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How hard is it to find extreme Nash equilibria in network congestion - - PowerPoint PPT Presentation

Nov 27, 2022 242 likes •815 views

How hard is it to find extreme Nash equilibria in network congestion games? E. Gassner J. Hatzl Graz University of Technology, Austria S.O. Krumke H. Sperber University of Kaiserslautern, Germany G. Woeginger Eindhoven University of

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How hard is it to find extreme Nash equilibria in network congestion games?

  • E. Gassner
  • J. Hatzl

Graz University of Technology, Austria S.O. Krumke

  • H. Sperber

University of Kaiserslautern, Germany

  • G. Woeginger

Eindhoven University of Technology, The Netherlands

13th Combinatorial Optimization Workshop Aussois, January 2009

Hatzl (TUG) Network congestion games January 2009 1 / 29

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Talk Outline

1 Problem Formulation 2 Preliminary Results 3 Complexity Results for Worst Nash Equilibria 4 Complexity Results for Best Nash Equilibria Hatzl (TUG) Network congestion games January 2009 2 / 29

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The model

A directed graph G(V , E) with multiple edges A source s and a sink t Non-decreasing latency functions ℓe : N0 → R+ N users, each routing the same amount of unsplittable flow Strategy set for all users: P — set of all simple s-t-paths

Hatzl (TUG) Network congestion games January 2009 3 / 29

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The model

A directed graph G(V , E) with multiple edges A source s and a sink t Non-decreasing latency functions ℓe : N0 → R+ N users, each routing the same amount of unsplittable flow Strategy set for all users: P — set of all simple s-t-paths s u t x 2x 1.5x 2x 1.5x

Hatzl (TUG) Network congestion games January 2009 3 / 29

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The model

A flow is a function f : P → N0 . The latency on a path P ∈ P is the sum

  • f the latencies on its edges, i.e.,

ℓP(f ) :=

  • e∈P

ℓe

  • P∈P: e∈P

fP

  • Given a flow f the social cost are given by

Cmax(f ) := max

P∈P:fP>0 ℓP(f ).

s u t x 2x 1.5x 2x 1.5x Cmax(f ) = max{1 + 3, 2 + 3, 1.5} = 5

Hatzl (TUG) Network congestion games January 2009 4 / 29

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Nash Equilibrium

Definition (Nash Equilibrium)

A flow f is a Nash equilibrium, iff for all paths P1, P2 with fP1 > 0 we have ℓP1(f ) ≤ ℓP2(˜ f ) with ˜ fP =      fP − 1 if P = P1 fP + 1 if P = P2 fP

  • therwise

. s u t x 2x 1.5x 2x 1.5x s u t x 2x 1.5x 2x 1.5x

Hatzl (TUG) Network congestion games January 2009 5 / 29

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Roughgarden Model

Network Congestion Game Roughgarden single-commoditiy multicommodity unsplittable, unweighted splittable makespan sum

Hatzl (TUG) Network congestion games January 2009 6 / 29

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Existence of Nash equilibria

Hatzl (TUG) Network congestion games January 2009 7 / 29

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Existence of Nash equilibria

Theorem (Roughgarden and Tardos (2002))

The Nash flows of an instance are precisely the optima of a non-linear convex programming problem. If f and ˜ f are Nash flows then ℓe(f ) = ℓe(˜ f ) for all e ∈ E. Hence, all Nash equilibria have the same social cost.

Hatzl (TUG) Network congestion games January 2009 7 / 29

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Existence of Nash equilibria

Theorem (Roughgarden and Tardos (2002))

The Nash flows of an instance are precisely the optima of a non-linear convex programming problem. If f and ˜ f are Nash flows then ℓe(f ) = ℓe(˜ f ) for all e ∈ E. Hence, all Nash equilibria have the same social cost.

Theorem (Fabrikant et al. (2004))

Given a network congestion game the optimal solution of the following min-cost flow problem MCF(G) yields a Nash equilibrium: For every edge e ∈ E we need N copies with costs cei = ℓe(i), i = 1, . . . , N. The capacities of all edges are 1 and we send N units of flow from s to t.

