Flow shop scheduling problems with decreasing linear deterioration under dominant machines

Computers & Operations Research 34 (2007) 2043–2058www.elsevier.com/locate/cor

Flow shop scheduling problems with decreasing lineardeterioration under dominant machines

Ji-Bo Wang

Department of Science, Shenyang Institute of Aeronautical Engineering, Shenyang 110034, People’s Republic of China

Available online 8 September 2005

Abstract

This paper considers the general, no-wait and no-idle flow shop scheduling problems with deteriorating jobs,respectively. By a deteriorating job, we mean that the processing time is a decreasing function of its execution startingtime. The normal processing time proportional to its decreasing rate is assumed and some dominant relationshipsbetween machines can be satisfied. It is shown that for the problems to minimize the makespan or the weightedsum of completion time, polynomial algorithms still exist, although these problems are more complicated thanthe classical ones. When the objective is to minimize the maximum lateness, the solutions of a classical versionmay not hold.� 2005 Elsevier Ltd. All rights reserved.

Keywords: Scheduling; Flow shop; Linear deterioration; Dominant machines

1. Introduction

In the classical scheduling theory, job processing times are considered to be constant. In practice, how-ever, we often encounter settings in which processing times increase or decrease over time. Researchershave formulated this phenomenon into different models and solved different problems for various crite-ria. Extensive surveys of different models and problems concerning start time-dependent job processingtimes can be found in [1,2]. Generally, there are two types of models describing this kind of processes.The first type is devoted to the problems in which the job processing time is characterized by a non-decreasing function, and the second type concerns problems in which the job processing time is given

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2044 J.-B. Wang / Computers & Operations Research 34 (2007) 2043–2058

by a non-increasing function. Applications of these models can be found, among others, in fire fighting,emergency medicine, police, machine maintenance, and computer science.

Some common examples of the problem in which the job processing time is an increasing start time-dependent function can be found in the areas of scheduling maintenance, cleaning assignments or met-allurgy, in which any delay often implies additional effort (or time) to accomplish the job. On the otherhand, an example considering the so called “learning effect” can be described by a non-increasing starttime-dependent function. Assume that a worker has to assemble a large number of similar products. Thetime required by the worker to assemble one product depends on his knowledge, skills, organization ofhis working place and others. The worker learns how to produce. After some time, he is better skilled, hisworking place is better organized and his knowledge has increased. As a result of his learning, the timerequired to assemble one product decreases. Another example is the process by which aerial threats are tobe recognized by a radar station [3]. In this case, a radar station has detected some objects approaching it.The time required to recognize the objects decreases as the objects get closer. Thus, the later the objectsare detected, the less time needed for their recognition.

Browne and Yechiali [4] consider a scheduling problem in which the processing times of the jobs arenot constant over time. n jobs have to be processed on a single machine to minimize the makespan. JobJi is characterized by: (1) a “basic” processing time ai , the length of time required to complete the jobif it were scheduled first, i.e., at t = 0, and (2) a parameter bi that jointly with ai determines the job’s(actual) processing time at t > 0; bi can be interpreted as the growth rate of the processing time of job Ji .Assuming linear deterioration, i.e., the processing time of the job increases linearly with its starting timet, the actual processing time is pi(t)= ai + bit . This problem can be solved optimally by scheduling jobsin an increasing order of ai/bi , the ratio of the basic processing time to the growth rate. Mosheiov [5]considers the problem that all jobs are characterized by a common positive basic processing time. Usingthis basic assumption, Mosheiov proves that the optimal schedule to minimize flowtime is symmetricand has a V-shaped property with respect to the increasing rates. Mosheiov [6] considers the followingobjective functions: makespan, total flow time, total weighted completion time, total lateness, maximumlateness and maximum tardiness, and number of tardy jobs. When the values of the normal processingtime equal zero, i.e., pi(t)= bit , all these problems can be solved polynomially. Bachman and Janiak [7]consider other variants of the linear model of start time-dependent job processing times. They assume thatfor all jobs, the increasing rates are k times greater than the basic processing times aj , i.e., bj =kaj , k > 0.Cheng and Ding [8] consider the scheduling model in which each job has a normal processing time whichdeteriorates as a step function if its starting time is beyond a given deterioration rate. They show that theflow time problem with identical task deteriorating dates is NP-complete and suggest a pseudo-polynomialalgorithm for the makespan problem. They also introduce a general method of solution for the flow timeproblems. Mosheiov [9] considers the makespan minimization on the complexity of flow shop, open shopand job shop problems. Mosheiov introduces a polynomial-time algorithm for the two-machine flowshop and proves NP-hardness when an arbitrary number of machines (three and above) is assumed. Zhaoet al. [10] consider a special type of the actual processing time, which is pi(t)=ai(a +bt), where a and bare positive constants. They prove that the single-machine scheduling problems of minimizing makespan,sum of weighted completion times, maximum lateness and maximum cost are polynomially solvable, andthe two-machine flow shop scheduling to minimize the makespan can be solved by Johnson’s rule. Wangand Xia [11] consider no-wait or no-idle flow shop scheduling problems with processing times dependenton starting time. In these problems job processing time is a simple linear function of the job’s starting timeand some dominating relationships between machines can be satisfied. They showed that for the problems

