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Apportionment, Divisor Methods, and Quota Violations

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  • jimrtex4192
    Warren Smith did some apportionment simulations for the United States House of Representatives to give an estimate of how frequent quota violations would be,
    Message 1 of 5 , Aug 1, 2010
      Warren Smith did some apportionment simulations for the United States House of Representatives to give an estimate of how frequent quota violations would be, using the various divisors (harmonic, geometric, and arithmetic means).

      The British government has proposed using the arithmetic mean in apportioning constituencies among the 4 parts of the UK (England, Northern Ireland, Scotland, and Wales). How feasible would it be to repeat the simulation for the UK?

      Intuitively, I would expect quota violations to be more common in the UK, given that the apportionment is among only 4 areas, and England has such a dominant population share (5/6). With only 4 areas, it is more likely that the fractions will be unequally distributed (3/4 or all with fractions greater than 1/2; or less than 1/2, compared to the United States where it would be quite unlikely that 3/4 of the States would have a fraction greater than 1/2).

      I used the 2009 electorates for the 4 parts, and found a quota violation for apportionment of 31, 88, 207, 257, 289, 294, 326, 370, 376, 408, 413, 495, and 614 constituencies. This is 2.2% of the distributions from 20-649*.

      *The British government proposes a fixed apportionment of 598 constituencies (there are two additional constituencies which are totally excluded from apportionment on the basis of electorate, to bring the total to 600). I just scrolled my spreadsheet a couple of screens to get to 649. 20 is the number of UK constituencies necessary for Northern Ireland to be entitled to more than 1/2 of a constituency. There is a quota violation for all distributions less than 13 constituencies, because of the 1 seat guarantee to Scotland, Wales, and Northern Ireland.

      Was Warren Smith's estimate of 1/160 for a quota violation for an apportionment of 435 representatives? Is this equivalent to an expectation of an average of 1 quota violation each census for apportionment in the range of 355 to 514 (a range of 160 representatives)? That is, is the probability of a quota violation for a given population distribution, independent of the number of seats apportioned?

      The 2009 UK electorates have some particular relationships which might make an estimate based on one population distribution particularly bad. For example, the ratio of England:Scotland is almost exactly 10:1 (9.996:1), Wales has very close to 1/20 of the UK electorate (1/20.06), and England is fairly close to 5/6 of the UK electorate (5.04/6).

      What, if anything, is wrong with simply using independent rounding? That is divide the UK electorate by 598, and then divide that electoral quota into the electorate for each part, and then round based on whether the fraction is greater than 1/2?

      For the current UK electorate distribution, this would result in one fewer constituency being apportioned in 102 of 630 instance, and one additional constituency being apportioned in 104 of 630 instances. In all cases, but one**, the sum of the absolute deviation of the average electorate in each part from the UK electorate* is less than using Saint-Lague. In the other 424 cases, Saint-Lague and independent rounding produce the same apportionment.

      *When additional seats are added or subtracted, the UK electorate is adjusted. In general, independent rounding improves the deviation for the electorates in the part of the UK where the independent rounding adjusts the apportionment, that outweighs any change in other parts due to the change of the UK electorate.

      **The one instance where independent rounding was worse was when the number of UK constituencies was 50. In that instance, Wales was entitled to 2.492 constituencies, and would have received 3 under Saint-Lague. But this is simply a case where use of the harmonic mean (2.4) would have better for the independent rounding.

      In the case of the UK for actual apportionment numbers, use of the harmonic mean, geometric mean, and arithmetic make little difference. Northern Ireland is entitled to 15.301 of 598 constituencies. The values of the three means for (15,16) are: 15.484; 15.492; and 15.500.
    • Raph Frank
      ... You potentially get the wrong number of seats. However, with 4 parts, that is probably not an issue. Since the errors must add up to zero, and the max
      Message 2 of 5 , Aug 4, 2010
        On Sun, Aug 1, 2010 at 7:41 PM, jimrtex4192 <jimrtex@...> wrote:
        > What, if anything, is wrong with simply using independent rounding? That is divide the UK electorate by
        > 598, and then divide that electoral quota into the electorate for each part, and then round based on
        > whether the fraction is greater than 1/2?

