Chapter 14 Exercises

Chapter 14 Exercises

14.1 The Apportionment Problem

Question 14.31

1. Jane has decided to track her daily expenses and finds them to be as listed in the table. Express these as percentages. If rounded to whole numbers, do the percentages add up to 100 percent?

Jane’s Expenses
Rent $31
Food 16
Transportation 7
Gym 12
Miscellaneous 5

1.

Jane’s Expenses
Category Percentage Rounded
Rent 43.66% 44%
Food 22.54% 23%
Transportation 9.86% 10%
Gym 16.90% 17%
Miscellaneous 7.04% 7%
Totals 100.00% 101%

The sum of the rounded percentages is not 100%.

Question 14.32

2. A mathematics department uses 20 teaching assistants to aid in its four-semester calculus course. The number of teaching assistants assigned to each level of the course depends on enrollment. Here are the fall enrollments:

Calculus I 500
Calculus II 100
Calculus III 350
Calculus IV 175
Total 1125

How many teaching assistants should be assigned to each level of the course?

Question 14.33

3. Should the mathematics department in Exercise 2 revise the assignments for its TAs? Grades have been posted for the previous semester, and some students need to repeat the previous level of the course. A total of 45 students move from Calculus II to Calculus I, 41 students move from Calculus III to Calculus II, and 12 students move from Calculus IV to Calculus III.

614

3.

No, although if the quotas are rounded in the usual way, there will be 21 sections. One must reduce the apportionment of one course, and a logical way to do so will lead to the original apportionment.

Question 14.34

4. Here is a typical apportionment problem. Round the numbers in the following sum to whole numbers:

The rounded numbers must add up to 40. How would you approach this?

Question 14.35

image 5. How would you round the numbers in the following sum to whole numbers? The rounded numbers must add up to 60.

5.

If you round the numbers to the nearest whole number, the sum will be 62. One approach to solving this problem would be to round the two numbers whose fractional parts are less than 0.6 down: round 14.53 to 14, and round 16.55 to 16. The rounded sum would be

14.2 The Hamilton Method

Question 14.36

6. Use the Hamilton method to round the numbers in the following sum to whole numbers. The sum of the rounded numbers must be the same as the original sum.

Question 14.37

7. Repeat Exercise 6 with the following sum:

7.

Question 14.38

8. The 37th pearl. Three friends have bought a bag guaranteed to contain 36 high-quality pearls for $14,900 at an auction. Abe contributed $5900, Beth’s contribution was $7600, and Charles supplied the remaining $1400. After taking the bag to your house, they pour the 36 pearls from the bag onto the kitchen table.

  1. How many should each friend get if the Hamilton method is used to apportion the pearls according to the size of the contributions?
  2. Charles has noticed that the bag isn’t empty! Another pearl comes out, so recalculate the apportionment.
  3. image How do you explain the result to Charles?

Question 14.39

9. A country has three political parties, and it allots seats in its 102-seat parliament by the Hare (Hamilton) method proportionately to the number of votes each receives. In a recent election, the Pro-UFO Party received 254,000 votes, the Anti-UFO Party got 153,000 votes, and the Who Cares Party polled 103,000 votes. Show that two of the parties are tied.

9.

The sum of the lower quotas is 100, leaving two seats to be given to the parties with the largest fractions. The fractional parts of the quotas are 0.8, 0.6, and 0.6. Therefore the Anti-UFO and Who Cares parties are tied for the 102nd seat.

Question 14.40

10. A small high school has one mathematics teacher who can teach a total of five sections. The subjects that she teaches, and their enrollments, are as follows: Geometry, 52; Algebra, 33; and Calculus, 12. Use the Hamilton method to apportion sections to the subjects.

Question 14.41

11. Repeat Exercise 10 using the following enrollments: Geometry, 77; Algebra, 18; and Calculus, 20.

11.

Geometry, 3 sections; algebra, 1 section; calculus, 1 section

Question 14.42

image 12. Use the Hamilton method to express the summands of the following expression as whole number percentages of the total:

Repeat the calculation for the following sum:

Do you see a paradox?

