This document presents a model of interdependent scheduling games (ISG) where multiple players each have tasks to schedule independently but the tasks may have dependencies on each other across players. The document provides an analysis of computational problems related to ISGs including welfare maximization, computing best responses, Nash equilibria existence, and complexity results. Key results are that welfare maximization is NP-hard even for uniform rewards and two tasks per player, but can be solved in polynomial time for a single player. Best responses can also be computed efficiently in some cases but are hard in others.

1. Interdependent Scheduling Games
Andr´es Abeliuk1,2
, Haris Aziz1,3
, Gerardo Berbeglia1,2
, Serge Gaspers1,3
, Petr Kalina4
,
Simon Mackenzie1,3
, Nicholas Mattei1,3
, Paul Stursberg5
, Pascal Van Hentenryck1,6
and
Toby Walsh1,3
1
National ICT Australia (NICTA)
2
University of Melbourne, Australia
3
University of New South Wales, Australia
4
Czech Technical University, Czech Republic
5
Technische Universit¨at M¨unchen, Germany
6
Australian National University, Australia
July 25, 2015
Abstract
We propose a model of interdependent scheduling games in which each player has a set
of tasks that they schedule independently. Each of these tasks only begin to accrue reward
for the player when all predecessor tasks, which may or may not be in the set of tasks for the
player, have been completed. This model, where players have interdependent tasks, is motivated
by the problems faced in planning and coordinating large-scale infrastructures, e.g., restoring
electricity and gas to residents after a natural disaster or providing medical care and supplies
after a disaster. We undertake a detailed game-theoretic analysis of this setting, in particular
consider the issues of welfare maximization, computing best responses, Nash dynamics, and
existence and computation of Nash equilibria.
1 Introduction
In many large, complex systems there are often multiple stakeholders each looking to maximize
their own utility. In some cases, the objectives of the individual stakeholders may be at odds
with some global measure of utility or social welfare. Our primary motivation for this work is
drawn from situations where companies and governments need to coordinate to restore coordinated
infrastructure after major disruption due to disasters and other outside forces (e.g. Cavdaroglu et
al. [2013]; Coffrin et al. [2012]). In many of these cases, one utility may be attempting to maximize
a private objective function that is at odds with the global objective: to restore all services to the
communities affected as quickly as possible. Other examples of these interdependent scheduling
games (ISG) are coordinating multiple providers for humanitarian logistics over multiple states
or regions and the coordination of interdependent supply chain networks which may involve ports,
inland terminals, railway, and truck operators Van Hentenryck et al. [2010]; Simon et al. [2012].
Consider, for instance, the joint restoration of an electrical power network and a gas network
after a natural disaster. These networks typically feature interdependencies: the gas supply may
be needed to run the electrical plant for a whole town. From the community or government’s
perspective, the overall goal is to reduce the size of the (gas and electrical power) blackout. However,
1

2. Complexity Reference
ISG Welfare
NP-complete for uniform rewards, two tasks each Theorem 1
in P for one player Theorem 4
ISG Best Response
in P for uniform rewards Theorem 5
NP-complete for conflict-free schedules Theorem 6
Table 1: Summary of results for interdependent scheduling games (ISG).
each operator may have their own objective of reducing the blackout size in its own network; without
regard for the global view Coffrin et al. [2012].
Informally we have a set of players, each of which has a set of tasks they want to achieve and
each task provides its own reward. The individual players tasks may have dependencies among
their own set of tasks and among the set of tasks of all other players, of which the player is aware.
The distinguishing feature of the games is that dependency of some task t on another task t does
not mean that t cannot be completed without completing t , but that the player can only start
accruing reward for the completion of task t once task t has been completed. In contrast, in most
traditional scheduling games, a task cannot be scheduled unless it is in the right time window
and when its dependencies have been fulfilled. In the games we consider, once a task is activated,
the player continues to gather reward for every time step in which the task is active. These soft
constraints make the scheduling problems easier in some cases and more difficult in others. Our
model captures, for example, settings in power restoration where interdependencies exist between
the gas and electrical network. The electric company may be able to restore the lines to individual
homes, but until the gas company can supply gas to the main generator, no electricity will flow.
Once the power is flowing, the electric company receives its reward (income) from those customers
who are now receiving power.
Contributions We present a scheduling model which represents dependencies among tasks and
has utility functions which capture scenarios in power restoration after natural disasters. The model
raises intriguing mechanism design questions regarding the efficiency of the system.