J.-B. Wang / Computers & Operations Research 34 (2007) 2043–2058 2045

to minimize makespan or minimize weighted sum of completion time polynomial algorithms still exist.When the objective is to minimize maximum lateness, the solutions of a classical version may not hold.Wang and Xia [12] consider general, no-wait or no-idle flow shop scheduling problems with processingtimes dependent on starting time. In these problems some dominating relationships between machinescan be satisfied. The job processing time is a special linear function of the job’s starting time, i.e., normalprocessing time is proportional to increasing rate. They show that for the problems to minimize makespanor minimize weighted sum of completion time polynomial algorithms still exist. When the objective is tominimize maximum lateness, the solutions of a classical version may not hold.

Apart from the increasing linear model for the job processing times, there is also a decreasing linearmodel. This model is introduced by Ho et al. [3]. It also contains two parts, but in contrast to the increasingmodel, the start time-dependent part is described by a decreasing rate. Ho et al. [3] consider the problemof solution feasibility with deadline restrictions. Cheng and Ding [13] consider some problems with adecreasing linear model of the job processing times, but with ready time and deadline restrictions. Theyidentify some interesting relationships between the linear models with decreasing and increasing starttime-dependent parts. Ng et al. [14] also consider three scheduling problems with a decreasing linearmodel of the job processing times, where the objective function is to minimize the total completion time,and two of the problems are solved optimally. A pseudo-polynomial time algorithm is constructed tosolve the third problem using dynamic programming. Some interesting relationships between the linearmodel with decreasing and increasing start time-dependent parts have also been presented by Ng et al.[14]. Bachman et al. [15] consider the single-machine scheduling problem with start time-dependent jobprocessing times. They prove that the problem of minimizing the total weighted completion time is NP-hard. They also consider some special cases. Wang and Xia [16] consider the scheduling problems undera special type of decreasing linear deterioration. They present optimal algorithms for single-machinescheduling of minimizing the makespan, maximum lateness, maximum cost and number of late jobs,respectively. For two-machine flow shop scheduling problem to minimize the makespan, they prove thatthe optimal schedule can be obtained by Johnson’s rule. If the processing times of operations are equalfor each job, they prove that the flow shop scheduling problems can be transformed into single-machinescheduling problems.

For the classical work on the flow shop scheduling in an environment of a series of dominatingmachines, the reader is referred to references Cepek et al. [17], Ho and Gupta [18], Monma and Rinnooy[19], Nouweland et al. [20] and Xiang et al. [21]. For the classical work on the no-wait or no-idle flowshop scheduling, the reader is referred to references Adiri and Pohoryles [22], Rock [23] and Wang andXia [24].

In this paper, we consider the general, no-wait and no-idle flow shop scheduling problem with de-creasing linear deterioration under dominant machines, respectively. Deterioration of a job means that itsprocessing time is a function of its execution start time. The “no-wait” constraint means that each job,once started, has to be processed without interruption until all of its operations are completed. In practice,this requirement may arise out of certain job characteristics (such as in the “hot ingot” problem, wheremetal has to be processed at a continuously high temperature) or out of the unavailability of intermediatestorage between machines. The “no-idle” constraint means that each machine, once it commences itswork, has to process all operations assigned to it without any interruption. Such a restriction may bevery natural in some real life situations, e.g., if machines represent expensive pieces of equipment whichhave to be rented only for the duration between the start of the first operation and the completion ofthe last one.


