Single machine scheduling with a restricted rate-modifying activity

Single Machine Scheduling with a Restricted Rate-Modifying Activity*

Yong He,1 Min Ji,1 T. C. E. Cheng2

1 Department of Mathematics, State Key Lab of CAD & CG, Zhejiang University, Hangzhou 310027,People’s Republic of China

2 Department of Logistics, The Hong Kong Polytechnic University, Kowloon, Hong Kong,People’s Republic of China

Received 14 November 2003; revised 26 January 2005; accepted 16 February 2005DOI 10.1002/nav.20083

Published online 22 March 2005 in Wiley InterScience (www.interscience.wiley.com).

Abstract: In this paper we consider the problem of scheduling a set of jobs on a single machine on which a rate-modifyingactivity may be performed. The rate-modifying activity is an activity that changes the production rate of the machine. So theprocessing time of a job is a variable, which depends on whether it is scheduled before or after the rate-modifying activity. Weassume that the rate-modifying activity can take place only at certain predetermined time points, which is a constrained case ofa similar problem discussed in the literature. The decisions under consideration are whether and when to schedule therate-modifying activity, and how to sequence the jobs in order to minimize some objectives. We study the problems of minimizingmakespan and total completion time. We first analyze the computational complexity of both problems for most of the possibleversions. The analysis shows that the problems are NP-hard even for some special cases. Furthermore, for the NP-hard cases ofthe makespan problem, we present a pseudo-polynomial time optimal algorithm and a fully polynomial time approximationscheme. For the total completion time problem, we provide a pseudo-polynomial time optimal algorithm for the case withagreeable modifying rates. © 2005 Wiley Periodicals, Inc. Naval Research Logistics 52: 361–369, 2005.

Keywords: scheduling; approximation algorithms; rate-modifying activity; computational complexity

1. INTRODUCTION

In a recent paper, Lee and Leon [9] proposed a problemof scheduling jobs on a single machine on which a rate-modifying activity may be performed. The rate-modify-ing activity is an activity that occupies the machine for agiven duration and changes the production rate of themachine. No job can be processed while the rate-modi-fying activity is being performed. Hence, the processingtime of a job is a variable, which depends on whether itis processed before or after the rate-modifying activity.The decisions under consideration are whether and when

to schedule the rate-modifying activity, and how to se-quence the jobs in order to minimize some regular per-formance measures.

This problem has important applications in electronicassembly lines, on which routine maintenance activitiesare performed in order to sustain production quality andefficiency. The maintenance work can be modelled as arate-modifying activity. In [9], Lee and Leon assumedthat the rate-modifying activity can take place as andwhen it is needed. They considered the problem withvarious performance measures including makespan, totalcompletion time, total weighted completion time, andmaximum lateness. They analyzed the computationalcomplexity of these problems, and provided polynomialor pseudo-polynomial time algorithms to solve them. Insome applications, the assumption that the rate-modify-ing activity can take place at any time may not be alwaysvalid because performing the rate-modifying activityneeds resources (e.g., operators, equipment, etc.), whichmay not be readily available. So we consider the problemwith a restricted rate-modifying activity, i.e., the rate-

* This research was supported by the Teaching and ResearchAward Program for Outstanding Young Teachers in Higher Edu-cation Institutions of the MOE, China, and the National NaturalScience Foundation of China (10271110, 60021201). The thirdauthor was supported in part by The Hong Kong PolytechnicUniversity under a grant from the Area of Strategic Developmentin China Business Services.

Correspondence to: T.C.E. Cheng ([email protected]);Y. He ([email protected]); M. Ji ([email protected])

© 2005 Wiley Periodicals, Inc.

modifying activity can take place only at certain prede-termined time points.

Formally, the single-machine scheduling problem with arestricted rate-modifying activity can be described as fol-lows: We are given n independent jobs � � { J1, J2, . . . ,Jn} to be processed on a single machine on which a rate-modifying activity r of duration t may be performed. Therate-modifying activity r can only take place at one of mpredetermined time points 0 � s1 � s2 � . . . � sm ��. If the rate-modifying activity must be performed, wesay that it is mandatory. Otherwise, it is called optional ifwe can choose not to perform the rate-modifying activity.Denote � � {s1, s2, . . . , sm}, and � is called the permit-ted start-time set of the rate-modifying activity. No preemp-tion is allowed for any jobs. Associated with each job Ji isa processing time pi � 0 if Ji is processed before therate-modifying activity. On the other hand, if Ji is processedafter the rate-modifying activity, then its processing time ismodified to be �ipi, where �i � 0 is the modifying rate. Ingeneral, no restrictions are imposed on the value of �i.However, the problem has useful applications for the specialcase with �i � 1 for all i [9]. For a given schedule, let Ci

denote the completion time of job Ji and Cmax �maxi�1,2,. . .,n{Ci} the makespan. The objective functionsconsidered in this paper are the makespan and total com-pletion time, ¥i�1

n Ci. Using the three-field notation todescribe scheduling problems [5], we denote the makespanproblem as 1/rm, �, man/Cmax for the mandatory case and1/rm, �, opt/Cmax for the optional case. Similarly, wedenote the mandatory and optional cases of the total com-pletion time problem as 1/rm, �, man/¥ Ci and 1/rm, �,opt/¥ Ci, respectively. In this paper both of the mandatoryand optional cases will be studied.

