GROUP DECISION MAKING UNDER MULTIPLE CRITERIA SOCIAL CHOICE THEORY Content Arrow's conditions for social walfare functions Criteria to evaluate the election methods Single transferable vote The Coombs method Arrow's conditions for social walfare functions Axiom 1: For all x and y, either x R y or y R x. [connected ] Axiom 2: For all x, y, and z, x R y and y R z imply x R z. [transitive] ­ x R y means «x is preferred or indifferent to y» Conditions: 1. domain properties: ­ The number of alternatives in A is greater than or equal to three. ­ The social welfare function f is defined for all possible profiles of individual orderings. ­ There are at least two individuals. 2. Positive Association of Social and Individual Values 3. Independence of Irrelevant Alternatives 4. Citizen's Sovereignty 5. Non-dictatorship Arrow's conditions POSITIVE ASSOCIATION OF SOCIAL AND INDIVIDUAL VALUES If the welfare function asserts that x is preferred to y for a given profile of individual preferences, it shall assert the same when the profile is modified as follows: ­ (a) The individual paired comparisons between alternatives other than x are not changed, ­ (b) Each individual paired comparison between x and any other alternative either remains unchanged or it is modified in x's favor. Monotonocity failure!* ­ Upward monotonicity failure: Given the use of voting method and a ballot profile in which candidate X is the winner, X may nevertheless lose if some voters change their preference in the favor of X. ­ Downward monotonicity failure: Given the use of voting method and a ballot profile B in which candidate X is a loser, X may nevertheless win if some voters change their preference against the favor of X. *Felsenthal and Tideman, (2014) 64 D.S. Felsenthal, N. Tideman / Mathematical Social Sciences 67 (2014) 57–66 64 D.S. Felsenthal, N. Tideman / Mathematical Social Sciences 67 (2014) 57–66 VoterVoter No. No. Ranking Ranking abcdefabcdefSum Sum 1 1 a b ad cb ed f c e f a – 9 7 10a 6 10– 42 9 7 10 6 10 42 2 a d c f b e b 7 – 7 10 10 8 42 2 a d c f b e b 7 – 7 10 10 8 42 3 3 b d bc ed ac f e a f c 9 9 – 5c 9 99 41 9 – 5 9 9 41 4 b d e c a f d 6 6 11 – 8 9 40 4 b d e c a f d 6 6 11 – 8 9 40 5 b e a f d c e 10 6 7 8 – 9 40 6 5 c b be ae da f f d c f 6 8 7 7e 7 –10 35 6 7 8 – 9 40 7 6 c d ca fb be e a d f Totalf 6 240 8 7 7 7 – 35 8 d a c b e f 7 c d a f b e Total 240 9 d c b e a f As the average Borda score remains 40 (240/6), candidates d, e, 8 d a c b e f 10 e a b d c f and f are eliminated and a revised matrix of paired comparisons 11 9 e f db dc c b a e a f among the remaining threeAs the candidates average is Borda computed: score remains 40 (240/6), candidates d, e, 12 10 f a ec ea bb d d c f and f are eliminated and a revised matrix of paired comparisons 13 f c a b d e abcSum 11 e f b d c a among the remaining three candidates is computed: 14 f e a d c b 12 f a c e b d a – 9 7 16 15 f e b c a d b 7 – 7 14 16 13 f e fd c aa b b d e abcSum c 9 9 – 18 14 f e a d c b Total 48 a – 9 7 16 In view of these15 rankings, thef Bordae scoreb of eachc a of thed six b 7 – 7 14 Example candidates is equal16 to the sum of hisf respectivee d row inc thea followingb Here the average Borda score of the three voters a–c is 16 c 9 9 – 18 matrix of paired comparisons: (48/3), so all candidates except c are eliminated and c becomes the winner. Note that, on the basis of his or her originalTotal ranking, voter 48 In view of theseabcdef rankings, the BordaSum score of each#7 of benefited the six from the upward move of a up in his or her ranking candidatesa is equal– to 9 the 7 sum 9 of 6 his 10 respective 41 row in thebecause, followingceteris paribusHere, s/he obtainedthe average as