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Hamilton’s Method of Apportionment -A bundle of contradictions??
 Hamilton’s Method
 Examples
 Problems
1
Hamilton’s Method:
Example #1
 We will begin these applications by creating a chart
with 6 columns
 Label the columns as seen here...
State
population
SQ
LQ
Rank
Apmt
The ORANGE type
indicates that these
columns will be labeled
apportionment
appropriately for the
given application.
 The number of rows will depend on the number
of “states”.
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Hamilton’s Method:
Vocabulary & Example
APPLICATION: Mathland is a small country
that consists of 3 states; Algebra,
Geometry, and Trigonometry. The
populations of the 3 states will be given in
the chart.
Suppose that one year, their government’s
representative body allows 176 seats to be
filled.
The number of seats awarded to each state
will be determined using Hamilton’s Method
of apportionment.
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Hamilton’s Method:
State
Algebra
Geometry
Trig
population
9230
8231
139
TOTAL:
#seats
SD:
176
*handout
Vocabulary & Example
SQ
LQ
Rank
#reps
4
Hamilton’s Method:
State
Use the
Algebra
given
information Geometry
to calculate
Trig
the TOTAL
population
TOTAL:
and the
#seats
“standard
divisor”
SD:
Vocabulary & Example
population
9230
8231
SQ
LQ Rank
#reps
139
17600
176
100
 The Standard Divisor is the “number of people per
representative” [calculated by dividing TOTAL by #seats]
 Here, the SD means that for every 100 people a state will
receive 1 representative
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Hamilton’s Method: Vocabulary & Example
The
State
STANDARD
Algebra
QUOTA
Geometry
is the exact
Trig
quotient
upon
dividing a
TOTAL:
state’s
#seats
population
SD:
by the SD
population
9230
8231
139
SQ
LQ Rank
#reps
17600
176
100
 The SQ is the exact # of seats that a state would be
allowed if fractional seats could be awarded.
 Notice that the “# of seats awarded” can also be thought of
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as the “# of representatives” for a state.
Hamilton’s Method: Vocabulary & Example
The
STANDARD
QUOTA
is the exact
quotient upon
dividing a
state’s
population by
the SD
State
Algebra
Geometry
Trig
population
9230
8231
139
TOTAL:
#seats
17600
176
SD:
100
SQ
LQ Rank
#reps
92.3
82.31
1.39
 For example…

The SQ for Algebra is: 9230/100 = 92.3

The SQ for Geometry is: 8231/100 = 82.31 etc…
 You will note that, very often, the values upon division must be rounded.
 There is a danger in rounding to too few decimal places, as the decimal
values will be used to determine which states are awarded the
7
remaining seats.
Hamilton’s Method: Vocabulary & Example
State
The LOWER Algebra
QUOTA
Geometry
is the integer
Trig
part of the
SQ
TOTAL:
#seats
SD:
population
9230
8231
139
SQ
92.3
82.31
1.39
LQ Rank
92
82
1
#reps
17600
176
100
 For example…
 The LQ for Algebra is: 92 etc…
 It may be thought of as “rounding down” to the lower of the
two integers the SQ lies between.
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Hamilton’s Method: Vocabulary & Example
If each state
State
was
Algebra
awarded its
Geometry
LQ,
Trig
What would
the total # of
seats
TOTAL:
apportioned
#seats
be?
SD:
population
9230
8231
139
17600
176
SQ
92.3
82.31
1.39
LQ Rank
92
82
1
#reps
175
100
 That leaves 1 seat empty!
 And we certainly can’t have that!
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Hamilton’s Method: Vocabulary & Example
The 1 empty
seat will be
awarded by
ranking the
decimal
portions of
the SQ.
State
Algebra
Geometry
Trig
population
9230
8231
139
TOTAL:
#seats
17600
176
SD:
100
SQ
92.3
82.31
1.39
LQ Rank
92
82
1
1st
#reps
175
 Since we only have 1 seat to fill, we need only rank the
highest decimal portion.
