1. Chapter 14: Apportionment Lesson Plan
The Apportionment Problem
The Hamilton Method
Divisor Methods
The Jefferson Method
The Webster Method
The Hill-Huntington Method
Which Divisor Method Is Best? 1

2. The Apportionment Problem
We are dividing and assigning things.
We are doing this on a proportional basis and in a planned, organized fashion. 2

3. The Apportionment Problem
Example: Percentage of season’s results of a field hockey team recorded and rounded to whole numbers.
HS Field Hockey Team: 2004–2005 Season
Percentage
Round
Games won 18
78.26%
78%
Games lost 4
17.39%
17%
Games tied 1
4.35% 4%
Games played 23
100.00%
99% 3

4. The Apportionment Problem
The sum does not reach 100%, and this becomes an apportionment problem.
The coach may want to give the extra percent to the games won—“to make it look good for the team.”
This would be biased. 4

5. An Apportionment Problem
In the U.S. Constitution, the seats of the House of Representatives shall be apportioned according to the state’s population.
An apportionment problem occurred in 1790: 5

6. An Apportionment Problem
The House of Representatives was to have 105 members.
The country’s total population of the 15 states was 3,615,920.
The average congressional district should have a population of 3,615,920 ÷ 105 = 34,437
(standard divisor). 6

7. The Apportionment Problem
To find the fair share (Quota) of House seats, we can divide each state’s population by this average, or standard divisor.
Virginia = 630,560/34,437 = and was only given 18 seats.
Virginia’s Quota was 18 7

8. The Apportionment Problem
President Washington came from Virginia.
Furthermore, Virginia’s apportionment was not rounded up to 19, and he may have been biased.
Hence, he did not agree this method was fair to all states.
When Congress passed the apportionment proposed bill and sent it to President Washington, he vetoed it! 8

9. First step: find Standard Divisor
Find the total population, p, and divide it by the house size, h (or M). p h
The standard divisor is calculated first, then you will be able to find the fair amount of shares to give to each class or house.
s = 9

10. Find the Quota To find the state, or class, quota, qi :
Divide its population, pi, by the standard divisor, s pi s
The quota is the exact share that would be allocated if a whole number were not required.
qi = 10

11. The Hamilton Method
An apportionment method was developed by Alexander Hamilton, also called the Method of Largest Fractions.
He realized that when rounding each quota to the nearest whole number, sometimes the apportionments do not add up correctly. 11

12. The Hamilton Method
Originally, he wrote the congressional apportionment bill that was vetoed by President Washington in 1792, but later adopted from 1850–1900.
Alexander Hamilton 12

13. The Hamilton Method
After the quotas are calculated, round down to the lower quota L. (Simply truncate (chop off) the decimal portion.)
Start with the state with the largest fraction (largest decimal portion), and round up this state’s quota to its upper quota U. Keep assigning the states their upper quotas in order of largest fractions until the total seats in the house is reached.
Alexander Hamilton 13

14. General Steps for All Methods:
First add up the total population of each state, p = ∑pi .
Calculate the standard divisor, s = p /h (total population ÷ house).
Calculate each state’s quota (or class), qi = pi /s. 14

15. Steps for Hamilton Method:
1. Tentatively assign to each state its lower quota L.
2. Give the upper quota U to the states whose quota has the largest fractional (decimal) portion until the house is filled. 15

16. Example Hamilton Method:
Three states: A, B, C
A has population of 37,000
B has population of 22,000
C has population of 61,000
How would 12 seats be allocated using Hamilton Method? 16

17. Example Hamilton Method:
Standard Divisor:
S = 37, , ,000 12
s = 10,000 17

18. Example Hamilton Method:
Standard Quotas:
qA = 37,000 = 3.7
10,000
qB = 22,000 = 2.2
qC = 61,000 = 6.1 18

19. Example Hamilton Method:
qA = A gets 3 seats
qB = B gets 2 seats
qC = C gets 6 seats
There is one seat still available.
A has the largest decimal of .7, so A gets one more seat.
Final allocation:
A gets 4 seats
B gets 2 seats
C gets 6 seats 19

20. Lets look at an example with 250 seats available
State
Population A
1,646,000 B
6,936,000 C
154,000 D
2,091,000 E
685,000 F
988,000
Total
12,500,000
Standard Divisor:
( ) / = 50,000