We also present computational and equilibrium existence and non-existence results. We show
that welfare maximization is NP-hard even when the rewards are uniform, i.e. the reward for all
tasks for all players is the same, and each player has two tasks. On the other hand, for one player,
finding a welfare maximizing schedule can be solved in polynomial time. Therefore, for multiple
players, even when rewards are uniform, the structure of dependencies already induces significant
complexity to the problem of welfare maximization.
We also consider the problem where a player may have an incentive to change its intended
schedule so as to get better total utility. For uniform rewards, a best response can be computed
in linear time. For general rewards, and only inter-player dependencies, a best response can be
computed in polynomial time. On the other hand, finding a conflict free best response schedule
in which no task are inactive is NP-hard. We show that for uniform rewards and two players,
best responses can cycle. Despite the possibility of cycles, we show that a pure Nash equilibrium
always exists in this setting. On the other hand, if the rewards are not uniform, then a pure Nash
equilibrium may not exist. Some of our computational results are summarized in Table 1.
2

3. 2 Related Work
We consider the problem of welfare maximization in ISG’s, which is equivalent to finding the
optimal global schedule. This is a well-studied problem for classic scheduling models. A review
of the scheduling literature can be found in Brucker and Brucker [2007]. We also consider game-
theoretic issues that are not common in scheduling theory such as best response dynamics and
Nash equilibria. These concepts are uniquely applicable in our setting as the players may or may
not have incentives to cooperate towards the optimal global welfare.
There is a vast literature in scheduling problems across a large number of domains. Most of
these problems are focused on allocating scarce resources to multiple tasks in order to maximize an
objective function or minimize total time. This has been a key area of study in computer science
over the last several decades. Unfortunately, many practical problems in this area are NP-hard Lee
et al. [1997]. Scheduling situations in which agents compete for common processing resources were
introduced by Agnetis et al. Agnetis et al. [2000, 2004] and Baker and Smith Baker and Smith
[2003]. The most traditional approach in multi-agent scheduling is to consider a single centralized
agent optimizing the whole domain.
It is only recently that decentralized scheduling mechanisms have been proposed. Agnetis et
al. Agnetis et al. [2007] considers auction and bargaining models, which are useful when several
agents have to negotiate for processing resources on the basis of their scheduling performance.
Scheduling auctions typically divide the schedule horizon into time slots, and these time slots are
auctioned among the players. The bargaining approach considers two agents that have to negotiate
upon possible schedules. Abeliuk et al. Abeliuk et al. [2015] considers a two player bargaining
mechanism for any setting where the utility of one player does not depend on the actions taken
by the other player. Thus, their results are applicable in ISG’s with two players and inter-player
dependencies are only one way. For further discussion of literature on mechanism design for non-
cooperative scheduling games see the surveys Heydenreich et al. [2007]; Christodoulou et al. [2004];
Angel et al. [2006].
In our setting specifically, we do not deal with shared resources between the agents, the most
related literature we find is on multi-agent project-scheduling. Each project is composed of a set of
activities, with precedence relations between the activities and each one requiring specific amounts
of local and shared (among projects) resources. The aim is to complete all the project activities
minimizing each project schedule length. A first mechanism design approach in multi-agent project-
scheduling proposed by Confessore et al. Confessore et al. [2007], uses a decentralized mechanism
using combinatorial auctions. Recently, Briand et al. Briand and Billaut [2011] take a first step in
analyzing game theoretical properties such as the existence of and computing Nash equilibrium as
well as detailing the price of anarchy for multi-agent project-scheduling.
3 Setup
A directed graph G is a pair (V, E), where V is a finite set of vertices and E ⊆ V × V is a set of
directed arcs where (u, v) ∈ V means there is a directed edge from u to v in G. We will always
assume that G is acyclic, i.e. there is no set of edges {(v1, v2), (v2, v3), . . . , (vn, v1)} ⊆ E. We say
that G is transitive if (u, v), (v, w) ∈ E implies that (u, w) ∈ E. The transitive closure of a graph
G = (V, E) is a graph G = (V, E ) such that ∀u, v ∈ V, (u, v) ∈ E whenever there exists a path
from u to v in G. The in-neighborhood of a vertex v is defined as N−
G (v) = {u ∈ V : (u, v) ∈ E},
the set of vertices with edges directed at v. The out-neighborhood of v are defined accordingly as
N+
G (v) = {u ∈ V : (v, u) ∈ E}, the set of vertices with edges directed from v. The neighborhood of
3

4. v is defined as NG(v) = N−
G (v) ∪ N+
G (v).
