§14.3, Apportionment Methods

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§14.3, Apportionment Methods


Learning Targets

I will find standard divisors and standard quotas.

I will understand the apportionment problem.

I will use Hamilton’s method, Jefferson’s method, Adam’s method, and Webster’s method to apportion.

I will understand the quota rule.


Standard Divisors and Standard QuotasStandard 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.

total populationStandard divisor =

number of allocated items

population of a particular groupStandard quota =

standard divisor


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 Total

Population 275 383 465 767 1890

Standard Quota


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.



• Standard quota for state A

• Standard quota for state B

• Standard quota for state C

• Standard quota for state D


population of state A 275= 4.37

standard divisor 63

population of state B 383= 6.08

standard divisor 63

population of state C 465= 7.38

standard divisor 63

population of state D 767= 12.17

standard divisor 63


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.



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.


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.

Apportionment Problem


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.


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.

Hamilton’s Method


• 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.

The Quota Rule


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.


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.

Example 3: Using Jefferson’s Method

Route A B C D E F Total

Average Number of Passengers

4360 5130 7080 10,245 15,535 22650 65,000


Solution: Using d = 486, the table below illustrates Jefferson’s method.

Example 3: Using Jefferson’s Method


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.


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.

Example 4: Using Adam’s Method



4360 5130 7080 10,245 15,535 22,650 65,000


Solution: We begin by guessing at a possible modified divisor, d, that we hope will work. We guess d = 512.



• 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.



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.


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.

Example 5: Using Webster’s Method



4360 5130 7080 10,245 15,535 22650 65,000


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.

Example 5: Using Webster’s Method

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Pg 801 – 803, #1 – 22.

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§14.3, Apportionment Methods