<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title></title>
</head>
<body>
Apologies Participants,<br>
The penny has finally dropped that I have been misunderstanding Woodall's
"Quota-Limited Trickle-Down" (QLTD) method.<br>
So my remark that it is "obviously easier to hand-count than Bucklin" is
wrong, but other than that I have not mis-represented<br>
its properties. In common with Bucklin and Highest Median Rank (HMR), it
fails Clone Independence and Mono-add-top.<br>
<br>
But I did get one result wrong in my last post.<br>
<blockquote type="cite">40: A>B>C <br>
25: B>A>C <br>
35: C>B>A <br>
100 ballots. B is the CW. <br>
<br>
</blockquote>
I wrongly wrote that A is the QLTD winner. In fact the QLTD winner is B.<br>
The "quota" is 50 (%). To reach this, A needs (10/25 = .4) of his second
preferences, whereas B needs (25/75 = .33333)<br>
of his second preferences to reach the quota. C needs (15/40 = .375). B's
fraction is the smallest, so B wins.<br>
<br>
My Wed. Apr. 21 post (with the error corrected):<br>
<br>
<div class="moz-text-flowed"
style="font-family: -moz-fixed; font-size: 13px;" lang="x-western"> Mike,
Dave, Forest, Kevin, anyone interested, <br>
<br>
<br>
I previously wrote (Sun.Apr.11): <br>
<br>
<blockquote type="cite">Plain WMA, as I have defined it, is descended
from an earlier version (from Joe Weinstein, Forest tells us) in which each
ballot <br>
approves as many of the highest-ranked candidates as possible without
their combined weight exceeding half the total weight, <br>
and then only approves the next ranked candidate if the weight of candidates
ranked below this (pivot) candidate is greater than <br>
the weight of candidates ranked above it. If the two weights are equal,
then the ballot half-approves that candidate. <br>
The problem with this is that it fails 3-candidate Condorcet. To distinguish
it, this earlier version could perhaps be called <br>
"Above Median Weighted Approval" (AMWA). </blockquote>
<br>
I had in mind this relatively simple type of example: <br>
<br>
40: A>B>C <br>
25: B>A>C <br>
35: C>B>A <br>
100 ballots. B is the CW. <br>
<br>
Weights A:40 B:25 C:35 <br>
AMWA approvals <br>
40: A (not B as in WMA, because A has a greater weight than C.) <br>
25: BA (because C has a greater weight than B) <br>
35: CB <br>
AMWA final scores A: 65 B: 60 C: 35, A wins. <br>
<br>
WMA easily elects B (with 100% "approval"). Bucklin, "Highest Median Rank"
(HMR), and QLTD also<br>
elect B. <br>
<br>
However I was wrong to imply that WMA meets 3-candidate Condorcet. <br>
An example adapted from one of Woodall's: <br>
300: A <br>
200: A>C>B <br>
300: B>C>A <br>
200: B>A>C <br>
300: C>A>B <br>
199: C>B>A <br>
1499 ballots. C is CW. <br>
<br>
WMA final scores A: 1000 B: 699 C: 999, A wins. (AMWA is the
same, except that C scores 799). <br>
WMA-STV eliminates B, and then elects C. <br>
Bucklin and its close relatives elect A. <br>
I adapted this from Woodall's proof that Condorcet is incompatible with
Later-no-help (and also mono-raise-random, <br>
mono-raise-delete, mono-sub-top, mono-sub-plump) in his paper "Monotonicity
and Single-Seat Election Rules", <br>
page 13, "Election 6". <br>
<a class="moz-txt-link-freetext"
href="http://groups.yahoo.com/group/election-methods-list/files/wood1996.pdf">http://groups.yahoo.com/group/election-methods-list/files/wood1996.pdf</a>
<br>
So WMA, in common with Bucklin, QLTD, and HMR fail 3-candidate Condorcet.
<br>
I can't see how WMA-STV can fail to meet it. <br>
<br>
This is "Election 2" from the same paper (page 11): <br>
12: A>B>C>D>E>F <br>
11: C>A>B>D>E>F <br>
10: B>C>A>D>E>F <br>
27: D>E>F <br>
60 ballots. Smith set comprises ABC. <br>
<br>
Bucklin, QLTD, HMR agree with WMA (and RP and BP) in electing A. WMA-STV
elects C. <br>
<br>
Weights A:12 B:10 C:11 D:27 E:0 F:0 <br>
WMA approvals <br>
12: ABC <br>
11: CAB <br>
10: BCA <br>
27: DEF <br>
<br>
ABC are all tied on 33, but we break the tie in favour of the candidate with
the greatest "weight", A. <br>
Also using weights to break ties to fix the WMA-STV elimination schedule,
we eliminate first F and E and then D and then B, <br>
and then C wins (C>A, 21-12). <br>
<br>
Now we add 6 A>D ballots: <br>
12: A>B>C>D>E>F <br>
06:A>D <br>
11: C>A>B>D>E>F <br>
10: B>C>A>D>E>F <br>
27: D>E>F <br>
66 ballots. Smith set is ABCD, Schwartz set is ABC. <br>
<br>
Now Bucklin, HMR, and QLTD all elect D. Adding ballots all with A ranked
first, causing A to lose, demonstrates that those methods fail Mono-add-top.
<br>
<br>
Weights A:18 B:10 C:11 D:27 E:0 F:0 <br>
WMA approvals <br>
12: ABC <br>
06: AD <br>
11: CAB <br>
10: BCA <br>
27: DEF <br>
WMA final scores A :39 B:33 C:33 D:33 E:27 F:27, A
wins. <br>
WMA-STV eliminates F and E, and then B (the "lightest" 33), and then C,
and then A wins (A>D, 39-27). (Ranked Pairs and Beat Path also pick
A.)<br>
<br>
It seems to me that WMA and WMA-STV meet Mono-add-top. <br>
<br>
Chris Benham <br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
</div>
<br>
</body>
</html>