Scheduling problems have traditionally been studied from a centralized point of view in which the goal is to find an assignment of jobs to machines so as to minimize some global objective function. Two of the classical results are that Smith’s rule, i.e., schedule jobs in decreasing order according to their ratio of weight over processing time, is optimal for single machine scheduling with the sum of weighted completion times objective , and list scheduling, i.e., greedily assign the job with the highest priority to a free machine, yields a -approximation for identical machines with the minimum makespan objective 
. Many modern systems provide service to multiple strategic users, whose individual payoff is affected by the decisions made by others. As a result, non-cooperative game theory has become an essential tool in the analysis of job-scheduling applications. The jobs are controlled by selfish users who independently choose which resources to use. The resulting job-scheduling games have by now been widely studied and many results regarding the efficiency of equilibria in different settings are known.
A particular focus has been placed on finding coordination mechanisms , i.e., local scheduling policies, that induce a good system performance.In these works it is common to assume that ties are broken in a consistent manner (see, e.g., Immorlica et al. ), or that there are no ties at all (see, e.g., Cole et al. ). In practice, there is no real justification for this assumption, except that it avoids subtle difficulties in the analysis. In this paper we relax this restrictive assumption and consider the more general setting in which machines have arbitrary individual priority lists. That is, each machine schedules those jobs that have chosen it according to its priority list. The priority lists are publicly known to the jobs.
In this paper we analyze the effect of having machine-dependent priority lists on the corresponding job-scheduling game. We study the existence of Nash equilibrium, the complexity of identifying whether a NE profile exists, the complexity of calculating a NE, in particular a good one, and the equilibrium inefficiency.
A second motivating application for this paper is traffic networks. A common way to model a traffic network is by means of a congestion game . However, this model has no element of time dependence and assumes that all users on a resource experience the same latency. Though in a road network, a vehicle can only be delayed by the vehicles ahead of it. Farzad et al.  introduced a simple modification that allows to incorporate some element of time dependence. They proposed a model in which the total cost of a resource is given by the integral under the cost function. The idea is that the area under the cost function can be divided among the players so that earlier players are associated with smaller latencies. For atomic players this idea can be implemented by means of priorities on resources: each player will only delay other players on the resource that have a lower priority.
1.1 The Model
An instance of a scheduling game with machine-dependent priority lists is given by a tuple where is a finite set jobs, is a finite set of machines, is the weight of job , is the processing delay of machine , and is the priority list of machine . In the literature, it is common to characterize the jobs by their processing time and the machines by their speed. We prefer to refer to weight instead of to processing time, and to delay, which is the inverse of speed, in order to be consistent with the general definition of congestion games.
A strategy profile assigns a machine to every job . Given a strategy profile , the jobs are processed according to their order in the machines’ priority lists. The set of jobs that delay in is . Note that job itself also belongs to . Let . The cost of job is equal to its completion time in , given by
A more general model is that of a congestion game with resource-dependent priority lists, in which the strategy space of a player consists of subsets of resources. Formally, an instance of the general game is given by a tuple , where is a finite set players, is a finite set of resources, is the set of feasible strategies for player , is the weight of player , is the cost coefficient of resource , and is the priority list of resource that defines its preference over the players using it.
Scheduling games are symmetric singleton congestion games in which the strategy space of each job is the set of all machines. For the general setting, the players’ costs are defined as follows. Given a strategy profile , for every player , and resource , let , and define . The cost of a player is given by,
Notice that for general congestion games, we assume that players’ costs are multiplied by their weight, whereas we do not make that assumption for scheduling games. This has no implications for the existence of Nash equilibria, but only affects the efficiency result.
Each job chooses a strategy so as to minimize its cost. A strategy profile is a pure Nash equilibrium (NE) if for all and all , we have . Let denote the set of Nash equilibria for a given instance . Notice that may be empty.
