Apportionment Schemes Dan Villarreal MATH 490-02 Tuesday, Sept. 15, 2009.

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Presentation transcript:

Apportionment Schemes Dan Villarreal MATH Tuesday, Sept. 15, 2009

Sept. 15, 2009 Apportionment Schemes But first…a quick PSA

Sept. 15, 2009 Apportionment Schemes What is Apportionment? The apportionment problem is to round a set of fractions so that their sum is maintained at original value. The rounding procedure must not be an arbitrary one, but one that can be applied constantly. Any such rounding procedure is called an apportionment method.

Sept. 15, 2009 Apportionment Schemes Example In the NHL season, the Stanley Cup Champion Philadelphia Flyers won 51 games, lost 18 games, and tied 11 games. Won:51  63.75%  64% Lost:18  22.5%  23% Tied:11  13.75%  14% But this adds up to 101%, an impossibility!

Sept. 15, 2009 Apportionment Schemes Dramatis Personae George Washington Alexander Hamilton Thomas Jefferson Daniel Webster Delaware Virginia

Sept. 15, 2009 Apportionment Schemes The Constitution Amendment 14, Section 2:  Article I, Section 2: “The actual Enumeration shall be made within three Years after the first Meeting of the Congress of the United States, and within every subsequent Term of ten Years, in such Manner as they shall by Law direct. The Number of Representatives shall not exceed one for every thirty Thousand, but each State shall have at Least one Representative” “Representatives shall be apportioned among the several States according to their respective numbers, counting the whole number of persons in each State”

Sept. 15, 2009 Apportionment Schemes The First Apportionment For the third session of Congress ( ) House of Representatives set at states U.S. Population: 3,615,920 3,615, = 34,437 people/district

Sept. 15, 2009 Apportionment Schemes Standard Divisors 34,437 is our standard divisor for More generally, SD t = Where HS t is the size of the House of Representatives (or whatever overall body) for year t. Pop total HS t

Sept. 15, 2009 Apportionment Schemes Quotas The number of Congressional districts a state should get is its quota: Q i = Take Delaware, for example… Pop i SD t

Sept. 15, 2009 Apportionment Schemes If only it were that easy… THERE’S NO SUCH THING AS.613 CONGRESSPERSONS. Hence the need for apportionment schemes, a way to map the quotas in R onto apportionments in Z.

Sept. 15, 2009 Apportionment Schemes If only it were that easy… StatePopulationQuota Virginia630, Massachusetts475, Pennsylvania432, North Carolina353, New York331, Maryland278, Connecticut236, South Carolina206, New Jersey179, New Hampshire141, Vermont85, Georgia70, Kentucky68, Rhode Island68, Delaware55, Totals3,615,920105

Sept. 15, 2009 Apportionment Schemes More on quotas The lower quota is the quota rounded down (or the integer part of the quota): LQ i = ⌊ Q i ⌋ The upper quota is the quota rounded up: UQ i = ⌈ Q i ⌉ = ⌊ Q i ⌋ + 1

Sept. 15, 2009 Apportionment Schemes If only it were that easy… StatePopulationQuotaLQUQ Virginia630, Massachusetts475, Pennsylvania432, North Carolina353, New York331, Maryland278, Connecticut236, South Carolina206, New Jersey179, New Hampshire141, Vermont85, Georgia70, Kentucky68, Rhode Island68, Delaware55, Totals3,615,

Sept. 15, 2009 Apportionment Schemes Alexander Hamilton One author of Federalist Papers First Secretary of the Treasury Most importantly for our purposes, devised the Hamilton Method for apportioning Congressional districts to states

Sept. 15, 2009 Apportionment Schemes The Hamilton Method State i receives either its lower quota or upper quota in districts; those states that receive their upper quota are those with the greatest fractional parts

Sept. 15, 2009 Apportionment Schemes Back to 1790 StatePopulationQuotaFrac. Part Virginia630, Massachusetts475, Pennsylvania432, North Carolina353, New York331, Maryland278, Connecticut236, South Carolina206, New Jersey179, New Hampshire141, Vermont85, Georgia70, Kentucky68, Rhode Island68, Delaware55, StateFrac. Part Kentucky.995 South Carolina.989 Rhode Island.988 Connecticut.878 Massachusetts.803 New York.629 Delaware.613 Pennsylvania.570 Vermont.484 Virginia.310 North Carolina.266 New Jersey.214 New Hampshire.118 Maryland.088 Georgia.057

Sept. 15, 2009 Apportionment Schemes If only it were that easy… StatePopulationQuotaLQUQApportionment Virginia630, Massachusetts475, Pennsylvania432, North Carolina353, New York331, Maryland278, Connecticut236, South Carolina206, New Jersey179, New Hampshire141, Vermont85, Georgia70, Kentucky68, Rhode Island68, Delaware55, Totals3,615,

