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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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Properties of SPT schedules

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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

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• Example:

• Cj = completion time of task j. (e.g. C3=4)

• Main quality criteria:

• Makespan -> (P||Cmax)

• Sum of completion times ( ∑Cj )-> (P|| ∑Cj )

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### SPT schedules

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• 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:

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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

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• 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 ∑xA x = ∑xB 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

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• Transformation:

• Partition: C={x1, x2, . . . , xn}

• Min. Max ∑Cj: k= ½ Min ∑Cj ;2 machines;2n tasks

• Example: C={ x1, x2, x3 }

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.

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### 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

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• 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

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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={ , , }

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• 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

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• 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.