Ann Oper Res (2011) 186:345–356DOI 10.1007/s10479-011-0835-1

Single-machine scheduling problems with timeand position dependent processing times

Xingong Zhang · Guangle Yan · Wanzhen Huang ·Guochun Tang

Published online: 15 January 2011© Springer Science+Business Media, LLC 2011

Abstract We consider single-machine scheduling problems with time and position depen-dent job processing times. In many industrial settings, the processing time of a job changesdue to either job deterioration over time or machine/worker’s learning through experiences.In the models we study, each job has its normal processing time. However, a job’s actualprocessing time depends on when its processing starts and how many jobs have completedbefore its start. We prove that the classical SPT (Shortest Processing Time) rule remainsoptimal when we minimize the makespan or the total completion time. For problems ofminimizing the total weighted completion time, the maximum lateness, and the discountedtotal weighted completion time, we present heuristic sequencing rules and analyze the worst-case bounds for performance ratios. We also show that these heuristic rules can be optimalunder some agreeable conditions between the normal processing times and job due dates orweights.

Keywords Single-machine scheduling · Learning effect · Performance ratio · Worst-caseerror

1 Introductions and literature review

In classical scheduling problems, the processing times of jobs are assumed to be constantvalues. However, there are situations where the processing times of jobs may be subject tochange due to deterioration and/or learning phenomena (Pinedo 2008). Machine scheduling

X. Zhang (�) · G. YanBusiness School, University of Shanghai for Science and Technology, Shanghai 200093, P.R. Chinae-mail: zhangxg7980@yahoo.com.cn

W. HuangDepartment of Mathematical Sciences, Lakehead University, Ontario, Canada

G. TangInstitute of Management Engineering, Shanghai Second Polytechnic University, Shanghai 201209,P.R. China

346 Ann Oper Res (2011) 186:345–356

problems with deteriorating jobs and or learning effect have been paid more attention inrecent years.

Learning effect has been investigated in a variety of industries, but is relatively unex-plored in the scheduling fields until recently. Biskup (1999) began to bring the concept oflearning effect into scheduling problems and proved that single-machine scheduling prob-lems to minimize the sum of job flow times and the total deviations of job completion timesfrom a common due date are polynomially solvable. In the past decade, more and moreresearchers have made contributions in this area. Mosheiov and Sidney (2003) considereda case of the job-dependent learning curve where the learning in the product process ofsome jobs is faster than of others. Mosheiov and Sidney (2005) introduced a polynomialtime solution for the single-machine scheduling problems to minimize the number of tardyjobs with general non-increasing job-dependent learning curves and a common due date.More recently, Biskup (2008) provided an extensive review of scheduling with learning ef-fects. Lee and Wu (2009) proposed a learning effect scheduling model which the actualprocessing time of a job depends not only on the total normal processing times of the jobsalready processed, but also on its scheduled position. They showed that both the position-based learning and the sum-of-processing-time-based learning models in the literature arespecial cases of the proposed model. They then provided some solvable cases for the single-machine and multiple-machine flowshop. Wang and Xia (2005) and Wang (2005) extendedtheir results to the model: Pegels’ learning curve (Pegels 1969), i.e., if job Jj is sched-uled in position k in a sequence, its actual processing time is pjk = pj [αak−1 + β], whereα, a and β are parameters obtained empirically. Wang et al. (2009) considered the sin-gle machine scheduling problem with exponential time-dependent learning effect and past-sequence-dependent setup times, where the actual processing time a job Jj at position k in

sequence π is pjk = pj (αa∑k−1

i=1 pπ(i) + β), where α ≥ 0, β ≥ 0, 0 < a ≤ 1 with α + β = 1and

∑k−1i=1 pπ(i) calculates the total normal processing times of jobs processed before Jj in

sequence π . Wang et al. (2010) considered the single machine scheduling problems with ex-ponential sum-of-logarithm-processing-times based learning effect. They the smallest (nor-mal) processing time first (SPT) rule minimizes the makespan Cmax, the total completiontime

∑Cj and the sum of the quadratic job completion times

∑C2

j . They also showed thatthe total weighted completion time and the maximum lateness minimization problems canbe solved in polynomial time under certain conditions.

