Kombinatorische Optimierung Matroids and Polymatroids in der - - PowerPoint PPT Presentation

Tobias Harks Augsburg University WINE Tutorial, 8.12.2015

Kombinatorische Optimierung in der Logistik und im Verkehr Matroids and Polymatroids in Congestion Games

Outline

◮ Part I: Congestion Games

◮ Existence of Equilibria ◮ Computation of Equilibria ◮ Matroids

◮ Part II: Integral Splittable Congestion Games

◮ Existence and Computation of Equilibria ◮ Integral Polymatroids

◮ Part III: Nonatomic Congestion Games

◮ Efficiency of Equilibria ◮ The Braess Paradox ◮ Matroids are Immune to Braess Paradox

Strategic Games

◮ Strategic game G = (N, X, π) ◮ N = {1, . . . , n} set of players ◮ X = ×i∈NXi set of pure strategies ◮ x = (x1, . . . , xN) strategy profile ◮ πi(x) : X → R, i ∈ N private cost/utility

Strategic Games

◮ Strategic game G = (N, X, π) ◮ N = {1, . . . , n} set of players ◮ X = ×i∈NXi set of pure strategies ◮ x = (x1, . . . , xN) strategy profile ◮ πi(x) : X → R, i ∈ N private cost/utility ◮ Mixed strategy: for each player a probability distribution over

pure strategies

Solution Concept

Definition

Pure Nash equilibrium (PNE): no player has an incentive to unilaterally deviate.

Definition

Mixed Nash equilibrium (MNE): no player has an incentive to unilaterally change her mixed strategy. ”Minimax Theorem” John von Neumann (1928), Nash (1950)

Motivation: Party Affiliation Games

◮ each strategy set Xi = {1, −1} ◮ Weight wi,j measures relationship between i and j ◮ payoff ui(x) = j∈N xi xj wi j → max

4 5 1 3 2 1 −1 u4(x) = 0 u1(x) = 1 u3(x) = 0 u2(x) = 0 u5(x) = 5 1 2 −1 1 −3 3 2

Solution Concept: Pure Nash Equilibrium

Definition

Pure Nash equilibrium (PNE): no player has an incentive to unilaterally change his pure strategy.

Definition

Mixed Nash equilibrium (MNE): no player has an incentive to unilaterally change his mixed strategy.

Motivation: Party Affiliation Games

◮ each strategy set Xi = {1, −1} ◮ Weight wi,j measures relationship between i and j ◮ payoff ui(x) = j∈N xi xj wi j → max

4 5 1 3 2 1 −1 u4(x) = 0 u1(x) = 1 u3(x) = 0 u2(x) = 0 u5(x) = 5 1 2 −1 1 −3 3 2

Existence of Nash Equilibria

Nash (1951)

Theorem

Every finite game possesses a mixed Nash equilibrium. Pure Nash Equilibrium need not exist! Example: Assymmetric Party Affiliation Game 1 2 1 −1 In the mixed Nash equilibrium, each player chooses each party with probability 1/2.

Part I Congestion Games

Congestion Games

Modell

◮ N = {1, . . . , n} set of players ◮ R = {r1, . . . , rm} set of resources ◮ X = ×i∈NXi set of strategy profiles with Xi ⊆ 2R ◮ Strategy profile x = (x1, . . . , xn) ∈ X ◮ Load of a resource xr = |{i ∈ N : r ∈ xi}| for x ∈ X ◮ Cost functions cr : N → R nondecreasing (convex) ◮ private cost: πi(x) = r∈xi cr(xr)

Example

Example

Example

Fundamental Questions

1 When do pure Nash equilibria exist ? 2 How do players find them ? 3 How difficult is it to compute them ?

Potential Functions

Definition (Exact potential function)

P : X1 × · · · × Xn → R If a player changes his action, the change in the potential function value is equal to the change in her payoff. ui(xi, x−i) − ui(yi, x−i) = P(xi, x−i) − P(yi, x−i)

Potential Functions

Definition (Exact potential function)

P : X1 × · · · × Xn → R If a player changes his action, the change in the potential function value is equal to the change in her payoff. ui(xi, x−i) − ui(yi, x−i) = P(xi, x−i) − P(yi, x−i) b–Potential ui(xi, x−i) − ui(yi, x−i) = bi (P(xi, x−i) − P(yi, x−i)) Monderer and Shapley (1996)

Merits of Potential Games

Theorem

Every finite exact potential game (with potential P)

◮ possesses a PNE ◮ every sequence of improving moves is finite (FIP) ◮ every local minimum of P is a PNE

Monderer and Shapley (1996)

Proof

◮ path γ = (x0, x1, . . . , ) sequence of unilateral moves ◮ improvement path γ = (x0, x1, . . . , ) sequence of unilateral

improving moves γ = x0, x1, . . . improvement path ⇒ P(x0) > P(x1) > · · · > must be finite.

