According to the 2010 census, the apportionment population of California was 37,341,989 and the state was apportioned 53 seats in the House of Representatives. Montana, with apportionment population 994,416, was apportioned just one seat. The district population of California was 704,565. Determine the district populations that would result if a seat were transferred from California to Montana. Would the difference be less if this transfer took place?

In 1991, the Census Bureau reported the apportionment of congressional seats resulting from the 1990 census. The states colored red on the above map lost representatives, and those in blue gained representatives. New York lost three representatives, and Ohio and Pennsylvania lost two apiece. Montana, whose apportionment decreased from 2 to 1, sustained the greatest percentage loss, and Montana sued to restore the lost seat. As precedents, Montana referred to the two famous cases, Baker v. Carr and Wesberry v. Sanders, in which the U.S. Supreme Court required state legislative and congressional district boundaries to be drawn so as to make district populations equal. Montana argued that the correct apportionment would be the one that met the Baker and Wesberry criterion of having district populations as nearly equal as possible, and asked the court to require the Census Bureau to recompute the apportionments using the Dean method. This would have resulted in the transfer of a congressional seat from Washington to Montana.

The Montana case proceeded concurrently with another federal lawsuit, Massachusetts v. Mosbacher, which asked the Federal District Court to order the apportionment to be calculated by the Webster method. if Massachusetts had won this suit, a seat would have been transferred to it from Oklahoma (see Exercise 42 on page 617). The apportionments of Montana and Washington would not have changed.

In U.S. Department of Commerce v. Montana, the Supreme Court unanimously rejected Montana’s claim. The opinion of the court, written by Justice John Paul Stevens, pointed out that intrastate districts, which were the subject of the Baker and Wesberry cases, could be equalized in population by drawing district boundaries correctly. Because congressional districts can’t cross state lines, some inequity is inevitable in congressional apportionment. The opinion conceded that there were alternative apportionment methods but concluded that the choice of apportionment method must be left to Congress.

Representative Share DEFINITION

Let be the apportionment given to a state whose population is . The quotient is called the representative share for that state. It represents the share of a congressional seat given to each citizen of the state.

Webster Method and Representative Share THEOREM

If a seat is transferred from one state to another in an apportionment that was calculated by the Webster method, the absolute difference between their representative shares will not decrease.

EXAMPLE 17 Inequity in the 113th Congress?

The 2010 census reported the following apportionment populations: North Carolina, 9,565,781, and Rhode Island, 1,055,247. The states were apportioned 13 seats and 2 seats in Congress, respectively. Would it have been fairer to give North Carolina 14 of the 15 seats between them? We will calculate the district populations and representative shares for both apportionments and make a comparison.

With 13 seats to North Carolina, the district populations are for North Carolina and for Rhode Island. The representative share for the states are for North Carolina and for Rhode Island. With a smaller district population and a larger representative share, Rhode Island has the advantage with this apportionment. The bottom line: Rhode Island’s district populations are 208,195 smaller than North Carolina’s, and Rhode Island’s representative share is 0. 536 seats per million larger than North Carolina’s.

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Table 14.17 summarizes what we have calculated so far and compares the results with those that would be obtained if North Carolina had been apportioned 14 seats to Rhode Island’s 1. With the 14-1 apportionment, North Carolina would have the advantage. Its district size would have been 371,977 less than Rhode Island’s. This would be a greater discrepancy than exists with the 13-2 apportionment. On the other hand, North Carolina’s representative share would have been 0.516 seats per million greater than Rhode Island’s—a smaller discrepancy than exists with the 13-2 apportionment!

Absolute and Percentage Difference DEFINITION

Given two positive numbers and , with , the absolute difference is equal to and the percentage difference is equal to the quotient . In computing the percentage difference between two numbers, the lesser of the two numbers goes in the denominator.

The world record for the men’s 100-meter race, 9.6 seconds, was set in 2009 by Usain Bolt. A century earlier, the record was 10.6 seconds, held by Knut Lindberg.

EXAMPLE 18 Percentage Inequity in the 113th Congress

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Let’s refer to Table 14.17 again and recalculate the “bottom line” as percentage differences. For example, for the percentage difference in district population with the “13-2” apportionment, we would divide the absolute difference, 208,195, by the lesser district population, which is Rhode Island’s, 527,634. The result is 39.46%. You can calculate the remaining entries and obtain the following.

You can see that the percentage differences in district population and representative share are the same, as Professor Huntington said that they would be, and that in terms of these, the “13-2” apportionment, where North Carolina gets 13 seats and Rhode Island gets 2, is preferred.

Geometric Mean DEFINITION

The geometric mean of two positive numbers and is equal to the square root of their product, .

Find the geometric mean of 5 and 20.

Hill-Huntington Rounding DEFINITION

Given a number that is not negative, let be the geometric mean of and ; that is,

This is called the Hill-Huntington rounding point for . The Hill-Huntington rounding of a number is equal to if but is equal to if .

EXAMPLE 19 Hill-Huntington Rounding

Suppose that . Jefferson and Webster would round 7.485 down to 7. Because and , the rounding point is , approximately. Since , Hill-Huntington rounds up to get 8.

Hill-Huntington Apportionment PROCEDURE

No Zero Apportionments THEOREM

The Hill-Huntington method is incapable of apportioning zero seats to any state.

