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

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My Hidden Co-Authors Wife Natasha Daughter
Linda Anastasia
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The Problem of Interest
Target
Integer release times
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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

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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.

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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
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Three-Macine Example of Unbounded Fractionality
The Sauer-Stone Theorem 1987 For any fixed mgt2
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
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Two-Machine Example of Unbounded Fractionality
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3/2 schedule
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The Example Description

• 3n3 jobs in triplets Aj, Bj, Cj,
• j 1,2,,n,n1
• Release times 0, 0, 0 for j 1,
• Release times 2j-3, 2j-2, 2j-2 for 1lt jltn1
• Relaese times 2j-3, 2j-2, 2j-1 for j n1
• Precedence constraints
• Aj ? Aj1, Bj ? Aj1, Cj ? Aj1, jltn1
  An1 ? Bn1 ? Cn1

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Schedules Comparison
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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
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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.

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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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Thank you

• v.timkovsky_at_econ.usyd.edu.au