The remaining part of the paper is organized as follows. In the next section, a precise formulation ofthe problem with decreasing job processing times is given. The problem of minimizing the makespan isgiven in the third section. The fourth section is devoted to minimizing the weighted sum of completiontime. The problem of minimizing the maximum lateness is given in the fifth section. The last sectioncontains some conclusions.

2. Basic notation, definition and observation

The flow shop scheduling problems consists of scheduling n jobs J1, J2, . . . , Jn on m machinesM1, M2, . . . , Mm. The processing time pij of job Jj (j = 1, 2, . . . , n) on machine Mi (i = 1, 2, . . . , m)

is given as a decreasing linear function of its execution starting time t:

pij (t) = aij − bij t , (1)

where aij > 0 denotes the normal processing time of job Jj on machine Mi and bij denotes its decreasingrate. It is assumed that the decreasing rates satisfy the following conditions:

0 < bij < 1 and bij

⎛⎝ m∑

i=1

n∑j=1

aij − aij

⎞⎠< aij .

The first condition ensures that the decrease of each job processing time is less than one unit for everyunit delay in its starting moment. The second one ensures that all job processing times are positive in afeasible schedule (see also [3] for detailed explanations). All jobs are available for processing at time 0.In this paper we consider a special case: aij is proportional to bij , bij = uaij , for some u > 0, that is

pij (t) = aij (1 − ut),

where u(∑n

i=1∑n

j=1 aij − amin) < 1, (amin = mini=1,2,...,m;j=1,2,...,n aij ). The objectives are to minimizethe makespan, the weighted sum of completion time, the maximum lateness and the maximum tardi-ness. We assume unlimited intermediate storage between successive machines for the general flow shopscheduling problem. We also restrict ourselves to permutation schedules only. The problems consideredin the paper are denoted according to the three-filed notation �|�|� introduced by Graham et al. [25]

We define the following notation:

dj = due date of job Jj ;wj = weight of job Jj ;Cij = completion time of job Jj on machine Mi in a given permutation;Cj = Cmj = completion time of job Jj in a given permutation;Lmax = max{Cj − dj |j = 1, 2, . . . , n}, maximum lateness of a given permutation;Tmax = max{0, Lmax}, maximum tardiness of a given permutation;Cmax = max{Cj |j = 1, 2, . . . , n}, makespan of a given permutation;� = ([1], [2], . . . , [n]), a permutation of (1, 2, . . . , n), where [j ] = i means job Ji is the jth oneto be processed.


J1

J1

J2

J2

J3

J3

J1 J2 J3

M1

M2

M3

Fig. 1. A sample of Fm|idm|� and Fm|no − idle, idm|�, n = 3, m = 3.

J1

J1

J2

J2

J3

J3

J1 J2 J3

M1

M2

M3

Fig. 2. A sample of Fm|no − wait, idm|�, n = 3, m = 3.

Definition 1. Mr is dominated by Mk , or Mk dominates Mr iff max{arj |j = 1, 2, . . . , n}� min{akj |j =1, 2, . . . , n}.

In an abbreviated notation, it is denoted as Mk > Mr .

Definition 2. The machines form an increasing series of dominating machines (idm) iff M1 < M2< · · · < Mm.

Definition 3. The machines form an decreasing series of dominating machines (ddm) iff M1 > M2> · · · > Mm.

Definition 4. The machines form an increasing–decreasing series of dominating machines (idm-ddm) iffM1 < M2 < · · · < Mh > · · · > Mm, where 1�h�m.

Definition 5. The machines form an decreasing–increasing series of dominating machines (ddm-idm) iffM1 > M2 > · · · > Mh < · · · < Mm, where 1�h�m.

Observation 1. For a given permutation s = ([1], [2], . . . , [n]) for the problems Fm|aij (1 − ut), idm|�,Fm|aij (1 − ut), no − wait, idm|� and Fm|aij (1 − ukt), no − idle, idm|�, the completion time C[j ] of jobJ[j ], if jobs are processed at the earliest possible time, is as follows (see Figs. 1 and 2):

C[j ] = 1

u

⎛⎝1 −

m∏i=1

(1 − uai[1])j∏

k=2

(1 − uam[k])

⎞⎠ . (2)

Proof. A formal proof can be found in Appendix A.

Using the similar method of Observation 1, we can obtain:


J1J1

J2J2

J3J3J1 J2 J3

M1

M2

M3

Fig. 3. A sample of Fm|no − idle, ddm|�, n = 3, m = 3.

