Methods of elimination in quota preferential STV

[Bart Ingles]

I haven't had time to think about multi-winner methods much, but I have
been leaning toward a modified cumulative voting method with
elimination.  I'm pretty sure the following method has been shown to
violate monotonicity & participation (see below), but I suspect the
violations may be less severe than STV's. 

Voters simply vote for multiple candidates, as they would with approval
voting, except that each choice gets an equal fraction of the vote (e.g.
if you vote for five candidates, each gets 1/5 of your vote).  You then
eliminate the weakest candidate and recount, so that the remaining
candidates get a larger share (if one candidate was eliminated from your
ballot, the remaining candidates now each get 1/4 of a vote).  Continue
eliminating candidates in the same fashion until the required number
remain.

No quotas are necessary since each voter always has the same cumulative
vote.  The fact that voters must weigh compromise choices rather than
simply rank them should yield higher overall utilities, and the method
is certainly simpler than STV.

I have the following references regarding the method (I haven't seen
them):

Bolger, E. M. (1983), "Proportional representation" in: S. J. Brams, W.
F. Lucas and P. D. Straffin, Jr., eds., Modules in Applied Mathematics,
Vol. 2 (Springer-Verlag, New York) 19-31.

Bolger, E. M. (1985), "Monotonicity and other paradoxes in some
proportional representation schemes," SIAM Journal on Algebraic and
Discrete Methods 6: 283-291.

-Bart



LAYTON Craig wrote:
> 
> The problem with these types of systems is that they require electors to
> rank ALL candidates.  While this is perfecly reasonable for single-winner
> elections, in PR the number of candidates make it impracticable & innacurate
> as the number of informals is too high.  In our state PR elections for 6
> members, we always have well over 50 candidates.  Implementation of any PR
> systems requiring the ranking of all candidates would prove difficult, and
> unpopular with voters.
> 
> -----Original Message-----
> From: DEMOREP1 at aol.com [mailto:DEMOREP1 at aol.com]
> Sent: Thursday, 5 October 2000 10:32
> To: election-methods-list at eskimo.com
> Subject: RE: Methods of elimination in quota preferential STV
> 
> A *democratic* legislative body exists only because the Electors (voters)
> cannot generally appear in person.
> 
> The legislators are the agents (i.e. representatives) of the Electors
> (voters).
> 
> The notion that a legislator must have only 1 vote in a legislative body is
> one of many political fictions.  A legislator should have a voting power in
> the legislative body equal to the number of voters that he/she represents.
> 
> The general case of choosing M winners among N choices in proportional
> representation elections is somewhat complex.
> 
> Example - elect 3 of 10
> 
> Vote YES on each acceptable choice and use number votes.
> 
> All of the combinations of the following would be done (i.e. an expanded
> head
> to head matrix) using number votes (1, 2, etc.) ---
> 
> Any 3 test winners     versus 1 test loser       (6 other test losers)
> 
> A first choice vote for each of the other test losers gets transferred to 1
> of the 3 test winners or the test loser.
> 
> Obviously a test winner may or may not win in ALL of his/her matches.
> 
> Example
> 
>        YES  Place votes
>      1    2     3     4    etc.
> 
> A  A1  A2  A3  A4
> B  B1  B2  B3  B4
> C  C1  C2  C3  C4
> D  D1  D2  D3  D4
> E  E1  E2   E3  E4
> etc.
> 
> Assume that D is a Condorcet winner (using both YES and NO votes).
> 
> One tiebreaker is to accumulate YES place votes to get the earliest Droop
> Quota (or the highest votes with truncated votes).
> 
> D2 (or D3, etc.) votes (YES or NO) would go to D (taking away first (second,
> 
> etc.) YES choice votes from the others).
> 
> Assume B gets a Droop Quota by summing the 1st to 3rd place B YES votes
> (reducing the first choice YES votes for the other candidates).
> 
> Assume E gets the highest YES votes (due to lots of truncated votes) (with D
> 
> and B getting more votes).
> 
> Another simpler (but somewhat inaccurate) tiebreaker would be simple IRV--
> the candidate with the lowest number of first choice (plus transferred)
> votes
> loses. Repeat until 3 candidates remain.
> 
> Another simpler tiebreaker would be to drop the choice with the lowest YES
> votes (repeatedly) and transfer the votes to the voter's next YES choice (if
> 
> any).
> 
> Votes from losers would transfer to an elected choice using the number votes
> 
> (whether or not a voter voted YES or NO for that elected choice).
> 
> Each winner should have a voting power equal to the final number of votes
> that he/she receives.
> 
> The complexity of the above is a reason why single winner math is only
> somewhat less complex.