Approval Voting for Committees: Threshold Approaches.
Peter Fishburn Saša Pekeč
Approval Voting for Committees: Threshold Approaches.
Saša PekečDecision Sciences
The Fuqua School of BusinessDuke University
[email protected]://faculty.fuqua.duke.edu/~pekec
(S)ELECTING A COMMITTEE
choosing a subset S from the set of m available alternatives
choosing a feasible (admissible) subset S
social choice
voting
multi-criteria decision-making
consumer choice (?)
CHOSING A SINGLE ALTERNATIVE
Information requirement on voters’ preferences
SWF rankings
plurality top choice
scoring rules constrained cardinal utility (IIA???)
approval voting subset choice
CHOSING A SINGLE ALTERNATIVE
V1 V2 V3 V4 V5 V6 V7 V8 V9
1 2 8 3 6 1 7 8 5 5
2 1 7 8 1 3 1 7 1 6
3 7 1 5 4 2 8 5 2 2
4 5 4 1 8 8 3 4 8 7
5 4 3 6 3 7 4 3 3 4
6 6 5 7 2 6 5 1 4 8
7 8 2 2 7 5 6 2 6 1
8 3 6 4 5 4 2 6 7 3
CHOSING A SINGLE ALTERNATIVE
V1 V2 V3 V4 V5 V6 V7 V8 V9
1 2 8 3 6 1 7 8 5 5
2 1 7 8 1 3 1 7 1 6
3 7 1 5 4 2 8 5 2 2
4 5 4 1 8 8 3 4 8 7
5 4 3 6 3 7 4 3 3 4
6 6 5 7 2 6 5 1 4 8
7 8 2 2 7 5 6 2 6 1
8 3 6 4 5 4 2 6 7 3
CHOSING A SINGLE ALTERNATIVE
BORDAV1 V2 V3 V4 V5 V6 V7 V8 V9
1 2 8 3 6 1 7 8 5 5
2 1 7 8 1 3 1 7 1 6
3 7 1 5 4 2 8 5 2 2
4 5 4 1 8 8 3 4 8 7
5 4 3 6 3 7 4 3 3 4
6 6 5 7 2 6 5 1 4 8
7 8 2 2 7 5 6 2 6 1
8 3 6 4 5 4 2 6 7 3
CHOSING A SINGLE ALTERNATIVE
BORDAV1 V2 V3 V4 V5 V6 V7 V8 V9
1 2 8 3 6 1 7 8 5 5 45
2 1 7 8 1 3 1 7 1 6 35
3 7 1 5 4 2 8 5 2 2 36
4 5 4 1 8 8 3 4 8 7 48
5 4 3 6 3 7 4 3 3 4 37
6 6 5 7 2 6 5 1 4 8 44
7 8 2 2 7 5 6 2 6 1 39
8 3 6 4 5 4 2 6 7 3 40
CHOSING A SINGLE ALTERNATIVE
V1 V2 V3 V4 V5 V6 V7 V8 V9
1 2 8 3 6 1 7 8 5 5
2 1 7 8 1 3 1 7 1 6
3 7 1 5 4 2 8 5 2 2
4 5 4 1 8 8 3 4 8 7
5 4 3 6 3 7 4 3 3 4
6 6 5 7 2 6 5 1 4 8
7 8 2 2 7 5 6 2 6 1
8 3 6 4 5 4 2 6 7 3
CHOSING A SINGLE ALTERNATIVE
PLURALITYV1 V2 V3 V4 V5 V6 V7 V8 V9
1 8 8
2 8
3 8
4 8 8 8
5
6 8
7 8
8
CHOSING A SINGLE ALTERNATIVE
PLURALITYV1 V2 V3 V4 V5 V6 V7 V8 V9
1
2
3
4
5
6
7
8
CHOSING A SINGLE ALTERNATIVE
PLURALITYV1 V2 V3 V4 V5 V6 V7 V8 V9
1 2
2 1
3 1
4 3
5 0
6 1
7 1
8 0
CHOSING A SINGLE ALTERNATIVE
V1 V2 V3 V4 V5 V6 V7 V8 V9
1 2 8 3 6 1 7 8 5 5
2 1 7 8 1 3 1 7 1 6
3 7 1 5 4 2 8 5 2 2