Congestion Games are Potential Games

Theorem (Rosenthal ’73)

Every congestion game

◮ admits an exact potential function ◮ possesses a PNE ◮ possesses the Finite Improvement Property, that is, every

sequence of improving moves is finite.

Proof

Rosenthal’s exact potential function P : X1 × · · · × Xn → R is defined as P(x) :=

• r∈R

xr

• k=1

cr(k). (1) Let x ∈ X and yi = xi be a unilateral deviation of i. ui(x−i, yi) − ui(x) =

• r∈yi

r / ∈xi

cr(xr + 1) −

• r∈xi

r / ∈yi

cr(xr).

The potential of (x−i, yi) is given by: P(x−i, yi) =

• r∈R

xr

• k=1

cr(k) +

• r∈yi

r / ∈xi

cr(xr + 1) −

• r∈xi

r / ∈yi

cr(xr)

• =ui (x−i ,yi )−ui (x)

= P(x) + ui(x−i, yi) − ui(x).

Complexity of Computing PNE

Theorem (Fabrikant et al. ’04, Ackermann et al. ’08)

It is PLS-complete to compute a PNE even for symmetric congestion games with affine costs.

Complexity of Computing PNE

Theorem (Fabrikant et al. ’04, Ackermann et al. ’08)

It is PLS-complete to compute a PNE even for symmetric congestion games with affine costs.

Theorem (Fabrikant et al. ’04)

For symmetric network congestion games, there is a polynomial time algorithm to compute a PNE. Subdivide each arc e into n parallel arcs with capacity 1 each and assign costs cei = ce(i) for i ∈ {1, . . . , n}.

Complexity of Computing PNE

Theorem (Fabrikant et al. ’04, Ackermann et al. ’08)

It is PLS-complete to compute a PNE even for symmetric congestion games with affine costs.

Theorem (Fabrikant et al. ’04)

For symmetric network congestion games, there is a polynomial time algorithm to compute a PNE. Subdivide each arc e into n parallel arcs with capacity 1 each and assign costs cei = ce(i) for i ∈ {1, . . . , n}.

Remark (Ackermann et al. ’08)

There are instances on which every best response dynamic needs exponential convergence time.

Complexity of Computing PNE

Theorem (Fabrikant et al. ’04, Ackermann et al. ’08)

It is PLS-complete to compute a PNE even for symmetric congestion games with affine costs.

Theorem (Fabrikant et al. ’04)

For symmetric network congestion games, there is a polynomial time algorithm to compute a PNE. Subdivide each arc e into n parallel arcs with capacity 1 each and assign costs cei = ce(i) for i ∈ {1, . . . , n}.

Remark (Ackermann et al. ’08)

There are instances on which every best response dynamic needs exponential convergence time. Are there other set systems Xi with efficiently comp. PNE?

Introduction Matroids

Definition (Matroid)

A matroid is a pair M = (R, I) where R is a set of resources, and I is a family of subsets of S such that:

1 ∅ ∈ I. 2 If I ⊂ J and J ∈ I, then I ∈ I. 3 Let I, J ∈ I and |I| < |J|, then there exists an x ∈ J \ I such

that I + x ∈ I. A set system R, I that only satisfies (1) and (2) is called an independence system.

Introduction Matroids

Definition (Matroid)

A matroid is a pair M = (R, I) where R is a set of resources, and I is a family of subsets of S such that:

1 ∅ ∈ I. 2 If I ⊂ J and J ∈ I, then I ∈ I. 3 Let I, J ∈ I and |I| < |J|, then there exists an x ∈ J \ I such

that I + x ∈ I. A set system R, I that only satisfies (1) and (2) is called an independence system. Bases are sets in I of maximal cardinality, denoted by B

Example: Uniform Matroids

The independent sets of a k-uniform matroid are the sets that contain at most k elements.

