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MISTA 2005. Properties of SPT schedules. Eric Angel, Evripidis Bampis, Fanny Pascual LaMI, university of Evry, France. Outline. Definition of an SPT schedule Quality of SPT schedules on these criteria: Min. Max ∑Cj: minimization of the maximum sum of completion times per machine.
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MISTA 2005 Properties of SPT schedules Eric Angel, Evripidis Bampis, Fanny Pascual LaMI, university of Evry, France
Outline • Definition of an SPT schedule • Quality of SPT schedules on these criteria: • Min. Max ∑Cj: minimization of the maximum sum of completion times per machine. • Fairness measure • Conclusion
Model 0 1 2 3 4 5 6 7 8 9 10 • Example: • Cj = completion time of task j. (e.g. C3=4) • Main quality criteria: • Makespan -> (P||Cmax) • Sum of completion times ( ∑Cj )-> (P|| ∑Cj ) 1 5 6 P1 n tasks m machines 2 4 P2 P3 3 time
SPT schedules 1 4 7 2 5 8 3 6 • SPT= Shortest Processing Time first Smith’s rule: SPTgreedy • Sort tasks in order of increasing lengths. • Schedule them as soon as a machine is available. • Algo which minimizes ∑Cj . • Class of the schedules which minimize ∑Cj : [Bruno et al]: Algorithms for minimizing mean flow time
Rank 1: Rank 2: Rank 3: 1 1 1 1 1 1 4 4 4 4 4 7 7 7 7 7 7 2 2 2 2 2 2 5 5 5 5 5 8 8 8 8 8 8 3 3 6 6 6 6 6 6 3 3 3 1 4 8 1 4 7 2 3 6 2 5 6 5 7 3 8 SPT schedules • [Bruno et al]: notion of rank. • A schedule minimizes ∑Cj iff it is an SPT schedule. The tasks of rank i are counted i times in the ∑Cj : ∑Cj= C1 + C4 + C7 + C2 + C5 + C8 + … = l(1) + (l(1)+ l(4) ) + ( l(1)+l(4)+ l(7) ) + … = 3 l(1) + 2 l(4) + l(7) + … machine 1
Outline • Definition of an SPT schedule • Quality of SPT schedules on these criteria: • Min. Max ∑Cj: minimization of the maximum sum of completion times per machine. • Example • NP-complete problem • Analysis of SPTgreedy • Fairness measure • Conclusion
Minimization of Max ∑Cj i {1,…,m} j on Pi 6 3 1 1 P1 P1 5 1 5 P2 7 P2 1 1 1 5 • Pb = Minimization of Max∑ Cj • To minimize Max ∑Cj To minimize ∑Cj Max ∑Cj = 7 Max ∑Cj = 6 ∑Cj = 10 ∑Cj = 11 • NP-complete problem.
To minimize Max ∑Cj is an NP-complete problem • We reduce the partition problem into a Min. Max ∑Cj problem. • Partition: Let C={ x1, x2, . . . , xn } be a set of numbers. Does there exist a partition (A,B) of C such that ∑xA x = ∑xB x ? • Min. Max ∑Cj: Let n tasks and m machines, and let k be a number. Does there exist a schedule such that Max ∑Cj= k ?
Min Max ∑Cj is NP-complete , , , , , x2 x x2 x2 x x x2 x x x x x2 x2 x x x 1 1 1 1 1 1 1 1 1 1 + + + + + + + + x x 3 3 3 3 3 2 3 3 2 2 3 3 3 2 3 2 2 3 x2 x3 x1 P1 P2 • Transformation: • Partition: C={x1, x2, . . . , xn} • Min. Max ∑Cj: k= ½ Min ∑Cj ;2 machines;2n tasks • Example: C={ x1, x2, x3 } Tasks = Claim: Solution of (Min. Max ∑Cj) ∑Cj = ∑Cj = k = ½ Min ∑Cj P1 P2 ≠ce contrib ∑Cj = + x3 x1 + x2 = x3 + x1 + x2
Min Max ∑Cj is NP-complete • transformation: • Partition: C={x1, x2, . . . , xn}. • Min. Max ∑Cj: k= ½ Min ∑Cj ; 2 machines; 2n tasks. n n - 1 n - 2 ... 1
Min. Max ∑Cj : analysis of SPTgreedy • Theorem 1 : • The approx. ratio of SPTgreedy is ≤ 3 – 3/m + 1/m2 . • Theorem 2 : • The approx. ratio of SPTgreedy is ≥ 2 – 2/(m2 + m).
Min. Max ∑Cj : analysis of SPTgreedy 1 1 6 1 1 1 1 1 1 1 1 1 1 6 • Theorem 2 : • The approx. ratio of SPTgreedy is ≥ 2 – 2/(m2 + m). ( example: for m=3, ratio ≥11/6 ) • Proof: • m(m-1) tasks of length 1 • A task of length B= m(m+1)/2 • Example for m=3: Max ∑Cj = 6 Max ∑Cj = 11
Outline • Definition of an SPT schedule • Quality of SPT schedules on these criteria: • Min. Max ∑Cj. • Fairness measure. • Conclusion
Fairness measure 1 2 4 Vector X = (1, 3, 4) • [Kumar, Kleinberg]: Fairness Measures For Ressources Allocation (FOCS 2000) • Definition: global approx ratio of a schedule S: • Max. ratio between the completion time of the ith task of S, and the min. completion time of the ith task of any other schedule. • I = instance; X = (sorted) vector of completion times • C(X) = min s.t. X Y Y= feasible schedule of I • C*(I) = min C(X) s.t. X = feasible schedule of I • C*= max C*(I)
Fairness measure I={ , , } 1 2 3 1 3 2 2 1 1 2 3 3 • Possible vectors: X + • (1, 2, 5) • (1, 3, 3) • (1, 3, 5) • (2, 3, 3) • (2, 3, 4) • (1, 3, 6) • (1, 4, 6) • (2, 3, 6) • (2, 5, 6) • (3, 4, 6) • (3, 5, 6) • Min = (1, 2, 3) • Example: Vector X = (1, 2, 4) C(X) = 4/3 C*(I) = 4/3
Fairness measure 1 2 1 • Theorem 1: • C(XSPTgreedy) ≤ 2 – 1/m. ( example: for m=2, C(XSPTgreedy) ≤ 3/2 ) • Theorem 2: • C*= 3/2 when m=2. • Proof of theorem 2: C(I) = C* = 3/2 Vector X = (1, 1, 3)
Conclusion – Future work • Conclusion • Minimization ofMax ∑Cj = NP-complete pb. • SPTgreedy between 2 – 2/(m2 + m) and 3 – 3/m + 1/m2 for Min. Max ∑Cj. • Good fairness measure for SPTgreedy. • Future work • A better bound for SPTgreedy for Min. Max ∑Cj. • Study of fairness measure on other problems.