If we set �� 1, and �i � 1 for all i, the mandatorycases of our problems reduce to the single machine sched-uling problems with a machine availability constraint,which have been considered in [1, 7, 10] and are denoted by1/nr � a/Cmax and 1/nr � a/¥ Ci, respectively. Themachine availability constraint stipulates a time window[s1, s1 � t] during which the machine is unavailable toprocess jobs when preventive maintenance is being per-formed. Lee [7] has studied the single machine scheduling

problem with a machine availability constraint with differ-ent performance measures. The problem has been extendedto different machine environments, such as parallel ma-chines and flowshops [3, 6, 7, 8, 11, 12]. Furthermore, ifwe assume �� 1 for the problems 1/rm, �, man/Cmax

and 1/rm, �, man/¥ Ci, they model the following appli-cation: Due to the need to perform preventive mainte-nance, the machine has unavailable time windows, butone can decide on the time from a given time set � toperform the maintenance activity. After maintenance, theprocessing time of a job will be changed according to itsmodifying rate.

The quality of an approximation algorithm is usuallymeasured by its worst-case ratio. Specifically, for themakespan problem, let CA(I) (or briefly CA) denote themakespan produced by an approximation algorithm A, andCOPT(I) (or briefly COPT) denote the makespan produced byan optimal off-line algorithm. Then the worst-case ratio ofalgorithm A is defined as the smallest number c such that forany instance I, CA(I) � cCOPT(I).

In this paper we analyze the computational complexity ofseveral related cases of both problems. For most of thesecases, we either provide their polynomial time algorithms,or prove their NP-hardness. The NP-hardness results arelisted in Table 1. In the table, we omit the case �� 2 ands1 � 0 of the problems, since all are NP-hard trivially fromthe NP-hardness of the case �� 1 and s1 � 0. Comparedto the results that 1/rm/Cmax and 1/rm/¥ Ci are polynomi-ally solvable [9], the problems under consideration areNP-hard even for some special cases. Furthermore, for themakespan problem, we present a pseudo-polynomial timealgorithm and a fully polynomial time approximationscheme (FPTAS) for all the NP-hard cases. For the totalcompletion time problem, we provide a pseudo-polynomialtime algorithm for the special case where pi � pj implies�ipi � �jpj for all i and j. The algorithmic results for theNP-hard problems are listed in Table 2.

This paper is organized as follows. Section 2 considersthe makespan problem. Section 3 considers the total com-pletion time problem. Finally, Section 4 presents someconcluding remarks.

Table 1. Summary of NP-hardness results.

�� 1

�� 2 and s1 � 0s1 � 0 s1 � 0

1/rm, �, opt/Cmax P [9] NP-hard if all �i � 1 P if all �i � 1, and NP-hard else1/rm, �, man/Cmax P trivially NP-hard if all �i � 1 P if all �i � 1, and NP-hard else

1/rm, �, opt/¥ Ci P [9] NP-hard if all �i � 1 [10]P if all �i � 1, and NP-hard elseOpen if all �i � 1

1/rm, �, man/¥ Ci P trivially NP-hard if all �i � 1 [1, 10]P if all �i � 1, and NP-hard elseOpen if all �i � 1

362 Naval Research Logistics, Vol. 52 (2005)

2. MINIMIZING THE MAKESPAN

2.1. NP-Hardness

We first consider the optional case. It is clear that if � �{s1 � 0}, then the problem 1/rm, �, opt/Cmax is easilysolved by a method similar to that in [9]. In fact, if t � ¥i�1

n

�ipi � ¥i�1n pi, then it is optimal to sequence the jobs

arbitrarily without the rate-modifying activity. Otherwise,perform the rate-modifying activity at time zero and thenprocess the jobs in an arbitrary order. In the followingTheorem 1 and Proposition 2, we show that the optionalcase of the problem becomes intractable if there is one moreitem in �, or � � {s1 � 0} and all the modifying rates areless than 1.

THEOREM 1: If � � {s1 � 0, s2 � 0}, the problem1/rm, �, opt/Cmax is NP-hard.

