Operations Research Letters 38 (2010) 123–126

Contents lists available at ScienceDirect

Operations Research Letters

journal homepage: www.elsevier.com/locate/orl

Complexity of a scheduling problem with controllable processing timesByung-Cheon Choi a,∗, Joseph Y.-T. Leung b, Michael L. Pinedo ca Department of Business, Chungnam National University, 79 Daehakro, Yuseong-gu Daejeon 305-764, Republic of Koreab Department of Computer Science, New Jersey Institute of Technology, Newark, NJ 07102, USAc Department of Information, Operations & Management Sciences, Stern School of Business, New York University, 44 West 4th Street, New York, NY 10012-1126, USA

a r t i c l e i n f o

Article history:Received 15 May 2008Accepted 7 October 2009Available online 24 October 2009

Keywords:Total weighted completion timeControllable processing timeSingle machineNP-hard

a b s t r a c t

We consider the problem of scheduling a set of independent jobs on a single machine so as to minimizethe total weighted completion time, subject to the constraint that the total compression cost is less thanor equal to a fixed amount. The complexity of this problem is mentioned as an open problem. In this notewe show that the problem is NP-hard.

© 2009 Elsevier B.V. All rights reserved.

1. Introduction

Scheduling problems with controllable processing times havebeen studied extensively; see [4,5] for the most recent surveys. Inthis note we consider a particular scheduling problem with con-trollable processing times that can be stated as follows. Supposewe have a set of n independent jobs {1, 2, . . . , n} to be scheduledon a single machine, where job processing times are compressible.Associated with each job j is a normal processing time p0j , a weightwj and a compression rate bj. If job j is compressed by spending anamount uj, then its actual processing time becomes pj = p0j − bjuj.Given a budget R, our problem is to find a schedule with minimum∑wjCj, subject to the constraint that the total amount spent on

compression,∑uj, does not exceed the fixed budget R. The com-

plexity of this problem is stated as open in [3,4]. In this note weshow that it is NP-hard. Our result implies also that the problemof minimizing

∑uj, subject to the constraint that

∑wjCj does not

exceed a given threshold, is also NP-hard.Wan et al. [6] and Hoogeveen and Woeginger [2] studied a

related problem in which the cost function∑wjCj +

∑cjuj has

to be minimized. They showed, independently from one another,that their problem is NP-hard [6,2]. On the surface, it may appearthat our problem is somewhat easier to solve than their problem.However, this note shows that our problem is NP-hard as well.

2. Problem reduction

In this section we will show that the decision version of ourproblem is NP-complete. We denote our decision problem as

∗ Corresponding author.E-mail addresses: polytime@cnu.ac.kr (B.-C. Choi), leung@cis.njit.edu

(J.Y.-T. Leung), mpinedo@stern.nyu.edu (M.L. Pinedo).

0167-6377/$ – see front matter© 2009 Elsevier B.V. All rights reserved.doi:10.1016/j.orl.2009.10.011

the Weighted Completion Time problem with Controllable Cost(WCTCC).WCTCC:Given two thresholdsR andWC , a set of n jobs {1, 2, . . . , n}with the initial processing time of job j equal to p0j , the weight ofjob j equal to wj, and the compression rate of job j equal to bj, isthere a schedule such that

∑wjCj ≤ WC and

∑uj ≤ R?

We will reduce the Equal Cardinality Partition problem to theWCTCC problem. The Equal Cardinality Partition problem is knownto beNP-complete; see [1]. The Equal Cardinality Partition problemcan be stated as follows.Equal Cardinality Partition: Given 2n integers a1, a2, . . . , a2n suchthat

∑2nj=1 aj = A, is there a subset I ⊆ {1, 2, . . . , 2n} such that

|I| = n and∑j∈I aj =

A2?