Hatzl (TUG) Network congestion games January 2009 7 / 29

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Extreme Nash equilibria

Consider the following instance with N = 2: s u1 u2 u3 t 2x 3x 2x 3x 2x 5x 2x 5x

Hatzl (TUG) Network congestion games January 2009 8 / 29

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Extreme Nash equilibria

Consider the following instance with N = 2: s u1 u2 u3 t 2x 3x 2x 3x 2x 5x 2x 5x The solution with minimum social cost of 12 is given by s u1 u2 u3 t 2x 3x 2x 3x 2x 5x 2x 5x

Hatzl (TUG) Network congestion games January 2009 8 / 29

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Extreme Nash equilibria

Consider the following instance with N = 2: s u1 u2 u3 t 2x 3x 2x 3x 2x 5x 2x 5x A Nash equlibirum with social cost of 13 is given by s u1 u2 u3 t 2x 3x 2x 3x 2x 5x 2x 5x

Hatzl (TUG) Network congestion games January 2009 8 / 29

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Extreme Nash equilibria

Consider the following instance with N = 2: s u1 u2 u3 t 2x 3x 2x 3x 2x 5x 2x 5x A Nash equlibirum with social cost of 14 is given by s u1 u2 u3 t 2x 3x 2x 3x 2x 5x 2x 5x

Hatzl (TUG) Network congestion games January 2009 8 / 29

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Extreme Nash equilibria

Worst Nash Equilibrium (W-NE for short): Given: Network congestion game (G = (V , E), (ℓe)e∈E, s ∈ V , t ∈ V , N) and a number K > 0 Question: Does there exist a Nash equilibrium f such that Cmax(f ) ≥ K? Best Nash Equilibrium (B-NE for short): Given: Network congestion game (G = (V , E), (ℓe)e∈E, s ∈ V , t ∈ V , N) and a number K > 0 Question: Does there exist a Nash equilibrium f such that Cmax(f ) ≤ K? Unfortunately, it can be shown that in general neither a best nor a worst Nash equilibrium is an optimal solution of MCF(G).

Hatzl (TUG) Network congestion games January 2009 9 / 29

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Extreme Nash equilibria

Theorem (Fotakis(2002), Gairing(2005))

If the users have different weights and the graph G has only parallel links W-NE and B-NE are NP-hard even for linear latency functions.

Hatzl (TUG) Network congestion games January 2009 10 / 29

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Nash equilibria in series-parallel graphs

The series composition G = S(G1, G2):

P1 Q1 Q2 Q3 P2 P3

Lemma

Let fi be a flow in Gi (i = 1, 2). Let f ∈ f1 ⊗ f2 then f is a Nash equilibrium in S(G1, G2) if and only if fi are Nash equilibria in Gi (i = 1, 2).

Hatzl (TUG) Network congestion games January 2009 11 / 29

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Nash equilibria in series-parallel graphs

The parallel composition G = S(G1, G2):

Lemma

Let fi be a flow in Gi (i = 1, 2). Then f = f1 ∪ f2 is a Nash equilibrium in P(G1, G2) if and only if fi are Nash equilibria in Gi (i = 1, 2) and Cmax(f2) ≤ L+

G1(f1) and Cmax(f1) ≤ L+ G2(f2).

Hatzl (TUG) Network congestion games January 2009 12 / 29

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Worst Nash equilibrium

Worst Nash Equilibrium (W-NE for short): Given: Network congestion game (G = (V , E), (ℓe)e∈E, s ∈ V , t ∈ V , N) and a number K > 0 Question: Does there exist a Nash equilibrium f such that Cmax(f ) ≥ K?

Hatzl (TUG) Network congestion games January 2009 13 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1. end do;

Hatzl (TUG) Network congestion games January 2009 14 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1. end do;

4x(0) 6x(0) 6x(0) x(0)

Hatzl (TUG) Network congestion games January 2009 14 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1. end do;

4x(0) 6x(0) 6x(0) x(0) 4 6 6 1

current makespan of user 1 = 5

Hatzl (TUG) Network congestion games January 2009 14 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1. end do;

6x(0) 6x(0) 4x(1) x(1) 4 6 6 1

current makespan of user 1 = 5

Hatzl (TUG) Network congestion games January 2009 14 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1. end do;

6x(0) 6x(0) 4x(1) x(1) 8 2 6 6

current makespan of user 1 = 5 current makespan of user 2 = 6

Hatzl (TUG) Network congestion games January 2009 14 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1. end do;

6x(0) 4x(1) x(1) 6x(1) 8 2 6 6

current makespan of user 1 = 5 current makespan of user 2 = 6

Hatzl (TUG) Network congestion games January 2009 14 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1. end do;

6x(0) 4x(1) x(1) 6x(1) 8 2 12 6

current makespan of user 1 = 6 current makespan of user 2 = 6 current makespan of user 3 = 8

Hatzl (TUG) Network congestion games January 2009 14 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1. end do;

4x(1) 6x(1) 6x(1) x(2) 8 2 12 6

current makespan of user 1 = 6 current makespan of user 2 = 6 current makespan of user 3 = 8 The last user yields the maximum makespan!

Hatzl (TUG) Network congestion games January 2009 14 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1 end do;

Hatzl (TUG) Network congestion games January 2009 15 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1 end do;

Theorem (Fotakis (2006))

Greedy Best Response yields a Nash equilibrium in series-parallel graphs.