PROOF: We show the result by reducing the PartitionProblem to this problem. Given an instance I of the PartitionProblem with a set of positive integers H � {h1, h2, . . . ,hn} and B � 1

2¥hi�H hi, where B � 1, we construct an

instance II of the problem 1/rm, �, opt/Cmax as follows:The permitted start-time set of the rate-modifying activity is� � {s1 � 0, s2 � B} and the duration of the rate-modifying activity r is t � 1. Associated with each hi, i �1, . . . , n, is job Ji with pi � hi and �i � B, i � 1, 2, . . . ,n. In addition, there is a job Jn�1 with pn�1 � B � B2 �3 and �n�1 � 1/(B � B2 � 3). Define the threshold Y �B � B2 � 2. We prove that instance I has a solution if andonly if instance II has a schedule with makespan no greaterthan Y.

If I has a solution, then there exist two subsets H1 and H2

of H such that H1 � H2 � H, H1 � H2 � A and ¥hi�H1

hi � ¥hi�H2hi � B. We can select s2 as the start-time of

the rate-modifying activity r. We process the jobs in { Ji �hi � H1} before r (recall that ¥hi�H2

pi � B � s2), andprocess job Jn�1 immediately after r, followed by the jobsin { Ji � hi � H2}. As a result, we obtain a feasible schedulewith makespan

Cmax � s2 � t � �n�1pn�1 � �hi�H2

�ipi

� B � 1 � 1 � B2 � Y.

Hence, we obtain a solution for II.Next, suppose II has a schedule with makespan no greater

than Y. It is clear that the rate-modifying activity must takeplace and job Jn�1 must be processed after the rate-modi-fying activity. Otherwise, we have

Cmax � pn�1 � B � B2 � 3 � Y,

contradicting the assumption that Cmax � Y. Furthermore,if the rate-modifying activity starts at time s1 � 0, we have

Cmax � s1 � t � �i�1

n�1

�ipi � 2B2 � 2 � Y,

since B � 1. Hence, we conclude that r must start at times2 � B. Then there exists a schedule in the form of (R1, r,R2) with Cmax � Y. R1 and R2 are subsets of the jobs suchthat the jobs in R1 are completed before r while those in R2

are processed after r. So, R1 � R2 � { J1, . . . , Jn, Jn�1},and recall that Jn�1 � R2. It is clear that ¥Ji�R1

pi � s2 �B. If ¥Ji�R1

pi � B, then ¥Ji�R2�{ Jn�1} pi � B. Thisimplies that

Cmax � s2 � t � �Ji�R2��Jn�1�

�ipi � �n�1pn�1 � B

� B2 � 2 � Y,

which is a contradiction to the assumption that Cmax � Y.So, we see that ¥Ji�R1

pi � ¥Ji�R2�{ Jn�1} pi � B, and,hence, we obtain a solution for instance I since hi �pi. �

Note that if all the modifying rates �i � 1 and � �{s1 � 0, s2 � 0, . . . , sm � 0}, the problem 1/rm, �,opt/Cmax is still easily solved by the method given in [9]. Infact, since all the modifying rates are no greater than 1, therate-modifying activity must take place at time s1 � 0 ifnecessary. Note that the total marginal gain by performingthe rate-modifying activity at time zero is ¥i�1

n (1 � �i) pi.Hence, if ¥i�1

n (1 � �i) pi � t, then it is optimal tosequence the jobs arbitrarily without the rate-modifying

Table 2. Summary of the algorithms.

�� 1

�� 2 and s1 � 0s1 � 0 s1 � 0

1/rm, �, opt/Cmax O(n) [9] Pseudo-polynomially solvable in O(nmsm) time,and an FPTAS running in O(mn2/�) exists

Polynomially solvable in O(n) timeif all �i � 11/rm, �, man/Cmax O(n)

1/rm, �, opt/¥ Ci O(n) [9] Open in general, and pseudo-polynomially1/rm, �, man/¥ Ci O(n) solvable for the agreeable rate case in

O(nmsm(¥i�1n �ipi)) time

363He, Ji, Cheng: Single-Machine Scheduling with a Restricted Rate-Modifying Activity

activity; otherwise, perform the rate-modifying activity attime zero and then process the jobs in an arbitrary order.

PROPOSITION 2: If � � {s1 � 0}, the problem 1/rm,�, opt/Cmax is NP-hard even when all the modifying ratesare less than 1.

PROOF: We again show the result by reducing the Par-tition problem to this problem. Given an instance I of thePartition problem with a set of positive integers H � {h1,h2, . . . , hn} and B � 1

2¥hi�H hi � 1, we construct an

instance II of the problem 1/rm, �, opt/Cmax as follows:The permitted start-time set of the rate-modifying activity is� � {s1 � 2B} and the duration of the rate-modifyingactivity is t � 1. Associated with each hi, i � 1, . . . , n,the job Ji with pi � 2hi and �i � 1/ 2, i � 1, 2, . . . , n.Define the threshold Y � 3B � 1. It can easily be shownthat instance I has a solution if and only if instance II has aschedule with makespan no greater than Y. �

Now, we analyze the mandatory case. It is trivial that theproblem 1/rm, �, man/Cmax with � � {s1 � 0} is poly-nomially solvable. Lee [7] has shown that 1/nr � a/Cmax isNP-hard, so we conclude that the problem 1/rm, �, man/Cmax with � � {s1 � 0} is NP-hard even when all themodifying rates are no greater than 1. However, using thesame method of reduction as in the proof of Proposition 2,we obtain the following proposition:

PROPOSITION 3: If � � {s1} with s1 � 0, the problem1/rm, �, man/Cmax is NP-hard even when all the modifyingrates are less than 1.

