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

ARTICLES

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

Jacek Blazewicz^I; Klaus Ecker^II; Tamás Kis^III; Michal Tanas^I

^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 T_i consists of two operations T_i1and T_i2 and a gap between them. During the gap, another task can be processed. Let p_i1 and p_i2 denote the processing times of operations T_i1 and T_i2, respectively.

The gap is exact when operation T_i2 has to start exactly h_iunits of time after the end of operation T_i1, where h_i denotes a length of the gap. In this paper, the only cases considered are those where all h_i are equal, i.e. h_i = h, i = l, 2,..., n.

Precedence constraints of coupled tasks can be strict or weak. T_iT_jmeans that T_i2

T_j1 if precedence constraints are strict, and T_i1

T_j1 L T_i2

T_J2if 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 = p_i1, p₂ = p_i2, h = h_ifor 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|C_max, which means:

• There is one processor in a system.

• Tasks are coupled and identical, with processing times

  p_i

  ₁

  = p_i

  ₂

  = 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

  C_max = max{

  t_j

  ₂}, where

  t_j

  ₂is a completion time of

  T_j(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 ⁺, ''_eAsuch that B/4 < s(aj) < B/2 and such that S Î_A s (a_j) = rB.

Question: Can A be partitioned into r disjoint sets AI, . . . ,A_rsuch that for 1 <i < r, S s(a_j) = B (note that each A_imust 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 = 3

  r

  ² +

  r B

• l = r

• For each item

  a_jÎ

  A, 1

  <

  j

  < 3

  r let us construct graph

  G_jin the following way:

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

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

  3. Create all possible edges (v_x, v_y) such that v_x Î K ,v_x¹ V_jand v_y Î D.

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

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

• G =

  

It is clear that G _jcan 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 v_j.

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 V_jvertices of K, so all vertices of D have to be coloured with the same colour as vertex V_j. So, the set of vertices of any D has to be monochromatic.

Let A₁, . . . , A_jbe a solution for the given instance of 3-Partition problem and let Ai = {a_i(1),a_i(2),a_i(3)}. Let us denote S_i = G_i(1) U G_i(2) U G_i(3) . G can be coloured such that sets D_i and D_jshare the same colour if and only if both D_iand D_jare in the same S_i. 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,...,S_r be a disjoint sets of vertices of graph G, such that V(G) =

S_i, each vertex in S_i is coloured with i-th colour and ''_i |S_i| = 3r + B. Let S be a subset of Si that contains only vertices belonging to D subgraphs. S_i 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<rD Î S_i. Because ''_1<j<3rB/4 < s(a_j) < B/2 each S_i has to contain exactly 3 subgraphs D, J = 1,2,3. So, we candenote A_i = { : 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 | C_max.

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|C_maxproblem in the following way:

• n = q

• h =

  q/l - 1

• For each vertex

  Vi of graph

  G define the coupled task

  T_Í.

• For each arc (

  v_i,V_j) of graph

  G define an arc

  T_i

  

  T_j of precedence constraints in the scheduling problem.

• y = C_max = 2

  n.

Let us assume that a balanced colouring of G exists. Let S₁, S₂, ... , S_Ibe subsets of V(G) such that S_i = V(G), and all vertices that belong to S_iare coloured with the i-thcolour and ''_1<i<I |S_i| = q/I. Sets S_i are partially ordered because vertices of G are partially ordered and G contains all transitive arcs. Schedule sets S_iin accordance to the partial order using the following algorithm:

Algorithm 1

begin

s:=0

repeat

Get a task T_j Î S_i.

Let s be the starting time of the task T_j.

S_i := S_i\T_j

s_i :=s + 1

until S_i¹

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|C_maxis 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.

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*

Partially supported by the KBN grant.

Publication Dates

• Publication in this collection
  16 Dec 2003
• Date of issue
  2001