Example

4 resources: {1, 2, 3, 4}

Example: Uniform Matroids

The independent sets of a k-uniform matroid are the sets that contain at most k elements.

Example

4 resources: {1, 2, 3, 4} Independent sets of the 3-uniform matroid: I = {∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}}

Example: Uniform Matroids

The independent sets of a k-uniform matroid are the sets that contain at most k elements.

Example

4 resources: {1, 2, 3, 4} Independent sets of the 3-uniform matroid: I = {∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}} Bases: B = {{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}}

Graphic Matroid (Cycle Matroid)

1 2 3 4 5 6

Figure : K4 with two bases B1 (red), B2 (blue).

Fundamental Results

Theorem

Let (R, I) be an independence system. The, the following is equivalent: M1 M = (R, I) is a matroid. M2 I, J ∈ I with |I| = |J| + 1 ⇒ there is x ∈ I \ J with J + x ∈ I. M3 For every I ⊆ R, every basis of I has the same cardinality (a basis of I ⊆ E is an inclusion-wise maximal set in I).

Proof.

(M1) ⇔ (M2) is trivial. (M1) ⇒ (M3) ist trivial. (M3) ⇒ (M1): Let I, J ∈ I with |I| > |J|. With (M3) we have that J is no base

• f I ∪ J. Hence, there is x ∈ (I ∪ J) \ J = I \ J ∈ I with

J + x ∈ I.

Fundamental Results

Theorem (Basis exchange theorem)

Let (R, I) be a matroid with basis system B. Then,

1 B = ∅ 2 For every B1, B2 ∈ B and x ∈ B1 \ B2 there is y ∈ B2 \ B1

such that B1 − x + y ∈ B.

Proof.

The bases set of (R, I) satisfies (1) since ∅ ∈ I. For condition (2) let B1, B2 ∈ B and x ∈ B1 \ B2. Since B1 − x ∈ I we can use (M2): |B1 − x| + 1 = |B2| hence there is y ∈ B2 \ (B1 − x) with B1 − x + y ∈ I. As all bases have the same cardinality (see (M3)) we get B1 − x + y ∈ B.

1 2 3 4 5 1 2 5 1 3 4

Figure : Two bases B (red) and B′ (blue) of a graphic matroid

1 2 3 4 5 1 2 5 1 3 4

Figure : Two bases B (red) and B′ (blue) of a graphic matroid

B \ B′ 2 5 B′ \ B 3 4

Figure : Bipartite graph G(B△B′) = (B△B′, R).

(x, y) ∈ R ⇔ B − x + y ∈ B.

1 2 3 4 5 1 2 5 1 3 4

Figure : Two bases B (red) and B′ (blue) of a graphic matroid

B \ B′ 2 5 B′ \ B 3 4

Figure : Bipartite graph G(B△B′) = (B△B′, R).

(x, y) ∈ R ⇔ B − x + y ∈ B.

1 2 3 4 5 1 2 5 1 3 4

Figure : Two bases B (red) and B′ (blue) of a graphic matroid

B \ B′ 2 5 B′ \ B 3 4

Figure : Bipartite graph G(B△B′) = (B△B′, R).

(x, y) ∈ R ⇔ B − x + y ∈ B.

1 2 3 4 5 1 2 5 1 3 4

Figure : Two bases B (red) and B′ (blue) of a graphic matroid

B \ B′ 2 5 B′ \ B 3 4

Figure : Bipartite graph G(B△B′) = (B△B′, R).

(x, y) ∈ R ⇔ B − x + y ∈ B.

Perfect matching in G(B△B′)

Theorem (Matching Property)

Let M = (R, I) be a matroid. Then, for every pair (B, B′) of bases of M there exists a perfect matching W (B, B′) in G(B△B′).

Matroid Congestion Games

Matroid Congestion Games

◮ N = {1, . . . , n} set of players ◮ R = {r1, . . . , rm} set of resources ◮ Xi ⊆ 2R with Xi = Bi for Mi = (R, Ii) ◮ private cost for B = (B1, . . . , Bn) ∈ B :

πi(B) =

• r∈Bi

cr(xr)

Best-Response - Polynomial Time

Theorem (Ackermann, R¨

• glin, V¨
• cking ’08)

For matroid congestion games, the best-response dynamic converges after at most n2 · m · maxi∈N rki ≤ n2m2 steps.