From the paragraph immediately after Theorem 1 and theproof of Theorem 1, it is easy to see that the problem 1/rm,�, man/Cmax with � � {s1 � 0, s2 � 0} is polynomialtime solvable if all �i � 1 and NP-hard else by omitting theconsideration of whether we perform the rate-modifyingactivity.

2.2. Pseudo-Polynomial Time Algorithm

Denote C0 � ¥i�1n pi. From now on, we assume that

C0 � sm. Otherwise, all the jobs can be completed by sm

and the rate-modifying activity will definitely not take placeat that time. Then we can delete sm from �. By repeatingthis argument, the assumption is valid. Next, we construct apseudo-polynomial time algorithm based on dynamical pro-gramming for both the mandatory and optional cases of themakespan problem. Hence, we conclude that the aboveNP-hard cases of the makespan problem are in fact onlyNP-hard in the ordinary sense.

If the rate-modifying activity r takes place at sj, wedefine fj(i, u) as the minimum total actual processing time

of the jobs processed after r if (i) we have assigned jobs J1,J2, . . . , Ji and (ii) the total processing time of the jobsassigned before r is u. Given fj(i � 1, u) for u, 0 � u �sj, we can schedule Ji either before or after r. In the formercase, the total actual processing time of the jobs processedafter r does not change, but u increases by pi. In the lattercase, the makespan increases by �ipi while u remainsunchanged. We have the following initial conditions:

fji, u � � 0, if i � 0, u � 0,��, if i � 0, u 0,

and the recursion for i � 1, . . . , n, and u � 0, . . . , sj is

fji, u � min�fji 1, u pi, fji 1, u � �ipi�.

The optimal makespan for this case is then determined as

Cj � sj � t � min0�u�sj

fjn, u.

Finally, the overall optimal makespan for the mandatorycase is

COPT � min�C1, C2, . . . , Cm�,

and for the optional case is

COPT � min�C0, C1, C2, . . . , Cm�.

It is clear that this algorithm requires at most O(nmsm)time. Hence, the makespan problem can be solved in pseu-do-polynomial time.

2.3. Approximation Algorithm

In this subsection we give an FPTAS for both the prob-lems 1/rm, �, man/Cmax and 1/rm, �, opt/Cmax. We willapply the FPTAS for the classical 0-1 Minimum Knapsackproblem as a subprocedure. Recall that for any instance ofthe 0-1 Minimum Knapsack problem, we are given n items,each with a profit ci and a weight aj, and a knapsack withcapacity C. We wish to put items into the knapsack suchthat the total weight of the selected items is not greater thanC while the total profit of the unselected items is minimized.For this problem, Babat [2] presented an FPTAS with a timecomplexity O(n4/�), and Gens and Levner [4] proposed anFPTAS with a time complexity O(n2/�).

To construct an FPTAS for the makespan problem, it iscrucial to determine which jobs are processed after therate-modifying activity. For any instance I, if an algorithmperforms the rate-modifying activity at time sj � �, then


we construct an instance II of the Minimum Knapsackproblem in the following way: For each job Ji, i � 1, . . . ,n, define an item with profit �ipi and weight pi, and set thecapacity of the knapsack as sj. By applying any FPTAS toinstance II, we obtain a partial solution for instance I.Namely, for every item put into the knapsack by the FP-TAS, we schedule the corresponding job before the rate-modifying activity, and schedule all the remaining jobs afterthe rate-modifying activity. Finally, we choose the best onefrom the m (m � 1) solutions as the output for the man-datory (optional) case.

Algorithm SRM:

1. For every j � 1, 2, . . . , m, perform the rate-modifying activity r at time sj � �. Determine thejobs processed after r by applying the FPTAS forthe Minimum Knapsack problem as above. Denoteby �SRM

j the set consisting of all the jobs processedafter r, and denote by �SRM

j the remaining jobs.Then, the resulting makespan is Cj � sj � t� ¥Ji��SRM

j �ipi if �SRMj � A, and Cj

� ¥Ji��SRMj pi otherwise.

2. Determine the index j0 such that Cj0 � min{C1,C2, . . . , Cm}, output Cj0 for the mandatory caseand min{Cj0, C0} for the optional case, whereC0 � ¥i�1

n pi as above.

