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A note on the lower bound for the Price of Anarchy of scheduling games on unrelated machines
Discrete Applied Mathematics 186 (2015) 295–300
Contents lists available at ScienceDirect
Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
Note
A note on the lower bound for the Price of Anarchy of scheduling games on unrelated machines Yujie Yan a , Zhihao Ding a,b , Zhiyi Tan a,∗ a
Department of Mathematics, Zhejiang University, Hangzhou 310027, PR China
b
H. Milton Stewart School of Industrial and Systems Engineering, Georgia Institute of Technology, USA
article
info
Article history: Received 8 February 2014 Received in revised form 1 January 2015 Accepted 7 January 2015 Available online 11 February 2015 Keywords: Scheduling game Coordination mechanism Price of Anarchy Worst-case ratio
abstract This note presents a lower bound for the Strong Price of Anarchy (SPoA) of coordination mechanisms for unrelated parallel machine scheduling games with social cost of minimizing the makespan. The SPoA of any set of non-preemptive strongly local policies satisfying the IIA property is at least m, the number of machines. Combining with the upper bound of the worst-case ratio of Shortest First algorithm for unrelated parallel machine scheduling problem with objective of minimizing the makespan (Ibarra and Kim, 1977), the SPoA of SPT policy, as well as the worst-case ratio of Shortest First algorithm, is exactly m. © 2015 Elsevier B.V. All rights reserved.
1. Introduction Scheduling games have been popular recently with the development of Internet and network economics. The most common model has a parallel machine setting. Given a job set J = {J1 , J2 , . . . , Jn } and a machine set M = {M1 , M2 , . . . , Mm }. Each job is processed on one of the machines, and the processing time needed for Jj on Mi is pji . Unlike classical scheduling problems where jobs are assigned by a central decision maker, each job can individually choose a machine for processing. The choices of all jobs constitute a schedule. Jobs are independent selfish players, each aiming at minimizing its individual cost, which is its own completion time on the selected machine in the schedule. A schedule is a Nash Equilibrium (NE) if no job can reduce its cost by moving to a different machine [8]. A schedule is a Strong Equilibrium (SE) if there is no coalition of jobs such that the cost of each job of the coalition will be reduced by migration simultaneously [1]. An SE is also an NE, but the reverse is not always true. Apart from individual cost of each job, the social cost which reflects utilities of the overall system is also of interest. In this note, we define social cost as the makespan of the schedule, i.e., the maximum completion time of all the jobs, which is natural and common in both theory and application. Due to lack of central coordination, an NE or SE may not be optimal in terms of the social cost. To measure the inefficiency of an equilibrium, the notions of PoA and SPoA are introduced. The Price of Anarchy (PoA) (res., Strong Price of Anarchy (SPoA)) is the worst-case ratio between the social cost of any NE (res., SE) and that of a social optimum [8,1]. Obviously, SPoA ≤ PoA by definition. To reduce the inefficiency of equilibria while not to impose centralized control on the jobs, one can design a scheduling policy for each machine [3]. Each job has the privilege to select the machine for processing. However, once all the jobs finish selecting machines, jobs on the same machine are processed according to the policy of that machine. More specifically, let the set of jobs selecting Mi be Ji and the scheduling policy of Mi be Pi . Once Ji is determined, Pi takes Ji as input and outputs the
∗
Corresponding author. E-mail address: [email protected] (Z. Tan).
http://dx.doi.org/10.1016/j.dam.2015.01.013 0166-218X/© 2015 Elsevier B.V. All rights reserved.
296
Y. Yan et al. / Discrete Applied Mathematics 186 (2015) 295–300
completion times of all the jobs in Ji . The set of scheduling policies of each machine forms a coordination mechanism [3]. The central problem on coordination mechanism is to analyze properties of common policies and design effective mechanism with the smallest possible PoA. Scheduling policies can be classified into different types in terms of how much information they need to determine the completion times [2]. Policy Pi is local if it only uses the information of the processing times of Ji . Policy Pi is strongly local if it only uses the information of the processing times of Ji on Mi . A policy is non-preemptive if it processes each job in a continuous fashion without any idle time. Policy Pi satisfies the property of the independence of irrelevant alternatives (IIA) if the precedence order of any Jj1 , Jj2 ∈ Ji is not influenced by the availability of any other job in Ji . Policy Pi is an ordering policy if Mi gives a global order of all the jobs, and jobs selecting Mi are processed according to it. A non-preemptive policy satisfying IIA is always an ordering policy [2]. In [2] and [4], it is proved that the PoA and SPoA of any set of non-preemptive strongly local policies satisfying the IIA property are at least m , respectively. 2 Another line of research concentrates on the inefficiency of certain scheduling policies [7]. Suppose that Ji = {Jj1 , Jj2 , . . . , Jjl } and pj1 ,i ≤ pj2 ,i ≤ · · · ≤ pjl ,i . In the MAKESPAN policy, all jobs of Ji are processed in parallel, and thus the completion
l
time of each job is k=1 pjk ,i . Policy SPT (res., LPT ) sequences jobs of Ji in non-decreasing (res., non-increasing) order of their processing times on Mi , and processes them one by one non-preemptively. Thus the completion time of Jjs under SPT
s
l
(res., LPT ) policy is k=1 pjk ,i (res., k=s pjk ,i ). Clearly, both SPT and LPT are non-preemptive strongly local policies. We summarize below some main results on the PoA and SPoA of certain coordination mechanisms. More results for the cases under certain special machine environments can be found in [7]. For notation simplicity, the PoA (res., SPoA) of scheduling policy P is in fact the PoA (res., SPoA) of coordination mechanism where the policy of each machine is P. It has been shown that NE always exists under all policies mentioned above [7]. The PoA of MAKESPAN policy is unbounded and the SPoA of MAKESPAN policy is exactly m [9,1,5]. The PoA of SPT policy is obtained in a different way. For the classical unrelated machine scheduling, there is an algorithm called Shortest First (Algorithm D of [6]). Shortest First selects the job among all unassigned jobs that can be completed the earliest if it is assigned to the appropriate machine. The worst-case ratio of Shortest First for the unrelated machine scheduling problem with the objective of minimizing the makespan is at most m [6]. In [7], it is proved that the set of NE for the SPT policy is precisely the set of schedules that can be generated by the Shortest First algorithm. Thus the PoA of SPT policy is the same as the worst-case ratio of Shortest First. In this note, we show that the SPoA of any set of non-preemptive strongly local policies satisfying the IIA property is at least m. Combining the upper bound of Shortest First [6], the PoA and SPoA of SPT policy are both m, and the worst-case ratio of Shortest First algorithm for unrelated machine scheduling with objective to minimize the makespan is exactly m. Thus the long-standing gap between the lower and upper bounds on the worst-case ratio of Shortest First algorithm, as well as the PoA of SPT policy, is eliminated. 2. Lower bound In this section, we present a lower bound on the SPoA of any set of non-preemptive strongly local policies satisfying the IIA property. The proof uses the same idea in [2,4] but a new instance, which results in a tight bound. Suppose that Pi is the policy of Mi such that Pi is non-preemptive, strongly local and satisfies the IIA property, i = 1, . . . , m. Thus Pi is an ordering policy. We will construct a job set J whose SPoA is m. For a subset S ⊆ J , let Jk,i (S ) be the job at the kth position among jobs of S under the policy Pi , k = 1, . . . , |S |, i = 1, . . . , m. Let x be a multiple of m!. Define ni =
(x+1)(m−1)!xm−i−1 , (i−1)!
1 ≤ i ≤ m − 1 and nm = 1. Let w0 = 1, wi =
i
l =1
nl for any i =
1, . . . , m, and |J | = wm = l=1 nl . We partition J into m disjoint subsets Si , i = 1, . . . , m with |Si | = ni = wi − wi−1 as follows (see Fig. 1). First let Sm = {Jn,m (J )}. Then recursively define J i and Si for i = m − 1, . . . , 1 as follows. Let J i be the complement set of the union of the m − i subsets Si+1 , . . . , Sm , i.e.,
m
Ji = J\
m
Sl .
l=i+1
Note that
|J i | = |J | −
m l=i+1
|Sl | = wm −
m
nl = w i .
l=i+1
The subset Si consists of the ni jobs which are ordered last among jobs of J i under policy Pi , i.e., Si = {Jwi−1 +1,i (J i ), Jwi−1 +2,i (J i ), . . . , Jwi ,i (J i )}. Now we define the processing time of all jobs. For the first m − 1 subsets S1 , S2 , . . . , Sm−1 , jobs of Si can be processed on Mi and Mi+1 . The job in the last subset Sm can only be processed on Mm . All the jobs that can be processed on Mi have the same processing time on Mi equaling to pi = pi pi+1
=
1 ix
,
i = 1, . . . , m − 1.
(i−1)!xi−m . (m−1)!
It is easy to verify that (1)
Y. Yan et al. / Discrete Applied Mathematics 186 (2015) 295–300
297
Fig. 1. Construction of Si .
Fig. 2. Partitioning of Si .
Consequently, for any i = 2, . . . , m, only jobs of Si−1 and Si can be processed on Mi . Recall that Si−1 ⊆ J i . By the definition of Si , we have the following observation. Observation 1. Any job of Si−1 is processed before any job of Si on Mi . In order to describe a schedule which is an SE precisely, the set Si is further partitioned (see Fig. 2). For i = 1, . . . , m − 1, let ui =
(m − 1)!(i − 1)xm−i−1 i−1 = i! pi+1
(2)
and ti =
ni − ui ix + 1
=
(m − 1)!xm−i−1 1 1 = = . i! pi+1 ixpi
(3)
Then Si is partitioned into disjoint union
Si =
ti
(i,j)
Mi
j =1
∪
ti
(i,j) Mi+1
∪ Mi(+i,01) ,
i = 1, . . . , m − 1,
j =1
(i,j)
(i,0)
(i,j)
(i,0)
where Mi+1 , Mi and Mi+1 , j = 1, . . . , ti will be defined formally in the following. Let Mi+1 consist of the ui jobs ordered first in Si under policy Pi+1 , i.e., (i,0)
Mi+1 = {J1,i+1 (Si ), J2,i+1 (Si ), . . . , Jui ,i+1 (Si )}. (i,j)
Mi
(i,j)
(i,0)
and Mi+1 , j = 1, . . . , ti are defined recursively as follows. Set Si1 = Mi+1 and j +1
Si
= Sij ∪ Mi(i,j) ∪ Mi(+i,j1) = Mi(+i,01) ∪
j (Mi(i,l) ∪ Mi(+i,l1) ) , l =1
j = 1, . . . , ti − 1.
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Y. Yan et al. / Discrete Applied Mathematics 186 (2015) 295–300
Fig. 3. SE of J .
(i,j)
Let Mi
j
(i,j)
consist of the ix jobs ordered first among jobs of Si \Si under policy Pi , and Mi+1 consist of the single job which is j Si
ordered first among the jobs of Si \( (i,j)
j J1,i Si Si
={
Mi
( \ ),
(i,j)
∪ Mi
) under policy Pi+1 . More precisely,
j J2,i Si Si
( \ ), . . . , Jix,i (Si \Sij )}
and (i,j)
(i,j)
j
Mi+1 = {J1,i+1 (Si \(Si ∪ Mi
))}.
By above definitions, we have the following observation (see Fig. 2). (i,j)
(i,j)
j
Observation 2. (i) For any 1 ≤ j ≤ ti , Mi+1 ⊆ Si \(Si ∪ Mi (i,j)
Mi
) ⊆ Si \Sij , and thus the job of Mi(+i,j1) is ordered after any job of
under policy Pi .
(ii) For any 1 ≤ j < ti , Mi(i,j+1) ⊆ Si \Sij+1 ⊆ Si \Sij , and thus any job of Mi(i,j+1) is ordered after any job of Mi(i,j) under
policy Pi . (iii) For any 0 ≤ j < ti , Mi(i,j+1) ⊆ Si \Sij+1 ⊆ Si \(Sij ∪ Mi(i,j) ), and thus any job of Mi(i,j+1) is ordered after the job of Mi(+i,j1) under policy Pi+1 . (iv) For any 0 ≤ j < ti , Mi(+i,j1+1) ⊆ Si \(Sij+1 ∪ Mi(i,j+1) ) ⊆ Si \(Sij ∪ Mi(i,j) ), and thus any job of Mi(+i,j1+1) is ordered after the (i,j)
job of Mi+1 under policy Pi+1 . We are ready to give a schedule σ by giving the jobs choosing Mi , i = 1, . . . , m, that is, t1
J1 =
(1,j)
M1
,
j =1
Ji =
t i−1
(i−1,j)
Mi
∪
j =0
ti
(i,j)
Mi
,
i = 2, . . . , m − 1,
j =1
and tm−1
Jm =
(m−1,j) Mm ∪ Sm .
j =0
The processing order of these jobs is given by Observations 1, 2. (See Fig. 3.) (i,j) (i,j) (i,0) Recall that |Mi+1 | = ui , |Mi | = ix and |Mi+1 | = 1 for any 1 ≤ j ≤ ti . By Observations 1, 2(ii)(iv) and (2), (3), we have the following observation. (i−1,0)
Observation 3. (i) The completion time of the last job of Mi
on Mi is ui−1 pi = i − 2, i = 2, . . . , m. (ii) The completion time of the job of Mi on Mi is ui−1 pi + jpi = i − 2 + jpi , j = 1, . . . , ti−1 , i = 2, . . . , m. (iii) The completion time of the last job of Mi(i,j) on Mi is (i−1,j)
ui−1 pi + ti−1 pi + jixpi = i − 1 + jixpi , (1,j)
The completion time of the last job of M1
j = 1, . . . , ti , i = 2, . . . , m.
on M1 is jxp1 , j = 1, . . . , ti .
By Observation 3(iii) and (1), (3), the completion time of the last job choosing Mi , i = 1, . . . , m − 1 is i − 1 + ti ixpi = i, and the completion time of the last job choosing Mm is m − 2 + tm−1 pm + pm = m. Thus the makespan of σ is m. In the optimal schedule σ ∗ , jobs of Si are assigned to Mi . The completion time of the last job on Mi , i = 1, . . . , m − 1 is ni pi = x+x 1 ,
and the completion time of the last job on Mm is pm = 1. Hence the makespan of σ ∗ is x+x 1 . Therefore, the ratio between the makespan of σ and σ ∗ is xm +1 → m(x → ∞). x
Now we are left to show that σ is an SE. On the contrary, suppose that σ is not an SE. There is a coalition of jobs that has incentive to deviate jointly.
Y. Yan et al. / Discrete Applied Mathematics 186 (2015) 295–300
299
Lemma 1. No job of S1 will join the coalition.
(1,j)
M1
(1,j)
t1
Proof. Recall that u1 = 0 and thus
M1
j =1
∪ M2(1,j) is a disjoint union of S1 . We prove by induction on j that jobs of
∪ M2(1,j) will not join the coalition. First consider the base step of j = 1. By Observation 3(iii)(ii), the completion time (1,1)
of the last job of M1
(1,1)
on M1 is xp1 , and the completion time of the job of M2
(1,1)
M2 , the new completion time will be at least p2 = xp1 by (1). If the job of M2 (1,1)
M2 no earlier than any job of M1 (1,1)
(1,1)
on M2 is p2 . If any job of M1
moves to
moves to M1 , it will be started to process on
by Observation 2(i), and thus the new completion time will be at least (x + 1)p1 > p2 .
∪ M2(1,1) will join the coalition. j−1 (1,l) (1,l) Assume that the result is true for any l ≤ j − 1, i.e., jobs of l=1 (M1 ∪ M2 ) will not join the coalition. Note that
Therefore, no job of M1
(1,j)
(1,j)
the completion time of the last job of M1 on M1 is jxp1 , and the completion time of the job of M2 on M2 is jp2 by Observation 3(ii)(iii). (1,j) (1,j−1) By Observation 2(iii)(i), if any job of M1 moves to M2 , it will be started to process no earlier than the job of M2 , (1,j)
thus the new completion time will be at least jp2 = jxp1 by (1). Similarly, if the job of M2 (1,j)
to process no earlier than any job of M1 (1,j)
no job of M1
moves to M1 , it will be started
, and thus the new completion time will be at least (jx + 1)p1 > jp2 . Therefore,
∪ M2(1,j) will join the coalition. Accordingly the assertion also holds for j.
Now we prove by induction on i that no job of Si will join the coalition. The base step follows from Lemma 1. Suppose i−1 that the result is true for any l ≤ i − 1, i.e., no job of l=1 Sl will join the coalition. We will prove the result for Si . Recall that Si =
ti
j=1
(i,j)
Mi
ti (i,j) (i,0) (i,0) ∪ j=1 Mi+1 ∪ Mi+1 . We first show that no job of Mi+1 will join the coalition. By Observations 1 and (i−1,ti−1 )
(i,0)
3(i)(ii), if any job of Mi+1 moves to Mi , it will be started after any job of Mi
. Thus the new completion time is at least (i,0)
i − 2 + ti−1 pi + pi−1 = i − 1 + pi−1 , while its completion time in σ is i − 1. It follows that no job of Mi+1 will join the coalition. (i,j)
Next we prove by induction on j that jobs in Mi
∪ Mi(+i,j1)
will not join the coalition. First consider the base step of j = 1. (i,1)
By Observation 3(iii)(ii), the completion time of the last job of Mi
(i,1) Mi+1
(i,1)
on Mi+1 is i − 1 + pi+1 . If any job of Mi
Thus the completion time will be at least i − 1 + pi+1 = (i,1)
process no earlier than any job of Mi
on Mi is i − 1 + ixpi , and the completion time of the job of (i,0)
moves to Mi+1 , it has to be started after any job in Mi+1 by Observation 2(iii). (i,1) i − 1 + ixp1 by (1). If the job of Mi+1
moves to Mi , it will be started to
by Observation 2(i). Hence, the new completion time will be at least i − 1 + ixp1 + p1 = (i,1)
i − 1 + p2 + p1 , while its completion time in σ is i − 1 + p2 . Therefore, no job of Mi tion 3(iii)(ii), the completion time of the last job of Mi
(i,l)
(i,l) Mi+1 )
∪ Mi(+i,11) will join the coalition.
(Mi ∪ will join the coalition. By Observa(i,1) on Mi is i − 1 + jixpi , and the completion time of the job of Mi+1
Assume that the claim is true for any l ≤ j − 1, i.e., no job of (i,1)
j−1 l=1
(i,j) moves to Mi+1 , it will be started to process no earlier than (i,j−1) the job of Mi+1 . Thus the new completion time will be at least i − 1 + (j − 1)pi+1 + pi+1 = i − 1 + jixpi . Similarly, if the (i,j) (i,j) job of Mi+1 moves to Mi , it will be started to process no earlier than any job of Mi . Thus the new completion time will be (i,j) (i,j) at least i − 1 + jixpi + pi > i − 1 + jpi+1 . Therefore, no jobs of Mi ∪ Mi+1 will join the coalition.
on Mi+1 is i − 1 + jpi+1 . By Observation 2(i)(iii), if any job of Mi
Up to now, we have proved that no job will join the coalition, which is a contradiction. Hence, σ is an SE, leading to the following main result of the note. Proposition 1. For any set of non-preemptive strongly local policies satisfying the IIA property, the SPoA is at least m.
Combining with the relation between NE and algorithm Shortest First [7], together with the upper bound of Shortest First [6], we have the following corollary. Corollary 1. The worst-case ratio of Shortest First for unrelated machine scheduling with objective of minimizing the makespan is exactly m. Acknowledgments The authors would like to acknowledge the anonymous referee for constructive comments which have improved the presentation of the paper. The third author was supported by the National Natural Science Foundation of China (11271324), Zhejiang Provincial Natural Science Foundation of China (LR12A01001). References [1] N. Andelman, M. Feldman, Y. Mansour, Strong Price of Anarchy, Games Econom. Behav. 65 (2009) 289–317. [2] Y. Azar, K. Jain, V.S. Mirrokni, (Almost) optimal coordination mechanisms for unrelated machine scheduling, in: Proceedings of the 19th Annual ACMSIAM Symposium on Discrete Algorithms, 2008, pp. 323–332. [3] G. Christodoulou, E. Koutsoupias, A. Nanavati, Coordination mechanisms, Theoret. Comput. Sci. 410 (2009) 3327–3336.
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[4] M. Feldman, M. Rosenthal, O. Sivan, Makespan minimization through coordination mechanisms. Working paper, 2013. [5] A. Fiat, H. Kaplan, M. Levy, S. Olonetsky, Strong Price of Anarchy for machine load balancing, in: Proceeesing of the 34th International Colloquium on Automata, Languages and Programming, in: Lecture Notes in Computer Science, 2007, pp. 583–594. [6] O.H. Ibarra, C.E. Kim, Heuristic algorithms for scheduling independent tasks on nonidentical processors, J. ACM 24 (1977) 280–289. [7] N. Immorlica, L. Li, V.S. Mirrokni, A. Schulz, Coordination mechanisms for selfish scheduling, Theoret. Comput. Sci. 410 (2009) 1589–1598. [8] E. Koutsoupias, C.H. Papadimitriou, Worst-case equilibria, Comput. Sci. Rev. 3 (2009) 65–69. [9] P. Schuurman, T. Vredeveld, Performance guarantees of local search for multiprocessor scheduling, INFORMS J. Comput. 19 (2007) 52–63.
Contents lists available at ScienceDirect
Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
Note
A note on the lower bound for the Price of Anarchy of scheduling games on unrelated machines Yujie Yan a , Zhihao Ding a,b , Zhiyi Tan a,∗ a
Department of Mathematics, Zhejiang University, Hangzhou 310027, PR China
b
H. Milton Stewart School of Industrial and Systems Engineering, Georgia Institute of Technology, USA
article
info
Article history: Received 8 February 2014 Received in revised form 1 January 2015 Accepted 7 January 2015 Available online 11 February 2015 Keywords: Scheduling game Coordination mechanism Price of Anarchy Worst-case ratio
abstract This note presents a lower bound for the Strong Price of Anarchy (SPoA) of coordination mechanisms for unrelated parallel machine scheduling games with social cost of minimizing the makespan. The SPoA of any set of non-preemptive strongly local policies satisfying the IIA property is at least m, the number of machines. Combining with the upper bound of the worst-case ratio of Shortest First algorithm for unrelated parallel machine scheduling problem with objective of minimizing the makespan (Ibarra and Kim, 1977), the SPoA of SPT policy, as well as the worst-case ratio of Shortest First algorithm, is exactly m. © 2015 Elsevier B.V. All rights reserved.
1. Introduction Scheduling games have been popular recently with the development of Internet and network economics. The most common model has a parallel machine setting. Given a job set J = {J1 , J2 , . . . , Jn } and a machine set M = {M1 , M2 , . . . , Mm }. Each job is processed on one of the machines, and the processing time needed for Jj on Mi is pji . Unlike classical scheduling problems where jobs are assigned by a central decision maker, each job can individually choose a machine for processing. The choices of all jobs constitute a schedule. Jobs are independent selfish players, each aiming at minimizing its individual cost, which is its own completion time on the selected machine in the schedule. A schedule is a Nash Equilibrium (NE) if no job can reduce its cost by moving to a different machine [8]. A schedule is a Strong Equilibrium (SE) if there is no coalition of jobs such that the cost of each job of the coalition will be reduced by migration simultaneously [1]. An SE is also an NE, but the reverse is not always true. Apart from individual cost of each job, the social cost which reflects utilities of the overall system is also of interest. In this note, we define social cost as the makespan of the schedule, i.e., the maximum completion time of all the jobs, which is natural and common in both theory and application. Due to lack of central coordination, an NE or SE may not be optimal in terms of the social cost. To measure the inefficiency of an equilibrium, the notions of PoA and SPoA are introduced. The Price of Anarchy (PoA) (res., Strong Price of Anarchy (SPoA)) is the worst-case ratio between the social cost of any NE (res., SE) and that of a social optimum [8,1]. Obviously, SPoA ≤ PoA by definition. To reduce the inefficiency of equilibria while not to impose centralized control on the jobs, one can design a scheduling policy for each machine [3]. Each job has the privilege to select the machine for processing. However, once all the jobs finish selecting machines, jobs on the same machine are processed according to the policy of that machine. More specifically, let the set of jobs selecting Mi be Ji and the scheduling policy of Mi be Pi . Once Ji is determined, Pi takes Ji as input and outputs the
∗
Corresponding author. E-mail address: [email protected] (Z. Tan).
http://dx.doi.org/10.1016/j.dam.2015.01.013 0166-218X/© 2015 Elsevier B.V. All rights reserved.
296
Y. Yan et al. / Discrete Applied Mathematics 186 (2015) 295–300
completion times of all the jobs in Ji . The set of scheduling policies of each machine forms a coordination mechanism [3]. The central problem on coordination mechanism is to analyze properties of common policies and design effective mechanism with the smallest possible PoA. Scheduling policies can be classified into different types in terms of how much information they need to determine the completion times [2]. Policy Pi is local if it only uses the information of the processing times of Ji . Policy Pi is strongly local if it only uses the information of the processing times of Ji on Mi . A policy is non-preemptive if it processes each job in a continuous fashion without any idle time. Policy Pi satisfies the property of the independence of irrelevant alternatives (IIA) if the precedence order of any Jj1 , Jj2 ∈ Ji is not influenced by the availability of any other job in Ji . Policy Pi is an ordering policy if Mi gives a global order of all the jobs, and jobs selecting Mi are processed according to it. A non-preemptive policy satisfying IIA is always an ordering policy [2]. In [2] and [4], it is proved that the PoA and SPoA of any set of non-preemptive strongly local policies satisfying the IIA property are at least m , respectively. 2 Another line of research concentrates on the inefficiency of certain scheduling policies [7]. Suppose that Ji = {Jj1 , Jj2 , . . . , Jjl } and pj1 ,i ≤ pj2 ,i ≤ · · · ≤ pjl ,i . In the MAKESPAN policy, all jobs of Ji are processed in parallel, and thus the completion
l
time of each job is k=1 pjk ,i . Policy SPT (res., LPT ) sequences jobs of Ji in non-decreasing (res., non-increasing) order of their processing times on Mi , and processes them one by one non-preemptively. Thus the completion time of Jjs under SPT
s
l
(res., LPT ) policy is k=1 pjk ,i (res., k=s pjk ,i ). Clearly, both SPT and LPT are non-preemptive strongly local policies. We summarize below some main results on the PoA and SPoA of certain coordination mechanisms. More results for the cases under certain special machine environments can be found in [7]. For notation simplicity, the PoA (res., SPoA) of scheduling policy P is in fact the PoA (res., SPoA) of coordination mechanism where the policy of each machine is P. It has been shown that NE always exists under all policies mentioned above [7]. The PoA of MAKESPAN policy is unbounded and the SPoA of MAKESPAN policy is exactly m [9,1,5]. The PoA of SPT policy is obtained in a different way. For the classical unrelated machine scheduling, there is an algorithm called Shortest First (Algorithm D of [6]). Shortest First selects the job among all unassigned jobs that can be completed the earliest if it is assigned to the appropriate machine. The worst-case ratio of Shortest First for the unrelated machine scheduling problem with the objective of minimizing the makespan is at most m [6]. In [7], it is proved that the set of NE for the SPT policy is precisely the set of schedules that can be generated by the Shortest First algorithm. Thus the PoA of SPT policy is the same as the worst-case ratio of Shortest First. In this note, we show that the SPoA of any set of non-preemptive strongly local policies satisfying the IIA property is at least m. Combining the upper bound of Shortest First [6], the PoA and SPoA of SPT policy are both m, and the worst-case ratio of Shortest First algorithm for unrelated machine scheduling with objective to minimize the makespan is exactly m. Thus the long-standing gap between the lower and upper bounds on the worst-case ratio of Shortest First algorithm, as well as the PoA of SPT policy, is eliminated. 2. Lower bound In this section, we present a lower bound on the SPoA of any set of non-preemptive strongly local policies satisfying the IIA property. The proof uses the same idea in [2,4] but a new instance, which results in a tight bound. Suppose that Pi is the policy of Mi such that Pi is non-preemptive, strongly local and satisfies the IIA property, i = 1, . . . , m. Thus Pi is an ordering policy. We will construct a job set J whose SPoA is m. For a subset S ⊆ J , let Jk,i (S ) be the job at the kth position among jobs of S under the policy Pi , k = 1, . . . , |S |, i = 1, . . . , m. Let x be a multiple of m!. Define ni =
(x+1)(m−1)!xm−i−1 , (i−1)!
1 ≤ i ≤ m − 1 and nm = 1. Let w0 = 1, wi =
i
l =1
nl for any i =
1, . . . , m, and |J | = wm = l=1 nl . We partition J into m disjoint subsets Si , i = 1, . . . , m with |Si | = ni = wi − wi−1 as follows (see Fig. 1). First let Sm = {Jn,m (J )}. Then recursively define J i and Si for i = m − 1, . . . , 1 as follows. Let J i be the complement set of the union of the m − i subsets Si+1 , . . . , Sm , i.e.,
m
Ji = J\
m
Sl .
l=i+1
Note that
|J i | = |J | −
m l=i+1
|Sl | = wm −
m
nl = w i .
l=i+1
The subset Si consists of the ni jobs which are ordered last among jobs of J i under policy Pi , i.e., Si = {Jwi−1 +1,i (J i ), Jwi−1 +2,i (J i ), . . . , Jwi ,i (J i )}. Now we define the processing time of all jobs. For the first m − 1 subsets S1 , S2 , . . . , Sm−1 , jobs of Si can be processed on Mi and Mi+1 . The job in the last subset Sm can only be processed on Mm . All the jobs that can be processed on Mi have the same processing time on Mi equaling to pi = pi pi+1
=
1 ix
,
i = 1, . . . , m − 1.
(i−1)!xi−m . (m−1)!
It is easy to verify that (1)
Y. Yan et al. / Discrete Applied Mathematics 186 (2015) 295–300
297
Fig. 1. Construction of Si .
Fig. 2. Partitioning of Si .
Consequently, for any i = 2, . . . , m, only jobs of Si−1 and Si can be processed on Mi . Recall that Si−1 ⊆ J i . By the definition of Si , we have the following observation. Observation 1. Any job of Si−1 is processed before any job of Si on Mi . In order to describe a schedule which is an SE precisely, the set Si is further partitioned (see Fig. 2). For i = 1, . . . , m − 1, let ui =
(m − 1)!(i − 1)xm−i−1 i−1 = i! pi+1
(2)
and ti =
ni − ui ix + 1
=
(m − 1)!xm−i−1 1 1 = = . i! pi+1 ixpi
(3)
Then Si is partitioned into disjoint union
Si =
ti
(i,j)
Mi
j =1
∪
ti
(i,j) Mi+1
∪ Mi(+i,01) ,
i = 1, . . . , m − 1,
j =1
(i,j)
(i,0)
(i,j)
(i,0)
where Mi+1 , Mi and Mi+1 , j = 1, . . . , ti will be defined formally in the following. Let Mi+1 consist of the ui jobs ordered first in Si under policy Pi+1 , i.e., (i,0)
Mi+1 = {J1,i+1 (Si ), J2,i+1 (Si ), . . . , Jui ,i+1 (Si )}. (i,j)
Mi
(i,j)
(i,0)
and Mi+1 , j = 1, . . . , ti are defined recursively as follows. Set Si1 = Mi+1 and j +1
Si
= Sij ∪ Mi(i,j) ∪ Mi(+i,j1) = Mi(+i,01) ∪
j (Mi(i,l) ∪ Mi(+i,l1) ) , l =1
j = 1, . . . , ti − 1.
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Y. Yan et al. / Discrete Applied Mathematics 186 (2015) 295–300
Fig. 3. SE of J .
(i,j)
Let Mi
j
(i,j)
consist of the ix jobs ordered first among jobs of Si \Si under policy Pi , and Mi+1 consist of the single job which is j Si
ordered first among the jobs of Si \( (i,j)
j J1,i Si Si
={
Mi
( \ ),
(i,j)
∪ Mi
) under policy Pi+1 . More precisely,
j J2,i Si Si
( \ ), . . . , Jix,i (Si \Sij )}
and (i,j)
(i,j)
j
Mi+1 = {J1,i+1 (Si \(Si ∪ Mi
))}.
By above definitions, we have the following observation (see Fig. 2). (i,j)
(i,j)
j
Observation 2. (i) For any 1 ≤ j ≤ ti , Mi+1 ⊆ Si \(Si ∪ Mi (i,j)
Mi
) ⊆ Si \Sij , and thus the job of Mi(+i,j1) is ordered after any job of
under policy Pi .
(ii) For any 1 ≤ j < ti , Mi(i,j+1) ⊆ Si \Sij+1 ⊆ Si \Sij , and thus any job of Mi(i,j+1) is ordered after any job of Mi(i,j) under
policy Pi . (iii) For any 0 ≤ j < ti , Mi(i,j+1) ⊆ Si \Sij+1 ⊆ Si \(Sij ∪ Mi(i,j) ), and thus any job of Mi(i,j+1) is ordered after the job of Mi(+i,j1) under policy Pi+1 . (iv) For any 0 ≤ j < ti , Mi(+i,j1+1) ⊆ Si \(Sij+1 ∪ Mi(i,j+1) ) ⊆ Si \(Sij ∪ Mi(i,j) ), and thus any job of Mi(+i,j1+1) is ordered after the (i,j)
job of Mi+1 under policy Pi+1 . We are ready to give a schedule σ by giving the jobs choosing Mi , i = 1, . . . , m, that is, t1
J1 =
(1,j)
M1
,
j =1
Ji =
t i−1
(i−1,j)
Mi
∪
j =0
ti
(i,j)
Mi
,
i = 2, . . . , m − 1,
j =1
and tm−1
Jm =
(m−1,j) Mm ∪ Sm .
j =0
The processing order of these jobs is given by Observations 1, 2. (See Fig. 3.) (i,j) (i,j) (i,0) Recall that |Mi+1 | = ui , |Mi | = ix and |Mi+1 | = 1 for any 1 ≤ j ≤ ti . By Observations 1, 2(ii)(iv) and (2), (3), we have the following observation. (i−1,0)
Observation 3. (i) The completion time of the last job of Mi
on Mi is ui−1 pi = i − 2, i = 2, . . . , m. (ii) The completion time of the job of Mi on Mi is ui−1 pi + jpi = i − 2 + jpi , j = 1, . . . , ti−1 , i = 2, . . . , m. (iii) The completion time of the last job of Mi(i,j) on Mi is (i−1,j)
ui−1 pi + ti−1 pi + jixpi = i − 1 + jixpi , (1,j)
The completion time of the last job of M1
j = 1, . . . , ti , i = 2, . . . , m.
on M1 is jxp1 , j = 1, . . . , ti .
By Observation 3(iii) and (1), (3), the completion time of the last job choosing Mi , i = 1, . . . , m − 1 is i − 1 + ti ixpi = i, and the completion time of the last job choosing Mm is m − 2 + tm−1 pm + pm = m. Thus the makespan of σ is m. In the optimal schedule σ ∗ , jobs of Si are assigned to Mi . The completion time of the last job on Mi , i = 1, . . . , m − 1 is ni pi = x+x 1 ,
and the completion time of the last job on Mm is pm = 1. Hence the makespan of σ ∗ is x+x 1 . Therefore, the ratio between the makespan of σ and σ ∗ is xm +1 → m(x → ∞). x
Now we are left to show that σ is an SE. On the contrary, suppose that σ is not an SE. There is a coalition of jobs that has incentive to deviate jointly.
Y. Yan et al. / Discrete Applied Mathematics 186 (2015) 295–300
299
Lemma 1. No job of S1 will join the coalition.
(1,j)
M1
(1,j)
t1
Proof. Recall that u1 = 0 and thus
M1
j =1
∪ M2(1,j) is a disjoint union of S1 . We prove by induction on j that jobs of
∪ M2(1,j) will not join the coalition. First consider the base step of j = 1. By Observation 3(iii)(ii), the completion time (1,1)
of the last job of M1
(1,1)
on M1 is xp1 , and the completion time of the job of M2
(1,1)
M2 , the new completion time will be at least p2 = xp1 by (1). If the job of M2 (1,1)
M2 no earlier than any job of M1 (1,1)
(1,1)
on M2 is p2 . If any job of M1
moves to
moves to M1 , it will be started to process on
by Observation 2(i), and thus the new completion time will be at least (x + 1)p1 > p2 .
