9.1: Apportionment - Jefferson’s, Adam’s, and Webster’s Methods
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Apportionment can be thought of as dividing a group of people (or other resources) and assigning them to different places.
Tom is moving to a new apartment. On moving day, four of his friends come to help and stay until the job is done since Tom promised they will split a case of beer afterwards. It sounds like a fairly simple job to split the case of beer between the five friends until Tom realizes that 24 is not evenly divisible by five. He could start by giving each of them (including himself) four beers. The question is how to divide the four remaining beers among the five friends assuming they only get whole beers. Apportionment methods can help Tom come up with an equitable solution
Basic Concepts of Apportionment
The apportionment methods we will look at in this chapter were all created as a way to divide the seats in the U.S. House of Representatives among the states based on the size of the population for each state. The terminology we use in apportionment reflects this history. An important concept is that the number of seats a state has is proportional to the population of the state. In other words, states with large populations get lots of seats and states with small populations only get a few seats.
- The seats are the people or items that are to be shared equally.
- The states are the parties that will receive a proportional share of the seats.
The first step in any apportionment problem is to calculate the standard divisor. This is the ratio of the total population to the number of seats. It tells us how many people are represented by each seat. The standard divisor is:
\[\mathrm{SD}=\frac{\text { total population }}{\# \text { seats }} \label{sd} \]
The next step is to find the standard quota for each state. This is the exact number of seats that should be allocated to each state if decimal values were possible. The standard quota is:
\[\mathrm{SQ}=\frac{\text { state population }}{\text { standard divisor }} \label{sq} \]
Hamiltonia, a small country consisting of six states is governed by a senate with 25 members. The number of senators for each state is proportional to the population of the state. The following table shows the population of each state as of the last census.
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | Total |
|---|---|---|---|---|---|---|---|
| Population | 24,000 | 56,000 | 28,000 | 17,000 | 65,000 | 47,000 | 237,000 |
Find the standard divisor and the standard quotas for each of the states of Hamiltonia.
Standard Divisor (Equation \ref{SD}):
\[\mathrm{SD}=\frac{\text { total population }}{\# \text { seats }}=\frac{237,000}{25}=9480 \nonumber \]
This means that each seat in the senate corresponds to a population of 9480 people.
Standard Quotas (Equation \ref{SQ}):
- Alpha: \[\mathrm{SQ}=\frac{\text { state population }}{\text { standard divisor }}=\frac{24,000}{9480}=2.532 \nonumber \]
- Beta:\[\mathrm{SQ}=\frac{\text { state population }}{\text { standard divisor }}=\frac{56,000}{9480}=5.907 \nonumber \]
If fractional seats were possible, Alpha would get 2.532 seats and Beta would get 5.907 seats.
Use similar calculations for the other states.
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | Total |
|---|---|---|---|---|---|---|---|
| Population | 24,000 | 56,000 | 28,000 | 17,000 | 65,000 | 47,000 | 237,000 |
| Standard Quota | 2.532 | 5.907 | 2.954 | 1.793 | 6.857 | 4.958 | 25.001 |
Notice that the sum of the standard quotas is 25.001, the total number of seats. This is a good way to check your arithmetic.
Note: Do not worry about the 0.001. That is due to rounding and is negligible.
The standard quota for each state is usually a decimal number but in real life the number of seats allocated to each state must be a whole number. Rounding off the standard quota by the usual method of rounding does not always work. Sometimes the total number of seats allocated is too high and other times it is too low. In Example \(\PageIndex{2}\) the total number of seats allocated would be 26 if we used the usual rounding rule.
When we round off the standard quota for a state the result should be the whole number just below the standard quota or the whole number just above the standard quota. These values are called the lower and upper quotas, respectively. In the extremely rare case that the standard quota is a whole number, use the standard quota for the lower quota and the next higher integer for the upper quota.
The lower quota is the standard quota rounded down. The upper quota is the standard quota rounded up.
Find the lower and upper quotas for each of the states in Hamiltonia.
