Single-machine scheduling with sum-of-logarithm-processing-times-based learning considerations

Information Sciences 179 (2009) 3127–3135

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Information Sciences

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Single-machine scheduling with sum-of-logarithm-processing-times-basedlearning considerations

T.C.E. Cheng a,*, Peng-Jen Lai b, Chin-Chia Wu c, Wen-Chiung Lee c

a Department of Logistics and Maritime Studies, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kongb Department of Mathematics, National Kaohsiung Normal University, Kaohsiung, Taiwanc Department of Statistics, Feng Chia University, Taichung, Taiwan

a r t i c l e i n f o a b s t r a c t

Article history:Received 15 July 2008Received in revised form 21 February 2009Accepted 4 May 2009

Keywords:SchedulingLearning effectSingle-machine

0020-0255/$ - see front matter � 2009 Elsevier Incdoi:10.1016/j.ins.2009.05.002

* Corresponding author. Tel.: +852 2766 5216; faE-mail address: [email protected] (T.C.E. C

Scheduling with learning effects has attracted growing attention of the scheduling researchcommunity. A recent survey classifies the learning models in scheduling into two types,namely position-based learning and sum-of-processing-times-based learning. However,the actual processing time of a given job drops to zero precipitously as the number of jobsincreases in the first model and when the normal job processing times are large in the sec-ond model. Motivated by this observation, we propose a new learning model where theactual job processing time is a function of the sum of the logarithm of the processing timesof the jobs already processed. The use of the logarithm function is to model the phenom-enon that learning as a human activity is subject to the law of diminishing return. Underthe proposed learning model, we show that the scheduling problems to minimize themakespan and total completion time can be solved in polynomial time. We further showthat the problems to minimize the maximum lateness, maximum tardiness, weightedsum of completion times and total tardiness have polynomial-time solutions under someagreeable conditions on the problem parameters.

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1. Introduction

A huge amount of research has been done in the field of deterministic scheduling, in which the job processing times areassumed to be fixed and known from the first job to be processed to the last job to be completed [18,23,26,27,29]. However,in many realistic situations, the efficiency of the production facility (e.g., a machine or a worker) improves continuously overtime [9,10,22,28,32]. As a result, the production time of a given product is shorter if it is scheduled (and so processed) later.For instance, Biskup [2] claimed that the repeated processing of similar tasks improves worker skills because workers areable to perform setups, deal with machine operations and software or handle raw materials and components at a faster pace.This phenomenon is known as the learning effect in the literature.

Biskup [2] and Cheng and Wang [4] were among the pioneers that introduced the learning effect to the scheduling field.Since then, many researchers have devoted large amounts of effort to this relatively young but vivid area of scheduling re-search. For example, Mosheiov and Sidney [25] considered a model in which the learning effects gained from doing somejobs are stronger than those from the other jobs, i.e., the so-called job-dependent learning model. Lin [24] presented thecomplexity results for a single-machine scheduling problem to minimize the number of late jobs where the processing timesof the jobs are defined by positional learning effects. Janiak and Rudek [13] showed that the single-machine scheduling prob-lem to minimize the number of late jobs with a positional learning effect is strongly NP-hard. Lee et al. [21] considered a

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3128 T.C.E. Cheng et al. / Information Sciences 179 (2009) 3127–3135

bicriterion single-machine scheduling problem, and Lee and Wu [20] considered the problem of minimizing the total com-pletion time in a two-machine flowshop. Wang [30] studied a model in which the job processing times are functions of theirstarting times and positions in the sequence. Koulamas and Kyparisis [16] introduced a general sum-of-job-processing-times-based learning effect model for scheduling, in which employees learn more if they perform a job with a longer pro-cessing time. Kuo and Yang [19] considered a sum-of-job-processing-times-based learning effect model. They providedthe optimal solution of the total completion time problem. Recently, Biskup [3] provided a comprehensive review of sched-uling research with learning considerations. In particular, he classified the learning models into two types, namely position-based learning and sum-of-processing-times-based learning. He further claimed that position-based learning assumes thatlearning takes place by processing time-independent operations like setting up machines. This seems to be a realisticassumption for the case where the actual processing time of a job is mainly machine-driven and has no (or near to zero)human interference. The sum-of-processing-times-based approach takes into account the experience that workers havegained from producing the jobs. This might, e.g., be the case for offset printing, where running the press itself is a highlycomplicated and error-prone process.

