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Coordination mechanism for selfish scheduling under a grade of service provision
Information Processing Letters 113 (2013) 251–254
Contents lists available at SciVerse ScienceDirect
Information Processing Letters www.elsevier.com/locate/ipl
Coordination mechanism for selfish scheduling under a grade of service provision ✩ Li Guan, Jianping Li ∗ Department of Mathematics, Yunnan University, Kunming 650091, PR China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 11 November 2011 Received in revised form 17 January 2013 Accepted 18 January 2013 Available online 23 January 2013 Communicated by J. Xu
In this paper, we study the problem of selfish scheduling game under a grade of service provision, where all machines and all jobs are labeled with the different grade of service (GoS) levels such that a job J can be assigned to execute on machine M only when the GoS level of machine M is not higher than the GoS level of job J . We consider two coordination mechanisms for this selfish scheduling game: the makespan policy and the LG-LPT policy. For the first mechanism, we show that the price of anarchy is exactly 32 for two machines
Keywords: Scheduling Selfish scheduling Grade of service Coordination mechanism Makespan Price of anarchy
and Θ( log log m ) for m ( 3) machines, respectively. For the second mechanism, we point
log m
1 out that the price of anarchy is 54 for two machines and 2 − m− for m ( 3) machines, 1 respectively, and we finally analyze the convergence to a Nash equilibrium of the induced game. © 2013 Elsevier B.V. All rights reserved.
1. Introduction The game theory is to study some situations concerning selfish agents who are interested in achieving their individual goals, as well as in opposing to obtain a global optimum. The agents act selfishly until reaching some equilibria. In general, the social optimum is not typically obtained. Quantifying the efficiency loss due to selfish behavior is an important research interest in such settings. The price of anarchy (POA, for short), which was first proposed by Koutsoupias and Papadimitriou [8], is the most popular measure to quantify the inefficiency of equilibrium. Precisely, the price of anarchy of a game is defined as the ratio between the worst objective function value of an equilibrium of the game and the objective function
✩ Supported by the National Natural Science Foundation of China [Nos. 10861012, 61063011], the Project of the First 100 High-level Overseas Talents of Yunnan Province IRTSTYN. Corresponding author. Tel.: +86 871 65033701; fax: +86 871 65033700. E-mail addresses: [email protected] (L. Guan), [email protected] (J. Li).
*
0020-0190/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.ipl.2013.01.014
value of an optimal outcome. We only consider the pure Nash equilibrium in this paper. How can we reduce the inefficiency of equilibrium in a game? An important approach is defined as a coordination mechanism, which is a local policy that assigns an outcome to each strategy s, where the outcome of such a strategy s is a function of the agents who have chosen the strategy s. The purpose is to lead the independent and selfish choices of the agents to obtain a better result in a socially desired outcome. Obviously, the primary goal of a coordination mechanism for the designer is to guarantee the existence of pure Nash equilibria for the induced game. The selfish scheduling game problem has been studied extensively in the literatures [1–4,7], which is defined as follows. There are n jobs owned by some independent agents, say J 1 , J 2 , . . . , J n , and m machines, say M 1 , M 2 , . . . , M m , and some processing times p i j , where p i j indicates the processing time for the job J i to be executed on the machine M j , each agent may select a machine to minimize his own completion time. The social objective is to minimize the makespan, i.e., the maximum completion time of m machines. A pure Nash equilibrium is an assignment of these n jobs to be executed on m machines such that no job has an unilateral incentive to switch to
252
L. Guan, J. Li / Information Processing Letters 113 (2013) 251–254
another machine. A coordination mechanism for this game is a set of scheduling policies, one for each machine, that determines the way to schedule the jobs on that machine. The price of anarchy of a selfish scheduling game is the ratio between the maximum makespan in all pure Nash equilibria and the minimum makespan in all available assignments. In service industry, the service providers often provide differentiated services to the special customers who are more valued than the regular customers. One simple scheme for providing differentiated services is to label all machines and all jobs with the different grade of service (GoS, for short) levels. A job J can be assigned to execute on a machine M only when the GoS level of machine M is not higher than the GoS level of job J . Thus, when we label relatively higher GoS levels on the jobs of more valued customers, we can provide a better service to them. The model of a selfish scheduling game under a GoS provision is defined as follows. There are a set of m machines, say M = { M 1 , M 2 , . . . , M m }, and a set of n jobs, say J = { J 1 , J 2 , . . . , J n }. Each job J j owned by a selfish agent of the game has a processing time p ( J j ), and it is labeled by a GoS level g ( J j ). Each machine M i is also labeled by a GoS level g ( M i ). The job J j is allowed to be executed on the machine M i only when g ( J j ) g ( M i ). We may assume that all jobs can be scheduled on these m machines. In this paper, we study the selfish scheduling game under a GoS provision, and we obtain main two results: (1) For the makespan policy, we prove that the price of anlog m archy is exactly 32 for two machines and Θ( log log m ) for m ( 3) machines, respectively; (2) For the LG-LPT policy, we point out that the price of anarchy is 54 for two machines 1 and 2 − m− for m ( 3) machines, respectively, and we fi1 nally analyze the convergence to a Nash equilibrium of the induced game. This paper is organized as follows. In Section 2, we consider the makespan policy and then obtain the results in (1); In Section 3, we consider the LG-LPT policy and then obtain the results in (2).