An interdependent scheduling game (ISG) is a tuple ((T1, . . . , Tk), G, r). Here, we have k players
(alternatively called agents) and each of them has a disjoint set of tasks Ti, 1 ≤ i ≤ k. We denote the
set of all tasks by T := k
i=1 Ti. We assume without loss of generality, that |T1| = · · · = |Tk| =: q,
and for all the problems analyzed in this paper we assume that each task has a duration of one unit
of time. The dependency relation is represented by the transitive directed graph G = (T, E). The
explicit dependencies graph GE is a possibly non-transitive directed acyclic graph used to express
the dependencies for specific classes of games. In case GE is given then the dependency relation
G corresponds to the transitive closure of GE. For each task v ∈ T, there is a reward r(v) ≥ 0.
Occasionally we will consider the more restrictive case of uniform rewards where each task has a
reward value of 1, i.e., for all v ∈ T, r(v) = 1.
To schedule their tasks, each player i selects a permutation πi of Ti, the set of all permutations
available to player i is denoted by Πi. For a given schedule and a given time step, a task v is active
during that time step if all tasks in N−
G [v] are scheduled at or before that time step. Given task v
we denote by t(v) the time step when v is scheduled and by a(v) ≥ t(v) the time when v becomes
active, i.e., t(v) = π−1(v) and a(v) = min{t : ∀w ∈ N−
G [v] t(w) ≤ t}. At each time step, all active
tasks v generate the reward r(v). Thus, for a schedule π = (π1, . . . , πk), the total reward of player
i is
Ri(π) =
|Ti|
j=1 v∈acti(j,π)
r(v).
Here acti(j, π) = {v ∈ Ti : a(v) ≤ j}, thus the welfare of a schedule π is k
i=1 Ri(π).
Certain subsets of the set Πi of permutations available to player i will be called schedule con-
figurations. This notation will be convenient is some of the proofs, since it allows us to denote the
set of all schedules with a subset of specific tasks being scheduled to specific time-slots. A schedule
configuration for player i, denoted by c = (c1, ..., cq) ∈ ({∗} ∪ Ti)q, is defined as follows:
c = (c1, ..., cq) := {πi ∈ Πi|∀cj = ∗ : πi(j) = cj} ,
where πi(j) = ∗ represents that slot j’s tasks is unspecified. We will be slightly imprecise at times
and use a permutation πi and the complete schedule configuration with cj = πi(j) for all j ∈ [q]
interchangeably, although the latter is in fact the singleton set containing only πi.
We often represent the games in a graphical form as shown in Figure 1. Player i’s set of tasks
Ti corresponds to the set of tasks (nodes) in the i-th row. The numbers in the nodes indicate
the reward extracted from the individual tasks. For ease of presentation, we often omit arrows
that correspond to the implicit dependencies (the transitive closure) of the tasks. It is therefore
important to remember that dependencies are transitive, i.e. a task will only activate when all its
predecessors are active.
Note that this representation is unambiguous. Task v can be uniquely identified by r(v) and
the edges in NG(v). Tasks v, w with r(v) = r(w) and NG(v) = NG(w) can be freely interchanged
within any particular solution to the game without affecting the utilities of the other players.
4

5. Player 1 10 1 1
Player 2 1 100 100
Player 1 1 1 10
Player 2100 100 1
Figure 1: An example of an ISG with two players and 3 tasks each. Both of the two 100 utility tasks
belonging to Player 2 are dependent upon a task belonging to Player 1. For the depicted schedule on
the left, Player 1 gets reward 3(10)+2(1)+1 = 33 while Player 2 gets reward 3(1)+2(100)+100 =
303. For the depicted schedule on the right, Player 1 gets reward 3(1)+2(1)+10 = 15 while Player
2 gets reward 3(100) + 2(100) + 1 = 501.
4 Utilitarian welfare maximization
We first consider the problem of utilitarian welfare maximization. To be clear, when speaking of
individual welfare we refer to the utility associated to a player; when considering social welfare, or
simply welfare, the notion is associated to all players as a whole. We can now formally define the
problem.
ISG Welfare
Input: An ISG ((T1, . . . , Tk), G, r) and an integer w.