Job deterioration in scheduling was first studied by Gupta and Gupta (1988) andBrowne and Yechiali (1990), and later studied extensively from a variety of perspectives.In Mosheiov’s (1991) model, all jobs are characterized by a common positive normal pro-cessing time, and proved that the minimum flow time schedule is symmetric and has theV-shaped property with respect to the increasing rates of deterioration. Mosheiov (1994)studied single-machine scheduling problems with objectives including makespan, total flowtime, total weighted completion time, total lateness, maximum lateness and maximum tardi-ness, and number of tardy jobs. When the values of the normal processing times equal zero,all these problems can be solved polynomially. Gawiejnowicz (2008) gave a detailed reviewon single-machine scheduling problems with time-dependent processing times. Generally,there are two types of time-dependencies under studies: the actual processing times are char-acterized by either non-decreasing or non-increasing functions (Ho et al. 1993). Researchworks regarding linear decreasing deterioration include Bachman et al. (2002), Wang andLiu (2009), and Yang (2010).

More recently, job deteriorations and learning effects are simultaneously considered insome scheduling problems because the phenomena can be found in many real-life situa-tions. Lee (2004) considered single-machine scheduling problems with both deterioration

Ann Oper Res (2011) 186:345–356 347

and learning effects. They proposed the following two models. In the first model, the actualprocessing time of job Jj is pjk(t) = αitk

a if it is scheduled in position k in a sequence,where αi , a and t are the deterioration rate, learning rate and starting time of job Ji , respec-tively. In the second model, the actual processing time is pjk(t) = (p0 +αit)k

a , where p0 isthe normal processing time. In Wang and Cheng’s (2007) model, the actual processing timeof job Jj is pjk = αj (b+ct)ka where aj is a deterioration rate, b, c (>0) are constants. Yangand Kuo (2010) considered some scheduling problems with deteriorating jobs and learningeffects using objective functions including the makespan, the total completion times, andthe total absolute differences in completion times.

The learning effect model we use is based on Wang and Xia’s (2005) exponential modelpjk = pjα

k−1, where 0 < α ≤ 1 is the learning factor. In reality, α closes to one that meansthe slow learning. If human interactions have a significant impact during the processing ofthe job, the processing time will add to the workers’ experience and cause learning effects.Under the proposed model, the actual processing time of a job depends on both the startingtime of the job and its scheduled position.

The remaining part of the paper is organized as follows: In Sect. 2, we formulate ourmodel. From Sect. 3 to Sect. 6, we study the problems that minimize various objectivesincluding the makespan, the total completion time, the total weighted conclusion time, themaximum lateness, and the discounted total weighted completion time. Conclusion is givenin Sect. 7.

2 Problem formulation

We have n independent and non-preemptive jobs J1, J2, . . . , Jn that are immediately avail-able for processing on one machine, one at a time. Associated with each job Jj , we have theweight wj , the due-date dj , and the normal processing time pj that is the processing timeif Jj is processed first at time zero. The actual processing time of job Jj depends on thestarting time of its processing as well as the position of the job in the sequence. If job Jj isscheduled at the kth position and starts the processing at time t , then the actual processingtime is

pjk(t) = pj (1 − βt)αk−1,

where α and β are parameters such that 0 < α ≤ 1, 0 < β < 1. The values of these param-eters can be estimated empirically. In practical setting, the value of α should not be verysmall so the processing times of jobs decrease rather slowly along the sequence. We alsoreasonably assume that each job has a positive processing time, so we have the requirementβ(

∑n

j=1 pj − min1≤j≤n pj ) < 1, which implies that β is fairly small number.Since we have only one machine, for all regular objective functions, a solution can be

identified simply by a sequence of jobs. Let π = [π(1), . . . , π(n)] be a permutation of[1,2, . . . , n], with π(k) representing the job index scheduled at kth position in π . To mini-mize the use of notations, we also use π to denote the job sequence (and therefore the sched-ule or solution of the problem) that based on the permutation π . Let Cj = Cj(π) representthe completion time of job Jj in sequence π when there is no confusion, j = 1,2, . . . , n.Following the traditional notation, we use Cmax = max{Cj |j = 1,2, . . . , n}, ∑Cj ,