2. Makespan policy In the makespan policy, each machine processes the jobs assigned in parallel. If a job J j is assigned to a machine M i , the completion time of job J j is equal to the completion time of machine M i . For any makespan policy, the pure Nash equilibrium of the induced selfish scheduling game always exists [7]. This fact is also true for the selfish scheduling game under a GoS provision since the latter is a special version of selfish scheduling game. We first consider POA for two machines, and we obtain the first result as follows. Theorem 1. For two machines, the price of anarchy of the makespan policy is exactly 32 for the selfish scheduling game under a grade of service provision. Proof. If the GoS levels of two machines are same or the GoS levels of all jobs are same, our scheduling problem becomes the ordinary identical parallel machine scheduling
problem P C max [5]. From the earlier work [7], we know the fact that the price of anarchy of the makespan policy for P C max is 43 for these two machines, which is less than 32 . Now we consider the state that the two machines are labeled with two different GoS levels meanwhile all jobs are labeled with two different GoS levels, respectively. The set of two machines is M = { M 1 , M 2 }, and for convenience, we may assume that g ( M 1 ) = 1, , . . . , J n }. Let g ( M 2 ) = 2. The set of n jobs is J = { J 1 , J 2 Ji = { J j | g ( J j ) = i , 1 j n} and P i = J j ∈J i p ( J j ) , where i = 1, 2. By distinguishing the following two cases, we shall prove that POA is at most 32 .
Case 1. P 1 > P 2 In this case, the induced game has only one Nash equilibrium, where all jobs in J1 are assigned to execute on M 1 and all jobs in J2 are assigned to execute on M 2 . This equilibrium schedule is also an optimum schedule. Thus, POA is equal to 1. Case 2. P 1 P 2 Let OPT denote the optimal value, i.e., the minimum makespan. We have
OPT max
max p ( J j ), P 1 ,
1 j n
P1 + P2 2
.
Let μ be any Nash equilibrium and li the completion time of the machine M i in μ, where i = 1, 2. Let cost(μ) be the makespan of μ, i.e., cost(μ) = max{l1 , l2 }. We consider the two subcases. (1) cost(μ) = l1 In this subcase, without loss of generality, we may assume that there is at least one job in J2 assigned to execute on machine M 1 , otherwise we obtain cost(μ) = OPT and the conclusion follows trivially. Let J ∗ be any job in J2 assigned to execute on machine M 1 . Suppose that if the job J ∗ changes its strategy, by moving from machine M 1 to machine M 2 , then the completion time of J ∗ should not be decreased, by the fact that μ is a Nash equilibrium. It follows l1 l2 + p ( J ∗ ). Thus, we obtain
2 cost(μ) l1 + l2 + p J ∗ = P 1 + P 2 + p J ∗ 3OPT which implies cost(μ) 32 OPT. (2) cost(μ) = l2 In this subcase, let J ∗ be any job assigned to M 2 . Obviously, we have g ( J ∗ ) = 2. Since μ is a Nash equilibrium, it follows l2 l1 + p ( J ∗ ). Thus, we get
2 cost(μ) l2 + l1 + p J ∗ = P 1 + P 2 + p J ∗ 3OPT which also implies cost(μ) 32 OPT.