Question 14.43

13. Abe, Beth, Charles, and David have decided to invest in rare coins. A dealer has offered to sell them a parcel containing 100 identical coins for $10,000. Each person invests all that he or she can afford, but there is not quite enough money, so Charles asks his Aunt Esther to join the group. The coins will be apportioned by the Hamilton method. Here are the amounts invested:

Investments
Abe $3,619
Beth 1,862
Charles 2,258
David 2,010
Esther 251
Total $10,000
image
  1. How should the coins be apportioned among the five contributors?
  2. After the coins are distributed, the dealer mentions that there will be $50 in excise tax. Everyone empties his or her wallet: Abe finds $16 more, Beth has $2, Charles has $1, and David finds $32. This adds up to $51, so $1 is returned to Aunt Esther. The apportionment is recalculated, and one of the coins changes hands. Who has to give a coin to whom?
  3. image Explain what happened.

To see how this situation works out with a different apportionment method, refer to Exercise 31 on page 616.

13.

(a) Abe, 36 coins; Beth, 19 coins; Charles, 23 coins; David, 20 coins; Esther, 2 coins

(b) Charles must give Esther a coin.

(c) Blame the population paradox.

615

Question 14.44

image 14. The new states paradox. The census of 1900 recorded the following apportionment populations:

Maine 694,466
New York 7,264,183
United States 74,562,608

The house size was 386. Apportionment was by the Hamilton method.

  1. Determine the quotas of New York and Maine. Given that New York was the last state to receive its upper quota, determine the numbers of seats that were apportioned to these two states.
  2. In 1907, Oklahoma became a state. Its population was stipulated to be 1,000,000 for the purpose of apportionment. Using the standard divisor that you found in part (a), determine the apportionment for Oklahoma.
  3. Add the stipulated population of Oklahoma to the 1900 population of the United States, and the number of seats that you found in part (b) to the house size before Oklahoma became a state. With these data, find out if the numbers of seats apportioned to New York and Maine changed. (New York and Maine actually retained their 1902 apportionments.)

Question 14.45

15. A country has five political parties. Here are the numbers of votes each received in a recent election: 5,576,330; 1,387,342; 3,334,241; 7,512,860; and 310,968. Seats in its parliament are apportioned by the Hare (Hamilton) method. Calculate the apportionments for house sizes of 89, 90, and 91. Does the Alabama paradox occur?

15.

As the following table shows, the Alabama paradox occurred. Party lost a seat when the house size increased from 89 to 90 and did not recover the seat when the house size increased to 91. (If you like, you can see what happens when the house size is 92.)

Party Population Seats
5,576,330 27 28 28
1,387,342 7 7 7
3,334,241 16 17 17
7,512,860 37 37 38
310,968 2 1 1
Totals 18,121,741 89 90 91

14.3 Divisor Methods

Question 14.46

16. Explain why the tentative Webster apportionment of a state with quota is .

Question 14.47

17. Reapportion the classes in Exercise 11 (page 614), using the Jefferson method.

17.

Geometry gets 4 sections, calculus gets 1 section, and there is no algebra class.

Question 14.48

18. Reapportion the classes in Exercise 10 (page 614), using the Webster method.

Question 14.49

image 19. The three friends who bought the pearls (see Exercise 8 on page 614) ask you to suggest a different apportionment method to distribute their purchase. Before answering, determine the apportionments given by the Jefferson and Webster methods for the 36- and 37-pearl house sizes, and then make your suggestion.

19.

The Jefferson and Webster methods give identical apportionments for 36 or 37 pearls: With 36 pearls, Abe gets 14, Beth gets 19, and Charles gets 3. The 37th pearl is awarded to Abe. Although the Jefferson and Webster methods yield the same result, that does not mean that they are right. If there is a principle on which to choose a method, it would probably be to choose the method by which the cost per pearl is as close as possible to the same for each of the friends. The cost per pearl is the district size, so they should use the Dean method (see Exercise 48 and Writing Project 2), which minimizes absolute differences in district size. Charles might want to study up on it, because it allocates the 37th pearl to him!

Question 14.50

image 20. The three friends in Exercise 8 have bought a lot of 36 identical diamonds, at a total cost of $36,000; Abe’s investment was $15,500, Beth’s was $10,500, and Charles’s was $10,000. They decided to apportion the diamonds using the Webster method, and they can’t make it work out. Can you help?

Question 14.51

21. A country has a 20-seat parliament. Seats are apportioned to parties by the d’Hondt method. The following table displays the results of a recent election. Make a d’Hondt table and determine the number of seats allocated to each party.