Recently, Eren [7] proposed a non-linear mathematical programming model for a single-machine scheduling problemwith unequal release dates and learning effect. Wang et al. [31] studied some scheduling problems with a time-dependentlearning effect. They provided several examples to show that the classical scheduling rules do not yield optimal solutions forthe problems to minimize the weighted sum of completion times, maximum lateness and number of tardy jobs. They alsoanalyzed their worst-case error bounds for the classical scheduling rules. Eren and Güner [6] considered a two-machineflowshop with position-based learning where the objective is to minimize a weighted sum of the total completion timeand makespan. They applied integer programming to solve problems up to 30 jobs, and utilized a heuristic algorithm anda tabu search based heuristic algorithm to handle large-sized problems. Janiak and Rudek [11] proposed an experience-basedlearning effect. They proved that the problem to minimize the makespan under the Bachman and Janiak [1] model remainspolynomially solvable when the experience-based approach is applied, but the same problem under the more general Chengand Wang [4] model becomes strongly NP-hard in the presence of the new learning effect. Janiak and Rudek [12] introduceda new model of learning into the scheduling field that relaxes one of the rigid constraints by assuming that each job providesa different experience to the processor. They formulated the shape of the learning curve as a non-increasing k-stepwise func-tion. Furthermore, they proved that the makespan problem is polynomially solvable if every job provides the same experi-ence to the processor, and it becomes NP-hard if the experiences are different. Moreover, Janiak and Rudek [14] proposed anew experience-based learning model, where the job processing times are described by S-shaped functions dependent onthe experience of the processor. They proved that the problem to minimize the makespan on a single processor is NP-hardor strongly NP-hard with most of the commonly considered learning models. Janiak et al. [15] investigated the single-processor problem to minimize the makespan with an S-shaped learning model. They proved that the problem is stronglyNP-hard even if the experience provided by each job is equal to its normal processing time. They constructed a branch-and-bound algorithm and some fast heuristic methods to find the optimal and near-optimal solutions, respectively.

Cheng et al. [5] introduced a new scheduling model with learning effects in which the actual processing time of a job is afunction of the sum of the normal processing times of the jobs already processed and of the job’s scheduled position. In addi-tion, Wu and Lee [33] studied the impact of the learning effect on the problem to minimize the total completion time in anm-machine permutation flowshop. Eren [8] developed a mathematical programming model for a parallel-machine schedul-ing problem with a learning effect. Koulamas and Kyparisis [17] introduced some type of learning effect on setup timeswhere the setup times can theoretically grow substantially when the batch size n is large.

However, the actual processing time of a given job drops to zero precipitously as the number of jobs increases in the po-sition-based learning model and when the normal job processing times are large in the sum-of-processing-times-basedlearning model classified in Biskup [3]. Motivated by this observation, we propose a new learning model where the actualjob processing time is a function of the sum of the logarithm of the processing times of the jobs already processed. The use ofthe logarithm function can be justified on the grounds that learning, like other human activities, is subject to the law ofdiminishing return. The remainder of this paper is organized as follows. We present the problem formulation in the nextsection. In Section 3, we provide polynomial-time solution algorithms for some single-machine problems. Finally, we con-clude the findings in the last section.

2. Problem formulation

Formulation of the proposed learning model in the single-machine case is as follows. There are n jobs ready to be pro-cessed on a single-machine. Each job j has a normal processing time pj with ln pj P 1, a weight wj and a due date dj. Dueto the learning effect, the actual processing time of job j is

pj½r� ¼ pj 1þXr�1

l¼1

ln p½l�

!a

;

if it is scheduled in the rth position in a sequence, where p½l� is the normal processing time of the job scheduled in the lthposition in the sequence, and a 6 0 is the learning index. The proposed learning model is a special case of the experi-ence-based learning model of Janiak and Rudek [14] with ej ¼ ln pj for j ¼ 1;2; . . . ;n.