To sum up, we prove that POA is at most 32 for two machines. The following example shows that this upper bound 32 on POA is tight. Suppose that we have M = { M 1 , M 2 } and J = { J 1 , J 2 , J 3 }, where g ( M 1 ) = 1, g ( M 2 ) = 2, g ( J 1 ) = 1, g ( J 2 ) = g ( J 3 ) = 2, p ( J 1 ) = p ( J 2 ) = 1 and p ( J 3 ) = 2. An optimal schedule assigns two jobs J 1 and J 2 to be executed
L. Guan, J. Li / Information Processing Letters 113 (2013) 251–254
on machine M 1 , job J 3 to be executed on machine M 2 , respectively, then we obtain OPT = 2. Now, we consider a schedule μ that assigns two jobs J 1 and J 3 to be executed on machine M 1 , job J 2 to be executed on machine M 2 , respectively. Obviously, μ is a Nash equilibrium with cost(μ) = 3. Observe the fact that any schedule to yield a larger makespan than 3 is not a Nash equilibrium, then we obtain POA = 32 . Thus, it shows that the price of anarchy of the induced game for two machines is exactly 32 . 2 In the reminder of this section, we present a tight log m bound Θ( log log m ) on the price of anarchy for m ( 3) machines. For the convenience, the Gamma function (n) is defined by (n) = n · (n − 1) · · · 2 · 1 (= n!) for any positive integer n. We shall use the fact that −1 (n) = log n (1 + o(1)). log log n Theorem 2. For m ( 3) machines, the price of anarchy of the makespan policy for the selfish scheduling game under a grade log m of service provision is Θ( log log m ). Proof. The scheduling problem under a GoS provision is actually a special version of the scheduling problem for the restricted assignment in [5], denoted by B C max , where each job can be scheduled on a subset of identical machines. Then POA for the scheduling game under a GoS provision is no more than POA for B C max . Gairing et al. [4] pointed out that POA of the makespan policy for B C max log m is Θ( log log m ). Thus, we have the upper bound of POA as log m
O ( log log m ). Now, we present the following example to show the log m tightness of this bound O ( log log m ). The machine set is M = { M 1 , M 2 , . . . , M m } and the job set is J = { J 1 , J 2 , . . . , J n }. Let Ji = { J j | g ( J j ) = i , 1 j n} and Mi = { M j | g ( M j ) = i , 1 j m}. We may recursively choose |M1 | = 1, |Mi | = (k − i + 2)|Mi −1 | for i = 2, 3 . . . , k + 1, and |J1 | = 0, |Ji | = |Mi | for i = 2, 3, . . . , k + 1, respectively. All jobs have a unit processing time. The schedule μ is defined as follows. We assign k − i + 2 jobs in Ji to execute on each machine in Mi −1 for i = 2, 3, . . . , k + 1. The machines in Mk+1 remain empty. Obviously, μ is a pure Nash equilibrium. The makespan of μ is k. Meanwhile the optimal schedule assigns each job in Ji to be executed on each machine in Mi for i = 2, 3, . . . , k + 1, we obtain a makespan with minimum value 1. Thus, the price of anarchy is at least k. Thus, we obtain
m = |M1 | + |M2 | + · · · + |Mk+1 | (k + 1)! = (k + 2) implying k −1 (m) − 2. This establishes the conclusion. 2 The example in Theorem 2 shows that, when the processing time of all jobs are 1, the price of anarchy for m ( 3) machines is still Θ( logloglogmm ).
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3. LG-LPT policy Instead of an arbitrary ordering, we consider the ordering of jobs where a job J j precedes a job J k when either g ( J j ) < g ( J k ), or g ( J j ) = g ( J k ) and p ( J j ) p ( J k ). We denote such a precedence by J j J k , and we call it LG-LPT order. Hwang et al. [6] designed an algorithm, denoted by LG-LPT, to solve the scheduling problem under a GoS provision. The strategy of the algorithm LG-LPT is as follows: (1) Sort the jobs in LG-LPT order; (2) Assign the first one job from the list of unassigned jobs to the least-loaded eligible machine until all jobs are scheduled. In the LG-LPT policy, the jobs assigned to a machine are executed in LG-LPT order, i.e., the job J j is executed before the job J k when J j J k holds. Theorem 3. The set of Nash equilibrium for the LG-LPT policy of the induced game is precisely the set of solutions that can be generated by the greedy algorithm LG-LPT. Proof. Consider a schedule μ produced by the algorithm LG-LPT [6]. On each machine, jobs are executed in LG-LPT order, i.e., the LG-LPT policy is effective. Suppose that it is beneficial for the job J i to switch from its current machine to another machine, let M j be the machine on which J i would be executed completely at first. Then the algorithm LG-LPT would schedule the job J i to be executed on machine M j by its greedy rule, which leads a contradiction. Thus, no job has an incentive to switch to another machine. It implies that μ is a Nash equilibrium. We prove the other direction by induction on the number of jobs. For n = 1, there is nothing to prove. Let μ be a Nash equilibrium scheduling with n + 1 jobs, and J i the job whose completion time is equal to the makespan of μ. When we remove the job J i from μ, by the fact that the job J i is last one, the remaining schedule, denoted by μ , is still a Nash equilibrium. According to the induction assumption, μ can be constructed by using the algorithm LG-LPT. Because μ is a Nash equilibrium, the job J i has no incentive to switch to another machine. Then the strategy of J i can be gained by the greedy algorithm LGLPT. It follows that μ can be interpreted as an output of the algorithm LG-LPT. 2 According to Theorem 3, the approximation factor of the greedy algorithm LG-LPT is the upper bound on the price of anarchy of the corresponding coordination mechanism. Lemma 1. (See [6].) For the scheduling problem under a grade of service provision, the greedy algorithm LG-LPT has the approx1 imation factor 54 for two machines and 2 − m− for m ( 3) 1 machines, respectively. And the greedy algorithm LG-LPT is tight. Now, we can obtain the following result as a consequence of Theorem 3 and Lemma 1. Corollary 1. The price of anarchy of the LG-LPT policy for the selfish scheduling game under a grade of service provision is 54
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for two machines and 2 − tively.