THEOREM 4: Algorithm SRM is an FPTAS for theproblem 1/rm, �, man/Cmax and 1/rm, �, opt/Cmax, i.e., forany positive number � � 0, we have CSRM/COPT � 1 � �.

PROOF: It is clear that if the rate-modifying activity doesnot take place in an optimal schedule, then algorithm SRMproduces an optimal solution. Hence, we assume that therate-modifying activity takes place at si in an optimal sched-ule. Denote �OPT � {the jobs processed after the rate-modifying activity in an optimal schedule} and BOPT �¥Ji��OPT

�ipi. Let BKNAPi denote the optimal value of the

constructed instance of the Minimum Knapsack problem(with capacity sj), then we have BKNAP

j � BOPT. Since SRMapplies the FPTAS for the Minimum Knapsack problem, wehave

�Ji��SRM

j

�ipi � 1 � �BKNAPj � 1 � �BOPT. (1)

Hence,

Cj

COPT�

sj � t � ¥Ji��SRMj �ipi

sj � t � BOPT

�sj � t � 1 � �BOPT

sj � t � BOPT� 1 � �. (2)

Therefore, we obtain

CSRM

COPT� 1 � �, (3)

since CSRM � Cj. �

It is clear that algorithm SRM has a time complexityO(mn2/�) by employing the FPTAS for the MinimumKnapsack problem presented by Gens and Levner [4]. Notethat for 1/nr � a/Cmax, Lee [9] showed that LPT has aworst-case ratio of 4/3. The above FPTAS can be applied tothis problem with a worst-case ratio of 1 � �.

3. MINIMIZING THE TOTALCOMPLETION TIME

3.1. NP-Hardness

We again first consider the optimal case. It is clear that if� � {s1 � 0}, then the problem 1/rm, �, opt/¥ Ci is easilysolved [9]. In fact, regardless of whether the rate-modifyingactivity takes place at time zero, processing the jobs in theShortest Processing Time (SPT) order minimizes the totalcompletion time. Hence, by comparing the objective valuesof the two schedules of performing and not performing therate-modifying activity at time zero, we can obtain an op-timal solution. Similarly, it also implies that the problem issolvable if all the modifying rates �i � 1 and � � {s1 �0, s2 � 0, . . . , sm � 0} for the optional case. Theorem 5shows that the problem becomes intractable for � � {s1 �0, s2 � 0} for the optional case. Theorem 8 shows that theproblem 1/rm, �, opt/¥ Ci with � � {s1 � 0} is alsointractable even for all �i � 1.

THEOREM 5: The problem 1/rm, �, opt/¥ Ci is NP-hardif � � {s1 � 0, s2 � 0}.

PROOF: We reduce the Partition problem to this prob-lem. Given an instance I of the Partition problem with a setof positive integers H � {h1, h2, . . . , hn} and B � 1

2¥hi�H hi, we construct an instance II of 1/rm, �, opt/¥ Ci

as follows:Associated with each hi, i � 1, 2, . . . , n, are two jobs

J1,i and J2,i. In addition, there is a job J0. Their processingtimes are as follows:

p1,i � 2n�1B � 2iB � hi, p2,i � 2n�1B � 2iB,

i � 1, 2, . . . , n,

and


p0 � Y � 1,

where Y is the threshold to be defined below. The permittedstart-time set of the rate-modifying activity is � � {s1 �0, s2 � (n � 1)2n�1B � B}, and the duration of therate-modifying activity is t � 1.

Define E � ¥i�1n (n � 1 � i) p2,i � 2nB2 � 2nB and

F � E � E ¥i�1n (n � 1 � i)ni. The modifying rates of

the jobs are as follows:

�1,i �

F � niE �2nBhi

n � 1 i

p1,i, �2,i �

F � niE

p2,i,

i � 1, 2, . . . , n,

and

�0 �1

p0.

The threshold is

Y � �i�1

n

n � 1 ip2,i � �i�1

n

n � 1 i�2,ip2,i � n � 1

� s2 � 2 � 2nB2 � 2nB

� F � F �i�1

n

n � 1 i � n � 1s2 � 2

� F �nn � 1

2F � n � 1s2 � 2. (4)

It is clear that the construction of instance II can be per-formed in polynomial time. We first discuss the structure ofthe solution for II.

Given a sequence, let [i] denote the ith position of thesequence. Let R1,i denote an element of the set {A, { J1,i},{ J2,i}, { J1,i, J2,i}} and R2,i � { J1,i, J2,i}�R1,i for 1 �i � n, where A is the empty set. A schedule in the form of(R1,1, R1,2, . . . , R1,n, r, J0, R2,1, R2,2, . . . , R2,n) iscalled a basic schedule.

LEMMA 6: If there exists a solution for II, then thereexists a basic schedule for II, in which there are exactly njobs in the partial schedule (R1,1, R1,2, . . . , R1,n).