Sept. 15, 2009 Apportionment Schemes If only it were that easy… StatePopulationQuotaLQUQApportionment Virginia630, Massachusetts475, Pennsylvania432, North Carolina353, New York331, Maryland278, Connecticut236, South Carolina206, New Jersey179, New Hampshire141, Vermont85, Georgia70, Kentucky68, Rhode Island68, Delaware55, Totals3,615,

Sept. 15, 2009 Apportionment Schemes

Sept. 15, 2009 Apportionment Schemes Back to Square One President Washington vetoed the Apportionment Bill because he believed, following the counsel of Edmund Randolph and Thomas Jefferson, that it was unconstitutional: A DE 2 1 Pop DE 55,54030,000 =<=<

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox The Hamilton Method was Congress’s preferred method of apportionment from 1850 to In 1881, the Alabama Paradox was first discovered. The Census Bureau, as a matter of course, calculated apportionments for a range of House sizes; in this case, Something interesting and very weird happened between the tables for HS = 299 and 300…

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 Population Alabama1,262,794 Illinois3,078,769 Texas1,592,574

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 HS = 299 PopulationLQFrac.A Alabama1,262, Illinois3,078, Texas1,592,

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 HS = 299 PopulationLQFrac.A Alabama1,262, Illinois3,078, Texas1,592,

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 HS = 299HS = 300 PopulationLQFrac.ALQFrac.A Alabama1,262, Illinois3,078, Texas1,592,

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 HS = 299HS = 300 PopulationLQFrac.ALQFrac.A Alabama1,262, Illinois3,078, Texas1,592,

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 HS = 299HS = 300 PopulationLQFrac.ALQFrac.A Alabama1,262, Illinois3,078, Texas1,592,

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 HS = 299HS = 300 PopulationLQFrac.ALQFrac.A Alabama1,262, Illinois3,078, Texas1,592,

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 HS = 299HS = 300 PopulationLQFrac.ALQFrac.A Alabama1,262, Illinois3,078, Texas1,592,

Sept. 15, 2009 Apportionment Schemes The Alabama Paradox US population in 1880 was 49,369,595 For HS = 299, SD = 165,116 For HS = 300, SD = 164,565 HS = 299HS = 300 PopulationLQFrac.ALQFrac.A Alabama1,262, Illinois3,078, Texas1,592,

Sept. 15, 2009 Apportionment Schemes Back to 1793… This particular issue with the Hamilton Method was not discovered until 1881, but the Constitutional constraints meant that it could not be used in A new method was proposed by Thomas Jefferson: the Jefferson Method.

Sept. 15, 2009 Apportionment Schemes Thomas Jefferson Biographical Information: You know this all already… Had the good fortune never to take a class in Morton Hall

Sept. 15, 2009 Apportionment Schemes The Jefferson Method Rather than use the standard divisor SD, the Jefferson Method uses the population of the smallest district, d. Each state receives an adjusted quota; this will need to be rounded down the actual apportionment: A i = ⌊ ⌋ Pop i d

Sept. 15, 2009 Apportionment Schemes The Jefferson Method In 1793, Jefferson used d = 33,000, so A VA = ⌊ 630,560 / 33,000 ⌋ = ⌊ ⌋ = 19 A DE = ⌊ 55,540 / 33,000 ⌋ = ⌊ ⌋ = 1 But how do we determine d in the first place?

Sept. 15, 2009 Apportionment Schemes Finding the Critical Divisor Start with the lower quota of each state; this is its tentative apportionment, n i. Next, find the critical divisor for each state: Pop i n i + 1 For example, d VA = 630,560 / (18 + 1) = 33,187 d DE = 55,540 / (1 + 1) = 27,770 d i =

Sept. 15, 2009 Apportionment Schemes The Critical Divisor The critical divisor for each state is the divisor for which the state will be entitled to n i + 1 seats. For example, if d > 27,770, Delaware gets only 1 seat, but for d ≤ 27,770, Delaware gets 2. But then Virginia gets ⌊ 630,560 / 27,770 ⌋ = ⌊ ⌋ = 22 seats. This will surely result in an overfull House Thus, d will need to be greater than 27,770

Sept. 15, 2009 Apportionment Schemes The Jefferson Method Step 1: Assume a tentative apportionment of the lower quota for each state: n i = LQ i Step 2: Determine the critical divisor d i for each state and rank by d i Step 3: If any seats remain to be filled, grant one to the state with the highest d i ; recompute d i for this state since its n i has now increased by 1. Step 4: Iterate Step 3 until the House is filled.