Question: Is there a schedule π for the ISG such that
k
i=1 Ri(π) ≥ w?
Firstly, we show that welfare maximization is NP-hard for the surprisingly restricted case where
agents have uniform rewards.
Theorem 1. ISG Welfare is NP-hard, even when the rewards are uniform, r(v) = 1 for all
v ∈ T, and each player has two tasks.
Proof. We give a reduction from the NP-hard Min 2SAT problem [Kohli et al., 1994].
Min 2SAT
Input: A propositional 2CNF formula F where each clause
has exactly two literals, and an integer k.
Question: Is there an assignment to the variables of F such that
at most k clauses are satisfied?
For each variable x in F, create a player Px with tasks Tx = {x, ¬x}. For each clause c in F,
create a player Pc with tasks Tc = {c1, c2}. For each clause c = ( 1 ∨ 2), the precedence graph
contains arcs (c1, c2), ( 1, c1), and ( 2, c1). The rewards are uniform, and we set w = 3n + 3m − k,
where n and m are the number of variables and clauses of F, respectively.
It remains to prove that F has an assignment satisfying at most k clauses if and only if the
ISG has a schedule generating a reward of at least w. For the forward direction, suppose F has an
assignment α : var(F) → {0, 1} satisfying at most k clauses. Consider the schedule where, for each
variable x, the player Px schedules first the literal of x that is set to false by α, i.e., x is scheduled
before ¬x iff α(x) = 0. Additionally, for each clause c, the task c1 is scheduled before c2. This
schedule generates a reward of 3 for each variable: a reward of 1 at the first time step and a reward
of 2 at the second time step. For a satisfied clause c, the schedule generates a reward of 2: at
the first time step no reward is generated since the literal satisfying the clause is scheduled at the
second time step and there is an arc from that literal to c1, and a reward of 2 is generated at the
5

6. second time step. For an unsatisfied clause c, the schedule generates a reward of 3: since neither
literal satisfies the clause, both literals are scheduled at the first time step. Thus, the total reward
generated for this schedule is at least 3n + 3m − k.
For the reverse direction, let π be a schedule generating a reward of at least w. Consider the
assignment α : var(F) → {0, 1} with α(x) = 0 iff player Px schedules x at the first time step. Note
that at the second time step, each player generates a reward of 2. Also, each player corresponding
to a variable generates an additional reward of 1 at the first time step since his tasks have in-degree
0. So, at least 3n + 3m − k − (3n + 2m) = m − k additional clause players generate a reward of 1 at
the first time step. But, for each such clause c, c1 is scheduled before c2 and both literals occurring
in c are scheduled at the first time step, which means that the assignment α sets these literals to
false. Therefore, α does not satisfy c. We conclude that α satisfies at most k clauses.
Definition 1. A schedule is conflict-free if no task is inactive (agent is not getting a reward from
the task) when the task has been scheduled.
For the uniform rewards case, it easily follows that any conflict-free schedule is a welfare maxi-
mizing schedule. This property does not hold in the case of non uniform rewards. An example is
shown in the following theorem.
Theorem 2. Even if a conflict-free schedule exists, the welfare maximizing schedule might not be
conflict-free.
Proof. Consider the figure below. The schedule in the left sub-figure is conflict free while the
schedule on the right has a conflict. Despite the conflict, the right schedule has a higher social
welfare as the two r = 100 tasks become active simultaneously in step two, providing more total
reward.
Player 1 1 1 1
Player 2 1 100 100
Welfare = 309.
Player 1 1 1 1
Player 2 1100 100
Welfare = 407.
For the right sub-figure, the schedule is indeed welfare maximizing: Both tasks with reward 100
require two tasks of player 1 to become active. Hence, none of the two tasks can become active
before timestep 2. The schedule shown is the only schedule by which both tasks with reward 100
become active in timestep 2, therefore it is welfare-maximizing.
Additionally, if one wants to compute the optimal schedule among the conflict-free schedules
with not necessarily uniform rewards, the constraint of conflict-freeness makes the problem of
welfare maximization NP-hard even for one agent. Note that dependency constraints in ISG can
be unmet. However, when there is only one player, the welfare maximizing schedule must be a
conflict-free schedule. Below we generalized this observation.
Proposition 1. For one player and non uniform rewards, the welfare maximizing schedule must
be a conflict-free schedule.
Proof. Assume this is not the case, then there must be a schedule π such that π(j) π(k) and
j > k. Given the precedence constraint, task π(k) can only became active after task π(j) is
scheduled, so swapping them can only increase the utility of the player.