PROOF: For any solution S for II, it is clear that job J0

must be processed after the rate-modifying activity r sincep0 � Y. In addition, r must take place at time s2 � (n �

1)2n�1B � B; otherwise, we have ¥i�12n�1 Ci � ¥i�1

2n

(2n � 1 � i)(F � niE) � 2(2n � 1) � Y. Denote by S1

the partial schedule consisting of the jobs processed beforer, and by S2 the partial schedule containing the remainingjobs after r. Then J0 � S2. Rescheduling S1 and S2 innondecreasing order of the actual processing times, respec-tively, we obtain two partial schedules S�1 and S�2. Denote byS� the new schedule by combining S�1 and S�2. From the SPTrule and the structure of II, it is easy to see that S� is a basicschedule.

We assume that S� � (R1,1, R1,2, . . . , R1,n, r, J0, R2,1,R2,2, . . . , R2,n) and there are w jobs in (R1,1, R1,2, . . . ,R1,n). If w � n � 1, then we have ¥i�1

w p[i] � (n �1)2n�1B � s2, which contradicts the definition of S�1. Ifw � n � 1, then �[i]p[i] � F for w � 2 � n � 1 � 2n �1, and we thus have

�i�1

2n�1

C i� � n � 2s2 � t � n � 2�0p0

� �i�n�1

2n�1

2n � 2 i� i�p i� � n � 2s2 � 1 � n � 2

� �i�n�1

2n�1

2n � 2 iF � n � 2s2 � 2

�n � 1n � 2

2F � Y, (5)

which is a contradiction to the assumption that ¥i�12n�1

C[i] � Y. Therefore, we have w � n. �

A schedule in the form of ( Jk1,1, Jk2,2, . . . , Jkn,n, r, J0,J3�k1,1, J3�k2,2, . . . , J3�kn,n) for 1 � i � n and ki � 1or 2, is called a canonical schedule.

LEMMA 7: If there exists a solution for II, then thereexists a canonical schedule for II.

PROOF: From Lemma 6, there exists a basic schedule inthe form of

R1,1, R1,2, . . . , R1,n, r, J0, R2,1, R2,2, . . . , R2,n,

and there are exactly n jobs in the partial schedule (R1,1,R1,2, . . . , R1,n).

We first show that there is exactly one job in R1,n. IfR1,n � { J1,n, J2,n}, then we have ¥i�1

n p[i] � (n �2)2n�1B � p1,n � p2,n � s2, contradicting the structure ofthe basic schedule. If R2,n � { J1,n, J2,n}, then, similar to(5), we have


�i�1

2n�1

C i� � n � 1s2 � 2 � �i�n�2

2n�1

2n � 2 iF

� 2� 2n�p 2n� F � � 2n�1�p 2n�1� F

� n � 1s2 � 2 �nn � 1

2F � 2F � nnE F

� F � nnE � 2nBhn F � Y � 3nnE F � Y,

(6)

which is a contradiction to the assumption that ¥i�12n�1

C[i] � Y. Using the same method, by induction, we canobtain that there is exactly one job in each R1,i, 1 � i � n.Therefore, we obtain a canonical schedule for II. �

Now, returning to the proof of Theorem 5, we show thatinstance I has a solution if and only if instance II has asolution.

If II has a solution, then from Lemma 7, there exists acanonical schedule

Jk1,1, Jk2,2, . . . , Jkn,n, r, J0, J3�k1,1, J3�k2,2, . . . , J3�kn,n.

We construct a corresponding subset of I as H1 � {hi �ki � 1, 1 � i � n}. If ¥hi�H1

hi � B, then we have

�i�1

n

p i� � n2n�1B � �i�1

n

2iB � �hi�H1

hi � n2n�1B

� 2n�1B 2B � �hi�H1

hi � n � 12n�1B B � s2,

(7)

which contradicts the structure of the canonical schedule.So, ¥hi�H1

hi � B. We suppose ¥hi�H1hi � B � 1.

From the definition of Y and the structure of II, we have

Y � �i�1

n

n � 1 ip2,i � �i�1

n

n � 1 i�2,ip2,i � n � 1

� s2 � 2 � 2nB2 � 2nB. (8)

On the other hand, the total completion time is

�i�1

2n�1

C i� � �i�1

n

n � 1 ip i� � �i�n�2

2n�1

2n � 2 i� i�p i�

� n � 1s2 � 2 � �i�1

n

n � 1 ip i�

� �i�1

n

n � 1 i� n�1�i�p n�1�i� � n � 1s2 � 2. (9)

Combining (8) and (9), we have

�i�1

2n�1

C i� Y � �hi�H1

n � 1 ihi � �hi�H1

2nBhi 2nB2

2nB � �hi�H1

hi � �hi�H1

2nBhi 2nB2 2nB � �hi�H1

hi

� 2nB� �hi�H1

hi B� 2nB. (10)

