Hamilton Method PROCEDURE

In Self Check 1, you found that the administrative assistant spent 41.67% of his time on his duties, 9.72% of his time on commuting, and 48.61% of his time on personal activities. Using the Hamilton method, round the percentages to whole numbers that add up to 100%.

EXAMPLE 4 The High School Teacher’s Dilemma

Let us use the Hamilton method to determine how many sections of geometry, precalculus, and calculus the high school teacher in Example 2 (page 577) should teach. Table 14.3 (page 577) displays the quotas for the three subjects,. 2.60, 1.65, and 0.75, respectively. Table 14.4 shows how to obtain the apportionment by the Hamilton method.

The lower quotas are the tentative apportionments, and their sum is 3, leaving two sections still to be apportioned. These go to precalculus and calculus, because the quotas for these courses have the largest fractional parts, 0.65 and 0.75, respectively.

Determine the “district population” for each of the subjects in Example 4. Explain, in familiar terms, what the district populations represent in this context.

EXAMPLE 5 Sharing Diamonds by the Hamilton Method

In Table 14.1 (page 572), we saw that Winnie’s, Louise’s, and Tim’s quotas for the 100 diamonds were 78.26, 17.39, and 4.35, respectively. We will round these quotas so that they sum to 100. To start, we apportion diamonds to Winnie, diamonds to Louise, and diamonds to Tim. This leaves one diamond, which must go to Louise because her fraction, 0.39, is the largest. The final apportionment is 78 diamonds for Winnie, 18 diamonds for Louise, and 4 diamonds for Tim. Winnie will want to veto this apportionment.

EXAMPLE 6 The Vetoed Apportionment Bill by Hamilton’s Method

The congressional apportionment bill that President Washington vetoed involved 15 states, and the House had 120 seats. According to the 1790 census, the U.S. population was 3,615,920. The standard divisor, , represents the population of the average congressional district.

Table 14.5 displays the apportionment, as calculated by Hamilton’s method. Each quota shown in the table was calculated by dividing the state’s population by this standard divisor. Adding the lower quotas, we find that their sum, 111, leaves 9 seats to be apportioned. These go to the 9 states whose quotas had the largest fractional parts.

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The apportionments following the censuses of 1790 through 1840 used a method proposed by Thomas Jefferson.

A divisor, representing the least population to be allowed per representative, was chosen. A state’s apportionment was obtained by dividing its apportionment population by that divisor, and rounding the quotient down to get a whole number. Following the censuses of 1790 and 1800, the divisor was 33,000. Thereafter, the divisor increased, to 35,000 in 1810, 40,000 in 1820, and 47,700 in 1830. in 1840, Webster’s method was used; that is, each state’s apportionment population was divided by a divisor (70,680 for the 1840 census), and the quotient rounded to the nearest whole number. (This is the same procedure that may have been used, with 30,000 as the divisor, for the vetoed apportionment bill—although Webster, then 10 years old, was obviously not involved.)

In 1850, the legislation that provided for the census also specified, for the first time, the house size (233 for the 1852 apportionment). The Hamilton method of apportionment was also specified. The apportionment acts of 1862 through 1892 were debated and enacted after the census counts were known. From 1882 on, house sizes where the Hamilton and Webster methods gave different results were avoided, so it could be claimed that either method was used. For an account of the apportionment following the 1870 census—the first after the Civil War—see Spotlight 14.2 on page 597.

EXAMPLE 7 The Apportionment of 1882

In 1880, the apportionment population of the United States was 49,373,329. When the house size was set at 299, the standard divisor was . Alabama’s apportionment population was 1,262,505. Thus, Alabama’s quota was . Alabama received its upper quota, so if the house size of 299 had been chosen, Alabama would have been apportioned 8 seats. All states with quotas having fractional parts less than Alabama’s were apportioned their lower quotas. The states following Alabama when comparing fractions were Illinois and Texas, with apportionment populations of 3,077,871 and 1,591,729, respectively. Their quotas were 18.63930 and 9.63936, respectively, and they were apportioned their lower quotas.