6

7. Theorem 3. For one player and non uniform rewards, welfare maximization is NP-hard.
Proof. We give a reduction from the NP-hard Single machine weighted completion time
problem [Lenstra and Rinnooy Kan, 1978].
Single machine weighted completion time
Input: A set of jobs Ji ∈ J. Each job has a weight wi and a
processing time pi = 1. Precedence constraints, such
that if i j, then Jj cannot be scheduled before Ji.
An integer k.
Question: Is there an ordering of the jobs such that i∈J wiCi ≤
k?, where Ci is the completion time of Ji.
For each job Ji ∈ J, create task ti with reward ri = wi and consider the same precedence graph
as the one given for jobs. We set w = (|J| + 1) i∈J wi − k.
By Proposition 1, without loss of generality, we can assume that any schedule for ISG’s with
one player are conflict-free schedules. It remains to prove that there is an ordering π of jobs with
a weighted completion time of at most k if and only if the ISG has a conflict-free schedule π
generating a reward of at least w.
Let π = π , then Ci is the completion time of both, job Ji and task ti given ordering π. Given that
π is a conflict-free schedule, the contribution of ti to the objective function is (|T| + 1 − Ci) ri. Thus,
R(π) = i∈T (|T| + 1 − Ci) ri = (|T|+1) i∈T ri− i∈T riCi. But, i∈T riCi = i∈J wiCi, which
corresponds to the weighted completion time of ordering π. Therefore, R(π) ≥ w ⇔ i∈J wiCi ≤ k,
which concludes the proof.
On the other hand, when restricting our attention to uniform rewards, we show that a welfare
maximization schedule can be found in polynomial time when there is only one player.
Theorem 4. For one player ISG and uniform rewards, welfare maximization can be solved in
O(|T| log |T|) time.
Proof. The problem can be solved as follows. Consider any topological ordering of the transitive
graph. Take the leaf node with the lowest reward v and place it at the end. Repeat the process.
We now argue that the algorithm results in a welfare maximizing schedule. Assume that there
exists a schedule π with a higher welfare and where a leaf node with the lowest reward in not
placed at the end. Then either the node placed at the end is a leaf node v with a higher reward
or a node v which is not a leaf node. In the first case, the welfare of π would increase if the
leaf node v with the higher award is swapped with the leaf node v with the lowest reward. In the
second case, v is not a leaf node and is placed at the end, which means that there is an arc going
backwards to some node v∗. Now consider the modification of π in which v∗ is inserted after v .
Task v∗ and all its successors did not become active before the time when v was scheduled (v∗’s
new position) as they depended on v by transitivity, hence their rewards remain unchanged. On
the other hand, all tasks between the original positions of v∗ and v are now scheduled one time
step earlier, thus their rewards can only increase.
5 Best Responses
We now consider the problem of computing best responses, where players care about their own
utility. We can now formally define the Best Response problem.
7

8. ISG Best Response
Input: An ISG ((T1, . . . , Tk), G, r), schedule profile π−i and
target utility W.
Question: Is there a schedule πi for the ISG such that
Ri(π−i, πi) ≥ W?
Theorem 5. For ISG with uniform rewards, there exists a linear-time algorithm to compute the
best response.
Proof. Dependencies between tasks of the other players are irrelevant, as their only influence on
decisions on the responder side is already captured in the transitive closure of the dependency
structure.
We give a greedy algorithm that solves the problem optimally. Starting from the first time
step, the algorithm successively schedules a maximal task according to the dependency structure,
minimizing the time that a task remains inactive after it was scheduled.
We observe the following: (i) At no point can it be beneficial to schedule a non-maximal
task. (ii) At each point, scheduling the maximal task with smallest time-to-activation is the best
response. Where by time-to-activation of a task v at time step t, we mean max{0, a(v) − t}.
Regarding (i) , suppose a non-maximal task is scheduled prior to its predecessor. Swapping the
two tasks can only increase the responders utility (by the same argument as in Theorem 4).
