1. Hamilton’s Method of Apportionment -- A bundle of contradictions??
Examples
Problems

2. 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 appropriately for the given application.
apportionment
The number of rows will depend on the number of “states”.

3. 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.

4. Hamilton’s Method: Vocabulary & Example
*handout
Hamilton’s Method: Vocabulary & Example
SD:
TOTAL:
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
176

5. Hamilton’s Method: Vocabulary & Example
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
176
Use the given information to calculate the TOTAL population
and the “standard divisor”
TOTAL:
17600
SD:
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

6. Hamilton’s Method: Vocabulary & Example
The STANDARD QUOTA
is the exact quotient upon dividing a state’s population by the SD
SD:
100
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
176
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 as the “# of representatives” for a state.

7. Hamilton’s Method: Vocabulary & Example
SD:
100
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
176
The STANDARD QUOTA
is the exact quotient upon dividing a state’s population by the SD
92.3
82.31
1.39
For example…
The SQ for Algebra is: 9230/100 = 92.3
The SQ for Geometry is: 8231/100 = 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 remaining seats.

8. Hamilton’s Method: Vocabulary & Example
92.3
82.31
1.39
SD:
100
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
176
The LOWER QUOTA
is the integer part of the SQ 92 82 1
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.

9. Hamilton’s Method: Vocabulary & Example
If each state was awarded its LQ,
What would the total # of seats apportioned be? 92 82 1
92.3
82.31
1.39
SD:
100
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
176
175
That leaves 1 seat empty!
And we certainly can’t have that!

10. Hamilton’s Method: Vocabulary & Example
The 1 empty seat will be awarded by ranking the decimal portions of the SQ. 92 82 1
92.3
82.31
1.39
SD:
100
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
176
1st
175
Since we only have 1 seat to fill, we need only rank the highest decimal portion.

11. Hamilton’s Method: Vocabulary & Example
So, in this application, the state with the highest decimal portion is awarded its UPPER QUOTA 92 82 1
92.3
82.31
1.39
SD:
100
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
176 92 82
1st 2
175
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!

12. Try the next one on your own!
HAMILTON’S METHOD
Try the next one on your own!

14. State
Population SQ LQ
Rank
Rep A
45,300
45.3 45 B
31,070
31.07 31 C
20,490
20.49 20
2nd 21 D
14,160
14.16 14 E
10,260
10.26 10 F
8720
8.72 8
1st 9
TOTAL
130,000
# Seats
130
128
2 extra seats SD
1000

15. HAMILTON’S METHOD
The ALABAMA PARADOX

16. 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.

17. Hamilton’s Method seems simple enough...
APPARENT CONTRADICTIONS...
So, with populations unchanged, a re-apportionment is done.
What results would make sense???
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
Some state would gain a seat, while all others remain the same.

18. Hamilton’s Method seems simple enough...
APPARENT CONTRADICTIONS...
A contradiction to the logical solution of a problem is called a PARADOX.
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 re-apportionment, the state of Alabama actually LOST a seat, even though its population had not been recalculated!1
1. For All Practical Purposes, 4th ed., COMAP, W.H. Freeman, 1997

19. Hamilton’s Method: ALABAMA PARADOX example
*handout
SD:
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
177
Reapportionment of Mathland representative body
with change in # seats.

20. Hamilton’s Method: ALABAMA PARADOX example
SD:
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
177
92.82 92
1st 93
82.77 82
2nd 83
1.40 1 1
Used to have 2 reps!
175
177
99.44
Remember:
SD = Total population/ #seats
SQ = state population/ SD
Values were rounded to fit in the chart after determining that it would not change the rankings.

21. Hamilton’s Method: ALABAMA PARADOX example
SD:
TOTAL:
17600
State
population SQ LQ
Rank
#reps
Algebra
9230
Geometry
8231
Trig
139
#seats
177
92.82 92
1st 93
82.77 82
2nd 83
1.40 1
99.44
175
Used to 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

22. The POPULATION PARADOX
HAMILTON’S METHOD
The POPULATION PARADOX

23. 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...

24. Hamilton’s Method: Population Paradox
*handout#1
Hamilton’s Method: Population Paradox
total#sect.: 25
Course
Proj. Enrol. SQ LQ
Rank
#sect.
Math A
400
Math B 90
Math C
225
Math D
200
TOTAL:
SD:
Mathland U. Graduate course apportionment
for Fall Semester

25. Hamilton’s Method: Population Paradox
*solution#1
Hamilton’s Method: Population Paradox
Course
Proj. Enrol. SQ LQ
Rank
#sect.
Math A
400
10.929 10
1st 11
Math B 90
2.459 2 2
Math C
225
6.148 6 6
Math D
200
5.464 5
2nd 6
TOTAL:
915
total#sect.: 25 23 25
SD:
36.6
Mathland U. Graduate course apportionment
for Fall Semester

26. 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...

27. Hamilton’s Method: Population Paradox
*handout#2
Hamilton’s Method: Population Paradox
total#sect.: 25
Course
Proj. Enrol. SQ LQ
Rank
#sect.
Math A
400
Math B 99
Math C
225
Math D
211
TOTAL:
SD:
Mathland U. Graduate course apportionment
for Spring Semester

28. Hamilton’s Method: Population Paradox
*solution#2
Hamilton’s Method: Population Paradox
Course
Proj. Enrol. SQ LQ
Rank
#sect.
Math A
400
10.695 10
1st 11
Math B 99
2.647 2
2nd 3
Math C
225
6.106 6 6
Math D
211
5.642 5 5
Gained the most students, but lost a section!
TOTAL:
935
total#sect.: 25 23 25
SD:
37.4
Mathland U. Graduate course apportionment
for SPRING Semester

29. Hamilton’s Method: Population Paradox
*solution#2
Hamilton’s Method: Population Paradox
Course
Proj. Enrol. SQ LQ
Rank
#sect.
Math A
400
10.695 10
1st 11
Math B 99
2.647 2
2nd 3
Math C
225
6.106 6 6
Math D
211
5.642 5 5
TOTAL:
935
total#sect.: 25 23 25
SD:
37.4
The POPULATION PARADOX occurs when:
The “population” of a “state” increases, but it loses a seat upon re-apportionment.

30. The NEW-STATES PARADOX
HAMILTON’S METHOD
The NEW-STATES PARADOX

31. 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...

32. Hamilton’s Method: New-States Paradox
*handout#3
Hamilton’s Method: New-States Paradox
Math E
120
Course
Proj. Enrol. SQ LQ
Rank
#sect.
Math A
400
Math B 90
Math C
225
Math D
200
total#sect.: 28
TOTAL:
SD:
Mathland U. Graduate course apportionment
for New Fall Semester

33. Hamilton’s Method: New-States Paradox
*solution#3
Hamilton’s Method: New-States Paradox
Math E
120
Course
Proj. Enrol. SQ LQ
Rank
#sect.
Math A
400
Math B 90
Math C
225
Math D
200
total#sect.: 28
TOTAL:
SD:
10.821 10
1st 11
Used to have 2 sections
2.435 2
2nd 3
6.087 6 6
Used to have 6 sections
5.411 5 5
3.246 3 3
1035 26 28
36.964
Mathland U. Graduate course apportionment
for New Fall Semester

34. 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 re-apportionment
and/or another original state gains a seat upon re-apportionment.

35. 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.