T.C.E. Cheng et al. / Information Sciences 179 (2009) 3127–3135 3129

For a given sequence S of processing the n jobs on the machine, let CjðSÞ, LjðSÞ ¼ CjðSÞ � dj and TjðSÞ ¼maxf0;CjðSÞ � djgdenote the completion time, lateness, and tardiness of job j under sequence S, respectively. Moreover, letCmax ¼max16j6nfCjg, Lmax ¼max16j6nfLjg and Tmax ¼max16j6nfTjg represent the makespan, maximum lateness, and maxi-mum tardiness, respectively. The objective is to find a sequence that minimizes one of the following criteria:Cmax;

PðwjÞCj; Lmax; Tmax, and

PTj.

3. Some solvable single-machine problems

In this section we show that some single-machine scheduling problems remain polynomially solvable under the proposedmodel. Before presenting the results, we first present several lemmas as follows (the proofs of the lemmas are given in theAppendix):

Lemma 1 (Kuo and Yang [19]). Let HðyÞ ¼ 1þ ayð1þ yÞa�1 � ð1þ yÞa, then HðyÞP 0 for 0 < y < 1, and a < 0.

Lemma 2. Let FðxÞ ¼ 1þ c0að1þ c0xÞa�1 � ð1þ c0xÞa, then FðxÞP 0 for x P 1, 0 < c0 < 1, and a < 0.

Lemma 3. Let GðkÞ ¼ ðk� 1Þ þ ð1þ c0 ln kþ c0xÞa � kð1þ c0xÞa, then GðkÞP 0 for k P 1, 0 < c0 < 1, x P 1, and a < 0.

Lemma 4. Let HðyÞ ¼ c1½1� ð1þ yÞa� � 1c1½1� ð1þ yþ y ln c1Þa�, then HðyÞ > 0 for 0 < y < 1, c1 > 1, and a < 0.

Lemma 5. Let FðxÞ ¼ c1½1� ð1þ c0xÞa� � 1c1½1� ð1þ c0xþ c0 ln c1Þa�, then FðxÞP 0 for x P 1, c1 P 1, 0 < c0 < 1, and a < 0.

Lemma 6. Let HðyÞ ¼ 1þ c1½1� ð1þ yÞa� þ ayc1ð1þ yþ y ln c1Þa�1, then HðyÞP 0 for 0 < y < 1, c1 > 1, and a < 0.

Lemma 7. Let FðxÞ ¼ 1þ c1½1� ð1þ c0xÞa� þ ac0c1ð1þ c0xþ c0 ln c1Þa�1, then FðxÞP 0 for x P 1, c1 P 1, 0 < c0 < 1, and a < 0.

Lemma 8. Let GðkÞ ¼ ðk� 1Þ þ c1k½1� ð1þ c0xÞa� � 1c1½1� ð1þ c0 ln kþ c0xþ c0 ln c1Þa�, then GðkÞP 0, for k P 1; c1 P 1; 0 <

c0 < 1; x P 1 and a < 0.

We prove the properties of the optimal solutions for some single-machine problems using the pairwise job interchangetechnique. Let S and S0 be two job schedules where the difference between S and S0 is a pairwise interchange of two adjacentjobs i and j, i.e., S ¼ ðp; i; j;p0Þ and S0 ¼ ðp; j; i;p0Þ, where p and p0 each denote a partial sequence. Furthermore, we assumethat there are r � 1 scheduled jobs in p and A is the completion time of the last job in p. Under S, the completion timesof jobs i and j are, respectively,

CiðSÞ ¼ Aþ pi 1þXr�1

l¼1

ln p½l�

!a

ð1Þ

and

CjðSÞ ¼ Aþ pi 1þXr�1

l¼1

ln p½l�

!a

þ pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

: ð2Þ

Similarly, the completion times of jobs j and i in S0 are, respectively,

CjðS0Þ ¼ Aþ pj 1þXr�1

l¼1

ln p½l�

!a

ð3Þ

and

CiðS0Þ ¼ Aþ pj 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

: ð4Þ

Property 1. For the 1=pj½r� ¼ pj 1þPr�1

l¼1 ln p½l�� a

=Cmax problem, the optimal schedule is obtained by sequencing jobs in the

shortest processing time (SPT) order.

