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.

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Properties of SPT schedules

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Properties of spt schedules

MISTA 2005

Properties of SPT schedules

Eric Angel, Evripidis Bampis, Fanny Pascual

LaMI, university of Evry, France


Outline

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

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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5

6

P1

n tasks

m machines

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4

P2

P3

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time


Spt schedules

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


Properties of spt schedules

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


Outline1

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

Minimization of Max ∑Cj

i  {1,…,m}

j on Pi

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3

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1

P1

P1

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P2

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P2

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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

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

Min Max ∑Cj is NP-complete

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,

,

,

,

x2

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x2

x2

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x2

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x2

x2

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x2

x3

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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 }

    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 complete1

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 spt greedy

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 spt greedy1

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


Outline2

Outline

  • Definition of an SPT schedule

  • Quality of SPT schedules on these criteria:

  • Min. Max ∑Cj.

  • Fairness measure.

  • Conclusion


Fairness measure

Fairness measure

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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 measure1

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 measure2

Fairness measure

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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 – 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.


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