For a profile , let denote the cost of . The cost is defined with respect to some objective. For example, the total players’ cost or the maximum cost of a player. It is well known that decentralized decision-making may lead to sub-optimal solutions from the point of view of the society as a whole. For a game , let be the set of feasible profiles of . We denote by the cost of a social optimal (SO) solution; i.e., . We quantify the inefficiency incurred due to self-interested behavior according to the price of anarchy (PoA)  and price of stability (PoS)  measures. The PoA is the worst-case inefficiency of a pure Nash equilibrium, while the PoS measures the best-case inefficiency of a pure Nash equilibrium.
Let be a family of games, and let be a game in . Let be the set of pure Nash equilibria of the game . Assume that .
The price of anarchy of is the ratio between the maximal cost of a NE and the social optimum of . That is, . The price of anarchy of the family of games is .
The price of stability of is the ratio between the minimal cost of a NE and the social optimum of . That is, . The price of stability of the family of games is .
1.2 Our Contribution
We start by studying scheduling games, i.e., each job has to choose one machine to be processed on, and then based on the choices of the jobs, each machine schedules the jobs according to its individual priority list. We first show that a Nash equilibrium in general need not exist, and use this example to show that it is NP-complete to decide whether a particular game has a Nash equilibrium. We then extend known results in order to provide a characterization of instances in which a pure Nash equilibrium is guaranteed to exist. Specifically, existence is guaranteed if the game belongs to at least one of the following four classes: all jobs have the same weight, there are two machines, all machines have the same processing delay (shown in ), and all machines have the same priority list (shown in ). For all four of these classes, there is a polynomial time algorithm that computes a Nash equilibrium. In fact, if jobs are unweighted, better-response dynamics converge to a Nash equilibrium in polynomial time. This characterization is tight in a sense that our inexistence example disobeys it in a minimal way: it describes a game on three machines, two of them having the same processing delay and the same priority list.
We analyze the efficiency of Nash equilibria by means of two different measures of efficiency: the makespan, i.e., the maximum completion time of a job, and the sum of completion times. For all four classes of games with a guaranteed Nash equilibrium we provide tight bounds for the price of anarchy and the price of stability with respect to both measures. Our results are summarized in Table 1.
|Instance class Objective||Makespan||Sum of Completion Times|
|Global priority list|
If jobs are unweighted, we show that the price of anarchy is equal to , which means that selfish behavior is optimal. For two machines with processing delays and , we prove that the PoA and the PoS are at most if , and if . Our analysis is tight for all . The maximal inefficiency, listed in Table 1, is achieved with . In case the sum of completion times is considered as an objective, the price of anarchy can grow linearly in the number of jobs. If machines have identical speeds, but potentially different priority lists, the price of anarchy with respect to the makespan is equal to . The upper bound follows because every Nash equilibrium can be seen as an outcome of Graham’s List-Scheduling algorithm. The lower bound example shows the bound is tight, even with respect to the price of stability. For the sum of completion times objective, we show that the price of anarchy is at most , and provide a lower bound example for which the price of stability grows in the order of . If there is a global priority list, and machines have arbitrary processing delays, we show that the -approximation of List-Scheduling carry over for the makespan inefficiency, and the results of carry over for the sum of completion times.
In terms of computational complexity, we prove that it is NP-hard to approximate the best NE within a factor of for all if machines have identical processing delays and the minimum makespan objective is considered. Recall that is the price of anarchy for these instances. In particular, this implies that the simple greedy algorithm that computes a Nash equilibrium by letting each machine with the current least load get the most preferred unassigned job, is the best one can hope for.
We finally generalize the model to allow for arbitrary strategy sets. We show that in general, even with unweighted jobs, a Nash equilibrium need not exist by making use of the famous Condorcet paradox . We then use this example to prove that the question whether a Nash equilibrium exists is NP-hard, even with unweighted jobs. We lastly study the price of anarchy with respect to the sum of weighted costs and show that the upper bound of proven by Cole et al.  for unrelated machine scheduling with Smith’s rule also extends to congestion games with resource-dependent priority lists. This ratio is smaller than the price of anarchy of the atomic game with priorities defined by Farzad et al. .