∑wjCj ,

Lmax = max{Cj − dj |j = 1,2, . . . , n} and∑

wj(1 − e−rCj ) (0 < r < 1) represent themakespan, the total completion time, the total weighted completion time, the maximum late-ness, and the discounted total weighted completion time of a given schedule π , respectively.In the remaining part of the paper, all problems considered in this paper will be denotedusing the three-field notation scheme α|β|γ introduced by Graham et al. (1979).

348 Ann Oper Res (2011) 186:345–356

3 Problem 1|pjk(t) = pj(1 − βt)αk−1|Cmax

We first study the makespan problem 1|pjk(t) = pj (1−βt)αk−1|Cmax. For any given sched-ule π , we first give a closed form of the makespan Cmax in the following lemma.

Lemma 1 For any given schedule π = [π(1),π(2), . . . , π(n)], with the first job Jπ(1) start-ing at time t0, the completion time of Jπ(j) (j = 1,2, . . . , n) in schedule π is

Cπ(j)(π) = t0

j∏

k=1

(1 − βpπ(k)αk−1) +

j∑

k=1

pπ(k)αk−1

j∏

i=k+1

(1 − βpiαi−1).

Proof Let Cπ(j)(π) be the completion time of job Jπ(j) in schedule π , j = 1,2, . . . , n. Themakespan is the completion time of the last job Jπ(n), or Cmax(π) = Cπ(n)(π). Note that thestarting time of the kth job Jπ(k) in the sequence π is Cπ(k−1), for k = 2,3, . . . , n, we havethe following recursion formula for calculating the makespan.

Cπ(1)(π) = t0 + pπ(1)1(t0)

= t0 + pπ(1)(1 − βt0) = t0(1 − pπ(1)β) + pπ(1),

Cπ(k)(π) = Cπ(k−1)(π) + pπ(k)k(Cπ(k−1)(π))

= Cπ(k−1)(π) + pπ(k)(1 − βC(k−1)(π))αk−1

= pπ(k)αk−1 + (1 − pπ(k)βαk−1)Cπ(k−1)(π), k = 2,3, . . . , n.

We now use mathematical induction to obtain the closed form for Cmax(π). First for k = 2we have

Cπ(2)(π) = pπ(2)α + (1 − pπ(2)βα)Cπ(1)(π)

= pπ(2)α + (1 − pπ(2)βα)(t0(1 − pπ(1)β) + pπ(1))

= t0(1 − pπ(1)β)(1 − pπ(2)βα) + pπ(1)(1 − pπ(2)βα) + pπ(2)α

= t0

2∏

l=1

(1 − βpπ(l)αl−1) +

2∑

l=1

pπ(l)αl−1

2∏

i=l+1

(1 − βpπ(i)αi−1).

Suppose for all 2 ≤ k ≤ n − 1, we have Cπ(k)(π) = t0∏k

l=1(1 − βpπ(l)αl−1) +

∑k

l=1 pπ(l)αl−1

∏k

i=l+1(1 − βpπ(i)αi−1). Then for the (k + 1)th job in the sequence,

Cπ(k+1)(π) = pπ(k+1)αk + (1 − pπ(k+1)βαk)Cπ(k)(π)

= pπ(k+1)αk + (1 − pπ(k+1)βαk)

(

t0

k∏

l=1

(1 − βpπ(l)αl−1)

+k∑

l=1

pπ(l)αl−1

k∏

i=l+1

(1 − βpπ(i)αi−1)

)

= t0

k+1∏

l=1

(1 − βpπ(l)αl−1) +

k+1∑

l=1

pπ(l)αl−1

k+1∏

i=l+1

(1 − βpπ(i)αi−1).