Regarding (ii) , suppose that, among the maximal tasks, a task a with with larger time-
to-activation is scheduled instead of another task b with smaller time-to-activation. If task b is
scheduled before a’s activation time and since b will become active before a, then swapping the
two tasks will increase utility by at least the difference in their activation times. In the case that
task b is scheduled after a’s activation time, if there are any successors of a scheduled before task
b and after a’s activation time, then swapping tasks a and b will make a’s successors inactive, thus
possible deteriorating the utility. To avoid this, an increase of the utility can be obtained with the
following transformation: (1) move b to a’s position, (2) move task a to the position corresponding
to a’s activation time and (3) shift towards the end by one all tasks between a’s activation time
and b’s original position. Tasks shifted are now scheduled one time step later, but since task a is
scheduled in its activation time, then the utility can not decrease.
Thus, an algorithm that always schedules the maximal element with smallest time-to-activation
computes the best response.
In the more general setting, with non-uniform rewards, let us consider only conflict-free best
response schedules. Lead by the observation that dependencies from the non-responder side induce
release dates on the tasks of the responding player, we can show NP-hardness of the problem.
Theorem 6. Computing the best conflict-free response is NP-hard.
Proof. A reduction from the single machine weighted completion time with release dates can be
obtained by adapting Theorem 3. Noticing that conflict-freenes is attained if and only if release
date constraints are met, the procedure can be carried out in a straightforward way and lead to
essentially the same construction.
However, as already seen in Theorem 2, best responses not necessarily satisfy all exterior de-
pendencies, even though there could exist a conflict-free schedule. So, it remains an open question
whether best responses in the general setting can be computed in polynomial time.
We show that even for ISG with uniform rewards, best responses can cycle.
8

9. Theorem 7. Even for ISG with uniform rewards, best responses can cycle.
Proof. Figures 3 shows that best responses can cycle.
c a d b
d a c b
(a) Player 2 best response
d b c a
d a c b
(b) Player 1 responds
d b c a
c d b a
(c) Player 2 responds
c a d b
c d b a
(d) Player 1 responds
Figure 3: A best response cycle.
The above example can be suitably modified to show the following:
Theorem 8. Even for ISG with strictly increasing rewards, better responses can cycle.
6 Nash equilibria
In this section we consider pure Nash equilibria in ISGs. We first show that ISGs with uniform
rewards always admit a pure Nash equilibrium (PNE). In addition we show that a social welfare
optimal schedule is also a PNE and we provide bounds for the price of anarchy of ISGs with uniform
rewards. For the general setting of ISGs, we show that a PNE is no longer guaranteed to exist.
6.1 ISGs with uniform rewards
Theorem 9. For any ISG with uniform rewards, a maximum welfare schedule profile is in PNE.
Proof. In the case of uniform rewards, maximizing the utility of a player is equivalent to minimizing
the time of tasks not active after they are scheduled. So, if all players have a conflict-free schedule,
then the profile is already in PNE.
If at least one of the schedules has a non active task scheduled, we will show this is the best
that player can do. By Theorem 5 we consider schedules where all the tasks of the same player
preserve the ordering induced by the transitive dependencies. Suppose player i has at least one
non-active task v in πi. Consider the following transformation: postponing task v until it becomes
active and shifting all other tasks earlier. Postponing the inactive task has no effects on the other
players, and tasks that where scheduled earlier can only activate earlier tasks of the other players.
Thus, this transformation can only improve the utility of all players. Therefore, task v cannot be
postponed or we would have found a better social welfare, contradicting our assumption. Finally,
permutations of active tasks cannot cause any improvement as they all have the same reward, so
player i has no strictly better response other than πi.
Corollary 1. For any ISG with uniform rewards, there is always a PNE profile.
9

10. There may be more than one PNE profile in ISG’s with uniform rewards. It is thus natural to
ask how bad the price of anarchy can be.
Proposition 2. For any ISG with uniform rewards, a PNE profile does not necessarily maximize
welfare.
Proof. Figure 4 shows a PNE which is not maximizing welfare solution. By bringing forward the
last task of player 1, we get a PNE that achieves the maximal welfare.
Player 1 1 1 1
Player 2 1 1 1
Figure 4: Game with a pure Nash equilibrium which does not maximize welfare.
Theorem 10. The price of anarchy of ISGs with uniform rewards is at least k(n+1)
n+2k−1, given k
players with n tasks each.
Proof. We extend the idea of Figure 4 to multiple players and tasks. Consider the tasks of all
players, except player 1, are dependent on a task t, belonging to player 1. Thus, as in Figure 4,
the worst PNE is obtained by scheduling task t at the end, as opposed to the PNE achieved when
task t is at the beginning, which results in the maximizing welfare schedule. The ratio between the
best and the worst schedule, given k players with n tasks each, is given by
kn(n+1)
2
n(n+1)
2 + (k − 1)n
=
k(n + 1)
n + 2k − 1
.
