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

Thank you

v.timkovsky@econ.usyd.edu.au