The Election Results
Demopublicans 44,856
Repocrats 34,944
Greenocrats 20,004
Greenicans 19,002
Independents 9,804

21.

See table below.

Question 14.52

22. Referring to the voting data in Exercise 21, make a Sainte-Laguë table to apportion the parliament.

Exercises 23–25 refer to the parliament in Example 9 on page 584. We will use only the data for the first election.

Question 14.53

23. Make a Sainte-Laguë table and apportion the first 30 seats.

23.

See table on next page

Question 14.54

24. In Example 14 (page 593), we used a d’Hondt table to apportion the first 30 seats. The Centrists did not get any of the first 30 seats. Which of seats 31-100 will be the first one that the Centrists receive?

Question 14.55

25. Which will be the first seat that the Centrists receive if the Sainte-Laguë table is used to apportion the seats?

25.

The 34th seat

Question 14.56

image image 26. Explain why the population paradox cannot occur if a divisor method of apportionment is used. (If one state receives an increased apportionment even though its population decreased, how did the divisor used in the second apportionment differ from the divisor used in the first one?)

Question 14.57

image image 27. Explain why the new states paradox (see page 585) cannot occur if a divisor method of apportionment is used. (Would the divisor used in the apportionment after the new state joined the union have to be different from the divisor used before?)

27.

Let be the divisor used to apportion before the new state is added. Let be the population of the new state. The new state should be apportioned , where square brackets indicate the method of rounding that is specified for the particular divisor method in use. This is the formula that would be used for each of the states before and after the new state joined; therefore, none of the original apportionments would change.

Seats Demopublicans Repocrats Greenocrats Greenicans Independents
Priority Seat Priority Seat Priority Seat Priority Seat Priority Seat
1 44,856 1 34944 2 20,004 4 19,002 5 9,804 11
2 22,428 3 17,472 6 10,002 10 9,501 12 4,902
3 14,952 7 11,648 8 6,668 17 6,334 19 3,268
4 11,214 9 8,736 14 5,001 4,751 2,451
5 8,971 13 6,989 16 4,001 3,800 1,961
6 7,476 15 5,824 20 3,334 3,167 1,634
7 6,408 18 4,992 2,858 2,715 1,401
8 5,607 4,368 2,501 2,375 1,226

The Demopublicans get 7 seats, the Repocrats get 6, the Greenocrats and the Greenicans each get 3, and the Independents get 1 seat.

616

Question 14.58

image 28. Round the following to whole percentages using the Hamilton, Jefferson, and Webster methods:

Do any of these apportionments show a violation of the quota condition?

Question 14.59

29. Round the following percentages to whole numbers, using the methods of Hamilton, Jefferson, and Webster.

Do any of these apportionments show a violation of the quota condition?

29.

Hamilton: . The quota condition is satisfied.

Jefferson: . Because 92.15 is rounded up to 95, the quota condition is violated.

Webster: . Again, the quota condition is violated: This time 92.15 is rounded down to 90.

Question 14.60

30. Use the Webster method to apportion the House of Representatives, based on the census of 1790. Choose a divisor such that each state’s district population is greater than 30,000, and as many seats as possible are apportioned.

Question 14.61

31. Recalculate the apportionment of the coins in Exercise 13 (on page 614) by the Webster method. Again, after the excise tax is paid, a coin changes hands. Who gives it to whom?

31.

Before Tax After Tax
Abe 36 36
Beth 19 19
Charles 22 23
David 20 20
Esther 3 2
Total 100 100

Now it is Esther who must give a coin to Charles!

Question 14.62

32. A country has two political parties, the Liberals and the Tories. The seats in its 99-seat parliament are apportioned to the parties according to the number of votes they receive in the election. If the Liberals receive 49% of the vote, how many seats do the Liberals get with the Hamilton (Hare) method? With the Webster (Sainte-Laguë) method? With the Jefferson (d’Hondt) method?

Question 14.63

image 33. A country with a parliamentary government has two parties that capture 100% of the vote between them. Each party is awarded seats in proportion to the number of votes received.

  1. Explain why the Webster (Sainte-Laguë) and Hamilton (Hare) methods will always give the same apportionment in this two-party situation.
  2. Explain how to use the result of part (a) to show that the Alabama and population paradoxes cannot occur when the Hamilton method is used to apportion seats between two parties or states.
  3. Explain why the result of part (a) implies that the Webster method satisfies the quota condition when the seats are apportioned between two parties or states.
  4. Will the Jefferson and Hill-Huntington methods also yield the same apportionments as the Hamilton method?

