The Price Of Stability for Network
Design with Fair Cost Allocation
Elliot Anshelevich, Anirban Dasgupta, Jon Kleinberg,
Eva Tardos, Tom Wexler, Tim Roughgarden
Presented by : Kobi Yablonka
(most of the slides are taken from Elliot Anshelevich’s presentation “The
price of Stability for Network Design” – www.cs.princeton.edu/~eanshele)
The context
• A network Design Game
• Number of independent agents
• Each agent try to minimize the cost
The Model
• Directed garph G=(V,E) , with each edge e having
a nonegative cost ce
• Each player i has a set of terminal nodes Ti that he
wants to connect
• Strategy for player i is Si  E s.t Si connects all
node in Ti
• All players using an edge split up the cost of the
edge equally
The Model cont.
S  (S1 , S2 ,...Sk )
vector of the players strategies
• xe - the number of players using edge e
Ce
• The cost to player i is Ci ( S )  
eS X e
n
• The total edge cost of the network is
 C (S )   C
i 1
i
e Si
i
• The cost to a player is affected from the strategies of
other players
e
Nash Equilibrium
• A Nash Equilibrium (NE) is a set of payments for players
such that no player wants to deviate.
• When considering deviations, player i assumes that other
player payments are fixed.
• Given a solution consisting of a vector of strategies S
'
there is no strategy Si for player i s.t
Ci ( S1 ,..., Si 1 , Si' , Si 1...Sk )  Ci (S1 ,..., Si 1 , Si , Si 1...Sk )
Price of stability
• The best Nash equilibrium relative to the
global optimum
• Stands in contrasts to the “price of anarchy”
which is the ratio of the worst Nash
equilibrium to the optimum
Congestion Games
This is a congestion game!
• Usual congestion games have latency/delay/load:
cost per player increases as the number of players sharing an
edge increases.
• Fair Connection Game has edge costs:
cost per player decreases as the number of players sharing an
edge increases.
Related Work
Shapley value cost sharing
[Feigenbaum, Papadimitriou, Shenker; Herzog, Shenker, Estrin]
Price of anarchy in routing and congestion games
[Roughgarden, Tardos]
Potential games
[Monderer, Shapley]
Example: High Price of Stability
t
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k
Example: High Price of Stability
cost(OPT) = 1+ε
t
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Example: High Price of Stability
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cost(OPT) = 1+ε
…but not a NE:
player k
pays (1+ε)/k,
k
could pay 1/k
Example: High Price of Stability
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1+
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k
so player k
would deviate
Example: High Price of Stability
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0
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k-1
...
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k
now player k-1
pays (1+ε)/(k-1),
could pay 1/(k-1)
Example: High Price of Stability
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k-1
...
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k
so player k-1
deviates too
Example: High Price of Stability
Continuing this
process, all
players defect.
t
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k-1
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Price of Stability is Hk = Θ(log k)!
k
This is a NE!
(the only Nash)
1
1
cost = 1 + 2 + … + k
To Show
• The Hk Price of Stability is worst case possible.
• Proof uses the idea of a Potential Game
[Monderer and Shapley].
• Extend results to many natural generalizations of
the Fair Connection Game.
Potential Games
A game is a potential game if there exists a
function Ф(S) mapping the current game state S
to a real value s.t.
If player i moves, i’s improvement = change in Ф(S).
Such games have pure NE: just do Best Response!
The Fair Connection Game is a potential game!
We extend analysis to bound Price of Stability.
A Potential Function
Define Фe(S) = ce[1+ 1/2 + 1/3 + … 1/ke]
where ke is # players using e in S.
Hk
Let Ф(S) = Σ Фe(S)
eєS
Consider some solution S (set of edges for each player).
Suppose player i is unhappy and decides to deviate.
What happens to Ф(S)?
Tracking Player Happiness
Фe(S) = ce[1+ 1/2 + 1/3 + … 1/ke]
Suppose player i’s new path includes e.
i pays ce/(ke+1) to use e.
Фe(S) increases by the same amount.
Likewise, if player i leaves an edge e’,
Фe’(S) exactly reflects the change
in i’s payment.
ce[1+ 1/2 +… +1/ke]
e
i
e’
ce’[1+ 1/2 +… +1/ke’]
Tracking Player Happiness
Фe(S) = ce[1+ 1/2 + 1/3 + … 1/ke]
Suppose player i’s new path includes e.
i pays ce/(ke+1) to use e.
Фe(S) increases by the same amount.
ce[1+ 1/2 +… +1/ke]+ce/(ke+1)
e
i
e’
Likewise, if player i leaves an edge e’,
Фe’(S) exactly reflects the change
in i’s payment.
ce’[1+ 1/2 +… +1/ke’] -ce’/ke’
Bounding Price of Stability
Consider starting from OPT (central optimum).
From OPT, players will settle on some Nash NE.
Bounding Price of Stability
Consider starting from OPT (central optimum).
From OPT, players will settle on some Nash NE.
_
Ф(NE) < Ф(OPT)
Bounding Price of Stability
Consider starting from OPT (central optimum).
From OPT, players will settle on some Nash NE.
_
Ф(NE) < Ф(OPT)
for any S,
cost(S) < Ф(S) < Hk cost(S).
_
_