4 5 4 1 8 8 3 4 8 7
5 4 3 6 3 7 4 3 3 4
6 6 5 7 2 6 5 1 4 8
7 8 2 2 7 5 6 2 6 1
8 3 6 4 5 4 2 6 7 3
CHOSING A SINGLE ALTERNATIVE
APPROVALV1 V2 V3 V4 V5 V6 V7 V8 V9
1 2 8 3 6 1 7 8 5 5
2 1 7 8 1 3 1 7 1 6
3 7 1 5 4 2 8 5 2 2
4 5 4 1 8 8 3 4 8 7
5 4 3 6 3 7 4 3 3 4
6 6 5 7 2 6 5 1 4 8
7 8 2 2 7 5 6 2 6 1
8 3 6 4 5 4 2 6 7 3
CHOSING A SINGLE ALTERNATIVE
APPROVALV1 V2 V3 V4 V5 V6 V7 V8 V9
1
2
3
4
5
6
7
8
CHOSING A SINGLE ALTERNATIVE
APPROVALV1 V2 V3 V4 V5 V6 V7 V8 V9
1 0 1 0 0 0 1 1 0 0
2 0 1 1 0 0 0 1 0 0
3 1 0 0 0 0 1 0 0 0
4 0 0 0 1 1 0 0 1 1
5 0 0 0 0 1 0 0 0 0
6 0 1 0 0 0 0 0 0 1
7 1 0 0 1 0 0 0 0 0
8 0 1 0 0 0 0 0 1 0
CHOSING A SINGLE ALTERNATIVE
APPROVALV1 V2 V3 V4 V5 V6 V7 V8 V9
1 0 1 0 0 0 1 1 0 0 3
2 0 1 1 0 0 0 1 0 0 3
3 1 0 0 0 0 1 0 0 0 2
4 0 0 0 1 1 0 0 1 1 4
5 0 0 0 0 1 0 0 0 0 1
6 0 1 0 0 0 0 0 0 1 2
7 1 0 0 1 0 0 0 0 0 2
8 0 1 0 0 0 0 0 1 0 2
APPROVAL VOTE PROFILEV1 V2 V3 V4 V5 V6 V7 V8 V9
1
2
3
4
5
6
7
8
37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46
APPROVAL VOTE PROFILE
V1 V2 V3 V4 V5 V6 V7 V8 V9
V=( 37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46 )
CHOSING A SUBSET
Information requirement on voters’ preferences
SWF rankings on all 2m subsets
plurality top choice among all 2m subsets
scoring rules constrained card. utility on all 2m subsets
approval voting subset choice on all 2m subsets
CHOSING A SUBSET
using “consensus” ranking of alternatives
but
for all voters: 1>2>3 or 2>1>3
AND
13>23>12 or 23>13>12
divide and conquer:
break into several separate singleton choices
proportional representation
IGNORING INTERDEPENDANCIES(substitutability and complementarity)
CHOSING A SUBSET
Barbera et al. (ECA91): impossibility
A manageable scheme that accounts for interdependencies?
Proposal: Approval Voting with modified subset count.
Threshold Approach:
- define t(S) for every feasible S
- ACt(S)= # of voters i such that |Vi S| t(S)
AV THRESHOLD APPROACH
Define t(S) for every feasible S
ACt(S)= # of voters i such that ViS = |Vi S| t(S)
Threshold functions (TF):
- t(S)=1 (favors small committees)
- t(S) = |S|/2 (majority)
- t(S) = (S|+1)/2 (strict majority)
- t(S) = |S| (favors large committees)
….