Best-Response - Polynomial Time

Theorem (Ackermann, R¨

• glin, V¨
• cking ’08)

For matroid congestion games, the best-response dynamic converges after at most n2 · m · maxi∈N rki ≤ n2m2 steps. Let L be a list of all cost values cr(i), r ∈ R = {r1, . . . , rm}, i ∈ N = {1, . . . , n} sorted in non-decreasing order. Define alternative cost function c′

r : N → {1, . . . , n · m},

where i ∈ [0, n] is mapped to list position of cr(i)in L (same costs are mapped to same position).

Lemma

Let B∗

i be a best-response w.r.t. B ∈ B with B∗ i = Bi. Then B∗ i

decreases also the private costs w.r.t. c′.

Lemma

Let B∗

i be a best-response w.r.t. B ∈ B with B∗ i = Bi. Then B∗ i

decreases also the private costs w.r.t. c′. Consider G(B∗

i ∆Bi) with perfect matching N. Let (u, v) ∈ N, i.e.,

v ∈ B∗

i \ Bi, u ∈ Bi \ B∗ i and B∗ i − v + u ∈ Bi.

For B∗ = (B∗

i , B−i) with load vector x∗ we have

cv(x∗

v ) ≤ cu(x∗ u + 1) for all (u, v) ∈ N,

as else B′

i := B∗ i − v + u would give less costs than B∗ i . There

must be (v, u) ∈ N with cv(x∗

v ) < cu(x∗ u + 1),

since B∗

i strictly decreases the private cost for i. Thus, also the

private costs under c′ must decrease.

Proof

Consider P′ w.r.t. c′. We get c′

r(i) ≤ n · m for all r ∈ R and all 1 ≤ i ≤ n.

Proof

Consider P′ w.r.t. c′. We get c′

r(i) ≤ n · m for all r ∈ R and all 1 ≤ i ≤ n.

Hence, P′(B) =

• r∈R

xr

• i=1

c′

r(i) ≤

• r∈R

xr

• i=1

n · m ≤ n2 · m · max

i∈N rki.

Proof

Consider P′ w.r.t. c′. We get c′

r(i) ≤ n · m for all r ∈ R and all 1 ≤ i ≤ n.

Hence, P′(B) =

• r∈R

xr

• i=1

c′

r(i) ≤

• r∈R

xr

• i=1

n · m ≤ n2 · m · max

i∈N rki.

Remark (Ackermann, R¨

• glin, V¨
• cking ’08)

For every non-matroid set system Xi, i ∈ N, there are isomorphic instances with exponentially long best-response sequences.

Part II Integral Splittable Congestion Games

Integral Splittable Congestion Games

Integral Congestion Games

◮ N = {1, . . . , n} set of players ◮ R = {r1, . . . , rm} set of resources ◮ Xi ⊆ 2R set of allowable subsets ◮ Demands di ∈ N ◮ Strategy corresponds to integral distribution of di among the

sets in Xi

◮ cost of a resource cr(x) = cr( i∈N xi,r) ◮ private cost πi(x) = r∈R cr(x)xi,r

A Counter-Example

Rosenthal (1973b)

◮ One player from A to C with

demand 2 (”two taxicaps”)

◮ One player from A to B with

demand 1

Positive Results

Theorem (Tran-Thanh et al. (2011))

◮ If Xi consists of singletons, then there is a PNE.

Maximal Structures

What are maximal structures of Xi such that the existence of PNE is guaranteed?

Maximal Structures

What are maximal structures of Xi such that the existence of PNE is guaranteed?

Theorem (H, Klimm, Peis (2014))

Polymatroids are the maximal structure.

Submodular Functions

Definition (integral, submodular, monotone, normalized)

◮ f : 2R → N is submodular if

f (U) + f (V ) ≥ f (U ∪ V ) + f (U ∩ V ) for all U, V ∈ 2R.

◮ f is monotone, if U ⊆ V ⇒ f (U) ≤ f (V ) ◮ f is normalized, if f (∅) = 0.

Submodular Functions

Definition (integral, submodular, monotone, normalized)

◮ f : 2R → N is submodular if

f (U) + f (V ) ≥ f (U ∪ V ) + f (U ∩ V ) for all U, V ∈ 2R.

