5-3 Election Theory

1. 5-3 Election Theory Apportionment Methods

2. WHAT YOU WILL LEARN • Standard quotas and standard divisors • Apportionment methods • Flaws of apportionment methods

3. Apportionment • The goal of apportionment is to determine a method to allocate the total number of items to be apportioned in a fair manner. • Four Methods • Hamilton’s method • Jefferson’s method • Webster’s method • Adam’s method

4. Definitions

5. Example • A Graduate school wishes to apportion 15 graduate assistantships among the colleges of education, business and chemistry based on their undergraduate enrollments. Using the table on the next slide, find the standard quotas for the schools.

6. Education Business Chemistry Total Population 3200 2940 1880 8020 Standard quota 14.999 Example (continued)

7. Hamilton’s Method 1. Calculate the standard divisor for the set of data. 2. Calculate each group’s standard quota. 3. Round each standard quota down to the nearest integer (the lower quota). Initially, each group receives its lower quota. 4. Distribute any leftover items to the groups with the largest fractional parts until all items are distributed.

8. Education Business Chemistry Total Population 3200 2940 1880 8020 Standard quota 14.999 Lower quota 5 5 3 13 Hamilton’s 6 5 4 15 Example: Apportion the 15 graduate assistantships

9. The Quota Rule • An apportionment for every group under consideration should always be either the upper quota or the lower quota.

10. Jefferson’s Method 1. Determine a modified divisor, d, such that when each group’s modified quota is rounded down to the nearest integer, the total of the integers is the exact number of items to be apportioned. We will refer to the modified quotas that are rounded down as modified lower quotas. 2. Apportion to each group its modified lower quota.

11. Education Business Chemistry Total Population 3200 2940 1880 8020 Standard quota 14.999 Modified quota 6.67 6.125 3.9167 Jefferson 6 6 3 15 Modified divisor = 480

12. Webster’s Method 1. Determine a modified divisor, d, such that when each group’s modified quota is rounded to the nearest integer, the total of the integers is the exact number of items to be apportioned. We will refer to the modified quotas that are rounded to the nearest integer as modified rounded quotas. 2. Apportion to each group its modified rounded quota.

13. Education Business Chemistry Total Population 3200 2940 1880 8020 Standard quota 14.999 Modified quota 5.98 5.49 3.51 Webster 6 5 4 15 Modified divisor = 535

14. Adams’s Method 1. Determine a modified divisor, d, such that when each group’s modified quota is rounded up to the nearest integer, the total of the integers is the exact number of items to be apportioned. We will refer to the modified quotas that are rounded up as modified upper quotas. 2. Apportion to each group its modified upper quota.

15. Education Business Chemistry Total Population 3200 2940 1880 8020 Standard quota 14.999 Modified quota 5.42 4.98 3.18 Adams 6 5 4 15 Modified divisor = 590

16. Three Flaws of Hamilton’s Method • The three flaws of Hamilton’s method are: the Alabama paradox, the population paradox, and the new-states paradox. • These flaws apply only to Hamilton’s method and do not apply to Jefferson’s method, Webster’s method, or Adam’s method. • In 1980 the Balinski and Young’s Impossibility Theorem stated that there is no perfect apportionment method that satisfies the quota rule and avoids any paradoxes.

17. Alabama Paradox • The Alabama paradox occurs when an increase in the total number of items to be apportioned results in a loss of an item for a group.

18. Office 1 2 3 Total Employees 161 250 489 900 Example: Demonstrating the Alabama Paradox A large company, with branches in three cities, must distribute 30 cell phones to the three offices. The cell phones will be apportioned based on the number of employees in each office shown in the table below.

19. Example: Demonstrating the Alabama Paradox (continued) • Apportion the cell phones using Hamilton’s method. • Does the Alabama paradox occur using Hamilton’s method if the number of new cell phones increased from 30 to 31? Explain.

