Best Response Dynamics in Multicast Cost Sharing

1
Best Response Dynamics in Multicast Cost Sharing

• Seffi Naor
• Microsoft Research and Technion

Based on papers with C. Chekuri, J. Chuzhoy, L.
Lewin-Eytan, and A. Orda EC 2006 M. Charikar,
C. Mattieu, H. Karloff, M. Saks 2007
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Motivation

• Traditional networks single entity, single
  control objective.
• Modern networking many entities, different
  parties.
• Users act selfishly, maximizing their objective
  function.
• Decisions of each user are based on the state of
  the network, which depends on the behavior of the
  other users.
• ? non-cooperative network games.

3
Our Framework

• A network shared by a finite number of users.
• Each edge has a fixed cost.
• Cost sharing method defines the rules of the
  game determines the mutual influence between
  players.
• Performance of a user is total payment
• sum of payments for all the edges it
  uses.
• Two fundamental models
• The congestion model.
• The cost sharing model.

4

• In both models
• Each user routes its traffic over a minimum-cost
  path.
• Splittable routing model vs. unsplittable
  routing model.

5
The Multicast Game

• A special root node r, and a set of n receivers
  (players).
• A players strategy is a routing decision the
  choice of a single path to r.
• Egalitarian (Shapley( cost sharing mechanism the
  cost of each edge is evenly split among the
  players using it.Each player on edge e with ne
  players pays ce / ne
• The goal of the players is to connect to the root
  by making a routing decision minimizing their
  payment.

6
The Multicast Game

• Two different models
• The integral model each player connects to the
  root through a single path.
• The fractional model each player is allowed to
  split its connection to the root into several
  paths (fractions add up to 1).

7
Nash Equilibrium

• Players are rational.
• Each player knows the rules of the underlying
  game.
• Nash Equilibrium no player can unilaterally
  improve its cost by changing its path to the
  root. Cost of a path takes into account cost
  sharing.
• Nash equilibrium solutions are stable operating
  points.

8
The Price of Anarchy

• Nash equilibrium outcomes do not necessarily
  optimize the overall network performance.
• Price of Anarchy The ratio between the cost of
  the worst Nash equilibrium and the (social)
  optimum.
• Quantifies the penalty incurred by lack of
  cooperation.

9
The Integral Multicast Game

• Potential function F of a solution T Rosenthal
  73
• Exact potential change in cost of a connection
  of player i to the root is equal to the change in
  the potential F.
• If edge e is deleted from T F F -
  ce / ne(T)
• If edge e is added to T F F
  ce / (ne(T)1)

10
The Integral Multicast Game

• Finite strategy space ? F has an optimal value.
• F ? Nash equilibrium existence.
• Global / Local optima of F correspond to a NE.
• A Nash equilibrium solution is a tree rooted at r
  spanning the players.
• Special case of a congestion game.

11
Price of Anarchy vs. Price of Stability

• Price of anarchy can be as bad as ?(n).
• Price of stability ratio between the cost of
  best Nash solution to the cost of OPT.
• Outcome of scenarios in the middle ground
  between centrally enforced solutions and
  non-cooperative games.
• E.g. central entity can enforce the initial
  operating point.
• Anshelevich et al., FOCS 2004
• Directed graphs - price of stability is ?(log
  n).
• Undirected graphs upper bound on the price of
  stability is O(log n). PoS can be reached from a
  2-approximate Steiner tree configuration
• C(TNash) ? F(TNash) ? F(T2-Seiner) ? log n C
  (T2-Steiner)

r
t
12
Best Response Dynamics

• Each player, in its turn, selects a path
  minimizing its cost (best response).
• Eventually, an equilibrium point is reached.
• PoA strongly depends on the choice of the initial
  configuration.
• Starting from a near-optimal solution may be hard
  to enforce requires relying on a central
  trusted authority.
• Question What happens if we start from an empty
  configuration? Chekuri, Chuzhoy, Lewin-Eytan,
  Naor, and Orda, EC 2006
• Round 1 Players arrive one by one, each player
  plays best response.
• Round 2 Best response dynamics continue in
  arbitrary order till NE.
• Question Can a good equilibrium always be
  achieved as a consequence of best-response
  dynamics in this model?

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Cost of user 1c (r, x, 1) 1e c (r , 1) 1
r
r
r
r
r
r
r
r
r
r
Greedy cost of 3, ,n 1
Cost of user 2c (r , x, 2) 1e c (r, 1, x, 2
) 12e c (r, 2) 1
1
1
1
3/4
1
1
1
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
¼ e
¼ e
¼ e
¼ e

2
1
3
n
1
1
1
1
1
1
2
1
2
1
2
1
2
1
3
2
1
3
2
1
3
2
1
3
2
1
3
2
1
3
2
1
n-2
n-1
Price of anarchy 4
Can a good equilibrium be achieved as a
consequence of best-response dynamics?
14
Results

• The integral multicast game for undirected
  graphs
• Upper bound of O(log3n) on the PoA of
  best-response dynamics in the two-round game
  starting from an empty configuration.
  (Improving over the bound of
  CCLNO-EC06 of .)
• Upper bound of O(log2? n) on the cost of the
  solution at the end of the first round.
• Lower bound of ?(log n) on the PoA of this
  game.
• Computing a Nash equilibrium minimizing
  Rosenthals potential function is NP-hard.

