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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

- 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

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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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P1

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m machines

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time

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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

- 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

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.

- 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 ?

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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

- transformation:
- Partition: C={x1, x2, . . . , xn}.
- Min. Max ∑Cj: k= ½ Min ∑Cj ; 2 machines; 2n tasks.

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- Theorem 1 :
- The approx. ratio of SPTgreedy is
≤ 3 – 3/m + 1/m2 .

- The approx. ratio of SPTgreedy is
- Theorem 2 :
- The approx. ratio of SPTgreedy is
≥ 2 – 2/(m2 + m).

- The approx. ratio of SPTgreedy is

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- Theorem 2 :
- The approx. ratio of SPTgreedy is ≥ 2 – 2/(m2 + m).
( example: for m=3, ratio ≥11/6 )

- The approx. ratio of SPTgreedy is ≥ 2 – 2/(m2 + m).
- 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

- Definition of an SPT schedule
- Quality of SPT schedules on these criteria:
- Min. Max ∑Cj.
- Fairness measure.
- Conclusion

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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)

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

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- Theorem 1:
- C(XSPTgreedy) ≤ 2 – 1/m.
( example: for m=2, C(XSPTgreedy) ≤ 3/2 )

- C(XSPTgreedy) ≤ 2 – 1/m.
- Theorem 2:
- C*= 3/2 when m=2.

- Proof of theorem 2:

C(I) = C* = 3/2

Vector X = (1, 1, 3)

- 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.