Proof. Suppose that pi 6 pj. To show that S dominates S0, it suffices to show that CjðSÞ 6 CiðS0Þ. Taking the difference between(2) and (4), we have

CiðS0Þ � CjðSÞ ¼ ðpj � piÞ 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

: ð5Þ


Substituting k ¼ pj

pi, x ¼ ln pi, and c0 ¼ 1

1þPr�1

l¼1ln p½l�

into (5), we derive from Lemma 3 that

CiðS0Þ � CjðSÞ ¼ pi 1þXr�1

l¼1

ln p½l�

!a

½ðk� 1Þ þ ð1þ c0 ln kþ c0xÞa � kð1þ c0xÞa�P 0;

since k P 1, x P 1, 0 < c0 < 1, and a < 0. Thus, S dominates S0. Therefore, repeating this interchange argument for jobs notsequenced in the SPT order completes the proof of the property. h


l¼1 ln p½l�� a.P

Cj problem, the optimal schedule is obtained by sequencing jobs in the SPTorder.

Proof. The proof is omitted since it is similar to that of Property 1. h

Smith [29] showed that the weighted smallest processing time (WSPT) rule yields an optimal schedule for the classicalsingle-machine scheduling problem to minimize the total weighted completion time, i.e., sequencing jobs in non-decreasingorder of pj=wj, where wj is the weight of job j. However, this rule does not yield an optimal schedule under the proposedmodel, as shown in the following example.

Example 1. Let n ¼ 2, p1 ¼ 12, p2 ¼ 8, w1 ¼ 25, w2 ¼ 15, and a ¼ �0:5. The WSPT sequence (1,2) yields a total weightedcompletion time of 544.28, while the sequence (2,1) yields the optimal value of 490.96.



wjCj problem, the optimal schedule is obtained by sequencing jobs in non-

decreasing order of pi=wi if the processing times and the weights are agreeable, i.e., pi 6 pj implies wi P wj for all the jobs i and j.

Proof. Suppose that pj

piP wj

wiP 1. Since pi 6 pj, we have from Property 1 that CjðSÞ 6 CiðS0Þ. Thus, to show that S dominates S0,

it suffices to show that wiCiðSÞ þwjCjðSÞ 6 wjCjðS0Þ þwiCiðS0Þ. From (1)–(4), we have

½wjCjðS0Þ þwiCiðS0Þ� � ½wiCiðSÞ þwjCjðSÞ� ¼ ðwipj �wjpiÞ 1þXr�1

l¼1

ln p½l�

!a

þwjpj 1þXr�1

l¼1

ln p½l�

!a

� 1þXr�1

l¼1

ln p½l� þ ln pi

!a" #

�wipi 1þXr�1

l¼1

ln p½l�

!a

� 1þXr�1

l¼1

ln p½l� þ ln pj

!a" #: ð6Þ

Substituting k ¼ pj=wj

pi=wi; c1 ¼

wj

wi; c0 ¼ 1

1þPr�1

l¼1ln p½l�

, and x ¼ ln pi into (6), we have from Lemma 8 that !a

½wjCjðS0Þ þwiCiðS0Þ� � ½wiCiðSÞ þwjCjðSÞ� ¼ wjpi 1þXr�1

l¼1

ln p½l� ðk� 1Þ þ kc1½1� ð1þ c0xÞa��

� 1c1½1� ð1þ c0 ln kþ c0xþ c0 ln c1Þa�

�P 0;

since k P 1; c1 P 1; 0 < c0 < 1; x P 1 and a < 0. Thus, S dominates S0. Repeating this interchange argument for jobs not se-quenced in the WSPT order completes the proof of Property 3. h

In the following we show that the earliest due date (EDD) rule gives an optimal solution for the problems to minimize thetotal tardiness, maximum lateness and maximum tardiness if the job processing times and the due dates are agreeable, i.e.,di 6 dj implies pi 6 pj for all the jobs i and j.



Ti problem, the optimal schedule is obtained by sequencing jobs in non-

decreasing order of di if the job processing times and due dates are agreeable.

Proof. Suppose that di 6 dj, which implies pi 6 pj. The total tardiness of the first r � 1 jobs are the same since they are pro-cessed in the same order. Since the makespan is minimized by the SPT rule (Property 1), the total tardiness of partialsequence p0 in S will not be greater than that of the partial sequence p0 in S0. Thus, to prove that the total tardiness of Sis less than or equal to that of S0, it suffices to show that TiðSÞ þ TjðSÞ 6 TjðS0Þ þ TiðS0Þ.