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for m ( 3) machines, respec-
An agent is called to be satisfied if this agent cannot reduce his completion time by unilaterally moving his job to another machine. The best-response action means that the agent moves his job to the machine on which he can get minimum completion time. Theorem 3 implies that the Nash equilibria exists for the induced game. We show that the selfish agents will converge automatically from any given initial schedule to a Nash equilibrium in reasonable times in the following theorem. Theorem 4. For the LG-LPT policy, selfish agents converge to a Nash equilibrium from any given initial schedule after n rounds of the best-response actions, and in the kth round, there are at most n − k + 1 agents who may take one best-response moving. Proof. We sort these n jobs in the LG-LPT order. Without loss of generality, we may assume that J 1 J 2 · · · J n . For any given initial schedule, each unsatisfied agent plays his best-response action regarding that other agents stay on their current machines. We call the procedure, that all unsatisfied agents play their own best-response actions, as a round. Now, we can prove by an induction that each job J i does not move in the kth round, where i = 1, 2, . . . , k − 1, and that if the job J k moves to some machine, it will remains there in the subsequent rounds. For the given strategies of J 1 , J 2 , . . . , J k−1 in the (k − 1)th round, we may suppose that M j is the machine on which the job J k achieves its minimum completion time. Then a best-response action for the job J k is to move to the machine M j in the kth round. Depending on the induction assumption, the jobs J 1 , J 2 , . . . , J k−1 do not change their strategies in the kth round. And for each i = 2, 3, . . . , n, because the job J i −1 precedes the job J i , then the jobs J 1 , J 2 , . . . , J k are always executed before the
jobs J k+1 , . . . , J n on any machine by the LG-LPT policy. Thus, we obtain the result: whenever the best-response actions are taken by J k+1 , . . . , J n in the kth round, the job J k is still satisfied at the end of the kth round. It means that the job J k cannot reduce its completion time by moving to another machine in the (k + 1)th round and it remains on the machine M j in the subsequent rounds. This establishes the theorem. 2 Acknowledgement The authors are grateful to the anonymous reviewer whose comments and suggestions have led to a substantially improved presentation for the paper. References [1] Y. Azar, K. Jain, V.S. Mirrokni, (Almost) optimal coordination mechanisms for unrelated machine scheduling, in: Proceedings of the 19th ACM-SIAM Symposium on Discrete Algorithms, 2008, pp. 323–332. [2] I. Caragiannis, Efficient coordination mechanisms for unrelated machine scheduling, in: Proceedings of the 20th Symposium on Discrete Algorithms, 2009, pp. 815–824. [3] G. Christodoulou, E. Koutsoupias, A. Nanavati, Coordination mechanisms, in: Proceedings of the 31st International Colloquium on Automata, Languages and Programming, in: Lecture Notes in Computer Science, vol. 3142, 2004, pp. 345–357. [4] M. Gairing, T. Lucking, M. Mavronicolas, B. Monien, Computing Nash equilibria for scheduling on restricted parallel links, in: Proceedings of the 36th ACM Symposium on Theory of Computing, 2004, pp. 613– 622. [5] R.L. Graham, E.L. Lawler, J.K. Lenstra, A.R. Kan, Optimization and approximation in deterministic sequencing and scheduling: a survey, Annals of Discrete Mathematics 5 (1979) 287–326. [6] Hark-Chin Hwang, Soo Y. Chang, Kangbok Lee, Parallel machine scheduling under a grade of service provision, Computers and Operations Research 31 (2004) 2055–2061. [7] N. Immorlica, L. Li, V.S. Mirrokni, A.S. Schulz, Coordination mechanisms for selfish scheduling, Theoretical Computer Science 410 (2009) 1589–1598. [8] E. Koutsoupias, C.H. Papadimitriou, Worst-case equilibria, in: Proceedings of the 16th Symposium on Theoretical Aspects of Computing Science, 1999, pp. 404–413.
