Opening the book…
- Find the standard advisor and standard quota.
- Apply Hamilton's method to solve the apportionment problem.
- Apply Jefferson's method to solve the apportionment problem.
- Apply Adam's method to solve the apportionment problem.
- Apply Webster's method to solve the apportionment problem.
What is the Apportionment Problem?
Apportionment is the problem of dividing a fixed number of items among groups of varying sizes. In politics, this takes the form of allocating a limited number of representatives among 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 also arises in various non-political contexts.
An apportionment problem is a mathematical method used to decide how to fairly distribute a fixed number of items as a whole number (you can't assign 2.4 items) (usually seats in a legislature) among groups (usually states or regions) based on their population.
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Assigning seats in the U.S. House of Representatives
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Dividing representatives in the student government
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Allocating funding to departments based on population or size
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Assigning voting power in committees
Standard Divisor and Standard Quota
The standard divisor is a concept used in apportionment problems, particularly when dividing the number of allocated items (seats or quotas) fairly among different groups based on population. The standard divisor is a key value used in apportionment problems to help decide how many seats each group should receive.
\[\text{Standard Divisor (SD)}=\frac{\text{Total Population}}{\text{Number of quota}}\nonumber\]
The standard quota for a particular group can be found by using the following formula
\[\text{Standard Quota for Particular Group}=\frac{\text{Population of that group}}{\text{Standard Divisor (SD)}}\nonumber\]
Note: If we round the standard quota DOWN to the nearest whole number, we call it standard lower quota or (Just lower quota), and if we round the standard quota UP to the nearest whole number, we call it standard upper quota or (Just upper quota).
Example \(\PageIndex{1}\): Finding Standard Divisor and Standard Quota
Delaware has three counties: Kent, New Castle, and Sussex. The Delaware State House of Representatives has \(41\) members. Find the standard divisor, standard quota for each county, upper quota, and lower quota.
\(\begin{array}{lr}
\text { County } & \text { Population } \\
\hline
\text { Kent } & 162,310 \\
\text { New Castle } & 538,479 \\
\text { Sussex } & 197,145 \\
\hline
\textbf{ Total } & \bf{ 897,934 }\\
\hline\end{array}\)
- Answer
-
First, we determine the standard divisor = \(\frac{897,934}{41} = 21,900.82927\)
Determine each country’s standard quota by dividing the country's population by the standard divisor, and find the upper and lower quotas.
\(\begin{array}{lr}
\text { County } & \text { Population } & \text{ Quota } & \text{ Lower Qutoa }& \text{Upper Quota}\\
\hline
\text { Kent } & 162,310 & 7.4111 & 7& 8 \\
\text { New Castle } & 538,479 & 24.5872 & 24& 25\\
\text { Sussex } & 197,145 & 9.0017 & 9& 10\\
\hline
\textbf{ Total } & \bf{ 897,934 } & &40& 43\\
\hline
\end{array}\)
As we see in example \(\PageIndex{1},\), rounding the standard quota to the upper and lower quotas does not give the total number of representatives in the Delaware State House. Now, we will discuss how to solve the apportionment problem using various methods. The first one is the Hamilton Method.
Hamilton's Method
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 can determine how many representatives each state should get. He follows these steps:
There are five steps we follow when applying Hamilton’s Method of apportionment:
Step 1: Find the standard divisor.
Step 2: Find each state’s standard quota.
Step 3: Give each state the state’s lower quota (with each state receiving at least one seat).
Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.
Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size from step \(4\).
Note on rounding: Today, we have technological advantages that Hamilton (and the others) couldn’t have imagined. Please take advantage of them, and keep them in several decimal places.
Example \(\PageIndex{2}\): Apportionment Using Hamilton's Method
Use Hamilton’s method to apportion the \(75\) seats of Rhode Island’s House of Representatives among its five counties.
