preemptive scheduling of intrees on two processors
DESCRIPTION
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:. - PowerPoint PPT PresentationTRANSCRIPT
Preemptive Scheduling of Intrees on Two
Processors
Coffman, E. G., Jr., Columbia UniversityMatsypura, D., Oron, D., Timkovsky, V. G.,
University of Sydney
Marseille, CIRM, May 12-16, 2008
My Hidden Co-Authors
Wife: NatashaDaughter: Linda Anastasia
The Problem of Interest
jjj CprP |1,,precpmtn,|
Target:
jjj CprP |1,,precpmtn,|2
Integer release times
Previous Results
jjj CprP |1,,outtree,pmtn|O(n2), Baptiste-Timkovsky, 2001O(n2), Brucker-Hurink-Knust, 2002; O(n log n), Huo-Leung, 2005
jjj CpprP |,,outtreepmtn,|2 O(n2), Lushchakova, 2006
jj CpP |1,precpmtn,|22
1O(n2), Coffman-Seturaman-Timkovsky, 2003
max2
1 |1,prec,pmtn|2 CpP j O(n2), Muntz-Coffman, 1969Half-Integrality Proof, Sauer-Stone, 1987
3/2 schedule
1
2 1
3
2
1
FRAGMENTS•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
max|1,precpmtn,| CpPm j
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
3,|1,precpmtn,| max mCpPm j
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
0
01A
Two-Machine Example of Unbounded Fractionality
33A
43B
43C
54A
64B
74C
01A
01B
01C
01B
01C
1 2 3 4 5 6 7 8
22B
22C
12A
22B
22C
12A
2
11
8
15
4
13
43B
33A
43C
54A
74C
8
16
8
17
8
18
01A
64B
12
122
132
1
releasetimes
3/2 schedule
These threejobs in the trunkare needed toprevent a delay of all precedingjobs
M1
M2
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
nonpreemptive
half-integer
preemptive
Schedules Comparison
3
2
2 3
2
2
11
2
11
5
4
4 5
8
15
2
15
6 7
6
8
16
6
2
16
7
7
8
17
2
17
4
4
13
2
13
32
12
4
12
52
14
8
14
0 1
1
0 1
0
2
1
2
1
8
8
8
18
8
2
18
9
Minimum Maximum and Total Completion Times for the Example
nnn
2
1563 2
nn
2
122
Minimum Maximum and Total Completion Time: Unbounded Fractionality
Minimum Maximum and Total Completion Time: Half-Integer
Minimum Maximum and Total Completion Time: Nonpreemptive
2
122 n
2
1563 2 nn
673 2 nn32 n
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.
||
max|1,prec,pmtn|3 CpP j
max|1,,intree,pmtn|2 CprP jj
Half-Integer Solution
jjj CprP |1,,precpmtn,|22
1
This problem reduces to serching a minimum pathin a directed graph and has an O(n15) algorithm.
What about
?|1,,precpmtn,|23
1 jjj CprP
?|1,,precpmtn,|24
1 jjj CprP
?|1,,precpmtn,|28
1 jjj CprP