Contents lists available at SciVerse ScienceDirect
Information Processing Letters www.elsevier.com/locate/ipl
Coordination mechanism for selfish scheduling under a grade of service provision ✩ Li Guan, Jianping Li ∗ Department of Mathematics, Yunnan University, Kunming 650091, PR China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 11 November 2011 Received in revised form 17 January 2013 Accepted 18 January 2013 Available online 23 January 2013 Communicated by J. Xu
In this paper, we study the problem of selfish scheduling game under a grade of service provision, where all machines and all jobs are labeled with the different grade of service (GoS) levels such that a job J can be assigned to execute on machine M only when the GoS level of machine M is not higher than the GoS level of job J . We consider two coordination mechanisms for this selfish scheduling game: the makespan policy and the LG-LPT policy. For the first mechanism, we show that the price of anarchy is exactly 32 for two machines
Keywords: Scheduling Selfish scheduling Grade of service Coordination mechanism Makespan Price of anarchy
and Θ( log log m ) for m ( 3) machines, respectively. For the second mechanism, we point
log m
1 out that the price of anarchy is 54 for two machines and 2 − m− for m ( 3) machines, 1 respectively, and we finally analyze the convergence to a Nash equilibrium of the induced game. © 2013 Elsevier B.V. All rights reserved.
1. Introduction The game theory is to study some situations concerning selfish agents who are interested in achieving their individual goals, as well as in opposing to obtain a global optimum. The agents act selfishly until reaching some equilibria. In general, the social optimum is not typically obtained. Quantifying the efficiency loss due to selfish behavior is an important research interest in such settings. The price of anarchy (POA, for short), which was first proposed by Koutsoupias and Papadimitriou [8], is the most popular measure to quantify the inefficiency of equilibrium. Precisely, the price of anarchy of a game is defined as the ratio between the worst objective function value of an equilibrium of the game and the objective function
✩ Supported by the National Natural Science Foundation of China [Nos. 10861012, 61063011], the Project of the First 100 High-level Overseas Talents of Yunnan Province IRTSTYN. Corresponding author. Tel.: +86 871 65033701; fax: +86 871 65033700. E-mail addresses: [email protected] (L. Guan), [email protected] (J. Li).
*
0020-0190/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.ipl.2013.01.014
value of an optimal outcome. We only consider the pure Nash equilibrium in this paper. How can we reduce the inefficiency of equilibrium in a game? An important approach is defined as a coordination mechanism, which is a local policy that assigns an outcome to each strategy s, where the outcome of such a strategy s is a function of the agents who have chosen the strategy s. The purpose is to lead the independent and selfish choices of the agents to obtain a better result in a socially desired outcome. Obviously, the primary goal of a coordination mechanism for the designer is to guarantee the existence of pure Nash equilibria for the induced game. The selfish scheduling game problem has been studied extensively in the literatures [1–4,7], which is defined as follows. There are n jobs owned by some independent agents, say J 1 , J 2 , . . . , J n , and m machines, say M 1 , M 2 , . . . , M m , and some processing times p i j , where p i j indicates the processing time for the job J i to be executed on the machine M j , each agent may select a machine to minimize his own completion time. The social objective is to minimize the makespan, i.e., the maximum completion time of m machines. A pure Nash equilibrium is an assignment of these n jobs to be executed on m machines such that no job has an unilateral incentive to switch to
252
L. Guan, J. Li / Information Processing Letters 113 (2013) 251–254
another machine. A coordination mechanism for this game is a set of scheduling policies, one for each machine, that determines the way to schedule the jobs on that machine. The price of anarchy of a selfish scheduling game is the ratio between the maximum makespan in all pure Nash equilibria and the minimum makespan in all available assignments. In service industry, the service providers often provide differentiated services to the special customers who are more valued than the regular customers. One simple scheme for providing differentiated services is to label all machines and all jobs with the different grade of service (GoS, for short) levels. A job J can be assigned to execute on a machine M only when the GoS level of machine M is not higher than the GoS level of job J . Thus, when we label relatively higher GoS levels on the jobs of more valued customers, we can provide a better service to them. The model of a selfish scheduling game under a GoS provision is defined as follows. There are a set of m machines, say M = { M 1 , M 2 , . . . , M m }, and a set of n jobs, say J = { J 1 , J 2 , . . . , J n }. Each job J j owned by a selfish agent of the game has a processing time p ( J j ), and it is labeled by a GoS level g ( J j ). Each machine M i is also labeled by a GoS level g ( M i ). The job J j is allowed to be executed on the machine M i only when g ( J j ) g ( M i ). We may assume that all jobs can be scheduled on these m machines. In this paper, we study the selfish scheduling game under a GoS provision, and we obtain main two results: (1) For the makespan policy, we prove that the price of anlog m archy is exactly 32 for two machines and Θ( log log m ) for m ( 3) machines, respectively; (2) For the LG-LPT policy, we point out that the price of anarchy is 54 for two machines 1 and 2 − m− for m ( 3) machines, respectively, and we fi1 nally analyze the convergence to a Nash equilibrium of the induced game. This paper is organized as follows. In Section 2, we consider the makespan policy and then obtain the results in (1); In Section 3, we consider the LG-LPT policy and then obtain the results in (2).