\(\begin{array}{lr}
\text { County } & \text { Population }\\
\hline \text { Bristol } & 49,875\\
\text { Kent } & 166,158\\
\text { Newport } & 82,888\\
\text { Providence } & 626,667\\
\text { Washington } & 126,979\\
\hline
\textbf{ Total } & \bf{ 1,052,567}\\
\hline
\end{array}\)
- Answer
-
1. The Standard divisor = \(\frac{1,052,567}{75} = 14,034.22667\)
2. Determine each county’s quota by dividing its population by the divisor:
\(\begin{array}{lrr}
\text { County } & \text { Population } & \text{ Standard Quota }\\
\hline \text { Bristol } & 49,875 & 3.5538\\
\text { Kent } & 166,158 & 11.8395\\
\text { Newport } & 82,888 & 5.9061\\
\text { Providence } & 626,667 & 44.6528\\
\text { Washington } & 126,979 & 9.0478\\
\hline
\textbf{ Total } & \bf{ 1,052,567 } & \\
\hline
\end{array}\)3. Remove the decimal part of each quota (That is, find the lower quota):
\(\begin{array}{lrrc}
\text { County } & \text { Population } & \text{ Standard Quota } & \text{ Initial (Lower Quota) }\\
\hline \text { Bristol } & 49,875 & 3.5538 & 3 \\
\text { Kent } & 166,158 & 11.8395 & 11 \\
\text { Newport } & 82,888 & 5.9061 & 5 \\
\text { Providence } & 626,667 & 44.6528 & 44\\
\text { Washington } & 126,979 & 9.0478 & 9\\
\hline
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 72 }\\
\hline
\end{array}\)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:
\(\begin{array}{lrrcc}
\text { County } & \text { Population } & \text{ Standard Quota } & \text{ Initial (Lower Quota) } & \text{ Final Quota } \\
\hline \text { Bristol } & 49,875 & 3.5538 & 3 & 3 \\
\text { Kent } & 166,158 & 11.8395 & 11 &11+1= 12 \\
\text { Newport } & 82,888 & 5.9061 & 5 & 5+1=6 \\
\text { Providence } & 626,667 & 44.6528 & 44 & 44+1=45 \\
\text { Washington } & 126,979 & 9.0478 & 9 & 9 \\
\hline
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 72 } & \bf{ 75 }\\
\hline
\end{array}\)Note that even though Bristol County’s decimal part is greater than \(0.5,\) it isn’t big enough to get an additional representative, because three other counties have greater decimal parts.
Hamilton’s method obeys something called the Quota Rule. The Quota Rule isn’t a law of any sort, but just an idea that some people, including Hamilton, think is a good one.
The Quota Rule says that the final number of representatives a state gets should be within one of that state’s quota. Since we’re dealing with whole numbers for our final answers, each state should either go up to the next whole number above its quota or down to the next whole number below its quota.
For example, let’s say there are \(100\) seats to distribute and \(5\) states. If State A's population justifies exactly \(9.4\) seats, the Quota Rule says: State A should get either \(9\) or \(10\) seats. Getting less than nine or more than \(10\) would violate the rule. The Hamilton Method always satisfies the quota rule.
Jefferson's Method
Thomas Jefferson proposed a different method for apportionment. After Washington vetoed Hamilton’s method, Jefferson’s method was adopted and used in Congress from \(1791\) through \(1842.\) Jefferson had political reasons for wanting his method to be used rather than Hamilton’s. Primarily, his method favors larger states, and his own home state of Virginia was the largest in the country at the time. He would also argue that it’s the ratio of people to representatives that is the critical thing, and apportionment methods should be based on that. However, the paradoxes we saw also provide mathematical reasons for concluding that Hamilton’s method isn’t so good, and while Jefferson’s method might or might not be the best one to replace it, at least we should look for other possibilities.
The first steps of Jefferson’s method are the same as Hamilton’s method. He finds the same divisor and the same quota, and cuts off the decimal parts in the same way, giving a total number of representatives that is less than the required total. The difference is in how Jefferson resolves that difference. He says that since we ended up with a too small answer, our divisor must have been too big. He changes the divisor by making it smaller, finding new quotas with the new divisor, cutting off the decimal parts, and looking at the new total, until he finds a divisor that produces the required total.
Step 1: Find the standard divisor.
Step 2: Choose a modified divisor slightly less than the standard divisor.
Step 3: Find the states’ or group lower quota and their sum (round down to the nearest whole number). Add all the lower quota numbers.
Step 4: If the total lower quota (after rounding) is more than available, increase the divisor (making quotas smaller). If the total is less, decrease the divisor (making quotas larger). Repeat until the sum of the lower quotas equals the total quota.
Step 5: Each group quota from step \(4\) is the final apportionment (or final quota).
In Jefferson's method, a modified divisor is less than the standard divisor.
Example \(\PageIndex{3}\): Jefferson's Method
Use Jefferson’s method to apportion the \(25\) seats (senators) among six states of the small country.
| State | A | B | C | D | E | F | Total |
|---|---|---|---|---|---|---|---|
| Population | \(24,000\) | \(56,000\) | \(28,000\) | \(17,000\) | \(65,000\) | \(47,000\) | \(237,000\) |
- Answer
-
Standard divisor is \(\frac{237,000}{25}=9480\).
