[EM] Proportional election method needed for the Czech Green party - Council elections

[Raph Frank]

On Tue, May 4, 2010 at 8:12 AM, Peter Zbornik <pzbornik at gmail.com> wrote:
> I am affraid that this is not possible. First we have mostly odd-numbered
> council sizes, and secondly the gender rule does not require that half of
> the men should be men and the other half women.
> Our current gender rule goes as following: "for every three members of the
> body, there has to be one person of each sex". A five member council thus
> has to have one woman and one man. For seven members it is two men and two
> women.

For elimination based PR-STV, I think my suggestion would be the most
reasonable.

- Set the threshold at one larger than the required number
- protect from elimination members of a gender if elimination would
reduce their number below the threshold
- prohibit from election members of a gender if that election would
leave less than a threshold for the other gender
- on the last round, remove remove the restrictions

In a 5 person council, that means that there must be at least 1 man and 1 woman.

If the candidates were

Men:
M1
M2
M3
M4
M5

and

Women
W1
W2
W3

then an election might go something like

Round 1

M1: 20
M2: 15
M3: 15
M4: 10
M5: 10
W1: 10
W2: 8
W3: 12

Total: 100
Quota: 17

M1 gets elected + 3 are distributed

Round 2
M1: 17*
M2: 16 (+1)
M3: 15
M4: 10
M5: 11 (+1)
W1: 11 (+1)
W2: 8
W3: 12

W2 is lowest, so is eliminated, +8 are distributed

Round 3
M1: 17*
M2: 16
M3: 15
M4: 13 (+3)
M5: 14 (+3)
W1: 13 (+2)
W2: 0
W3: 12

W3 is lowest.

However, eliminating W3 would reduce the number of women below 2, so
the lowest man is eliminated.

M4 is eliminated + 13 are distributed

Round 4
M1: 17*
M2: 17 (+1)
M3: 17 (+2)
M4: 0
M5: 16 (+2)
W1: 17 (+4)
W2: 0
W3: 16 (+4)

M2, M3 and W1 all meet the quota, so all are elected, but no surplus
is distributed.

Round 5
M1: 17*
M2: 17*
M3: 17*
M4: 0
M5: 16
W1: 17*
W2: 0
W3: 16

If this had been a previous round, W3 would be protected from
elimination, as there are only 2 women left.

However, since this is the last round, (only 1 seat left to fill and 2
candidates for the seat), the restriction is lifted.

Both W3 and M5 have 16 votes, so a tie break rule (say coin toss),
would decide which one is eliminated.

If M5 is eliminated, then the results are:

M1+M2+M3+W1+W3

if W3 is eliminated, then the results are

M1+M2+M3+M5+W1

In both cases, the requirement for at least 1 man and 1 woman is met.

>> Methods like Schulze STV work by comparing possible councils to determine
>> which are best. Thus, it may be possible to limit them to only consider
>> "balanced" councils. I'm not sure how to do this in ordinary STV, however,
>> since it doesn't work that way, and in any case, this would be untested.

Yes, you can.  The software would just need to be updated.

A council with 5 men and 0 women would be considered to lose to a
council of 4 men and 1 woman.