Sept. 15, 2009 Apportionment Schemes The Jefferson Method This method actually was used for the 1793 apportionment, and it resulted in Virginia receiving 19 seats to Delaware’s one. Used until about 1840 Not subject to the Alabama paradox But fails to satisfy the quota condition…

Sept. 15, 2009 Apportionment Schemes The Quota Condition The quota condition is twofold:  1. No state may receive fewer seats than its lower quota  2. No state may receive more seats than its upper quota The Jefferson Method does just fine with 1, but not 2

Sept. 15, 2009 Apportionment Schemes Example U.S. population in 1820 was 8,969,878, with a House size of 213, so SD = 8,969,878 / 213 = 42,112 New York had a population of 1,368,775: Q NY = 1,368,775 / 42,112 = So if the quota condition was satisfied, New York’s delegation should be either 32 or 33 Using the Jefferson Method and d = 39,900, we actually get 34 seats for New York

Sept. 15, 2009 Apportionment Schemes What’s the Problem? The Jefferson Method always skews in favor of the large states. Let u i = p i / d be the state’s adjusted quota. Then A i = ⌊ u i ⌋. Now compare u i with the state’s quota: M = = / = × = Then u i = M * Q i => a i = ⌊ M * Q i ⌋ The rich only get richer… u i Pop i Pop i Pop i SD SD Q i d SD d Pop i d

Sept. 15, 2009 Apportionment Schemes The Webster Method Daniel Webster devised an apportionment method that was similar in nature to Jefferson’s, but that did not unconditionally favor large states. Used for reapportionments, then

Sept. 15, 2009 Apportionment Schemes The Webster Method Step 1: Determine SD, and find the quota Q i for each state i. Step 2: Round each quota up or down and let this be the tentative apportionment n i for each state. Step 3: Determine the total apportionment at this point. 3 cases:  1. The total apportionment equals HS  2. The total apportionment is greater than HS  3. The total apportionment is less than HS

Sept. 15, 2009 Apportionment Schemes Adjusting the Apportionment If we have an overfill, at least one or more seats needs to be pared off. Let the critical divisor be d i - = p i / (n i - 1/2). The state with the smallest d i - will be the next to lose a seat. Conversely, if we have an underfill, we need to add more seats. Let the critical divisor be d i + = p i / (n i + 1/2). The state with the smallest d i + will be the next to gain a seat. Iterate either process until done.

Sept. 15, 2009 Apportionment Schemes Large State Bias How does the Webster Method avoid susceptibility to the large-state bias exhibited by the Jefferson Method? We get a similar expression for M: M = SD/d

Sept. 15, 2009 Apportionment Schemes Large State Bias M > 1 when there is an underfill, thus in this circumstance, the larger states are more likely to receive another seat But when there is an overfill and we must subtract, M < 1, and the larger states are more likely to get a seat subtracted Equally likely to get an overfill or underfill Thus, equally likely that the Webster Method will favor neither large nor small

Sept. 15, 2009 Apportionment Schemes Timeline Present Jefferson Method Webster Method Hamilton Method Webster Method Hill-Huntington Method

Sept. 15, 2009 Apportionment Schemes Hill-Huntington Method Step 1: Start with assumption that each state gets 1 seat (i.e., set n i = 1 for all i) Step 2: Calculate the priority value for each state Pop i (n i (n i + 1)) 1/2 Step 3: The state with the greatest PV i is granted the next seat, increasing its tentative apportionment n i by 1; recalculate this state’s PV i Step 4: Iterate Step 3 until the House is filled. PV i,n =

Sept. 15, 2009 Apportionment Schemes Hill-Huntington in 2000 US population in 2000: 281,421, seats in the House California population: 33,871,648 Texas population: 20,851,820 PV CA,1 = 33,871,648 / (2) 1/2 = 23,992,697 PV TX,1 = 20,851,820 / (2) 1/2 = 14,781,356 PV CA,2 = 33,871,648 / (6) 1/2 = 13,852,190 ortionment/00pvalues.txt ortionment/00pvalues.txt

Sept. 15, 2009 Apportionment Schemes The Population Paradox The Hill-Huntington Method is immune to the Alabama paradox, but may violate the quota condition. In the 1970s, two mathematicians attempted to devise a method that was immune to both violations, and they did…but another paradox popped up: the population paradox. This paradox occurs when the population of one state increases at a greater rate than others, but fails to gain a seat.

