4.5: Huntington-Hill Method

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Mike Kenyon & David Lippman
Pierce College via The OpenTextBookStore

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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.

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 \(n\). Compute \(\sqrt{n(n+1)}\), 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 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 7

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

\(\begin{array}{lrrccc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Lower Quota } & \text{ Geom Mean } & \text{ Initial } \\
\hline \text { Kent } & 162,310 & 7.4111 & 7 & 7.48 & 7\\
\text { New Castle } & 538,479 & 24.5872 & 24 & 24.49 & 25\\
\text { Sussex } & 197,145 & 9.0017 & 9 & 9.49 & 9\\
\textbf{ Total } & \bf{ 897,934 } & & & & \bf{ 41 }\end{array}\)

Solution

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

Example 8

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

\(\begin{array}{lrrccc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Lower Quota } & \text{ Geom Mean } & \text{ Initial } \\
\hline \text { Bristol } & 49,875 & 3.5538 & 3 & 3.46 & 4\\
\text { Kent } & 166,158 & 11.8395 & 11 & 11.49 & 12 \\
\text { Newport } & 82,888 & 5.9061 & 5 & 5.48 & 6\\
\text { Providence } & 626,667 & 44.6528 & 44 & 44.50 & 45 \\
\text { Washington } & 126,979 & 9.0478 & 9 & 9.49 & 9\\
\textbf{ Total } & \bf{ 1,052,567 } & & & & \bf{ 76 }\end{array}\)

Solution

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

\(\begin{array}{lrrccc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Lower Quota } & \text{ Geom Mean } & \text{ Initial } \\
\hline \text { Bristol } & 49,875 & 3.5372 & 3 & 3.46 & 4 \\
\text { Kent } & 166,158 & 11.7843 & 11 & 11.49 & 12 \\
\text { Newport } & 82,888 & 5.8786 & 5 & 5.48 & 6 \\
\text { Providence } & 626,667 & 5.8786 & 5 & 44.50 & 44 \\
\text { Washington } & 126,979 & 9.0056 & 9 & 9.49 & 9\\
\textbf{ Total } & \bf{ 1,052,567 } & & & & \bf{ 75 }\end{array}\)

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 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.

\(\begin{array}{lr}
\text { State } & \text { Population } \\
\hline \mathrm{A} & 300,500 \\
\mathrm{B} & 200,000 \\
\mathrm{C} & 50,000 \\
\mathrm{D} & 38,000 \\
\mathrm{E} & 21,500
\end{array}\)

Solution

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.

\(\begin{array}{lrr}
\text { State } & \text { Population } & \text { Quota } \\
\hline \text { A } & 300,500 & 34.48361 \\
\text { B } & 200,000 & 22.95082 \\
\text { C } & 50,000 & 5.737705 \\
\text { D } & 38,000 & 4.360656 \\
\text { E } & 21,500 & 2.467213
\end{array}\)

Webster’s Method

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

\(\begin{array}{lrrr}
\text { State } & \text { Population } & \text { Quota } & \text { Initial } \\
\hline \mathrm{A} & 300,500 & 34.48361 & 34 \\
\mathrm{B} & 200,000 & 22.95082 & 23 \\
\mathrm{C} & 50,000 & 5.737705 & 6 \\
\mathrm{D} & 38,000 & 4.360656 & 4 \\
\mathrm{E} & 21,500 & 2.467213 & 2
\end{array}\)

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

\(\begin{array}{lrrr}
\text { State } & \text { Population } & \text { Quota } & \text { Initial } \\
\hline \mathrm{A} & 300,500 & 34.54023 & 35 \\
\mathrm{B} & 200,000 & 22.98851 & 23 \\
\mathrm{C} & 50,000 & 5.747126 & 6 \\
\mathrm{D} & 38,000 & 4.367816 & 4 \\
\mathrm{E} & 21,500 & 2.471264 & 2
\end{array}\)

Huntington-Hill Method

3. 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.

\(\begin{array}{lrrrrr}
\text { State } & \text { Population } & \text { Quota } & \begin{array}{r}
\text { Lower } \\
\text { Quota }
\end{array} & \begin{array}{r}
\text { Geometric } \\
\text { Mean }
\end{array} & \text { Initial } \\
\hline \mathrm{A} & 300,500 & 34.48361 & 34 & 34.49638 & 34 \\
\mathrm{B} & 200,000 & 22.95082 & 22 & 22.49444 & 23 \\
\mathrm{C} & 50,000 & 5.737705 & 5 & 5.477226 & 6 \\
\mathrm{D} & 38,000 & 4.360656 & 4 & 4.472136 & 4 \\
\mathrm{E} & 21,500 & 2.467213 & 2 & 2.44949 & 3
\end{array}\)

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.