Flaws of the Apportionment Methods Section 14.4 Flaws of the Apportionment Methods

What You Will Learn Upon completion of this section, you will be able to: Determine if a given apportionment demonstrates the Alabama paradox. Determine if a given apportionment demonstrates the population paradox. Determine if a given apportionment demonstrates the new-states paradox.

Three Flaws of Hamilton’s Method The three flaws of Hamilton’s method are: the Alabama paradox the population paradox the new-states paradox.

Three Flaws of Hamilton’s Method 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.

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.

Example 1: Demonstrating the Alabama Paradox Consider Stanhope, a small country with a population of 15,000 people and three states A, B, and C. There are 150 seats in the legislature that must be apportioned among the three states, according to their population. Show that the Alabama paradox occurs if the number of seats is increased to 151.

Example 1: Demonstrating the Alabama Paradox Round standard divisors and standard quotas to the nearest hundredth.

Example 1: Demonstrating the Alabama Paradox Solution With 150 seats, the standard divisor is 15,000 ÷ 150 = 100 The standard quotas for each state and the apportionment for each state are shown in the following table.

Example 1: Demonstrating the Alabama Paradox Solution

Example 1: Demonstrating the Alabama Paradox Solution With 151 seats, the standard divisor is 15,000 ÷ 151 ≈ 99.34 The standard quotas for each state and the apportionment for each state are shown in the following table.

Example 1: Demonstrating the Alabama Paradox Solution

Example 1: Demonstrating the Alabama Paradox Solution When the number of seats increased from 150 to 151, state A’s apportionment actually decreased, from 8 to 7. This example illustrates the Alabama paradox.

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.

Example 2: Demonstrating the Population Paradox Consider Alexandria, a small country with a population of 100,000 and three states A, B, and C. There are 100 seats in the legislature that must be apportioned among the three states. Using Hamilton’s method, the apportionment is shown in the table.

Example 2: Demonstrating the Population Paradox

Example 2: Demonstrating the Population Paradox Suppose that the population increases according to the table below and that the 100 seats are reapportioned. Show that the population paradox occurs when Hamilton’s method is used.

Example 2: Demonstrating the Population Paradox Solution Calculate the percent increases. State A has an increase of 399 people. Therefore, state A has a percent increase of

Example 2: Demonstrating the Population Paradox Solution State B has an increase of 100 people. Therefore, state B has a percent increase of

Example 2: Demonstrating the Population Paradox Solution State C has an increase of 185 people. Therefore, state C has a percent increase of

Example 2: Demonstrating the Population Paradox Solution All three states had an increase in their population, but state B is increasing at a faster rate than state A and state C. The standard divisor using the new population is

Example 2: Demonstrating the Population Paradox Solution The table shows the reapportionment, using Hamilton’s method using the standard divisor 1006.84.

Example 2: Demonstrating the Population Paradox Solution State B has lost a seat to state A even though state B’s population grew at a faster rate than state A’s. As a result, we have an example of the population paradox.

New-States Paradox The new-states paradox occurs when the addition of a new group, and additional items to be apportioned, reduces the previous apportionment of another group.

Example 3: Demonstrating the New-States Paradox The Oklahoma Public Library System has received a grant to purchase 100 laptop computers to be distributed between two libraries A and B. The 100 laptops will be apportioned based on the population served by each library. The apportionment using Hamilton’s method is shown in the table.

Example 3: Demonstrating the New-States Paradox The standard divisor is

Example 3: Demonstrating the New-States Paradox Suppose that an anonymous donor decides to donate money to purchase six more laptops provided that a third library, C, that serves a population of 625, is included in the apportionment. Show that the new-states paradox occurs when the laptops are reapportioned.

Example 3: Demonstrating the New-States Paradox Solution Total population is 10,000 + 625 = 10,625 and the total number of laptops to be apportioned is 100 + 6 = 106. The new standard divisor is

Example 3: Demonstrating the New-States Paradox Solution The new standard quota and Hamilton’s apportionment are shown here.

Example 3: Demonstrating the New-States Paradox Solution Before library C was added, library B would receive 79 laptops. By adding a new library and increasing the total number of laptops to be apportioned, library B ended up losing a laptop to library A. Thus, we have a case of the new-states paradox.

Balinski and Young’s Impossibility Theorem There is no perfect apportionment method that satisfies the quota rule and avoids any paradoxes.

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