If we fix the number of players k, the highest value of the ratio is limn→∞
k(n+1)
n+2k−1 = k. Similarly,
when fixing the number of tasks n, then limk→∞
k(n+1)
n+2k−1 = n+1
2 . This motivates the following
theorem.
Theorem 11. The price of anarchy of ISG with uniform rewards is at most n+1
2 , with n the
number of tasks per player.
Proof. Any PNE profile can not be worse than if all tasks activate at time step n. Thus, NE ≥ kn.
Any schedule can not be better than if there is no precedence constraints. Thus, the optimum
schedule OPT ≤ kn(n+1)
2 . Therefore,
PoA =
OPT
NE
≤
kn(n+1)
2
kn
=
n + 1
2
.
6.2 ISGs with non uniform rewards
Finally, we show that an ISG with two players does not always admit a pure Nash equilibrium.
10

11. Player 1 1 4 3 2
Player 2 2 4 1 3
Figure 5: Game not admitting a Pure Nash Equilibrium.
Theorem 12. An ISG with two players and non uniform rewards does not always admit a pure
Nash equilibrium.
Proof. Consider the explicit game given by Figure 5. Let’s consider the game admits a PNE.
Notice that any best response of player 1 must satisfy that task 4, being the highest reward task,
is scheduled immediately after task 1. Therefore, any possible best response of player 1 has to
adopt one of the following schedule configurations: (i) π1 ∈ (1, 4, ∗, ∗), (ii) π1 ∈ (∗, 1, 4, ∗) or (iii)
π1 ∈ (∗, ∗, 1, 4).
In a similar way, task 4 of player 2, for any best response of player 2, must be scheduled as
soon as possible. These observations narrow the number of possible PNE configurations, which are
explored below.
• Case (i) Player’s 2 best response, given any schedule of the form π1 ∈ (1, 4, ∗, ∗) is π2 =
(2, 4, 1, 3). However, such a schedule triggers a player 1’s best response of π1 = (3, 1, 4, 2),
which take us to case (ii).
• Case (ii) Player’s 2 best response, given any schedule of the form π1 ∈ (∗, 1, 4, ∗) is π2 =
(1, 3, 4, 2). However, such a schedule triggers a player 1’s best response of π1 = (1, 4, 2, 3),
which is an instance of case (i). This leads to a cycle of best responses.
• Case (iii) Player’s 2 best response, given any schedule of the form π1 ∈ (∗, ∗, 1, 4) is π2 ∈
{(2, 1, 3, 4), (1, 3, 2, 4)}. However, such schedules triggers a player 1’s best response of π1 =
(3, 1, 4, 2) if π2 = (2, 1, 3, 4), or π1 = (1, 4, 3, 2) in the other case. Both schedules being an
instance of case (ii) or (i), respectively.
Therefore, for any schedule π1, there is no schedule π2, such that (π1, π2) is a PNE.
We conjecture that the above example is minimal, in the sense that no other instance with
less than 4 tasks per player, does not admit a PNE. The same goes when considering the number
of explicit dependencies. We leave as an open problem the computational complexity of checking
whether there exists a pure Nash equilibrium schedule profile.
7 Conclusions
This paper introduced a class of interdependent scheduling games which were motivated by appli-
cations in large-scale power restoration, humanitarian logistics, and integrated supply-chains. We
addressed computational complexity issues of the centralized social maximizing welfare problem, as
well as in the decentralized, non cooperative game version of the framework. For the latter setting,
we focused our attention to the Nash equilibrium solution concept.
The model we have presented leads to intriguing mechanism design issues where we want to
incentivize players to report their valuations for their tasks truthfully and for the mechanism
to satisfy desirable welfare properties. Some of our NP-hardness results even hold under sever
11

12. restrictions of the model. It will be interesting to undertake a parametrized complexity analysis of
the problems as well as to consider approximation algorithms. Known approximation algorithms for
traditional interdependent scheduling games (with hard dependencies and non-accruing rewards)
cannot be directly applied to the model we have presented. It is also important to generalize
interdependent scheduling games to integer time durations.
Acknowledgments
NICTA is funded by the Australian Government through the Department of Communications and
the Australian Research Council through the ICT Centre of Excellence Program.
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