1.3 Related Work
Scheduling games. The existence and efficiency of Nash equilibria in scheduling games gained lots of attention over the recent years. We refer to Vöcking  for a recent overview. For existence, Immorlica et al  proved that for unrelated machines, i.e., different machines can have different processing times for jobs, and priority lists based on shortest processing time first with consistent tie-breaking, the set a Nash equilibria is always non-empty and corresponds to the set of solutions of the Ibarra-Kim algorithm. This result is also proven in Heydenreich et al. . A Nash equilibrium is in fact reached after rounds of best-responses. It remains open whether a Nash equilibrium exists for unrelated machines and priority lists based on either longest processing time first with consistent tie-breaking or a random order. Dürr and Nguyen  proved that a potential and thus a Nash equilibrium exists for two machines with a random order and balanced jobs, whereas no potential exists for three or more machines. Azar et al.  showed that for unrelated machines with priorities based on the ratio of a job’s processing time to its faster processing time, there need not be a Nash equilibrium. Lu and Yu  proposed the so-called group-makespan mechanism that guarantees the existence of a Nash equilibrium.
The standard measure for the efficiency of Nash equilibria is the price of anarchy . This measure has been widely studied for different measures of efficiency. Most attention has been addressed on minimizing the makespan. Czumaj and Vöcking  gave tight bounds for related machines that grow as the number of machines grows, whereas Awerbuch et al.  and Gairing et al.  provided tight bounds for restricted machine settings. An alternative measure of efficiency is utilitarian social welfare, that is, the sum of weighted completion times. Correa and Queyranne  proved a tight upper bound of 4 for restricted related machines with priority lists derived from Smith’s rule. Cole et al.  generalized the bound of 4 to unrelated machines with Smith’s rule. Hoeksma and Uetz  gave a tighter bound for the more restricted setting in which jobs have unit weights and machines are related.
Christodoulou et al. 
introduced the idea of using coordination mechanisms, i.e., local functions that specify the costs of the jobs as a function of the vector of weights, to improve the performance of these games. Immorlica et al. studied coordination mechanisms based on shortest and longest processing times first. Kollias  considered non-preemptive coordination mechanism and shows that preemption is necessary to improve lower bounds on the price of anarchy lower bounds. In fact, Caragiannis et al.  proposed a framework that uses price of anarchy results of Nash equilibria in scheduling games to come up with combinatorial approximation algorithms for the centralized problem.
Congestion games. Rosenthal  proved that congestion games are potential games and thus have a pure Nash equilibrium. Awerbuch et al.  and Christodoulou and Koutsoupias  proved a tight bound of for affine congestion games, and 2.618 for weighted affine congestion games. Fotakis et al.  showed that weighted congestion games need not have a pure Nash equilibrium, unless we restrict cost functions to be linear. Ackermann et al.  proved that as long as strategy spaces are restricted to be matroidal, then a Nash equilibrium always exists.
Congestion games with priorities. Rosenthal  proved that congestion games are potential games and thus have a pure Nash equilibrium. Ackermann et al.  were the first to study a congestion game with priorities. They proposed a model in which users with higher priority on a resource displace users with lower priorities such that the latter incur infinite cost. Farzad et al.  studied priority based selfish routing for non-atomic and atomic users and analyzed the inefficiency of equilibrium. Gourvès et al.  studied capacitated congestion games to characterize the existence of pure Nash equilibria and computation of an equilibrium when they exist. Piliouras et al.  assumed that the priority lists are unknown to the players a priori and consider different risk attitudes towards having a uniform at random ordering.
2 Equilibrium Existence and Computation
In this section we give a precise characterization of instances that are guaranteed to have a NE. The conditions that we provide are sufficient but not necessary. A natural question is to decide whether a given game instance that does not fulfill any of the conditions has a NE. We show that answering this question is an NP-complete problem.
We first show that a NE may not exist, even with only three machines, two of which have the same delay and the same priority list.
Consider the game with jobs, , and three machines, , with , and . The first machine has delay while the two other machines have delay . The job weights are , and , where but small.