From (1)–(4), we derive that the tardiness of jobs i and j in S are

TiðSÞ ¼max Aþ pi 1þXr�1

l¼1

ln p½l�

!a

� di;0

( )


and

TjðSÞ ¼max Aþ pi 1þXr�1

l¼1

ln p½l�

!a

þ pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

� dj;0

( ):

Similarly, the tardiness of jobs i and j in S0 are

TjðS0Þ ¼ max Aþ pj 1þXr�1

l¼1

ln p½l�

!a

� dj;0

( )

and

TiðS0Þ ¼ max Aþ pj 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� di;0

( ):

To compare the total tardiness of jobs i and j in S and in S0, we consider two cases. In the first case where

Aþ pj 1þPr�1


6 dj, the total tardiness of jobs i and j in S and in S0 are

TiðSÞ þ TjðSÞ ¼max Aþ pi 1þXr�1

l¼1

ln p½l�

!a

� di;0

( )

þmax Aþ pi 1þXr�1

l¼1

ln p½l�

!a

þ pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

� dj;0

( )

and

TjðS0Þ þ TiðS0Þ ¼max Aþ pj 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� di;0

( ):

Suppose that neither TiðSÞ nor TjðSÞ is zero. Note that this is the most restrictive case since it comprises the case that eitherone or both of TiðSÞ and TjðSÞ are zero. From Property 1 and di 6 dj, we have

fTjðS0Þ þ TiðS0Þg � fTiðSÞ þ TjðSÞg ¼ ðpj � piÞ 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

þ dj � pi 1þXr�1

l¼1

ln p½l�

!a

P 0:

Thus, fTjðS0Þ þ TiðS0Þg � fTiðSÞ þ TjðSÞgP 0 in the first case. In the second case where Aþ pj 1þPr�1


> dj, the totaltardiness of jobs i and j in S and in S0 are

TiðSÞ þ TjðSÞ ¼max Aþ pi 1þXr�1

l¼1

ln p½l�

!a

� di;0

( )

þmax Aþ pi 1þXr�1

l¼1

ln p½l�

!a

þ pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

� dj;0

( )

and

TjðS0Þ þ TiðS0Þ ¼ 2Aþ 2pj 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� di � dj:

Suppose that neither TiðSÞ nor TjðSÞ is zero. From Property 1, di 6 dj and pi 6 pj, we have

fTjðS0Þ þ TiðS0Þg � fTiðSÞ þ TjðSÞg ¼ 2ðpj � piÞ 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

P 0:

Thus, fTjðS0Þ þ TiðS0Þg � fTiðSÞ þ TjðSÞgP 0 in the second case. This completes the proof of Property 4. h


The earliest due date rule provides an optimal schedule for the classical single-machine scheduling problem to minimizethe maximum lateness. However, it does not yield an optimal schedule under the proposed model, as shown in the followingexample.

Example 2. Let n ¼ 2, p1 ¼ 100, p2 ¼ 60, d1 ¼ 101, d2 ¼ 102, and a ¼ �0:5. The EDD sequence (1,2) yields a maximumlateness of 23.34, while the sequence (2,1) yields the optimal value of 3.31.



=Lmax problem, the optimal schedule is obtained by sequencing jobs in non-


Proof. By definition, the lateness of jobs i and j in S and jobs j and i in S0 are, respectively,

LiðSÞ ¼ Aþ pi 1þXr�1

l¼1

ln p½l�

!a

� di;

LjðSÞ ¼ 2Aþ pi 1þXr�1

l¼1

ln p½l�

!a

þ pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

� dj;

LjðS0Þ ¼ Aþ pj 1þXr�1

l¼1

ln p½l�

!a

� dj;

and

LiðS0Þ ¼ 2Aþ pj 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� di:

Suppose that di 6 dj, which implies pi 6 pj. Interchanging jobs i and j has no impact on the maximum lateness of the jobs insubsequence p since they are processed in the same order. As the makespan is minimized by the SPT rule in Property 1, themaximum lateness of the jobs in subsequence p0 of S cannot be larger than that of the jobs in p0 of S0. To show that S dom-inates S0, it suffices to show that maxfLiðSÞ; LjðSÞg 6 LiðS0Þ.