◮ f is monotone, if U ⊆ V ⇒ f (U) ≤ f (V ) ◮ f is normalized, if f (∅) = 0.

Example

◮ rank function rk : 2R → N of a matroid M = (I, R) ◮ for F ⊆ R : rk(F) := max{|B|, B basis of F}

Integral Polymatroids

integral polymatroid Pf :=

• x ∈ NR |
• r∈U

xr ≤ f (U) for all U ⊆ R

• .

Integral Polymatroids

integral polymatroid Pf :=

• x ∈ NR |
• r∈U

xr ≤ f (U) for all U ⊆ R

• .

“truncated” integral polymatroid Pf (d) := {x ∈ NR |

• r∈U

xr ≤ f (U) for all U ⊆ R,

• r∈R

xr ≤ d}.

Integral Polymatroids

integral polymatroid Pf :=

• x ∈ NR |
• r∈U

xr ≤ f (U) for all U ⊆ R

• .

“truncated” integral polymatroid Pf (d) := {x ∈ NR |

• r∈U

xr ≤ f (U) for all U ⊆ R,

• r∈R

xr ≤ d}. integral polymatroid base polytope Bf (d) :=

• x ∈ NR |
• r∈U

xr ≤ f (U) for all U ⊆ R,

• r∈R

xr = d

• .

Congestion Games on Polymatroids

Strategic Game

◮ G = (N, X, π) ◮ Xi = Bf (i)(di) ◮ πi(x) = r∈R ci,r(x)xi,r

Bf (i)(di) =

• xi ∈ NR |
• r∈U

xi,r ≤ f (i)(U) for all U ⊆ R,

• r∈R

xi,r = di

Examples

Singletons

f (i)({r}) = di falls r ∈ Ri und f (i)({r}) = 0, sonst.

Bases of Matroids

f (i) is rank function of some matroid Mi = (Ri, Ii) and di = rki(Ri).

Spanning Trees

◮ R edges of a graph ◮ Ri ⊂ R connected subgraph ◮ Bases of Mi = (Ri, Ii) are spanning trees of Ri

Existence Proof

◮ Induction over d = i∈N di ◮ d = 1 trivial ◮ d → d + 1

◮ let x be a PNE for a game with demand d ◮ In step d → d + 1 exactly for one player i: di → di + 1.

Existence Proof (contd.)

Hamming Distance of x, y ∈ N|R|: H(y, x) =

r∈R |yr − xr|

Sensitivity Lemma - Increased Demand

Let x be given. There exists a best response yi ∈ argmin

• πi(yi, x−i) s.t. yi ∈ Bf (i)(di + 1)
• with H(xi, yi) = 1.

Existence Proof (contd.)

Hamming Distance of x, y ∈ N|R|: H(y, x) =

r∈R |yr − xr|

Sensitivity Lemma - Increased Demand

Let x be given. There exists a best response yi ∈ argmin

• πi(yi, x−i) s.t. yi ∈ Bf (i)(di + 1)
• with H(xi, yi) = 1.

Existence Proof (contd.)

Sensitivity Lemma - Increased Load

Let ar =

i=j xi,r, r ∈ R be given and xj be a best response to a.

If ar → ar + 1 for one r ∈ R, then there exists a best response yj ∈ argmin

• πj(yj, a) s.t. yj ∈ Bf (j)(dj)
• with H(xj, yj) ∈ {0, 2}.

Existence Proof (contd.)

Sensitivity Lemma - Increased Load

Let ar =

i=j xi,r, r ∈ R be given and xj be a best response to a.

If ar → ar + 1 for one r ∈ R, then there exists a best response yj ∈ argmin

• πj(yj, a) s.t. yj ∈ Bf (j)(dj)
• with H(xj, yj) ∈ {0, 2}.

Existence Proof (contd.)

Sensitivity Lemma - Increased Load

Let ar =

i=j xi,r, r ∈ R be given and xj be a best response to a.

If ar → ar + 1 for one r ∈ R, then there exists a best response yj ∈ argmin

• πj(yj, a) s.t. yj ∈ Bf (j)(dj)
• with H(xj, yj) ∈ {0, 2}.

Existence Proof (contd.)

Invariant

There is a sequence of best-responses yk with H(y0 = y, yk) =

r∈R |yr − xr| = 2.