EXAMPLE 20 Rounding Summands

Let’s use the Hill-Huntington method to round each summand in

Each summand is the quota, for the standard divisor is 1. The rounding points are as follows:

Each summand is greater than its corresponding rounding point, so all are rounded up, resulting in a too large sum: . We’ll need to choose a divisor larger than 1; let’s try 2. The apportionment quotients are

The first two apportionment quotients are between 0 and 1, and so they are rounded to 1; the third is below the rounding point between 1 and 2, so it too rounds to 1. The sum, 3, is too small.

Let’s try again with a smaller divisor, . The new apportionment quotients are

The first two apportionment quotients are still between 0 and 1, so they round to 1. The third is greater than the rounding point between 1 and 2, so it rounds to 2. Therefore the Hill-Huntington rounding of this sum is

Use the Hill-Huntington method to round the percentages:

(You may use 1.007 as the divisor.) Does the apportionment satisfy the quota condition?

Equity and the Hill-Huntington Method THEOREM

An apportionment is equitable by percentage differences if and only if it is the same as the apportionment that is produced by the Hill-Huntington method.

EXAMPLE 21 Percent Effort

Faculty members at a certain university must state the percentage of their time spent in several activities. Professor Worktorule has requisitioned five stopwatches to keep track of her activities. Table 14.18 shows, in its left columns, what she recorded over the course of one week.

The professor is too busy to convert the data into percentages—which the university requires in whole numbers with sum 100%—so we’ll do it, using the Hill–Huntington method. (She requested this method because she wanted the result to display a nonzero percentage for each of her activities.) As with any percentage apportionment problem, the house size is 100, so the standard divisor—one percentage unit—is the population, 3600, divided by 100, or 36 minutes. Table 14.18 shows in its right columns, the quotas, the rounding points, and the tentative apportionment, obtained by rounding the quotas up or down, depending on whether the quota is above the rounding point or not.

Because too many “seats” were awarded, we’ll try divisors larger than the standard divisor, 36. The results are shown in Table 14.19. The first divisor we tried, 36.5, produced 98 “seats” too few. The left column under that divisor shows the corresponding apportionment quotients (AQ), obtained by dividing the minutes devoted to each activity by that divisor. The middle column shows the Hill-Huntington rounding points (RP) for the apportionment quotients, and the right column displays the Hill-Huntington tentative apportionment (TA). To increase the number of “seats” apportioned, the next trial divisor must be closer to 36 (but more than 36). We tried the divisor 36.2 and found that 99 “seats” were apportioned. Our third trial divisor was 36.1 (not shown in the table), which apportioned 101 “seats”—too many. The divisor we needed was therefore between 36.1 and 36.2. We set the divisor equal to 36.15, and the table shows that exactly 100 “seats” were apportioned. The right column of Table 14.19 displays the percentages that Professor Worktorule should put into her effort report.

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EXAMPLE 22 The 435th Seat

The Webster and Hill-Huntington methods give almost the same apportionments, based on the 2010 census. The only difference is that the Webster method apportions the last seat to North Carolina, while the Hill-Huntington method gives it to Rhode Island.

The standard divisor is the apportionment population of the United States, 309,183,463, divided by the number of seats, 435. Thus, . The quotas, obtained by dividing the states’ populations by s, are for North Carolina and for Rhode Island. With the Webster method, both quotas would be rounded down and only 434 seats would be apportioned. If we reduce the divisor to 708,000, North Carolina’s quotient becomes 13.51, which will be rounded up, and Rhode Island’s quotient is 1.49, which will be rounded down. North Carolina will get 14 seats, and Rhode Island will get 1.

With the Hill-Huntington method, the rounding point for numbers between 13 and 14 is , which is greater than North Carolina’s quota. The rounding point for numbers between 1 and 2 is , which is less than Rhode Island’s quota. Therefore, North Carolina’s quota is rounded down to get its apportionment, 13, and Rhode Island’s quota is rounded up to get its apportionment, 2, with the Hill-Huntington method.

The first American to consider apportionment from a theoretical point of view was Walter Willcox (1861–1964), who strongly advocated the Webster method and had computed the apportionment of the 78th Congress in 1902. His arguments convinced Congress to use the Webster method again in 1912. In 1911, Joseph Hill, a statistician at the Census Bureau, proposed the Hill-Huntington method—with the endorsement of Edward V. Huntington, a mathematics professor at Harvard.

In 1920, the two methods were in competition. There were significant differences in the apportionments determined by the two methods, and the result was Washington gridlock: No apportionment bill passed during the decade, and the 1912 apportionment was retained throughout the 1920s. In preparation for the 1930 census results, the National Academy of Sciences formed a committee of distinguished mathematicians to study apportionment. In 1929, the committee endorsed the Hill-Huntington method.

The 1930 census was remarkable in that the apportionments calculated by the Webster method were the same as the Hill-Huntington apportionments. Therefore, the House of Representatives was reapportioned, but the method used could be claimed to be either one of the competing methods. The coincidence was almost repeated in the 1940 census, but there was one difference: The Hill-Huntington method gave the last seat to Arkansas, while Webster’s method gave it to Michigan. At the time, Michigan was a predominantly Republican state and Arkansas was in the Democratic column. The vote on the apportionment bill split strictly along party lines, with Democrats supporting the Hill-Huntington method and Republicans voting for the Webster method. Because the Democrats had the majority, the Hill-Huntington method became law.