In Jefferson’s method, the modified divisor is always smaller than the standard divisor. There is no formula to find the modified divisor, so we start by guessing.
Let’s try the modified divisor, \(d = 9000.\)
Table \(\PageIndex{2}:\) Quotas for \(d = 9000\) State A B C D E F Total Population \(24,000\) \(56,000\) \(28,000\) \(17,000\) \(65,000\) \(47,000\) \(237,000\) Modified Quota
\(d = 9000\)
\(2.67\) \(6.22\) \(3.11\) \(1.89\) \(7.22\) \(5.22\) Lower Quota \(2\) \(6\) \(3\) \(1\) \(7\) \(5\) \(24\) The sum \(24\) is smaller than the given quota \(25,\) so we must try again by making the modified divisor smaller than \(9000.\) Let’s try \(d = 8000.\)
Table \(\PageIndex{3}:\) Quotas for \(d = 8000\) State A B C D E F Total Population \(24,000\) \(56,000\) \(28,000\) \(17,000\) \(65,000\) \(47,000\) \(237,000\) Modified Quota \(d = 8000\) \(3.00\) \(7.00\) \(3.50\) \(2.13\) \(8.13\) \(5.88\) Lower Quota \(3\) \(7\) \(3\) \(2\) \(8\) \(5\) \(28\) This time, the sum of \(28\) is too big. Try again, making the modified divisor larger. We know the divisor must be between \(d = 8000\) and \(d = 9000\) so let’s try \(d = 8500.\)
Table \(\PageIndex{4}:\) Quotas for \(d = 8500\) State A B C D E F Total Population \(24,000\) \(56,000\) \(28,000\) \(17,000\) \(65,000\) \(47,000\) \(237,000\) Modified Quota \(d = 8500\) \(2.82\) \(6.59\) \(3.29\) \(2.00\) \(7.65\) \(5.53\) Lower Quota
Final apportion
\(2\) \(6\) \(3\) \(2\) \(7\) \(5\) \(25\) This time the sum is \(25,\) so we are done. A gets two senators, B gets six senators, C gets three senators, D gets two senators, E gets seven senators, and F gets five senators.
Any modified divisor between \(8,200\) and \(8,500\) should work in this problem.
Note: You can double-check the above calculation using the following Jefferson Method Calculator.
Example \(\PageIndex{4}\): Jefferson's Method
Use Jefferson’s method to apportion the \(75\) seats of Rhode Island’s House of Representatives among its five counties.
\(\begin{array}{lrrc}
\hline\text { County } & \text {Population} \\
\hline \text { Bristol } & 49,875 \\
\text { Kent } & 166,158 \\
\text { Newport } & 82,888 \\
\text { Providence } & 626,667 \\
\text { Washington } & 126,979 \\
\hline
\textbf{ Total } & \bf{ 1,052,567 } \\
\hline
\end{array}\)
- Answer
-
First, find the standard divisor. The Standard divisor = \(\frac{1,052,567}{75} = 14,034.22667\)
We need \(75\) representatives and only have \(75,\) so we need to use a smaller divisor. Let’s try \(13,500.\)
\(\begin{array}{lrrc}
\hline\text { County } & \text { Population } & \text{Modified Quota } & \text{ Lower quota } \\
\hline \text { Bristol } & 49,875 & 3.6944 & 3 \\
\text { Kent } & 166,158 & 12.3080 & 12 \\
\text { Newport } & 82,888 & 6.1399 & 6 \\
\text { Providence } & 626,667 & 46.4198 & 46 \\
\text { Washington } & 126,979 & 9.4059 & 9\\
\hline
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 76 }\\
\hline
\end{array}\)We’ve gone too far. We need a divisor that’s greater than \(13,500\) but less than \(14,034.22667.\) Let’s try \(13,700.\)
\(\begin{array}{lrrc}
\hline\text { County } & \text { Population } & \text{ Modified Quota } & \text{ Lower quota (Final apportion) } \\
\hline \text { Bristol } & 49,875 & 3.6405 & 3 \\
\text { Kent } & 166,158 & 12.1283 & 12 \\
\text { Newport } & 82,888 & 6.0502 & 6 \\
\text { Providence } & 626,667 & 45.7421 & 45 \\
\text { Washington } & 126,979 & 9.2685 & 9\\
\hline
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{75 }\\
\hline
\end{array}\)This works.
Any modified divisor between \(13,650\) and \(13,800\) should work in this problem.
Notice, in comparison to Hamilton’s method, that although the results were the same, they came about in a different way, and the outcome was almost different. Providence County (the largest) almost went up to \(46\) representatives before Kent (which is much smaller) got to \(12\). Although that didn’t happen here, it can. Divisor-adjusting methods, such as Jefferson’s, are not guaranteed to follow the quota rule.
Adam’s Method:
Adams’s method of apportionment is another method of apportionment that is based on a modified divisor. However, instead of basing the changes on the sum of the lower quotas, as Jefferson did, Adams used the upper quotas. Adam’s method of apportionment is very similar to Jefferson’s, but with a twist in how it rounds numbers. It is named after John Quincy Adams, the \(6\)th President of the U.S.
Step 1: Find the standard divisor.
Step 2: Choose a modified divisor slightly greater than the standard divisor.
Step 3: Find the states’ or group upper quota and their sum (round UP to the nearest whole number). Add all the upper quota numbers.
Step 4: If the total upper quota (after rounding) is more than available, increase the divisor (making quotas smaller). If the total is less, decrease the divisor (making quotas larger). Repeat until the sum of the upper quotas equals the total quota.
Step 5: Each group quota from step \(4\) is the final apportionment (or final quota).
Note: In Adam's method, a modified divisor is greater than the standard divisor.
Example \(\PageIndex{5}\): Using Adam's Method
In Adamstown, \(42\) new firefighters have just completed their training. They are to be assigned to the five firehouses in town in a manner proportional to the population in each fire district. The populations are listed in the following table.
| District | A | B | C | D | E | Total |
|---|---|---|---|---|---|---|
| Population | \(25,010\) | \(8,760\) | \(11,590\) | \(9,025\) | \(15,080\) | \(69,465\) |
Apportion the new firefighters to the firehouses using the Adams Method.
- Answer
-
\[\text{Standard divisor} = \frac{69,465}{42}=1653.92857\nonumber\]
Adams’s method always rounds up, making the sum of the upper quotas too large. First, choose the modified divisor slightly larger than the standard divisor to see if that works. The results are summarized below in Table \(\PageIndex{2}.\)
Guess #\(1:\) \( d = 1700.\) The total is too still too large (\(43\)) so make the modified divisor larger. (Does not work)
Guess #\(2:\) \( d = 1800.\) Now the total is too small (\(41\)) so make the modified divisor smaller. (Does not work)
Guess #\(3:\) \( d = 1750.\) The total is too large again (\(43\)) so make the modified divisor larger. (Does not work)
Guess #\(4:\) \( d = 1775.\) The sum is \(42 \) so we are done. (Works)
Table \(\PageIndex{2}:\) Apportion by Adams’s Method for Adamstown District A B C D E Total Population \(25,010\) \(8,760\) \(11,590\) \(9,025\) \(15,080\) \(69,465\) \(d = 1700\) \(14.712\) \(5.153\) \(6.818\) \(5.309\) \(8.871\) quota \(15\) \(6\) \(7\) \(6\) \(9\) \(43\) \(d = 1800\) \(13.894\) \(4.86\) \(6.439\) \(5.014\) \(8.378\) quota \(14\) \(5\) \(7\) \(6\) \(9\) \(41\) \(d = 1750\) \(14.291\) \(5.006\) \(6.623\) \(5.157\) \(8.617\) quota \(15\) \(6\) \(7\) \(6\) \(9\) \(43\) \(d = 1775\) \(14.090\) \(4.935\) \(6.530\) \(5.085\) \(8.496\) Final Allocation \(15\) \(5\) \(7\) \(6\) \(9\) \(42\) Any modified divisor between \(1755\) and \(1785\) should work in this case.
Webster's Method
This method is named after Daniel Webster, an American statesman in the \(19\)th century. Webster’s method of apportionment is another method of apportionment that is based on a modified divisor. However, instead of basing the changes on the sum of the lower quotas, as Jefferson did, or the sum of the upper quotas, as Adams did, Webster used traditional rounding.
Step 1: Find the standard divisor.
Step 2: Choose a modified divisor either slightly greater than or slightly less than the standard divisor. But you must start with a standard divisor. This could be a modified divisor.
Step 3: Find the states or group rounded quota (If the decimal is \(0.5\) or more, round UP. If it's less than \(0.5,\) round DOWN). Add all rounded quota numbers.
Step 4: If the number of quotas (after rounding) is more than what is available, increase the divisor (making quotas smaller). If the total is less, decrease the divisor (making quotas larger). Repeat until the sum of the rounded quotas equals the total seats or quotas.
Step 5: Each group quota from step \(4\) is the final apportionment (or final quota).
