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Katie Kilgore

on 7 October 2014

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Transcript of Apportionment

Rounding each number in a set of numbers that add up to a whole number. The sum of the rounded numbers must equal the original sum.
Standard Divisor
The total population, divided by the house size.
The Hamilton Method
Each state receives either its lower quota, its quota rounded down, or its upper quota, obtained by rounding up.
The states that receive their upper quotas are those quotas have the largest fractional parts
Jefferson's Method
Chapter 14

By: Katie Kilgore, Kyana Vizcarra, Alli Bell, Saeda Tecumseh
Apportionment Method
A Procedure for solving all apportionment problems without making arbitrary choices
Total population
Standard Divisor=
House Size
The fraction of the total number of seats a state would be entitled to if the seats were not indivisible.
State Population
Standard Quota=
Standard Divisor
The Websters Method
The Webster method is the divisor method that rounds the quota (adjusted if necessary) to the nearest whole number, rounding up when the fractional part is greater than or equal to ½, and rounding down when the fractional part is less than ½.
The Webster method is neutral, favoring neither large nor small states.
The Webster method rarely violates the quota condition by giving a state more than its upper quota, or fewer seats than its lower quota, and would not have done so in any of the 23 congressional apportionments that have occurred so far.
1.. Obtain the tentative apportionment by rounding each states quota.
2. Add the rounded quotas. If their sum is equal to the house size, the job is finished.
3. When the rounded quotas of the states don’t add up to the house size, then calculate adjusted quotas, using a trial divisor as with the Jefferson method. All trial methods must me larger than the standard divisor if the sum of the rounded quotas is greater than the house size, and smaller than the standard divisor if the sum of the quotas is less than house size.
4. Round the adjusted quotas from step 3. If there sum is more than the house size, try a larger divisor; if the sum is less, try a smaller divisor. If the sum is equal to the house size, the rounded adjusted quotas provide the correct apportionment.
Example Suppose a nation has 4 states, with populations shown below, and a representative body with 60 seats. How many seats should each state get?
State Population
A 74,285
B 63,614
C 42,355
D 27,383

Total 207,637
State Population Fair Share Rounded
A. 74,285 21.47 21
B. 63,614 18.38 18
C. 42,355 12.24 12
D. 27,383 7.91 8
Total 207,637 60 59

But we need the fair share to be a whole number, so we round the numbers up or down to become a whole number and the sum should equal the total seats available.
*Does not equal total number of seats (60)
The apportionment problem gets its name from Article 1 Section 2 of the U.S. Constitution requires that "representatives shall be apportioned among the several States according to their respective numbers, counting the whole number of persons in each State..." Our United States Congress is broken up into two parts the Senate And the House of Representatives. Each state is represented in the House proportioned to its population, each state is entitled to at least one representative. The fixed number of representatives is 435, although that has not always been the case. Throughout this presentation we are going to show you how to round fractional shares that can not be distributed much like our government has done with the House of Representatives.
A field hockey team is making a poster to displaying the percentage of their team record of 18 won, 4 lost, and 1 tied game out of 23 games total. Represented in whole numbers to keep it simple.

Winning percentage = 78.26.. rounding to 78
Losing percentage = 17.39.. rounding to 17
Tie percentage = 4.35.. rounding to 4
Because all percents are rounded down the total is only 99%

So how do we get at last 1%? This is where Apportionment method comes into play
Standard Quota
House size (because it is a percentage) is 100

Total population (the number of games played) is 23

Standard Divisor is 23
= 0.23
Total populations = = 23
Population of the category (wins, losses,and ties)
Standard Divisor = / 100
Wins: 18 / 0.23 = 78.26%
Losses: 4 / 0.23 = 17.39%
Ties: 1 / 0.23 = 4.25%
Change the divisor until you get the total number of seats available.
Since the number we got when rounded is less, we want to try a smaller divisor.
Modified divisor= 3450
State Population Fair Share Rounded
A. 74,285 21.47 21
B. 63,614 18.38 18
C. 42,355 12.24 12
D. 27,383 7.91 8

Total 207,637 60 59
State Population Modified Share Rounded
A. 74,285 21.53 22
B. 63,614 18.43 18
C. 42,355 12.27 12
D. 27,383 7.9 8