Since pi 6 pj, we have

LiðS0Þ � LiðSÞ ¼ Aþ ðpj � piÞ 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

> 0:

Since pi 6 pj and di 6 dj, we have from Property 1 that

LiðS0Þ � LjðSÞ ¼ ðpj � piÞ 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

� ðdi � djÞ > 0:

Thus, repeating this job interchange argument for all the jobs not sequenced in the EDD rule completes the proof of Property5. h



=Tmax problem, an optimal schedule is obtained by sequencing jobs in non-


Proof. The proof is omitted since it is similar to that of Property 5. h

4. Conclusions

Scheduling problems with learning effects have captured many scheduling researchers’ attention in recent years. However,the actual processing time of a given job drops to zero precipitously as the number of jobs increases in the position-based mod-el or when the normal job processing times are large in the sum-of-processing-times-based model. Motivated by this obser-vation, we proposed a new learning model where the actual job processing time is a function of the sum of the logarithm of theprocessing times of the jobs already processed. In particular, we showed that the problems to minimize the makespan and totalcompletion time on a single-machine are polynomially solvable under the proposed learning model. In addition, we showedthat the single-machine scheduling problems to minimize the total weighted completion time, total tardiness, maximum late-ness and maximum tardiness are polynomially solvable under some agreeable conditions on the problem parameters.

Acknowledgements

We are thankful to the Editor and the anonymous referees for their helpful comments on earlier versions of our paper.This paper was supported in part by the NSC under grant number NSC 97-2221-E-060-MY2.


Appendix

Lemma 1 (Kuo and Yang [19]). Let HðyÞ ¼ 1þ ayð1þ yÞa�1 � ð1þ yÞa, then HðyÞP 0 for 0 < y < 1, and a < 0.

Proof. Taking the first derivative of HðyÞ with respect to y, we have

H0ðyÞ ¼ aða� 1Þyð1þ yÞa�2:

Since a < 0 and 0 < y < 1, we have H0ðyÞ > 0. Thus, HðyÞ is an increasing function of y. Hence, HðyÞP Hð0Þ ¼ 0. h

Lemma 2. Let FðxÞ ¼ 1þ c0að1þ c0xÞa�1 � ð1þ c0xÞa, then FðxÞP 0 for x P 1, 0 < c0 < 1, and a < 0.

Proof. Taking the first derivative of FðxÞ with respect to x, we have

F 0ðxÞ ¼ c20aða� 1Þð1þ c0xÞa�2 � c0að1þ c0xÞa�1

:

Since a < 0, we have F 0ðxÞ > 0. This implies that FðxÞ is an increasing function of x, so we have from Lemma 1 thatFðxÞP Fð1ÞP 0. h

Lemma 3. Let GðkÞ ¼ ðk� 1Þ þ ð1þ c0 ln kþ c0xÞa � kð1þ c0xÞa, then GðkÞP 0 for k P 1, 0 < c0 < 1, x P 1, and a < 0.

Proof. Taking the first and the second derivatives of GðkÞ with respect to k, we have

G0ðkÞ ¼ 1þ ac0ð1þ c0 ln kþ c0xÞa�1

k� ð1þ c0xÞa

and

G00ðkÞ ¼ aða� 1Þc20ð1þ c0 ln kþ c0xÞa�2

k2 � ac0ð1þ c0 ln kþ c0xÞa�1

k2 :

Since a < 0 and 0 < c0 < 1, we have G00ðkÞ > 0. This implies that G0ðkÞ is an increasing function of k. Hence, from Lemma 2follows G0ðkÞP G0ð1ÞP 0. This implies that GðkÞ is also an increasing function of k. Thus, we haveGðkÞP Gð1Þ ¼ ð1þ c0xÞa � ð1þ c0xÞa ¼ 0. h

Lemma 4. Let HðyÞ ¼ c1½1� ð1þ yÞa� � 1c1½1� ð1þ yþ y ln c1Þa�, then HðyÞ > 0 for 0 < y < 1, c1 > 1, and a < 0.


