Multistage interval scheduling games

Abstract

We study a game theoretical model of multistage interval scheduling problems in which each job consists of exactly one task (interval) for each of t stages (machines). In the game theoretical model, the machine of each stage is controlled by a different selfish player who wants to maximize her total profit, where the profit for scheduling the task of a job j is a fraction of the weight of the job that is determined by the set of players that also schedule their corresponding task of job j. We provide criteria for the existence of pure Nash equilibria and prove bounds on the Price of Anarchy and the Price of Stability for different social welfare functions.

Appendix

1.1 Payoff matrices

We present the payoff matrices for the examples in Lemmas 1 and 2 (Tables 1, 2).

Table 1 The payoff matrix for the example in the proof of Lemma 1

Table 2 The payoff matrix for the example in the proof of Lemma 2

1.2 Proofs of Theorem 9 and Theorem 10

We now present the proofs of Theorems 9 and 10.

Proof of Theorem 9

Let $\varGamma =\varGamma (N,J,(\alpha _j)_j)$ be a MISG with job-dependent rewards. If $\vert J\vert = 1$, the statement is trivial. If $\vert J\vert \geqslant 2$, we construct a MISG $\overline{\varGamma }=\overline{\varGamma }(\overline{N},\overline{J},\overline{\alpha })$ with job-independent rewards as follows: For each job $j\in J$, we create a new player, i.e., $\overline{N} :=N \dot{\cup } J$ and $\overline{t} :=t + \vert J \vert$, and the set of jobs is given as $\overline{J}:=J \cup \{M\}$, where M denotes a job with such a large weight that it is always beneficial for every player to include it in her schedule. The weight of each job $j\in \overline{J} \;\setminus \{M\}$ is set to $\overline{w}_j :=1+\max _{A\subseteq N} \max _{i\in A} \alpha _j(i,A)$. The reward function $\overline{\alpha }$ of $\overline{\varGamma }$ is defined by setting

$$\begin{aligned} \overline{\alpha }(i,A) :={\left\{ \begin{array}{ll} 0, &{}i \notin A \\ 1, &{}i\in A\cap J \\ 0, &{}i\in A\cap N, A\cap J =\emptyset \\ \alpha _j(i,A\setminus \{j\}) \cdot \frac{1}{\overline{w}_j}, &{} i\in A\cap N, A\cap J = \{j\} \\ 1, &{} i\in A\cap N, \vert A \cap J \vert \geqslant 2. \end{array}\right. } \end{aligned}$$

Note that, by definition of the weights $\overline{w}_j$, $\overline{\alpha }$ takes values in [0, 1]. The intersections of all jobs for the players in N (the original players) remain the same, and the additional job M is presented such that its tasks do not intersect with any other tasks. The intersections for a new player $i\in \overline{N} \setminus N$ are as shown in Fig. 9, so that her optimal schedule is always $\overline{S}_i = \{i,M\}$.

We define a mapping $f:\pmb {S}\rightarrow \pmb {\overline{S}}$ from the set of schedules of $\varGamma$ to the set of schedules of $\overline{\varGamma }$ by setting

$$\begin{aligned}&f((S_1, \ldots , S_t)) \\&:=(S_1 \cup \{M\},\ldots , S_t \cup \{M\}, \{1,M\}, \ldots , \{n,M\}). \end{aligned}$$

Hence, the schedules of the original players are just complemented by the job M and do not change otherwise, and the schedule of each new player $i\in J= \overline{N}\setminus N$ is set to $\{i,M\}$. Thus, f is obviously injective.

Let $\text {NE}(\varGamma )$ and $\text {NE}(\overline{\varGamma })$ denote the sets of all pure NEs of $\varGamma$ and $\overline{\varGamma }$, respectively. We now show that the restriction $f\vert _{\text {NE}(\varGamma )}: \text {NE}(\varGamma )\rightarrow \text {NE}(\overline{\varGamma })$ is well-defined and surjective, which implies that $f\vert _{\text {NE}(\varGamma )}$ is a bijection and, thus, shows the claim.

If S is a NE of $\varGamma$, then f(S) is a NE of $\overline{\varGamma }$ because an original player clearly cannot improve her schedule and, as argued above, the schedule $\{i,M\}$ is optimal for a player $i\in J = \overline{N}\setminus N$.

Fig. 9

The tasks presented to player $i\in \overline{N}\setminus N$. The schedule $\overline{S} = \{i,M\}$ is a strictly dominant strategy for player i in $\overline{\varGamma }$

Now let $\overline{S}$ be a NE of $\overline{\varGamma }$. As argued above, each player $i\in J=\overline{N}\setminus N$ adopts the schedule $\overline{S}_i=\{i,M\}$. Hence, $\vert A_M^{\overline{S}} \vert \geqslant \vert J \vert \geqslant 2$, so $\overline{\alpha }(i,A_M^{\overline{S}}) = 1$ for all players $i\in N$. Thus, the job M must also be contained in every schedule $\overline{S}_i$ for $i\in N$. Consequently, the schedule S of $\varGamma$ defined by setting $S_i :=\overline{S}_i \setminus \{M\}$ for all $i\in N$ satisfies $f(S) = \overline{S}$ and is a NE of $\varGamma$. $\square$

Proof of Theorem 10

Given a MISG $\varGamma$ with restricted availabilities, we construct a MISG $\overline{\varGamma }$ with job-dependent rewards as follows: We just introduce a new job M with such a large weight M that it is always beneficial to include it in a schedule. Therefore, we set $\alpha _M(i,A) :=C$ for all $A\subseteq N$ and $i \in A$, where C is a large constant ($C> t\cdot \vert J \vert \cdot \max _{j\in J}\max _{A\subseteq N}\max _{i\in A}\alpha _j(i,A)$). In $\overline{\varGamma }$, for player i, all intersections of the tasks in $J_i$ remain the same as for the original instance. Moreover, we present job M such that it does not intersect with any job from $J_i$. Finally, we present all jobs in $J\setminus J_i$ directly above job M such that they are never included in a dominant schedule.

We define a mapping $g:\pmb {S} \rightarrow \pmb {\overline{S}}$ by setting

$$\begin{aligned} g((S_1, \ldots , S_t)) :=(S_1 \cup \{M\}, \ldots , S_t \cup \{M\}). \end{aligned}$$

In particular, the schedule $g(S)_i$ of player i in $\overline{\varGamma }$ is now $S_i \cup \{M\}$.

Now, we can argue analogously to the proof of Theorem 9 to show that the restriction $g_{\vert NE(\varGamma )}: \text {NE}(\varGamma ) \rightarrow \text {NE}(\overline{\varGamma })$ is a bijection. This follows since the intersections of the available tasks remain the same for each player and the utilities change only by an additive constant. $\square$