Note that Cmax(π) = Cπ(n)(π), by the induction we proved the lemma. �

Ann Oper Res (2011) 186:345–356 349

Since actual processing times of jobs depend on job positions in the sequence, and thejobs scheduled in the later positions of the sequence get the processing times reduced propor-tionally, the intuitive thought is that we schedule the long jobs later to maximize the learningdiscount in processing time. In the next theorem, we will prove that the SPT (the ShortestProcessing Time) rule is optimal for the makespan problem.

Theorem 1 For the problem 1|pjk(t) = pj (1 − βt)αk−1|Cmax, an optimal schedule can beobtained by sequencing the jobs in non-decreasing order of pj (the SPT rule).

Proof Let π and π ′ be two job schedules where the difference between the two is thepairwise interchange of two adjacent jobs Ji and Jj , that is, π = [S1, Ji, Jj , S2], π ′ =[S1, Jj , Ji, S2], where S1 and S2 are partial sequences, S1 or S2 may be empty. Furthermore,we assume that there are k − 1 jobs in S1. Thus, Ji and Jj are the kth and the (k + 1)th jobsin π , respectively. Let t0 = Cπ(k−1)(π) = Cπ ′(k−1)(π

′) denote the completion time of the lastjob in S1 in both π and π ′. Assume that pi ≤ pj , all we need to prove is that Cj(π) ≤ Ci(π

′).Under π , the completion times of jobs Ji and Jj are

Ci(π) = piαk−1 + t0(1 − βpiα

k−1), (1)

Cj(π) = pjαk + piα

k−1(1 − βpjαk) + t0(1 − βpiα

k−1)(1 − βpjαk). (2)

Whereas under π ′, they are

Cj(π′) = pjα

k−1 + t0(1 − βpjαk−1), (3)

Ci(π′) = piα

k + pjαk−1(1 − βpiα

k) + t0(1 − βpjαk−1)(1 − βpiα

k). (4)

The difference between (2) and (4) can be simplified to

Cj(π) − Ci(π′) = (1 − βt0)(pi − pj )(α

k−1 − αk).

Since 0 < α ≤ 1, 1 − βt0 > 0 due to the assumption β(∑n

j=1 pj − min1≤j≤n pj ) < 1 andpi ≤ pj , we have Cj(π) ≤ Ci(π

′). This completes the proof. �

4 Problem 1|pjk(t) = pj(1 − βt)αk−1|∑Cj(∑

wjCj)

In the following theorem, we show that SPT rule is also optimal for problem the total sumof completion time problem 1|pjk(t) = pj (1 − βt)αk−1|∑Cj .

Theorem 2 For the problem 1|pjk(t) = pj (1 −βt)αk−1|∑Cj , an optimal schedule can beobtained by sequencing the jobs in non-decreasing order of pj (the SPT rule).

Proof We use the same notations and the sequences π and π ′ as in the proof of Theorem 1.Under the same assumption pi ≤ pj , it suffices to show that (i) Cj(π) ≤ Ci(π

′) and (ii)Ci(π) + Cj(π) ≤ Ci(π

′) + Cj(π′).

The proof of part (i) is given in Theorem 1. In addition, from pi ≤ pj , we haveCi(π) ≤ Cj(π

′), hence Ci(π) + Cj(π) ≤ Ci(π′) + Cj(π

′). This completes the proof ofpart (ii) and thus of the theorem. �

For the classical total weight completion time with fixed job processing times, an optimalschedule can be obtained by the weighted shortest processing time first (WSPT) rule (Smith

350 Ann Oper Res (2011) 186:345–356

1956). However, Mosheiov (2001) showed that this polynomially time obtainable optimalsolution for the classical version failed when learning is considered. In the next theorem, weprove that the WSPT rule “almost” minimize the total weighted completion time by givinga bound for the worst-case performance ratio.

Theorem 3 Let π∗ be an optimal schedule and π be a WSPT schedule for the problem1|pjk(t) = pj (1 − βt)αk−1|∑wjCj , with the starting time of the first job t0 = 0, then

ρ1 =∑

wjCj (π)∑

wjCj (π∗)≤ 1

αn−1∏n

l=1(1 − βpl)

and the bound is tight.

