At 05:59 AM 1/25/2010, Juho wrote:
Here's one simple approach.
- all voters rank all the rooms
- use Borda like personal utility values => last room = 0 points,
one but last = 1 point etc. (also other than this kind of linear
scale could be used)
- find the room allocation that gives the highest sum of utilities
- if there is a tie one can use seniority to break it
- the utility values of each voters are multiplied by some
seniority factor and then summed up again
- the factors could be quite small if one just wants to break
the ties (e.g. 1.0001, 1.0002)
This tie breaking approach is intended to work so that if there is
for example some room that all consider to be the best then that
room would be given to the most senior voter.
Any chances to work?
Sure. But equal ranking must be allowed, otherwise noise is
introduced. Borda with equal ranking (and therefore empty ranks,
otherwise equal ranked votes are reduced in strength) is Range. Why
not just use Range, allowing greater precision. One could use a Range
method with N*R resolution, where the "Borda" version has N equal to 1.
The allocation problem as "highest sum of utilities" is NP-hard,
though, I believe. I suspect that an iterative method would be
better. Use an approval method, which will divide up voters into
blocks. Then use more sophisticated analysis on each block. If you
are the only one to approve a particular room, perhaps you get the
room! So you would not over-approve at the first iteration. Standard
approval strategy when there is iteration.
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