Observation 2. For a given permutation s = ([1], [2], . . . , [n]) for the problems Fm|aij (1 − ukt), ddm|�and Fm|aij (1 − ut), no − wait, ddm|�, the completion time C[j ] of job J[j ] is as follows:

C[j ] = 1

u

⎛⎝1 −

j∏i=1

(1 − ua1[i])m∏

k=2

(1 − uak[j ])

⎞⎠ . (3)

Observation 3. For a given permutation s = ([1], [2], . . . , [n]) for the problem Fm|aij (1 − ut),no − idle, ddm|�, the completion time C[j ] of job J[j ] is as follows (see Fig. 3):

C[j ] = 1

u

⎛⎝1 −

n∏k=1

(1 − ua1[k])m∏

i=2

(1 − uai[n])j∏

k=1

(1 − uam[k])/

n∏k=1

(1 − uam[k])

⎞⎠ . (4)

Observation 4. For a given permutation s = ([1], [2], . . . , [n]) for the problems Fm|aij (1 − ut), idm −ddm|� and Fm|aij (1 − ut), no − wait, idm − ddm|�, the completion time C[j ] of job J[j ] is as follows:

C[j ] = 1

u

⎛⎝1 −

h−1∏i=1

(1 − uai[1])j∏

k=1

(1 − uah[k])m∏

i=h+1

(1 − uai[j ])

⎞⎠ . (5)

Observation 5. For a given permutation s = ([1], [2], . . . , [n]) for the problem Fm|aij (1 − ut),no − idle, idm − ddm|�, the completion time C[j ] of job J[j ] is as follows:

C[j ] = 1

u

⎛⎝1 −

h−1∏i=1

(1 − uai[1])n∏

k=1

(1 − uah[k])m∏

i=h+1

(1 − uai[n])

×j∏

k=1

(1 − uam[k])/

n∏k=1

(1 − uam[k])

⎞⎠ . (6)

Observation 6. For a given permutation s = ([1], [2], . . . , [n]) for the problem Fm|aij (1 − ut),no − idle, ddm − idm|�, the completion time C[j ] of job J[j ] is as follows:

C[j ] = 1

u

⎛⎝1 −

n∏i=1

(1 − ua1[i])h∏

k=2

(1 − uak[n])m−1∏i=h

(1 − uai[1])

×j∏

k=1

(1 − uam[k])/

n∏i=1

(1 − uah[i])

⎞⎠ . (7)


3. Minimizing the makespan

We first study the case of idm − ddm.By (5), we know that

Cmax = C[n] = 1

u

(1 −

h−1∏i=1

(1 − uai[1])n∏

k=1

(1 − uah[k])m∏

i=h+1

(1 − uai[n]))

. (8)

By (6), we know that

Cmax = C[n] = 1

u

(1 −

h−1∏i=1

(1 − uai[1])n∏

k=1

(1 − uah[k])m∏

i=h+1

(1 − uai[n]))

. (9)

It is the same as (8), hence the problem Fm|aij (1 − ut), no − idle, idm − ddm|Cmax is equivalent to theproblems Fm|aij (1 − ut), idm − ddm|Cmax and Fm|aij (1 − ut), no − wait, idm − ddm|Cmax.

Since the term∏n

k=1 (1 − uah[k]) is independent of permutation, Cmax is minimized by placing in thefirst position the job which maximizes the term

∏h−1i=1 (1 − uai[1]) and placing in the last position the

job which maximizes the term∏m

i=h+1 (1 − uai[n]). Algorithm 1 below generates an optimal solutionfor the problems Fm|aij (1 − ut), idm − ddm|Cmax, Fm|aij (1 − ut), no − wait, idm − ddm|Cmax andFm|aij (1 − ut), no − idle, idm − ddm|Cmax.

Algorithm 1. Step 1: Let job Jv and job Jt satisfy the following Eqs. (10) and (11), respectively:

h−1∏i=1

(1 − uaiv) = max

{h−1∏i=1

(1 − uaij )|j = 1, 2, . . . , n

}, (10)

m∏i=h+1

(1 − uait ) = max

{m∏

i=h+1

(1 − uaij )|j = 1, 2, . . . , n

}. (11)

Step 2: If job Jv and job Jt are different, go to Step 3; otherwise, job Jv and job Jt are the same job,i.e., v = t , which is unique in satisfying Eqs. (10) and (11). Let job Jv′ and job Jt ′ satisfy the followingequations, respectively:

h−1∏i=1

(1 − uaiv′) = max

{h−1∏i=1

(1 − uaij )|j �= v and j = 1, 2, . . . , n

},

m∏i=h+1

(1 + bit ′) = max

{m∏

i=h+1

(1 + bij )|j �= t and j = 1, 2, . . . , n

}.