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | Total |
|---|---|---|---|---|---|---|---|
| Population | 24,000 | 56,000 | 28,000 | 17,000 | 65,000 | 47,000 | 237,000 |
| Standard Quota | 2.532 | 5.907 | 2.954 | 1.793 | 6.857 | 4.958 | 25.001 |
| Lower Quota | 2 | 5 | 2 | 1 | 6 | 4 | 20 |
| Upper Quota | 3 | 6 | 3 | 2 | 7 | 5 | 26 |
Note: The total of the lower quotas is 20 (below the number of seats to be allocated) and the total of the upper quotas is 26 (above the number of seats to be allocated).
Hamilton’s Method
The U.S. Constitution requires that the seats for the House of Representatives be apportioned among the states every ten years based on the sizes of the populations. Since 1792, five different apportionment methods have been proposed and four of these methods have been used to apportion the seats in the House of Representatives. The number of seats in the House has also changed many times. In many situations the five methods give the same results. However, in some situations, the results depend on the method used. As we will see in the next section, each of the methods has at least one weakness. Because it was important for a state to have as many representatives as possible, senators tended to pick the method that would give their state the most representatives. In 1941, the number of seats in the House was fixed at 435 and an official method was chosen. This took the politics out of apportionment and made it a purely mathematical process.
Alexander Hamilton proposed the first apportionment method to be approved by Congress. Unfortunately for Hamilton, President Washington vetoed its selection. This veto was the first presidential veto utilized in the new U.S. government. A different method proposed by Thomas Jefferson was used instead for the next 50 years. Later, Hamilton’s method was used off and on between 1852 and 1901.
Summary of Hamilton’s Method:
- Use the standard divisor to find the standard quota for each state.
- Temporarily allocate to each state its lower quota of seats. At this point, there should be some seats that were not allocated.
- Starting with the state that has the largest fractional part and working toward the state with the smallest fractional part, allocate one additional seat to each state until all the seats have been allocated.
Use Hamilton’s method to finish the allocation of seats in Hamiltonia.
Let’s use red numbers below in Table \(\PageIndex{4}\) to rank the fractional parts of the standard quotas from each state in order from largest to smallest. For example, Zeta’s standard quota, 4.958, has the largest fractional part, 0.958. Also find the sum of the lower quotas to determine how many seats still need to be allocated.
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | Total |
|---|---|---|---|---|---|---|---|
| Population | 24,000 | 56,000 | 28,000 | 17,000 | 65,000 | 47,000 | 237,000 |
| Standard Quota | 2.532(6) | 5.907(3) | 2.954(2) | 1.793(5) | 6.857(4) | 4.958(1) | 25.001 |
| Lower Quota | 2 | 5 | 2 | 1 | 6 | 4 | 20 |
Twenty of the 25 seats have been allocated so there are five remaining seats. Allocate the seats, in order, to Zeta, Gamma, Beta, Epsilon and Delta.
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | Total |
|---|---|---|---|---|---|---|---|
| Population | 24,000 | 56,000 | 28,000 | 17,000 | 65,000 | 47,000 | 237,000 |
| Standard Quota | 2.532 | 5.907 | 2.954 | 1.793 | 6.857 | 4.958 | 25.001 |
| Lower Quota | 2 | 5 | 2 | 1 | 6 | 4 | 20 |
| Final Allocation | 2 | 6 | 3 | 2 | 7 | 5 | 25 |
Overall, Alpha gets two senators, Beta gets six senators, Gamma gets three senators, Delta gets two senators, Epsilon gets seven, and Zeta gets five senators.