Given an instance of the Equal Cardinality Partition problem,we construct an instance of the WCTCC problem as follows. Therewill be 2n jobs. The initial processing time,weight and compressionrate of job j are given as follows.

p0j = B+ aj, wj = B+ aj and bj =B+ ajM + aj

, j = 1, . . . , 2n,

where B = nA3 and M = 2n+12n+2B =

n(2n+1)2n+2 A

3. All jobs can be

completely compressed, i.e., the upper bound of uj isp0jbj= M +

aj, j = 1, . . . , 2n. The thresholdsWC and R are given as

WC =12

((n2 + n)B2 + (n+ 1)AB

+

(2n∑j=1

aj

)2+

2n∑j=1

a2j

and R = nM +A2.

Note that since wj

p0j= 1, j = 1, . . . , 2n, the jobs that are not

compressed will always follow the compressed jobs in any order.

124 B.-C. Choi et al. / Operations Research Letters 38 (2010) 123–126

∆ ≤ ε(B+ ak)(−((|Nkl| + 2)B+ A1)(M + al)+ ((|Nkl| + 1)B+ A2)(M + ak)

(M + ak)(M + al)

)= ε(B+ ak)

(−BM + (|Nkl| + 1)akB− (|Nkl| + 2)alB− akM + akA2 − alA1

(M + ak)(M + al)

)Box I.

Lemma 1. If there is a solution to the Equal Cardinality Partitionproblem, then there is a solution to the WCTCC problem.

Proof. Let I ⊆ {1, 2, . . . , 2n} be such that |I| = n and∑j∈I aj =

A2 .

Let I ′ = {1, 2, . . . , 2n} − I . We fully compress every job in I .The upper bound of uj is M + aj for every job in I . Therefore,∑j∈I uj = nM +

A2 = R. The jobs in I

′ are not compressed at alland they are scheduled in any order.Without loss of generality, suppose that the jobs in I ′ are

scheduled in the order (σ (1), . . . , σ (n)). Then2n∑j=1

wjCj =∑j∈I ′wjCj

=

n∑k=1

wσ(k)

k∑j=1

pσ(j) =n∑k=1

(B+ aσ(k))k∑j=1

(B+ aσ(j))

=12

(∑j∈I ′(B+ aj)

)2+

∑j∈I ′(B+ aj)2

.Since

|I ′| = n,∑j∈I ′aj =

A2and

A2

4+

∑j∈I ′a2j <

(2n∑j=1

aj

)2+

2n∑j=1

a2j ,

we have2n∑j=1

wjCj =12

((nB+

A2

)2+ nB2 + AB+

∑j∈I ′a2j

)

=12

((n2 + n)B2 + (n+ 1)AB+

A2

4+

∑j∈I ′a2j

)< WC .

�

We now show that if there is a solution to the WCTCC problem,then there is a solution to the Equal Cardinality Partition problem.Let S be a schedule such that

∑2nj=1wjCj ≤ WC and

∑2nj=1 uj ≤ R.

Without loss of generality, we may assume that the entire budgetR is used for the compression of jobs, i.e.,

∑2nj=1 uj = R.

Lemma 2. In the reduced problem, the objective value of the schedulewith at least two partially compressed jobs is larger than the objectivevalue of the schedule with at most one partially compressed job.

Proof. Consider a schedule S with at least two partially com-pressed jobs. The jobs in S can be put into three categories: (1)those that are fully compressed, (2) those that are partially com-pressed, and (3) those that are not compressed at all. For jobsin the first category, their actual processing times have been re-duced to zero. Thus, we only need to consider in schedule S jobsin the second and third categories. Without loss of generality, wemay assume that these jobs are scheduled in WSPT order and that∑2nj=1 uj = R, where WSPT order implies a decreasing order of

wjpj.

Clearly, the jobs in the second category are scheduled before thejobs in the third category, in WSPT order. Let jobs k and l be thefirst and second job, respectively, in schedule S. Let Nkl be all the

jobs that follow job l in schedule S. We construct a new sched-ule S by compressing more of job k and ‘decompressing’ job l;i.e., uk = uk+ ε and ul = ul− ε, where ε > 0 and uk and ul are thenew amounts spent on the compression of jobs k and l. Clearly, thenew schedule is feasible, since

∑2nj=1 uj = R. Moreover,

pk = pk −B+ akM + ak

ε and pl = pl +B+ alM + al

ε.

