• Aucun résultat trouvé

Proof of NP-completeness for a scheduling problem with coupled-tasks and compatibility graph

N/A
N/A
Protected

Academic year: 2021

Partager "Proof of NP-completeness for a scheduling problem with coupled-tasks and compatibility graph"

Copied!
3
0
0

Texte intégral

(1)

HAL Id: lirmm-00262286

https://hal-lirmm.ccsd.cnrs.fr/lirmm-00262286

Submitted on 11 Mar 2008

HAL is a multi-disciplinary open access

archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Proof of NP-completeness for a scheduling problem with

coupled-tasks and compatibility graph

Gilles Simonin

To cite this version:

Gilles Simonin. Proof of NP-completeness for a scheduling problem with coupled-tasks and compati-bility graph. [Research Report] RR-08007, LIRMM. 2008. �lirmm-00262286�

(2)

In this technical rapport, we investigate a scheduling problem with coupled-tasks (denoted ac-quisition tasks) and a compatibility graph. We will show that the problem 1|prec, coupled − task,(pai= pbi= 1, Li= α) ∪ (pTi, pmtn), Gc|Cmax is N P-complete, with α ≥ 3. This problem is

denoted by Π.

Theorem 1 Let n be the acquisition tasks number, the problem to decide if an instance of the problem Π has a scheduling length Cmax = 2n +PTi∈T pTi, is N P-complete.

Proof

Our approach is similar to the proof of Lenstra and Rinnoy Kan [1] for the problem P|prec; pj = 1|Cmax. This demonstration is based on the Clique decision problem (see

Garey and Johnson GT19 [2]):

INSTANCE:A graph G = (V, E) where |V | = n, and an integer K. QUESTION:Can we find a clique of size K in G?

Cmax= 2×5+2×5 = 20 K = α + 1 = 3 pTi = α Li= α = 2 n= 5 A1 A1 A2 A2 A3 A3 A4 A4 A5 A5 T1 T1 T2 T2 T3 T3 T4 T4 T5 T5 a1 b1 a2 a3 a4 b2 b3 b4 a5 b5 The set Gp A graph G

The compatibility Graph Gc

0 1 1 2 2 3 3 4 4 5 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Figure 1: Illustration of polynomial-time transformation Clique∝ Π

Our proof is based on the polynomial-time transformation Clique ∝ Π. It is easy to see that the problem Π is in N P.

Let I∗ an instance of Clique, we will construct an instance I of Π with Cmax= 2n +

X

Ti∈T

pTi in

the following way:

Let G = (V, E) a graph in the instance I, with |V | = n:

• ∀v ∈ V , an acquisition task Av is introduced, composed of two sub-tasks av and a′v with

processing time pav = pa′v = 1 and with a latency time, between these two sub-tasks, of

length α = (K − 1), called slot.

• For each edge e = (v, w) ∈ E, there is a compatibility relation between the two acquisition tasks Av and Aw.

• For each task Av, we introduce a treatment task Tv which is its successor.

(3)

• Each Tv has a processing time noted pTv = α. Thus, the treatment tasks will replace all

the inactivity slot of all the Av after the clique.

• We suppose that there is a clique of length K = (α + 1) in the graph G. Let us show that there is a scheduling in Cmax = (2n +PTi∈T pTi) units of time. For that, consider the

following scheduling:

– From t = 0 to t = α, we schedule the K = (α + 1) tasks which represent the vertices of the clique of size K.

– From t = (2α + 2), we schedule the (n − K) remaining tasks Av.

– In each slot from these (n − K) tasks Av, we schedule the tasks Tv. Since each Tv has

as a value pTv = α, by scheduling (n − K) tasks Tv, we will fill each slot of length α

of the (n − K) tasks Av.

– Remeaning treatment tasks are scheduled at the end of the schedule.

With this allocation, we fill all the slots and we give a valid scheduling in (2n +PTi∈T pTi)

units of time.

• Reciprocally, let us suppose that there is a scheduling in (2n +PTi∈T pTi) units of time

without inactivity time, then let us show that the graph G contains a clique of size K = (α + 1).

From these suppositions, we make essential comments:

– With the precedence constraints between the tasks Av and Tv, it is easy to see that

we can schedule only tasks Av at t = 0, ∀v ∈ V . Thus, the first treatment task could

be scheduled only starting from t = (α + 2).

– Let ap1 be the first sub-task of acquisition scheduled at t = 0, with a slot of length α.

We need a clique of size (α + 1) to obtain a scheduling without inactivity slot. Thus we have (α + 1) acquisition tasks which are compatibles. And in the compatibility graph Gc, we will have an edge between each couple of these tasks Av. Consequently, the tasks

Ap1, Ap2, . . . , Apα, associated to the vertices of the graph G, form a clique of size K = (α + 1).

This concludes our proof of Theorem 1.

 Bibliographie

1. R.L. Graham, E.L. Lawler, J.K. Lenstra, and A.H.G. Rinnooy Kan. Optimization and approximation in deterministic sequencing and scheduling: a survey. Annals of Discrete Mathematics, 5:287–326, 1979.

2. M.R. Garey and D.S. Johnson. Computers and Intractability; A Guide to the Theory of NP-Completeness. W. H. Freeman & Co., New York, NY, USA, 1990.

Figure

Figure 1: Illustration of polynomial-time transformation Clique ∝ Π

Références

Documents relatifs

To test whether the vesicular pool of Atat1 promotes the acetyl- ation of -tubulin in MTs, we isolated subcellular fractions from newborn mouse cortices and then assessed

Néanmoins, la dualité des acides (Lewis et Bronsted) est un système dispendieux, dont le recyclage est une opération complexe et par conséquent difficilement applicable à

Cette mutation familiale du gène MME est une substitution d’une base guanine par une base adenine sur le chromosome 3q25.2, ce qui induit un remplacement d’un acide aminé cystéine

En ouvrant cette page avec Netscape composer, vous verrez que le cadre prévu pour accueillir le panoramique a une taille déterminée, choisie par les concepteurs des hyperpaysages

Chaque séance durera deux heures, mais dans la seconde, seule la première heure sera consacrée à l'expérimentation décrite ici ; durant la seconde, les élèves travailleront sur

A time-varying respiratory elastance model is developed with a negative elastic component (E demand ), to describe the driving pressure generated during a patient initiated

The aim of this study was to assess, in three experimental fields representative of the various topoclimatological zones of Luxembourg, the impact of timing of fungicide

Attention to a relation ontology [...] refocuses security discourses to better reflect and appreciate three forms of interconnection that are not sufficiently attended to