170 likes | 287 Views
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.
E N D
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.