So, ¥i�12n�1 C[i] � Y � 0 since ¥hi�H1

hi � B � 1, acontradiction. Thus, we obtain ¥hi�H1

hi � ¥hi�H1hi � B,

which is a solution for I.If I has a solution, i.e., there is a subset H1 � {hi1

,hi2

, . . . , hiw} with ¥hi�H1

� B. We can construct a cor-responding schedule S as

Jk1,1, Jk2,2, . . . , Jkn,n, r, J0, J3�k1,1, J3�k2,2, . . . , J3�kn,n,

where ki � 1 for i � {i1, i2, . . . , iw}, and ki � 2 else.Similar to (7), we have ¥i�1

n p[i] � n2n�1B � ¥i�1n 2iB �

¥hi�H1hi � n2n�1B � (2n�1B � 2B) � ¥hi�H1

hi �(n � 1)2n�1B � B � s2, and thus the schedule S is acanonical schedule. Then, similar to (10), we have

�i�1

2n�1

C i� Y � �hi�H1

n � 1 ihi 2nB � 0.

Therefore, we obtain a solution for II. This completes theproof of Theorem 5. �

THEOREM 8: If � � {s1} with s1 � 0, the problem1/rm, �, opt/¥ Ci is NP-hard even when all the modifyingrates are no greater than 1.

PROOF: We prove this theorem by modifying the NP-hardness proof of the problem 1/nr � a/¥ Ci (i.e., 1/rm, �,man/¥ Ci with �� 1 and all �i � 1). In [10], Lee andLiman presented a proof by reducing the Even-Odd Parti-tion problem to the problem 1/nr � a/¥ Ci.

The Even-Odd Partition problem: Given n � Z� and a setX � { x1, x2, . . . , x2n} of positive integers, where xi �xi�1 for 1 � i � 2n, does there exist a partition of X intosubsets X1 and X2 such that ¥x�X1

x � ¥x�X2x and such

that for each i, 1 � i � n, X1 (and, hence, X2) containsexactly one of { x2i�1, x2i}?

For any instance of the Even-Odd Partition problem, thecorresponding instance of the problem 1/rm, �, man/¥ Ci

with �� 1 and all �i � 1 was constructed as follows:


Number of jobs: 2n � 1,Processing times: pi � M � xi, i � 1, 2, . . . ,2n, p2n�1 � P,Start time of the rate-modifying activity: s1 � nM �Z,Duration of the rate-modifying activity: t � M,Threshold: y � ¥i�1

n (n � i � 1)( x2i�1 � x2i) �2n(n � 1) M � nZ � {2(nM � Z) � M} � P,

where Z � (¥i�1n xi)/ 2, M � 2nZ, and P � ¥i�1

n

(n � i � 1)( x2i�1 � x2i) � 2n(n � 1) M � nZ �{2(nM � Z) � M}.

To prove our theorem, we only modify it to the instanceof our problem 1/rm, �, opt/¥ Ci with � � {s1 � 0} asfollows: We add one more job J0 with p0 � y� � 1 and�0 � 1/p0, where y� � y � (nM � Z � 1) � (n � 1)is the new threshold. The processing times and modifyingratios of the other jobs, s1 and t are unchanged.

Note that for the constructed instance of our problem1/rm, �, opt/¥ Ci with � � {s1 � 0}, the rate-modifyingactivity must be performed. Otherwise, we have ¥i�1

n Ci �p0 � y�. Furthermore, �0p0 � 1 � pi, i � 1, . . . , 2n �1, job J0 must be processed right after the rate-modifyingactivity. It can be deduced that the contribution of J0 to theobjective is (nM � Z � 1) � (n � 1) � y� � y.Therefore, using the proof in [10], we conclude that thereexists a solution for the instance of the Even-Odd Partitionproblem if and only if there exists a schedule for theconstructed instance of our problem with total completiontime y�. �

Whether the problem 1/rm, �, opt/¥ Ci with �� {s1 � 0} and all �i � 1 is NP-hard is open.

Now we consider the mandatory case. Similar to themakespan problem, it is trivial that the problem 1/rm, �,man/¥ Ci with � � {s1 � 0} is polynomially solvable.Since 1/nr � a/¥ Ci is NP-hard [1, 10], we obtain thefollowing proposition:

PROPOSITION 9: If � � {s1 � 0}, the problem 1/rm,�, man/¥ Ci is NP-hard even when all the modifying ratesare no greater than 1.

Whether the problem 1/rm, �, man/¥ Ci with �� {s1 � 0} and all �i � 1 is NP-hard is open. From theparagraph right before Theorem 5 and the proof of Theorem5, it is easy to see that the problem 1/rm, �, man/¥ Ci with� � {s1 � 0, s2 � 0} is polynomially solvable if all �i �1 and NP-hard otherwise, by omitting the consideration ofwhether we perform the rate-modifying activity.

