La tarjeta ciudadana de Zaragoza.

We conclude this chapter with results regarding equilibria existence on cost- sharing in the splittable version [24, 46, 51, 65, 74] of congestion games with

Figure 4.1: The network congestion game in Corollary 5. sp sp′ sq′ sq tp tp′ tq′ tq x3 _{(1− ǫ) · x}3 5· x3 2· x3 _{5· x}3 5· x3

Table 4.2: Players’ costs in example of Corollary 5.

P1 P3 P2 20 + 8 5 + (13− 9) 20 + 8 (16_{− 9) + (13 − 9)} P4 20 + 7 5 + 13 20 + (2_{− )} (17_{− 4) + (5 − 4)} P3 20 + (11− 4) 5 + (37− 9) 20 + (5_{− 4)} (37_{− 9) + (10 − 9)}

multi-commodity players. In the splittable version of such games, the weight wq of a commodity q∈ Q can be split among its strategies in Pq; i.e., a frac- tional strategy of commodity q _{∈ Q is a vector P}q = (wq,P)P ∈Pq ∈ R|P

q_|

≥0 with

P

P ∈Pqwq,P = wq. For the unsplittable version, vector Pqhas only one non-zero and equal to wq component, which is not necessarily the case for the splittable games. For the single-commodity per player model, it is known that the propor- tional sharing method, having players paying a cost share proportional to their flows on each resource, guarantees existence of a pure Nash equilibrium [65].

Below we present two results with respect to the pure Nash equilibrium existence property of cost-sharing methods.

Theorem 4.2.3. Using the Shapley value to share player costs in splittable congestion games with multi-commodity players yields a potential game. Proof. We prove this theorem by showing that these games are potential games. We use the potential function that was used for the unsplittable case

Φ(P ) =X e∈E

X

i∈N

χ(i, fe≤i,π(P ), Ce).

The only difference for the splittable case, is that any fraction of the weight of a commodity can deviate to another strategy. In our main model, Shapley value is applied on the total flows players assign on a resource, fi

e(P ), which equals to the sum of the weight-fractions of the commodities of each player on the resource, i.e., P

q∈Qi,e∈Piwq,P, as opposed to the

P

q∈Qi,e∈Piwq (for

the unsplittable case). But this difference has no affect in the proof steps of Theorem 4.2.1, therefore the same proof steps as in Theorem 4.2.1 apply on splittable games.

Theorem 4.2.4. For a weighted Shapley value with parameter γ large enough, there exists a splittable congestion game with single-commodity per player ad- mitting no PNE.

Proof. We provide a sketch of the proof. For simplicity of the proof, we only proof the case of γ =_{∞. This case captures the main idea of our proof. However,} our instance also provides a counter-example if γ is finite, but large enough.

Consider two players 1 and 2 who control one commodity each, q _{∈ Q}1 and q0 _{∈ Q}2 with wq = wq0 = 1. There is a set of three parallel resources

E =_{{1, 2, 3} with associated cost functions C}1(x) = C3(x) = x and C2(x) = x3. The strategy set of commodity q and q0is_Pq ₌

{{1}, {2}} and Pq0 ₌

{{2}, {3}}, respectively. Each commodity splits her flow of 1 between the available re- sources. Since each player controls only one commodity, for the rest of the proof we refer to players instead of the commodities.

First, consider the case where players assign different flows on the common resource 2. Let y and z be these flows for player 1 and player 2, respectively. Assume that both players are in a pure Nash equilibrium and, without loss of generality, player 1 assigns more flow than player 2 on resource 2, that is, y > z. Since γ = _{∞, player 1 is certain to come last in the Shapley value ordering.} This gives the following costs for the players:

χ1,1+ χ1,2 = 1− y + (y + z)3− z3

= 1_{− y + y}3+ 3_{· z}2_{· y + 3 · z · y}2
, (4.10)

χ2,3+ χ2,2 = 1− z + z3. (4.11)

Note that the cost share of player 2 depends only on her flow z and observe that (4.11) is minimized for z = √1

3. Thus, either y≤ 1 √

3, in which case player 2 can improve by slightly increasing x, or y > √1

3 in which case we must have z = _√1

3 by the Nash equilibrium condition. Substituting z = 1 √

3 in (4.10), we get y3+√3_{· y}2, which is increasing in y. Thus, in this case, player 1 can improve by slightly reducing y. Therefore there is no equilibrium when players assign different flows on resource 2, which completes the first part of the proof. Focus now on the case where players assign the same flow (y = z) on resource 2. Then their cost shares of both players are the same and equal to

χ1,1+ χ1,2= 1− y + 1

2 · (2 · y) 3

= 1_{− y + 4 · y}3. (4.12) We claim that there is at least one player who can improve her cost decreasing her flow on resource 2 by an  small. Let player 1 be the player who deviates. Then she comes first in the ordering and her cost becomes

χ1,1+ χ1,2= 1− y +  + (y − )3,

which for small enough  is less than her intial cost (4.12). This completes the second part of the proof.