Existence Proof (contd.)

Invariant

There is a sequence of best-responses yk with H(y0 = y, yk) =

r∈R |yr − xr| = 2.

Convergence

The sequence yk is finite.

◮ Define “marginal costs” of every unit of every player ◮ Sorted vector of marginal costs decreases lexicographically

Remark

For “non-polymatroid” Xi, there are counterexamples.

Part III Nonatomic Congestion Games - The Braess Paradox

Braess Paradox via Cost Reductions

s

x 1

t

1 x

s

x 1

t

1 x Left equilibrium: C(x) = ( 1

2 · 1 2 + 1 2 · 1) · 2 = 3/2

Right equilibrium: C(x) = 1 · 1 + 1 · 1 = 2

Braess Paradox via Cost Reductions

s

x 1

t

1 x

s

x 1

t

1 x Left equilibrium: C(x) = ( 1

2 · 1 2 + 1 2 · 1) · 2 = 3/2

Right equilibrium: C(x) = 1 · 1 + 1 · 1 = 2 G is immune to BP-free iff G is series-parallel [Milchtaich, ’06, Chen et al. ’15].

Braess Paradox via Cost Reductions

s

x 1

t

1 x

s

x 1

t

1 x Left equilibrium: C(x) = ( 1

2 · 1 2 + 1 2 · 1) · 2 = 3/2

Right equilibrium: C(x) = 1 · 1 + 1 · 1 = 2 G is immune to BP-free iff G is series-parallel [Milchtaich, ’06, Chen et al. ’15].

What about other strategy spaces:

◮ tours in a graph ◮ scheduling jobs on machines ◮ spanning trees ◮ Steiner trees

Braess Paradox via Cost Reductions

s

x 1

t

1 x

s

x 1

t

1 x Left equilibrium: C(x) = ( 1

2 · 1 2 + 1 2 · 1) · 2 = 3/2

Right equilibrium: C(x) = 1 · 1 + 1 · 1 = 2 G is immune to BP-free iff G is series-parallel [Milchtaich, ’06, Chen et al. ’15].

What about other strategy spaces:

◮ tours in a graph ◮ scheduling jobs on machines ◮ spanning trees ◮ Steiner trees

Q: What is the maximal combinatorial structure of strategy spaces so that there is no Braess paradox?

Nonatomic Congestion Model

Model M

◮ N = {1, . . . , n} populations, represented by [0, di], i ∈ N ◮ R = {r1, . . . , rm} resources ◮ Strategies Si ⊆ 2R with S = ∪i∈NSi ◮ Strategy distribution (xS)S∈Si with S∈Si xS = di ◮ Load of overall distribution x: xr = S∈S:r∈S xS ◮ Cost function cr : R+ → R+ nondecreasing

Nonatomic Congestion Model

Model M

◮ N = {1, . . . , n} populations, represented by [0, di], i ∈ N ◮ R = {r1, . . . , rm} resources ◮ Strategies Si ⊆ 2R with S = ∪i∈NSi ◮ Strategy distribution (xS)S∈Si with S∈Si xS = di ◮ Load of overall distribution x: xr = S∈S:r∈S xS ◮ Cost function cr : R+ → R+ nondecreasing

Example

◮ Si corresponds to paths from si to ti in a graph ◮ tours in a graph ◮ machines ...

Nonatomic Congestion Games

If player from population i picks S she gets disutility: πi,S(x) =

r∈S cr(xr).

Definition (Wardrop Equilibrium)

πi(x) :=

• r∈Si

cr(xr) ≤

• r∈S′

i

cr(xr) for any Si, S′

i ∈ Si with xSi > 0, for all i ∈ N.

Braess Paradox via Cost Reductions

s

x 1 ∞

t

1 x

s

x 1

t

1 x

Figure : Example of the Braess paradox.

Braess Paradox via Demand Reductions

s1, s2 t2, s3

x

t1, t3

2 c(x) x c(x) M M + 1 1 d1 = 1, d2 = 2, d3 = M with cost C(x∗) = 1 · 2 + 2 · 2 + M · 0 = 6 Reducing d2 → 0: C(xnew) = M + 2

Weak/Strong Braess Paradox

Let x and ¯ x be Wardrop equilibria before and after a cost/demand reduction, resp.

