4.1 Apportionment Problems Ch 4: Apportionment 4.1 Apportionment Problems

Basic Elements States: the players involved in the apportionment Seats: the set of M identical, indivisible objects that are being divided among N states; M>N but not every state is guaranteed a seat Populations: set of positive numbers which are used as the basis for the apportionment of the seats to the states (actual population, minutes studied, number of patients, etc)

Divisors & Quotas Standard Divisor (SD): the ratio of population to seats (total population/total # of seats) Standard Quotas: exact fractional number of seats each state would get if fractional seats were allowed (population/SD) Lower Quota: the quota rounded down (the LQ of 2.62 would be 2) Upper Quota: the quota rounded up (the UQ of 2.62 would be 3)

Pg. 150 #1 The Bandana Republic is a small country consisting of four states (Apure, Barinas, Carabobo, and Dolores). There are M=160 seats in the Bandana Congress. The populations of each state (in millions) are given in the following table. Find the standard divisor, standard quota, and lower and upper quotas. State A B C D Total Population 3.31 2.67 1.33 0.69 SQ Lower Quota Upper Quota

Pg. 150 #3 The SMARTS operates 6 bus routes (A, B, C, D, E, and F) and 130 buses. The buses are apportioned among the routes based on the average number of daily passengers per route, given in the following table. Describe the “states” and the “seats”. Find the SD and explain what it represents in this problem. Route A B C D E F Average # of passengers 45,300 31,070 20,490 14,160 10,260 8,720

Pg. 151 #5 The total population is 23.8 million. The following table shows each state’s standard quota. Find the number of seats, the SD, and each state’s population. State A B C D E SQ 40.50 29.70 23.65 14.60 10.55

Pg. 151 #7 & #9 7) Texas had 7.43% of the US population in 2000. Find their standard quota of the 435 seats in the House of Representatives. 9) If there are 200 seats apportioned among six planets, find the SD and SQ for each planet listed in the following table. Planet A B C D E F % of Population 40.50 29.70 23.65 14.60 10.55 SQ

4.2 Hamilton’s Method

Hamilton’s Method Step 1: Calculate each states standard quota. Step 2: Give each state its lower quota. Step 3: Give the surplus seats to the states with the largest decimal.

Hamilton’s Method Example: State A has a population of 940, state B has a population of 9,030, and state C has a population of 10,030. There are 200 seats to be divided among the three states using Hamilton’s Method. State Pop SQ LQ HQ A 940 B 9030 C 10030 Total

Hamilton’s Method Example: There are 50 seats to divide among 5 states. A has a population of 150, B has 78, C has 173, D has 204, and E has 295. Find each states apportionment of seats using Hamilton’s Method. State Pop SQ LQ HQ A 150 B 78 C 173 D 204 E 295 Total

Quota Violations

The Quota Rule The Quota Rule: Each state must receive either their lower quota or their upper quota.

Lower Quota Violation When a state ends up with an apportionment smaller than it’s lower quota.

Upper Quota Violation When a state ends up with an apportionment larger than it’s upper quota.

4.3 Paradoxes

Alabama Paradox This occurs when an increase in the total number of seats being apportioned forces a state to lose one of its seats.

Population Paradox This occurs when State A has a higher percentage of increase in population than State B, but State A loses one of its seats to State B.

New-States Paradox This occurs when the addition of a new state with its fair share of seats affects the apportionments of the other states.

4.4 Jefferson’s Method

Jefferson’s Method Find a modified divisor (D) so that each states lower quotas add to be the exact number of seats.

Jefferson’s Method Example: State A has a population of 940, state B has a population of 9,030, and state C has a population of 10,030. There are 200 seats to be divided among the three states using Jefferson’s Method. State Pop SQ LQ JQ A 940 B 9030 C 10030 Total

Jefferson’s Method Example: There are 50 seats to divide among 5 states. A has a population of 150, B has 78, C has 173, D has 204, and E has 295. Find each states apportionment of seats using Jefferson’s Method. State Pop SQ LQ JQ A 150 B 78 C 173 D 204 E 295 Total

4.5 Adam’s Method

Adam’s Method Find a modified divisor (D) so that each states upper quotas add to be the exact number of seats. Start by using a modified divisor that is bigger than the standard divisor.

Adam’s Method Example: State A has a population of 940, state B has a population of 9,030, and state C has a population of 10,030. There are 200 seats to be divided among the three states using Adam’s Method. State Pop SQ UQ AQ A 940 B 9030 C 10030 Total

Adam’s Method Example: There are 50 seats to divide among 5 states. A has a population of 150, B has 78, C has 173, D has 204, and E has 295. Find each states apportionment of seats using Adam’s Method. State Pop SQ UQ AQ A 150 B 78 C 173 D 204 E 295 Total

4.6 Webster’s Method

Webster’s Method Find a modified divisor (D) so that each states standard quotas rounded conventionally add to be the exact number of seats.

Webster’s Method Example: State A has a population of 940, state B has a population of 9,030, and state C has a population of 10,030. There are 200 seats to be divided among the three states using Webster’s Method.

Webster’s Method Example: There are 50 seats to divide among 5 states. A has a population of 150, B has 78, C has 173, D has 204, and E has 295. Find each states apportionment of seats using Webster’s Method.