H0ðyÞ ¼ �ac1ð1þ yÞa�1 þ að1þ ln c1Þc1

ð1þ yþ y ln c1Þa�1:

Since ð1þ yÞa�1> ð1þ yþ y ln c1Þa�1, and jac1j > að1þln c1Þ

c1

�� for a < 0, 0 < y < 1, and c1 > 1, we have H0ðyÞ > 0. This implies that

HðyÞ is an increasing function of y. Thus, we have HðyÞP Hð0Þ ¼ 0. h

Lemma 5. Let FðxÞ ¼ c1½1� ð1þ c0xÞa� � 1c1½1� ð1þ c0xþ c0 ln c1Þa�, then FðxÞP 0 for x P 1, c1 P 1, 0 < c0 < 1, and a < 0.


F 0ðxÞ ¼ �ac0c1ð1þ c0xÞa�1 þ ac0

c1ð1þ c0xþ c0 ln c1Þa�1

:

Since ð1þ c0xÞa�1> ð1þ c0xþ c0 ln c1Þa�1 and jac0c1j > ac0

c1

�� for c1 P 1, 0 < c0 < 1, x P 1, and a < 0, we have F 0ðxÞ > 0. Thisimplies that F(x) is an increasing function of x. Hence, we have from Lemma 4 that

FðxÞP Fð1Þ ¼ c1½1� ð1þ c0Þa� �1c1½1� ð1þ c0 þ c0 ln c1Þa�P 0: �

Lemma 6. Let HðyÞ ¼ 1þ c1½1� ð1þ yÞa� þ ayc1ð1þ yþ y ln c1Þa�1, then HðyÞP 0 for 0 < y < 1, c1 > 1, and a < 0.


H0ðyÞ ¼ �ac1ð1þ yÞa�1 þ ac1ð1þ yþ y ln c1Þa�1 þ aða� 1Þð1þ ln c1Þy

c1ð1þ yþ y ln c1Þa�2

:

Since jac1j > ac1

�� and ð1þ yÞa�1 > ð1þ yþ y ln c1Þa�1 for 0 < y < 1, c1 > 1, and a < 0, we have H0ðyÞ > 0. This implies that HðyÞis an increasing function of y. Hence, HðyÞP Hð0Þ ¼ 1. h


Lemma 7. Let FðxÞ ¼ 1þ c1½1� ð1þ c0xÞa� þ ac0c1ð1þ c0xþ c0 ln c1Þa�1, then FðxÞP 0 for x P 1, c1 P 1, 0 < c0 < 1, and a < 0.


F 0ðxÞ ¼ �ac0c1ð1þ c0xÞa�1 þ aða� 1Þc20

c1ð1þ c0xþ c0 ln c1Þa�2

:

Since c1 P 1, 0 < c0 < 1, x P 1, and a < 0, we have F 0ðxÞ > 0. This implies that F(x) is an increasing function of x. From Lemma6, we have

FðxÞP Fð1Þ ¼ 1þ c1½1� ð1þ c0Þa� þac0

c1ð1þ c0 þ c0 ln c1Þa�1 P 0: �

Lemma 8. Let GðkÞ ¼ ðk� 1Þ þ c1k½1� ð1þ c0xÞa� � 1c1½1� ð1þ c0 ln kþ c0xþ c0 ln c1Þa�, then GðkÞP 0, for k P 1; c1 P 1; 0 <

c0 < 1; x P 1, and a < 0.

Proof. Taking the first and the second derivatives of GðkÞ with respect to k, we have

G0ðkÞ ¼ 1þ c1½1� ð1þ c0xÞa� þ c0að1þ c0 ln kþ c0xþ c0 ln c1Þa�1

kc1

and

G00ðkÞ ¼ aða� 1Þc20ð1þ c0 ln kþ c0xþ c0 ln c1Þa�2

c1k2 � ac0ð1þ c0 ln kþ c0xþ c0 ln c1Þa�1

c1k2 :

Since k P 1; c1 P 1; 0 < c0 < 1; x P 1, and a < 0, we have G00ðkÞP 0. This implies that G0ðkÞ is an increasing function of k.From Lemma 7, we have G0ðkÞP G0ð1ÞP 0. This implies that GðkÞ is an increasing function of k. Hence, we have from Lemma5 that

GðkÞP Gð1Þ ¼ c1½1� ð1þ c0kÞa� �1c1½1� ð1þ c0xþ c0 ln c1Þa�P 0: �

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Single-machine scheduling with sum-of-logarithm-processing-times-based learning considerations