Definition

(Si)i∈N admits the

1 weak BP, if there exists a resource r ∈ R with ¯

cr(¯ xr) > cr(xr).

2 strong BP, if there exists a population i ∈ N with

πi(¯ x) > πi(x).

Main Result

Q: What is the maximal combinatorial structure of (Si)i∈N which is immune to the weak or strong Braess paradox?

Main Result

Q: What is the maximal combinatorial structure of (Si)i∈N which is immune to the weak or strong Braess paradox? A: Bases of matroids (and only matroids!) (jointly with S. Fujishige, M. Goemans, B. Peis and R. Zenklusen)

Main Result

Q: What is the maximal combinatorial structure of (Si)i∈N which is immune to the weak or strong Braess paradox? A: Bases of matroids (and only matroids!) (jointly with S. Fujishige, M. Goemans, B. Peis and R. Zenklusen)

Theorem

Let (Si)i∈N be a family of set systems. Then, the following statements are equivalent. (I) For all i ∈ N, the clutter cl(Si) consists of the base sets of a matroid Mi = (R, Ii). (II) (Si)i∈N is immune to the weak (and strong) Braess paradox.

Main Result

Q: What is the maximal combinatorial structure of (Si)i∈N which is immune to the weak or strong Braess paradox? A: Bases of matroids (and only matroids!) (jointly with S. Fujishige, M. Goemans, B. Peis and R. Zenklusen)

Theorem

Let (Si)i∈N be a family of set systems. Then, the following statements are equivalent. (I) For all i ∈ N, the clutter cl(Si) consists of the base sets of a matroid Mi = (R, Ii). (II) (Si)i∈N is immune to the weak (and strong) Braess paradox. Proof via Sensitivity Theory for Polymatroids.

Compact Representations of Strategies

P(M) :=   x ∈ RS

≥0

• S∈Si

xS = di for all i ∈ N    .

Compact Representations of Strategies

P(M) :=   x ∈ RS

≥0

• S∈Si

xS = di for all i ∈ N    . Resource-based polytope ˜ P(M) ⊆ RR

≥0

˜ P(M) :=   

• i∈N
• S∈Si

xS · χS

• x ∈ P(M)

   , χS ∈ {0, 1}R for S ⊆ R is the characteristic vector of S.

Compact Representations of Strategies

P(M) :=   x ∈ RS

≥0

• S∈Si

xS = di for all i ∈ N    . Resource-based polytope ˜ P(M) ⊆ RR

≥0

˜ P(M) :=   

• i∈N
• S∈Si

xS · χS

• x ∈ P(M)

   , χS ∈ {0, 1}R for S ⊆ R is the characteristic vector of S.

Theorem (Beckmann et al. ’56)

x is a Wardrop equilibrium if and only if it solves min

x∈˜ P(M)

  Φ(x) :=

• r∈R

xr

• cr(t) dt

   . (2) We call Φ the Beckmann potential.

Polymatroids and Submodular Functions

Definition (submodular, monotone, normalized)

◮ f : 2R → R is submodular if

f (U) + f (V ) ≥ f (U ∪ V ) + f (U ∩ V ) for all U, V ∈ 2R.

◮ f is monotone, if U ⊆ V ⇒ f (U) ≤ f (V ) ◮ f is normalized, if f (∅) = 0.

Ph :=

• x ∈ RR

+ | x(U) ≤ h(U) ∀U ⊆ R, x(R) = h(R)

• ,

for U ⊆ R, x(U) :=

r∈U xr.

Polymatroids and Submodular Functions

Definition (submodular, monotone, normalized)

◮ f : 2R → R is submodular if

f (U) + f (V ) ≥ f (U ∪ V ) + f (U ∩ V ) for all U, V ∈ 2R.

◮ f is monotone, if U ⊆ V ⇒ f (U) ≤ f (V ) ◮ f is normalized, if f (∅) = 0.

Ph :=

• x ∈ RR

+ | x(U) ≤ h(U) ∀U ⊆ R, x(R) = h(R)

• ,

for U ⊆ R, x(U) :=

r∈U xr.

