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Computers & Operations Research 34 (2007) 19882000www.elsevier.com/locate/cor

Single machine scheduling problems with resource dependentrelease times

Byung-Cheon Choia, Suk-HunYoonb,, Sung-Jin ChungaaDepartment of Industrial Engineering, Seoul National University, Seoul, Republic of Korea

bDepartment of Industrial and Information Systems Engineering, Soongsil University, Seoul, Republic of KoreaAvailable online 6 October 2005

Abstract

Weconsider two singlemachine scheduling problemswith resource dependent release times that can be controlledby a non-increasing convex resource consumption function. In the rst problem, the objective is to minimize thetotal resource consumption with a constraint on the sum of job completion times.We show that a recognition versionof the problem is NP-complete. In the second problem, the objective is to minimize the weighted total resourceconsumption and sum of job completion times with an initial release time greater than the total processing times.We provide some optimality conditions and show that the problem is polynomially solvable. 2005 Elsevier Ltd. All rights reserved.

Keywords: Single machine scheduling; Resource dependent release time; NP-completeness

1. Introduction

In single machine scheduling problems, it is generally assumed that job release times are known andconstant. However, in reality the release times can be varied such as in steel plants where jobs need tobe preprocessed before they undergo processing. These preprocessing times can be considered as jobrelease times, and they depend on the resource consumed for the preprocessing treatment. Ingots must bepreheated by gas in soaking pits to the required temperature, before they can be hot-rolled by a bloomingmill. The preheating time of an ingots is a non-increasing function of the amount of gas consumed.

Corresponding author. Tel.: +82 2 820 0687; fax: +82 2 825 1094.E-mail address: yoon@ssu.ac.kr (S.-H.Yoon).

0305-0548/$ - see front matter 2005 Elsevier Ltd. All rights reserved.doi:10.1016/j.cor.2005.06.021

B.-C. Choi et al. / Computers & Operations Research 34 (2007) 19882000 1989

Thus, the preheating time of an ingot may be treated as the release time at which the ingot is availablefor the jobs of ingot rolling [2,46,10].

Scheduling problems with resource dependent release times have been studied mainly in single ma-chine environments [1,2]. Janiak [3] considered a single machine scheduling problem in which theobjective was to minimize the maximal job completion time (makespan) subject to the total re-source consumption. He assumed that all jobs had a common resource consumption function andshowed that the problem could be efciently solved by ordering jobs according to non-increasingprocessing times.

Cheng and Janiak [4] considered a problem that exchanged the objective functionwith the resource con-straint in the Janiak problem [3]. Janiak [5] generalized the research of Janiak [3] by considering differentresource consumption functions, and showed that the problem was strongly NP-hard. For the problemof Janiak [5], Janiak [6] presented some polynomially solvable cases and proposed some approximationalgorithms with good worst-case performance ratios.

Janiak and Li [7] considered a single machine scheduling problem with release times that dependedon the convex function of the amount of resources consumed. The objective was to minimize the totalweighted completion time subjective to the total resource consumed. They showed that the problem wasstronglyNP-hard. Li [8] considered a singlemachine scheduling problemwith release times that dependedon a non-increasing function of the amount of resources consumed. The objective was to minimize thetotal resource consumption with a constraint on the sum of job completion times. He showed that theproblem was NP-hard. However, he remarked that when the resource consumption function was convex,the computational complexity remained open. For the problem of Li [8], Vasilev and Foote [9] providedseveral optimality properties.

Li [10] considered a single machine scheduling problem of minimizing the total resource consumptionplus the sum of job completion times, and showed that the problem was NP-hard even for an identicalpiecewise-linear resource consumption function.

In this paper, we consider two single machine scheduling problems with release times that dependon a non-increasing convex function of resources consumed. The objective of the rst problem is tominimize the total resource consumption with a constraint on the sum of job completion times. Theobjective of the second problem is to minimize the weighted total resource consumption and sum of jobcompletion times.

The rest of this paper is organized as follows. In Section 2, our two problems are dened formally. InSection 3, we show that the recognition version of the rst problem is NP-complete. In Section 4, weprovide some optimality properties for the second problem and show that the problem can be solved inpolynomial time. Finally, we provide the summary and concluding remarks.

