14.4 14.3 Divisor Methods

The Jefferson Method

Following President Washington’s veto of the first apportionment bill in 1792, Congress moved quickly to pass a new apportionment bill. Each state’s apportionment population was divided by 33,000, and the resulting quotient was rounded down to get a whole number, which was the state’s apportionment. Thus, Virginia, with apportionment population 630,560, received

Delaware’s apportionment was seat. Because the number of seats in the House was not specified in advance, the apportionment problem was I avoided. (It turned out that 105 seats were apportioned.)

If a state with population was apportioned seats, its district population would be Since

it follows that the district population

In words, the state’s district population would be at least 33,000.

image
Thomas Jefferson favored a method of apportionment biased in favor of states with large populations.
National Portrait Gallery, Smithsonian Institution; gift of the Regents of the Smithsonian Institution, the Thomas Jefferson Memorial Foundation, and the Enid and Crosby Kemper Foundation/Art Resource, NY

Self Check 9

Referring to Table 14.2 on page 574, determine the number of seats that would have been apportioned if the divisor had been set to be 30,000, and explain why this was the maximum house size that would conform to the requirement that each state’s district population must be at least 30,000.

  • The following table is the apportionment. In any apportionment in which no state receives more than one representative per 30,000 population, the maximum apportionment permitted is

    These are the apportionments shown in the table. Thus, the maximum possible house size is 112.

    State Population Pop/30,000 Apportionment
    Virginia 630,560 21.02 21
    Massachusetts 475,327 15.84 15
    Pennsylvania 432,879 14.43 14
    North Carolina 353,523 11.78 11
    New York 331,589 11.05 11
    Maryland 278,514 9.28 9
    Connecticut 236,841 7.89 7
    South Carolina 206,236 6.87 6
    New Jersey 179,570 5.99 5
    New Hampshire 141,822 4.73 4
    Vermont 85,533 2.85 2
    Georgia 70,835 2.36 2
    Kentucky 68,705 2.29 2
    Rhode Island 68,446 2.28 2
    Delaware 55,540 1.85 1
    Totals 3,615,920 120.53 112

The method just described was suggested by Thomas Jefferson, and it is called the Jefferson method.

With this method, a divisor is specified. Using , rather than the standard divisor , as the divisor, an apportionment quotient, rather than a quota, is obtained for each state. A state’s apportionment quotient is rounded down to obtain its apportionment.

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Apportionment Quotient DEFINITION

The result of dividing a state’s population by a divisor is the state’s apportionment quotient.

Jefferson Method DEFINITION

A divisor is chosen, and, using that divisor, each state’s apportionment quotient is determined. The apportionment quotients are rounded down to obtain the apportionments for the states.

The Jefferson method is one of a family of apportionment methods called divisor methods.

Divisor Method DEFINITION

A divisor method of apportionment determines each state’s apportionment by selecting an appropriate divisor and rounding the resulting apportionment quotients, using a specified rounding rule. Divisor methods differ in the rule used to round the apportionment quotient.

In an apportionment problem, there is a specified house size. When a divisor method is used in an apportionment problem, the challenge is to choose the divisor that achieves the intended house size. Select a divisor by trial and error, with the insight that to reduce the number of seats apportioned the divisor must be increased, and to increase the number of seats, a smaller divisor must be used.

The Jefferson Method is the divisor method associated with the rounding rule that rounds each apportionment quotient down. In this case, the divisor must be less than the standard divisor, because if the standard divisor were used, each state would be apportioned its lower quota—and we know that the lower quotas sum to less than the house size (except in the very unusual case where all the quotas turn out to be whole numbers). Let’s see how apportioning the diamonds (Example 5 on page 579) works with the Jefferson method.

EXAMPLE 10 Sharing Diamonds by the Jefferson Method

Winnie, Louise, and Tim have bought 18, 4, and 1 lottery tickets, respectively, and have won 100 diamonds. The Jefferson method can be used to share the diamonds. The house size is 100 diamonds, and the total population is the 23 lottery tickets. The standard divisor is 0.23. Table 14.11 shows what happens with a few trial divisors, starting with 0.22, which turns out to be too small, as it apportions a total of 103 diamonds. We gradually increase the divisor until 100 are apportioned. For each trial divisor, there are two columns: the apportionment quotients on the left, and the same quotients rounded down on the right. The sums of the rounded apportionment quotients are shown, and we see that the divisor 0.227 gets the correct house size, 100. (There is no need to add the apportionment quotients, as their sums play no role in determining the apportionment.)