APPROVAL TOP 3-SET
V1 V2 V3 V4 V5 V6 V7 V8 V9
1 0 1 0 0 0 1 1 0 0 3
2 0 1 1 0 0 0 1 0 0 3
3 1 0 0 0 0 1 0 0 0 2
4 0 0 0 1 1 0 0 1 1 4
5 0 0 0 0 1 0 0 0 0 1
6 0 1 0 0 0 0 0 0 1 2
7 1 0 0 1 0 0 0 0 0 2
8 0 1 0 0 0 0 0 1 0 2
S=124 gets 10 votes total.
CHOOSING 3-SET, t(S) 1V1 V2 V3 V4 V5 V6 V7 V8 V9
1
2
3
4
5
6
7
8
37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46
CHOOSING 3-SET, t(S) 1V1 V2 V3 V4 V5 V6 V7 V8 V9
1
2
3
4
5
6
7
8
37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46
S=234 is the only 3-set approved by all voters
CHOOSING 3-SET, t(S) 2V1 V2 V3 V4 V5 V6 V7 V8 V9
1
2
3
4
5
6
7
8
37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46
CHOOSING 3-SET, t(S) 2V1 V2 V3 V4 V5 V6 V7 V8 V9
1
2
3
4
5
6
7
8
37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46
CHOOSING 3-SET, t(S) 2V1 V2 V3 V4 V5 V6 V7 V8 V9
1
2
3
4
5
6
7
8
37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46
S=123 is the only 3-set approved by at least three voters
COMPLEXITY of AVCT
If X= the set of all feasible subsets, is part of the input thencomputing AVCT winner is polynomial in mn+|X|
Theorem. If X is predetermined (not part of the input), then computing AVCT winner is NP-complete at best.
COMPLEXITY of AVCT
If X= the set of all feasible subsets, is part of the input thencomputing AVCT winner is polynomial in mn+|X|
Theorem. If X is predetermined (not part of the input), then computing AVCT winner is NP-complete at best.
Proof: choosing a k-set, t1. Suppose |Vi|=2 for all i.
Note: alternatives ~ vertices of a graphVi ~ edges of a graph
k-set approved by all voters ~ vertex cover of size kVertex Cover is a fundamental NP-complete problem.
COMPLEXITY cont’d
not as problematic as it seems.
Theorem. (Garey-Johnson)
If X is predetermined (not part of the input), then computing
maxS inX sumi inS score(i)
is NP-complete.
CHOSING A SINGLE ALTERNATIVE
BORDAV1 V2 V3 V4 V5 V6 V7 V8 V9
1 2 8 3 6 1 7 8 5 5 45
2 1 7 8 1 3 1 7 1 6 35
3 7 1 5 4 2 8 5 2 2 36
4 5 4 1 8 8 3 4 8 7 48
5 4 3 6 3 7 4 3 3 4 37
6 6 5 7 2 6 5 1 4 8 44
7 8 2 2 7 5 6 2 6 1 39
8 3 6 4 5 4 2 6 7 3 40
CHOSING A SINGLE ALTERNATIVE
PLURALITYV1 V2 V3 V4 V5 V6 V7 V8 V9
1 2
2 1
3 1
4 3
5 0
6 1
7 1
8 0
CHOSING A SINGLE ALTERNATIVE
APPROVALV1 V2 V3 V4 V5 V6 V7 V8 V9
1 0 1 0 0 0 1 1 0 0 3
2 0 1 1 0 0 0 1 0 0 3
3 1 0 0 0 0 1 0 0 0 2
4 0 0 0 1 1 0 0 1 1 4
5 0 0 0 0 1 0 0 0 0 1
6 0 1 0 0 0 0 0 0 1 2
7 1 0 0 1 0 0 0 0 0 2
8 0 1 0 0 0 0 0 1 0 2
LARGER IS NOT BETTER
Example: m=8, n=12, strict majority TF: t(S)=(|S|+1)/2
V= (123,15,1578,16,278,23,24,34,347,46,567,568)
1-set (AC): 1,2,3,4,5,6,7 all approved by 4 voters (8 is approved by 3 voters)
LARGER IS NOT BETTER
Example: m=8, n=12, strict majority TF: t(S)=(|S|+1)/2
V= (123,15,1578,16,278,23,24,34,347,46,567,568)
1-set (AC): 1,2,3,4,5,6,7 all approved by 4 voters (8 is approved by 3 voters)
2-set: 15,23,34,56,57,58,78 all approved by 2 voters
LARGER IS NOT BETTER
Example: m=8, n=12, strict majority TF: t(S)=(|S|+1)/2
V= (123,15,1578,16,278,23,24,34,347,46,567,568)