2. Notation and problem denition

Let Jj represent job j. For job j, let pj be the processing time, Cj the completion time, and rj theactual release time. Let v be an initial release time of jobs and f (rj ) a resource consumption functionof rj . Let and be the weight of the total resource consumption function and the sum of job com-pletion times, respectively. Let = {(1), . . . , (n)} be a job sequence, where (j) = k implies thatjob k is positioned jth in the sequence. Two problems are considered in this paper. They are denedas below.

1990 B.-C. Choi et al. / Computers & Operations Research 34 (2007) 19882000

Fig. 1. Non-increasing convex resource consumption function.

Problem P1. In an n-job, single machine with resource dependent release times, determine (r, )such that

Minn

j=1f (r(j)) =

nj=1

max{v r(j), 0}

s.t.n

j=1(r(j) + p(j))E,

where f (rj ) is the non-increasing convex function shown in Fig. 1 and E is a threshold value.

Problem P2. In an n-job, single machine with resource dependent release times, determine (r, ) thatminimizes K(r, ), dened as

K(r, ) = n

j=1max{v r(j), 0} +

nj=1

(r(j) + p(j)),

where vn

j=1 pj .

Note that Problem P2 is NP-hard if there is no limitation for v, an initial release time of jobs [10]. InProblemP1 andP2,

nj=1 (r(j)+p(j)) is the sumof job completion times. Sincef (rj ) is non-increasing,

we assume that each job starts as soon as it is released.

3. NP-completeness of Problem P1

In this section, we dene the signed knapsack problem and permutation integer problem, and provethat both problems are NP-complete. We show that the recognition version of Problem P1 is NP-

B.-C. Choi et al. / Computers & Operations Research 34 (2007) 19882000 1991

complete. We will prove signed knapsack problem by reducing it from knapsack problem which is NP-complete [11].

Knapsack Problem (KP): Given positive integers aj for j=1, . . . , n andA, are there integers xj {0, 1}such that

nj=1 ajxj = A?

Signed Knapsack Problem (SKP): Given positive integers bj for j = 1, . . . , n and B, are there integersyj {1, 1} such thatnj=1 bjyj = B?Theorem 1. SKP is NP-complete.

Proof. It is clear that SKP is in NP. Given an arbitrary instance of KP, we construct an instance of SKPas follows : Let bj = aj for j = 1, . . . , n and B = 2Anj=1 aj . This can be constructed in polynomialtime. We show that the answer to KP is yes if and only if the answer to SKP is yes.

Suppose that there is a solution x such thatn

j=1 aj xj = A, where xj {0, 1}. If xj = 0, let yj = 1and otherwise, let yj = 1. It implies that yj = 2xj 1.

nj=1

bj yj =n

j=1aj (2xj 1) = 2

nj=1

aj xj n

j=1aj = 2A

nj=1

aj = B.

Thus, y is a solution of SKP.Suppose that there is a solution y such that

nj=1 bj yj =B, where yj {1, 1}. If yj =1, let xj =0

and otherwise, let xj = 1. It implies that xj = (yj + 1)/2.

nj=1

aj xj =n

j=1aj

(yj + 1)2

= 12

n

j=1bj yj +

nj=1

aj

= 1

2

B + n

j=1aj

= A.

Thus, x is a solution of KP. This completes proof.

Denition 1. A permutation solution is an integer vector x = (x1, . . . , xn)T such that 1xi = xj n,i = j .

Permutation integer problem (PIP): Given positive integers dj for j = 1, . . . , m and D, is there apermutation solution x such that

mj=1 djxj = D?

Theorem 2. PIP is NP-complete.

Proof. It is clear that PIP is in NP. Given an arbitrary instance of SKP, we construct an instance of PIPas follows: there are positive integers dj for j = 1, . . . , m, m = 2n such that d2j1 = Mj bj andd2j =Mj +bj for j =1, . . . , n, andD=nj=1 (4j 1)Mj +B, whereM =Bn. This can be constructedin polynomial time. We show that the answer to PIP is yes if and only if the answer to SKP is yes.

Suppose that there is a solution y such thatn

j=1 bj yj = B, where yj {1, 1}. If yj = 1, letx2j1 = 2j 1 and x2j = 2j and otherwise, let x2j1 = 2j and x2j = 2j 1, for j = 1, . . . , n.

