General discussion about Congressional Apportionment (user search)
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  General discussion about Congressional Apportionment (search mode)
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Author Topic: General discussion about Congressional Apportionment  (Read 16144 times)
True Federalist (진정한 연방 주의자)
Ernest
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« on: February 19, 2010, 03:53:23 PM »

I think the arithmetic mean would be fairer as divisor than the geometric one. The current system clearly advantages small States (a State whose population is equivalent to 1.41 seats has as much chances to get its second seat as one of 52.5 to get its 53th). This is particularly senseless, considering that anyways small States get overrepresented thanks to the Senate.

Depends on what one considers "fair".  Using the geometric mean keeps the range of people per representative smaller.

For sake of example, assume we have a population of 435 million to apportion among 435 seats for an average of 1,000,000 per seat.

Assuming no State is getting its first representative because of the floor of 1 seat, using the arithmetic mean gives a range in potential district sizes of 750,000 to 1,500,000 while the geometric mean gives a range of 707,107 to 1,414,214.  Hence the geometric mean gives a 5.72% smaller variance in district sizes than the arithmetic mean does.

Another method of reducing the variance is to increase the number of seats, and not just because of the smaller averages.  For example, if we had an average district size of 1,000,000 but each State had enough population to be guaranteed two districts, then AM has a range of 833,333 to 1,250,000 (416,667) and GM has a range of 816,497 to 1,224,745 (408,248) both of which are considerably smaller ranges than if having 1 seat is a possibility.  GM still has an advantage, but it is a smaller advantage (2.02%).  No matter how many seats is the minimum of honestly earned seats, GM will always be have the smaller variance in district sizes, tho the advantage will be smaller as the number of seats increases.

BTW, the reason I oppose statehood for the Virgin Islands, American Samoa, Guam and the Northern Mariana Islands is that they don't have enough population to warrant a Representative of their own without the floor.  (Even if Guam and the NMI were to unite as the Mariana Islands, they'd be too small population-wise.)
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #1 on: February 19, 2010, 07:34:34 PM »

I'm in favor of the cube root rule myself, which by the Census Bureau's estimates would give the US House 676 seats. and UN if it had a body apportioned by population 1,895 seats.
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #2 on: February 20, 2010, 07:18:49 PM »

BTW, the reason I oppose statehood for the Virgin Islands, American Samoa, Guam and the Northern Mariana Islands is that they don't have enough population to warrant a Representative of their own without the floor.  (Even if Guam and the NMI were to unite as the Mariana Islands, they'd be too small population-wise.)
You oppose statehood for Wyoming also? It'd be a lot fairer (though also completely absurd) if we could gerrymander a couple hundred thousand Montana voters into Wyoming's district.
No.  If we added one more seat to the HoR and doubled Wyoming's population in 2000, it would have legitimately earned 2 seats under the current apportionment formula (its 2nd seat would be the 387th seat under that scenario), so there's no reason for Wyoming to not be a State.  For the 2000 Census the cutoff point for Statehood under my criteria would be 456,567 (half of the minimum population needed to get a second Representative in a House with 436 members). Wyoming was above  that with an apportionment population 495,304.

The population of all four minor territories combined in 2000 was only 389,929 so even together they didn't meet the target. Possibly the combined Marianas (NMI + Guam) might be able to grow into Statehood, especially if we ever increased the size of the House to use the cube root formula as both Guam and NMI are growing at a faster rate than the U.S. With 676 seats in 2010, the target number would be approximately 322 thousand and the combined Marianas will have approximately 270 thousand.  (Kept at 435 seats and the target number for 2010 will be approximately 500 thousand, and Wyoming will again be comfortably over that number.)
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #3 on: February 21, 2010, 02:38:46 PM »

As for the geometric mean that jimrtex omitted in in his previous post, it's justification is as follows :

If the population were the geometric mean of 200,000 and 300,000 (244,949 if rounded to the nearest integer) then if you had two representatives you would have 100,000*sqrt(1.5) people per representative and if you had three representatives you would have  100,000/sqrt(1.5) people per representative.

(Conversely with two representatives you have 1/sqrt(1.5) per 100,000 people and with three you have sqrt(1.5) per 100,000 people.)

Hence by the arguments jimrtex put forth concerning whether people per representative or representatives per people is the criterium to judge by, the geometric mean is equally fair to both criteria.  (What using the geometric mean minimizes is the difference in the logarithm of both ratios from the ideal value.)

The geometric mean, m,  of any two numbers a and b is the number such that m/a = b/m, which is why the multiplier and the divisor are the same above.  Insert n for a and n+1 for b and then solve for m and you get m=sqrt(n*(n+1)).

The geometric mean above is thus 100,000*sqrt(6).

100,000*sqrt(6)/2=100,000*sqrt(6)/sqrt(4)=100,000*sqrt(6/4)=100,000*sqrt(1.5)
100,000*sqrt(6)/3=100,000*sqrt(6)/sqrt(9)=100,000*sqrt(6/9)=100,000*sqrt(9/6)=100,000/sqrt(1.5)



The arithmetic mean does have one advantage over the geometric and harmonic means, which, while not relevant to apportionment where every contender for seats is guaranteed at least one seat, pertains to allotting seats in a proportional representation election.  Namely that that there is a natural cutoff between having 0 seats and having 1 seat.  This is because both the harmonic and geometric mean of 0 and any other number is 0, so if an unmodified geometric or harmonic mean were used to allocate seats, every party that got at least one vote would be guaranteed 1 seat.
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #4 on: February 21, 2010, 07:52:08 PM »

Also, Nevada wouldn't have had a single representative in the U.S. House under the methods used at the time with no floor for much of its time as a state, stretching through 1930s reapportionment and I believe through the 1940s reaportionment if the Congressional Dems hadn't switched from major fractions to equal proportions in 1941 to prevent then-solidly Democratic Arkansas from losing a seat to then-Republican-leaning Michigan.

