Single machine scheduling problems with resource dependent release times

Computers & Operations Research 34 (2007) 1988–2000www.elsevier.com/locate/cor

Single machine scheduling problems with resource dependentrelease times

Byung-Cheon Choia, Suk-Hun Yoonb,∗, Sung-Jin Chunga

aDepartment of Industrial Engineering, Seoul National University, Seoul, Republic of KoreabDepartment of Industrial and Information Systems Engineering, Soongsil University, Seoul, Republic of Korea

Available online 6 October 2005

Abstract

We consider two single machine scheduling problems with resource dependent release times that can be controlledby a non-increasing convex resource consumption function. In the first 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: [email protected] (S.-H. Yoon).

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

http://www.elsevier.com/locate/cor

mailto:[email protected]

B.-C. Choi et al. / Computers & Operations Research 34 (2007) 1988–2000 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 [2,4–6,10].

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 efficiently solved by ordering jobs according to non-increasingprocessing times.

Cheng and Janiak [4] considered a problem that exchanged the objective function with 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 wasstrongly NP-hard. Li [8] considered a single machine scheduling problem with 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 first 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 defined formally. InSection 3, we show that the recognition version of the first 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 definition

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 definedas below.

1990 B.-C. Choi et al. / Computers & Operations Research 34 (2007) 1988–2000

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=1

f (r�(j)) =n∑

j=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, �), defined as

K(r, �) = �n∑

j=1

max{v − r�(j), 0} + �n∑

j=1

(r�(j) + p�(j)),

where v�∑n

j=1 pj .

Note that Problem P2 is NP-hard if there is no limitation for v, an initial release time of jobs [10]. InProblem P1 and P2,

∑nj=1 (r�(j)+p�(j)) is the sum of job completion times. Since f (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 define 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-


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 and A, 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 that

∑nj=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 = 2A −∑n

j=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 that∑n

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.

n∑j=1

bj yj =n∑

j=1

aj (2xj − 1) = 2n∑

j=1

aj xj −n∑

j=1

aj = 2A −n∑

j=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.

n∑j=1

aj xj =n∑

j=1

aj

(yj + 1)

2= 1

2

⎛⎝ n∑

j=1

bj yj +n∑

j=1

aj

⎞⎠= 1

2

⎛⎝B +

n∑j=1

aj

⎞⎠= A.

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

Definition 1. A permutation solution is an integer vector x = (x1, . . . , xn)T such that 1�xi �= 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 d2j−1 = Mj − bj andd2j =Mj +bj for j =1, . . . , n, and D=∑n

j=1 (4j −1)Mj +B, where M =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 that∑n

j=1 bj yj = B, where yj ∈ {−1, 1}. If yj = 1, letx2j−1 = 2j − 1 and x2j = 2j and otherwise, let x2j−1 = 2j and x2j = 2j − 1, for j = 1, . . . , n.


Thus, −x2j−1 + x2j = yj and x2j−1 + x2j = 4j − 1.

m∑j=1

dj xj =n∑

j=1

[(Mj − bj )x2j−1 + (Mj + bj )x2j ]

=n∑

j=1

(x2j−1 + x2j )Mj +

n∑j=1

(−x2j−1 + x2j )bj

=n∑

j=1

(4j − 1)Mj +n∑

j=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.

2n∑j=1

dj xj =n∑

j=1

[(Mj − bj )x2j−1 + (Mj + bj )x2j ]

=n∑

j=1

(x2j−1 + x2j )Mj +

n∑j=1

(−x2j−1 + x2j )bj =n∑

j=1

(4j − 1)Mj + B.