1-set (AC): 1,2,3,4,5,6,7 all approved by 4 voters (8 is approved by 3 voters)
2-set: 15,23,34,56,57,58,78 all approved by 2 voters
3-set: 234 approved by 5 voters
4-set: 5678 approved by 3 voters
5-set: 15678 approved by 4 voters
LARGER IS NOT BETTER
Example: m=8, n=12, strict majority TF: t(S)=(|S|+1)/2
V= (123,15,1578,16,278,23,24,34,347,46,567,568)
1-set (AC): 1,2,3,4,5,6,7 all approved by 4 voters (8 is approved by 3 voters)
2-set: 15,23,34,56,57,58,78 all approved by 2 voters
3-set: 234 approved by 5 voters
4-set: 5678 approved by 3 voters
5-set: 15678 approved by 4 voters
TOP INDIVIDUAL NOT IN A TOP TEAM
Example: m=5, n=6, majority TF: t(S)=|S|/2
V= (123,124,135,145,25,34)
TOP INDIVIDUAL NOT IN A TOP TEAM
Example: m=5, n=6, majority TF: t(S)=|S|/2
V= (123,124,135,145,25,34)
Top individual: 1 approved by 4 voters (all other alternatives approved by 3 voters)
TOP INDIVIDUAL NOT IN A TOP TEAM
Example: m=5, n=6, majority TF: t(S)=|S|/2
V= (123,124,135,145,25,34)
Top individual: 1 approved by 4 voters (all other alternatives approved by 3 voters)
Top team
2345 is the only team approved by all 5 voters
TOP INDIVIDUAL NOT IN A TOP TEAM
Example: m=5, n=6, majority TF: t(S)=|S|/2
V= (123,124,135,145,25,34)
Top individual: 1 approved by 4 voters (all other alternatives approved by 3 voters)
Top team
2345 is the only team approved by all 5 voters
- could generalize examples for almost any TF
- could generalize to top k individuals
THRESHOLD SENSITIVITY
Theorem
For any K>1, there exist n,m and a corresponding V such that AVCT winner Sk (where X is the set of all K-sets), k=1,…,K are mutually disjoint.
ANY GOOD PROPERTIES?
P1. Nullity. If every vote is the empty set, any choice is good.
P2. Anonymity. If U is a permutation of V, the choices for U and V are identical.
P3. Partition Consistency. If S is chosen in two voter disjoint elections, then S would be chosen in the joint election.)
P4. Partition Inclusivity. If no S is chosen by a single voter and in an election of the remaining n-1 voters, then any choice would also be chosen in an election w/o one of the voters.
SINGLE VOTER PROPERTIES
(S) = min{AS: S is a choice for A}
P5. For every choice S, there exists votes A and B such that A is a choice for S but not for B.
P6. Let S be a choice for vote A that does not choose everyone. If BS>AS then S is a choice for B
P7. For every S, there is an A such that AS= (S) -1
P8. Suppose vote B chooses every committee. For all A1, A2 and for all choices S, T: If A1S= (S), A2T= (T), then BS>A1S implies BT>A2T
THE LAST THEOREM OF FISHBURN
Theorem.
If P1-8 hold, then the subset choice function is the AVCT.
AV THRESHOLD APPROACH
- low informational burden
- simplicity
- takes into account subset preferences
Results:- properties of TFs, axiomatic characterization- complexity- robustness properties: theorems show what is possible and not what is probable
Need: -Comparison with other methods, data validation- strategic considerations
Approval Voting for Committees: Threshold Approaches.
