14 3 Apportionment Methods 2010 Pearson Prentice Hall

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14. 3 Apportionment Methods © 2010 Pearson Prentice Hall. All rights reserved. 1

14. 3 Apportionment Methods © 2010 Pearson Prentice Hall. All rights reserved. 1

1. 2. 3. 4. 5. 6. 7. Objectives Find standard divisors and standard quotas.

1. 2. 3. 4. 5. 6. 7. Objectives Find standard divisors and standard quotas. Understand the apportionment problem. Use Hamilton’s method. Understand the quota rule. Use Jefferson’s method. Use Adam’s method. Use Webster’s method. © 2010 Pearson Prentice Hall. All rights reserved. 2

Standard Divisors and Standard Quotas Standard Divisors The standard divisor is found by dividing

Standard Divisors and Standard Quotas Standard Divisors The standard divisor is found by dividing the total population under consideration by the number of items to be allocated. The standard quota for a particular group is found by dividing that group’s population by the standard divisor. © 2010 Pearson Prentice Hall. All rights reserved. 3

Example 1: Finding Standard Quotas The Republic of Margaritaville is composed of four states,

Example 1: Finding Standard Quotas The Republic of Margaritaville is composed of four states, A, B, C, and D. The table of each state’s population (in thousands) is given below. State A B C D Population 275 383 465 767 Total 1890 Standard Quota © 2010 Pearson Prentice Hall. All rights reserved. 4

Example 1: Finding Standard Quotas The standard quotas are obtained by dividing each state’s

Example 1: Finding Standard Quotas The standard quotas are obtained by dividing each state’s population by the standard divisor. We previously computed the standard divisor and found it to be 63. © 2010 Pearson Prentice Hall. All rights reserved. 5

Example 1: Finding Standard Quotas • Standard quota for state A • Standard quota

Example 1: Finding Standard Quotas • Standard quota for state A • Standard quota for state B • Standard quota for state C • Standard quota for state D © 2010 Pearson Prentice Hall. All rights reserved. 6

Example 1: Finding Standard Quotas State A B C D Total Population 275 383

Example 1: Finding Standard Quotas State A B C D Total Population 275 383 465 767 1890 Standard Quota 4. 37 6. 08 7. 38 12. 17 30 Notice that the sum of all the standard quotas is 30, the total number of seats in the congress. © 2010 Pearson Prentice Hall. All rights reserved. 7

Apportionment Problem • The standard quotas represent each state’s exact fair share of the

Apportionment Problem • The standard quotas represent each state’s exact fair share of the 30 seats for the congress of Margaritaville. – Can state A have 4. 37 seats in congress? • The apportionment problem is to determine a method for rounding standard quota into whole numbers so that the sum of the numbers is the total number of allocated items. – Can we round standard quotas up or down to the nearest whole number and solve this problem? • The lower quota is the standard quota rounded down to the nearest whole number. • The upper quota is the standard quota rounded up to the nearest whole number. © 2010 Pearson Prentice Hall. All rights reserved. 8

Apportionment Problem Here’s the apportionment problem for Margaritaville. There are four different apportionment methods

Apportionment Problem Here’s the apportionment problem for Margaritaville. There are four different apportionment methods called Hamilton’s method, Jefferson’s method, Adam’s method, and Webster’s method. © 2010 Pearson Prentice Hall. All rights reserved. 9

Hamilton’s Method 1. Calculate each group’s standard quota. 2. Round each standard quota down

Hamilton’s Method 1. Calculate each group’s standard quota. 2. Round each standard quota down to the nearest whole number, thereby finding the lower quota. Initially, give to each group its lower quota. 3. Give the surplus items, one at a time, to the groups with the largest decimal parts until there are no more surplus items. © 2010 Pearson Prentice Hall. All rights reserved. 10

Hamilton’s Method Example: Consider the lower quotas for Margaritaville. We have a surplus of

Hamilton’s Method Example: Consider the lower quotas for Margaritaville. We have a surplus of 1 seat. The surplus seat goes to the state with the greatest decimal part. The greatest decimal part is 0. 38 for state C. Thus, state C receives the additional seat, and is assigned 8 seats in congress. © 2010 Pearson Prentice Hall. All rights reserved. 11

The Quota Rule • A group’s apportionment should be either its upper quota or

The Quota Rule • A group’s apportionment should be either its upper quota or its lower quota. An apportionment method that guarantees that this will always occur is said to satisfy the quota rule. • A group’s final apportionment should either be the nearest upper or lower whole number of its standard quota. © 2010 Pearson Prentice Hall. All rights reserved. 12