Job is clearly on in every NE. It is easy to see that in every NE at least one of and is on . Therefore, job is first on or . Since these two machines have the same priority list and the same delay function, we can assume w.l.o.g., that if a NE exists, then there exists a NE in which job is on . We show that no NE exists by considering the three possible strategies of job .
is on : If is not on or , then prefers to . If is on , then is on (since ). As a result, prefers (since ), so prefers . Finally, given that is on , is not on .
is on : job prefers , where it completes at time , while after on it completes at time . Now prefers , (since ). So prefers .
is on : Being after , job prefers .
Thus, the game has no pure Nash equilibrium.
We can use the above example to show that deciding whether a game instance has a NE is NP-complete by using a reduction from 3-bounded 3-dimensional matching.
Given an instance of a scheduling game, it is NP-complete to decide whether the game has a NE.
Proof: Given a game and a profile , verifying whether is a NE can be done by checking for every job whether its current assignment is also its best-response, therefore the problem is in NP.
The hardness proof is by a reduction from -bounded -dimensional matching (DM-). The input to the DM- problem is a set of triplets , where . The number of occurrences of every element of in is at most . The number of triplets is . The goal is to decide whether has a -dimensional matching of size , i.e., there exists a subset , such that , and every element in appears exactly once in . DM- is known to be NP-hard .
Given an instance of DM- matching and , we construct the following scheduling game. The set of jobs consists of:
The jobs from the game in Example 2.1.
A single dummy job, , of weight .
A set of dummy jobs of weight .
A set of dummy jobs of weight .
jobs of weight - one for each element in .
There are machines, . All the machines except for and have delay . For and , .
The heart of the reduction lies in determining the priority lists. The first three machines will mimic the no-NE game from Example 2.1. Note that if job is missing from then there exists a NE of on . The idea is that if a DM- matching exists, then job would prefer and leave the first three machines for . However, if there is no DM-, then some job originated from the elements in will precede job on , and ’s best-response would be on or - where it is guaranteed to have completion time , and the no-NE game would come to life. The dummy jobs in are long enough to guarantee that each of the jobs prefers the first three machines over the last machines.
The priority lists are defined as follows. When the list includes a set, it means that the set elements appear in arbitrary order. For the first machine, . For the second and third machines, . For the forth machine, . The remaining machines are triplet-machines. For every , the priority list of the triplet-machine corresponding to is .
Observe that in any NE, the dummy job of weight is assigned as the first jobs on . Also, the dummy jobs in have the highest priority on the triplet-machines, thus, in every NE, there are triplet-machines on which the first job is from . Finally, it is easy to see that in every NE there is exactly one dummy job from on each of the last machines.
Figure 1 provides an example for and .
If a D-matching of size exists, then the game has a NE.
Proof: Let be a matching of size . Assign the jobs of on the triplet-machines corresponding to and the jobs of on the remaining triplet-machines. Assign and on . Also, assign a single job from on all but the first machines. We are left with the jobs that are assigned on the first three machines: and on , on and on . It is easy to verify that the resulting assignment is a NE. The crucial observation is that all the jobs originated from completes at time at most , and therefore have no incentive to select . Thus, job completes at time on and therefore, has no incentive to join the no-NE game on the first three machines.
If a D-matching of size does not exist, then the game has no NE.
Proof: Since a matching does not exist, at least one job from , is not assigned on its triplet machine, and thus prefers , where its completion time is . Thus, job prefers to be first on or , where its completion time is . The long dummy jobs guarantee that machines and attracts exactly the jobs and the no-NE game is played on the first three machines.
By combining the two claims, the reduction proof is complete.
Our next results are positive. When combined with known results regarding equilibrium existence, and our example above, we get a tight characterization of classes of instances with a guaranteed Nash equilibrium.
The following algorithm is intended for instances in the class , that is, for all , . It assigns the jobs greedily, where in each step, a job is added on a machine on which the cost of the next job is minimal.
If for all jobs , then Algorithm 1 calculates a NE.