2. Makespan policy In the makespan policy, each machine processes the jobs assigned in parallel. If a job J j is assigned to a machine M i , the completion time of job J j is equal to the completion time of machine M i . For any makespan policy, the pure Nash equilibrium of the induced selfish scheduling game always exists [7]. This fact is also true for the selfish scheduling game under a GoS provision since the latter is a special version of selfish scheduling game. We first consider POA for two machines, and we obtain the first result as follows. Theorem 1. For two machines, the price of anarchy of the makespan policy is exactly 32 for the selfish scheduling game under a grade of service provision. Proof. If the GoS levels of two machines are same or the GoS levels of all jobs are same, our scheduling problem becomes the ordinary identical parallel machine scheduling
problem P C max [5]. From the earlier work [7], we know the fact that the price of anarchy of the makespan policy for P C max is 43 for these two machines, which is less than 32 . Now we consider the state that the two machines are labeled with two different GoS levels meanwhile all jobs are labeled with two different GoS levels, respectively. The set of two machines is M = { M 1 , M 2 }, and for convenience, we may assume that g ( M 1 ) = 1, , . . . , J n }. Let g ( M 2 ) = 2. The set of n jobs is J = { J 1 , J 2 Ji = { J j | g ( J j ) = i , 1 j n} and P i = J j ∈J i p ( J j ) , where i = 1, 2. By distinguishing the following two cases, we shall prove that POA is at most 32 .
Case 1. P 1 > P 2 In this case, the induced game has only one Nash equilibrium, where all jobs in J1 are assigned to execute on M 1 and all jobs in J2 are assigned to execute on M 2 . This equilibrium schedule is also an optimum schedule. Thus, POA is equal to 1. Case 2. P 1 P 2 Let OPT denote the optimal value, i.e., the minimum makespan. We have
OPT max
max p ( J j ), P 1 ,
1 j n
P1 + P2 2
.
Let μ be any Nash equilibrium and li the completion time of the machine M i in μ, where i = 1, 2. Let cost(μ) be the makespan of μ, i.e., cost(μ) = max{l1 , l2 }. We consider the two subcases. (1) cost(μ) = l1 In this subcase, without loss of generality, we may assume that there is at least one job in J2 assigned to execute on machine M 1 , otherwise we obtain cost(μ) = OPT and the conclusion follows trivially. Let J ∗ be any job in J2 assigned to execute on machine M 1 . Suppose that if the job J ∗ changes its strategy, by moving from machine M 1 to machine M 2 , then the completion time of J ∗ should not be decreased, by the fact that μ is a Nash equilibrium. It follows l1 l2 + p ( J ∗ ). Thus, we obtain
2 cost(μ) l1 + l2 + p J ∗ = P 1 + P 2 + p J ∗ 3OPT which implies cost(μ) 32 OPT. (2) cost(μ) = l2 In this subcase, let J ∗ be any job assigned to M 2 . Obviously, we have g ( J ∗ ) = 2. Since μ is a Nash equilibrium, it follows l2 l1 + p ( J ∗ ). Thus, we get
2 cost(μ) l2 + l1 + p J ∗ = P 1 + P 2 + p J ∗ 3OPT which also implies cost(μ) 32 OPT.
