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Section 3.4 Huntington-Hill Method

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 3.4.1 Huntington-Hill Method
  1. 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.

  2. 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.

  3. Cut off the decimal part of the quota to obtain the lower quota, which we’ll call \(n\text{.}\) Compute \(\sqrt{n(n+1)}\text{,}\) which is the geometric mean of the lower quota and one value higher.

  4. 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.

  5. If the total from Step 4 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 4 was larger than the total number of representatives, increase the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 4 is equal to the total number of representatives. The divisor we end up using is called the modified divisor or adjusted divisor.

Example 3.4.2

Again, Delaware, with an initial divisor of 21,900.82927:

Figure 3.4.3 Video solution by David Lippman
County Population Quota Lower Quota Geom Mean Initial
Kent 162,310 7.4111 7 7.48 7
New Castle 538,479 24.5872 24 24.49 25
Sussex 197,145 9.0017 9 9.49 9
Total 897,934       41

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

Example 3.4.4
Figure 3.4.5 Video solution by David Lippman

Again, Rhode Island, with an initial divisor of 14,034.22667:

County Population Quota Lower Quota Geom Mean Initial
Bristol 49,875 3.5538 3 3.46 4
Kent 166,158 11.8395 11 11.49 12
Newport 82,888 5.9061 5 5.48 6
Providence 626,667 44.6528 44 44.50 45
Washington 126,979 9.0478 9 9.49 9
Total 1,052,567       76

This is too many, so we need to increase the divisor. Let’s try 14,100:

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 3.4.6
Figure 3.4.7 Video solution by David Lippman

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.

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

1. The total population is 610,000. Dividing this by the 70 representatives gives the divisor: 8714.286

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

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

Webster’s Method

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

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

4. 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

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

Huntington-Hill Method

5. 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.

State Population Quota Lower Quota Geometric Mean Initial
A 300,500 34.48361 34 34.49638 34
B 200,000 22.95082 22 22.49444 23
C 50,000 5.737705 5 5.477226 6
D 38,000 4.360656 4 4.472136 4
E 21,500 2.467213 2 2.44949 3

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

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.