10.1: Hamilton’s Method
Apportionment is the problem of dividing up a fixed number of things among groups of different sizes. In politics, this takes the form of allocating a limited number of representatives amongst voters. This problem, presumably, is older than the United States, but the best-known ways to solve it have their origins in the problem of assigning each state an appropriate number of representatives in the new Congress when the country was formed. States also face this apportionment problem in defining how to draw districts for state representatives. The apportionment problem comes up in a variety of non-political areas too, though. We face several restrictions in this process:
- The things being divided up can exist only in whole numbers.
- We must use all of the things being divided up, and we cannot use any more.
- Each group must get at least one of the things being divided up.
- The number of things assigned to each group should be at least approximately proportional to the population of the group. (Exact proportionality isn’t possible because of the whole number requirement, but we should try to be close, and in any case, if Group A is larger than Group B, then Group B shouldn’t get more of the things than Group A does.)
In terms of the apportionment of the United States House of Representatives, these rules imply:
- We can only have whole representatives (a state can’t have 3.4 representatives)
- We can only use the (currently) 435 representatives available. If one state gets another representative, another state has to lose one.
- Every state gets at least one representative
- The number of representatives each state gets should be approximately proportional to the state population. This way, the number of constituents each representative has should be approximately equal.
Alexander Hamilton proposed the method that now bears his name. His method was approved by Congress in 1791, but was vetoed by President Washington. It was later adopted in 1852 and used through 1911. He begins by determining, to several decimal places, how many things each group should get. Since he was interested in the question of Congressional representation, we’ll use the language of states and representatives, so he determines how many representatives each state should get. He follows these steps:
- 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 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 . Since we can only allocate whole representatives, Hamilton resolves the whole number problem, as follows:
- Cut off all the decimal parts of all the quotas (but don’t forget what the decimals were). These are called the lower quotas . Add up the remaining whole numbers. This answer will always be less than or equal to the total number of representatives (and the “or equal to” part happens only in very specific circumstances that are incredibly unlikely to turn up).
- Assuming that the total from Step 3 was less than the total number of representatives, assign the remaining representatives, one each, to the states whose decimal parts of the quota were largest, until the desired total is reached.
Make sure that each state ends up with at least one representative!
Note on rounding: Today we have technological advantages that Hamilton (and the others) couldn’t even have imagined. Take advantage of them, and keep several decimal places.
The state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has 41 members. If Delaware wants to divide this representation along county lines (which is not required, but let’s pretend they do), let’s use Hamilton’s method to apportion them. The populations of the counties are as follows (from the 2010 Census):
\(\begin{array}{lr}
\text { County } & \text { Population } \\
\hline
\text { Kent } & 162,310 \\
\text { New Castle } & 538,479 \\
\text { Sussex } & 197,145 \\
\textbf{ Total } & \bf{ 897,934 }\end{array}\)
Solution
1. First, we determine the divisor: \(897,934/41 = 21,900.82927\)
2. Now we determine each county’s quota by dividing the county’s population by the divisor:
\(\begin{array}{lr}
\text { County } & \text { Population } & \text{ Quota } \\
\hline
\text { Kent } & 162,310 & 7.4111 \\
\text { New Castle } & 538,479 & 24.5872 \\
\text { Sussex } & 197,145 & 9.0017 \\
\textbf{ Total } & \bf{ 897,934 } & \end{array}\)
3. Removing the decimal parts of the quotas gives:
\(\begin{array}{lrrc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial }\\
\hline
\text { Kent } & 162,310 & 7.4111 & 7\\
\text { New Castle } & 538,479 & 24.5872 & 24\\
\text { Sussex } & 197,145 & 9.0017 & 9\\
\textbf{ Total } & \bf{ 897,934 } & & \bf{ 40 }\end{array}\)
4. We need 41 representatives and this only gives 40. The remaining one goes to the county with the largest decimal part, which is New Castle:
\(\begin{array}{lrrcc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial } & \text{ Final } \\
\hline
\text { Kent } & 162,310 & 7.4111 & 7 & 7 \\
\text { New Castle } & 538,479 & 24.5872 & 24 & 25 \\
\text { Sussex } & 197,145 & 9.0017 & 9 & 9 \\
\textbf{ Total } & \bf{ 897,934 } & & \bf{ 40 } & \bf { 41 } \end{array}\)
Use Hamilton’s method to apportion the 75 seats of Rhode Island’s House of Representatives among its five counties.
