DM02_Final.qxp:DM02.qxp 5/9/14 2:43 PM Page 72 Lesson 2.3 Apportionment Models The problem of dividing an estate fairly involves discrete objects, but also involves cash. When a fair division problem is strictly discrete, it can be impossible to solve in a way that treats all parties fairly. A variety of situations involve fair allocation of discrete objects. For example, your school s administrators must decide a fair way to allocate teaching positions to the school s various departments and equipment such as computers to classrooms. One of the most politically charged fair allocation problems in the United States involves the apportionment of seats in the U.S. House of Representatives among the states. (The House is reapportioned every ten years after a new census is completed.) Unlike the discrete objects in estate division situations, the value of a seat in the U.S. House is not subjective. Thus, fairness has a simple definition, one that is mandated by the Constitution: that the seats be distributed among the states according to population. Although the definition of fairness that is used to apportion seats in the U.S. House is uncontroversial, the method of apportionment can be. The first veto by a U.S. president occurred in 1792 when George Washington rejected an apportionment bill advocated by Alexander Hamilton in favor of a method championed by Thomas Jefferson. This lesson considers the two models for apportioning seats in a governmental body that were at the center of the Hamilton-Jefferson dispute. Alexander Hamilton.
Lesson 2.3 Apportionment Models 73 Since the apportioning of the U.S. House of Representatives involves 435 seats and 50 states, you will feel more comfortable starting with a simpler situation. (Although some of this lesson s examples may seem artificial, they are designed to reflect the large population differences among the 50 states.) Central High School has sophomore, junior, and senior classes of 464, 240, and 196 students, respectively. The 20 seats on the school s student council are divided among the classes according to population. Since there are 900 students in the school and since 900 20 = 45, ideally each member of the Thomas Jefferson. council would represent 45 students. In other words, the ideal ratio of students to seats is 45. Total population Ideal ratio = Number of seats In cases of political representation, the ideal ratio is often called the ideal district size. If, for example, the population of the United States is 280 million, then the ideal district size is 280,000,000 435, or about 645,000. Ideally, each member of the U.S. House would represent 645,000 people. This ideal cannot be achieved because district boundaries cannot cross state lines. Because Central High s sophomore class has 464 members, it deserves 464 45 = 10.31 seats. Accordingly, 10.31 is called the sophomore class quota. Similarly, the junior quota is 5.33, and the senior quota is 4.36. The Quotas Sophomores: 10.31 Juniors: 5.33 Seniors: 4.36 Class size Quota = Ideal ratio In the case of the U.S. House, a state s quota is determined by dividing its population by the ideal district size. For example, a state with 4 million people deserves 4,000,000 645,000, or about 6.2 seats.
74 Chapter 2 Fair Division It isn t possible to split a seat in Central High s council and give 0.36 of it to the seniors, 0.33 of it to the juniors, and 0.31 of it to the sophomores. The school must decide a fair way to award this seat to one of the classes. The model favored by Hamilton has something in common with the one favored by Jefferson: Each begins by ignoring the quota s decimal part and assigning a number of seats equal to the whole number part. Regardless of whether the quota is 10.31 or 10.91, both Hamilton and Jefferson begin by awarding 10 seats. Ignoring the decimal part of a number in this way is called truncating. The total of the truncated quotas is 10 + 5 + 4 = 19 seats. The difference between the Hamilton and Jefferson models is in the way they award the remaining seat. The Hamilton model gives the remaining seat to the class whose quota has the largest decimal part. Since the decimal part of the senior quota, 0.36, is larger than either of the other decimal parts, the senior class gets the extra seat. The Hamilton results are summarized in the following table. Class Size Quota Hamilton Apportionment 464 10.31 10 240 5.33 5 196 4.36 5 The Hamilton model seems reasonable. Perhaps some members of your class proposed a similar model in Lesson 2.1. However, the Hamilton model has fallen out of favor in the United States for reasons you will consider in this lesson s exercises, after you have examined Jefferson s approach. You might think of the Jefferson model as conducting a race to see which quota can get to the next whole number first. That is, whether the sophomore quota can increase to 11 or the junior quota can increase to 6 or the senior quota can increase to 5 first. Here is how this race is conducted.
