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Apportionment
Ch. 14 Finite Math
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The Apportionment Problem
An apportionment problem is to round a set of
fractions so that their sum is maintained at its
original value. The rounding procedure must not
be an arbitrary one, but one that can be applied
constantly. Any such rounding procedure is
called an apportionment method.
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Terminology
 States:
 House
The categories
size:
In the problem of rounding
percentages the house size is 100
 Populations: The numbers in each category
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Standard Divisor
The quotient of the total population, p, divided by
the house size is called the standard divisor. If h
denotes the house size and s is the standard
divisor, then:
p
s
h

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Quota
In an apportionment problem, the quota is the
exact share that would be allocated if a whole
number were not required. To obtain a state’s
quota, divide its population by the standard
divisor.
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Example: The High School
Mathematics Teacher
 Courses

to be taught correspond to states.
Geometry, Pre-calculus. and Calculus
 The
number of students enrolled in each course
correspond to populations.

52,33, and 15 respectively
 The
house size is the total number of sections to be
scheduled.

One teacher has time for 5 sections
 The

average section will have 20 students
100 students/5 sections=20 This is our standard divisor
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The quota is each section’s
population divided by the
standard divisor
Course
Population
Quota
Rounded
Geometry
52
52/20=2.6
3
Precalculus
33
33/20=1.65
2
Calculus
15
15/20=0.75
1
Totals
100
5
6
The purpose of an apportionment method is to
find an equitable way to round a set of numbers
such as these quotas without increasing or
decreasing the original sum.

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Example- California’s Quota
The Census Bureau recorded the apportionment
population of the US as of April 1st, 2000 to be
281,424,177. There are 435 seats in the House of
Representatives.; therefore, the standard divisor
and California’s quota are:
281,424,177
s
 646,952
435
California’s population
33,930,798
q
 52.447 seats
646,952
California’s apportionment, which must be a whole
number, was set at 53
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14.2 The Hamilton Method
Each state receives either
its lower quota, which is its
quota rounded down, or its
upper quota obtained by
rounding the quota up. The
states that receive their
upper quotas are those
whose quotas have the
largest fractional parts.
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3 steps for the Hamilton Method
1) Calculate each state’s quota
2)
Tentatively assign 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.
3) Allot the remaining seats, one each, to the
states whose quotas have the largest fractional
parts, until the house is filled.
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Example- Math Teacher
Course
Population
Quota
Rounded
Geometry
52
52/20=2.6
3
Precalculus
33
33/20=1.65
2
Calculus
15
15/20=0.75
1
Totals
100
5
6
1) Done above
2) Geo-2/Pre-1/Calc-0
3) Geo-2/Pre-2/Calc-1
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Paradoxes of the Hamilton Method
A paradox is a fact that seems obviously false.
The Alabama Paradox
When a state loses a seat as a result of an
increase in the house size
State
House Size
299
Inc. in Quota
300
Alabama
7.646
7.671
0.025
Illinois
18.640
18.702
0.062
Texas
9.460
9.672
0.032
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The Population Paradox
When one state’s population
increases, and its
apportionment decreases,
while simultaneously
another states population
increases proportionally
les, or decreases, and its
apportionment increases.
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Apportioning Seats in Parliament
Party
First Election
Second Election
Whigs
5,525,381
5,657,564
Tories
3,470,152
3,507,464
Liberals
3,864,226
3,885,693
Centrists
201,203
201,049
Totals
13,060,962
13,251,770
Standard Divisor
130,609.62
132,517.70
The Tories
and Liberals
get the extra
votes:
42,27,30,1
Party
First Election
The Whigs
and Centrists
get the extra
votes:
43,26,29,2
Second Election
Whigs
42.3045
42.6929
Tories
26.5689
26.4679
Liberals
29.5861
29.3221
Centrists
1.5405
1.5171
Totals
100
100
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Practice Problems on page 515
#’s 2 and 4
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14.3 Divisor Methods
A divisor method of apportionment determines
each state’s apportionment by dividing its
population by a common divisor d and rounding
the resulting quotient.
Divisor methods differ in the rule used to round
the quotient.
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The Jefferson Method
The method proposed by
Jefferson to replace
Hamilton’s method.
This method favors the
states with large
populations.
He used an adjusted
quota which is always
rounded down at first.
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The Jefferson Method-examples

A district needed at least a population of 30,000 according to
the constitution.