Peter Fishburn Saša Pekeč
DIGRESSION
Subset Choice and Cooperative Games
APPROVAL VOTE
subset choice
alternative vote count:
i. For every S find:
u(S)= # voters whose approval set is S
ii. AC(j) = ΣS: j in S u(S)
APPROVAL VOTE PROFILEV1 V2 V3 V4 V5 V6 V7 V8 V9
1 3
2 3
3 2
4 4
V= (3,12,2,4,4,13,12,4,4)
u(4)=4,u(12)=2,u(2)=u(3)=u(13)=1; for all other S: u(S)=0
AC(j) = ΣS: j in S u(S)
e.g. AC(1)=u(12)+u(13)=2+1=3
APPROVAL VOTE
i. For every S find:
u(S)= # voters whose approval set is S
ii. AC(j) = ΣS: j in S u(S)
APPROVAL VOTE
For every S find:
u(S)= # voters whose approval set is S
Cooperative Game u
solution concepts for cooperative games
- core, nucleolus, Shapley Value ...
- define how to attribute subset values to individual alt’s.
- implicitly define rankings on alternatives
APPROVAL VOTE
For every S find:
u(S)= # voters whose approval set is S
Cooperative Game u
solution concepts for cooperative games
- core, nucleolus, Shapley Value ...
- define how to attribute subset values to individual alt’s.
- implicitly define rankings on alternatives
APPROVAL VOTE
i. For every S find:
u(S)= # voters whose approval set is S
ii. Use your “favorite” solution concept
to define a ranking on alternatives
Is there a solution concept that generates ranking identical to The Approval Count (AC)?
POWER INDICES
p(j)= c ΣS:j in S w(S,j) [u(S) – u(S\{j})]
Shapley-Shubik Index: w(S,j) = (|S|-1)!(m-|S|)!
Banzhaf-Coleman Index: w(S,j)=1
…
Proposition:
Banzhaf-Coleman Index pBC( ) is the only power index such that, for every u, the ranking of alternatives induced by pBC( ) is identical to the ranking induced by the Approval Vote Count AC( ).
Proposition: Banzhaf-Coleman Index pBC( ) is the only power index such that, for every u, the ranking of alternatives induced by pBC( ) is identical to the ranking induced by the Approval Vote Count AC( ).
Proof: AC(j) = ΣS: j in S u(S)
pBC(j) = ΣS:j in S [u(S) – u(S\{j})]
= ΣS:j in S u(S) – ΣS:j not in S u(S)
Note that ΣS u(S) = n, so
pBC(j) = 2 AC(j) – n
Converse is a bit tedious, constructing V to exploit differences in w(S,j) (Recall: p(j)= c ΣS:j inS w(S,j)[u(S)–u(S\{j})]).
AV AND COOPERATIVE GAMES
it is all about subset choice
demonstrated a link between AV and cooperative games
how to use large body of research in cooperative games?
- opens up possibilities for new aggregation methods
- social choice implications for power indices
PLAN OF ACTION
Motivation/Introduction
Subset Choice and Cooperative Games
Approval Voting: Threshold Approach(with Fishburn)
Balancing Teams (with Baucells)
BALANCING TEAMS
MBA student teams- N individuals divided into G groups/teams- Each individual i described by values aij of predefined characteristics j
BALANCING TEAMS
MBA student teams- N individuals divided into G groups/teams- Each individual i described by values aij of predefined characteristics j
Want as perfectly balanced team assignment as possible:
For any characteristic j and any value a*j, the difference across any two teams in the number of people with value a*j in characteristic j is at most one.
BALANCING TEAMS
MBA student teams- N individuals divided into G groups/teams- Each individual i described by values aij of predefined characteristics j
Want as perfectly balanced team assignment as possible:
For any characteristic j and any value a*j, the difference across any two teams in the number of people with value a*j in characteristic j is at most one.
other examples: consultants
showroom settings (cars, furniture)
BALANCING TEAMS
- INSEAD, Stern (Weitz and Jelassi, JORS 92)- Tuck (Baker et al., JORS 02,03)- Kelley (Cutshall et al., Interfaces 06)- Rotman (Krass and Ovchinnikov, Interfaces 2006). . .