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Table 14.12: TABLE 14.11 Apportioning 100 Diamonds by the Jefferson Method
Divisors
Participant Population 0.22 0.225 0.227
Winnie 18 81.82 81 80.00 80 79.30 79
Louise 4 18.18 18 17.78 17 17.62 17
Tim 1 4.55 4 4.44 4 4.41 4
Total 23 103 101 100

The apportionment is what Winnie wanted: 79 diamonds for her, 17 for Louise, and 4 for Tim.

Self Check 10

Without any calculations, explain why the divisor 0.226 would also produce the apportionment of 79 diamonds for Winnie, 17 for Louise, and 4 for Tim.

  • Table 14.11 shows that with the divisor 0.225, the apportionment quotient for Winnie was exactly 80. The apportionment quotients for Louise and Tim were 17.78 and 4.44, respectively. The divisor 0.227 reduced the apportionment quotient for Winnie to 79.30, while Louise and Tim had apportionment quotients of 17.62 and 4.41, respectively. Thus, the divisor 0.226 would produce apportionment quotients for Winnie between 79.30 and 80, for Louise between 17.62 and 17.78, and for Tim between 4.41 and 4.44. These quotas, rounded down, would be 79, 17, and 4 diamonds, respectively.

Apportioning with the Jefferson Method1 PROCEDURE

  1. Make a table with the “states” in the left column.
  2. Choose a divisor that is slightly less than the standard divisor.
  3. Divide each state’s population by the chosen divisor to get its apportionment quotient. Round down to a whole number to get its tentative apportionment. Thus, if the apportionment quotient is 9.999, round it to 9.
  4. Add the tentative apportionments. If the sum equals the house size, you have completed the apportionment.
  5. If the total in Step 4 is less than the house size, choose a smaller divisor and repeat Steps 3 and 4. If the total in Step 4 is larger than the house size, choose a larger divisor—but not larger than the standard divisor—and repeat Steps 3 and 4.

We will implement this procedure for our high school mathematics teacher. The Jefferson divisor in this context can be interpreted as a minimum class size because the number of classes allocated for each subject is obtained by dividing the number of students enrolled by , and discarding any fractional section in the quotient. Thus, each section will have at least students.

EXAMPLE 11 Apportioning Classes by the Jefferson Method

The teacher can be assigned five classes. There are 52 students enrolled in geometry, 33 in precalculus, and 15 in calculus. The calculations to determine her teaching assignment by the Jefferson method are shown in Table 14.12. We determined in Example 2 (page 577) that the standard divisor is 20, so we’ll use 18 as our first trial divisor.

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This time, our starting divisor was too large (only three sections were apportioned). We need to try a smaller divisor. With , 6 sections would be apportioned. Thus, the Jefferson divisor has to be more than 15 and less than 18. it turns out that works as a divisor, resulting in 3 geometry classes and 2 precalculus classes. There will be no calculus class because the 15 students enrolled are not enough for a class when the minimum class must have at least 16 students.

Table 14.13: Table 14.12 Apportioning High School Mathematics Classes
Divisors
Subject Population 18 15 16
Geometry 52 2.89 2 3.47 3 3.25 3
Precalculus 33 1.83 1 2.20 2 2.06 2
Calculus 15 0.83 0 1.00 1 0.94 0
Total 100 3 6 5

Comparing the results of Example 11 and Example 4 (page 579), where the Hamilton and Jefferson apportionment methods were applied to the same apportionment problem, we see that the results differ. Examples 10 and 5 also exhibit different apportionments from another apportionment problem. Now let us apportion the TAs in the scenario we encountered in Example 8 on page 582. Will the Alabama paradox occur with the Jefferson method? Recall that with the Hamilton method, Calculus III had 3 TAs when 30 TAs were available, but only 2 TAs when another TA was hired.

EXAMPLE 12 Apportioning TAs with the Jefferson Method

in Example 8 (page 582), a mathematics department was to apportion its TAs among five courses. The populations were the numbers of students enrolled in each course, the states were the courses, and the house size was the number of TAs that were available. Here are the enrollment data: College Algebra, 188; Calculus l, 142; Calculus ll, 138; Calculus ill, 64; and Contemporary Mathematics, 218. For a house size of 30, the standard divisor was 25. in Table 14.13, we will follow the procedure used in the previous two examples. The apportionment quotients are not shown, just the tentative apportionments that were obtained by rounding them.

