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Journal of the Brazilian Computer Society

Print version ISSN 0104-6500On-line version ISSN 1678-4804

J. Braz. Comp. Soc. vol.7 no.3 Campinas  2001

http://dx.doi.org/10.1590/S0104-65002001000200004 

ARTICLES

 

A note on the complexity of scheduling coupled tasks on a single processor*

 

 

Jacek BlazewiczI; Klaus EckerII; Tamás KisIII; Michal TanasI

IInstitute of Computing Science, PoznanUniversity of Technology, Poznan, Poland
IIInstitute of Computing Science, Technical University Clausthal, Germany
IIIUniversity of Siegen, Germany

 

 


ABSTRACT

This paper considers a problem of coupled task scheduling on one processor, where all processing times are equal to 1, the gap has exact length h, precedence constraints are strict and the criterion is to minimise the schedule length. This problem is introduced e.g. in systems controlling radar operations. We show that the general problem is NP-hard.

Keywords: Scheduling, coupled tasks, NP-hardness.


 

 

1 Introduction

A scheduling problem is, in general, a problem answering a question of how to allocate some resources over time in order to perform a given set of tasks [1]. In practical applications resources are processors, money, manpower, tools, etc. Tasks can be described by a wide range of parameters, like ready times, due dates, relative urgency factors, precedence constraints and many more. Different criteria can be applied to measure the quality of a schedule. The general formulation of scheduling problems and the commonly used notation can be found in books such as [5], [16], [2], [15], [3] and [4]. A survey of the most important results is given in [11].

One branch of scheduling theory is concerned with scheduling of coupled tasks. A task is called coupled if it contains two operations where the second has to be processed some time after a completion of the first one. This problem, described in [17] and [18], often appears in radar-like devices, where the first operation is the transmission of an electromagnetic pulse and the second is the reception of its echo. Several algorithms designed to solve the problem of radar pulse scheduling can be found in [9] and [12].

The complexity of various scheduling problems with coupled tasks has been studied in [13]. Although most of the cases are NP-hard [13], some polynomial algorithms were found in [14].

A coupled task scheduling problem with variable length gap is surveyed in [10] and [7]. NP-hardness of this case is proved in [19], where some interesting connections between coupled tasks and flow shops are also given.

In this note, we complement the above results by presenting the NP-completeness proof for the problem of scheduling coupled tasks on a single processor, with all processing times equal to 1, exact, integer gap length, general strict precedence constraints and the optimisation criteria of minimising the schedule length.

An organisation of the paper is as follows. The problem is formulated in Section 2. The NP-hardness proof is presented in Section 3. We conclude in Section 4.

 

2 Problem formulation

We consider the problem of scheduling n coupled tasks on a single machine, where each coupled task Ti consists of two operations Ti1and Ti2 and a gap between them. During the gap, another task can be processed. Let pi1 and pi2 denote the processing times of operations Ti1 and Ti2, respectively.

The gap is exact when operation Ti2 has to start exactly hi units of time after the end of operation Ti1, where hi denotes a length of the gap. In this paper, the only cases considered are those where all hi are equal, i.e. hi = h, i = l, 2,..., n.

Precedence constraints of coupled tasks can be strict or weak. Ti Tj means that Ti2 Tj1 if precedence constraints are strict, and Ti1Tj1 L Ti2 TJ2if they are weak.

The special case of a coupled task problem involves identical tasks. Commonly, such tasks are denoted by (p1, h,p2), where p1 = pi1, p2 = pi2, h = hi for all 1 < i < n.

Adapting the commonly accepted notation for scheduling problems [8], the scheduling problem considered in this paper can be denoted by 1|(1,h, 1) – coupled, strict prec, exact gap|Cmax, which means:

  • There is one processor in a system.
  • Tasks are coupled and identical, with processing times pi1 = pi2 = 1, ''1<i<n
  • Gaps are exact and have uniform length h.
  • Precedence constrains are strict.
  • The optimisation criterion is to minimise the schedule length Cmax = max{tj2}, where tj2is a completion time of Tj (its second operation).

 

3 NP-hardness of the general case

In case where precedence constraints are general the problem of scheduling coupled tasks on a single processor is NP-hard even for unit processing times. We will prove this by a series of lemmae showing NP-hardness of some intermediate problems.

Lemma 1 Problem of Balanced Colouring of Graphs with Partially Ordered Vertices is NP-hard.

Proof: The problem of Balanced Colouring of Graphs with Partially Ordered Vertices (BCGPOV for short) can be stated as follows: Instance: A directed, acyclic graph G = (V,E) where |V| = q. (It is clear that the arcs define a partial order in set V.)

Question: Can the vertices of G be coloured with l colours such that no pair of adjacent vertices shares the same colour and exactly q/l vertices are coloured with the same colour. (We will call such a colouring the balanced colouring.) Without loss of generality we can assume that q/l Î +.

Firstly, we prove that BCGPOV belongs to class NP. To verify a solution of BCGPOV is enough to verify that

No pair of adjacent vertices is monochromatic. Because each vertex has no more than (q– 1) adjacent vertices the complexity of this step is O(q2).

Exactly q/l vertices are coloured with each colour. Complexity of this step is O(q).

A solution of BCGPOV can be verified in polynomial time, which means that the problem BCGPOV belongs to class NP.

