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    Largest remainders methods: more remaining seats than parties?

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    • A
      akazukin5151 last edited by

      Hello,

      I've found a weird edge case in using the Droop quota for the largest remainders methods in proportional representation. Consider the following election results for 100 voters and 30 seats:

      Party Votes
      A 23
      B 26
      C 51

      The droop quota is 1 + (100 / (1 + 30)) = 4.23. There are various flooring strategies, but they all should yield 4.

      A total of 23 seats are automatically allocated:

      Party Formula Automatic seats
      A floor(23 / 4) 5
      B floor(26 / 4) 6
      C floor(51 / 4) 12

      There are still 7 seats remaining, but there are only 3 parties. It doesn't matter what the remainders are, there are only 3 remainders.

      The examples on Wikipedia assumes that the remaining number of seats are less than the number of parties. This election does works if there are 20 seats total:

      The quota is 1 + (100 / (1 + 20)) = 5.76 ~= 5. 19 seats are automatically allocated, the largest remainder is party A with 0.6, hence the result is A: 5, B: 5, C : 10 seats.

      My question is, what is supposed to happen in this situation? Will there just be unfilled seats in parliament? How can this be fixed or mitigated? Did I make any mistakes above? Thank you in advance


      PS: I'd post this to the proportional representation category but there was no new topic button

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      • T
        Toby Pereira @akazukin5151 last edited by

        @akazukin5151 First of all, I'd say that the Droop Quota isn't very good when the quota is so small. It's basically a rounding up of the Hagenbach-Bischoff Quota, which would be approx 3.23 in this case. So 4 is a massive difference from this (24% more I think). A more realistic quota for most elections would be in the thousands, so rounding up to the next integer would be a very small percentage difference. I think this is largely where the problem stems from.

        But if we're going with 4 as the quota anyway, I think we're just minimising the amount over quota they go. So we have the following seats "owed" to each party:

        A - 5.75
        B - 6.5
        C - 12.75

        You'd end up with:

        A - 7 (1.25 over)
        B - 8 (1.5 over)
        C - 14 (1.25 over)

        This gives 29 seats and then there would be one last seat to go to either A or C (a tie) putting them 2.25 over quota.

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        • A
          akazukin5151 @Toby Pereira last edited by

          I'd say that the Droop Quota isn't very good when the quota is so small.

          I agree, I'm not allocating seats in a real life election so this is just purely theoretical.

          A more realistic quota for most elections would be in the thousands, so rounding up to the next integer would be a very small percentage difference. I think this is largely where the problem stems from.

          So that's the core reason, makes sense to me.

          But if we're going with 4 as the quota anyway, I think we're just minimising the amount over quota they go. So we have the following seats "owed" to each party:

          That's pretty smart. This is the algorithm as I understand it:

          If there are still seats remaining to elect, award seats to the party with the smallest difference between seats won and "party quota" (seats_won - votes / quota). Repeat until all seats filled

          (For simplicity I'm breaking ties in favour of the first party)

          So my example would start from the automatic seats [5, 6, 12]. Award seats based on the largest remainders method as usual, stopping when all parties has been awarded a seat: [6, 6, 12], [6, 6, 13], [6, 7, 13].

          The remaining 4 seats would be filled like this

          [7, 7, 13], [7, 7, 14], [7, 8, 14], [8, 8, 14]

          Thank you very much for the help!

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