\(\begin{array}{lr}
\text { County } & \text { Population }\\
\hline \text { Bristol } & 49,875\\
\text { Kent } & 166,158\\
\text { Newport } & 82,888\\
\text { Providence } & 626,667\\
\text { Washington } & 126,979\\
\textbf{ Total } & \bf{ 1,052,567 }\end{array}\)
Solution
1. The divisor is \(1,052,567/75 = 14,034.22667\)
2. Determine each county’s quota by dividing its population by the divisor:
\(\begin{array}{lrr}
\text { County } & \text { Population } & \text{ Quota }\\
\hline \text { Bristol } & 49,875 & 3.5538\\
\text { Kent } & 166,158 & 11.8395\\
\text { Newport } & 82,888 & 5.9061\\
\text { Providence } & 626,667 & 44.6528\\
\text { Washington } & 126,979 & 9.0478\\
\textbf{ Total } & \bf{ 1,052,567 } & \end{array}\)
3. Remove the decimal part of each quota:
\(\begin{array}{lrrc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial }\\
\hline \text { Bristol } & 49,875 & 3.5538 & 3 \\
\text { Kent } & 166,158 & 11.8395 & 11 \\
\text { Newport } & 82,888 & 5.9061 & 5 \\
\text { Providence } & 626,667 & 44.6528 & 44\\
\text { Washington } & 126,979 & 9.0478 & 9\\
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 72 }\end{array}\)
4. We need 75 representatives and we only have 72, so we assign the remaining three, one each, to the three counties with the largest decimal parts, which are Newport, Kent, and Providence:
\(\begin{array}{lrrcc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial } & \text{ Final } \\
\hline \text { Bristol } & 49,875 & 3.5538 & 3 & 3 \\
\text { Kent } & 166,158 & 11.8395 & 11 & 12 \\
\text { Newport } & 82,888 & 5.9061 & 5 & 6 \\
\text { Providence } & 626,667 & 44.6528 & 44 & 45 \\
\text { Washington } & 126,979 & 9.0478 & 9 & 9 \\
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 72 } & \bf{ 75 }\end{array}\)
Note that even though Bristol County’s decimal part is greater than .5, it isn’t big enough to get an additional representative, because three other counties have greater decimal parts.
Controversy
After seeing Hamilton’s method, many people find that it makes sense, it’s not that difficult to use (or, at least, the difficulty comes from the numbers that are involved and the amount of computation that’s needed, not from the method), and they wonder why anyone would want another method. The problem is that Hamilton’s method is subject to several paradoxes. Three of them happened, on separate occasions, when Hamilton’s method was used to apportion the United States House of Representatives.
The Alabama Paradox is named for an incident that happened during the apportionment that took place after the 1880 census. (A similar incident happened ten years earlier involving the state of Rhode Island, but the paradox is named after Alabama.) The post-1880 apportionment had been completed, using Hamilton’s method and the new population numbers from the census. Then it was decided that because of the country’s growing population, the House of Representatives should be made larger. That meant that the apportionment would need to be done again, still using Hamilton’s method and the same 1880 census numbers, but with more representatives. The assumption was that some states would gain another representative and others would stay with the same number they already had (since there weren’t enough new representatives being added to give one more to every state). The paradox is that Alabama ended up losing a representative in the process, even though no populations were changed and the total number of representatives increased.
The New States Paradox happened when Oklahoma became a state in 1907. Oklahoma had enough population to qualify for five representatives in Congress. Those five representatives would need to come from somewhere, though, so five states, presumably, would lose one representative each. That happened, but another thing also happened: Maine gained a representative (from New York).
The Population Paradox happened between the apportionments after the census of 1900 and of 1910. In those ten years, Virginia’s population grew at an average annual rate of 1.07%, while Maine’s grew at an average annual rate of 0.67%. Virginia started with more people, grew at a faster rate, grew by more people, and ended up with more people than Maine. By itself, that doesn’t mean that Virginia should gain representatives or Maine shouldn’t, because there are lots of other states involved. But Virginia ended up losing a representative to Maine .