Since D =∑nj=1 (4j − 1)Mj + B and M is sufficiently large, we can infer from the above relations that

the following equations hold:

n∑j=1

(x2j−1 + x2j )Mj =

n∑j=1

(4j − 1)Mj , (1)

n∑j=1

(−x2j−1 + x2j )bj = B. (2)

From (1), for j = n, x2n−1 = 2n − 1, x2n = 2n or x2n−1 = 2n, x2n = 2n − 1. Likewise, for j �n − 1,x2j−1 = 2j − 1, x2j = 2j or x2j−1 = 2j , x2j = 2j − 1. It implies that −x2j−1 + x2j = ±1. Letyj = −x2j−1 + 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, �)=∑n

j=1 f (r�(j)) the total resource consumption of solution (r, �). The recognition version of Problem P1can be stated as follows:

Recognition version of Problem P1: Given processing timespj for j=1, . . . , n, a non-increasing convexfunction 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 P1as 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{∑n

j=1 dj −rk, 0} for k=1, . . . , n+1.E =∑n

j=1 dj + D + H and G = (n + 1)∑n

j=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.


Lemma 1. If there is a permutation solution x such that∑n

j=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+1∑k=1

k∑j=1

p�(j) =n∑

k=1

kp�(n−k+1) +n+1∑k=1

p�(k).

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

C(r, �) =n∑

j=1

dj xj +n∑

j=1

dj + H = D +n∑

j=1

dj + H = E. (3)

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

n∑k=1

k∑j=1

p�(j) = D.

Since r�(j) <∑n

j=1 dj for j = 1, . . . , n and r�(n+1) =∑nj=1 dj ,

F(r, �) =n+1∑k=1

max

⎧⎨⎩

n∑j=1

dj − r�(k), 0

⎫⎬⎭=

n∑k=1

⎛⎝ n∑

j=1

dj − r�(k)

⎞⎠

=n∑

k=1

⎛⎝ n∑

j=1

dj −k−1∑j=1

p�(k)

⎞⎠=

n∑k=1

⎛⎝ n∑

j=1

dj −⎛⎝ k∑

j=1

p�(j) − p�(k)

⎞⎠⎞⎠

=n∑

k=1

n∑j=1

dj −n∑

k=1

k∑j=1

p�(j) +n∑

k=1

p�(k) = (n + 1)

n∑j=1

dj − D = G.

Thus, F(r, �) = G. This completes proof. �

Lemma 2. If there is a solution (r, �) such that C(r, �)�E and F(r, �)�G, then there is a permutationsolution x such that

∑nj=1 dj xj = D.

Proof.

Claim 1. Job (n + 1) should be positioned last in �.


Proof. Suppose that job (n + 1) is positioned the lth in �, l �= n + 1. Since∑n+1

j=1 p�(j) > p�(l) =H >

∑nj=1 dj + D,

C(r, �)�n+1∑k=1

k∑j=1

p�(j)�p�(l) +n+1∑j=1

p�(j) > H +n∑

j=1

dj + 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+1

j=1 p�(j) =∑nj=1 dj + H and C(r, �)�E = D +∑n

j=1 dj + H ,

C(r, �) =n+1∑k=1

k∑j=1

(�j + p�(j))

=n+1∑k=1

k∑j=1

�j +n∑

k=1

k∑j=1

p�(j) +n∑

j=1

dj + H �D +n∑

j=1

dj + H . (4)

Since∑n+1

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

n∑k=1

k∑j=1

(�j + p�(j)) < D. (5)

Since r�(n+1) >∑n

j=1 dj ,

F(r, �) =n+1∑k=1

max

⎧⎨⎩

n∑j=1

dj − r�(k), 0

⎫⎬⎭

=n∑

k=1

max

⎧⎨⎩

n∑j=1

dj − r�(k), 0

⎫⎬⎭ �

n∑k=1

⎛⎝ n∑

j=1

dj − r�(k)

⎞⎠

=n∑

k=1

⎛⎝ n∑

j=1

dj −k∑

j=1

(�j + p�(j)) + p�(k)

⎞⎠

= (n + 1)

n∑j=1

dj −n∑

k=1

k∑j=1

(�j + p�(j)).