To sum up, we prove that POA is at most 32 for two machines. The following example shows that this upper bound 32 on POA is tight. Suppose that we have M = { M 1 , M 2 } and J = { J 1 , J 2 , J 3 }, where g ( M 1 ) = 1, g ( M 2 ) = 2, g ( J 1 ) = 1, g ( J 2 ) = g ( J 3 ) = 2, p ( J 1 ) = p ( J 2 ) = 1 and p ( J 3 ) = 2. An optimal schedule assigns two jobs J 1 and J 2 to be executed
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on machine M 1 , job J 3 to be executed on machine M 2 , respectively, then we obtain OPT = 2. Now, we consider a schedule μ that assigns two jobs J 1 and J 3 to be executed on machine M 1 , job J 2 to be executed on machine M 2 , respectively. Obviously, μ is a Nash equilibrium with cost(μ) = 3. Observe the fact that any schedule to yield a larger makespan than 3 is not a Nash equilibrium, then we obtain POA = 32 . Thus, it shows that the price of anarchy of the induced game for two machines is exactly 32 . 2 In the reminder of this section, we present a tight log m bound Θ( log log m ) on the price of anarchy for m ( 3) machines. For the convenience, the Gamma function (n) is defined by (n) = n · (n − 1) · · · 2 · 1 (= n!) for any positive integer n. We shall use the fact that −1 (n) = log n (1 + o(1)). log log n Theorem 2. For m ( 3) machines, the price of anarchy of the makespan policy for the selfish scheduling game under a grade log m of service provision is Θ( log log m ). Proof. The scheduling problem under a GoS provision is actually a special version of the scheduling problem for the restricted assignment in [5], denoted by B C max , where each job can be scheduled on a subset of identical machines. Then POA for the scheduling game under a GoS provision is no more than POA for B C max . Gairing et al. [4] pointed out that POA of the makespan policy for B C max log m is Θ( log log m ). Thus, we have the upper bound of POA as log m
O ( log log m ). Now, we present the following example to show the log m tightness of this bound O ( log log m ). The machine set is M = { M 1 , M 2 , . . . , M m } and the job set is J = { J 1 , J 2 , . . . , J n }. Let Ji = { J j | g ( J j ) = i , 1 j n} and Mi = { M j | g ( M j ) = i , 1 j m}. We may recursively choose |M1 | = 1, |Mi | = (k − i + 2)|Mi −1 | for i = 2, 3 . . . , k + 1, and |J1 | = 0, |Ji | = |Mi | for i = 2, 3, . . . , k + 1, respectively. All jobs have a unit processing time. The schedule μ is defined as follows. We assign k − i + 2 jobs in Ji to execute on each machine in Mi −1 for i = 2, 3, . . . , k + 1. The machines in Mk+1 remain empty. Obviously, μ is a pure Nash equilibrium. The makespan of μ is k. Meanwhile the optimal schedule assigns each job in Ji to be executed on each machine in Mi for i = 2, 3, . . . , k + 1, we obtain a makespan with minimum value 1. Thus, the price of anarchy is at least k. Thus, we obtain
m = |M1 | + |M2 | + · · · + |Mk+1 | (k + 1)! = (k + 2) implying k −1 (m) − 2. This establishes the conclusion. 2 The example in Theorem 2 shows that, when the processing time of all jobs are 1, the price of anarchy for m ( 3) machines is still Θ( logloglogmm ).
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3. LG-LPT policy Instead of an arbitrary ordering, we consider the ordering of jobs where a job J j precedes a job J k when either g ( J j ) < g ( J k ), or g ( J j ) = g ( J k ) and p ( J j ) p ( J k ). We denote such a precedence by J j J k , and we call it LG-LPT order. Hwang et al. [6] designed an algorithm, denoted by LG-LPT, to solve the scheduling problem under a GoS provision. The strategy of the algorithm LG-LPT is as follows: (1) Sort the jobs in LG-LPT order; (2) Assign the first one job from the list of unassigned jobs to the least-loaded eligible machine until all jobs are scheduled. In the LG-LPT policy, the jobs assigned to a machine are executed in LG-LPT order, i.e., the job J j is executed before the job J k when J j J k holds. Theorem 3. The set of Nash equilibrium for the LG-LPT policy of the induced game is precisely the set of solutions that can be generated by the greedy algorithm LG-LPT. Proof. Consider a schedule μ produced by the algorithm LG-LPT [6]. On each machine, jobs are executed in LG-LPT order, i.e., the LG-LPT policy is effective. Suppose that it is beneficial for the job J i to switch from its current machine to another machine, let M j be the machine on which J i would be executed completely at first. Then the algorithm LG-LPT would schedule the job J i to be executed on machine M j by its greedy rule, which leads a contradiction. Thus, no job has an incentive to switch to another machine. It implies that μ is a Nash equilibrium. We prove the other direction by induction on the number of jobs. For n = 1, there is nothing to prove. Let μ be a Nash equilibrium scheduling with n + 1 jobs, and J i the job whose completion time is equal to the makespan of μ. When we remove the job J i from μ, by the fact that the job J i is last one, the remaining schedule, denoted by μ , is still a