If∏h−1

i=1 (1 − uaiv′)∏m

i=h+1 (1 − uait )�∏h−1

i=1 (1 − uaiv)∏m

i=h+1 (1 − uait ′), set v = v′ and go toStep 3; otherwise, set t = t ′ and go to Step 3.


Step 3: An optimal solution is one in which jobs Jv and Jt are placed in the first and last position,respectively, and the other jobs are placed between them in any order.

Theorem 1. For the problems Fm|aij (1−ut), idm−ddm|Cmax, Fm|aij (1−ut), no−wait, idm−ddm|Cmaxand Fm|aij (1 − ut), no − idle, idm − ddm|Cmax, Algorithm 1 generates an optimal solution.

Proof. The result follows from (8)and (9) directly. �

When h = n, idm − ddm is idm, hence Algorithm 1 can be changed as follows:

Algorithm 1′. Step 1: Let job Jv satisfy the following Eqs. (12),

m−1∏i=1

(1 − uaiv) = max

{m−1∏i=1

(1 − uaij )|j = 1, 2, . . . , n

}. (12)

Step 2: An optimal solution is one in which job Jv is placed in the first position, and the other jobs areplaced in any order.

Theorem 2. For the problems Fm|aij (1 − ut), idm|Cmax, Fm|aij (1 − ut), no − wait, idm|Cmax andFm|aij (1 − ut), no − idle, idm|Cmax, Algorithm 1′ generates an optimal solution.

When h = 1, idm − ddm is ddm, hence Algorithm 1 can be changed as follows:

Algorithm 1′′. Step 1: Let job Jt satisfy the following Eqs. (13),

m∏i=2

(1 − uait ) = max

{m∏

i=2

(1 − uaij )|j = 1, 2, . . . , n

}. (13)

Step 2: An optimal solution is one in which job Jt is placed in the last position, and the other jobs areplaced in any order.

Theorem 3. For the problems Fm|aij (1 − ut), ddm|Cmax, Fm|aij (1 − ut), no − wait, ddm|Cmax andFm|aij (1 − ut), no − idle, ddm|Cmax, Algorithm 1′′ generates an optimal solution.

Now, we consider the case of no − idle, ddm − idm.By (7), we know that

Cmax = C[n] = 1

u

(1 −

n∏i=1

(1 − ua1[i])h∏

k=2

(1 − uak[n])m−1∏i=h

(1 − uai[1])

×n∏

k=1

(1 − uam[k])/

n∏i=1

(1 − uah[i]))

. (14)

Since the terms∏n

i=1 (1 − ua1[i]),∏n

k=1 (1 − uam[k]) and∏n

i=1 (1 − uah[i]) are independent of per-mutation, respectively. Cmax is minimized by placing in the first position the job which maximizes the


term∏m−1

i=h (1 − uai[1]) and placing in the last position the job which maximizes the term∏h

k=2 (1 −uak[n]). Algorithm 2 below generates an optimal solution for the problem Fm|aij (1 − ut), no − idle,ddm − idm|Cmax.

Algorithm 2. Step 1: Let job Jv and job Jt satisfy the following Eqs. (15) and (16), respectively:

m−1∏i=h

(1 − uaiv) = max

{m−1∏i=h

(1 − uaij )|j = 1, 2, . . . , n

}, (15)

h∏k=2

(1 − uakt ) = max

{h∏

k=2

(1 − uakj )|j = 1, 2, . . . , n

}. (16)

Step 2: If job Jv and job Jt are different, go to Step 3; otherwise, job Jv and job Jt are the same job,i.e., v = t , which is unique in satisfying Eqs. (15) and (16). Let job Jv′ and job Jt ′ satisfy the followingequations, respectively:

m−1∏i=h

(1 − uaiv′) = max

{m−1∏i=h

(1 − uaij )|j �= v and j = 1, 2, . . . , n

},

h∏k=2

(1 − uakt ′) = max

{h∏

k=2

(1 − uakj )|j �= t and j = 1, 2, . . . , n

}.