Jefferson’s Method 1. Find a modifier divisor, d, such that each group’s modified quota

Jefferson’s Method 1. Find a modifier divisor, d, such that each group’s modified quota is rounded down to the nearest whole number, the sum of the quotas is the number of items to be apportioned. • The modified quotients that are rounded down are called modified lower quotas. 2. Apportion to each group its modified lower quota. © 2010 Pearson Prentice Hall. All rights reserved. 13

Example 3: Using Jefferson’s Method A rapid transit service operates 130 buses along six

Example 3: Using Jefferson’s Method A rapid transit service operates 130 buses along six routes, A, B, C, D, E, and F. The number of buses assigned to each route is based on the average number of daily passengers per route. Use Jefferson’s method with d = 486 to apportion the buses. Route Average Number of Passengers A B C 4360 5130 7080 © 2010 Pearson Prentice Hall. All rights reserved. D E F Total 10, 245 15, 535 22650 65, 000 14

Example 3: Using Jefferson’s Method Solution: Using d = 486, the table below illustrates

Example 3: Using Jefferson’s Method Solution: Using d = 486, the table below illustrates Jefferson’s method. © 2010 Pearson Prentice Hall. All rights reserved. 15

Adam’s Method 1. Find a modifier divisor, d, such that when each group’s modified

Adam’s Method 1. Find a modifier divisor, d, such that when each group’s modified quota is rounded up to the nearest whole number, the sum of the quotas is the number of items to be apportioned. • The modified quotients that are rounded up are called modified upper quotas. 2. Apportion to each group its modified upper quota. © 2010 Pearson Prentice Hall. All rights reserved. 16

Example 4: Using Adam’s Method A rapid transit service operates 130 buses along six

Example 4: Using Adam’s Method A rapid transit service operates 130 buses along six routes, A, B, C, D, E, and F. The number of buses assigned to each route is based on the average number of daily passengers per route. Use Adam’s method to apportion the buses. Route Average Number of Passengers A B C 4360 5130 7080 © 2010 Pearson Prentice Hall. All rights reserved. D E F 10, 245 15, 535 22, 650 Total 65, 000 17

Example 4: Using Adam’s Method Solution: We begin by guessing at a possible modified

Example 4: Using Adam’s Method Solution: We begin by guessing at a possible modified divisor, d, that we hope will work. We guess d = 512. © 2010 Pearson Prentice Hall. All rights reserved. 18

Example 4: Using Adam’s Method • Because the sum of modified upper quotas is

Example 4: Using Adam’s Method • Because the sum of modified upper quotas is too high, we need to lower the modified quotas. • In order to do this, we must try a higher divisor. If we let d = 513, we obtain 129 for the sum of modified upper quotas. • Because this sum is too low, we decide to increase the divisor just a bit to d = 512. 7 instead. © 2010 Pearson Prentice Hall. All rights reserved. 19

Webster’s Method 1. Find a modifier divisor, d, such that when each group’s modified

Webster’s Method 1. Find a modifier divisor, d, such that when each group’s modified quota is rounded to the nearest whole number, the sum of the quotas is the number of items to be apportioned. • The modified quotients that are rounded down are called modified rounded quotas. 2. Apportion to each group its modified rounded quota. © 2010 Pearson Prentice Hall. All rights reserved. 20

Example 5: Using Webster’s Method A rapid transit service operates 130 buses along six

Example 5: Using Webster’s Method A rapid transit service operates 130 buses along six routes, A, B, C, D, E, and F. The number of buses assigned to each route is based on the average number of daily passengers per route. Use Webster’s method to apportion the buses. Route Average Number of Passengers A B C 4360 5130 7080 © 2010 Pearson Prentice Hall. All rights reserved. D E F Total 10, 245 15, 535 22650 65, 000 21

Example 5: Using Webster’s Method Solution: We begin by guessing at a possible modified

Example 5: Using Webster’s Method Solution: We begin by guessing at a possible modified divisor, d, that we hope will work. We guess d = 502, a divisor greater than the standard divisor, 500. • We would obtain 129 for the sum of modified rounded quotas. • Because 129 is too low, this suggests that d is too large. So, we guess a lower d than 500. Say, d = 498. © 2010 Pearson Prentice Hall. All rights reserved. 22

Pg 801: 1 – 39 odd © 2010 Pearson Prentice Hall. All rights reserved.

Pg 801: 1 – 39 odd © 2010 Pearson Prentice Hall. All rights reserved.