        You potentially get the wrong number of seats. However, with 4 parts,
        that is probably not an issue.

        Since the errors must add up to zero, and the max error is 0.5, you
        can't be more than one seat out.

        You could have something like

        50.49 -> 50
        50.49 -> 50
        50.49 -> 50
        48.53 -> 49

        If only 2 districts ended up with an excess, then you couldn't have
        the 2 other districts lose seats.
      • Raph Frank
        Forgot to add 50.49 - 50 50.49 - 50 49.51 - 50 49.51 - 50 However, you could have 50.49 - 50 50.49 - 50 49.49 - 49 49.51 - 50 So, one loss.
        Message 3 of 5 , Aug 4, 2010
          Forgot to add

          50.49 -> 50
          50.49 -> 50
          49.51 -> 50
          49.51 -> 50

          However, you could have

          50.49 -> 50
          50.49 -> 50
          49.49 -> 49
          49.51 -> 50

          So, one loss.
        • jimrtex4192
          ... Never more than an error of one for the total. If the fractions add to 3.0, either 3 or 4 of the fractions must be greater than 0.5. If there are 4, then
          Message 4 of 5 , Aug 5, 2010
            --- In RangeVoting@yahoogroups.com, Raph Frank <raphfrk@...> wrote:
            >
            > On Sun, Aug 1, 2010 at 7:41 PM, jimrtex4192 <jimrtex@...> wrote:
            > > What, if anything, is wrong with simply using independent rounding? That is divide the UK electorate by
            > > 598, and then divide that electoral quota into the electorate for each part, and then round based on
            > > whether the fraction is greater than 1/2?
            >
            > You potentially get the wrong number of seats. However, with 4 parts,
            > that is probably not an issue.

            Never more than an error of one for the total.

            If the fractions add to 3.0, either 3 or 4 of the fractions must be greater than 0.5. If there are 4, then a 601st MP would be added.

            If the fractions add to 2.0, either 1, 2, or 3 of the fractions are greater than 0.5 (assume no fractions are 0.5). If there are 3, then, a 601st MP would be added, if there are 1, then only 599 would be needed.

            If the fractions add to 1.0, either 0 or 1 of the fractions are greater than 0.5. If there is 0, then only 599 MPs would be added.

            Using current electorates and a range of total constituencies, about 1/6 of the time there would be an extra MP and 1/6 of the time there would be one less.

            The average electorate in the part that is independently rounded (to a different number than Sainte-Lague) would be closer to the UK electorate. Those of the other's might be slightly closer or further (due to the change of the denominator).
          • jimrtex4192
            ... Which is the same as Sainte-Lague ... I adjusted the example to avoid a tie, and so that they totaled to 200. In this case Sainte-Lague apportions a 51st
            Message 5 of 5 , Aug 5, 2010
              --- In RangeVoting@yahoogroups.com, Raph Frank <raphfrk@...> wrote:
              >
              > Forgot to add
              >
              > 50.49 -> 50
              > 50.49 -> 50
              > 49.51 -> 50
              > 49.51 -> 50

              Which is the same as Sainte-Lague

              > However, you could have
              >
              > 50.497 -> 50
              > 50.496 -> 50
              > 49.497 -> 49
              > 49.510 -> 50

              I adjusted the example to avoid a tie, and so that they totaled to 200.

              In this case Sainte-Lague apportions a 51st seat to the 1st part.

              If you independently round, and then compute the deviation from the average electorate:

              A: -.00987 .00489 -.00499 51/200 vs 50/199
              B: .00992 .00487 -.00512 50/200 vs 50/199
              C: .01014 .00509 -.00505 49/200 vs 49/199
              D: -.00980 -.01475 .00495 50/200 vs 50/199

              So the deviation for the first three is less, and for the last is larger.

              But why re-calculate the UK electoral quota, just because there is one less MP? If the ideal is that each MP represent 1/598 of the electorate, and this can be better achieved with 597 or 599 MPs, why not?
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