Raph Frank wrote:
On Tue, Nov 3, 2009 at 7:41 AM, Kristofer Munsterhjelm
<[email protected]> wrote:
Raph Frank wrote:
If there are 5 seats and you have 20%+ of the votes, you are
guaranteed to get 1 seat under both d'Hondt and Droop.
There is a typo there, I meant 4 seats and 20%+ (I replied in a different post).
How about Sainte-Lague/Webster's? Since it's a divisor method, it would
(seldomly) violate quota, and so a ballot-based version of it couldn't meet
the DPC. Yet, I would say that such a version would (absent other flaws) be
proportional - I just don't know how to actually construct it.
That might be possible by reducing the quota. However, doing that
could result in to many candidates winning a seat.
Webster's method adjusts the divisor until it gives the right result.
Perhaps something similar could be done with the quota? Adjust towards
Hare until there are too many seats, then back off a bit. However, the
quota doesn't use rounding, so the analogy to the divisor fails at that
point.
If there are 4 seats, then a party is entitled to get 1 seat if they
get between 0.5 and 1.5 "seat's worth of votes".
In a 2 party situation, where 1 party gets 12.5%+ of the vote, the
smaller party will still get 1 seat, even though it is only much lower
than the Droop quota.
This is why most jurisdictions don't use the standard version.
Instead of dividing by
1,3,5,7,9,...
they divide by
1.4,3,5,7,9,...
To my knowledge, this was actually a compromise between the largest
party and the smaller parties, at least here in Norway: the largest
party wanted Sainte-Laguë with the divisor at 1.5 whereas the smaller
parties wanted Hare (because it did not discriminate against them). The
former system was D'Hondt - and the compromise worked out to
Sainte-Laguë with the first divisor at 1.4 (in exchange for the parties'
cooperation regarding some other laws).
The effect is that it is harder for parties to get their first seat.
Parties with 2 or more seats are no affected.
Is that true? Consider a maximally unfair variant, something like
2.999, 3, 5, 7, 9...
Now the larger parties can get many seats before the intermediate and
small parties get in the running. This naturally decreases the number of
free seats that may be allocated to the small parties.
Applying that to PR-STV could be something like
-> candidates must designate what party they are members of
Initially, the Droop quota is used as Quota_single, but it might take
a bit of tweaking to find one that gives the right number of seats, in
any given election (like Webster's method).
When a party has some members elected, the quota is reduced for all
other members of the party (but max 1 candidate may be elected at a
time).
[snip]
STV has an advantage in that it doesn't need to care about parties. I'd
prefer to preserve that in any competing method. My "Setwise Highest
Average" method treats solid coalitions as parties (roughly speaking,
perhaps better is to consider them parts of a party with a tree
structure) - this might be a way to do what you propose, but without
explicit party information. On the other hand, Setwise Highest Average,
is severely nonmonotonic, and it's not house-monotone either.
Also, if the districts are only 5 or so seats in size, then it doesn't
really help that much at all, as only large parties will get more than
1 seat anyway, though it could help medium parties get a 2nd seat.
Yes. If complexity is not a problem, Schulze's MMP proposal could be
used to fix that. Norway has something like "party list MMP": a certain
number of seats are top-up and allocated to maximize proportionality
after the district seats are allocated, with proportionality presumably
being defined according to a national Modified Sainte-Laguë count.
If the limitations of apportionment methods are true for party-neutral
multiwinner methods as well, then it's impossible to have both population
pair monotonicity (what we usually call "monotonicity") and to always obey
quota.
Well, PR-STV doesn't meet the monotonicity criterion.
I am not sure if an alternative elimination ording could help there,
but probably not.
If my reasoning is correct, then no tinkering with the elimination
ordering would solve the problem completely, because the resulting
method would in any case still always obey quota, and we can't have both
quota and population-pair monotonicity.
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