Proof Without loss of generality, we can suppose that p1w1

≤ · · · ≤ pn

wn, and therefore π =

[1,2, . . . , n]. Since 0 < α ≤ 1 and β(∑n

j=1 pj − min1≤j≤n pj ) < 1, we have

∑wjCj (π) = w1p1 + w2(p2α + p1(1 − βp2α)) + · · ·

+ wn

(n∑

j=1

pjαj−1

n∏

i=j+1

(1 − βpiαi−1)

)

≤ w1p1 + w2(p2 + p1) + · · · + wn

(n∑

j=1

pj

)

=n∑

j=1

wj

(j∑

i=1

pi

)

,

where∑n

j=1 wj(∑j

i=1 pi) is the optimal value of the classical version of the problem.In the optimal schedule π∗ = [π∗(1),π∗(2), . . . , π∗(n)], we have

∑wjCj (π

∗) =∑

wπ∗(j)Cπ∗(j)(π∗)

=n∑

j=1

wπ∗(j)

(j∑

k=1

pπ∗(k)αk−1

j∏

l=k+1

(1 − βpπ∗(l)αl−1)

)

≥n∑

j=1

wπ∗(j)

(j∑

k=1

pπ∗(k)αn−1

n∏

l=1

(1 − βpπ∗(l)αl−1)

)

=n∑

j=1

wπ∗(j)

(j∑

k=1

pπ∗(k)

)

αn−1n∏

l=1

(1 − βpπ∗(l)αl−1)

≥n∑

j=1

wπ∗(j)

(j∑

k=1

pπ∗(k)

)

αn−1n∏

l=1

(1 − βpl).

Hence,

ρ1 =∑

wjCj (π)∑

wjCj (π∗)≤ 1

αn−1∏n

l=1(1 − βpl).

Ann Oper Res (2011) 186:345–356 351

It is not difficult to see that the bound is tight, since, if α = 1 and b → 0, we have ρ1 → 1.When α = 1 and β → 0, the WSPT schedule is optimal. �

In the next theorem, we prove that WSPT is optimal if the processing times and the jobweights satisfy certain agreeable condition, defined as below. We say the job weights {wj }and (normal) processing times {pj } are agreeable if when jobs have the WSPT order suchthat

p1

w1≤ p2

w2· · · ≤ pn

wn

,

then we have pi

pj≤ wi

wj≤ 1, for all i < j . In our three-field notation, we will use (pj ,wj ) to

denote this relationship between the job processing times and the weights. This conditionis stronger that the classical agreeable condition where we only require that pi < pj ⇒wi ≤ wi .

Theorem 4 For the problem 1|pjk(t) = pj (1 − βt)αk−1, (pj ,wj )|∑wjCj , an optimalschedule is obtained by sequencing jobs in non-decreasing order of

pj

wj, if the processing

times and the weights are agreeable in this order.

Proof We use the same notations as in the proofs of Theorems 1 and 2, and the structureof sequences π and π ′. Assume that i < j which implies pi

pj≤ wi

wj≤ 1. It suffices to show

that the r th and (r + 1)th jobs in π and π ′ satisfy the condition wiCi(π) + wjCj (π) ≤wjCj (π

′) + wiCi(π′). We have

wiCi(π) + wjCj (π) − wjCj (π′) − wiCi(π

′)

= wi[piαk−1 + t0(1 − βpiα

k−1)] + wj [pjαk + piα

k−1(1 − βpjαk)

+ t0(1 − βpiαk−1)(1 − βpjα

k)] − wj [pjαk−1 + t0(1 − βpjα

k−1)]− wi[pjα

k + piαk−1(1 − βpjα

k) + t0(1 − βpiαk−1)(1 − βpjα

k)]= (1 − βt0)[(wi − wj)pipjβαk−1αk + (wi + wj)(pi − pj )(α

k−1 − αk)

+ (wjpi − wipj )αk].