Lesson 2.3 Apportionment Models 75 Since a quota is found by dividing the class size by the ideal ratio, the quota becomes larger when the ideal ratio becomes smaller. For example, consider what happens if the ideal ratio is decreased from 45 students per seat to 40 students per seat. The results are summarized in the following table. Quota with Quota with Class Size Ideal Ratio of 45 Ideal Ratio of 40 464 464 45 = 10.31 464 40 = 11.6 240 240 45 = 5.33 240 40 = 6.0 196 196 45 = 4.36 196 40 = 4.9 Seats 10 + 5 + 4 = 19 11 + 6 + 4 = 21 For example, the sophomore class receives 10 seats when the ideal ratio is 45 and 11 seats when the ratio is 40. Therefore, there must be a ratio between 45 and 40 that causes the sophomore apportionment to be exactly 11. It can be found by dividing the sophomore class size by 11: 464 11 42.18. The number 42.18 is called a Jefferson adjusted ratio. Class size Jefferson adjusted ratio = Truncated quota + 1 Similarly, the junior quota passes 6 when the ratio drops below 240 6 = 40. The senior quota passes 5 when the ratio drops below 196 5 = 39.2. Proponents of the Jefferson model argue that since the ideal ratio does not apportion all of the seats, it should be replaced with a ratio that does give a complete apportionment and that this new ratio should be as close as possible to the ideal ratio. If the ratio gradually decreases from the ideal 45, it reaches a value at which the sophomores receive another seat before it reaches a value at which either of the other classes receives one, as shown in the following table. Ideal Ratio Sophomore Seats Junior Seats Senior Seats 45 10 5 4 42.18 11 5 4 40 11 6 4 39.2 11 6 5
76 Chapter 2 Fair Division Greater detail can be seen in the following table, in which the adjusted ratio is decreased in steps of 0.2. Note that the sophomore class quota passes the next integer before either the junior or the senior quota does. Adjusted Ratio Sophomore Junior Senior 45.00 10.31 5.33 4.36 44.80 10.36 5.36 4.38 44.60 10.40 5.38 4.39 44.40 10.45 5.41 4.41 44.20 10.50 5.43 4.43 44.00 10.55 5.45 4.45 43.80 10.59 5.48 4.47 43.60 10.64 5.50 4.50 43.40 10.69 5.53 4.52 43.20 10.74 5.56 4.54 43.00 10.79 5.58 4.56 42.80 10.84 5.61 4.58 42.60 10.89 5.63 4.60 42.40 10.94 5.66 4.62 42.20 11.00 5.69 4.64 42.00 11.05 5.71 4.67 41.80 11.10 5.74 4.69 41.60 11.15 5.77 4.71 41.40 11.21 5.80 4.73 41.20 11.26 5.83 4.76 41.00 11.32 5.85 4.78 40.80 11.37 5.88 4.80 40.60 11.43 5.91 4.83 40.40 11.49 5.94 4.85 40.20 11.54 5.97 4.88 40.00 11.60 6.00 4.90 39.80 11.66 6.03 4.92 39.60 11.72 6.06 4.95 39.40 11.78 6.09 4.97 39.20 11.84 6.12 5.00 39.00 11.90 6.15 5.03 The Jefferson model can be summarized as an algorithm: 1. Divide the total population by the number of seats to obtain the ideal ratio. 2. Divide the population of each class (state, district, etc.) by the ideal ratio to obtain the class quota.
Lesson 2.3 Apportionment Models 77 3. Assign a number of seats to each class that equals its truncated quota. 4. If the number of seats assigned matches the total number of seats to be apportioned, then stop. 5. If the number of seats assigned is smaller than the total number of seats to be apportioned, then divide the size of each class by one more than the number of seats assigned to it in step 3, to obtain an adjusted ratio. 6. Give an extra seat to the class with the largest adjusted ratio. (In other words, to the class whose adjusted ratio is closest to the ideal ratio.) This algorithm applies only to situations in which the total of the truncated quotas falls one short of the number of seats to be apportioned. In some cases, the shortfall after truncation is more than one seat. This lesson s exercises consider what to do in such cases. Exercises 1. The Central High council has had trouble deciding a number of issues because of disagreements between the sophomore representatives and the other representatives. The vote has been a 10-10 tie. In order to avoid future ties, the council decides to add one seat. a. On the basis of the data in this lesson, which class do you think should get the extra seat? Why? b. Find the new ideal ratio of students per seat for the 21-seat council. c. Use this ideal ratio to find the quota for each of the three classes. d. Use the Hamilton model to apportion the 21 seats on the new council among the three classes. e. Compare the Hamilton apportionment for the 21-seat council to that of the 20-seat council and explain why the results constitute a paradox.
78 Chapter 2 Fair Division 2. A senior council member who recently studied apportionment in the school s discrete mathematics course is unhappy about the 21- seat Hamilton apportionment and proposes the Jefferson model. a. Find an adjusted ratio for each class as described in the Jefferson model s algorithm of this lesson. b. Decrease the 21-seat ideal ratio until all 21 seats are apportioned. State the number of seats that the Jefferson model gives each class. c. Compare the 21-seat Jefferson apportionment with the 20-seat Jefferson apportionment. Does the Jefferson model produce a paradox similar to the one in Exercise 1? 3. Revise the Jefferson apportionment algorithm given in this lesson to account for situations in which more than one seat remains after truncation. 4. The paradox you observed in Exercise 1 occurs because increases in a divisor do not produce equal changes in all quotients. When the size of a representative assembly increases and the total population remains the same, the ideal ratio decreases. As an example, consider two classes with populations of 100 and 230. An increase in the size of the council causes the ideal ratio to decrease from 22 to 21. a. Complete the following table. Then explain why it could result in the shift of a council seat from one class to the other. Class Quota with Ideal Quota with Ideal Size Ratio of 22 Ratio of 21 100 230 b. Does the paradox in Exercise 1 result in the loss of a seat for a small class or a large class? Why? c. Why do you think Thomas Jefferson opposed the Hamilton model? (If you re not sure, look up the 1790 census. Jefferson was from Virginia.)