According to Ask.com, “a paradox is a statement that apparently contradicts itself and yet might be true.” (Ask.com, 2014) Hamilton’s method and the other apportionment methods discussed in section 9.2 are all subject to at least one paradox. None of the apportionment methods is perfect. The Alabama paradox was first noticed in 1881 when the seats in the U.S. House of Representatives were reapportioned after the 1880 census. At that time the U.S. Census Bureau created a table which showed the number of seats each state would have for various possible sizes of the House of Representatives. They did this for possible sizes of the House from 275 total seats to 350 total seats. This table showed a strange occurrence as the size of the House of Representatives increased from 299 to 300. With 299 total seats, Alabama would receive 8 seats. However, if the house size was increased to 300 total seats, Alabama would only receive 7 seats. Increasing the overall number of seats caused Alabama to lose a seat.
The Alabama paradox happens when an increase in the total number of seats results in a decrease in the number of seats for a given state.
A mother has an incentive program to get her five children to read more. She has 30 pieces of candy to divide among her children at the end of the week based on the number of minutes each of them spends reading. The minutes are listed below in Table \(\PageIndex{6}\).
| Child | Abby | Bobby | Charli | Dave | Ed | Total |
|---|---|---|---|---|---|---|
| Population | 188 | 142 | 138 | 64 | 218 | 750 |
Use Hamilton’s method to apportion the candy among the children. The standard divisor (Equation \ref{SD}) is
\[S D=\frac{750}{30}=25 \nonumber \]
After dividing each child’s time by the standard divisor, and finding the lower quotas for each child, there are three pieces of candy left over. They will go to Ed, Bobby, and Dave, in that order, since they have the largest fractional parts of their quotas.
| Child | Abby | Bobby | Charli | Dave | Ed | Total |
|---|---|---|---|---|---|---|
| Population | 188 | 142 | 138 | 64 | 218 | 750 |
| Standard Quota | 7.520 | 5.680 | 5.520 | 2.560 | 8.720 | 30.000 |
| Lower Quota | 7 | 5 | 5 | 2 | 8 | 27 |
| Final Allocation | 7 | 6 | 5 | 3 | 9 | 30 |
At the last minute, the mother finds another piece of candy and does the apportionment again. This time the standard divisor will be 24.19. Bobby, Abby, and Charli, in that order, will get the three left over pieces this time.
| Child | Abby | Bobby | Charli | Dave | Ed | Total |
|---|---|---|---|---|---|---|
| Population | 188 | 142 | 138 | 64 | 218 | 750 |
| Standard Quota | 7.772 | 5.870 | 5.705 | 2.646 | 9.012 | 31.005 |
| Lower Quota | 7 | 5 | 5 | 2 | 9 | 28 |
| Final Allocation | 8 | 6 | 6 | 2 | 9 | 31 |
Notice that adding another piece of candy (a seat) caused Dave to lose a piece while Abby and Charli gain a piece. This is an example of the Alabama paradox.
9.3: Huntington-Hill Method
The Huntington-Hill method is the method currently used to apportion the seats for the U.S. House of Representatives. As with the other apportionment methods, the method of rounding off the quotas is what distinguishes this method from the others. The Huntington-Hill method starts out similarly to Webster’s method since some quotas are rounded up and some quotas are rounded down. The difference is that the cut-off for rounding is not 0.5 anymore. Now the cut-off depends on the geometric mean between the lower and upper quotas.
The geometric mean \(G\) of two positive numbers \(A\) and \(B\) is
\[G = \sqrt{AB} \label{gm} \]
Example \(\PageIndex{1}\): Geometric Mean
Find the geometric mean between 5 and 6.
Note that the geometric mean between A and B must be a number between A and B. In this example the geometric mean of 5.477 is between 5 and 6.
Summary of the Huntington-Hill Method:
- Find the standard divisor,
. Use the standard divisor as the first modified divisor.
- Divide each state’s population by the modified divisor to get the modified quota.
- Round each modified quota to the nearest integer using the geometric mean as the cut off. If the quota is less than the geometric mean between the upper and lower quotas, round the quota down to the lower quota. If the quota is more than the geometric mean between the upper and lower quotas, round the quota up to the upper quota.
- Find the sum of the rounded quotas.