Table 14.14: Table 14.13 Apportioning TAs by the Jefferson Method
Divisors
Subject Population 23 24 23.5
College Algebra 188 8 7 8
Calculus I 142 6 5 6
Calculus II 138 6 5 5
Calculus III 64 2 2 2
Contemporary Math 218 9 9 9
Totals 750 31 28 30

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The first trial divisor was 23, less than the standard divisor. The number of sections apportioned was 31, which is what we will need when the additional TA arrives. Attempting to apportion 30 sections, we increase the divisor to 24, but that reduces the seats apportioned to 28. The divisor that we need is therefore between 23 and 24, so we will try 23.5, which indeed produces an apportionment of exactly 30 sections.

Self Check 11

The Mathematics Department has made a rule that all sections must have at least 24 students enrolled. Assuming that the department intends to open the maximum number of sections, how many TAs will be needed?

  • The maximum number of sections for a course would be the population for that course, divided by 24, and rounded down. This would be found by the Jefferson method, using the divisor 24. Referring to Table 14.13, where that divisor was employed, we see that 28 TAs are needed.

Table 14.13 shows that when the house size increases from 30 to 31, Calculus II gets the new TA, and no course loses a TA. In fact, the Jefferson method prevents the Alabama paradox, as the following theorem confirms.

Divisor Methods Avoid Paradoxes THEOREM

The Alabama paradox, the population paradox, and the new states paradox cannot occur with any divisor method of apportionment.

We will verify that the Alabama paradox is impossible if the Jefferson method is used.

Let be the divisor used to apportion a house with seats. For a house size of , the divisor must be less than . Consider a state with population . Its apportionment with house size will be , and its apportionment with house size will be .

Since , it follows that and hence

Therefore, an increase in the house size, with no change in any state’s population, cannot lead to a decrease in any state’s apportionment. This shows that the Jefferson method prevents the Alabama paradox.

Because divisor methods only differ by the rounding rule used, the same argument shows that any divisor method will prevent the Alabama paradox. The other two paradoxes also do not occur with divisor methods (see Exercises 26 and 27 on page 615).

Here is another distinction between the Hamilton and Jefferson methods— and this time it is in the Hamilton method’s favor. The Hamilton method always gives each state either its upper quota or its lower quota.

Self Check 12

Explain why it is that, since the Jefferson divisor must not be more than the standard divisor, the Jefferson method always apportions to each state at least its lower quota.

  • A state has population . Let s be the standard divisor and be the Jefferson divisor. The state’s apportionment quotient is and its quota is . Since ,

    It follows that rounding the apportionment quotient down results in a whole number not less than the lower quota, which we obtain by rounding the quota down.

Although the house size was ignored when the Jefferson method was used to apportion seats in the House of Representatives, when the apportionment was complete, it was possible to compare each state’s apportionment with its quota. In 1822, there was a surprise.

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EXAMPLE 13 A Problem with the 1822 Congressional Apportionment

The 1820 census reported that New York had an apportionment population of 1,368,775. The total apportionment population of the United States was 8,969,878. Using the divisor , a total of 213 seats were apportioned by the Jefferson method. With the 213 as the house size, the standard divisor was . New York’s quota was . The Hamilton method, with 213 seats in the House, would have apportioned to New York its upper quota, 33 seats. However, New York’s apportionment quotient was . The Jefferson method rounded the apportionment quotient down and apportioned 34 seats to New York, one more than New York’s upper quota.

Quota Condition DEFINITION

An apportionment method is said to satisfy the quota condition if in every situation, each state’s apportionment is equal to either its lower quota or its upper quota.

The Jefferson method never gives a state fewer seats than its lower quota, but it takes only one example like the 1822 apportionment to show that a state’s Jefferson apportionment may be more than its upper quota. Therefore the Jefferson method does not satisfy the quota condition. In fact, if the House had continued to use the Jefferson method, it would have given some state more than its upper quota in every apportionment since 1850. For example, the Jefferson apportionment of the House according to the 2010 census gives California 55 seats, although its quota is 52.54; and Texas, with quota 35.55, would get 37 seats.

Self Check 13

Use the Jefferson method to round to whole percentages:

(You may use the divisor 0.98 to do this.) Is the quota condition violated by this apportionment?

