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.
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.