a result Borda the election score of of c the three voters a–c is 16 which s/he ranked initially above a. 64 Apply Nanson’s functionD.S. Felsenthal,: N. Tidemanmatrix of / Mathematical pairedb 7 comparisons: – Social 7 10 Sciences 10 8 67 42 (2014) 57–66 (48/3), so all candidates except c are eliminated and c becomes the c 9 9 – 5 9 9 41 winner. Note that, on the basis of his or her original ranking, voter d 7 6 11 – 8 9 41 3.5.3. Upward monotonicity failure — dynamic voters worse off [UW] abcdef Sum #7 benefited from the upward move of a up in his or her ranking Voter No. Ranking e 10 6 7 8 – 9 40 abcdefSuppose instead that, ceterisSum paribus, voter #8 changes his or her f a 6 8– 7 9 7 7 7 – 9 35 6 10 41 ranking from d because,a c toceterisa d paribusc ,thereby s/he obtainedincreasing as a result the election of c Total 240 ··· ··· 1 a b d c e f b 7 – 7 10a 10 8– 42 9a’s 7 support. 10 So we 6which now 10 obtain s/he 42 the ranked same matrices initially and above final resultsa. as in Section 3.5.1, i.e., here too candidate c is elected. The only 2 a d c f b e c 9 9 – 5b 9 919 7 41 – 7 10 10 8 42 The sum of the Borda scores of all six candidates is 240, hence difference is that, on the basis of his or her original ranking, here 3 b d c e a f the average Bordad score is 407 (240 6/6). 11 According –c 8to the 9 Nanson9 41 9 –voter #8 5 is worse 93.5.3. off 9 as Upwarda result 41 of monotonicity the rise in the failure ranking — of dynamica voters worse off [UW] method, one eliminatese at the10 end 6 of every 7 counting 8 – round 9 those 40 because s/he obtains the election of c which s/he ranked originally 64 4 b d e c a D.S.f Felsenthal, N. Tideman / Mathematical Social Sciencesd 67 (2014)6 57–66 6 11 – 8Suppose 9 40 instead that, ceteris paribus, voter #8 changes his or her candidates whose Bordaf score6 does 8 not 7 exceed 7 the 7 average – Borda 35 lower than a. ranking from d a c to a d c thereby increasing 5 b e a f d c score of all the candidates participating in thate round.10 So here 6 7 8 – 9 40 ··· ··· Total 240 a’s support. So we now obtain the same matrices and final results 6 Voterc No.b Rankinge a d f candidates e and f are eliminated and a revisedf matrix of pairedabcdef6 83.5.4. 7 Downward 7 7 monotonicity –Sum 35 failure — dynamic voters better off comparisons among the remaining four candidates is computed: [DB] as in Section 3.5.1, i.e., here too candidate c is elected. The only 7 1 c d aa bf db c e e f Totala – 9 719 10 6 10 42 240 The sum of the Borda scores of all six candidates is 240, Ashence candidate bdifferencewas not elected is that, initially, on the suppose basis now of his that, or her original ranking, here 8 2 d a ac db c e f f b thee average Borda scoreabc is 40 d (240Sum/6).b According7 to – theceteris 7 Nanson 10 paribus, 10 votervoter 8 #5 downgrades #8 42 is worse candidate off asb by a result changing of his the rise in the ranking of a rank order from b e a to e b a . As a result, 9 3 d c bb de ca e f a method,f one eliminatesa – 9As at 7 the the 9 end average 25 of everyc Borda counting9 score 9 round – remains those 5 409 (240because 9/ 416), s/he··· candidates obtains the d election,···e, of c which s/he ranked originally ceteris paribus, the initial Borda score of b decreases from 42 to 41 4 b d e c a f b 7 – 7 10 24 d 6 6 11 – 8lower 9 than40 a. 10 e a b d c f candidates whose Bordaand scoref are does eliminated not exceed and the averagea revisedand the Borda initial matrix Borda of score paired of e increases comparisons from 40 to 41, while the 5 b e a f d c c 9 9 – 5 23 e 10 6 7 8 – 9 40 11 e f b d c a score of all thed candidates7among 6 11 participating –the 24 remaining in that three round.