- If the sum is the same as the number of seats to be apportioned, you are done. If the sum is too big, pick a new modified divisor that is larger than d. If the sum is too small, pick a new modified divisor that is smaller than d. Repeat steps two through five until the correct number of seats are apportioned.
Example \(\PageIndex{2}\): Huntington-Hill Method
Use the Huntington-Hill method to apportion the 25 seats in Hamiltonia from Example \(\PageIndex{2}\).
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | Total |
| Population | 24,000 | 56,000 | 28,000 | 17,000 | 65,000 | 47,000 | 237,000 |
The first step is to use the standard divisor as the first modified divisor. We also include a row for the geometric mean between the upper and lower quotas for each state.
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | 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 | |
| Lower Quota | 2 | 5 | 2 | 1 | 6 | 4 | |
| Upper Quota | 3 | 6 | 3 | 2 | 7 | 5 | |
| Calculation for G | |||||||
| Geometric Mean | 2.449 | 5.477 | 2.449 | 1.414 | 6.481 | 4.472 | |
| Rounded Quota | 3 | 6 | 3 | 2 | 7 | 5 | 26 |
The total number of seats, 26, is too big so we need to try again by making the modified divisor larger. Try d = 10,500.
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | 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 | |
| Geometric Mean | 2.449 | 5.477 | 2.449 | 1.414 | 6.481 | 4.472 | |
| Rounded Quota | 3 | 6 | 3 | 2 | 7 | 5 | 26 |
| d = 10,500 | 2.29 | 5.33 | 2.67 | 1.62 | 6.19 | 4.48 | |
| Geometric Mean | 2.449 | 5.477 | 2.449 | 1.414 | 6.481 | 4.472 | |
| Rounded Quota | 2 | 5 | 3 | 2 | 6 | 5 | 23 |
The total number of seats, 23 is too small. We need to try again with a modified divisor between 9480 and 10,500. Since 23 is further from 25 than 26 is, try a divisor closer to 9480. Try d = 9800.
| State | Alpha | Beta | Gamma | Delta | Epsilon | Zeta | 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 | |
| Geometric Mean | 2.449 | 5.477 | 2.450 | 1.414 | 6.481 | 4.472 | |
| Rounded Quota | 3 | 6 | 3 | 2 | 7 | 5 | 26 |
| d = 10,500 | 2.29 | 5.33 | 2.67 | 1.62 | 6.19 | 4.48 | |
| Geometric Mean | 2.449 | 5.480 | 2.450 | 1.410 | 6.480 | 4.470 | |
| Rounded Quota | 2 | 5 | 3 | 2 | 6 | 5 | 23 |
| d = 9800 | 2.4490 | 5.71 | 2.86 | 1.73 | 6.63 | 4.80 | |
| Geometric Mean | 2.4495 | 5.480 | 2.450 | 1.410 | 6.480 | 4.470 | |
| Rounded Quota | 2 | 6 | 3 | 2 | 7 | 5 | 25 |
Note: It was necessary to use more decimal places for Alpha’s quota than the other quotas in order to see which way to round off.
This is the same apportionment we got with most of the other methods.
Example \(\PageIndex{3}\): Comparison of all Apportionment Methods
In the city of 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 fire houses using Hamilton’s, Jefferson’s, Adams’s, Webster’s, and Huntington-Hill’s methods.
The standard divisor is
Hamilton’s Method:
Table \(\PageIndex{6}\): Hamilton’s Method for Adamstown
Start by dividing each population by the standard divisor and rounding each standard quota down.
| District | A | B | C | D | E | Total |
| Population | 25,010 | 8,760 | 11,590 | 9,025 | 15,080 | 69,465 |
| Standard Quota | 15.121 | 5.296 | 7.007 | 5.456 | 9.117 | 41.998 |
| Lower Quota | 15 | 5 | 7 | 5 | 9 | 41 |
| Final Allocation | 15 | 5 | 7 | 6 | 9 | 42 |
Using the lower quotas, there is one firefighter left over. Assign this firefighter to District D since D has the largest fractional part.