  • The sum of the lower quotas is 98%. Divide each quota by 0.98, as suggested, to get apportionment quotients 99.18%, 1.94%, 0.92%. Rounding down we have

    This violates the quota condition, because the apportionment to the first percentage, 97.2%, is greater than its upper quota, 98%.

The Hamilton method satisfies the quota condition. This was obvious to Congress in 1850, and that is why it switched to the Hamilton method in that year.2

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Congress has never used an apportionment method that satisfies the quota condition and avoids the paradoxes of the Hamilton method. In the 1970s, the mathematicians Michel L. Balinski and H. Peyton Young tried to devise such a method. Their research showed that the population paradox is possible for all apportionment methods except divisor methods. (The Hamilton method paradoxes are impossible with divisor methods.) However, no divisor method satisfies the quota condition! Thus, while Balinski and Young failed to meet their objective, they did prove that there is no apportionment method that satisfies the quota condition and prevents the population paradox. Their theorem is like Kenneth Arrow’s impossibility theorem, which tells us that there is no completely satisfactory way to decide multicandidate elections based on voter preference schedules (see Section 9.4, page 424).

The Jefferson method favors the larger states. It is not an accident that in every example that we have considered, the “state” with the largest population fared better with the Jefferson method than it did with the Hamilton method. Winnie, not Louise, received the extra diamond when the Jefferson method was used, as compared with the Hamilton method, and in the example of the high school math teacher, there were more sections of geometry and no sections of calculus when the Jefferson method was substituted for the Hamilton method.

Let’s see why the Jefferson method is biased in favor of larger states. Because the Jefferson method always rounds the apportionment quotients down, each state’s apportionment quotient must be greater than its quota (otherwise, each state would get its lower quota of seats or fewer). Therefore, the divisor used to obtain these apportionment quotients must be less than the standard divisor. If we compare the quotas to the apportionment quotients, we will find that their ratio is equal to the ratio of the standard divisor to the Jefferson divisor:

Denote this ratio by . Since the Jefferson divisor is less than the standard divisor, , so we will let . Then

and so the amount the state’s apportionment quotient exceeds the quota is

The larger the state’s quota is, the more it increases when the apportionment quotient is substituted. This is an advantage to the states with large populations.

Consider the congressional apportionment of 1822. The standard divisor is , and the Jefferson divisor is . The quotient is

thus . Now suppose that the quota for state is . The state’s apportionment quotient is obtained by increasing the state’s quota by . A state with a large quota will get a greater raise under these circumstances than a small state will.

To see how this works with numbers, consider a state with . The apportionment quotient is , or 1.003 greater than , which brings it to 20. The upper quota for is 19, but is awarded 20 seats! This violates the quota condition, and in fact, every state whose quota is 18.997 or more will be guaranteed to get at least its upper quota. If a state has a quota greater than , an identical calculation shows that it will receive at least its upper quota plus 1 seat. On the other hand, consider a small state whose lower quota is 1. To increase its apportionment to 2, its quota must be at least . Thus, a state with quota 18.997 gets more than its upper quota, and a state with quota 1.899 has to settle for its lower quota.

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When parliamentary seats are being apportioned, the Jefferson method’s bias favors major parties, which typically attract many votes, over less popular parties. The Jefferson method is used by many nations that prefer voters to select party lists rather than to vote for individual candidates for parliament. It is believed that this practice favors more stable government because there will be less need for a major party to be forced into a coalition with a small party.

For historical reasons, when the Jefferson method is used to apportion seats in parliament, it is called the d’Hondt method, after Victor d’Hondt (1841–1901). Countries that use the d’Hondt method to apportion seats in their parliaments include Argentina, Denmark, and Israel.

The procedure for apportioning seats by the d’Hondt method is not the same as the one that we have used for the Jefferson method, but it leads to the same apportionment. With the d’Hondt method, seats are apportioned one at a time until the house size is reached.

d’Hondt Method PROCEDURE

  1. Each party is given a priority number, which is reduced as the party accumulates seats. A party’s priority number is initially the number of votes that it received.
  2. After a party has received seats, its priority number is equal to the number of votes that it received divided by .
  3. Award the first seat to the party that has the largest number of votes; divide that party’s vote total by 2 to get its new priority number.
  4. Award the second seat to the party that now has the highest priority; recalculate its priority. Continue in this way until all the seats have been distributed.