21. State
Population
Step 1 Quota
Lower Quota A
1,646,000
50000
32.92 32 B
6,936,000
138.72
138 C
154,000
3.08 3 D
2,091,000
41.82 41 E
685,000
13.70 13 F
988,000
19.76 19

22. State
Population
Step 1 Quota
Lower Quota A
1,646,000
32.92 32 B
6,936,000
138.72
138 C
154,000
3.08 3 D
2,091,000
41.82 41 E
685,000
13.70 13 F
988,000
19.76 19
Total
250.00
246

23. This example there are 4 extra seats
What happens if there are more than one extra seats?
This example there are 4 extra seats
State
Step 1 Quota
Lower Quota A
32.92 32 B
138.72
138 C
3.08 3 D
41.82 41 E
13.70 13 F
19.76 19
Total
250.00
246

24. State
Step 1
Quota
Lower
Rank
Fractional
parts
Hamilton
method A
32.92 32
1st 33 B
138.72
138
4th C
3.08 3 D
41.82 41
2nd 42 E
13.70 13 F
19.76 19
3rd 20
Total
250.00
246
250

25. Alabama Paradox
Discovered in 1881, a study was done on several apportionments for different house sizes, and the following was discovered: 25

26. The Hamilton Method can create this paradox
Alabama Paradox
Occurs when a state loses a seat as the result of an increase in the house size.
The Hamilton Method can create this paradox 26

27. Alabama Paradox
With a 299-seat house, three state quotas and apportionments were:
Alabama q =  8,
Illinois q =  18,
Texas q =  9
With a 300-seat house, TX and IL gained seats, Alabama lost a seat , but the house added a seat:
Alabama q =  7,
Illinois q =  19,
Texas q =  10 27

28. The Population Paradox
Occurs when one state’s population increases and its apportionment decreases, while simultaneously another state’s population decreases, and its apportionment increases.
Another way of a stating it: when state A loses a seat to state B even though the population of A grew at a higher rate than the population of B.
The Hamilton Method can create this paradox also. 28

29. 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.
Thomas Jefferson 29

30. The Hill-Huntington Method
Divisor Methods
The Jefferson Method
The Webster Method
The Hill-Huntington Method
Thomas Jefferson
Daniel Webster
Edward V. Huntington 30

31. The Jefferson Method
Proposed by Thomas Jefferson to replace the Hamilton Method. 31

32. Steps for the Jefferson Method:
1. First add up the total population, p, of all the states, ∑pi .
2. Calculate the standard divisor,
s = p /h (total population ÷ house) .
3. Calculate each state’s quota,
qi = pi /s. 32

33. Steps for the Jefferson Method:
4. Use the lower quota L for a tentative apportionment.
5. Calculate the critical divisor for each state, di = pi /(L + 1).
The state with the largest critical divisor will get the next seat. 33

34. Steps for the Jefferson Method:
6. Re-compute the apportionments pi / d using this largest critical divisor, d.
7. List the new lower apportionments L for all states.
8. Continue this process until the house is filled. 34

35. Example Jefferson Method:
Three states: A, B, C
A has population of 37,000
B has population of 22,000
C has population of 61,000
How would 12 seats be allocated using Jefferson Method? 35

36. Example Jefferson Method:
Standard Divisor:
S = 37, , ,000 12
s = 10,000 36

37. Example Jefferson Method:
Standard Quotas:
qA = 37,000 = A gets 3 seats
10,000
qB = 22,000 = B gets 2 seats
qC = 61,000 = C gets 6 seats 37

38. Example Jefferson Method:
Compute critical divisor:
dA = 37,000 = 9250 4
dB = 22,000 = 7333 3
dC = 61,000 = 8714 7
Since A’s critical divisor is largest, it receives extra seat. 38

39. Example Jefferson Method:
A gets 4 seats
B gets 2 seats
C gets 6 seats 39

40. Jefferson Method:
Bad News- Jefferson’s method can produce upper-quota violations!
To make matters worse, the upper-quota violations tend to consistently favor the larger states. 40

41. The Webster Method
A divisor method invented by Representative Daniel Webster.
The Webster method is the divisor method that rounds the quota to the nearest whole number. Rounding fraction the usual way
Round up when greater than or equal to ½ .
Round down when less than ½ . 41