1992 B.-C. Choi et al. / Computers & Operations Research 34 (2007) 19882000

Thus, x2j1 + x2j = yj and x2j1 + x2j = 4j 1.m

j=1dj xj =

nj=1

[(Mj bj )x2j1 + (Mj + bj )x2j ]

=n

j=1(x2j1 + x2j )Mj +

nj=1

(x2j1 + x2j )bj

=n

j=1(4j 1)Mj +

nj=1

bj yj =n

j=1(4j 1)Mj + B = D.

Thus, x is the permutation solution of PIP.Suppose that there is a permutation solution x such that

2nj=1 dj xj = D.

2nj=1

dj xj =n

j=1[(Mj bj )x2j1 + (Mj + bj )x2j ]

=n

j=1(x2j1 + x2j )Mj +

nj=1

(x2j1 + x2j )bj =n

j=1(4j 1)Mj + B.

Since D =nj=1 (4j 1)Mj +B and M is sufciently large, we can infer from the above relations thatthe following equations hold:

nj=1

(x2j1 + x2j )Mj =n

j=1(4j 1)Mj , (1)

nj=1

(x2j1 + x2j )bj = B. (2)

From (1), for j = n, x2n1 = 2n 1, x2n = 2n or x2n1 = 2n, x2n = 2n 1. Likewise, for jn 1,x2j1 = 2j 1, x2j = 2j or x2j1 = 2j , x2j = 2j 1. It implies that x2j1 + x2j = 1. Letyj = x2j1 + x2j . Then y is a solution of SKP from (2). This completes proof.

Let C(r, )=nj=1 (r(j) +p(j)) be the sum of job completion times of solution (r, ) and F(r, )=nj=1 f (r(j)) the total resource consumption of solution (r, ). The recognition version of Problem P1

can be stated as follows:Recognition version ofProblemP1:Givenprocessing timespj for j=1, . . . , n, a non-increasing convex

function f (rj ), and integers E and G, is there a solution (r, ) such that C(r, )E and F(r, )G?Given an arbitrary instance of PIP, we construct an instance of the recognition version of Problem P1

as follows: there are n+ 1 jobs. The processing times are pj = dj for j = 1, . . . , n and pn+1 =H , whereH >

nj=1 dj +D.A non-increasing convex function f (rk)=max{

nj=1 dj rk, 0} for k=1, . . . , n+1.

E =nj=1 dj + D + H and G = (n + 1)nj=1 dj D.We show that the answer to PIP is yes if and only if there is a solution (r, ) such that C(r, )E and

F(r, )G for the instance of the recognition version of Problem P1.

B.-C. Choi et al. / Computers & Operations Research 34 (2007) 19882000 1993

Lemma 1. If there is a permutation solution x such thatnj=1 dj xj = D, then there is a solution (r, )such that C(r, )E and F(r, )G.

Proof. We construct = ((1), . . . , (n+1)) in which r(1) =0 and r(k+1) = r(k) +p(k), k=1, , n:If xj = k for j = 1, . . . , n, position job j the (n k + 1)th in the sequence and job n + 1 the last.

Then,

C(r, ) =n+1k=1

kj=1

p(j) =n

k=1kp(nk+1) +

n+1k=1

p(k).

Since pj = dj , pn+1 = H , (n k + 1) = j and xj = k,

C(r, ) =n

j=1dj xj +

nj=1

dj + H = D +n

j=1dj + H = E. (3)

Thus, C(r, ) = E. Also, from (3) the following can be derived:

nk=1

kj=1

p(j) = D.

Since r(j) p(l) =H >

nj=1 dj + D,

C(r, )n+1k=1

kj=1

p(j)p(l) +n+1j=1

p(j) >H +n

j=1dj + D = E.

This is a contradiction.

Claim 2. There should be no idle time in (r, ).