True, but Nevada was granted Statehood way too early so as to gin up three Union electoral votes in 1864.  Had it not been for concerns over the Confederacy and the Mormons, it would have been better to have split Nevada between Utah and California and then split California into a North and South California.  Of course had that happened, there never would have been a Las Vegas.
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #5 on: February 22, 2010, 05:58:34 PM »

Las Vegas might have happened anyway. Without the state of Nevada, the Clark county area was part of New Mexico territory. It wasn't even part of NV until 2 years after statehood. This is an 1857 map of NM terr:



There would be a village there, but there likely wouldn't be any casinos.  For a long time Nevada was the only State that allowed legal gambling on anything besides horseracing.
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #6 on: February 26, 2010, 09:01:48 PM »

Washington's veto seems quite legitimate, since the Constitution did indeed set a minimum population per representative

The Constitutional limit is "The Number of Representatives shall not exceed one for every thirty Thousand" not "The Number of Representatives for each State shall not exceed one for every thirty Thousand".

So long as the total number of Representatives apportioned as a result of the 1790 census was 120 or less, a method that exceeded that limit for a particular state would have been Constitutional.

Incidentally, if any of Webster, Hamilton, or Huntington-Hill been applied to the 1790 Census, Delaware would have had 2 Representatives instead of 1 and Virginia would have had 18 instead of 19.  Under Huntington-Hill an additional change is that Vermont would have had 3 instead of 2 while Pennsylvania would have had 12 instead of 13.
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #7 on: February 27, 2010, 07:16:18 PM »

Washington's veto seems quite legitimate, since the Constitution did indeed set a minimum population per representative
The Constitutional limit is "The Number of Representatives shall not exceed one for every thirty Thousand" not "The Number of Representatives for each State shall not exceed one for every thirty Thousand".
Washington's veto message says:

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Since 8 States were affected, they must have originally tried to apportion the maximum 120 possible Representatives in the bill Washington vetoed, but doing so would run afoul of Washington's stated objection no matter what method was used.  With Washington's stated objection the maximum number of Representatives could be 112 for the 1790 census.

So any objection to using Hamilton's method on Constitutional grounds is a sham as by appropriately reducing the size of the House, any method could meet the stated reason, and without a reduction, none could.

The maximum size for the House apportioned using the 1790 Census and Hamilton's method while keeping the Ratio in each State above 30K per Representative would have been 100, which would have given Virginia only 17 of 100 Representatives under Hamilton (or under Webster or Huntington-Hill, but the House would need to be even smaller under those two methods, at most 99 for Webster and 92 for Huntington-Hill).

Yet rather than increase the quota to be used for Hamilton's method, they switched to a different method that produced a bias for large States such as Virginia.  It could have been worse.  Using a quota that would have produced a House of 112 Jefferson's method would have given Virginia 21 of 112 seats for an even larger bias.

Note: All of the stated maximums above come from selecting a quota that produces as large a House as possible using the 1790 Census data, but does not necessarily work a priori, assuming the Representation limit applies per State without knowing what the numbers were.  Using a quota of 30,000 in Jefferson's method works with any set of data but results in there being a House of 105 members using the 1790 data. Webster's quota needs to be 40,000 (or more) which results in a House of 91 members. Huntington-Hill needs a priority value of at least 42,427 (30,000 * sqrt(2)) which results in a House of 86 members. The quota needed for Hamilton to work under all data sets is 60,000 which results in a House of 60 members.
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #8 on: March 04, 2010, 07:13:40 PM »

Actually the problem with the fairness rule that the divisor methods have results from choosing to have a fixed number of seats.  If one instead chooses a fixed priority value, then the divisor methods will always meet the fairness rule, but there might be extra or fewer seats compared to the ideal.  For example, in 2000, using the Hamilton-Hill method with a fixed priority value of 646,952 (which is 281,424,177 / 435) one ends up with a 433 seat House, with California and North Carolina both not getting a seat that they received due to the requirement for a fixed size of the house., but in 1990, one ends up with a 438 seat House, with Massachusetts, New Jersey, and New York getting the 436th, 437th, and 438th seats respectively.

There is no requirement that the size of the House be determined a priori.

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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #9 on: March 05, 2010, 04:27:50 PM »

Actually if you're going to talk fairness, one ought to consider the effect of removing the States that under any system will only get the mandated minimum of 1 from consideration.

If you do that and remove Alaska, North Dakota, Vermont, and Wyoming (the states with less than 1/435 of the apportionment population) you get slightly different values, tho no effect under your fairness criteria, which would still recommend shifting a seat from California to Utah.

However, in 1970 your method would recommend shifting a seat from South Dakota to Connecticut and a second from Montana to Oregon, whereas my refinement of only considering the 432 seats left after removing States that under any rational system get only 1 seat (North Dakota had more than 1/435 of the population in 1970) would have South Dakota's 2nd seat going to Oregon, and then Oregon handing the seat it just got from Oregon over to Connecticut.  (I had to go to four decimal places to see whether Montana or Oregon would send a seat to Connecticut.)
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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #10 on: March 05, 2010, 07:34:28 PM »

But we're not considering how to allocate 435 seats, but rather how to allocate 385 seats.
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