Approval Voting for Committees: Threshold Approaches. Peter Fishburn Saša Pekeč

Approval Voting for Committees: Threshold Approaches.

Peter Fishburn Saša Pekeč


Saša PekečDecision Sciences

The Fuqua School of BusinessDuke University

[email protected]://faculty.fuqua.duke.edu/~pekec

mailto:[email protected]

(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


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


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


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



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


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


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



1

2

3

4

5

6

7

8



1 2

2 1

3 1

4 3

5 0

6 1

7 1

8 0


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


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



1

2

3

4

5

6

7

8



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



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


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


1

2

3

4

5

6

7

8

37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46


1

2

3

4

5

6

7

8

37, 1268, 2 , 47 , 45 , 13 , 12 , 48 , 46


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.



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



1 2

2 1

3 1

4 3

5 0

6 1

7 1

8 0



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)



V= (123,15,1578,16,278,23,24,34,347,46,567,568)


2-set: 15,23,34,56,57,58,78 all approved by 2 voters



V= (123,15,1578,16,278,23,24,34,347,46,567,568)



3-set: 234 approved by 5 voters





V= (123,15,1578,16,278,23,24,34,347,46,567,568)






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)



V= (123,124,135,145,25,34)

Top individual: 1 approved by 4 voters (all other alternatives approved by 3 voters)



V= (123,124,135,145,25,34)


Top team

2345 is the only team approved by all 5 voters



V= (123,124,135,145,25,34)


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


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



ii. AC(j) = ΣS: j in S u(S)

APPROVAL VOTE

For every S find:


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:


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



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


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


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


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


Proof. Take [aij] with exactly two ones in each column.

Note: individual ~ vertex of a graph, characteristic ~ edge

Balanced team assignment ~ G-equicoloring.

COMPLEXITY





Claim. k-coloring and k-equicoloring are in the samecomplexity class.(Add (n-k)(k-1) independent vertices.)

COMPLEXITY





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





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


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

mailto:[email protected]

Documents

Approval Voting for Committees: Threshold Approaches. Peter Fishburn Saša Pekeč