Proof. Let = (1,2, . . . ,n+1) and = (1, (1), . . . ,j , (j), . . . ,n+1, (n + 1)), where j isthe idle time before job (j) starts. Suppose that = 0 is a non-negative vector. By Claim 1, job (n+ 1)is positioned last. Since

n+1j=1 p(j) =

nj=1 dj + H and C(r, )E = D +

nj=1 dj + H ,

C(r, ) =n+1k=1

kj=1

(j + p(j))

=n+1k=1

kj=1

j +n

k=1

kj=1

p(j) +n

j=1dj + H D +

nj=1

dj + H . (4)

Sincen+1

j=1 j > 0, we derive the following inequality from (4):

nk=1

kj=1

(j + p(j))n

j=1 dj ,

F(r, ) =n+1k=1

max

nj=1

dj r(k), 0

=n

k=1max

nj=1

dj r(k), 0

nk=1

n

j=1dj r(k)

=n

k=1

n

j=1dj

kj=1

(j + p(j)) + p(k)

= (n + 1)n

j=1dj

nk=1

kj=1

(j + p(j)).

B.-C. Choi et al. / Computers & Operations Research 34 (2007) 19882000 1995

From Eq. (5),

F()> (n + 1)n

j=1dj D = G.

This is a contradiction.

By Claim 1 and 2, in a solution (r, ) such that C(r, )E and F(r, )G, there are not intermittentidle times and job (n + 1) is positioned last. Thus,

0C(r, ) E =n

k=1

kj=1

p(j) +n

j=1dj + H

D + n

j=1dj + H

=n

k=1

kj=1

p(j) D.

Thus,n

k=1k

j=1 p(j)D.Also, by Claim 1 and 2,

0F(r, ) G =n+1k=1

max

nj=1

dj r(k), 0

(n + 1) n

j=1dj D

=n

k=1

n

j=1dj

k

j=1p(j) p(k)

(n + 1) n

j=1dj + D

= n

k=1

kj=1

p(j) + D.

Thus,n

k=1k

j=1 p(j)D.From the above two equations,

nk=1

kj=1

p(j) = D.

For l = 1, . . . , n, suppose job l is positioned (n k + 1)th in . Letting xl = k,n

j=1dj xj =

nk=1

kp(nk+1) =n

k=1

kj=1

p(j) = D.

Thus, there is a permutation solution x such thatn

j=1 djxj = D. This completes proof.

1996 B.-C. Choi et al. / Computers & Operations Research 34 (2007) 19882000

Theorem 3. Recognition version of Problem P1 is NP-complete.

Proof. It is clear that recognition version of Problem P1 is in NP. For any instance of PIP, we can constructan instance of the recognition version of Problem P1 in polynomial time. Theorem 3 immediately followsfrom Lemmas 1 and 2.

4. Optimality conditions for Problem P2

Next two lemmas discuss some optimality conditions for Problem P2.

Lemma 3. Intermittent idle times can be eliminated in an optimal solution without costs.

Proof. Suppose that there exists an optimal solution (r, ) such that r(k) + p(k) < r(k+1).Case 1: r(k+1)v.We can construct a new solution (r, ) by letting r(k) = r(k) + , and a new solution (r, ) by

letting r(k+1) = r(k+1) , where > 0:

K(r, ) = K(r, ) ( ), (6)

K(r, ) = K(r, ) + ( ). (7)

From (6) and (7), min{K(r, ),K(r, )}K(r, ). This implies that one of left and right-shiftingjobs cannot make a new solution worse until there are no intermittent idle times between the jobs.

Case 2: r(k+1) > v.We can make a new solution (r, ) by letting r(k+1) = r(k+1) , where > 0.

K(r, ) = K(r, ) . (8)

From (8), K(r, )

B.-C. Choi et al. / Computers & Operations Research 34 (2007) 19882000 1997

Fig. 2. Two solutions.

Proof. By Lemma 3, we only consider an optimal solution without idle times. Then, both conditions (a)and (b) cannot be true at a time. Suppose that there exists an optimal solution (r, ) for which conditions(a) and (b) are not true. Two cases can be considered as shown in Fig. 2.

Case 1: r(1) +n

i=1p(i) < v.In this case, the objective function K(r, ) can be rewritten as

K(r, ) = n

j=1(v r(j)) +

nr(1) +

ni=1

ij=1

p(j)

.

We can construct a new solution (r, ) by letting r(j) = r(j) + , j = 1, . . . , n and a new solution(r, ) by letting r(j) = r(j) , j = 1, , n, > 0, from the optimal solution (r, ):

K(r, ) =...