eS
ce

eS
ce H xe

eS
(ce H k )
So cost(NE) < Ф(NE) < Ф(OPT) < Hk cost(OPT).
_
_
_
Extensions
• Take a fair connection game with each edge
having a nondecreasing concave cost
function ce(x),where x is the number of
players using edge e. Then the price of
stability is at most Hk
• The proof is analogous to the previous
proof.
Extensions
• All results hold if edges have capacities.
• Incorporate distance:
cost to player i = ci(Pi) + length(Pi)
• Utility function of player i can depend on both cost and the
set Si picked by i e є Si
– cost to player i = Σ ce(ke)/ke + fi(Si)
– PoS is still within log(k) if ce is concave
More Questions
• Cost and Latency
• Only Latency
– Nash exist (same potential argument)
– Best NE costs at most OPT w/ twice as many players.
• Best Response Dynamics
– Can construct games with k players so that a certain
ordering of moves takes 2O(k) time.
• Weighted Game
Adding Latency
What if we want to model congestion?
…marginal cost increases, so not buy-at-bulk.
Every edge has increasing delay function de(ke).
Cost of edge e for player i is
ce(ke)/ke+de(ke).
Total cost of edge is
ce(ke) + kede(ke).
Cost + Latency
From earlier proof, we know that if for all S,
cost(S) < AФ(S) < ABcost(S),
then the price of stability is < AB.
if ce is concave, de is nondecreasing for all e, and x ed e( x e)  A x1 d e( x)
for all e and xe then the price of stability is at most AHk (separate cost
and latency)
xe
E.g. if ce is concave, de is polynomial with degree m,
then Price of Stability is < (m+1)log(k).
Latency
In this case Nash Equilibria can be computed.
d(x)
Convert all
edges
d(1)
d(2)
All edges
capacity 1
d(3)
…
Claim: A min cost flow corresponds to a NE.
Idea: Since d is increasing, flow will use d(1), then d(2), etc,
mirroring a potential function.
[Fabrikant, Papadimitriou, Talwar]
Latency
Results (with single source)
• Nash exist (same potential argument)
• Best NE costs at most OPT w/ twice as many players.
Best Response Dynamics
How long before players settle on a NE?
• In games with 2 players, O(n) time,
since shared segment grows monotonically.
• Can construct games with k players so that a certain
ordering of moves takes 2O(k) time.
• Can 3-player games run for exponential time?
• Can k-player games be scheduled to be polytime?
Weighted Game
If some player has more traffic, should pay more…
In a weighted game, player i has weight w(i).
Players pay for edges proportionally to their weight.
No potential function exists. Do NE always exist?
• Best Response converges for single commodity.
• Games with at most 2 players per edge have NE.
• If NE do exist, Price of Stability will be >> log(k)
Games with at most 2 players per
edge have NE
For edge e used by
players i and j

ce wi

ce w j


e (S )   (   wi w j )
ce wi w j
wi  w j


0
i use e in S
j use e in S
both i and j use e in S
otherwise
• For edge with one player F(s) = wici if I uses e 0 otherwise
( S )   e  e ( S )
• When player i moves the change in F(S) is equal to the
change in player's i payment scaled up by wi
Best Response converges for single
commodity
•
•
•
•
•
•
•
•
All players have the same source and sink
For every s-t path P the marginal weight is c( P)   eP c e
we
Order the paths of the players in tuple in according their marginal
weight(lexicographic order)
Show that in every step the lexicographic size of the tuple decrees
If the move is P1 -> P2
P* = group of paths that have edges in either P1 or P2
C’(P2) – the cost of P2 after the move
show that : c '( P 2)  min{c ( p ) | p
 P*}
If NE do exist, Price of Stability will
be >> log(k)
t
1/2
1
1
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2
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1
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0
k-1
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k
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wi = 2i-1
Thank you.