If∏m−1

i=h (1 − uaiv′)∏h

k=2 (1 − uakt )�∏m−1

i=h (1 − uaiv)∏h

k=2 (1 − uakt ′), set v = v′ and go to Step 3;otherwise, set t = t ′ and go to Step 3.

Step 3: An optimal solution is one in which jobs Jv and Jt are placed in the first and last position,respectively, and the other jobs are placed between them in any order.

Theorem 4. For the problem Fm|aij (1 − ut), no − idle, ddm − idm|Cmax, Algorithm 2 generates anoptimal solution.

Proof. The result follows from (14) directly. �

Algorithm 1 (Algorithm 1′,Algorithm 1′′ andAlgorithm 2) only needs O(mn) time to obtain an optimalsolution.

4. Minimizing the total weighted completion time

Lemma 1. The∑n

j=1 wj

∏jk=1 (1 − uak) is maximized when calculated over the permutation order by

non-decreasing values of aj/[wj(1 − uaj )].Proof. Using simple job interchanging technique, the result can be easily obtained. �


Theorem 5. For the problems Fm|aij (1 − ut), idm|∑wjCj , Fm|aij (1 − ut), no − wait, idm|∑wjCj

and Fm|aij (1 − ut), no − idle, idm|∑wjCj with a fixed job in the first sequence position, an opti-mal schedule can be obtained by sequencing the remaining n − 1 jobs in non-decreasing order ofamj/[wj(1 − uamj )].Proof. By (2), we know that

∑wjCj

=n∑

j=1

w[j ]

⎡⎣1

u

⎛⎝1 −

m∏i=1

(1 − uai[1])j∏

k=2

(1 − uam[k])

⎞⎠⎤⎦

= 1

u

n∑j=1

w[j ] − w[1]1

u

m∏i=1

(1 − uai[1]) − 1

u

m∏i=1

(1 − uai[1])

⎡⎣ n∑

j=2

w[j ]j∏

k=2

(1 − uam[k])

⎤⎦ ,

where the term (1/u)∑n

j=1 w[j ] is a constant. The terms w[1](1/u)∏m

i=1 (1−uai[1]) and (1/u)∏m

i=1 (1−uai[1]) are also constants (for a given first job). The term

∑nj=2 w[j ]

∏jk=2 (1 − uam[k]) is maximized by

the non-decreasing order of amj/[wj(1 − uamj )] (Lemma 1). Thus, for the considered problem with afixed job in the first sequence position, an optimal schedule can be obtained by sequencing the remainingn − 1 jobs in non-decreasing order of amj/[wj(1 − uamj )]. �

To solve the problems Fm|aij (1 − ut), idm|∑wjCj , Fm|aij (1 − ut), no − wait, idm|∑wjCj andFm|aij (1−ut), no−idle, idm|∑wjCj , therefore, all jobs can be considered at the first sequence positionto generate n schedules. The one with the minimum weighted sum of completion time among these nschedules is an optimal schedule.

Theorem 6. The problems Fm|aij (1 − ut), ddm|∑wjCj and Fm|aij (1 − ut), no − wait, ddm|∑wjCj

can be solved optimally by scheduling jobs in non-decreasing order of the ratio a1j /[wj

∏mk=1 (1−uakj )].

Proof. Assume that there are given two schedules � and �′. The schedule �′ is obtained from the sched-ule � by interchanging the jobs in the jth and the j + 1st positions. Assume also that the schedule �in which a1[j ]/[w[j ]

∏mk=1 (1 − uak[j ])] > a1[j+1]/[w[j+1]

∏mk=1 (1 − uak[j+1])] is optimal. Stem from

Observation 2, for both schedules only the completion times of the jobs in the jth and the j + 1st po-sitions are different. We now calculate the difference between the values of

∑wjCj obtained for both

schedules.

