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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

1

3

1

2

1

2

3/2 schedule

Previous ResultsO(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

jobs in the trunk

are needed to

prevent a delay

of all preceding

jobs

release

times

M1

M2

0

Two-Machine Example of Unbounded Fractionality1

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

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

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