With the d’Hondt method, it is unnecessary to compute the standard divisor or the quota for any party. The d’Hondt method provides the order in which members of parliament receive their seats. The first member to be seated is the first on the list of the party with the plurality of votes, and so on. However, it achieves exactly the same result as the Jefferson method. Think of the parties’ priority numbers as divisors. When a party receives the last seat (because its priority is highest), its priority number (before that seat is awarded to it) is the divisor that will produce the Jefferson apportionment. Here is why this is so: Suppose the party that is about to receive the last seat got votes and already has seats. Its priority number is . Thus, (exactly—no rounding needed), so with divisor , party will get seats. The seats awarded to the other parties were awarded with priority numbers larger than , so the divisor will award those seats to them as well. They will not get any additional seats, though, because their current priority numbers are less than .

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EXAMPLE 14 Using the d’Hondt Method

We will use the d’Hondt method to apportion the first 30 seats in the parliament of Example 9 (page 584), based on the data from the first election. The calculations are presented in Table 14.14, a d’Hondt table. There is a column in the table for each party. The first entry in each column is the number of votes that the party received—the party’s initial priority number. Running down each column, the entries are the number of votes for the parties divided by 2, 3, 4, and so on. When a seat is awarded, the priority number for the party that receives the seat is crossed out because it has been used. The next seat goes to the party with the highest priority that has not been crossed out.

Table 14.15: TABLE 14.14 A d’Hondt Table
Apportioned Whigs Tories Liberals Centrists
1 #1 5,525,381 #3 3,470,152 #2 3,864,226 201,203
2 #4 2,762,691 #7 1,735,076 #5 1,932,113 100,602
3 #6 1,841,794 #10 1,156,717? #9 1,288,075 67,068
4 #8 1,381,345 #14 867,538 #12 966,057 50,301
5 #11 1,105,076 #17 694,030 #16 772,845 40,241
6 #13 920,897 #21 578,359 #19 644,038 33,534
7 #15 789,340 #25 495,736 #23 552,032 28,743
8 #18 690,673 #28 433,769 #26 483,028 25,150
9 #20 613,931 385,572 #29 429,358 22,356
10 #22 552,538 347,015 386,423 20,120
11 #24 502,307 315,468 351,293 18,291
12 #27 460,448 289,179 322,019 16,767
13 #30 425,029 266,935 297,248 15,477
14 394,670 247,868 276,016 14,372

At any stage of the d’Hondt process, we can see how many seats have been assigned to a party: it is the number of entries in its column that have been crossed out. if a party has been assigned seats so far, the first remaining entry in its column is its population divided by .

As shown in the table, the Whigs get the first seat. The first priority number is marked #1 and gets crossed out because it has been used. The greatest remaining priority number is for the Liberals, who get the second seat. As seats are awarded, the priority numbers are numbered in sequence and crossed out, in decreasing order. The number of seats awarded to each party is equal to the number of numbers crossed out in its column. Of the first 30 seats, the Whigs get 13, the Tories get 8, the Liberals get 9, and the Centrists get none.

Self Check 14

Which party gets seat #31?

  • Compare the current priority numbers: Whigs, 394,670; Tories, 385,572; Liberals, 386,423; Centrists, 201,203. The Whigs have the highest priority and get seat #31.

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The Webster Method

A second divisor method was introduced in 1832 by Senator Daniel Webster of Massachusetts.

image
Statesman and orator Daniel Webster (1782–1852), who developed a divisor method for apportioning the U.S. House of Representatives.

The Webster Method DEFINITION

The Webster method is the divisor method that rounds the apportionment quotient to the nearest whole number, rounding up when the fractional part is greater than or equal to , and rounding down when the fractional part is less than .

The Webster and Jefferson methods are free of the paradoxes that we have seen with the Hamilton method, but neither satisfies the quota condition. However, the Jefferson method favors the large states, while the Webster method is neutral, favoring neither the large nor the small states. Furthermore, the Webster method rarely violates the quota condition by giving a state more than its upper quota, or fewer seats than its lower quota, and would not have done so in any of the 23 congressional apportionments that have occurred so far.