We conjecture that the above NP-hard problems might beNP-hard in the strong sense. Hence, it might be impossibleto find a pseudo-polynomial time algorithm to solve them

optimally. However, in the next subsection, we propose aspecial case that is pseudo-polynomially solvable (but it isstill NP-hard), and that has application in the real world.

3.2. A Special Pseudo-Polynomial Solvable Case

We now present a pseudo-polynomial time algorithm fora special case of the problems 1/rm, �, man/¥ Ci and 1/rm,�, opt/¥ Ci, where pi � pj implies �ipi � �jpj for all i andj (called the agreeable rate assumption). The agreeable rateassumption is quite realistic in real applications [9]. Forexample, it includes the case that all �i are the same. Recallthat the problem 1/nr � a/¥ Ci (i.e., 1/rm, �, man/¥ Ci

with �� 1 and all �i � 1) satisfies the agreeable rateassumption, and is NP-hard.

It is clear that in an optimal schedule, jobs assignedbefore and after the rate-modifying activity r should besequenced in the SPT order, respectively, where the SPTorder after the rate-modifying activity means that �[i]p[i] ��[i�1]p[i�1] for all i. In the remainder, we assume that thejobs are reindexed in the SPT order. The pseudo-polynomialtime algorithm is based on dynamic programming, which issimilar to the Dynamic Programming Algorithm 2 in [9] forthe nonrestricted problem 1/rm/¥ wiCi presented. We pro-vide it here for the sake of completeness.

Let Z0 � ¥i�1n (n � 1 � i) pi, which is the objective

value of the case where we do not perform the rate-modi-fying activity. Assuming that the rate-modifying activitytakes place at time sj � �, we assign jobs in the sequencefrom left to right. Define fj(i, u, v) as the minimum totalcompletion time if (i) we have assigned jobs J1, J2, . . . , Ji,(ii) the total actual processing time of the jobs assignedbefore the rate-modifying activity is u, and (iii) the totalactual processing time of the jobs assigned before the rate-modifying activity is v. Given fj(i � 1, u, v), we canschedule job Ji either before or after r. In the former case,the total completion time increases by u � pi, and uincreases by pi while v remains unchanged. In the lattercase, the total completion time increases by sj � t � v ��ipi, and v increases by �ipi while u remains unchanged.Let T � ¥i�1

n �ipi. We have the following initial condi-tions:

fji, u, v � � 0, if i � 0, u � 0, v � 0,��, if i � 0, u 0, or v 0

and the recursion for i � 1, . . . , n, u � 0, . . . , sj, and v� 0, . . . , T is

fji, u, v � min�fji 1, u pi, v

� u, fji 1, u, v �ipi � sj � t � v�.


The optimal value of this case is then determined as

Zj � minu�0,1,. . .,sj;v�0,1,. . .,T

fjn, u, v.

Finally, the overall optimal value for the mandatory case is

min�Z1, . . . , Zm�,

and for the optimal case is

min�Z0, Z1, . . . , Zm�.

It is clear that the algorithm requires at most O(nmsmT)time. Hence, the problems 1/rm, �, man/¥ Ci and 1/rm, �,opt/¥ Ci with agreeable modifying rates can be solved inpseudo-polynomial time.

4. CONCLUSION AND DISCUSSION

We studied the constrained single machine schedulingproblem with a rate-modifying activity. Two objectiveswere considered. We analyzed the computational complex-ity of several related cases. For most of these cases, weeither provide their polynomial time algorithms, or provetheir NP-hardness. Some results were obtained easily fromthose provided in [7, 9, 10], since theoretically, if we onlyallow a few fixed start-times for the rate-modifying activity,then similar approaches (in terms of the design of optimalalgorithms, as well as complexity analysis) used in solvingmachine scheduling problems with availability constraintscan be applied (we just need to try all possible allowablerate-modifying activity start-times, and find the best oneamong them); and if the number of permitted start-times ofrate-modifying activity is large, or it is optional to have nosuch activity, then the analysis may be similar to that in [9].However, other results cannot be so derived, as the con-straint of permitted start-times changes the complexity sta-tus. This can be seen from the fact that the problems underconsideration are NP-hard even for some special caseswhile the unrestricted problems 1/rm/Cmax and 1/rm/¥ Ci

are polynomially solvable [9].For the makespan objective, we further presented a pseu-

do-polynomial time algorithm and an FPTAS. For the total

completion time objective, we presented a pseudo-polyno-mial algorithm for the special case with agreeable modify-ing rates. The design of approximation algorithms for thisobjective is interesting. For future research, it is promisingto extend our results to parallel machines or flowshops.

ACKNOWLEDGMENTS

The authors wish to thank an associate editor and theanonymous referees for their helpful comments on earlierversions of this paper.

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Single machine scheduling with a restricted rate-modifying activity