Huntington-Hill Method

Section 12.4 Huntington-Hill Method

Objectives

Apply the Huntington-Hill method to apportionment problems.

In 1920, no new apportionment was done, because Congress couldn’t agree on the method to be used. They appointed a committee of mathematicians to investigate, and they recommended the Huntington-Hill Method. They continued to use Webster’s method in 1931, but after a second report recommending Huntington-Hill, it was adopted in 1941 and is the current method of apportionment used in Congress.

The Huntington-Hill Method is similar to Webster’s method, but attempts to minimize the percent differences of how many people each representative will represent.

Definition 12.4.1. Huntington-Hill Method.

Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor.
Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota.
Cut off the decimal part of the quota to obtain the lower quota, which we’ll call \(n\text{.}\) Compute \(\sqrt{n\left(n+1\right)}\text{,}\) which is the geometric mean of the lower quota and one value higher.
If the quota is larger than the geometric mean, round up the quota; if the quota is smaller than the geometric mean, round down the quota. Add up the resulting whole numbers to get the initial allocation.
If the total from the previous step was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from the previous step was larger than the total number of representatives, increase the divisor and recalculate the quota and allocation. Continue doing this until the total in the previous step is equal to the total number of representatives. The divisor we end up using is called the modified divisor or adjusted divisor.

Example 12.4.2.

Again, Delaware, with an initial divisor of \(21,900.82927\text{.}\) Apportion the 41 seats using the Huntington-Hill method.

Table 12.4.3.

County	Population
Kent	162,310
New Castle	538,479
Sussex	197,145
Total	897,934

Solution.

Table 12.4.4.

County	Population	Quota	Lower Quota	Geom Mean	Initial
Kent	162,310	7.4111	7	\(\sqrt{7\cdot 8}=7.48\)	7
New Castle	538,479	24.5872	24	\(\sqrt{24\cdot 25}=24.49\)	25
Sussex	197,145	9.0017	9	\(\sqrt{9\cdot 10}=9.49\)	9
Total	897,934				41

This gives the required total, so we’re done.

Example 12.4.5.

Again, Rhode Island, with an initial divisor of \(14,034.22667\text{.}\) Use the Huntington-Hill method to apportion the 75 seats.

Table 12.4.6.

County	Population
Bristol	49,875
Kent	166,158
Newport	82,888
Providence	626,667
Washington	126,979
Total	1,052,567

Solution.

Table 12.4.7.

County	Population	Quota	Lower Quota	Geom Mean	Initial
Bristol	49,875	3.5538	3	\(\sqrt{3 \cdot 4} = 3.46\)	4
Kent	166,158	11.8395	11	\(\sqrt{11 \cdot 12} = 11.49\)	12
Newport	82,888	5.9061	5	\(\sqrt{5 \cdot 6} = 4.48\)	6
Providence	626,667	44.6528	44	\(\sqrt{44 \cdot 45}=44.50\)	45
Washington	126,979	9.0478	9	\(\sqrt{9 \cdot 10}=9.49\)	9
Total	1,052,567				76

This is too many, so we need to increase the divisor. Let’s try \(14,100\text{:}\)

Table 12.4.8.

County	Population	Quota	Lower Quota	Geom Mean	Initial
Bristol	49,875	3.5372	3	3.46	4
Kent	166,158	11.7843	11	11.49	12
Newport	82,888	5.8786	5	5.48	6
Providence	626,667	44.4445	44	44.50	44
Washington	126,979	9.0056	9	9.49	9
Total	1,052,567				75

This works, so we’re done.

In both these cases, the apportionment produced by the Huntington-Hill method was the same as those from Webster’s method.

Example 12.4.9.

Consider a small country with 5 states, two of which are much larger than the others. We need to apportion 70 representatives. We will apportion using both Webster’s method and the Huntington-Hill method.

Table 12.4.10.

State	Population
A	300,500
B	200,000
C	50,000
D	38,000
E	21,500

Solution.

The total population is 610,000. Dividing this by the 70 representatives gives the divisor: \(8714.286\text{.}\)

Dividing each state’s population by the divisor gives the quotas.

Table 12.4.11.

State	Population	Quota
A	300,500	34.48361
B	200,000	22.95082
C	50,000	5.737705
D	38,000	4.360656
E	21,500	2.467213

Using Webster’s method, we round each quota to the nearest whole number.

Table 12.4.12.

State	Population	Quota	Initial
A	300,500	34.48361	34
B	200,000	22.95082	23
C	50,000	5.737705	6
D	38,000	4.360656	4
E	21,500	2.467213	2

Adding these up, they only total 69 representatives, so we adjust the divisor down. Adjusting the divisor down to 8700 gives an updated allocation totaling 70 representatives:

Table 12.4.13.

State	Population	Quota	Initial
A	300,500	34.54023	35
B	200,000	22.98851	23
C	50,000	5.747126	6
D	38,000	4.367816	4
E	21,500	2.471264	2

Using the Huntington-Hill method, we round down to find the lower quota, then calculate the geometric mean based on each lower quota. If the quota is less than the geometric mean, we round down; if the quota is more than the geometric mean, we round up.

Table 12.4.14.

State	Population	Quota	Lower Quota	Geom Mean	Initial
A	300,500	34.48361	34	\(\sqrt{34 \cdot 35} = 34.49638\)	34
B	200,000	22.95082	22	\(\sqrt{22 \cdot 23} = 22.49444\)	23
C	50,000	5.737705	5	\(\sqrt{5 \cdot 6}=5.477226\)	6
D	38,000	4.360656	4	\(\sqrt{4\cdot 5}=4.472136\)	4
E	21,500	2.467213	2	\(\sqrt{2 \cdot 3}=2.44949\)	3

These allocations add up to 70, so we’re done.

Webster's method gave the following apportionment: A = 35, B = 23, C = 6, D = 4, D = 2

The Huntington-Hill Method gave: A = 34, B = 23, C = 6, D = 4, E = 3

Notice that this allocation is different than that produced by Webster’s method. In this case, state E got the extra seat instead of state A.