CHAPTER 14

1. CHAPTER 14 Voting and Apportionment

2. 14.4 Flaws of Apportionment Methods

3. Objectives • Understand and illustrate the Alabama paradox. • Understand and illustrate the population paradox. • Understand and illustrate the new-states paradox. • Understand Balinski and Young’s Impossibility Theorem.

4. The Alabama Paradox An increase in the total number of items to be apportioned results in the loss of an item for a group.

5. Example 1: Illustrating the Alabama Paradox A small country with a population of 10,000 is composed of three states. According to the country’s constitution, the congress will have 200 seats. A table of the country’s population distribution is given below. Use Hamilton’s method and show that the Alabama paradox occurs if the number of seats is increased to 201.

6. Example 1: Illustrating the Alabama Paradox Solution: We begin with 200 seats and compute the standard divisor. Using this value, we obtain the apportionment for each state using Hamilton’s method.

7. Example 1: Illustrating the Alabama Paradox Next, we find each state’s apportionment with 201 congressional seats. Again, we compute the standard divisor.

8. Example 1: Illustrating the Alabama Paradox The final apportionments are summarized below. When the number of seats increased from 200 to 201, state C’s apportionment decreased. This is an example of Alabama’s Paradox. • States A and B benefit at state C’s expense.

9. The Population Paradox Group A loses items to group B, even though the population of group A grew at a faster rate than that of group B.

10. Example 2: Illustrating the Population Paradox A small country with a population 10,000 is composed of three states. There are 11 seats in congress, divided among the three states according to their respective populations. Using Hamilton’s method, the apportionment of congressional seats for each state is shown. a. Find the percent of increase in the population of each state. b. Use Hamilton’s method to show that the population paradox occurs.

11. Example 2: Illustrating the Population Paradox Solution: a. • From the percentages, state C is increasing at a faster rate than state A and B. • From the percentages, state B is increasing at a faster rate than state A.

12. Example 2: Illustrating the Population Paradox • We use the Hamilton’s method to find the apportionment for each state with its new population. First we compute the standard divisor.

13. Example 2: Illustrating the Population Paradox The final apportionments are shown below. Even though state A now has a congressional seat, state B has lost a seat to state A, but recall state B’s population grew faster than state A’s. This is an example of population paradox.

14. New-States Paradox The addition of a new group changes the apportionments of other groups.

15. Example 3: Illustrating the New-States Paradox A school district has two high schools: East High and West High. The school district has a counseling staff of 48 counselors. The standard divisor is

16. Example 3: Illustrating the New-States Paradox Suppose a new high school, North High, is added to the district with 1448 more students. Using the standard divisor of 200, the district hires 7 counselors.Showthat the new-states paradox occurs when the counselors are reapportioned. Solution: The new standard divisor is

17. Example 3: Illustrating the New-States Paradox Before North High, East High had 8 counselors and West High had 40. After we added North High, East High added 1 counselor, West High lost 1 counselor. Hence, the addition of another high school changed the apportionments of the other high schools. This is an example of new-states paradox.

18. Balinski and Young’s Impossibility Theorem Is there an ideal apportionment method? • There is no perfect apportionment method. • Any apportionment method that does not violate the quota rule must produce paradoxes. • Any apportionment method that does not produce paradoxes must violate the quota rule.