The Webster Method PROCEDURE

  1. Obtain the tentative apportionments by rounding each state’s quota to the nearest whole number. (Round down to if the fractional part of is less than 0.5; otherwise, round up to .)
  2. Add the rounded quotas. If their sum is equal to the house size, the job is finished. The tentative apportionments calculated in Step 1 are the final apportionments.
  3. When the rounded quotas of the states don’t add up to the house size, calculate apportionment quotients, using a trial divisor as with the Jefferson method. All trial divisors must be larger than the standard divisor if the sum of the rounded quotas is larger than the house size, and smaller than the standard divisor if the sum of the quotas is smaller than the house size.
  4. Round the apportionment quotients from Step 3. If their sum is more than the house size, increase the divisor; if the sum is less, decrease the divisor. If the sum is equal to the house size, the rounded apportionment quotients provide the correct apportionment.

It is possible that first use of the Webster method was before Webster invented it! The apportionment bill that was vetoed by President Washington was consistent with the Webster apportionment, with 30,000 as the divisor. The resulting apportionment was unconstitutional, but perhaps it would have not been vetoed if a large enough divisor had been used.

Let’s revisit the three friends who won 100 diamonds in a lottery (see Example 10 on page 586).

EXAMPLE 15 Sharing Diamonds by the Webster Method

Tim and Louise have convinced Winnie to go with the Webster method because the Jefferson method is biased in her favor. The quotas are 78.26 diamonds for Winnie, 17.39 for Louise, and 4.35 for Tim; these are rounded to get the tentative shares of 78, 17, and 4 diamonds, respectively, which add up to 99. The standard divisor was 0.23, and because we need to apportion one more seat, we will have to try a smaller divisor. Our trials are shown in Table 14.15.

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Table 14.16: TABLE 14.15 Apportioning Diamonds by Webster’s Method
Divisors
Participant Tickets 0.227 0.228 0.229
Winnie 18 79.2952 79 78.9474 79 78.6026 79
Louise 4 17.6211 18 17.5439 18 17.4672 17
Tim 1 4.4053 4 4.3860 4 4.3668 4
Total 23 101 101 100

The first divisor that we tried was 0.227. The apportionment quotients are in the left column under that divisor in the table, and to their right are the rounded values. Because the total number of “seats” with that divisor turned out to be 101 diamonds, we tried the divisor 0.228, with the same result. The divisor 0.229 produced our goal, 100 diamonds. We never tried any divisor greater than 0.23 because we knew from the start that 0.23 is too large. The result was Winnie’s preferred apportionment: 79 diamonds for her, 17 for Louise, and 4 for Tim.

Self Check 15

You saw in Self Check 13 (page 590) that rounding the percentages in

by the Jefferson method produced a violation of the quota condition. What happens with the Webster method?

  • With the Webster method, start by rounding each quota to the nearest whole number. This yields , which is the Webster apportionment. There is no violation of the quota condition.

EXAMPLE 16 Apportioning Classes by the Webster Method

Let us return to the case of the mathematics teacher who is to teach a total of five classes in geometry, precalculus, and calculus. The enrollments are 52 for geometry, 33 for precalculus, and 15 for calculus. With a total of 100 students enrolled, and a house size of 5, the standard divisor is 20. The quotas, determined by dividing the enrollments for the three subjects by the standard divisor, are 2.60, 1.65, and 0.75, respectively. The tentative apportionments are 3, 2, and 1; their total, 6, exceeds the house size. We therefore will try a divisor greater than 20. using 21 as the divisor, we find that the apportionment quotients are 2.48 for geometry, 1.57 for precalculus, and 0.71 for calculus. Rounded, these quotients become 2, 2, and 1, respectively, for a total of 5 classes. Thus, geometry and precalculus are each apportioned 2 classes, and calculus gets 1 class. This is the same apportionment that we obtained with the Hamilton method.

To see why the Webster method is not biased in favor of large states or small states, we will use the same analysis that we applied to Jefferson’s method. Let be the standard divisor, let be the divisor that is used in the Webster method, and let . Each state’s apportionment quotient is equal to

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as with the Jefferson method. When (this happens when the number of seats tentatively apportioned is less than the house size), the states all receive an across-the-board increase in their quotas; and just as in a company where the workers receive the same percentage raise, the larger states are favored. When , the reverse is true because a large number multiplied by will decrease more than a small number would. Thus, the Webster method favors neither large nor small states when the tentative apportionment exactly fills the house; it favors small states when the tentative apportionment must be reduced, and it favors large states when the tentative apportionment must be expanded. On balance, the Webster method is neutral because it is equally likely that the tentative apportionment will be less than the house size or that it will be greater than the house size.

The Webster Method Has No Population Bias THEOREM

Among all divisor methods, the Webster method alone shows no bias with regard to state population.