Lesson 2.3 Apportionment Models 79 5. The student council members at Central High, aware of the strange results that slight differences can make, decide to monitor the council s apportionment. At the end of the first quarter of the school year, the class numbers have changed somewhat: Sophomores Juniors Seniors 459 244 197 a. Use the Hamilton model to divide the council s 21 seats among the classes. At the end of the first semester the classes have changed again: Sophomores Juniors Seniors 460 274 196 At the council s first meeting of the new semester, the members are surprised when one of the representatives of the senior class, the only class that has decreased in size, demands a reapportionment. b. Use the Hamilton model to reapportion the council. c. Explain why the results constitute a paradox. d. Use an analysis similar to that of Exercise 4 to explain why this type of paradox occurs. Does it have an adverse effect on small classes or large classes? 6. The cartoon on page 74 has been described as an indiscreet attempt to apply a continuous division procedure to a discrete problem. Explain what this means. 7. Read the news article on the following page about the 2010 census results. In the 2010 census, the populations of Texas, New York, and California were 25,268,418, 19,421,055, and 37,341,989, respectively. a. Find the quota for each of these three states. b. Compare the quotas to the actual apportionment mentioned in the news article. c. Do you think the apportionment treated any of these three states unfairly? Explain.
80 Chapter 2 Fair Division 2010 Census: Slowest Growth Since Great Depression USA Today February 3, 2011 The U.S. population grew 9.7% in the past decade to 308,745,538, according to the first results of the 2010 Census the slowest growth since the Great Depression for a nation hard hit by a recession and housing bust. Census population counts taken every 10 years are used to distribute $400 billion a year in federal funds and reallocate seats in the U.S. House of Representatives. The shifting of House seats affects 18 states. Arizona, Georgia, Nevada, South Carolina, Utah and Washington all gain one seat. Florida gains two, and Texas gains four. Illinois, Iowa, Louisiana, Massachusetts, Michigan, Missouri, New Jersey and Pennsylvania all lose a seat. Both Ohio and New York lose two. Texas is the big winner. It gained more people than any other state for the first time, outpacing California, which had dominated for almost a century. Texas' four new seats will give it 36, still behind California's 53. New York's loss of two seats and Florida's gain of two will give each state 27. d. The 2010 census total for Louisiana was 4,553,962. Minnesota's total was 5,314,879. The apportionment gave Louisiana 6 seats and Minnesota 8. Louisiana felt that the apportionment was unfair and went to court, but lost. Do you think the apportionment was less fair to Louisiana than to Minnesota? Explain. 8. A county is composed of four districts: A, B, C, and D. Their populations are 210; 1,082; 311; and 284; respectively. The county commission has 18 seats. a. What is the ideal ratio? b. Find each district s quota. c. Find the Hamilton apportionment. d. Does the Jefferson model apportion all of the seats initially? Explain. e. Find the final Jefferson apportionment. f. This exercise shows why some do not like the Jefferson model. Explain.
Lesson 2.3 Apportionment Models 81 Computer/Calculator Explorations 9. Develop a spreadsheet to do the Jefferson apportionment for the three classes in this lesson s example. When finished, the spreadsheet should be similar to the following. A B C D 1 2 3 4 5 6 7 Sophomore Junior Senior Total Seats Ideal ratio Population 464 240 196 900 20 45 Quota 10.311111 5.333333 4.355556 Seats 10 5 4 19 The values in columns C and D and in cells B5 and B7 should be calculated with formulas. The formulas in column D require the use of your spreadsheet s truncation function. On many spreadsheets, this function is abbreviated TRUNC. For example, TRUNC(C2,0) truncates the value in cell C2 so it has 0 decimal places. When your spreadsheet is done, show how the proper apportionment can be found by changing the value in cell B7 (the ideal ratio). Projects 10. Research and report on models that have been used to apportion the U.S. House of Representatives and the controversies that have arisen. Why has the apportionment model been changed? Why has the size of the House been changed? Did paradoxes occur with any of the models? 11. The president of the United States is chosen in the Electoral College. The number of electoral votes a state has is determined by the size of its congressional delegation. Thus, apportionment can affect the Electoral College vote. Research and report on the impact of apportionment on the election of the president. Has the apportionment model ever made a difference in the person the Electoral College chose?