33.

(a) One quota will be rounded up and the other down to obtain the Webster apportionment. The quota that is rounded up will have a fractional part greater than 0.5 and will be greater than the fractional part of the quota that is rounded down. The Hamilton method will give the party whose quota has the larger fractional part an additional seat. Thus, the apportionments will be identical.

(b) These paradoxes never occur with the Webster method, which gives the same apportionment in this case.

(c) The Hamilton method, which always satisfies the quota condition, gives the same apportionment.

Seats Whigs Tories Liberals Centrists
Priority Seat # Priority Seat # Priority Seat # Priority Seat #
1 5,525,381 1 3,470,152 3 3,864,226 2 201,203
2 1,841,794 4 1,145,717 6 1,288,075 5 67,068
3 1,105,076 7 694,030 10 772,845 9 40,241
4 789,340 8 495,736 14 552,032 12 28,743
5 613,931 11 385,572 17 429,358 15 22,356
6 502,307 13 315,468 21 351,293 19 18,291
7 425,029 16 266,935 24 297,248 22 15,477
8 368,359 18 231,343 28 257,615 26 13,414
9 325,022 20 204,127 227,307 29 11,835
10 290,810 23 182,640 203,380 10,590
11 263,113 25 165,245 184,011 9,581
12 240,234 27 150,876 168,010 8,748
13 221,015 30 138,806 154,569 8,048
14 204,644 128,524 143,119 7,452
15 190,530 119,660 133,249 6,938

The Whigs get 13 of the first 30 seats, the Liberals get 9, the Tories get 8, and the Centrists get none of the first 30 seats.

(d) No, each of these methods is capable of producing a different apportionment. For example, suppose that one party receives 99.9% of the vote and the other receives 0.1%. If the house size is 100, the Jefferson method would apportion all 100 seats to the dominant party. (So would the Hamilton and Webster methods.) The Hill-Huntington method would apportion the dominant party 99 seats, and 1 seat to the other party. If the dominant party received 99.4% and the other party received 0.6%, the Jefferson method would still apportion all of the seats to the dominant party, but the Hamilton and Webster methods would give 1 seat to the other party.

14.4 Which Divisor Method Is Best?

Question 14.64

34. A barista uses 11 grams of coffee to make an espresso and 16 grams to make a doppio. Find the percentage difference in the amounts of coffee used for the two drinks.

Question 14.65

35. Jim is 72 inches tall and Alice is 65 inches tall. What is the percentage difference in their heights?

35.

10.77%

Question 14.66

36. Find the Hill-Huntington rounding points for numbers between 0 and 1; between 1 and 2; between 2 and 3; and between 3 and 4.

Question 14.67

37. A high school has one math teacher who can teach five sections. A total of 56 students have enrolled in the algebra class, 28 have signed up for geometry, and 7 students will take calculus. Use the Hill-Huntington method to decide how many sections of each course to schedule.

37.

The standard divisor is 18.2, but with this divisor the rounded quotas (rounding by Hill-Huntington) are 3, 2, and 1, so 6 sections are apportioned. We will choose a larger divisor; 20 is convenient. The apportionment quotients are 2.8 (which rounds to 3), 1.4, and 0.35. Remember that all numbers between 0 and are rounded to 1; therefore, there will be 3 sections of algebra and 1 section each for geometry and calcus.

Question 14.68

image 38. One year later, the high school described in Exercise 37 still has just one math teacher who teaches 5 sections. The enrollments are algebra, 36; geometry, 61; and calculus, 3. Apportion the classes by the Webster and Hill-Huntington methods. Which apportionment do you think the school principal would prefer?

Question 14.69

image image 39. In 2001, Utah sued to increase its apportionment (Utah v. Evans). Federal employees stationed abroad are counted in the apportionment population of the state of their residence, and Utah wanted to include in its apportionment population religious missionaries who were based in the state and serving abroad.