Hatzl (TUG) Network congestion games January 2009 15 / 29

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Worst Nash equilibria in SP-graphs

Greedy Best Response (GBR): For i = 1 to N do User i chooses a path with minimal latency with respect to load = current flow +1 end do;

Theorem (Fotakis (2006))

Greedy Best Response yields a Nash equilibrium in series-parallel graphs.

Theorem (GHKSW(2008))

Greedy Best Response yields a worst Nash equilibrium in series-parallel graphs.

Hatzl (TUG) Network congestion games January 2009 15 / 29

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Worst Nash equilibria in arbitrary graphs

Theorem (GHKSW (2008))

Determining a worst Nash equilibrium is strongly NP-hard even for two users on acyclic networks and with linear latency functions.

Hatzl (TUG) Network congestion games January 2009 16 / 29

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Worst Nash equilibria in arbitrary graphs

Blocking Path Problem: Given: Digraph G = (V , E) with source s ∈ V and sink t ∈ V . Question: Does there exist an s-t-path P ∈ P such that after deleting the edges of P there is no path from s to t?

Hatzl (TUG) Network congestion games January 2009 17 / 29

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Worst Nash equilibria in arbitrary graphs

Blocking Path Problem: Given: Digraph G = (V , E) with source s ∈ V and sink t ∈ V . Question: Does there exist an s-t-path P ∈ P such that after deleting the edges of P there is no path from s to t?

Theorem (GHKSW (2008))

The Blocking Path Problem is strongly NP-hard even on acyclic networks. Proof: Reduction from 3SAT.

Hatzl (TUG) Network congestion games January 2009 17 / 29

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Worst Nash equilibria in arbitrary graphs

s t

Hatzl (TUG) Network congestion games January 2009 18 / 29

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Worst Nash equilibria in arbitrary graphs

s t

construct positive and integral edge lengths ae such that every path from s to t has the same length L∗.

Hatzl (TUG) Network congestion games January 2009 18 / 29

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Worst Nash equilibria in arbitrary graphs

s t

construct positive and integral edge lengths ae such that every path from s to t has the same length L∗. ℓe(x) =

  • aex

if e ∈ E (L∗ + 1

2)(x)

if e = (s, t)

Hatzl (TUG) Network congestion games January 2009 18 / 29

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Worst Nash equilibria in arbitrary graphs

s t

construct positive and integral edge lengths ae such that every path from s to t has the same length L∗. ℓe(x) =

  • aex

if e ∈ E (L∗ + 1

2)(x)

if e = (s, t) ∃ blocking path P∗ ⇐ ⇒ ∃ Nash equilibrium f for two users with Cmax(f ) ≥ L∗ + 1

2.

Hatzl (TUG) Network congestion games January 2009 18 / 29

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Worst Nash equilibria in arbitrary graphs

s t

P2 P∗

construct positive and integral edge lengths ae such that every path from s to t has the same length L∗. ℓe(x) =

  • aex

if e ∈ E (L∗ + 1

2)(x)

if e = (s, t) ∃ blocking path P∗ ⇐ ⇒ ∃ Nash equilibrium f for two users with Cmax(f ) ≥ L∗ + 1

2.

Hatzl (TUG) Network congestion games January 2009 18 / 29

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Extreme Nash equilibria

series-parallel graph arbitrary graph Worst NE polynomially solvable (Greedy) strongly NP-hard Best NE

Hatzl (TUG) Network congestion games January 2009 19 / 29

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Best Nash equilibrium

Best Nash Equilibrium (B-NE for short): Given: Network congestion game (G = (V , E), (ℓe)e∈E, s ∈ V , t ∈ V , N) and a number K > 0 Question: Does there exist a Nash equilibrium f such that Cmax(f ) ≤ K?

Hatzl (TUG) Network congestion games January 2009 20 / 29

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Best Nash equilibrium: N is part of input

Theorem (GHKSW (2008))

Determining a best Nash equilibrium is strongly NP-hard even on series-parallel graphs and with linear latency functions if the number of users is part of the input.

Hatzl (TUG) Network congestion games January 2009 21 / 29

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Best Nash equilibrium: N is part of input

Numerical 3-Dimensional Matching: Given: Disjoint sets X, Y , Z, each containing m elements, a weight w(a) for all elements a ∈ X ∪ Y ∪ Z and a bound B ∈ Z+. Question: Does there exist a partition of X ∪ Y ∪ Z into m disjoint sets A1, . . . , Am such that each Aj contains exactly one element from each of X, Y and Z and

a∈Ai w(a) = B for all i.