From Eq. (5),

F(�) > (n + 1)

n∑j=1

dj − 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,

0�C(r, �) − E =n∑

k=1

k∑j=1

p�(j) +n∑

j=1

dj + H −⎛⎝D +

n∑j=1

dj + H

⎞⎠

=n∑

k=1

k∑j=1

p�(j) − D.

Thus,∑n

k=1∑k

j=1 p�(j)�D.Also, by Claim 1 and 2,

0�F(r, �) − G =n+1∑k=1

max

⎧⎨⎩

n∑j=1

dj − r�(k), 0

⎫⎬⎭−

⎛⎝(n + 1)

n∑j=1

dj − D

⎞⎠

=n∑

k=1

⎛⎝ n∑

j=1

dj −⎛⎝ k∑

j=1

p�(j) − p�(k)

⎞⎠⎞⎠− (n + 1)

n∑j=1

dj + D

= −n∑

k=1

k∑j=1

p�(j) + D.

Thus,∑n

k=1∑k

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

n∑k=1

k∑j=1

p�(j) = D.

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

n∑j=1

dj xj =n∑

k=1

kp�(n−k+1) =n∑

k=1

k∑j=1

p�(j) = D.

Thus, there is a permutation solution x such that∑n

j=1 djxj = D. This completes proof. �


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, �∗) byletting 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, �∗) < K(r∗, �∗). This is a contradiction. This completes proof. �

By Lemma 3, since r�(i) = r�(1) +∑i−1j=1 p�(j), the objective function K(r, �) can be rewritten as

K(r, �) = �n∑

j=1

max{v − r�(j), 0} + �

⎛⎝nr�(1) +

n∑i=1

i∑j=1

p�(j)

⎞⎠ .

Lemma 4. An optimal solution can be left-shifted up to time zero or right-shifted until an initial jobrelease time without any cost. Equivalently, one of the conditions holds:

(a) r�(1) = 0;(b) there is job �(j) such that r�(j) = v for j = 1, . . . , n.


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) +n∑

i=1

i∑j=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, �∗) = K(r∗, �∗) − (� − �)n�, (9)

K(r, �∗) = K(r∗, �∗) + (� − �)n�. (10)

From (9) and (10), min{K(r, �∗), K(r, �∗)}�K(r∗, �∗). This implies that we can left-shift the first jobstarting at time zero (condition (a)) or right-shift the last job starting at time v (condition (b)).

Case 2: v < r∗�∗(1) +∑n

i=1p�∗(i).Let job �(l) be the job that is being processed at time v. Then, the objective function K(r∗, �∗) can be

rewritten as

K(r∗, �∗) = �l∑

j=1

(v − r�∗(j)) + �

⎛⎝nr�∗(1) +

n∑i=1

i∑j=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 solution (r∗, �∗). Thus,

K(r, �∗) = K(r∗, �∗) − (�l − �n)�, (11)

K(r, �∗) = K(r∗, �∗) + (�l − �n)�. (12)

From (11) and (12), min{K(r, �∗), K(r, �∗)}�K(r∗, �∗). This implies that we can left-shift job �(l) tofinish at time v or right-shift job �(l) to start at time v. This completes proof. �


Assignment problem. Given n × n matrix Q = [qij ], determine a vector x = (x1, . . . , xn)T such that

minn∑

i=1

n∑j=1

qijxij

s.t.n∑

i=1

xij = 1, j = 1, . . . , n

n∑j=1

xij = 1, i = 1, . . . , n

xij ∈ {0, 1}, i = 1, . . . , n, j = 1, . . . , n.

Define Pk as a constrained problem from Problem P2 such that r�(k) = v where 1�k�n, and P0 canbe the case where r�(1) = 0.

Theorem 4. Pk can be reduced to the assignment problem in polynomial time.