  1. In the apportionment based on the 2000 census, North Carolina received the last seat. Its apportionment population was 8,067,673, and it was apportioned 13 seats. Find the largest divisor that would allow North Carolina 13 seats.
  2. Utah’s apportionment population (not counting missionaries) was 2,236,714, and Utah was apportioned 3 seats in the House of Representatives. Find the population that would be required for Utah to be apportioned 4 seats, with the same divisor that you found in part (a).
  3. To justify the transfer of a seat from North Carolina to Utah, by how much would Utah’s population have to increase? (In the suit, Utah claimed that its residents should include 11,000 missionaries.)

39.

(a) The rounding point between 12 and 13 is . Dividing North Carolina’s apportionment population by . we obtain . With a divisor greater than , North Carolina will receive fewer than 13 seats; with a divisor less than , North Carolina’s apportionment will be at least 13.

(b) With a population greater than , Utah would be apportioned 4 seats.

(c) 856

617

Question 14.70

40. (a) Show that for any positive numbers and , the geometric mean is less than the arithmetic mean,5 except when ; in that case, the two means are equal. (Hint: Show that the triangle in Figure 14.2 is a right triangle.)

5 The arithmetic mean of and is equal to .

image
Figure 14.2: Figure 14.2 Is this a right triangle?

(b) If is a number such that the Webster and Hill-Huntington roundings of differ, show that is greater than the Hill-Huntington rounding point and is less than the Webster rounding point . Conclude that, in this case, the Hill-Huntington rounding of is equal to and the Webster rounding is equal to .

(c) Explain why the fact established in part (b) implies that the Hill-Huntington method is more favorable to small states than the Webster method.

Question 14.71

image 41. A city has three districts with populations of 100,000, 600,000, and 700,000, respectively. Its council has 20 members, and seats on the council are apportioned by the Hill-Huntington method according to the district populations. Show that there is a tie. Would a tie occur with any of the other apportionment methods that we have considered?

41.

The quotas for the three districts are , and 10, respectively. The Hamilton method simply rounds the first two districts to 1 and 9, respectively, and the Webster and Jefferson methods give the same apportionment. Therefore, these methods do not lead to ties. The Hill-Huntington rounding points are . (It is important to notice that .) The Hill-Huntington roundings of the quotas are therefore 2, 9, and 10, respectively. The apportionment quotient for the first district will be less than (so that it will receive 1 seat), if the divisor is greater than . Using this divisor, the apportionment quotients would be for the first district and , the rounding point between 8 and 9, for the second district. Any divisor large enough to cause the first district’s quota to be rounded down will also cause the second district’s quota to be rounded down. The result would be 1 seat for the first district, 8 seats for the second district, and 10 seats for the third district—a total of

19 seats. The first and second districts are tied for the last seat.

Question 14.72

image 42. In a 1991 federal lawsuit, Massachusetts v. Mosbacher, Massachusetts claimed that the Hill-Huntington method of apportionment is unconstitutional because it does not reflect the “one person, one vote” principle as well as the Webster method does. Would Massachusetts have gained a seat from Oklahoma if the Webster method had been used to apportion the House of Representatives in 1991? In your calculation, use the following populations and Hill-Huntington apportionments: Massachusetts was apportioned 10 seats for a population of 6,029,051, and Oklahoma was apportioned 6 seats for a population of 3,157,604.

Question 14.73

image 43. In 1822, Congressman William Lowndes of South Carolina proposed an apportionment method, which was never used. Lowndes started, as Hamilton did, by giving each state its lower quota. But where Hamilton apportions the remaining seats to the states whose quotas have the largest fractional parts—in other words, the states for which the absolute difference between the quota and the lower quota is greatest—Lowndes gives the extra seats to the states where the percentage difference is greatest, increasing the apportionments of as many states as necessary to their upper quotas to fill the House.

  1. Compared with the Hamilton method, would this method be more beneficial to states with large populations or small populations?
  2. Does the Lowndes method satisfy the quota condition?
  3. Would there be any trouble with paradoxes with the Lowndes method?
  4. Use the method to apportion the 1790 House of Representatives with a 120-seat house. The populations and quotas resulting from the 1790 census are in Table 14.5 on page 580.

43.

(a) Lowndes favors small states.