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: StateOld CensusQA A5,525,381 B3,470,152 C3,864,226 D201,203 Tot13,060,962

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 StateOld CensusQA A5,525,381 B3,470,152 C3,864,226 D201,203 Tot13,060,962

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 SD = 13,060,962 / 100 = 130,610 StateOld CensusQA A5,525,381 B3,470,152 C3,864,226 D201,203 Tot13,060,962

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 SD = 13,060,962 / 100 = 130,610 StateOld CensusQA A5,525, B3,470, C3,864, D201, Tot13,060,962

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 SD = 13,060,962 / 100 = 130,610 StateOld CensusQA A5,525, B3,470, C3,864, D201, Tot13,060,962

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 SD = 13,060,962 / 100 = 130,610 StateOld CensusQA A5,525, B3,470, C3,864, D201, Tot13,060,962

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 StateOld CensusQANew CensusQA A5,525, ,657,564 B3,470, ,507,464 C3,864, ,885,693 D201, ,049 Tot13,060,96213,251,770

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 SD = 13,251,770 / 100 = 132,518 StateOld CensusQANew CensusQA A5,525, ,657,564 B3,470, ,507,464 C3,864, ,885,693 D201, ,049 Tot13,060,96213,251,770

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 SD = 13,251,770 / 100 = 132,518 StateOld CensusQANew CensusQA A5,525, ,657, B3,470, ,507, C3,864, ,885, D201, , Tot13,060,96213,251,770

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 SD = 13,251,770 / 100 = 132,518 StateOld CensusQANew CensusQA A5,525, ,657, B3,470, ,507, C3,864, ,885, D201, , Tot13,060,96213,251,770

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 SD = 13,251,770 / 100 = 132,518 StateOld CensusQANew CensusQA A5,525, ,657, B3,470, ,507, C3,864, ,885, D201, , Tot13,060,96213,251,770

Sept. 15, 2009 Apportionment Schemes The Population Paradox Exercise 10, COMAP page 535: House size set at 100 State D lost population, yet gained a seat! StateOld CensusQANew CensusQA A5,525, ,657, B3,470, ,507, C3,864, ,885, D201, , Tot13,060,96213,251,770

Sept. 15, 2009 Apportionment Schemes SO MANY PARADOXES!!! The apportionment methods that Congress has used have either violated the quota condition (Jefferson, Webster, Hill-Huntington) or the Alabama and population paradoxes (Hamilton) The quota method (never used by Congress) violates the population paradox Is this just another instance of that old joke?

Sept. 15, 2009 Apportionment Schemes SO MANY PARADOXES!!! It turns out that this is endemic to the situation Theorem “No apportionment method that satisfies the quota condition is free of paradoxes” (COMAP, p. 519) Proof The only methods that are free of paradoxes are the divisor methods (Jefferson, Webster, Hill-Huntington). But the divisor methods are all subject to violating the quota condition. Thus, we are basically screwed.

Sept. 15, 2009 Apportionment Schemes Hill-Huntington…in 1790! StatePopulationQuotaJ Virginia630, Massachusetts475, Pennsylvania432, North Carolina353, New York331, Maryland278, Connecticut236, South Carolina206, New Jersey179, New Hampshire141, Vermont85, Georgia70, Kentucky68, Rhode Island68, Delaware55, Totals3,615,920105

Sept. 15, 2009 Apportionment Schemes Hill-Huntington…in 1790! StatePopulationQuotaJH Virginia630, Massachusetts475, Pennsylvania432, North Carolina353, New York331, Maryland278, Connecticut236, South Carolina206, New Jersey179, New Hampshire141, Vermont85, Georgia70, Kentucky68, Rhode Island68, Delaware55, Totals3,615,920105

Sept. 15, 2009 Apportionment Schemes Hill-Huntington…in 1790! StatePopulationQuotaJHH-H Virginia630, Massachusetts475, Pennsylvania432, North Carolina353, New York331, Maryland278, Connecticut236, South Carolina206, New Jersey179, New Hampshire141, Vermont85, Georgia70, Kentucky68, Rhode Island68, Delaware55, Totals3,615,920105

Sept. 15, 2009 Apportionment Schemes The point of the story being… Delaware should’ve gotten 2 seats.

Sept. 15, 2009 Apportionment Schemes Selected Sources COMAP, For All Practical Purposes, 7 th ed. (2006), Chapter 14 Wikipedia (multiple pages) ts/ pdf ts/ pdf ion2.html ion2.html

Sept. 15, 2009 Apportionment Schemes Photo Credits Images: George Washington: Alexander Hamilton: amilton_portrait_by_John_Trumbull_1806.jpg amilton_portrait_by_John_Trumbull_1806.jpg Jefferson: jpg jpg Webster: Delaware: shirt.JPGhttp:// shirt.JPG Virginia: zone/usa/virginia/images/state-flag-virginia.jpghttp://wwp.greenwichmeantime.com/time- zone/usa/virginia/images/state-flag-virginia.jpg Constitution: fba/Globe_Photo/2008/08/15/we__ _8547.jpghttp://cache.boston.com/bonzai- fba/Globe_Photo/2008/08/15/we__ _8547.jpg Veto: