1. Scheduling tasks from selfish multi-tasks agents Johanne Cohen, CNRS, LRI, Orsay Fanny Pascual, Université Pierre et Marie Curie (UPMC), LIP6, Paris

2. Introduction Scheduling problem with independant (selfish) agents. - K agents, each agent having a set of tasks of different lengths. - Set of m decentralized shared machines. Each agent chooses on which machines he will schedule his tasks. His aim is to minimize the sum of completion times of his own tasks. Aim : Give policies to the machines in order to obtain schedules which – are stable, and - minimize the sum of the completion times of the tasks. 2

3. Coordination mechanism Introduced by Christodoulou et al. [ICALP 2004]. Coordination mechanism = set of scheduling policies, one for each machine. Each policy : –Gives the order of the tasks on the machine, and may introduce idle times. –Is local : it depends on the tasks scheduled on the machine only. A coordination mechanism is stable if, for each instance, there exists a Nash equilibrium: an assignment of the tasks to the machines such that no agent has incentive to move his tasks. 3

4. Example 2 agents A et B A has 2 tasks : and B has 2 tasks: Both machines schedule the tasks by increasing order of lengths (policy SPT). 4 14 23 1 42 3 Nash equilibrium M1M1 M2M2 time 1 4 2 3  Cj= 7,  Cj= 6  Cj= 6,  Cj= 7 B A A B

5. Our problem Aim: design stable coordination mechanim, and of good quality (sum of completion times). Price of anarchy = max Focus on natural coordination mechanism : - policies identical for all the machines - deterministic policies - the ID of the tasks are used only to break the ties 5 Sum completion times in the worst NE Sum completion times in OPT Instances

6. Outline of this talk 1.State of the art 2.The machines do not know the owners of the tasks -No natural mechanism is stable if it does not introduce idle times -Price of anarchy ≥ 2 3.The machines know the owners of the tasks EqualSPTprio coordination mechanism 4.Conclusion and future work 6

7. State of the art Coordination mechanism for agents owning one task: Numerous papers (different social costs, machine environments) [Christodoulou et al, ICALP 2004; Azar et al, SODA 2008; Immorlica et al, TCS 2009 ; Hoeksma et al, WAOA 2011 ; Cole et al, STOC 2011 ; Correa et al, NRL 2012 ; Caragianis, Algorithmica 2013 ; etc.]. Coordination mecha. for agents owning several tasks: [Abed, Correa, Huang, ESA 2014]: - Agent cost = ∑ w i C i, Social cost = ∑ agents costs - Superclass of Nash equilibrium: no agent decreases his cost by moving exactly one task to a different machine. - Policy = length/weight : price of anarchy = 4 7

8. Outline of this talk 1.State of the art 2.The machines do not know the owners of the tasks -No natural coord. mechanism is stable if it does not introduce idle times -Price of anarchy ≥ 2 3.The machines know the owners of the tasks EqualSPTprio coordination mechanism 4.Conclusion and future work 8

9. No natural mechanism is stable if it does not introduce idle times Property : If all the machines use the same deterministic policy which doesn’t use idle times, then there does not always exist a Nash equilibrium. Proof: (m=2) Let 3 tasks s.t. –If A 1 and B 1 are alone on a same machine : < –If B 1 and A 2 are alone on a same machine : < There is no Nash equilibrium. Agent A wants : Agent B wants : 9 B1B1 A2A2 A1A1 B1B1 B1B1 A2A2 M2M2 M1M1 A1A1 B1B1 A2A2 M2M2 M1M1 A1A1 B1B1 A2A2 A1A1

10. The price of anarchy of a natural mechanism is at least 2 Property : If all the machines have the same deterministic policy, then the price of anarchy is at least 2. Proof: m=2, 3 tasks of length 1 s.t. < < Idle times : - If there are 2 tasks of length 1 on the same machine : - If there is one task of length 1 alone on a machine : 10 A1A1 B1B1 B1B1 A2A2 i11i21 i31

11. We distinguish 4 cases: –Case 1: - Case 2: –Case 3: - Case 4: the 3 tasks are together 11 A 1 B1B1 i1i2 i3 A2A2 B 1 A2A2 i1i2 i3A1A1 A 1 A2A2 i1i2 i3 B1B1

12. - Case 1:  Cj= C B1 = i1+1+i2+1 = 2+i1+i2 If B 1 goes on M 2 :  Cj= i1+1 => This not a Nash equilibrium. 12 A 1 B1B1 i1i2 i3 A2A2 B 1 A2A2 i1i2 i3 A1A1 B B

13. We distinguish 4 cases: –Case 1: - Case 2: –Case 3: - Case 4: the 3 tasks are together 13 A 1 B1B1 i1i2 i3 A2A2 B 1 A2A2 i1i2 i3A1A1  Cj= i1+i2+2 If B 1 goes on M 2 :  Cj= i1+1  This not a Nash equilibrium.  Cj= i1+i2+i3+3 Exchange A 1  A 2 :  Cj= i1+i3+2  This not a Nash equilibrium. A 1 A2A2 i1i2 i3 B1B1  Cj= 2 * i1+i2+3 Nash equilibrium only if i1+i3+2  2 * i1+i2+3, i.e. if i3  1+i1+i2 If i3  1 then price of anarchy  2.  Price of anarchy  2

14. Outline of this talk 1.State of the art 2.The machines do not know the owners of the tasks -No natural mechanism is stable if it does not introduce idle times -Price of anarchy ≥ 2 3.The machines know the owners of the tasks EqualSPTprio coordination mechanism 4.Conclusion and future work 14

15. First attempt : PrioSPT PrioSPT coord. mechanism : each machine schedules first the tasks of Agent 1 (in SPT order), and then the tasks of Agent 2, etc. Example : 15 3 23 2 2 1 11 2 3 1 2 3 Agent A 1 : Agent A 2 : Agent A 3 :

16. First attempt : PrioSPT PrioSPT is stable. However, its price of anarchy is unbounded. 16 3 2 3 2 2 1 1 1 2 3 M1M1 M2M2 M3M3 M4M4 M5M5 M6M6 1 2 3 A 1,A 2,A 3

17. EqualPrioSPT coord. mechanism We assume that K ≤ m. EqualPrioSPT : divides the machines in K sets of size  m/K  (or  m/K  +1). The machines of the i-th set schedule - by priority the tasks of agent i (in SPT order), and then - the tasks of agent (i+1) mod m, - and then the tasks of agent (i+2) mod m, etc. 17

18. EqualPrioSPT: example m=6,K=3. 18 323 2 2 1 11 2 3 M1M1 M2M2 M3M3 M4M4 M5M5 M6M6 123 A 1,A 2,A 3 A 2,A 3,A 1 A 3,A 1,A 2 2 3 3 3

19. EqualPrioSPT: properties Prop: EqualPrioSPT is stable. Prop: Its price of anarchy is equal to m /  m/K  Sketch of the proof: 1)PoA ≥ m /  m/K  : example (m=6,K=3). 19 11111166 66 111111 66 66 66 66 11 11 11 66 66 11 11 11 66 66 66 66 Sol. obtained with EqualPrioSPTOptimal solution PoA= (36  2 +54  ) / (12  2 +54  ) 3  ∞ 66

20. EqualPrioSPT: properties 2)PoA ≤ m /  m/K  : cost of A i in a Nash eq. ≤ its cost in X. X = schedule where each agent schedules his tasks with the SPT list algorithm on q =  m/K  machines. Cost of A i in X ≤ (m/q) opt. cost of A i when A i owns m machines   Cost of X ≤ m/q OPT => Cost of NE ≤ m/q OPT 20 X : Lemma: let S j = ∑C i of a set of tasks scheduled with the SPT list algo. on j machines. Let q<m. S q ≤ (m/q) S m. ≤ (m/q) cost of A i in Opt. sol.

21. Conclusion and future work The machines do not know the owners of the tasks: coord. mecha shoud use randomized policies, or different policies, or idle times between tasks (in this case POA ≥ 2).  Does there exist a stable coordination mechanism ? The machines know the owners of the tasks: EqualPrioSPT coord. mechanism (PoA is about K).  Coordination mechanism with a better price of anarchy ? 21