Since pi

pj≤ wi

wj≤ 1, simplifying the above equation will lead to wiCi(π) + wjCj (π) ≤

wjCj (π′) + wiCi(π

′). This completes the proof of Theorem 4. �

5 Problem 1|pjk(t) = pj(1 − βt)αk−1|Lmax

Now we consider the problem with job due dates. For the classical due date problem, weknow that the earliest due-date (EDD) rule (Smith 1956) generates an optimal schedule for1|pj , dj |Lmax. However, Mosheiov (2001) also showed that the problem becomes polyno-mially unsolvable when learning is considered in the processing times.

Next we use the EDD rule as a heuristic for the problem 1|pjk(t) = pj (1−βt)αk−1|Lmax

and study its performance ratio. Since the optimal maximum lateness may be zero, whichcauses problem in the denominator of the regular definition of performance ratio, here weuse the modified performance ratio, as first suggested by Kise et al. (1979) and Cheng andWang (2000). The modified performance ratio is Lmax(π)+dmax

Lmax(π∗)+dmaxfor any heuristic schedule π ,

where π∗ denotes the optimal schedule, and dmax = max{dj |j = 1, . . . , n}. In the next theo-rem, we give an upper bound for the worst-case performance ratio for any EDD schedules.

352 Ann Oper Res (2011) 186:345–356

Theorem 5 Let π∗ be an optimal schedule and π be an EDD schedule for the problem1|pjk(t) = pj (1 − βt)αk−1|Lmax, with the starting time of the first job t0 = 0, then

ρ2 = Lmax(π) + dmax

Lmax(π∗) + dmax≤

∑n

j=1 pj

C∗max

,

where C∗max is the optimal makespan of the problem 1|pjk(t) = pj (1 − βt)αk−1|Cmax and

the bound is tight.

Proof Without loss of generality, we can suppose that d1 ≤ · · · ≤ dn, and therefore π =[1,2, . . . , n]. Since 0 < α ≤ 1 and β(

∑n

j=1 pj − min1≤j≤n pj ) < 1, we have

Lmax(π) = max

{j∑

i=1

piαi−1

j∏

l=i+1

(1 − βp[l]αl−1) − dj

}

≤ max

{j∑

i=1

pi − dj

}

,

and we see that max1≤j≤n{∑j

i=1 pi − dj } is the optimal value of the classical version of theproblem. For the optimal schedule π∗, we have

Lmax(π∗) = max

1≤j≤n

{j∑

i=1

pπ∗(i)αi−1

j∏

l=i+1

(1 − βpπ∗(l)αl−1) − dπ∗(j)

}

= max1≤j≤n

{j∑

i=1

pπ∗(i) − dπ∗(j) −j∑

i=1

pπ∗(i) +j∑

i=1

pπ∗(i)αi−1

j∏

l=i+1

(1 − βpπ∗(l)αl−1)

}

≥ max1≤j≤n

{j∑

i=1

pπ∗(i) − dπ∗(j)

}

−n∑

i=1

pi +n∑

i=1

pπ∗(i)αi−1

n∏

l=i+1

(1 − βpπ∗(l)αl−1)

≥ max1≤j≤n

{j∑

i=1

pπ∗(i) − dπ∗(j)

}

−n∑

i=1

pi + C∗max,

where C∗max can be obtained by the SPT rule (by Theorem 1), hence

Lmax(π) − Lmax(π∗) =

n∑

i=1

pi − C∗max,

therefore

ρ2 = Lmax(π) + dmax

Lmax(π∗) + dmax≤ Lmax(π

∗) + ∑n

i=1 pi − C∗max + dmax

Lmax(π∗) + dmax

≤ 1 +∑n

i=1 pi − C∗max

Lmax(π∗) + dmax≤ 1 +

∑n

i=1 pi − C∗max

C∗max

≤∑n

j=1 pj

C∗max

.

It is not difficult to see that the bound is tight, since, if α = 1 and b → 0, we have ρ1 → 1and C∗

max = ∑pj . When α = 1 and b → 0, the EDD schedule is optimal. �

Ann Oper Res (2011) 186:345–356 353

In the next theorem, we will show that the EDD sequence does provide an optimal sched-ule for problem the problem 1|pjk(t) = pj (1−βt)αk−1|Lmax when job processing times andthe due dates are agreeable in the sense that pi < pj implies di ≤ dj . We denoted the prob-lem by 1|pjk(t) = pj (1 − βt)αk−1, (pj , dj )|Lmax.

