Chapter 15 Apportionment Part 3 Jefferson s Method

Chapter 15: Apportionment Part 3: Jefferson’s Method

A little History • The first Congressional bill ever to be vetoed by the President of the United States was a bill in 1790 containing a new apportionment of the House based on Hamilton’s Method. • The reason for the veto may have been related to the following requirement stated in the U. S. Constitution (Article 1, section 2): The number of people per single seat in the House should be at least 30, 000. • Remember that the standard divisor represents the average number of people per seat in the nation as a whole. • In 1790, there were 15 states and 105 seats in the House. According to the 1790 census, the U. S. population was 3, 615, 920. Thus the standard divisor would have been 3, 615, 920/105 = 34, 437. • So why the veto?

A little History • It is likely that at least one reason George Washington vetoed the 1790 House apportionment as calculated by Alexander Hamilton is that under Hamilton’s apportionment, two seats were assigned to Delaware while the 1790 census indicated a population of 55, 540 for Delaware. Therefore, there were actually 55, 540/2 = 27, 770 people per seat for Delaware, a violation of the Constitution. • How did this happen? The answer comes from the fact that Hamilton’s method had awarded Delaware the extra seat by rounding up the quota for Delaware. That is, in 1790, Delaware’s quota was q = (state population)/(standard divisor) = (55, 540)/(34, 437) = 1. 613. • Following Hamilton’s Method and rounding quota’s, it so happened that Delaware was rounded up to get 2 seats.

A little History • Congress was unable to override the Presidential veto, and to avoid a stalemate, they turned to Thomas Jefferson, who had devised another means of apportionment … • Jefferson’s Method was the method actually used for the first apportionment of the House, which was finally done in 1792. This method was then used until 1840 when Hamilton’s Method made a return. • To use Jefferson’s Method we must find a modified divisor such that when dividing each state’s population by this divisor, and rounding down, the sum of the adjusted quotas (seats) is equal to the total number of seats. • One justification for this method is that it may seem more fair in the sense that every state’s quota will be rounded down – based on division by the same modified divisor.

Jefferson’s Method – Example #1 • Let’s consider a simple example. Suppose there are only three states: Texas, Alabama and Illinois and 100 seats in the House. Suppose the populations are as given in the table. Population We begin by calculating the standard divisor, which is 20, 000/100 = 200. Texas 10, 030 Then we determine the lower quota for each state using the standard divisor… Illinois 9, 030 Alabama 940 Total 20, 000

Jefferson’s Method – Example #1 • Now we compute the initial apportionment, as defined previously, using the standard divisor of 20, 000/100 = 200. Population Texas Lower quotas, using standard divisor of 200 10, 030 50 Illinois 9, 030 45 Alabama 940 4 Total 20, 000 99 (1 seat still available) Note that not all of the 100 seats have been apportioned

Jefferson’s Method – Example #1 • We haven’t apportioned all of the seats using the standard divisor of 200, so we will choose another modified divisor… We choose a modified divisor so that when we round all of the quotas down, they add to the required total. Population Lower quotas (with standard divisor) Texas 10, 030 50 Illinois 9, 030 45 Alabama 940 4 Total 20, 000 99 (so 1 seat still available)

Jefferson’s Method – Example #1 • We find the sum of the lower quotas is less than the required total of 100 seats. Therefore, we seek a modified divisor to replace the standard divisor. By trial and error (or as described below) we find that a modified divisor of d = 196. 6 will work. Population Lower quotas (with standard divisor) Texas 10, 030 50 Illinois 9, 030 45 Alabama 940 4 Total 20, 000 99 (so 1 seat still available) Is this ok? Is it ok to change the divisor? The answer is yes – there is no constitutional (or mathematical) requirement for using the standard divisor. In fact, we are determining a value representing the lowest ratio of people per seat for any of the states. In this case, it can be found by adding one to each state’s apportionment and dividing into the population and then comparing each of these resulting values and taking the largest, which is 196. 67