(b) Yes

(c) Yes

(d)

State Quota Lower Quota Priority Apportionment
DE 1.843 1 84.30% 2
VT 2.839 2 41.95% 3
NJ 5.959 5 19.18% 6
NH 4.707 4 17.68% 5
GA 2.351 2 17.55% 3
SC 6.844 6 14.07% 7
KY 2.280 2 14.00% 3
RI 2.271 2 13.55% 3
CT 7.860 7 12.29% 8
NC 11.732 11 6.65% 11
MA 15.774 15 5.16% 15
VA 20.926 20 4.63% 20
MD 9.243 9 2.70% 9
PA 14.366 14 2.61% 14
NY 11.004 11 0.04% 11
Totals 120 111 120

Question 14.74

image image 44. Let the populations of states and be and respectively. The apportionments will be and . Assuming that district populations for state are larger than district populations for state , show that the percentage difference in district populations is

Also show that same expression is equal to the percentage difference in representative share. Hence the percentage difference in district populations is equal to the percentage difference in representative shares.

Question 14.75

image 45. John Quincy Adams, the sixth president of the United States, proposed that the House of Representatives should be apportioned by a divisor method based on the rounding rule that rounds each fraction up to the next whole number.

  1. Is it likely that the initial tentative apportionment will be final?
  2. Will the divisor that produces the correct apportionment be greater than or less than the standard divisor?
  3. Does the method favor small states or large states?
  4. Can the Adams method apportion any state zero seats?
  5. The Adams method can violate the quota condition. Explain why it can never apportion to a state more than its upper quota.

45.

(a) No

(b) The divisor will have to be greater than the standard divisor.

(c) It favors small states.

(d) No, because an apportionment quotient between 0 and 1 must be rounded up to get .

(e) The Adams method always uses a divisor that is greater than the standard divisor. Therefore the apportionment quotients are less than the quotas that are obtained by dividing populations by the standard divisor. In other words, if is the quota for some state, and is the apportionment quotient for the same state, then . It follows that . But is the Adams apportionment and is the upper quota; hence the Adams method cannot exceed the upper quota for any state.

618

Question 14.76

image 46. The choice of a divisor method for ?4P apportioning classes to subjects according to enrollments, as in Example 2 (page 577), depends on what the school principal considers most important.

  1. The principal wants to set a minimum class size. For example, if the minimum class size is 20, and 39 students are signed up for English III, there would be one section, because there are not enough students for two sections with enrollment of at least 20. If there were 40 students, there would be two sections. The minimum class size is adjusted so that as many sections as possible are running. What apportionment method should she use?
  2. The principal prefers to set a maximum class size. For example, if the maximum class size is 33, and 67 students are taking History I, there will be three sections because there are too many students to fit in two 33-student sections. If there were only 66 students taking History I, there would be two sections. The maximum class size is adjusted so that as many sections as possible are running. What apportionment method should she use? (Hint: This divisor method is not described in the text but is mentioned in one of the previous exercises.)
  3. The principal wants to cancel any class that has an enrollment of just one student. Which apportionment methods should she avoid using?

Question 14.77

image image 47. Let be the quotas for states in an apportionment problem, and let the apportionments assigned by some apportionment method be denoted . The absolute deviation for state is defined to be ; it is a measure of the amount by which the state’s apportionment differs from its quota. The maximum absolute deviation is the largest of these numbers. Explain why the Hamilton method always gives the least possible maximum absolute deviation.

47.

Consider an apportionment problem where we will compare two methods, the Hamilton method and the method. (We do not specify what the method is, and we leave the apportionment problem also unspecified.) Our objective is to show that the maximum absolute deviation (MAD) for the apportionment is at least as large as the MAD for the Hamilton apportionment.

Since the Hamilton method satisfies the quota condition, we will assume that the method has given each state either its upper or lower quota; otherwise, its MAD would be greater than 1, and thus worse than Hamilton’s. If the methods give different apportionments, they must differ for more than one state; otherwise, they would not have the same house size.

Let be the state that has the largest absolute deviation with the Hamilton method, and let be its quota. The apportionment for is either the lower quota, , or the upper quota, , so the absolute deviation for (and the MAD for the Hamilton apportionment) is either .

If the apportionment for the state is the same as the Hamilton apportionment for , then the MAD for the apportionment is at least equal to the absolute deviation for , so it cannot be less than the MAD for the Hamilton apportionment.

