Chapter 15: Apportionment
Part 2: The Hamilton Method
The Hamilton Method
• The Hamilton Method, one of several methods of apportionment we
study, is named after Alexander Hamilton. It was first used to decide
the initial apportionment of the seats in the House of
Representatives in 1790. That apportionment was vetoed by
George Washington and the House was reapportioned that year
according to the Jefferson Method. The Hamilton Method did come
back into use in 1850 and was then used until 1900.
• To explain this and other methods, we must define certain terms…
• There are problems of apportionment other than those involving the
House of Representatives but because apportionment of the House
has been so historically significant to the development of these
methods in general, the terms we use reflect that connection.
The Hamilton Method
• Suppose the total population of all states is p and that are there are
h seats in the house (we call h the house size). That is, house size
is the total number of seats available.
• We define the standard divisor, s, as follows:
total population
standard divisor 
house size
or
p
s
h
The Hamilton Method
• Next, we define the term quota as follows:
population of state i
quota 
standard divisor
or
pi
q
s
• Thus, each state could have a different quota.
• In the case of apportioning seats in the House of Representatives,
the quota is the number of seats that a state would get if they could
have a fractional part of a seat. That is, it is a fractional part of the
whole before being rounded to an integer.
The Hamilton Method
• By definition,
population of state i
quota 
standard divisor
and
total population
standard divisor 
.
house size
That is,
pi
q
s
and
p
s .
h
pi
pi h pi
q
    h.
 p 1 p p
 
h
Therefore,
Conclusion:
pi
q  h
p
The Hamilton Method
Notice that because of the conclusion
pi
q  h
p
we can calculate a state’s quota without actually using the standard divisor.
This is what was done in a previous example…
Suppose house size is 200, Alabama has a population of 940 and the total
population is 20,000.
Then h = 200, pi=940 and p = 20,000. Thus, to find Alabama’s quota
q
940
200  0.047200  9.4 seats
20,000
The question then becomes, how do we deal with the fraction of a seat?
The Hamilton Method
• For the Hamilton method we will also need the following definitions:
 
• The lower quota, which is denoted by q , is the value that results
from rounding down the quota to the next lowest integer
 