Nash equilibrium. According to the induction assumption, μ can be constructed by using the algorithm LG-LPT. Because μ is a Nash equilibrium, the job J i has no incentive to switch to another machine. Then the strategy of J i can be gained by the greedy algorithm LGLPT. It follows that μ can be interpreted as an output of the algorithm LG-LPT. 2 According to Theorem 3, the approximation factor of the greedy algorithm LG-LPT is the upper bound on the price of anarchy of the corresponding coordination mechanism. Lemma 1. (See [6].) For the scheduling problem under a grade of service provision, the greedy algorithm LG-LPT has the approx1 imation factor 54 for two machines and 2 − m− for m ( 3) 1 machines, respectively. And the greedy algorithm LG-LPT is tight. Now, we can obtain the following result as a consequence of Theorem 3 and Lemma 1. Corollary 1. The price of anarchy of the LG-LPT policy for the selfish scheduling game under a grade of service provision is 54
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for m ( 3) machines, respec-
An agent is called to be satisfied if this agent cannot reduce his completion time by unilaterally moving his job to another machine. The best-response action means that the agent moves his job to the machine on which he can get minimum completion time. Theorem 3 implies that the Nash equilibria exists for the induced game. We show that the selfish agents will converge automatically from any given initial schedule to a Nash equilibrium in reasonable times in the following theorem. Theorem 4. For the LG-LPT policy, selfish agents converge to a Nash equilibrium from any given initial schedule after n rounds of the best-response actions, and in the kth round, there are at most n − k + 1 agents who may take one best-response moving. Proof. We sort these n jobs in the LG-LPT order. Without loss of generality, we may assume that J 1 J 2 · · · J n . For any given initial schedule, each unsatisfied agent plays his best-response action regarding that other agents stay on their current machines. We call the procedure, that all unsatisfied agents play their own best-response actions, as a round. Now, we can prove by an induction that each job J i does not move in the kth round, where i = 1, 2, . . . , k − 1, and that if the job J k moves to some machine, it will remains there in the subsequent rounds. For the given strategies of J 1 , J 2 , . . . , J k−1 in the (k − 1)th round, we may suppose that M j is the machine on which the job J k achieves its minimum completion time. Then a best-response action for the job J k is to move to the machine M j in the kth round. Depending on the induction assumption, the jobs J 1 , J 2 , . . . , J k−1 do not change their strategies in the kth round. And for each i = 2, 3, . . . , n, because the job J i −1 precedes the job J i , then the jobs J 1 , J 2 , . . . , J k are always executed before the
jobs J k+1 , . . . , J n on any machine by the LG-LPT policy. Thus, we obtain the result: whenever the best-response actions are taken by J k+1 , . . . , J n in the kth round, the job J k is still satisfied at the end of the kth round. It means that the job J k cannot reduce its completion time by moving to another machine in the (k + 1)th round and it remains on the machine M j in the subsequent rounds. This establishes the theorem. 2 Acknowledgement The authors are grateful to the anonymous reviewer whose comments and suggestions have led to a substantially improved presentation for the paper. References [1] Y. Azar, K. Jain, V.S. Mirrokni, (Almost) optimal coordination mechanisms for unrelated machine scheduling, in: Proceedings of the 19th ACM-SIAM Symposium on Discrete Algorithms, 2008, pp. 323–332. [2] I. Caragiannis, Efficient coordination mechanisms for unrelated machine scheduling, in: Proceedings of the 20th Symposium on Discrete Algorithms, 2009, pp. 815–824. [3] G. Christodoulou, E. Koutsoupias, A. Nanavati, Coordination mechanisms, in: Proceedings of the 31st International Colloquium on Automata, Languages and Programming, in: Lecture Notes in Computer Science, vol. 3142, 2004, pp. 345–357. [4] M. Gairing, T. Lucking, M. Mavronicolas, B. Monien, Computing Nash equilibria for scheduling on restricted parallel links, in: Proceedings of the 36th ACM Symposium on Theory of Computing, 2004, pp. 613– 622. [5] R.L. Graham, E.L. Lawler, J.K. Lenstra, A.R. Kan, Optimization and approximation in deterministic sequencing and scheduling: a survey, Annals of Discrete Mathematics 5 (1979) 287–326. [6] Hark-Chin Hwang, Soo Y. Chang, Kangbok Lee, Parallel machine scheduling under a grade of service provision, Computers and Operations Research 31 (2004) 2055–2061. [7] N. Immorlica, L. Li, V.S. Mirrokni, A.S. Schulz, Coordination mechanisms for selfish scheduling, Theoretical Computer Science 410 (2009) 1589–1598. [8] E. Koutsoupias, C.H. Papadimitriou, Worst-case equilibria, in: Proceedings of the 16th Symposium on Theoretical Aspects of Computing Science, 1999, pp. 404–413.