Clearly, compressing job k will cause every job in the scheduleto be shifted to the left (and hence completed earlier). Similarly,decompressing job l will cause every job in Nkl ∪ {l} to be shiftedto the right (and hence to be completed later). Therefore, the totalamount of change (∆) in

∑wjCj can be calculated as follows:

∆ = −

( ∑Nkl∪{k,l}

wj

)B+ akM + ak

ε +

( ∑j∈Nkl∪{l}

wj

)B+ alM + al

ε.

Since∑j∈Nkl∪{k,l}

wj =∑

j∈Nkl∪{k,l}

(B+ aj) = (|Nkl| + 2)B+ A1

and∑j∈Nkl∪{l}

wj =∑

j∈Nkl∪{l}

(B+ aj) = (|Nkl| + 1)B+ A2,

where A1 =∑j∈Nkl∪{k,l}

aj and A2 =∑j∈Nkl∪{l}

aj, we have

∆ = −((|Nkl| + 2)B+ A1)(B+ ak)

M + akε

+((|Nkl| + 1)B+ A2)(B+ al)

M + alε.

There are two cases to consider.Case 1: ak ≥ al.In this case, we have the equation in Box I.

Since −(|Nkl| + 2)alB − akM + akA2 − alA1 < 0, B = nA3,M = 2n+1

2n+2nA3, |Nkl| + 1 ≤ 2n and ak < A

2 , we have

∆ <ε(B+ ak)

(M + ak)(M + al)

(−n2(2n+ 1)2n+ 2

A6 + n2A4).

Since− 2n+12n+2A2+ 1 < 0, we have

∆ <ε(B+ ak)

(M + ak)(M + al)n2A4

(−2n+ 12n+ 2

A2 + 1)< 0.

Case 2: ak < al.In this case, we have

B+ akM + ak

−B+ alM + al

=(M − B)(ak − al)(M + ak)(M + al)

> 0.

Therefore, we have

∆ <(B+ ak)εM + ak

(−((|Nkl| + 2)B+ A1)+ ((|Nkl| + 1)B+ A2))

=(B+ ak)εM + ak

(−B− ak) < 0.

B.-C. Choi et al. / Operations Research Letters 38 (2010) 123–126 125

From Cases 1 and 2, we have ∆ < 0. This implies that we candecrease the objective value of the schedule by fully compressingjob k. By repeating this argument, we can obtain a schedule with atmost one partially compressed job. �

In S, let T be the set of all fully compressed jobs, let P be the setof partially compressed jobs and let U be the set of jobs that havenot undergone any compression. By Lemma 2,wemay assume that|P| ≤ 1. If |T | ≥ n+ 1, then we have∑j∈T

(M + aj) ≥ (n+ 1)M > R.

Thus, we may assume that |T | ≤ n. If |T | ≤ n− 2, then we have∑j∈T∪P

(M + aj) < (n− 1)M < R.

This violates our assumption that the entire budget has been usedfor the compression of jobs. By the above argument, there are onlythree cases we need to consider: (1) (|T |, |P|, |U|) = (n, 0, n), (2)(|T |, |P|, |U|) = (n, 1, n− 1), (3) (|T |, |P|, |U|) = (n− 1, 1, n).

Lemma 3. If there is a solution to the WCTCC problem, then there isa solution to the Equal Cardinality Partition problem.

Proof. Following the discussion above, we will consider the threecases separately.Case 1: (|T |, |P|, |U|) = (n, 0, n)In this case we have

∑j∈T aj =

∑j∈U aj =

A2 . The total allocated

resources can be calculated as follows:∑j∈T

uj =∑j∈T

(M + aj) = nM +A2= R.

The jobs in U are scheduled in an arbitrary order since wjp0j= 1 for

all j ∈ U . Therefore, the objective value is calculated as follows:

2n∑j=1

wjCj =∑j∈U

wjCj =12

(∑j∈U

(B+ aj)

)2+

∑j∈U

(B+ aj)2

=12

(∑j∈U

B+∑j∈U

aj

)2+

∑j∈U

(B2 + 2Baj + a2j )

.Since

|U| = n,∑j∈U

aj =A2and

A2

4+

∑j∈U

a2j <

(n∑j=1

aj

)2+

n∑j=1

a2j ,

we have2n∑j=1

wjCj =12

((n2 + n)B2 + (2n+ 2)

A2B+

A2

4+

∑j∈U

a2j

)< WC .