Matroid Base Polytope

Pdi·rki =

• xi ∈ RR

+ | xi(U) ≤ di · rki(U)∀U ⊆ R, xi(R) = di · rki(R)

• ˜

P(M) :=

i∈N Pdi·rki = P

i∈N di·rki

Sensitivity in Polymatroid Optimization

min

x∈Ph

  Φ(x) :=

• r∈R

xr

• cr(t) dt

   , (3) where Ph is a polymatroid base polytope with rank function h and for all r ∈ R, cr : R≥0 → R≥0, are non-decreasing and continuous functions.

Sensitivity in Polymatroid Optimization

min

x∈Ph

  Φ(x) :=

• r∈R

xr

• cr(t) dt

   , (3) where Ph is a polymatroid base polytope with rank function h and for all r ∈ R, cr : R≥0 → R≥0, are non-decreasing and continuous functions.

Optimality Conditions

x ∈ Ph is optimal if and only if ce(xe) ≤ cf (xf ) for all e, f ∈ R, xe > 0 with x′ := x + ǫ(χf − χe) ∈ Ph for some ǫ > 0.

Matroids are Immune to Braess Paradox

To show ¯ ce(¯ xe) ≤ ce(xe) for all e ∈ R.

Assume ¯ ce(¯ xe) > ce(xe) for some e ∈ R.

Matroids are Immune to Braess Paradox

To show ¯ ce(¯ xe) ≤ ce(xe) for all e ∈ R.

Assume ¯ ce(¯ xe) > ce(xe) for some e ∈ R. Then ¯ xe > xe.

Matroids are Immune to Braess Paradox

To show ¯ ce(¯ xe) ≤ ce(xe) for all e ∈ R.

Assume ¯ ce(¯ xe) > ce(xe) for some e ∈ R. Then ¯ xe > xe. Strong exchange property of polymatroids: ∃f ∈ R − e with ¯ xf < xf und ǫ > 0 such that x − ǫχf + ǫχe ∈ Ph and ¯ x − ǫχe + ǫχf ∈ Ph.

Matroids are Immune to Braess Paradox

To show ¯ ce(¯ xe) ≤ ce(xe) for all e ∈ R.

Assume ¯ ce(¯ xe) > ce(xe) for some e ∈ R. Then ¯ xe > xe. Strong exchange property of polymatroids: ∃f ∈ R − e with ¯ xf < xf und ǫ > 0 such that x − ǫχf + ǫχe ∈ Ph and ¯ x − ǫχe + ǫχf ∈ Ph. Wardrop equilibrium condition:

1 cf (xf ) ≤ ce(xe), as x WE for c,

Matroids are Immune to Braess Paradox

To show ¯ ce(¯ xe) ≤ ce(xe) for all e ∈ R.

Assume ¯ ce(¯ xe) > ce(xe) for some e ∈ R. Then ¯ xe > xe. Strong exchange property of polymatroids: ∃f ∈ R − e with ¯ xf < xf und ǫ > 0 such that x − ǫχf + ǫχe ∈ Ph and ¯ x − ǫχe + ǫχf ∈ Ph. Wardrop equilibrium condition:

1 cf (xf ) ≤ ce(xe), as x WE for c, 2 ¯

ce(¯ xe) ≤ ¯ cf (¯ xf ), as ¯ x WE for ¯ c.

Matroids are Immune to Braess Paradox

To show ¯ ce(¯ xe) ≤ ce(xe) for all e ∈ R.

Assume ¯ ce(¯ xe) > ce(xe) for some e ∈ R. Then ¯ xe > xe. Strong exchange property of polymatroids: ∃f ∈ R − e with ¯ xf < xf und ǫ > 0 such that x − ǫχf + ǫχe ∈ Ph and ¯ x − ǫχe + ǫχf ∈ Ph. Wardrop equilibrium condition:

1 cf (xf ) ≤ ce(xe), as x WE for c, 2 ¯

ce(¯ xe) ≤ ¯ cf (¯ xf ), as ¯ x WE for ¯ c. We get the following contradiction cf (xf ) ≤ ce(xe) < ¯ ce(¯ xe) ≤ ¯ cf (¯ xf ) ≤ cf (xf ).

Summary

Part I matroids lead to fast convergence of best-response in congestion games Part II polymatroids allow PNE in integral splittable congestion games Part III matroids are immune to Braess Paradox

Summary

Part I matroids lead to fast convergence of best-response in congestion games Part II polymatroids allow PNE in integral splittable congestion games Part III matroids are immune to Braess Paradox

◮ all results are tight (in some sense)!