Proof: Let be the schedule produced by Algorithm 1. We show that is a NE. Note that the jobs are assigned one after the other according to their completion time in . That is, if is assigned before then . Assume by contradiction that is not a NE, and let be a job that can migrate from its current machine to machine and reduce its completion time. Assume that if it migrates, then would be assigned as the -th job on machine . This contradicts the choice of the algorithm when the -th job on machine is assigned - since should have been selected. If no job is -th on machine , then we get a contradiction to the assignment of .
In fact, for the unweighted case, every sequence of better responses converges in polynomial time. Given a strategy profile , a strategy for job is a better response if .
If for all jobs , then jobs reach an equilibrium after polynomially many better response moves.
Proof: We analyze a potential function that is introduced by Gourvès et al. . Define the rank of job on machine by . Let be a strategy profile. For all , we define , we write , where is the job with the highest priority choosing machine and is the job with the lowest priority choosing machine (in particular, there are jobs choosing machine ), and we define for all . With each machine , we associate a multiset of elements , …, . We define .
For a value , define the multiset for a given strategy profile . Moreover, define . By the definition, is the number of elements corresponding to profile with costs less than and is the total sum of ranks of these elements. We define a partial order on strategy profiles. Two strategy profile satisfy if for the smallest such that , we have , or and . In other words, a strategy profile is smaller than strategy profile if for the smallest for which , we have that there are more elements in with costs less than , or there are an equal number of elements with costs less than , but then the sum of ranks is smaller.
Let and . Suppose that is a better response for given . We show that . We can restrict attention to machines and . Observe that the costs of the smallest element that is different between and are equal to . But then two cases could occur: either there is no lower priority job on and then an element with cost is added which is smaller than , or there is a lower priority job on , but then job has a better ranking than that job and thus we have in both cases that .
Now we bound the number of strategy changes to reach a Nash equilibrium from arbitrary profile in better responses. Let be an arbitrary strategy profile. By definition of order , there are at most values of that we have to consider. Moreover, for each , and . Hence, there are at most elements which are -different. Thus, from an arbitrary profile, better responses converges to a Nash equilibrium in at most strategy changes.
Our next result considers the number of machines and completes the picture. Since our inexistence example uses three machines, out of which two are identical (in both delay and priority list), we cannot hope for a wider positive result.
If , then a NE exists and can be calculated efficiently.
Proof: For a single machine, the priority list defines the only feasible schedule, which is clearly a NE. For , assume w.l.o.g., that and . Consider the following algorithm, which initially assigns all the jobs on the fast machine. Then, the jobs are considered according to their order in , and every job gets an opportunity to migrate to .
Denote by the schedule after the first step of the algorithm (where all the jobs are on ), and let denote the schedule after the algorithm terminates. We show that is a NE.
No job for which has a beneficial migration.
Proof: Assume by contradiction that job is assigned on and has a beneficial migration. Assume that . Job was offered to perform a migration in the -th iteration of step 2 of the algorithm, but chose to remain on . The only migrations that took place after the -th iteration are from to . Thus, if migrating is beneficial for after the algorithm completes, it should have been beneficial also during the algorithm, contradicting its choice to remain on .
No job for which has a beneficial migration.
Proof: Assume by contradiction that the claim is false and let be the first job on (first with respect to ) that may benefit from returning to . Let denote the schedule before job migrates to - during the second step of the algorithm. Recall that is the completion time of job on , and is its completion time on before its migration.
Since the jobs are activated according to in the -nd step of the algorithm, no jobs are added before job on . Job may be interested in returning to only if some jobs that were processed before it on , move to after its migration. Denote by the set of these jobs, and let be their total weight. Let be the last job from to complete its processing in . Since job performs its migration out of after job , and jobs do not join during step 2 of the algorithm, the completion time of when it performs the migration is at most . The migration from to is beneficial for , thus, .
The jobs in are all before job in and after job in . Therefore, , and . Finally, we assume that is not stable and would like to return to . By returning, its completion time would be . Given that the migration is beneficial for , and that is the first job who likes to return to , we have that .