Case 2: (|T |, |P|, |U|) = (n, 1, n− 1).In this case, let job k be the partially compressed job. Clearly, job kis scheduled in the first position in S. Since |T | = n and |P| = 1,we have

∑j∈T uj = nM +

∑j∈T aj < R = nM +

A2 . Thus, uk can be

calculated as follows:

uk = R−∑j∈T

(M + aj) =A2−

∑j∈T

aj =∑j∈U∪{k}

aj −A2.

In other words,∑j∈T∪{k}

uj = R.

The actual processing time of job k is

pk = B+ ak −∆,

where∆ = B+akM+ak

(∑j∈U∪{k} aj −

A2

). Thus, we have

2n∑j=1

wjCj =∑j∈U∪{k}

wjCj −∆∑j∈U∪{k}

wj

=12

( ∑j∈U∪{k}

(B+ aj)

)2+

∑j∈U∪{k}

(B+ aj)2

−∆ ∑j∈U∪{k}

wj

=12

( ∑j∈U∪{k}

B+∑j∈U∪{k}

aj

)2+

∑j∈U∪{k}

(B2 + 2Baj + a2j )

−∆

∑j∈U∪{k}

wj.

Since |U| = n− 1, we have

2n∑j=1

wjCj =12

((n2 + n)B2 + (2n+ 2)

( ∑j∈U∪{k}

aj

)B

+

( ∑j∈U∪{k}

aj

)2+

∑j∈U∪{k}

a2j

−∆ ∑j∈U∪{k}

wj.

Since∆ = B+akM+ak

(∑j∈U∪{k} aj −

A2

), we have

2n∑j=1

wjCj = WC + (n+ 1)

( ∑j∈U∪{k}

aj −A2

)B+ ε

−B+ akM + ak

( ∑j∈U∪{k}

aj −A2

) ∑j∈U∪{k}

(B+ aj),

where ε =12

((∑j∈U∪{k} aj

)2+∑j∈U∪{k} a

2j −

(∑2nj=1 aj

)2−∑2nj=1 a

2j

). Therefore,

2n∑j=1

wjCj = WC +

( ∑j∈U∪{k}

aj −A2

)

×

((n+ 1)B−

B+ akM + ak

(nB+

∑j∈U∪{k}

aj

))+ ε.

SinceM = 2n+12n+2B, ak <

A2 and

∑j∈U∪{k} aj < A, we have

(n+ 1)B−B+ akM + ak

(nB+

∑j∈U∪{k}

aj

)

=

B2 +

(2ak − 2

∑j∈U∪{k}

aj

)B− 2ak

∑j∈U∪{k}

aj

2( 2n+12n+2B+ ak

)>B2 − 2AB− A2

2B+ A.

Also, since∑2nj=1 aj = A and aj <

A2 , we have

ε > −12

(A2 +

n2A2)= −

12

(1+

n2

)A2 > −

n2A2.

126 B.-C. Choi et al. / Operations Research Letters 38 (2010) 123–126

Thus, since A−(2nA +

1n2A3+ 1+ 1

2nA2

)> 0, we have

2n∑j=1

wjCj > WC +B2 − 2AB− A2

2B+ A−n2A2

= WC +2n2A5(A− 2

nA −1n2A3− 1− 1

2nA2)

2(2B+ A)> WC .

In this case, the objective value of the schedule is larger thanWC .Case 3: (|T |, |P|, |U|) = (n− 1, 1, n).Let job k be the partially compressed job in S. Clearly, job k isthe first job scheduled in S. Since |T | = n − 1 and |P| = 1,∑j∈T uj = (n − 1)M +

∑j∈T aj < R = nM + A

2 . Therefore, ukcan be calculated as follows:

uk = R−∑j∈T

(M + aj) = M +A2−

∑j∈T

aj.