Proof. For P0, by Lemmas 3 and 4, the objective function K(r, �) can be written as follows:

K(r, �) = �n∑

j=1

(v − r�(j)) + �n∑

i=1

i∑j=1

p�(j).

Since r�(i) =∑i−1j=1 p�(j),

K(r, �) = �nv − �n−1∑i=1

i∑j=1

p�(j) + �n∑

i=1

i∑j=1

p�(j)

= �nv − �n−1∑j=1

(n − j)p�(j) + �n∑

j=1

(n + 1 − j)p�(j)

= �nv +n∑

j=1

(�(n + 1 − j) − �(n − j))p�(j).

For i = 1, . . . , n, j = 1, . . . , n, by letting qij = (�(n + 1 − j) − �(n − j))pi , where pi = p�(j), P0 canbe reduced to the assignment problem with Q.

For Pk , k�1, the objective function of Pk can be written as follows:

K(r, �) = �k−1∑j=1

(v − r�(j)) + �

⎛⎝nr�(1) +

n∑i=1

i∑j=1

p�(j)

⎞⎠ .


Table 1Results of the numerical example

Problem Optimal sequence Objective value

P0 (J3, J1, J2) 57P1 (J2, J1, J3) 41

P2(J2, J1, J3)

(J1, J2, J3)40

P3 (J1, J2, J3) 41

Since r�(i) = v −∑k−1j=i p�(j),

K(r, �) = �k−1∑i=1

k−1∑j=i

p�(j) + �n

⎛⎝v −

k−1∑j=1

p�(j)

⎞⎠+ �

n∑j=1

(n − j + 1)p�(j).

Since �∑n

j=1 (n − j + 1)p�(j) = �∑k−1

j=1 (n − j + 1)p�(j) + �∑n

j=k (n − j + 1)p�(j),

K(r, �) = �nv +k−1∑j=1

(�j − �(j − 1))p�(j) +n∑

j=k

�(n − j + 1)p�(j).

For i=1, . . . , n, by letting qij =(�j −�(j −1))pi for j =1, . . . , k−1, and by letting qij =�(n−j +1)pi

for j = k, . . . , n, where pi =p�(j), Pk can be reduced to the assignment problem with Q. This completesproof. �

Theorem 5. Problem P2 can be solved in O(n4).

Proof. An optimal solution of Problem P2 can be obtained by minimizing Pk , 0�k�n. Since each Pk

can be solved in O(n3) by the Hungarian method, Problem P2 can be solved in O(n4). �

Numerical example. Consider a 3-job single machine scheduling problem in which v=10, �=2, �=1,p1 = 2, p2 = 1 and p3 = 4. We can divide this instance into P0, P1, P2 and P3. Let Qi be the matrix forPi , i = 0, . . . , 3. Thus, by Theorem 4, Qi can be constructed as follows:

Q0 =(−2 0 2

−1 0 1−4 0 4

), Q1 =

( 6 4 23 2 1

12 8 4

), Q2 =

(4 4 22 2 18 8 4

), Q3 =

(4 6 22 3 18 12 4

).

By using the Hungarian method, we can obtain the result provided in Table 1.Let (�∗, r∗) be an optimal solution. Then, an optimal schedule with no intermittent idle time is �∗ =

(J2, J1, J3) or (J1, J2, J3) with r�∗(2) = 10.

5. Summary and conclusion

We consider single machine scheduling problems with release times that depend on a non-increasingconvex resource consumption function. The first objective is the minimization of the total amount of


resource consumed with a constraint on the sum of job completion times. We show that the recognitionversion of the problem is NP-complete by reducing it from the permutation integer problem that we proveto be NP-complete. The second objective is the minimization of the weighted total resource consumptionand sum of job completion times with an unrestricted initial release time. We provide some optimalityconditions and show that the problem can be solved in polynomial time.

Acknowledgements

This work was supported by the Soongsil University Research Fund.

References

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Single machine scheduling problems with resource dependent release times