When the house size increased to 300, the standard divisor was less, . Dividing the three states’ apportionment populations by this smaller divisor yielded the following quotas: Alabama, 7.67117; Illinois, 18.70159; and Texas, 9.67158. One additional seat was available for apportionment. With this house size, the quota for Illinois had the largest fractional part of the three, followed by Texas; Alabama was in third place. The seat that had been Alabama’s, and the 300th seat, went to Illinois and Texas; both states were apportioned their upper quotas. Alabama received its lower quota. Table 14.6 summarizes the result of these calculations.

Alabama Paradox DEFINITION

The Alabama paradox occurs when a state loses a seat as the result of an increase in the house size, or gains a seat due to a reduction of the house size, with no change in any state’s population.

EXAMPLE 8 A Mathematics Department Meets the Alabama Paradox

A mathematics department has 30 teaching assistants (TAs) to cover recitation sections for College Algebra, Calculus I, Calculus II, Calculus III, and Contemporary Mathematics. The enrollments of these courses are given in Table 14.7. The department will use the Hamilton method to apportion the TAs to the five subjects. in this problem, the house size is 30 (the number of TAs) and the population is the number of students, 750. The states are the five courses to be offered. The standard divisor is , which represents the average number of students per recitation section. Each quota shown in the table was determined by dividing the enrollment of the course by this divisor.

The lower quotas add up to 27, so the three courses whose quotas have the largest fractional parts, Calculus I and III and Contemporary Mathematics, were given their upper quotas.

After the TAs were given their teaching assignments, the graduate school authorized the department to hire an additional TA. To determine which course should get the new TA, the department had to recalculate the apportionment. With 31 TAs, the standard divisor was . The new quotas, determined by dividing each population by this new divisor, are shown in Table 14.8. Now the lower quotas add up to 28, so again three additional TAs go to the subjects whose quotas have the largest fractions. The Calculus III fraction, which had been larger than the College Algebra fraction when there were just 30 teaching assistants, has been surpassed. The new TA was placed in College Algebra, and one of the Calculus III TAs had to be reassigned to Calculus II.

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Suppose that the reverse had occurred. The department started with 31 TAs, and just before the semester began, one of the TAs was awarded a research position and released from teaching duties, so that only 30 TAs were available. Is there a paradox in this case?

Population Paradox DEFINITION

Two apportionments are made by the same apportionment method, based on censuses taken at different times. The house size is the same for both apportionments, but the populations for each of the states have changed. The population paradox occurs if there is a pair of states and such that when the first census results are compared with those of the second, it is found that

EXAMPLE 9 Apportioning Seats in Parliament

A country has four political parties. its parliament has 100 members, and seats are apportioned by the Hare method after each election so that the number of seats that each party is awarded is as close as possible to being proportional to the number of votes that the party receives.

An election is held, but the parties are unable to form a government, so there is a repeat election. Table 14.9 shows the results of the two elections. The three major parties—Whigs, Tories, and Liberals—all received more votes in the second election, but the Centrists received fewer. The quotas for each party, shown in Table 14.10, were determined by dividing each party’s votes by the standard divisors

for the first election, and 132,517.70 for the second election.

The lower quotas for the results of the first election were 42, 26, 29, and 1, with a sum of 98; thus, the Tories and the Liberals, with the largest fractions, get extra seats. The apportionment after the first election was Whigs, 42; Tories, 27; Liberals, 30; and Centrists, 1.

For the repeat election, the lower quotas were the same, but now the largest fractions belong to the Whigs and the Centrists. Therefore, the new apportionment is Whigs, 43; Tories, 26; Liberals, 29; and Centrists, 2.

The Centrists have gained a seat, although they received fewer votes in the repeat election, while the Tories and the Liberals each lost a seat, even though their vote totals increased in the repeat election. This is an instance of the population paradox. This has a disturbing implication: A group of voters might have unintentionally caused the Centrist Party to gain a seat by switching their votes to the Liberal Party.