Jefferson’s Method
Jefferson’s method always rounds down making the sum of the lower quotas too small. Make the standard divisor smaller to get the first modified divisor. The results are summarized below in Table \(\PageIndex{7}\).
Guess #1: d = 1600. The sum of 41 is still too small so make the modified divisor smaller.
Guess #2: d = 1550. The sum is 42 so we are done.
| District | A | B | C | D | E | Total |
| Population | 25,010 | 8,760 | 11,590 | 9,025 | 15,080 | 69,465 |
| d = 1600 | 15.631 | 5.475 | 7.244 | 5.641 | 9.425 | |
| quota | 15 | 5 | 7 | 5 | 9 | 41 |
| d = 1550 | 16.135 | 5.652 | 7.477 | 5.823 | 9.729 | |
| Final Allocation | 16 | 5 | 7 | 5 | 9 | 42 |
Adams’s Method:
Adams’s method always rounds up making the sum of the upper quotas too large. Make the standard divisor larger to get the first modified divisor. The results are summarized below in Table \(\PageIndex{8}\).
Guess #1: d = 1700. The total is too still too large so make the modified divisor larger.
Guess #2: d = 1900. Now the total is too small so make the modified divisor smaller.
Guess #3: d = 1750. The total is too large again so make the modified divisor larger.
Guess #4: d = 1775. The sum is 42 so we are done.
| 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 | 4.867 | 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 |
Webster’s Method:
Webster’s method rounds the usual way so we cannot tell if the sum is too large or too small right away. Try the standard divisor as the first modified divisor. The results are summarized below in Table \(\PageIndex{9}\).
Guess #1: d = 1654. The sum of 41 is too small so make the modified divisor smaller.
Guess #2: d = 1600. The sum of 43 is too large so make the modified divisor larger.
Guess #3: d = 1625. The sum is 42 so we are done.
| District | A | B | C | D | E | Total |
| Population | 25,010 | 8,760 | 11,590 | 9,025 | 15,080 | 69,465 |
| d = 1654 | 15.121 | 5.296 | 7.007 | 5.456 | 9.117 | |
| quota | 15 | 5 | 7 | 5 | 9 | 41 |
| d = 1600 | 15.631 | 5.475 | 7.244 | 5.641 | 9.425 | |
| quota | 16 | 5 | 7 | 6 | 9 | 43 |
| d = 1625 | 15.391 | 5.391 | 7.132 | 5.554 | 9.280 | |
| Final Allocation | 15 | 5 | 7 | 6 | 9 | 42 |
Huntington-Hill’s Method:
Huntington-Hill’s method rounds off according to the geometric mean. Use the standard divisor as the first modified divisor. The results are summarized below in Table \(\PageIndex{10}\).
Guess #1: d = 1654. The sum of 41 is too small so make the modified divisor smaller. Look at District D. It was really close to being rounded up rather than rounded down so we do not need to change the modified divisor by very much.
Guess #2: d = 1625. The sum is 42 so we are done.
| District | A | B | C | D | E | Total |
| Population | 25,010 | 8,760 | 11,590 | 9,025 | 15,080 | 69,465 |
| d = 1654 | 15.121 | 5.296 | 7.007 | 5.456 | 9.117 | |
| Geometric mean | 15.492 | 5.477 | 7.483 | 5.477 | 9.487 | |
| quota | 15 | 5 | 7 | 5 | 9 | 41 |
| d = 1625 | 15.391 | 5.391 | 7.132 | 5.554 | 9.280 | |
| Geometric mean | 15.492 | 5.477 | 7.483 | 5.477 | 9.487 | |
| Final Allocation | 15 | 5 | 7 | 6 | 9 | 42 |
Hamilton’s, Adams’s, Webster’s, and Huntington-Hill’s methods all gave the same apportionment: 15 firefighters to District A, five to District B, seven to District C, six to District D, and nine to District E.
Jefferson’s method gave a different apportionment: 16 firefighters to District A, five to District B, seven to District C, five to District D, and nine to District E.