In order to prove NP-completeness of the problem BCGPOV we will use the 3-Partition problem, which is NP-complete in the strong sense according to [6] . The 3-Partition problem is defined as follows: Instance: A collection A of 3r items, bound B Î +, and size s(aj) Î +, ''eA such that B/4 < s(aj) < B/2 and such that S ÎA s (aj) = rB.

Question: Can A be partitioned into r disjoint sets AI, . . . ,Ar such that for 1 < i < r, S s(aj) = B (note that each Ai must contain exactly three elements from A)?

The transformation: For any instance of the 3-Partition problem let us define the corresponding instance of the BCGPOV problem as follows:

  • q = 3r2 + r B
  • l = r
  • For each item aj ÎA, 1< j < 3r let us construct graph Gj in the following way:

    1. Construct complete graph K . on r vertices. Denote one vertex of K . by Vj.

    2. Construct set D of s(aj) independent vertices.

    3. Create all possible edges (vx, vy) such that vx Î K ,vx¹ Vj and vy Î D.

It is clear that edges of graph Gj can be directed to define a partial order in the set of vertices of Gj.

An example of such a graph is shown in Fig 1.

  • G =

 

 

It is clear that G j can be coloured with r colours only in the following way: each vertex from Ka04img05.gif has a different colour and all vertices from D are coloured with the same colour as vertex vj .

No two vertices of any subgraph K, 1 < j < 3r can share the same colour, so after colouring of all K, exactly 3r vertices are coloured with the same colour. All vertices of D are connected with all but Vj vertices of K, so all vertices of D have to be coloured with the same colour as vertex Vj. So, the set of vertices of any D has to be monochromatic.

Let A1, . . . , Aj be a solution for the given instance of 3-Partition problem and let Ai = {ai(1),ai(2),ai(3)}. Let us denote Si = Gi(1) U Gi(2) U Gi(3) . G can be coloured such that sets Di and Dj share the same colour if and only if both Di and Dj are in the same Si. It means that for all i exactly s = B vertices of sets D, so exactly 3r + B vertices of graph G, are coloured with each colour.

On the other hand, let S1,S2,...,Sr be a disjoint sets of vertices of graph G, such that V(G) = Si, each vertex in Si is coloured with i-th colour and ''i |Si| = 3r + B. Let S be a subset of Si that contains only vertices belonging to D subgraphs. Si has to contain 3r vertices belonging to K, so |S| – B for each i. For each feasible colouring each D is monochromatic, so ''1<j<3r $1<i<r D Î Si. Because ''1<j<3r B/4 < s(aj) < B/2 each Si has to contain exactly 3 subgraphs D, J = 1,2,3. So, we candenote Ai = { : j = 1,2,3}, i = 1, 2,..., r and this is the solution of the 3-Partition problem.

The complexity of this transformation is O(r2 + rB). Let us note that this is a pseudopolynomial transformation. On the other hand, 3-Partition is NP-complete, thus, we have the desired result.

Now, we will show that proplem BCGPOV polynomially transforms to our scheduling problem.

Lemma 2 The problem of Balanced Colouring of Graphs with Partially Ordered Vertices polynomially transforms to 1ô (1,h,1) - coupled, strict prec, exact gap | Cmax.

Proof: Let G = (V, E) be an instance of problem BCGPOV. Let it contain all transitive arcs. Let us define a corresponding instance of 1|(1, h, 1) – coupled, strict prec, exact gap|Cmax problem in the following way:

  • n = q
  • h = q/l - 1
  • For each vertex Vi of graph G define the coupled task TÍ.
  • For each arc (vi,Vj) of graph G define an arc Ti Tj of precedence constraints in the scheduling problem.
  • y = Cmax = 2n.

Let us assume that a balanced colouring of G exists. Let S1, S2, ... , SI be subsets of V(G) such that Si = V(G), and all vertices that belong to Si are coloured with the i-thcolour and ''1<i<I |Si| = q/I. Sets Si are partially ordered because vertices of G are partially ordered and G contains all transitive arcs. Schedule sets Si in accordance to the partial order using the following algorithm:

Algorithm 1

begin

s:=0

repeat

Get a task Tj Î Si.

Let s be the starting time of the task Tj.

Si := Si\Tj

si :=s + 1

until Si ¹

end;

The schedule generated in such way is shown in Fig. 2.

 

 

This procedure guarantees that no precedence constraint will be violated. The schedule contains no idle intervals, so its length is y.

On the other hand, let us assume that a schedule of length y for the given instance of the coupled tasks problem exists. The schedule does not contain idle intervals, so it has to be a sequence of segments as shown in Fig. 2. Each segment contains h+1 independent tasks, which means that the corresponding vertices in G are also independent. So, the vertices from one segment can be coloured with the same colour, which means G can be coloured with n/(h + 1) colours such that exactly h + 1 vertices shares the same colour.

Now we can conclude.

Theorem 1 Problem 1 | (1, h, 1) _ coupled,strict prec, exact gap|Cmax is NP-hard.

Proof: Follows immediately Lemmae 1 and 2.

 

4 Conclusions

In the paper, scheduling of coupled tasks has been considered. General precedence constraints resulted in a strong NP-hardness of the problem, even for unit processing times and equal gap lengths for all the tasks. The other cases, especially where precedence constraints are chain-like or tree-shaped are still open.

 

References

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* Partially supported by the KBN grant.

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