When seats are apportioned to parties after an election, a method that is equivalent to the Webster method can be used. This method, called the Sainte-Laguë method (after André Sainte-Laguë, 1878–1950), involves a process that is analogous to the d’Hondt method. A Sainte-Laguë table is constructed with a column for each party. The number of votes received is at the top of each party’s column, and below that the number of votes is divided by successive odd numbers, 3, 5, 7, …. Seats are apportioned to the parties by treating the entries in the table as a priority list. The party with the highest priority gets a seat, and then that priority number is crossed out. The next seat goes to the party with the largest remaining priority in the Sainte-Laguë table. Germany and New Zealand use the Sainte-Laguë method to apportion seats to their parliaments.

If you are curious to know why the Sainte-Laguë method produces the same apportionment as the Webster method, refer again to the party that is about to get the last seat. Let be the number of votes that it received, and suppose that it has seats already, before the last seat is awarded. Its priority, , is the highest remaining in the table. [The odd number is .] Multiply this priority by 2 to get a number . Using this as the divisor, we see that

(The s cancel each other out, as we divide the quotients by inverting the divisor and multiplying.) This quotient can be rounded up to get seats. The other parties have lower priority, and so they get no new seats with as the divisor.

Self Check 16

Referring again to the high school teacher who was to teach five classes, where 52 students were enrolled in geometry, 33 students were enrolled in precalculus, and 15 students were enrolled in calculus, make a Sainte-Laguë table to determine the Webster apportionment.

  • Geometry Precalculus Calculus
    #1 52 #2 33 #4 15
    #3 17.33 5 11 5
    10.4 6.6 3

    Geometry and precalculus get two sections, and calculus gets one section.

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So far, three of the four methods that have been used to apportion seats in the U.S. House of Representatives have been introduced. The fourth is the Hill-Huntington method, to be described in Section 14.4. This divisor method was adopted by statute following the 1940 census and is still used. In Table 14.16, the properties of the four apportionment methods are described.

Table 14.18: TABLE 14.16 Apportionment Methods Used by Congress
Hamilton Jefferson Webster Hill-Huntington
Censuses 1850;
1860*–1900*
1790*–1830* 1840*,
1910, 1930
1940–present
Divisor method No Yes Yes Yes
Paradoxes3 Yes No No No
Bias for None Large states None Small states
Quota
condition
Satisfied Often exceeds
upper quota
Not satisfied,
rarely violates
Not satisfied,
rarely violates
Table 14.18: In years marked with an asterisk (*), the house size was determined after the apportionment was calculated. In asterisk years when a divisor method was in use, the divisor was chosen and the house size was determined by the calculation. For the Hamilton method, Congress was presented with apportionments for a range of house sizes and chose the one that suited.

A Puzzling Apportionment of the U.S. House of Representatives Spotlight 14.2

The strangest apportionment of seats in the U.S. House of Representatives followed the census of 1870. This was the first apportionment to be based on the Fourteenth Amendment to the U.S. Constitution, which redefined the apportionment populations. On February 2, 1872, President Grant signed an apportionment bill specifying 283 seats. With this house size, the Hamilton and Webster methods provided the same apportionment (see Spotlight 14.1 on page 580).

Later that year, the Congress passed a bill awarding one additional seat to each of nine states. The bill was signed by the president on May 30, bringing the house size to 292. With this house size, the Hamilton and Webster methods gave different apportionments, and the new apportionment did not agree with either of them! Circumstantial evidence indicates that the supplemental apportionment was not due to partisan considerations: the House was controlled by the Democrats, while the president and the Senate were Republican. Five of the nine states that received additional seats voted Republican, and four voted Democratic in the 1876 presidential election. Northern and southern states were among those favored with extra seats.

The supplemental apportionment may have influenced the outcome of the controversial presidential election in 1876, in which Rutherford B. Hayes lost the popular vote but defeated Samuel J. Tilden, 185 to 184 electoral votes. (Because Colorado was admitted to the Union on August 1, 1876, with one seat in the House, the House size at the time of the election was 293.) Without the extra seats from the bill on May 30, 1872, the election would have been tied with 180 electoral votes for each candidate. The House of Representatives would have decided the election, under the provision of Article ll, Section 1, of the U.S. Constitution, in which each state’s delegation has one vote. it is probable, but not certain, that the House would have elected Tilden, 20 states to 18.