Because some quotas will be rounded up and others will be rounded down, we cannot immediately determine whether the total number of seats is too large or too small. Unlike Jefferson’s and Adam’s methods, we do not know how to adjust the modified divisor. This forces us to use the standard divisor as the first modified divisor. So, the modified divisor in this method can be less than, greater than, or equal to the standard divisor.
The modified divisor in the Webster's Method can be less than, greater than, or equal to the standard divisor.
Example \(\PageIndex{6}\): Webester's Method
Use Webster’s method to apportion the \(25\) seats among \(6\) states of the small country.
| State | A | B | C | D | E | F | Total |
|---|---|---|---|---|---|---|---|
| Population | \(24,000\) | \(56,000\) | \(28,000\) | \(17,000\) | \(65,000\) | \(47,000\) | \(237,00\) |
- Answer
-
First, find the standard divisor = \(\frac{237,000}{25}= 9480.\) and use it as a modified divisor to see if we are lucky.
Table \(\PageIndex{2}:\) Quotas for \(d = 9480\) State A B C D E F Total Population \(24,000\) \(56,000\) \(28,000\) \(17,000\) \(65,000\) \(47,000\) \(237,000\) \(d = 9480\) \(2.53\) \(5.91\) \(2.95\) \(1.79\) \(6.86\) \(4.96\) Rounded Quota \(3\) \(6\) \(3\) \(2\) \(7\) \(5\) \(26\) Since the total of \(26\) seats is too big, we need to make the modified divisor larger. Try \(d = 11,000.\)
Table \(\PageIndex{3}:\) Quotas for \(d = 11,000\) State A B C D E F Total Population \(24,000\) \(56,000\) \(28,000\) \(17,000\) \(65,000\) \(47,000\) \(237,000\) \(d = 11,000\) \(2.18\) \(5.09\) \(2.55\) \(1.55\) \(5.91\) \(4.27\) Rounded Quota \(2\) \(5\) \(3\) \(2\) \(6\) \(4\) \(22\) The total number of seats \(22\) is smaller than we hoped (\(25.\)) That means that \(d = 11,000\) is much too big. We need to pick a new modified divisor between \(9480\) and \(11,000.\) Try a divisor closer to \(9480\) such as \( d = 10,000.\)
Table \(\PageIndex{4}:\) Quotas for \(d = 10,000\) State A B C D E F Total Population \(24,000\) \(56,000\) \(28,000\) \(17,000\) \(65,000\) \(47,000\) \(237,000\) \(d = 10,000\) \(2.40\) \(5.60\) \(2.80\) \(1.70\) \(6.50\) \(4.70\) Rounded Quota \(2\) \(6\) \(3\) \(2\) \(7\) \(5\) \(25\) Any modified divisor between \(9,650\) and \(10,000\) works here.
Note: You can double-check the above calculation using the following Webster's Method Calculator.
Example \(\PageIndex{7}\): Comparing all \(4\) Apportionment Methods
The following table summarizes the results of the Hamilton, Jefferson, Adams, and Webster methods when applied to the apportionment of \(616\) schools among five Hawaiian counties.
| Hawaii | Honolulu | Kalawao | Kauai | Maui | |
|---|---|---|---|---|---|
| Under 19 years old | \(46,310\) | \(224,230\) | \(20\) | \(16,560\) | \(38,450\) |
| Hamilton | \(87\) | \(424\) | \(1\) | \(31\) | \(73\) |
| Jefferson | \(87\) | \(425\) | \(1\) | \(31\) | \(72\) |
| Adams | \(88\) | \(422\) | \(1\) | \(31\) | \(73\) |
| Webster | \(87\) | \(424\) | \(1\) | \(31\) | \(73\) |
- Do any of the apportionment methods result in the same apportionment? If so, which ones?
- Which apportionment method would the citizens of the largest county likely favor most and least? Justify your answer.
- As a group, which apportionment method would the citizens of the other four counties likely favor most and least? Justify your answer.
- Answer
-
- Yes, the Hamilton and Webster methods result in the same apportionment.
- The largest county is Honolulu. The citizens would likely favor the Jefferson method of apportionment most since they received the most seats by that method. They would likely favor the Adams method of apportionment least because they received the least number of seats by that method.
- As a group, the other four counties received \(192\) seats by either the Hamilton or Webster method, \(193\) seats by the Adams method, and \(191\) seats by the Jefferson method. They would likely favor the Adams method the most and favor the Jefferson method the least.
This page titled 6.1: Apportionment Methods was last modified on Thu, 13 Aug 2026 15:37:47 GMT and is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by Mike Kenyon & David Lippman via source content that was edited to the style and standards of the LibreTexts platform.
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