Total 207,637 60

Wins 18
Losses 4
Ties 1
Total 23
Category Populations
Category Pop x0.23 x0.227 x0.228 x0.229

Wins: 18 78.26= 78 79.29 = 79 78.94 = 79 78.60 = 79

Losses: 4 17.39 = 17 17.62 = 18 17.54 = 18 17.46 = 17

Ties: 1 4.35 = 4 4.40 = 4 4.38 = 4 4.36 = 4

Total: 23 99 101 101 100
Using the prior information about the field hockey team express the record in the form of a whole percent. Make sure the total sum equals 100% when rounded
Things To Know:
1.Divide the population by the number of seats to get the standard divisor.
2.The divide the states population by the standard divisor.

There are tow paradoxes in the Hamilton method:
*House monotone- It is not possible for any state to receive fewer seats after the house size than it had before the increase, provided that there is no change in any states population
*Alabama paradox- when a stat loses a seat as the result of an increase in the size of House of Representatives, with no change in any states population

Alabama 8 7
Illinois 18 19
States House Size
1. Tentatively assign each to each state its lower quota of representatives. Each state whose quota is not a whole number loses a fraction of a seat at this stage, so the total number of seats assigned at this point will be less than the house size. This leaves additional seats to be apportioned.

2. Allot the remaining seats, one each, to the states whose quotas have the largest fractional parts, until the house is filled

Kentucky 1.995 1^ 2
S. Carolina 5.989 5^ 5
Rhode Island 1.998 1^ 2
Connecticut 6.878 6^ 7
Massachusetts 13.803 13^ 14
Delaware 1.613 1^ 2
Pennsylvania 12.570 12^ 13
Vermont 2.484 2 2
Virginia 18.310 18 18
N. Carolina 10.266 10 10
New Jersey 5.214 5 5
N. Hampshire 4.118 4 4
Maryland 8.088 8 8
Georgia 2.057 2 2
Totals 105 97 105
States Quota Lower Quota State Apportioned

College Algebra: 188 7.52 7
Calculus: 142 5.68 5
Calculus I: 138 5.52 5
Calculus II: 64 2.56 2
Totals: 750 30.00 27
218 8.72 8
Course Enrollment Quota Lower Quota
Find the Apportionment using the Hamilton method.

College Algebra: 188 7.52 7 6
Calculus: 142 5.68 5 6
Calculus I: 138 5.52 5 5
Calculus II: 64 2.56 2 3
Totals: 750 30.00 27 30
218 8.72 8 9
Course Enrollment Quota Lower Quota Apportionment
First find the standard divisor:
(total/available seats)
Standard divisor: 207,637/60= 3460.6166
Fair Share: Population/standard divisor

Find the standard divisor and standard quota for the following question. Round to the nearest tenth.

A high school has one mathematics teacher who teaches all geometry, pre-calculus, and calculus classes. She teaches a total of 5 sections, and 100 students are enrolled as follows: 52 for geometry, 33 pre-calculus, and 15 calculus
Population(Total Students)=100 Geometry=52
States(Courses)=3 Pre-Calculus=33
House size(Number of sections taught)=5 Calculus= 15

Standard Divisor = =20

Standard Quota:
Geometry= =2.60

Pre-Cal= =1.65

Calculus= =0.75

Adjusted Quota: The result of dividing a states population by a divisor that is not the standard divisor
Replace the standard divisor with the smallest district in the nation ( )
Jefferson uses the adjusted quota of 33,000 because at that time it was the smallest district.
State Populations
Adjusted Quota
= Apportionment for the state
Virginia's populations = 630560
= 19.108
After rounding down to the nearest whole number, Virginia's receives 19 seats.
Category Pop x0.22 x0.225 x0.227

Wins: 18 81.82= 81 80.00 = 80 79.30 = 79

Losses: 4 18.18 = 18 17.78 = 17 17.62 = 17

Ties: 1 4.55 = 4 4.44 = 4 4.41 = 4

Total: 23 103 101 100
Given the prior information about the field hockey team, use the Jefferson method to apportion the problem evenly.
18 wins, 4 losses, 1 tie, an 23 games total.
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