Thus, we have∑j∈T∪{k}

uj = R.

The actual processing time of job k can be calculated as follows:

pk = B+ ak −B+ akM + ak

(M +

A2−

∑j∈T

aj

)

=B+ akM + ak

( ∑j∈T∪{k}

aj −A2

).

Since∑j∈T∪{k} aj −

A2 =

A2 −

∑j∈U aj, we have

pk =B+ akM + ak

(A2−

∑j∈U

aj

).

Since job k is scheduled first in S, we have2n∑j=1

wjCj =∑j∈U

wjCj + pk∑j∈U∪{k}

wj

=12

(∑j∈U

(B+ aj)

)2+

∑j∈U

(B+ aj)2

+ pk ∑j∈U∪{k}

wj.

Since |U| = n, we have

2n∑j=1

wjCj =12

((n2 + n)B2 + (2n+ 2)

(∑j∈U

aj

)B

+

(∑j∈U

aj

)2+

∑j∈U

a2j

+ pk ∑j∈U∪{k}

wj.

Since∑j∈U aj <

A2 , we have

2n∑j=1

wjCj = WC − (n+ 1)

(A2−

∑j∈U

aj

)B+ ε

+B+ akM + ak

(A2−

∑j∈U

aj

) ∑j∈U∪{k}

(B+ aj),

where ε = 12

((∑j∈U aj

)2+∑j∈U a

2j −

(∑2nj=1 aj

)2−∑2nj=1 a

2j

).

Therefore, we have2n∑j=1

= WC +

(A2−

∑j∈U

aj

)

×

(B+ akM + ak

∑j∈U∪{k}

(B+ aj)− (n+ 1)B

)+ ε.

SinceM = 2n+12n+2B and ak <

A2 , we have

B+ akM + ak

((n+ 1)B+

∑j∈U∪{k}

aj

)− (n+ 1)B

=

B2 + 2B∑

j∈U∪{k}ak + 2ak

∑j∈U∪{k}

aj

2( 2n+12n+2B+ ak

) >B2

2B+ A.

Also, as we showed in Case 2,

ε > −n2A2.

Thus, since A− 1− 12nA2

> 0 and B = nA3, we have

2n∑j=1

wjCj > WC +B2

2B+ A−n2A2 = WC +

2n2A5(A− 1− 1

2nA2

)2(2B+ A)

> WC .

Again, the objective value of the schedule is larger thanWC .Summarizing the above three cases, the only schedule with

objective value smaller than WC is the one with (|T |, |P|, |U|) =(n, 0, n). All other schedules have objective values larger thanWC .Therefore, if the WCTCC problem has a solution, then the EqualCardinality Partition problem has a solution as well. �

By Lemmas 1 and 3, we have the following theorem.

Theorem 1. The WCTCC problem is NP-complete.

Acknowledgements

Byung-Cheon Choi was supported in part by the Korea ResearchFoundation Grant KRF-2008-357-D00289. Joseph Y-T. Leung wassupported in part by the NSF Grant DMI-0556010. Michael L.Pinedo was supported in part by the NSF Grant DMI-0555999.

References

[1] M.R. Garey, D.S. Johnson, Computers and Intractability: A Guide to the Theoryof NP-Completeness, W.H. Freeman, New York, 1979.

[2] H. Hoogeveen, G.J. Woeginger, Some comments on sequencing with control-lable processing times, Computing 68 (2002) 181–192.

[3] A. Janiak, Scheduling in computer and manufacturing systems, WroclawUniversity of Technology, Wroclaw, Poland, 2006.

[4] A. Janiak, W. Janiak, M. Lichtenstein, Resource management in machinescheduling problems: A survey, DecisionMaking inManufacturing and Services1 (2007) 59–89.

[5] D. Shabtay, G. Steiner, A survey of scheduling with controllable processingtimes, Discrete Applied Mathematics 155 (2007) 1643–1666.

[6] G.Wan, B. Yen, C-L. Li, Singlemachine scheduling tominimize total compressionplus weighted flow cost is NP-hard, Information Processing Letters 79 (2001)273–280.