Preemptive scheduling of intrees on two processors
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Preemptive Scheduling of Intrees on Two Processors. Coffman, E. G., Jr., Columbia University Matsypura, D., Oron, D., Timkovsky, V. G., University of Sydney Marseille, CIRM, May 12-16, 2008. My Hidden Co-Authors Wife : Natasha Daughter : Linda Anastasia. The Problem of Interest. Target:.

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Preemptive Scheduling of Intrees on Two Processors

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Preemptive Scheduling of Intrees on Two Processors

Coffman, E. G., Jr., Columbia University

Matsypura, D., Oron, D., Timkovsky, V. G., University of Sydney

Marseille, CIRM, May 12-16, 2008


My Hidden Co-Authors

Wife: Natasha

Daughter: Linda Anastasia


The Problem of Interest

Target:

Integer release times


FRAGMENTS

1

3

1

2

1

2

3/2 schedule

Previous Results

O(n2), Baptiste-Timkovsky, 2001

O(n2), Brucker-Hurink-Knust, 2002;

O(n log n), Huo-Leung, 2005

O(n2), Lushchakova, 2006

O(n2), Coffman-Seturaman-Timkovsky, 2003

O(n2), Muntz-Coffman, 1969

Half-Integrality Proof, Sauer-Stone, 1987

  • Any optimal schedule is a concatenation

  • of these fragments

  • Each job is preempted at most once

  • in the middle


Schedule Fractionality

  • The fractionality of a preemptive schedule is the greatest reciprocal 1/k such that the interval between every two event times (i.e., start times, completion times, or preemption times) in the schedule is a multiple of 1/k .

  • Preemptive schedules of fractionality 1/2 are simply half-integer schedules.

  • We say that a preemptive scheduling problem has a bounded fractionality if there exist constant k and optimal preemptive schedules of fractionality 1/k for all its instances, or an unbounded fractionality otherwise.


Fractionality Conjecture

Weak Conjecture [mid 80s]:

This problem has bounded fractionality 1/m

Strong Conjecture [mid 80s]:

This problem has bounded fractionality 1/p(m),

where p is a polynom

BOTH ARE NOT TRUE


Three-Macine Example of Unbounded Fractionality

The Sauer-Stone Theorem [1987]: For any fixed m>2 there exists an instance of this problem, i.e., precedence constraints, whose all optimal schedules are of fractionality 1/mn.

Sauer, N. W., Stone, M. G., Rational preemptive scheduling, Order 4 (1987) 195-206


These three

jobs in the trunk

are needed to

prevent a delay

of all preceding

jobs

release

times

M1

M2

0

Two-Machine Example of Unbounded Fractionality

1

2

3

4

5

6

7

8

3/2 schedule


The Example Description

  • 3n+3 jobs in triplets: Aj, Bj, Cj,

    j = 1,2,…,n,n+1

  • Release times: 0, 0, 0 for j = 1,

  • Release times: 2j-3, 2j-2, 2j-2 for 1< j<n+1

  • Relaese times: 2j-3, 2j-2, 2j-1 for j = n+1

  • Precedence constraints:

    Aj Aj+1, Bj Aj+1, Cj Aj+1, j<n+1 An+1 Bn+1 Cn+1


preemptive

0

1

2

3

4

5

6

7

8

half-integer

0

1

2

3

4

5

6

7

8

nonpreemptive

0

1

2

3

4

5

6

7

8

9

Schedules Comparison


Minimum Maximum and Total Completion Times for the Example

Minimum Maximum and Total Completion Time: Unbounded Fractionality

Minimum Maximum and Total Completion Time: Half-Integer

Minimum Maximum and Total Completion Time: Nonpreemptive


NP Preemption Hypothesis

  • Recognition versions of preemptive problems in the

    classification belong to NP.

  • In other words, there exist solutions to these problems that can be checked in polynomial (in problem size) time.

  • IS THIS TRUE?

  • The problems

    are strong candidates in finding counterexamples to the hypothesis.


Half-Integer Solution

This problem reduces to serching a minimum path

in a directed graph and has an O(n15) algorithm.

What about


Thank you

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