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Hamilton’s Method: Vocabulary & Example
So, in this
application,
the state
with the
highest
decimal
portion is
awarded its
UPPER
QUOTA
State
Algebra
Geometry
Trig
population
9230
8231
139
TOTAL:
#seats
17600
176
SD:
100
SQ
92.3
82.31
1.39
LQ Rank
92
82
1
1st
175
#reps
92
82
2
176
 The Upper Quota is the first integer that is higher that the
SQ.
 It may be thought of as “rounding up” to the higher of the
two integers the SQ lies between.
 And now the TOTAL # reps is equal to the allowable #seats!
11
HAMILTON’S METHOD
Try the next one on
your own!
12
13
State
Population
SQ
LQ
Rank
A
45,300
45.3
45
45
B
31,070
31.07
31
31
C
20,490
20.49
20
D
14,160
14.16
14
14
E
10,260
10.26
10
10
F
8720
8.72
8
1st
TOTAL
130,000
# Seats
130
128
2 extra seats
SD
1000
2nd
Rep
21
9
14
HAMILTON’S METHOD
The ALABAMA
PARADOX
15
Hamilton’s Method seems simple enough...
So, what’s the problem?
 The only confusion might occur if two decimal values
are the same and both can not be ranked.
 In that case, usually the state with the higher integer
value would be awarded the extra seat..
 Problems may occur when Hamilton’s Method is
used repeatedly in an application which has certain
changes (in # of seats, populations, and/or # of
states) over a period of time.
16
Hamilton’s Method seems simple enough...
APPARENT CONTRADICTIONS...
 Think about this…
 As our country was
continuing to grow in
the 18th & 19th
centuries, the number
of SEATS in the USHR
continued to increase.
 Suppose that one
election year, though a
new census had not
been taken, Congress
decided to ADD a seat
in the HR
 So, with populations
unchanged, a reapportionment is done.
 What results would
make sense???
Some state would gain
a seat, while all others
remain the same.
17
Hamilton’s Method seems simple enough...
APPARENT CONTRADICTIONS...
 Of course, that would
make perfect sense.
 And much of the time,
that’s what occurs…
 However, in 1881, this
very thing happened
 But... upon reapportionment, the
state of Alabama
actually LOST a seat,
even though its
population had not been
recalculated!1
 A contradiction to the
logical solution of a
problem is called a
PARADOX.
1. For All Practical Purposes, 4th ed.,
COMAP, W.H. Freeman, 1997
18
Hamilton’s Method: ALABAMA PARADOX example
*handout
State
Algebra
population
9230
Geometry
Trig
8231
139
TOTAL:
#seats
SD:
17600
177
SQ
LQ Rank
#reps
Reapportionment of Mathland representative body
with change in # seats.
19
Hamilton’s Method: ALABAMA PARADOX example
State
Algebra
population
9230
Geometry
Trig
8231
139
TOTAL:
#seats
SD:
17600
177
99.44
 Remember:


SD = Total population/ #seats
SQ = state population/ SD
SQ
92.82
82.77
1.40
LQ Rank
92
1st
82
1
175
2nd
#reps
93
83
1
177 to
Used
have 2
reps!
Values were rounded to fit
in the chart after
determining that it would
not change the rankings. 20
Hamilton’s Method: ALABAMA PARADOX example
State
Algebra
Geometry
population
9230
8231
SQ
92.82
Trig
139
1.40
TOTAL:
#seats
17600
177
SD:
99.44
82.77
LQ Rank
92
1st
82 2nd
1
175
#reps
93
83
1
Used to
177
have 2
reps!
 The ALABAMA PARADOX occurs when:
 The number of seats is increased, and,
 although there has been no re-calculation of
populations,
 a state loses a seat upon re-apportionment
21
HAMILTON’S METHOD
The POPULATION
PARADOX
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Hamilton’s Method
POPULATION PARADOX EXAMPLE...
 EXAMPLE part 1: The Mathematics Department at
Mathland U. plans to offer 4 graduate courses during
the Fall semester.
 They are allowed to schedule 25 class sections,
 and wish to apportion these sections using Hamilton’s
Method.
 The projected enrollments are given in the
chart on the following slide...
23
Hamilton’s Method: Population Paradox
Course
Math A
Math B
Proj. Enrol.
400
90
Math C
Math D
225
200
SQ
LQ Rank
*handout#1
#sect.
