ALGOS TRUTH JUSTICE
Fair Division V: Indivisible Goods
Teachers: Ariel Procaccia (this time) and Alex Psomas
INDIVISIBLE GOODS Set of goods Each good is indivisible Players - - PowerPoint PPT Presentation
T RUTH J USTICE A LGOS Fair Division V: Indivisible Goods Teachers: Ariel Procaccia (this time) and Alex Psomas INDIVISIBLE GOODS Set of goods Each good is indivisible Players = 1, , have valuations
Teachers: Ariel Procaccia (this time) and Alex Psomas
π for
bundles of goods
π π = Οπβπ» π π π
denoted π© = (π΅1, β¦ , π΅π)
infeasible!
$50 $30 $3 $2 $5 $5 $5 Total: $50 Total: $30 Total: $20
$50 $30 $3 $2 $5 $5 $5 Total: $50 Total: $30 Total: $20 $3 $2 $5 $40 $10 $20 $20 Total: $40 Total: $30 Total: $30
π1,β¦,ππ min π
π(π π)
π(π΅π) is at
1 1 1 1 1 1 1 1 1 1 1 1 17 25 12 1 2 22 3 28 11 0 21 23
3 -1 -1 -1 0 0 0 0 0 0 0 0 3 -1 0 0
3 0 -1 0 0 0 -1 0 0 0 0 -1
Player 1 Player 2 Player 3
π π β€ π π(π)
π1 π2 π3 π4 π2 π3 π1 π4
π πΆ π = π π π΅π β₯ π π π΅π = π π πΆπ β ππ
π π΅π β₯ π π π΅π β π
Complexity of π-EF in the RW model?
1 π
π π2
1 π2
π2 π
Poll 1
π π΅π β² β₯ π π π΅π for all π β π, and
π π΅π β² > π π π΅π for some π β π
product of values NW π© = ΰ·
πβπ
π
π(π΅π)
chooses an allocation that maximizes the Nash welfare
NW π© = 0 for all π©
additive valuations, the MNW solution is EF1 and efficient
π(π)/π π(π)
π π΅π = π π(π΅π β² ) for all π β π, π,
π π΅π β² = π π π΅π + π π πβ , and
π π΅π β² = π π π΅π β π π πβ
NW π΅ > 1 β 1 β
ππ πβ ππ π΅π
1 +
ππ πβ ππ π΅π
> 1 β
ππ πβ ππ πβ
π
π π΅π + π π πβ
< π
π π΅π
π
π πβ
π
π πβ β€
Οπβπ΅π π
π π
Οπβπ΅π π
π π = π π π΅π
π
π π΅π
π
π π΅π + π π πβ < π π π΅π
5 10 15 20 25 30 5 10 15 20 25 30 35 40 45 50
Time (s) Number of players
[Caragiannis et al., 2016]