Largest remainders methods: more remaining seats than parties?

akazukin5151

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

Toby Pereira

@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.

akazukin5151

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!