20. Office 1 2 3 Total Employees 161 250 489 900 Standard Quota 5.37 8.33 16.3 Lower Quota 5 8 16 29 Hamilton’s apportionment 6 8 16 30 Example: Demonstrating the Alabama Paradox (continued) • Based on 30 cell phones, the table is as follows: (Note: standard divisor = 900/30 = 30)

21. Office 1 2 3 Total Employees 161 250 489 900 Standard Quota 5.55 8.61 16.84 Lower Quota 5 8 16 29 Hamilton’s apportionment 5 9 17 31 Example: Demonstrating the Alabama Paradox (continued) • Based on 31 cell phones, the table is as follows: (Note: standard divisor = 900/31 ≈ 29.03)

22. Example: Demonstrating the Alabama Paradox (continued) • When the number of cell phones increased from 30 to 31, office one actually lost a cell phone, while the other two offices actually gained a cell phone under Hamilton’s apportionment.

23. Population Paradox • The Population Paradox occurs when group A loses items to group B, even though group A’s population grew at a faster rate than group B’s.

24. School A B C D E Total Population in 2003 733 1538 933 1133 1063 5400 Population in 2005 733 1539 933 1133 1112 5450 Example: Demonstrating Population Paradox A school district with five elementary schools has funds for 54 scholarships. The student population for each school is shown in the table below.

25. Example: Demonstrating Population Paradox (continued) • Apportion the scholarships using Hamilton’s method. • If the school wishes to give the same number of scholarships two years later, does a population paradox occur?

26. School A B C D E Total Population in 2003 733 1538 933 1133 1063 5400 Standard Quota 7.33 15.38 9.33 11.33 10.63 Lower Quota 7 15 9 11 10 52 Hamilton’s apportionment 7 16 9 11 11 54 Solution • Based on the population in 2003, the table is as follows: (Note: standard divisor = 5400/54 = 100)

27. School A B C D E Total Population in 2005 733 1539 933 1133 1112 5450 Standard Quota 7.26 15.25 9.24 11.23 11.02 Lower Quota 7 15 9 11 11 53 Hamilton’s apportionment 8 15 9 11 11 54 Solution (continued) • Based on the population in 2005, the table is as follows: (Note: standard divisor = 5450/54 ≈ 100.93)

28. Solution (continued) • In the school district in 2005, school B actually gives one of its scholarships to school A, even though the population in school B actually grew by 1 student and the population in School A remained the same.

29. New-States Paradox • The new-states paradox occurs when the addition of a new group reduces the apportionment of another group.

30. Apportionment Method Hamilton Jefferson Adams Webster May violate the quota rule No Yes Yes Yes May produce the Alabama paradox Yes No No No May produce the population paradox Yes No No No Yes No No No May produce the new-states paradox Appointment method favors Large states Large states Small states Small states Summary

31. A country has four states and 40 seats in the legislature. The population of each state is shown below. Determine each state’s apportionment using Hamilton’s method. a. 16, 8, 9, 7 b. 15, 8, 10, 7 c. 14, 9, 9, 8 d. 15, 9, 9, 7

32. A country has four states and 40 seats in the legislature. The population of each state is shown below. Determine each state’s apportionment using Hamilton’s method. a. 16, 8, 9, 7 b. 15, 8, 10, 7 c. 14, 9, 9, 8 d. 15, 9, 9, 7

33. A country has four states and 40 seats in the legislature. The population of each state is shown below. Determine each state’s apportionment using Jefferson’s method. a. 16, 8, 9, 7 b. 15, 8, 10, 7 c. 14, 9, 9, 8 d. 15, 9, 9, 7

34. A country has four states and 40 seats in the legislature. The population of each state is shown below. Determine each state’s apportionment using Jefferson’s method. a. 16, 8, 9, 7 b. 15, 8, 10, 7 c. 14, 9, 9, 8 d. 15, 9, 9, 7

35. Practice Problems