Jefferson’s Method – Example #1 • We find the sum of the lower quotas is less than the required total of 100 seats. Therefore, we seek a modified divisor to replace the standard divisor. By trial and error (or as described below) we find that a modified divisor of d = 196. 6 will work. Population Lower quotas (with standard divisor) Texas 10, 030 50 Illinois 9, 030 45 Alabama 940 4 Total 20, 000 99 (so 1 seat still available) By finding the value of 196. 6 we are essentially solving a problem of optimization – to find the maximize value of the minimum number of people per seat in any state. This is what we call a maximin later in game theory. This was Jefferson’s problem – he wanted to make sure to satisfy the Constitutional requirement that there were at least 30, 000 people per seat in the apportionment of the House. So he was looking at the resulting ratios of people per seat in every state and sought to maximize this value – to make sure it was always more than 30, 000.

Jefferson’s Method – Example #1 • Now, using the modified divisor, we calculator lower modified quotas for each state… Population Initial lower quotas Texas 10, 030 50 Illinois 9, 030 45 Alabama 940 4 Total 20, 000 99 (so 1 seat still available) Lower quotas (with modified divisor of 196. 6) Total apportionment = 100

Jefferson’s Method – Example #1 • Now, using the modified divisor, we calculator lower modified quotas for each state… Population Initial lower quotas Texas 10, 030 50 51 Illinois 9, 030 45 45 Alabama 940 4 4 Total 20, 000 99 (1 seat still available) Lower quota using modified divisor of 196. 6 100 Final apportionment 100

Jefferson’s Method – Example #2 • Let’s consider another example of Jefferson’s Method. Suppose a country consists of six different states: A, B, C, D, E, F. Suppose the populations of these states are different and the country has a House of Representatives with 250 seats. • How should these 250 seats be apportioned? Let’s use Jefferson’s Method to answer that question. Here are the population figures from that country’s most recent census: A population B C D E F total 1, 646 6, 936 154 2, 091 685 988 12, 500 • To use Jefferson’s Method, we first need s, the standard divisor. In this case it’s s = 12, 500/250 = 50. (That is, on average, throughout the country, there should be about 50 people per Congressional district – or equivalently, about 50 people per Congressional seat. )

Jefferson’s Method – Example #2 population initial app. (divide and round down) A 1646/50 = 32 B 6936/50 =138 C 154/50 = 3 D 2091/50 = 41 E 685/50 = 13 F 988/50 = 19 total 12, 500 246 (so 4 seats remain) Initial apportionments are found by dividing population for each state by the standard divisor.

Jefferson’s Method – Example #2 population initial app. (divide and round down) A 1646/50 = 32 B 6936/50 =138 C 154/50 = 3 D 2091/50 = 41 E 685/50 = 13 F 988/50 = 19 total 12, 500 246 (so 4 seats remain) Initial apportionments are found by dividing population for each state by the standard divisor. Because there are still 4 seats to be assigned we will need to increase the apportionments of some of the states… We do this by searching for a new modified divisor…

Jefferson’s Method – Example #2 To find a modified divisor, we seek to maximize the minimum number of people per seat in any given state. population initial apportionment (after rounding down) A 1646/50 = 32 B 6936/50 =138 C 154/50 = 3 D 2091/50 = 41 E 685/50 = 13 F 988/50 = 19 total 12, 500 246 (so 4 seats remain) Standard Unfortunately, this example is quotas complicated by the fact that we must add a total of 4 seats to reach the required total of 250. We could find the necessary modified divisor by experimentation or by the following procedure… For each of the four remaining seats, we can temporarily add 1 to all of the apportionments – divide that result into the population – and determine the largest ratio. The state with the largest ratio will get the next seat. We then continue that process for each of the remaining seats.

Jefferson’s Method – Example #2 By experimentation, we find a modified divisor of 49. 5 will work. The final result is in the last column and represents the final apportionment… population initial apportionment (after rounding down) A 1646 B Final apportionment Modified quotas Modified lower quotas 1646/50 = 32 33. 25 33 33 6936/50 =138 140. 12 140 C 154/50 = 3 3. 11 3 3 D 2091/50 = 41 42. 24 42 42 E 685/50 = 13 13. 84 13 13 F 988/50 = 19 19. 95 19 19 total 12, 500 246 (so 4 seats remain) 252. 5 250