∪ M2(1,1) will join the coalition. j−1 (1,l) (1,l) Assume that the result is true for any l ≤ j − 1, i.e., jobs of l=1 (M1 ∪ M2 ) will not join the coalition. Note that
Therefore, no job of M1
(1,j)
(1,j)
the completion time of the last job of M1 on M1 is jxp1 , and the completion time of the job of M2 on M2 is jp2 by Observation 3(ii)(iii). (1,j) (1,j−1) By Observation 2(iii)(i), if any job of M1 moves to M2 , it will be started to process no earlier than the job of M2 , (1,j)
thus the new completion time will be at least jp2 = jxp1 by (1). Similarly, if the job of M2 (1,j)
to process no earlier than any job of M1 (1,j)
no job of M1
moves to M1 , it will be started
, and thus the new completion time will be at least (jx + 1)p1 > jp2 . Therefore,
∪ M2(1,j) will join the coalition. Accordingly the assertion also holds for j.
Now we prove by induction on i that no job of Si will join the coalition. The base step follows from Lemma 1. Suppose i−1 that the result is true for any l ≤ i − 1, i.e., no job of l=1 Sl will join the coalition. We will prove the result for Si . Recall that Si =
ti
j=1
(i,j)
Mi
ti (i,j) (i,0) (i,0) ∪ j=1 Mi+1 ∪ Mi+1 . We first show that no job of Mi+1 will join the coalition. By Observations 1 and (i−1,ti−1 )
(i,0)
3(i)(ii), if any job of Mi+1 moves to Mi , it will be started after any job of Mi
. Thus the new completion time is at least (i,0)
i − 2 + ti−1 pi + pi−1 = i − 1 + pi−1 , while its completion time in σ is i − 1. It follows that no job of Mi+1 will join the coalition. (i,j)
Next we prove by induction on j that jobs in Mi
∪ Mi(+i,j1)
will not join the coalition. First consider the base step of j = 1. (i,1)
By Observation 3(iii)(ii), the completion time of the last job of Mi
(i,1) Mi+1
(i,1)
on Mi+1 is i − 1 + pi+1 . If any job of Mi
Thus the completion time will be at least i − 1 + pi+1 = (i,1)
process no earlier than any job of Mi
on Mi is i − 1 + ixpi , and the completion time of the job of (i,0)
moves to Mi+1 , it has to be started after any job in Mi+1 by Observation 2(iii). (i,1) i − 1 + ixp1 by (1). If the job of Mi+1
moves to Mi , it will be started to
by Observation 2(i). Hence, the new completion time will be at least i − 1 + ixp1 + p1 = (i,1)
i − 1 + p2 + p1 , while its completion time in σ is i − 1 + p2 . Therefore, no job of Mi tion 3(iii)(ii), the completion time of the last job of Mi
(i,l)
(i,l) Mi+1 )
∪ Mi(+i,11) will join the coalition.
(Mi ∪ will join the coalition. By Observa(i,1) on Mi is i − 1 + jixpi , and the completion time of the job of Mi+1
Assume that the claim is true for any l ≤ j − 1, i.e., no job of (i,1)
j−1 l=1
(i,j) moves to Mi+1 , it will be started to process no earlier than (i,j−1) the job of Mi+1 . Thus the new completion time will be at least i − 1 + (j − 1)pi+1 + pi+1 = i − 1 + jixpi . Similarly, if the (i,j) (i,j) job of Mi+1 moves to Mi , it will be started to process no earlier than any job of Mi . Thus the new completion time will be (i,j) (i,j) at least i − 1 + jixpi + pi > i − 1 + jpi+1 . Therefore, no jobs of Mi ∪ Mi+1 will join the coalition.
on Mi+1 is i − 1 + jpi+1 . By Observation 2(i)(iii), if any job of Mi
Up to now, we have proved that no job will join the coalition, which is a contradiction. Hence, σ is an SE, leading to the following main result of the note. Proposition 1. For any set of non-preemptive strongly local policies satisfying the IIA property, the SPoA is at least m.
Combining with the relation between NE and algorithm Shortest First [7], together with the upper bound of Shortest First [6], we have the following corollary. Corollary 1. The worst-case ratio of Shortest First for unrelated machine scheduling with objective of minimizing the makespan is exactly m. Acknowledgments The authors would like to acknowledge the anonymous referee for constructive comments which have improved the presentation of the paper. The third author was supported by the National Natural Science Foundation of China (11271324), Zhejiang Provincial Natural Science Foundation of China (LR12A01001). References [1] N. Andelman, M. Feldman, Y. Mansour, Strong Price of Anarchy, Games Econom. Behav. 65 (2009) 289–317. [2] Y. Azar, K. Jain, V.S. Mirrokni, (Almost) optimal coordination mechanisms for unrelated machine scheduling, in: Proceedings of the 19th Annual ACMSIAM Symposium on Discrete Algorithms, 2008, pp. 323–332. [3] G. Christodoulou, E. Koutsoupias, A. Nanavati, Coordination mechanisms, Theoret. Comput. Sci. 410 (2009) 3327–3336.
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