Chapter 14 Exercises
14.1 The Apportionment Problem
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%.
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?
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.
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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.
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?
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
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.
7. Repeat Exercise 6 with the following sum:
7.
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.
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.
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.
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
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?
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 |
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.
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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.
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
16. Explain why the tentative Webster apportionment of a state with quota is .
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.
18. Reapportion the classes in Exercise 10 (page 614), using the Webster method.
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!
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?
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.
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.
23. Make a Sainte-Laguë table and apportion the first 30 seats.
23.
See table on next page
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?
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
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?)
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.
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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?
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.
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.
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!
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?
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.
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?
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.
35. Jim is 72 inches tall and Alice is 65 inches tall. What is the percentage difference in their heights?
35.
10.77%
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.
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.
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?
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.
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
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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 .
(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.
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.
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.
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.
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 |
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.
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.
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.
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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.
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
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?
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.
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.
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.
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 |
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.
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.