Hatzl (TUG) Network congestion games January 2009 22 / 29

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Best Nash equilibrium: N is part of input

Numerical 3-Dimensional Matching: Given: Disjoint sets X, Y , Z, each containing m elements, a weight w(a) for all elements a ∈ X ∪ Y ∪ Z and a bound B ∈ Z+. Question: Does there exist a partition of X ∪ Y ∪ Z into m disjoint sets A1, . . . , Am such that each Aj contains exactly one element from each of X, Y and Z and

a∈Ai w(a) = B for all i.

Assume w.l.o.g. that w(a) ≤ 2w(b) and w(b) ≤ 2w(a) for all a, b ∈ X (Y , Z) holds.

Hatzl (TUG) Network congestion games January 2009 22 / 29

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Best Nash equilibrium: N is part of input

X Y Z w(a)x

Hatzl (TUG) Network congestion games January 2009 23 / 29

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Best Nash equilibrium: N is part of input

X Y Z w(a)x

∃ numerical 3-dimensional matching ⇐ ⇒ ∃ Nash equilibrium f for m users with Cmax(f ) ≤ B

Hatzl (TUG) Network congestion games January 2009 23 / 29

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Best Nash equilibrium: N is fixed

Theorem ([GHKSW (2008))

Determining a best Nash equilibrium is weakly NP-hard even for two users

  • n series-parallel graphs and with linear latency functions.

Proof: Reduction from Even-Odd Partition Problem.

Hatzl (TUG) Network congestion games January 2009 24 / 29

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Best Nash Equilibrium: N is fixed

A dynamic programming algorithm Let f be a Nash flow, then C(f ) denotes the set of latencies of the users with respect to f . C(f ) is called cost profile.

Hatzl (TUG) Network congestion games January 2009 25 / 29

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Best Nash Equilibrium: N is fixed

A dynamic programming algorithm Let f be a Nash flow, then C(f ) denotes the set of latencies of the users with respect to f . C(f ) is called cost profile. SG(C) . . . maximum latency for an additional user in a Nash flow in G with cost profile C.

Hatzl (TUG) Network congestion games January 2009 25 / 29

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Best Nash Equilibrium: N is fixed

A dynamic programming algorithm Let f be a Nash flow, then C(f ) denotes the set of latencies of the users with respect to f . C(f ) is called cost profile. SG(C) . . . maximum latency for an additional user in a Nash flow in G with cost profile C. Idea: Find best C such that SG(C) < ∞.

Hatzl (TUG) Network congestion games January 2009 25 / 29

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Best Nash Equilibrium: N is fixed

The series composition: SG(C) = max

C1⊗C2≤C{SG1(C1) + SG2(C2)}

Hatzl (TUG) Network congestion games January 2009 26 / 29

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Best Nash Equilibrium: N is fixed

The series composition: SG(C) = max

C1⊗C2≤C{SG1(C1) + SG2(C2)}

The parallel composition: SG(C) = max

C1∪C2=C C1≤SG2(C2) C1≤SG1(C1)

min{SG1(C1), SG2(C2)}

Hatzl (TUG) Network congestion games January 2009 26 / 29

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Best Nash Equilibrium: N is fixed

The series composition: SG(C) = max

C1⊗C2≤C{SG1(C1) + SG2(C2)}

The parallel composition: SG(C) = max

C1∪C2=C C1≤SG2(C2) C1≤SG1(C1)

min{SG1(C1), SG2(C2)}

  • There is a huge number multisets C!

O((|V | maxe∈N ℓe(N))N) = ⇒ pseudopolynomial-time algorithm for fixed N

Hatzl (TUG) Network congestion games January 2009 26 / 29

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Best Nash Equilibrium: N is fixed

The series composition: SG(C) = max

C1⊗C2≤C{SG1(C1) + SG2(C2)}

The parallel composition: SG(C) = max

C1∪C2=C C1≤SG2(C2) C1≤SG1(C1)

min{SG1(C1), SG2(C2)}

  • There is a huge number multisets C!

O((|V | maxe∈N ℓe(N))N) = ⇒ pseudopolynomial-time algorithm for fixed N

  • Result is best possible!

Hatzl (TUG) Network congestion games January 2009 26 / 29

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Extreme Nash equilibria

series-parallel graph arbitrary graph Worst NE polynomially solvable (Greedy) strongly NP-hard Best NE strongly NP-hard strongly NP-hard if N is part of input if N is part of input weakly (!) NP-hard weakly (?) NP-hard for fixed N for fixed N

Hatzl (TUG) Network congestion games January 2009 27 / 29

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Open Questions

Can we give a bound on the price of anarchy for the network congestion games if the graph is series-parallel? What can be said about the price of stability for the network congestion games if the graph is series-parallel?

Hatzl (TUG) Network congestion games January 2009 28 / 29

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The end Thank you for your attention!

Hatzl (TUG) Network congestion games January 2009 29 / 29