Hamilton’s method did not account for this so Jefferson used
the 33,000 as his divisor. This was the smallest population.
Virginia’s apportionment:
Population of Virginia  630,560
 
 19.108 19




  33,000 
33,000
Delaware’s apportionment:

Population of Delaware  55,540
 
 1.683 1




 33,000
33,000
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Critical Divisor for the Jefferson
method
A state’s critical divisor is the number that
can be divided into the state’s population to
produce a number just on the borderline
for changing the state’s apportionment.
Let N be the tentative apportionment obtained
as the lower quota from the Hamilton method
for a state X.
Population of X
N 1
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Apportionment of the extra seats
 Since
we always round down, some extra seats will
need to be allotted. Here, the states with the largest
critical divisors get the seats first.
 Only
states with a critical divisor equal to or
greater than the divisor will receive additional
seats.
 When
even more seats must be allotted, the states
who added seats must recalculate their critical
divisor.
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Example: The Field Hockey Team
A field hockey team had 18 wins, 4 losses,
and 1 tie last season. The coach wants to
display this on a poster in terms of
win/loss percentages.
House size is 100%
Total population is 23
games played
This makes the Quotas:
18
4
1
 78.26%
17.39%
 4.35%
23
23
23
The lower quotas are:
78,17,4
This does not add to
100%
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Example: The field hockey team
To decide who gets the extra percent we
must calculate the critical divisors:
18
4
1
 0.22785
 0.22222
 0.2
78 1
17 1
4 1
“Wins” gets the extra percent!
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The Quota Condition
If every state’s apportionment is equal to
either is upper or lower quota.
The Jefferson method does not satisfy this,
but the Hamilton method does.
But, the Hamilton method has the Alabama
paradox to face and the Jefferson method
does not.
No known method avoids all issues…
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The Webster Method
The Webster Method is the
divisor method that rounds the
quota to the nearest whole
number, rounding up when the
fractional part is greater than
or equal to 0.5.
This method also avoids
the paradoxes and is
neutral unlike the
Jefferson method.
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Procedure for the Webster method:
 1)
Determine the standard divisor and use it
to find each state’s quota
 2)
Round up if the fractional part is 0.5 or
above, if not round down to find tentative
apportionment.
 3)
If the rounded quotas equal the house size
you are finished. If not, use the same method
for allotting additional seats as used in the
Jefferson method, but with a different divisor.
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Critical divisors for the Webster
method:
If the rounded numbers fall short of the house
size we calculate the following to allot
additional seats:
V

d 
1
N
2
If the rounded numbers exceed the house size
we calculate the following to remove seats:
N is the tentative
V


d 
apportionment
1
N
V is the state’s population
2
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Practice Problems pg.516 #’s 6,7,8
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14.4 Which method is best?
Representative Share
If A is the apportionment given to a state of
population V, then A/V represents the share
of a congressional seat given to each
citizen. One would hope that this is the
same for all.
The Webster method is the method that
makes the representative share the most
even.
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District Population
The district population of state X is V/A.
This is the average population of a
congressional district in the state
Absolute and Relative Differences
Given two positive numbers A and B, with
A>B, the Absolute difference is A-B and the
relative difference is the quotient:
AB
100%
B
The relative difference in district population is
the same as the relative difference in
representative share. Therefore, either can be

used to compare methods.
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The Hill-Huntington Method
 Another
divisor method with a different way of
rounding related to geometric means.
Geometric Mean
The geometric mean of two positive
numbers A and B is equal to the square root
of their product: A  B
 Find
the Geometric mean of the higher and lower
quota and call it q*

 If
q<q* round down, if not round up
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Critical Divisors for the HillHuntington Method
N is the tentative apportionment, V is the state’s
population.

d 

V

d 
NN  1

V
NN 1
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Practice Problems pg. 516 #’s 12,
13, 14