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Price of Anarchy Lower Bound

• Theorem Price of anarchy of our game ?(logn).
• Proof Adaptation of lower bound proof for the
  online Steiner problem ImazeWaxman
  
• Online Steiner problem used edges have cost 0
• Take hard instance IW and replace each
  terminal by a star of n2 1 terminals at zero
  distance
• The n2 1 terminals choose the same path to root
• cost of used edge becomes negligible

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PoA in Undirected Graphs Upper Bound

• Analysis is performed in two steps
• Round 1 players connect one by one to the root
  via best response.
• Solution T is reached after a sequence of
  arrivals t1 , t2 ,, tn . We show
• 
  ? O(log3n ) c(TOPT )
• c(T) O(log2? n )
  c(TOPT )
• Round 2 players play in arbitrary order till NE
  is reached. c(TNash) ? F(TNash) ? F(T)

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The First Round

• Choose a threshold ? 2 (0,1).
• Terminal t is ?-good if cost of next terminal
  using the same path as t (1-?) cost(t).
• Otherwise, terminal t is ?-bad.
• Idea bound separately the contribution to F of
  the ?-good terminals and the ?-bad terminals.

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Charging ?-good Terminals
r

• Suppose there is a tree T spanning the ?-good
  terminals.
• Upon arrival t pays cost(t)
• Upon arrival, t pays at most
• cost(t) ? d (1-?) cost(t)

cost(t)

• t and t are ?-good terminals
• t arrives first, then t arrives

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Charging ?-good Terminals (contd.)

• Charges decay at an exponential rate along a root
  leaf path in T
• Upon arrival of ti it pays at most
• cost(ti) di
  (1-?)cost(ti-1)
• cost(tk) dk dk-1(1-?)
• dk-2(1-?)2 d1(1-?)k-1
  

r
d1
t1
d2
t2
d3
t3
d4
t4
dk
tk
The charge to each edge e 2 T d(e) ?i
(1-?)i ne(i)

• t1,, tk are ?-good terminals
• arrival order t1,, tk

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Auxiliary Tree

• TOPT is transformed to an auxiliary tree T
  defined on the set of ?-good terminals
• The descendants of terminal t in T are terminals
  which have arrived after t.
• c(T) O(1/? logn) c(TOPT )
• The depth of T O(1/? logn)

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Contribution of ?-good Terminals
Theorem The contribution of the ?-good terminals
to F in the first phase is bounded as
follows Proof Follows from the properties
of T together with the exponential decay of the
charges to the edges of T of the ?-good
terminals.
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Contribution of ?-bad Terminals

• Theorem The contribution of the ?-bad terminals
  to F in the first phase is bounded as follows
• Intuition The cost of the edges opened for
  the first time by ?-bad terminals constitutes
  only a small part of the sum of the costs of the
  ?-bad paths.
• Setting ? O(1/logn) O(log4n ) upper
  bound on PoA
• Setting ? cleverly O(log3n ) upper
  bound on PoA

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The Fractional Multicast Game

• Players split their connection to the source.
• A splittable multicast model.

r

• Flow on (r, y)
• ¼ unit of flow is shared by users 1 2.
• ½ unit of flow is used only by user 2.

• Flow on (r, x)
• ¼ unit of flow is shared by users 1 2.
• ½ unit of flow is used only by user 1.

3/4
3/4
1/4
1/4
x
y
1
2
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The Fractional Multicast Game

• The cost of each flow fraction is split evenly
  between its users.
• ?
  cefe,n_e is the total cost of edge e.

fe,1 1/4 fe,2 3/4 fe,3 7/8
1/4
3/4
7/8
Total cost of user 3 ce (1/12 1/4 1/8).
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The Fractional Multicast Game

• The potential function F of the fractional model
• F is an exact potential.
• A fractional flow configuration defining a local
  minimum of F corresponds to a NE.

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Results Fractional Multicast Game

• Nash equilibrium existence.
• NE - minimizing the potential function
• Can be computed in polynomial time (using LP).
• It is NP-hard in the case of an integral Nash.
• PoA of the computed NE is O(log n).

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Computing a Minimum Potential NE

• Create a new graph G (V, E) by replacing each
  edge e by n copies e1, e2, , en.
• Copy ej should be used if j players use edge
  e.
• The cost of a unit flow on copy j of edge e is ce
  / j.
• The undirected graph is replaced by a directed
  flow network.

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Computing a Minimum Potential NE

• The linear program
• Objective function potential function
• The capacity of edge ej is 0 ? xe_j ? 1.
• Variables of the LP
• Flows of the users on the edges ej?E.
• Capacities of the edges in E.
• Constraints
• Non-aggregating flow constraint (flow of user i
  on ej ) ? xe_j .
• Aggregating flow constraint (total flow on ej )
  j xe_j .

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The Linear Program

• Theorem 5 There exists an optimal solution to
  the linear program that corresponds to a
  fractional multicast flow.
• Heavily depends on the non-increasing property of
  the cost function.
• LP can be used
• For any cost sharing mechanism that is
  cross-monotonic.
• In case players are not restricted to have a
  common source.
• PoA of a minimum potential fractional NE is O(log
  n).

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Integral vs. Fractional Potential Minimization

• There exists an instance where the gap between
  the integral and fractional minimum potential
  solutions is a small constant.
• Finding an integral Nash equilibrium that
  minimizes the potential function is NP-hard.
• Building block a variation of the
  Lund-Yannakakis hardness proof of approximating
  the set cover problem.

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The Weighted Multicast Game

• Each player i has a positive weight wi .
• The cost share of each player is proportional to
  its weight.
• Integral cost share of player i ce (wi / We)
  
• (We weight of players
  currently using e)
• Fractional weighted sharing on each fraction.
• Theorem A NE always exists for the weighted
  fractional model.
• Note NE does not necessarily exists for the
  weighted integral model Chen-Roughgarden, SPAA
  06.

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• Thank You!