• The upper quota, which is denoted by q , is the value that results
from rounding up the quota to the next highest integer.
• If q is an integer, then
done.
q = q = q .
That is, no rounding is
The Hamilton Method
•
The Hamilton Method of apportionment is as follows:
1.
2.
3.
•
Calculate each state’s quota.
Temporarily assign each state it’s lower quota.
Starting with state’s having the largest fractional part in their
original quota, distribute any remaining seats, in order from
largest to smallest, until all remaining seats are distributed.
The Hamilton Method does not specify what to do in the case of a
tie – if two states had the same fractional part in their quota - but
this is unlikely to occur.
Example #1 - The Hamilton Method
•
Consider a country with 4 states and 30 seats in a congress and
populations distributed as in the table below.
Population
State A
275
State B
383
State C
465
State D
767
Total
1890
Example #1 - The Hamilton Method
•
Consider a country with 4 states and 30 seats in a congress and
populations distributed as in the table below.
Population
State A
275
State B
383
State C
465
State D
767
Total
1890
We find the standard divisor
to be 1890/30 = 63
Example #1 - The Hamilton Method
•
Consider a country with 4 states and 30 seats in a congress and
populations distributed as in the table below.
Population
Quota
Quota
Lower
quota
State A
275
275/63
4.37
4
State B
383
383/63
6.08
6
State C
465
465/63
7.38
7
State D
767
767/63
12.17
12
Total
1890
30
29
Add
remaining
seat
+1
Example #1 - The Hamilton Method
•
Consider a country with 4 states and 30 seats in a congress and
populations distributed as in the table below.
State A
Population
Quota
Quota
Lower
quota
275
275/63
4.37
4
State B
383
383/63
6.08
6
State C
465
465/63
7.38
7
State D
767
767/63
12.17
12
Total
1890
30
29
Add
remaining
seat
State C has the
largest fractional
part
+1
Example #1 - The Hamilton Method
•
Consider a country with 4 states and 30 seats in a congress and
populations distributed as in the table below.
Population
Quota
Quota
Lower
quota
Add
remaining
seat
State A
275
275/63
4.37
4
4
State B
383
383/63
6.08
6
6
State C
465
465/63
7.38
7
State D
767
767/63
12.17
12
12
Total
1890
30
29
30
+1
Final
apportionment
8
Example 2: The Hamilton Method
•
Let’s use the Hamilton Method to apportion students into classes.
•
Suppose a high-school math teacher can teach a maximum of five
classes in one day. Suppose that she teaches geometry, precalculus and calculus. Suppose that 52 students registered for
geometry, 33 for pre-calc and 15 for calculus.
•
How many of each class should the teacher be assigned?
•
We have a total population of students: 52+33+15 = 100 students.
These 100 students are divided among 3 subjects. We are also
dividing the 5 classes among the 3 subjects.
Example 2: The Hamilton Method
•
We can look at these apportionment problems in terms of groups
and items. There are several groups, like states which have
different populations and there are items, like seats to be divided
among those groups, based on the populations.
•
Here we have 3 groups, in this case the subjects – geometry, precalc and calculus and 5 items, the classes of each subject to be
taught. How do we distribute the available 5 classes among those
groups?
•
Notice how the groups are weighted differently by their populations
– in this case by their enrollments. We plan to distribute the items
among the groups based on their relative size.
Example 2: The Hamilton Method
•
Because we have 100 students and 5 classes we can see the
teacher will expect about 20 students per class. This is the
standard divisor. The total population from all of the groups is 100
and the number of items is 5. Thus (total population)/(number
items) = 100/5 =20. Of course, in other examples, this can be
written (total population)/(house size)
•
Now, each group (each subject) will get how many of the classes
(items)? The answer is given by the quota. In general,
quota = (group population)/(standard divisor)
•
•
•
Geometry: quota = 52/20 = 2.6 classes
Pre-Calc: quota = 33/20 = 1.65 classes
Calculus: quota = 15/20 = 0.75 classes
Example 2: The Hamilton Method
•
We have
Geometry: quota = 52/20 = 2.6 classes
Pre-Calc: quota = 33/20 = 1.65 classes
Calculus: quota = 15/20 = 0.75 classes
•
Notice that 2.6 + 1.65 + 0.75 = 5.
•
Notice that we are trying to divide up the whole total number of
items (available classes) among groups (different subjects) but with
different groups getting more or less than others. That is, the
division is not equal and the problem of apportionment is to round
the fractional parts in a way to maintain a sum equal to the whole.
•
In other words, obviously we can not have 2.6 geometry classes or
0.75 calculus classes.
Example 2: The Hamilton Method
•
We have
Geometry: quota = 52/20 = 2.6 classes
Pre-Calc: quota = 33/20 = 1.65 classes
Calculus: quota = 15/20 = 0.75 classes
•
The Hamilton method is one way to round the values of the sum
2.6 + 1.65 + .75 so that the terms are integer but the sum is the
same.
•
By the Hamilton method we’ll first round every quota to the lower
quota…
Geometry = 2 classes
Pre-Calc = 1 class
Calculus = 0 classes
Example 2: The Hamilton Method
•
We have
Geometry: quota = 52/20 = 2.6 classes -> 2 classes
Pre-Calc: quota = 33/20 = 1.65 classes -> 1 class
Calculus: quota = 15/20 = 0.75 classes -> 0 classes
•
When we round down to the lower quota, we will add any
remaining items to the groups with the largest fractional part.
•
Those are assigned in order from largest fractional part to smallest
until all classes are assigned.
•
In this case, we assign 1 class to calculus and 1 to pre-calc.
Example 2: The Hamilton Method
•
We have
Geometry: quota = 52/20 = 2.6 classes -> 2 classes
Pre-Calc: quota = 33/20 = 1.65 classes -> 1 class + 1
Calculus: quota = 15/20 = 0.75 classes -> 0 classes + 1
•
•
The final apportionment is:
–
Geometry: 2 classes
–
Pre-Calc: 2 classes
–
Calculus: 1 class
Notice how the average is 100/5 = 20 students per class but it may occur
that no class has 20 students. For example, the calculus class will have
15 students and geometry (with 52 total students) could have 26 students
each.