Theorem 6 For the problem 1|pjk(t) = pj (1−βt)αk−1, (pj , dj )|Lmax, an optimal schedulecan be obtained by sequencing jobs in non-decreasing order of dj (i.e. the EDD rule).

Proof Let π be the EDD schedule. Under π , the lateness of the jobs Ji and Jj areLi(π) = Ci(π) − di and Lj(π) = Cj(π) − dj . Let π ′ be the schedule obtained from π

by exchange the positions of jobs Ji and Jj . The lateness of the jobs Ji and Jj in π ′ areLi(π

′) = Ci(π′) − di and Lj(π

′) = Cj(π′) − dj . Assume that pi ≤ pj , then by the agree-

able assumption we have di ≤ dj . From Theorems 1 and 2, we have

Li(π) = Ci(π) − di ≤ Ci(π′) − di = Li(π

′),

Lj (π) = Cj(π) − dj ≤ Ci(π′) − dj ≤ Ci(π

′) − di = Li(π′).

Therefore we have max{Li(π),Lj (π)} ≤ max{Li(π′),Lj (π

′)}. This completes theproof of the theorem. �

When the objective is to minimize the total tardiness problem, we can similarly provethat EDD rule provides an optimal schedule if the processing times and the due dates areagreeable.

6 Problem 1|pjk(t) = pj(1 − βt)αk−1|∑wj(1 − e−kCj )

Pinedo (2008) considered the single-machine problem with the objective of discounted totalweighted completion time 1 ‖ ∑

wj(1 − e−rCj ) (with 0 < r < 1). He showed that for thisproblem, an optimal schedule can be obtained by the weighted discounted shortest process-

ing time first (WDSPT) rule, i.e., sequencing the jobs in non-decreasing order of 1−e−rpj

wj e−rpj

.

Unfortunately, the WDSPT sequences does not yield an optimal schedule for our problem1|pjk(t) = pj (1 − βt)αk−1|∑wj(1 − e−rCj ) in general. We may use WDSPT rule to gen-erate a heuristic schedule for the problem. Before presenting a worst-case upper bound forthe performance ratio, we first introduce a useful lemma.

Lemma 2 1 − e−ax ≥ x(1 − ea) for all 0 < x ≤ 1 and a ∈ R.

The proof of the Lemma 2 can be obtained by differentiation, so this proof is omitted.

Theorem 7 Let π∗ be an optimal schedule and π be an WDSPT schedule for the problem1|pjk(t) = pj (1−βt)αk−1|∑wj(1− e−rCj ), and the starting time of the first job be 0, then

ρ3 =∑

wj(1 − e−rCj (π))∑

wj(1 − e−rCj (π∗))≤ 1

αn−1∏n

l=1(1 − βpl),

and the bound is tight.

354 Ann Oper Res (2011) 186:345–356

Proof Without loss of generality, we can suppose that 1−e−rp1

w1e−rp1≤ 1−e−rp2

w2e−rp2≤ · · · ≤ 1−e−rpn

wne−rpn and

therefore π = [1,2, . . . , n]. Since 0 < α ≤ 1, 0 < r < 1 and β(∑n

j=1 pj −min1≤j≤n pj ) < 1,we have

∑wj(1 − e−rCj (π)) = w1(1 − e−rp1) + w2(1 − e−r(p1(1−βp2α)+p2α))

+ · · · + wn(1 − e−r(∑n

i=1 piαi−1 ∏n

l=i+1(1−βplαl−1)))

≤ w1(1 − e−rp1) + w2(1 − e−r(p1+p2)) + · · · + wn(1 − e−r∑n

i=1 pi )

=n∑

j=1

wj(1 − e−r∑j

i=1 pi ),

and we see that∑n

j=1 wj(1 − e−r∑j

i=1 pi ) is the optimal value of the classical ver-sion of the problem with fixed job processing times. For the optimal schedule π∗ =[π∗(1),π∗(2), . . . , π∗(n)], we have

