1. Copyright © 2009 Pearson Education, Inc. Chapter 15 Section 3 - Slide 1 5-3 Election Theory Apportionment Methods

2. Chapter 15 Section 3 - Slide 2 Copyright © 2009 Pearson Education, Inc. WHAT YOU WILL LEARN Standard quotas and standard divisors Apportionment methods Flaws of apportionment methods

3. Chapter 15 Section 3 - Slide 3 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 3 - Slide 4 Copyright © 2009 Pearson Education, Inc. Definitions

5. Chapter 15 Section 3 - Slide 5 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 3 - Slide 6 Copyright © 2009 Pearson Education, Inc. Example (continued) 14.999Standard quota 8020188029403200Population TotalChemistryBusinessEducation

7. Chapter 15 Section 3 - Slide 7 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 3 - Slide 8 Copyright © 2009 Pearson Education, Inc. Example: Apportion the 15 graduate assistantships 15456Hamilton’s 13355 Lower quota 14.999 Standard quota 8020188029403200Population TotalChemistryBusinessEducation

9. Chapter 15 Section 3 - Slide 9 Copyright © 2009 Pearson Education, Inc. The Quota Rule An apportionment for every group under consideration should always be either the upper quota or the lower quota.

10. Chapter 15 Section 3 - Slide 10 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 3 - Slide 11 Copyright © 2009 Pearson Education, Inc. Modified divisor = 480 15366Jefferson 3.91676.1256.67 Modified quota 14.999 Standard quota 8020188029403200Population TotalChemistryBusinessEducation

12. Chapter 15 Section 3 - Slide 12 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 3 - Slide 13 Copyright © 2009 Pearson Education, Inc. Modified divisor = 535 15456Webster 3.515.495.98 Modified quota 14.999 Standard quota 8020188029403200Population TotalChemistryBusinessEducation

14. Chapter 15 Section 3 - Slide 14 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 3 - Slide 15 Copyright © 2009 Pearson Education, Inc. Modified divisor = 590 15456Adams 3.184.985.42 Modified quota 14.999 Standard quota 8020188029403200Population TotalChemistryBusinessEducation

16. Chapter 15 Section 4 - Slide 16 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 4 - Slide 17 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 4 - Slide 18 Copyright © 2009 Pearson Education, Inc. 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. 900489250161Employees Total321Office

19. Chapter 15 Section 4 - Slide 19 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 4 - Slide 20 Copyright © 2009 Pearson Education, Inc. Example: Demonstrating the Alabama Paradox (continued) Based on 30 cell phones, the table is as follows: (Note: standard divisor = 900/30 = 30) 900489250161 Employees 291685 Lower Quota 301686 Hamilton’s apportionment 16.38.335.37 Standard Quota Total321 Office

21. Chapter 15 Section 4 - Slide 21 Copyright © 2009 Pearson Education, Inc. Example: Demonstrating the Alabama Paradox (continued) Based on 31 cell phones, the table is as follows: (Note: standard divisor = 900/31 ≈ 29.03) 900489250161 Employees 291685 Lower Quota 311795 Hamilton’s apportionment 16.848.615.55 Standard Quota Total321 Office

22. Chapter 15 Section 4 - Slide 22 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 4 - Slide 23 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 4 - Slide 24 Copyright © 2009 Pearson Education, Inc. 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. 5400106311339331538733 Population in 2003 5450111211339331539733 Population in 2005 DETotalCBASchool

25. Chapter 15 Section 4 - Slide 25 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 4 - Slide 26 Copyright © 2009 Pearson Education, Inc. Solution Based on the population in 2003, the table is as follows: (Note: standard divisor = 5400/54 = 100) 16 15 15.38 1538 B 9 9 9.33 933 C 11 11.33 1133 D 54 52 5400 Total 11 10 10.63 1063 E 733 Population in 2003 7 Lower Quota 7 Hamilton’s apportionment 7.33 Standard Quota A School

27. Chapter 15 Section 4 - Slide 27 Copyright © 2009 Pearson Education, Inc. Solution (continued) Based on the population in 2005, the table is as follows: (Note: standard divisor = 5450/54 ≈ 100.93) 15 15.25 1539 B 9 9 9.24 933 C 11 11.23 1133 D 54 53 5450 Total 11 11.02 1112 E 733 Population in 2005 7 Lower Quota 8 Hamilton’s apportionment 7.26 Standard Quota A School

28. Chapter 15 Section 4 - Slide 28 Copyright © 2009 Pearson Education, Inc. 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. Chapter 15 Section 4 - Slide 29 Copyright © 2009 Pearson Education, Inc. New-States Paradox The new-states paradox occurs when the addition of a new group reduces the apportionment of another group.

30. Chapter 15 Section 4 - Slide 30 Copyright © 2009 Pearson Education, Inc. Summary Small states Large states Appointment method favors No Yes May produce the new- states paradox No Yes May produce the population paradox No Yes May produce the Alabama paradox Yes No May violate the quota rule WebsterAdamsJeffersonHamilton Apportionment Method

31. Slide 15 - 31 Copyright © 2009 Pearson Education, Inc. 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, 7b.15, 8, 10, 7 c.14, 9, 9, 8d.15, 9, 9, 7 StateABCDTotal Population780045315122395521,408

32. Slide 15 - 32 Copyright © 2009 Pearson Education, Inc. 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, 7b.15, 8, 10, 7 c.14, 9, 9, 8d.15, 9, 9, 7 StateABCDTotal Population780045315122395521,408

33. Slide 15 - 33 Copyright © 2009 Pearson Education, Inc. 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, 7b.15, 8, 10, 7 c.14, 9, 9, 8d.15, 9, 9, 7 StateABCDTotal Population780045315122395521,408

34. Slide 15 - 34 Copyright © 2009 Pearson Education, Inc. 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, 7b.15, 8, 10, 7 c.14, 9, 9, 8d.15, 9, 9, 7 StateABCDTotal Population780045315122395521,408

35. Slide 15 - 35 Copyright © 2009 Pearson Education, Inc. Practice Problems

36. Slide 15 - 36 Copyright © 2009 Pearson Education, Inc.