42. The Webster Method
The Webster method minimizes differences of representative shares between states
(it does not favor the large states, as does Jefferson method). 42

43. Steps for the Webster Method:
First add up the total population, p, of all the states ∑pi .
2.Calculate the standard divisor,
s = p /h (total population ÷ house).
3.Calculate each state’s (or class) quota, qi = pi /s. 43

44. Steps for the Webster Method:
4. Round all quotas the usual way
5. If the sum of the quotas is enough to fill the house, we are finished. 44

45. Steps for the Webster Method:
6. If not…, we adjust the quotas
Adjusting the divisor:
If the quotas do not fill the house, the state which has the largest critical divisor where di = pi /(qi + ½) will receive a seat.
If the quotas overfill the house, the state which has the smallest critical divisor where di = pi /(qi − ½) will lose a seat. 45

46. Steps for the Webster Method:
7. Re-compute the quotas
qi = pi / d using this critical divisor, d.
8. Continue this process until the house is filled. 46

47. Example Webster Method:
Standard Quotas:
qA = 32,000 = A gets 3 seats
10,000
qB = 24,000 = B gets 2 seats
qC = 64,000 = C gets 6 seats 47

48. Example Webster Method:
The quotas do not fill the house, the state which has the smallest critical divisor where di = pi /(qi + ½) will receive a seat.
dA = 32,000 = A gets 4 seats
3.5
dB = 24,000 = B gets 2 seats
2.5
dC = 64,000 = C gets 6 seats
6.5 48

49. Which Divisor Method Is Best? (3 comparisons)
To find which of the two divisor methods is the fairest, compare the apportionments produced by the methods.
There are 3 measures of fairness. 49

50. Which Divisor Method Is Best? (3 comparisons)
If p is the state’s population and A is its apportionment.
Measures of Fairness:
1. Representative Share, A/p
2. District Population, p /A
3.Absolute and Relative Differences 50

51. 1. Representative Share, A/V
Represents the share of a congressional seat given to each citizen of the state.
In an ideal apportionment, every state would have the same representative share. Instead, we look for the least discrepancy.
The Webster method is the fairest when comparing representative shares. 51

52. 2. District Population, V /A
District population is the average population of a congressional district in the state.
This method surprisingly does not agree with Webster being the fairest. 52

53. 3. Absolute and Relative Differences
Given two positive numbers A and B, with A > B,
the absolute difference is A − B
the relative difference is the quotient:
(A − B)/B × 100%. 53

54. 3. Absolute and Relative Differences
For any two states, the relative difference in district population is equal to the relative difference in representative share. 54

55. 3. Absolute and Relative Differences
Therefore, an apportionment method that minimizes relative difference in representative shares will also minimize the relative difference in district population. 55

56. The Hill-Huntington Method
The Hill-Huntington method is a divisor method related to the geometric mean that has been used to apportion the U.S. House of Representatives since 1940. 56

57. Steps for the Hill-Huntington Method
First, add up the total population, p, of all the states, ∑pi .
Calculate the standard divisor,
s = p /h (total population ÷ house).
3.Next, calculate each state’s (or class’) quota, qi = pi / s.
4.Find the geometric mean
q* =  qi qi . 57

58. Steps for The Hill-Huntington Method
5.Round each state’s quota the Hill-Huntington way by comparing the quota to the geometric mean to obtain a first tentative apportionment.
The Hill-Huntington rounding of q is equal to:
qi if q < q*
qi if q > q*
where geometric mean q* =  qi qi 58

59. Steps for The Hill-Huntington Method
6.If the sum of the tentative apportionments is equal to the house size, the job is finished.
7.If not, a list of critical divisors must be constructed. 59

60. Steps for the Hill-Huntington Method (continued):
Step 7. (continued) Calculating the critical divisors, if needed …
If the tentative apportionments do not fill the house, then the critical divisor for state X with population V and tentative apportionment N is:
The state with the largest critical divisor is first in line to receive a seat. 60

61. Steps for the Hill-Huntington Method (continued):
Step 7. (continued) Calculating the critical divisors, if needed …
If the tentative apportionments overfill the house, then the critical divisor for state X with population V and tentative apportionment N is:
The state with the smallest critical divisor is first in line to lose a seat. 61