∑wj(1 − e−rCj (π∗)) = wπ∗(1)(1 − e−rpπ∗(1) ) + wπ∗(2)(1 − e−r(pπ∗(1)(1−βpπ∗(2)α)+pπ∗(2)α))

+ · · · + wπ∗(n)(1 − e−r(∑n

i=1 pπ∗(i)αi−1 ∏n

l=i+1(1−βpπ∗(l)αl−1)))

≥n∑

j=1

wπ∗(j)(1 − e−r∑j

i=1 pπ∗(i)αn−1 ∏n

l=1(1−βpπ∗(l)αl−1))

≥ αn−1n∏

l=1

(1 − βpπ∗(l)αl−1)

n∑

j=1

w[j ](1 − e−r∑j

i=1 pπ∗(i) ) (Lemma 2)

≥ αn−1n∏

l=1

(1 − βpl)

n∑

j=1

wπ∗(j)(1 − e−r∑j

i=1 pπ∗(i) ).

Hence,

ρ3 =∑

wj(1 − e−rCj (π))∑

wj(1 − e−rCj (π∗))≤ 1

αn−1∏n

l=1(1 − βpl).

This completes the proof for Theorem 7. We can also see that the bound is tight: when α = 1and b → 0, we have ρ1 → 1. When α = 1 and b → 0, the WDSPT schedule is optimal. �

In the next theorem, we can prove that the WDSPT sequence does provide an optimalschedule for problem 1|pjk(t) = pj (1 − βt)αk−1|∑wj(1 − e−rCj ), if the job processingtimes and the weights are anti-agreeable, in the sense that pi ≤ pj implies wi ≥ wj , and wedenote it by anti(pj ,wj ).

Theorem 8 For the problem 1|pjk(t) = pj (1 − βt)αk−1, anti(pj ,wj )|∑wj(1 − e−rCj ),there exists an optimal schedule in which the jobs are ordered according to the WDSPTrule.

Proof Let π be a schedule in which there are adjacent jobs Ji and job Jj such that Ji is

scheduled before Jj and 1−e−rpj

wj e−rpj

≥ 1−e−rpi

wi e−rpi

. Let π ′ be the schedule generated from π by

Ann Oper Res (2011) 186:345–356 355

exchange the position of jobs Ji and job Jj . We will show that π dominates π ′. It sufficesto show that

wi(1 − e−rCi (π)) + wj(1 − e−rCj (π)) ≤ wi(1 − e−rCi (π′)) + wj(1 − e−rCj (π ′)).

From Theorems 1 and 2, we have Cj(π) ≤ Ci(π′) and Ci(π) ≤ Cj(π

′). Since 0 < r < 1,we have

wi(1 − e−rCi (π)) + wj(1 − e−rCj (π)) − wi(1 − e−rCi (π′)) − wj(1 − e−rCj (π ′))

= wje−rCj (π ′) + wie

−rCi (π′) − wie

−rCi (π) − wje−rCj (π)

≤ wje−rCi (π) + wie

−rCj (π) − wie−rCi (π) − wje

−rCj (π)

= (wi − wj)(e−rCj (π) − e−rCi (π))

≤ 0.

This completes the proof. �

7 Conclusion

In this paper, we considered some single-scheduling problems with time-and position-dependent job processing times. We showed that the makespan and the total completiontime problems remain polynomially solvable. For the total weighted completion time, themaximum lateness and the discounted total weighted completion time problems, we stud-ied popular heuristic rules used in the corresponding problems with fixed processing times.Theorems are presented to give the worst-case error bounds of these heuristic. We furthershowed that under certain conditions, these heuristic rules do provide optimal schedules.

Our future research include investigating different time-and position-dependent job pro-cessing times, job dependent parameters αj and βj , and other scheduling settings such asmulti-machine and job-shop scheduling..

Acknowledgements This research was supported by the NSFC (10371071) (70731160015), the SHKDCP(S30504) and The Innovation Fund Project For Graduate Student of Shanghai (JWCXSL1001).

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