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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 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 • v.timkovsky@econ.usyd.edu.au