Combining the above inequalities, we get
This contradicts the fact that and .
By combining the two claims, we conclude that is a NE.
3 Equilibrium Inefficiency
Two common measures for evaluating the quality of a schedule are the makespan, given by , and the sum of completion times, given by . In this section we analyze the equilibrium inefficiency with respect to each of the two objectives, for each of the four classes for which a NE is guaranteed to exist.
We begin with , the class of instances with unweighted jobs. For this class we show that allowing arbitrary priority lists does not hurt the social cost, even on machines with different speeds.
PoA PoS for both the min-makespan and the sum of completion times objective.
Proof: Let be a schedule of unit-weight jobs. The quality of is characterized by the vector specifying the number of jobs on each machine. The makespan of is given by , and the sum of completion times in is .
Recall Algoirthm 1 that assigns the jobs greedily, where on each step a job is added on a machine on which the cost of the next job is minimized. Let denote the resulting schedule, and let be the sorted vector of jobs’ costs (that is, completion times) in . By the algorithm, this is also the order according to which the jobs are assigned. We show that every NE schedule has the same sorted cost vector as , and that this vector achieves both minimum makespan and minimum sum of completion times.
Let be a NE schedule with sorted cost vector and let , and assume by contradiction that it has a different cost vector. Let be the minimal index such that . Since and agree on the costs of the first jobs, and since assigns the -th job on a minimal-cost machine, it holds that . We get a contradiction to the stability of - since some job can reduce its cost to .
The following Lemma, whose proof is given in the Appendix, completes the proof.
The schedule is optimal with respect to makespan and sum of completion times objectives.
Proof: We show that achieves the minimum makespan. Assume that there exists a schedule such that . Let . It must be that . Since , there must be a machine such that . Thus, the last job on machine in can benefit from migrating to machine , as its cost will be at most . This contradicts the assumption that is a NE.
Next, we analyze the sum of completion times objective. For a schedule , the sum of completion times is . Using similar arguments, if is not optimal with respect to the sum of completion times, there exists a beneficial migration from a machine whose contribution to the sum is maximal, to a machine with a lower contribution.
In Theorem 2.6 it is shown that a NE exists for any instance on two related machines. We now analyze the equilibrium inefficiency of this class. Let denote the class of games played on two machines with delays and .
For the min-makespan objective, PoAPoS if , and PoAPoS if .
Proof: Let . Let be the total weight of all jobs. Assume first that . For the minimum makespan objective, . Also, for any NE , we have that , since every job can migrate to be last on the fast machine and have completion time at most . Thus, PoA .
Assume next that . Let job be a last job to complete in a worst Nash equilibrium , be the total weight of all jobs different from on machine , and be the total weight of all jobs different from on machine in . Then since is a Nash equilibrium, and . Combining these two inequalities yields
where for the inequality we use that and , and thus PoA.
For the PoS lower bound, assume first that . Consider an instance consisting of two jobs, and , where and . The priority lists are . The unique NE is that both jobs are on the fast machine. . For every , it holds that , therefore, job does not have a beneficial migration. An optimal schedule assigns job on the slow machine, and both jobs complete at time . The corresponding PoS is . 222For , by taking , the PoS approaches as .
Assume now that . Consider an instance consisting of three jobs, , and , where , and . The priority lists are . Note that for every . The unique NE is when jobs and are on the fast machine, and job on the slow machine. Indeed, job prefers being alone on the slow machine since . Job prefers joining on the fast machine since . The makespan is . In an optimal schedule, job is alone on the fast machine, and jobs and are on the slow machine. Both machines have the same completion time . The PoS is .
For the sum of completion times objective, PoA and PoS for all .
Proof: For the upper bound, note that and in every NE schedule , . This implies PoA . For the PoS lower bound, consider an instance consisting of a set of jobs of weight , and two jobs, and , where and . The priority lists are . Note that , therefore, in every NE, job is first on and job is first on . Thus, every -job has completion time at least . The total completion time is at least . An optimal schedule assigns and on and all the -jobs on . The total completion time is at most . For small enough , we get that the PoS is .