TOTAL:
total#sect.:
25
SD:
Mathland U. Graduate course apportionment
for Fall Semester
24
Hamilton’s Method: Population Paradox
Course
Math A
Math B
Proj. Enrol.
400
90
Math C
Math D
225
200
TOTAL:
915
25
36.6
total#sect.:
SD:
SQ
LQ Rank
10.929 10
1st
2.459
2
6.148
5.464
6
5
23
2nd
*solution#1
#sect.
11
2
6
6
25
Mathland U. Graduate course apportionment
for Fall Semester
25
Hamilton’s Method
POPULATION PARADOX EXAMPLE...
 EXAMPLE part 2: The Mathematics Department at
Mathland U. also plans to offer 4 graduate courses
during the Spring semester.
 Once again, they are allowed to schedule 25 class
sections,
 and wish to apportion these sections using Hamilton’s
Method.
 The Spring semester projected enrollments
are given in the chart on the following slide...
26
Hamilton’s Method: Population Paradox
Course
Math A
Math B
Proj. Enrol.
400
99
Math C
Math D
225
211
SQ
LQ Rank
*handout#2
#sect.
TOTAL:
total#sect.:
25
SD:
Mathland U. Graduate course apportionment
for Spring Semester
27
Hamilton’s Method: Population Paradox
Course
Math A
Math B
Proj. Enrol.
400
99
Math C
Math D
225
211
TOTAL:
935
25
37.4
total#sect.:
SQ
LQ Rank
10.695 10
1st
2.647
2
2nd
6.106
5.642
6
5
*solution#2
#sect.
11
3
6
5
23
25
Gained
SD:
the most
students,
Mathland U. Graduate course apportionment
but lost
for SPRING Semester
a
section!
28
Hamilton’s Method: Population Paradox
Course
Math A
Math B
Proj. Enrol.
400
99
Math C
Math D
225
211
TOTAL:
935
25
37.4
total#sect.:
SD:
SQ
LQ Rank
10.695 10
1st
2.647
2
2nd
6.106
5.642
*solution#2
#sect.
11
3
6
5
6
5
23
25
The POPULATION PARADOX occurs when:
The “population” of a “state” increases, but it loses a
seat upon re-apportionment.
29
HAMILTON’S METHOD
The NEW-STATES
PARADOX
30
Hamilton’s Method
NEW-STATES PARADOX EXAMPLE...
 EXAMPLE part 3: During the Fall semester following
the one mentioned in EXAMPLE part 1, the
Mathematics Department at Mathland U. plans to offer
5 graduate courses.
 They decide to schedule 28 class sections, due to
the addition of the new course,
 and, once again, wish to apportion these sections
using Hamilton’s Method.
 The projected enrollments for the 5 classes
are given in the chart on the following slide...
31
Hamilton’s Method: New-States Paradox
Course
Math A
Math B
Proj. Enrol.
400
90
Math C
Math D
Math E
225
200
120
SQ
LQ Rank
*handout#3
#sect.
TOTAL:
total#sect.:
28
SD:
Mathland U. Graduate course apportionment
for New Fall Semester
32
Hamilton’s Method: New-States Paradox
Course
Math A
Math B
Proj. Enrol.
400
90
Math C
Math D
Math E
225
200
120
TOTAL:
1035
28
36.964
total#sect.:
SD:
SQ
LQ Rank
10.821 10
1st
2.435
2
2nd
6.087
5.411
3.246
*solution#3
#sect.
11
3
6
5
3
6
5
3
26
28
Used to
have 2
sections
Used to
have 6
sections
Mathland U. Graduate course apportionment
for New Fall Semester
33
Hamilton’s Method: New-States Paradox
The NEW-STATES PARADOX occurs when:
The total “population” changes due to the addition
of a new “state”
The “# of seats” changes appropriately using the
old apportionment
And, though the populations of the original
“states” remain unchanged,
An original state loses a “seat” upon reapportionment
and/or another original state gains a seat upon reapportionment.
34
Hamilton’s Method… then what?
 Remember that Hamilton’s Method of
apportionment is very useful and easy.
 Even though these paradoxes can occur, they
happen infrequently and do not invalidate the
method.
 Next class we will explore two other methods of
apportionment that were proposed.
35