1. Chapter 15: Apportionment
Part 2: The Hamilton Method

2. 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 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.

3. 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: or

4. The Hamilton Method
Next, we define the term quota as follows: or
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.

5. The Hamilton Method
By definition,
and
That is,
and
Therefore,
Conclusion:

6. The Hamilton Method
Notice that because of the conclusion
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
The question then becomes, how do we deal with the fraction of a seat?

7. The Hamilton Method
For the Hamilton method we will also need the following definitions:
The lower quota, which is denoted by , is the value that results from rounding down the quota to the next lowest integer
The upper quota, which is denoted by , is the value that results from rounding up the quota to the next highest integer.
If q is an integer, then = = q . That is, no rounding is done.

8. The Hamilton Method
The Hamilton Method of apportionment is as follows:
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.

9. 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

10. 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

11. 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
Lower quota
Add remaining seat
State A
275
275/63
4.37 4
State B
383
383/63
6.08 6
State C
465
465/63
7.38 7 +1
State D
767
767/63
12.17 12
Total
1890 30 29

12. 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
Lower quota
Add remaining seat
State A
275
275/63
4.37 4
State B
383
383/63
6.08 6
State C
465
465/63
7.38 7 +1
State D
767
767/63
12.17 12
Total
1890 30 29
State C has the
largest fractional
part

13. 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
Lower quota
Add remaining seat
Final apportionment
State A
275
275/63
4.37 4
State B
383
383/63
6.08 6
State C
465
465/63
7.38 7 +1 8
State D
767
767/63
12.17 12
Total
1890 30 29

14. 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, pre-calculus 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: = 100 students. These 100 students are divided among 3 subjects. We are also dividing the 5 classes among the 3 subjects.

15. 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, pre-calc 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.

16. 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

17. 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 = 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.

18. 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 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

19. 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.

20. 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.