FEASIBILITY PROBLEM
Simplify to binary characteristics
Input: N,G and 0-1 matrix A= [aij]. (Let qj= Σi aij /G)
Feasibility problem:
1,0
...1,...1,
...1,//
...1,1
1
1
1
ig
jij
N
iigj
N
iig
G
gig
x
MjGgqaxq
GgGNxGN
Nix
COMPLEXITY
Theorem. Balancing Teams is NP-complete.
COMPLEXITY
Theorem. Balancing Teams is NP-complete.
Proof. Take [aij] with exactly two ones in each column.
Note: individual ~ vertex of a graph, characteristic ~ edge
Balanced team assignment ~ G-equicoloring.
COMPLEXITY
Theorem. Balancing Teams is NP-complete.
Proof. Take [aij] with exactly two ones in each column.
Note: individual ~ vertex of a graph, characteristic ~ edge
Balanced team assignment ~ G-equicoloring.
Claim. k-coloring and k-equicoloring are in the samecomplexity class.(Add (n-k)(k-1) independent vertices.)
COMPLEXITY
Theorem. Balancing Teams is NP-complete.
Proof. Take [aij] with exactly two ones in each column.
Note: individual ~ vertex of a graph, characteristic ~ edge
Balanced team assignment ~ G-equicoloring.
Claim. k-coloring and k-equicoloring are in the samecomplexity class.(Add (n-k)(k-1) independent vertices.)
Finally, Graph k-coloring is NP-complete for k>2.
COMPLEXITY
Theorem. Balancing Teams is NP-complete.
Proof. Take [aij] with exactly two ones in each column.
Note: individual ~ vertex of a graph, characteristic ~ edge
Balanced team assignment ~ G-equicoloring.
Claim. k-coloring and k-equicoloring are in the samecomplexity class. Add (n-k)(k-1) independent vertices.)
Finally, Graph k-coloring is NP-complete for k>2.
Theorem. Any reasonable approximate balancing is also NP complete. (Reduction to exact cover by 3sets.)
SIMULATION
2500+ instances using Solver Premium
3000+ instances using CPLEX (w/o preprocessing)
up to 40 binary categories used
Logistic regression model
variables: N, M, N/G, density, # tight constraints
Prob of Feasibility
0%
20%
40%
60%
80%
100%
0 10 20 30 40 50 60
Number of (binary) atribules K
N=20; q0=4
N=20; q0=6
N=20; q0=8
N=20; q0=10
N=20; q0=12
N=60; q0=4
N=60; q0=6
N=60; q0=8
N=60; q0=10
N=60; q0=12
N=100; q0=4
N=100; q0=6
N=100; q0=8
N=100; q0=10
N=100; q0=12
Q0=N/G
0%
20%
40%
60%
80%
100%
0 10 20 30 40 50 60
Number of binary attributes K
N=20; Int=7
N=20; Int=6
N=20; Int=5
N=20; Int=4
N=20; Int=3
N=60; Int=7
N=60; Int=6
N=60; Int=5
N=60; Int=4
N=60; Int=3
N=100; Int=7
N=100; Int=6
N=100; Int=5
N=100; Int=4
N=100; Int=3
e^Bx/(1+e^Bx)
1
10
100
1000
10000
-15 -10 -5 0 5 10 15
Linear Score Bx
Tim
e (S
ec.)
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
Feasible
Infeasible
Undecided
Pr(Feasible)
BALANCING TEAMS IS EASY
Example: N=72, G=12
- Easy for M<20,
- M=33 probability of success is ~0.5
Sources of hardness:- large K, small N- many tight constraints- density- large G, small N
Problem instances related to MBA programs are easy.
PLAN OF ACTION
Motivation/Introduction
Subset Choice and Cooperative Games
Approval Voting: Threshold Approach(with Fishburn)
Balancing Teams (with Baucells)
It’s all over but the crying.
ASSEMBLING TEAMS
Saša PekečDecision Sciences
The Fuqua School of BusinessDuke University
[email protected]://faculty.fuqua.duke.edu/~pekec
*thanks to Manel Baucells, Peter Fishburn