Suppose that the apportionment for differs from the Hamilton apportionment: Say Hamilton assigns its lower quota and assigns its upper quota. There must be another state to which Hamilton assigns the upper quota and assigns the lower quota. Since the Hamilton method rounds down and up, the fractional part of is greater than or equal to the fractional part of . The absolute deviation in the apportionment for state is the fractional part of , so the MAD for the apportionment, which must be at least the absolute deviation for state , is greater than or equal to the MAD for the Hamilton method in this apportionment problem.

If the Hamilton method awards state its upper quota, we can reason as before and reach the same conclusion—that the apportionment has a MAD no less than that of the Hamilton method.

Chapter Review

Question 14.78

48. The Legis County Board of Supervisors has 145 seats and a total population of 115,275. The county is divided into five townships, whose populations are as follows:

Township Population
Alpha 16,210
Beta 40,052
Gamma 8,284
Delta 48,018
Epsilon 2,711

Use the Webster method to determine the apportionment of seats on this Board, then repeat, using the Hamilton method. Are the results the same?

Question 14.79

49. The Legis County Board of Supervisors (see Exercise 48) has too many seats for its space in the County Administration building. It has decided to use the Jefferson method of apportionment, with 3000 as divisor.

  1. Calculate the apportionment and find the new house size.
  2. Why would Epsilon Township have a problem with this?
  3. Find the quota for each township based on the house size you determined in part (a).
  4. Would this apportionment satisfy the quota condition?

49.

(a) Divide each township’s population by 3000 and round the quotient down to get the apportionment.

Township Population Seats Quota
Alpha 16,210 5 5.062
Beta 40,052 13 12.508
Gamma 8,284 2 2.587
Delta 48,018 16 14.996
Epsilon 2,711 0 0.847

The house size is the total number of seats apportioned: 36.

(b) Epsilon Township is unrepresented on the Board, but if a divisor less than 2711 had been chosen, it would have a seat.

(c) The standard divisor is the county’s population, 115,275, divided by the house size, 36. This is . The quotas, obtained by dividing the populations by the standard divisor, are shown in the table displayed in the answer for part (a).

(d) No. The upper quota for Delta Township is 15 seats, but it is awarded 16 seats.

Question 14.80

50. Responding to objections raised against the apportionment scheme in Exercise 49, the Legis County Board of Supervisors has adopted the Webster method of apportionment, with 3000 as the divisor.

  1. Using the same divisor, compute the apportionment and find the house size.
  2. Find the quota for each township based on the house size you determined in part (a).
  3. Is the quota condition satisfied?

Question 14.81

51. A county has five townships and elects a 301-seat board of supervisors, using the Webster method of apportionment. The populations of the townships are as follows:

Township Population
109,050
55,920
67,770
61,260
7,500
  1. Determine the Webster apportionment.
  2. Determine the district population for each township. Call the township with the greatest district population the “disadvantaged township.”
  3. To make the apportionment more equitable in terms of district population, which township should give a seat to the disadvantaged township?

51.

(a) The total population of this county is 301,500, and the house size is 301. The standard divisor is therefore . The quotas, determined by dividing the township populations by the house size, are shown in the following table. The sum of the rounded quotas (using the Webster rounding) is equal to the house size. Thus, each township’s Webster apportionment is its rounded quota.

Township Population Quota Rounded Quota District Population
109,050 108.87 109 1000.46
55,920 55.83 56 998.57
67,770 67.66 68 996.62
61,260 61.16 61 1004.28
7,500 7.49 7 1071.43

(b) The district populations are shown in the table displayed in part (a). Township is disadvantaged, because its district population is greater than the district population of each of the others by approximately 70.

(c) The following table shows the absolute difference between the district populations of Township and each of the other townships, as well as the difference if a seat were transferred from the other township to Township .

Township Abs Diff Before Transfer Abs Diff After Transfer
70.97 72.22
72.86 79.23
74.81 73.99
67.17 83.50

The table shows that if a seat is taken from

Township and given to Township , the absolute difference in district population will be less.

Question 14.82

image 52. Considering the problem of apportioning goods to people according to financial contributions, as in Exercise 8 on page 614 (allocating pearls bought at auction) or Exercise 13 on page 614 (rare coins), list the pros and cons for each of the apportionment methods listed below. Considerthe appropriateness of the following standards of comparison: paradoxes, quota condition, bias favoring small or large states, district population equity, representative share equity, and percentage equity.

  1. Hamilton
  2. Jefferson
  3. Webster
  4. Dean
  5. Hill-Huntington