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

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
The Problem of Interest

Target:

Integer release times


Previous results

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
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
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
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


Two machine example of unbounded fractionality

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
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


Schedules comparison

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 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
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
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
Thank you

  • [email protected]


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