We turn to analyze the equilibrium inefficiency of the class , consisting of games played on identical-speed machines, having machine-based priority lists. The proof of the following theorem is based on the observation that every NE schedule is a possible outcome of Graham’s List-scheduling (LS) algorithm .
For the min-makespan objective, PoAPoS .
Proof: Let be a NE schedule. We claim that is a possible outcome of Graham’s List-scheduling algorithm . Indeed, assume that List-scheduling is performed and the jobs are considered according to their start time in . Every job selects its machine in , as otherwise, we get a contradiction to the stability of . Since List-scheduling provides a approximation to the makespan, we get the upper bound of the PoA.
For the lower bound, given , the following is an instance for which PoS. The instance consists of a single job of weight and unit-weight jobs. In all priority lists, the heavy job is last and the unit-weight jobs are prioritized arbitrarily. It is easy to verify that in every NE profile the unit-weight jobs are partitioned in a balanced way among the machines, and the heavy job is assigned as last on one of the machines. Thus, the completion time of the heavy job is . On the other hand, an optimal assignment assign the heavy job on a dedicated machine, and partition the unit-weight job in a balanced way among the remaining machines. In this profile, all the machines have load . The corresponding PoS is .
For the sum of completion times objective, PoA, and for every , PoS.
Proof: For the upper bound of the PoA, note that, independent of the number of machines, the total completion time is at least . Also, for every job , if is not assigned on any machine, then there exists a machine with load at most , therefore, in every NE profile, the completion time of job is at most . Summing this equation for all the jobs, we get that the total completion time of any NE is at most
We conclude that the PoA is at most .
For the PoS lower bound, given , let , and consider an instance with jobs, out of which, jobs have length and the other jobs have length . Assume that gives the highest priority to then to all the -jobs, and then to the other unit jobs.
In every NE, machine processes first the unit-job , followed by -jobs. Thus, every job has completion time at least . The sum of completion times is . On the other hand, an optimal solution assigns on machine a set of -jobs followed by one unit-job for , resulting in total completion time . The PoS tends to as decreases.
The last class of instances for which a NE is guaranteed to exist includes games with a global priority list, and is denoted by . It is easy to verify that for this class, the only NE profiles are those produced by List-Scheduling algorithm, where the jobs are considered according to their order in the priority list. Different NE may be produced by different tie-breaking rules. Thus, the equilibrium inefficiency is identical to the approximation ratio of LS . Since the analysis of LS is tight, this is also the PoS.
For the min-makespan objective, PoSPoA.
For the sum of completion times objective, we note that the proof of Theorem 3.4 for two related machines uses a global priority list. The analysis of the PoA is independent of the number and delays of machines.
For the sum of completion times objective, PoA and PoS.
3.1 Hardness of Approximating the Minimum Makespan NE
Correa and Queyranne  showed that if all the machines have the same speeds, but arbitrary priority lists, then a NE is guaranteed to exist, and can be calculated by a simple greedy algorithm. In Theorem 3.5, we have shown that the PoA is at most . In this subsection, we show that we cannot hope for a better algorithm than the simple greedy algorithm. More formally, we prove that it is NP-hard to approximate the best NE within a factor of for all .
If for all machines , then it is NP-hard to approximate the best NE w.r.t. the makespan objective within a factor of for all .
Proof: We show that for every , there is an instance on identical machines for which it is NP-hard to distinguish whether the game has a NE profile with makespan at most or at least .
The hardness proof is by a reduction from -bounded -dimensional matching (DM-), defined in the proof of Theorem 2.1.
Given an instance of DM- and , consider the following game on machines, . The set of jobs includes job of weight , job of weight , a set of dummy jobs of weight , two dummy jobs of weight , a set of unit-weight jobs, and jobs of weight - one for each element in .
We turn to describe the priority lists. When the list includes a set, it means that the set elements appear in arbitrary order. For the first machine, . For the second machine, . The right machines are triplet-machines. For every , the priority list of the triplet-machine corresponding to is