∑wjCj (�) −

∑wjCj (�

′)

= w[j ]1

u

⎛⎝1 −

j−1∏i=1

(1 − ua1[i])(1 − ua1[j ])m∏

k=2

(1 − uak[j ])

⎞⎠


+ w[j+1]1

u

⎛⎝1 −

j−1∏i=1

(1 − ua1[i])(1 − ua1[j ])(1 − ua1[j+1])m∏

k=2

(1 − uak[j+1])

⎞⎠

− w[j+1]1

u

⎛⎝1 −

j−1∏i=1

(1 − ua1[i])(1 − ua1[j+1])m∏

k=2

(1 − uak[j+1])

⎞⎠

− w[j ]1

u

⎛⎝1 −

j−1∏i=1

(1 − ua1[i])(1 − ua1[j+1])(1 − ua1[j ])m∏

k=2

(1 − uak[j ])

⎞⎠

=[w[j+1]a1[j ]

m∏k=1

(1 − uak[j+1]) − w[j ]a1[j+1]m∏

k=1

(1 − uak[j ])]

j−1∏i=1

(1 − ua1[i]).

The result obtained above is positive, because the expressions w[j+1]a1[j ]∏m

k=1 (1 − uak[j+1])−w[j ]a1[j+1]

∏mk=1 (1 − uak[j ]) > 0 and a1[j ]/[w[j ]

∏mk=1 (1 − uak[j ])] > a1[j+1]/[w[j+1]

∏mk=1 × (1 −

uak[j+1])] is equivalent. Thus, it has been shown that the schedule in which the jobs are arranged innon-decreasing order of the ratio a1j /[wj

∏mk=1 (1 − uakj )] is optimal. �

Theorem 7. For the problem Fm|aij (1−ut), no− idle, ddm|∑wjCj and a fixed job in the last sequenceposition, an optimal schedule can be obtained by sequencing the remaining n− 1 jobs in non-decreasingorder of amj/[wj(1 − uamj )].Proof. It is the same as Theorem 5. �

To solve the problem Fm|aij (1 − ut), no − idle, ddm|∑wjCj , therefore, all jobs can be considered atthe last sequence position to generate n schedules. The one with the minimum weighted sum of completiontime among these n schedules is an optimal schedule.

Theorem 8. For the problems Fm|aij (1−ut), idm−ddm|∑wjCj and Fm|aij (1−ut), no−wait, idm−ddm|∑wjCj with a fixed job in the first sequence position, an optimal schedule can be obtained bysequencing the remaining n − 1 jobs in non-decreasing order of amj/[wj

∏mk=h (1 − uakj )].

Proof. It stems from Theorems 5 and 6. �

To solve the problems Fm|aij (1 − ut), idm − ddm|∑wjCj and Fm|aij (1 − ut), no − wait, idm −ddm|∑wjCj , therefore, all jobs can be considered at the first sequence position to generate n schedules.The one with the minimum weighted sum of completion time among these n schedules is an optimalschedule.

Theorem 9. For the problems Fm|aij (1 − ut), no − idle, idm − ddm|∑wjCj and Fm|aij (1 − ut),no − idle, ddm − idm|∑wjCj with a fixed job in the first sequence position and a fixed job in thelast sequence position, an optimal schedule can be obtained by sequencing the remaining n − 2 jobs innon-decreasing order of amj/[wj(1 − uamj )].Proof. It is the same as Theorem 5. �


To solve the problems Fm|aij (1 − ut), no − idle, idm − ddm|∑wjCj and Fm|aij (1−ut), no−idle,ddm − idm|∑wjCj , therefore, all jobs can be considered at the first sequence position and for the fixedfirst job the remaining n−1 jobs can be considered at the last sequence position. Hence, there are n(n−1)

possible schedules. The one with the minimum weighted sum of completion time among these n(n − 1)

schedules is an optimal schedule.

5. Minimizing the maximum lateness (tardiness)

Lateness of a job is the deviation (negative or positive) of its completion time from its due date whiletardiness is the positive deviation of a job completion time from its due date. A schedule which minimizesLmax also minimizes Tmax [26]. Therefore, the developments in this section consider the minimization ofLmax only.

Theorem 10. For the problems Fm|aij (1 − ut), idm|Lmax, Fm|aij (1 − ut), no − wait, idm|Lmax andFm|aij (1 − ut), no − idle, idm|Lmax, with a fixed job in the first sequence position, an optimal schedulecan be obtained by sequencing the remaining n − 1 jobs in non-decreasing order of dj (EDD rule).

Proof. Consider a schedule �, suppose which is not EDD, is optimal. In this schedule there must be atleast two adjacent jobs, say J[i] and J[i+1] (2�i�n − 1), such that d[i] > d[i+1]. Schedule �′ is obtainedfrom schedule � by interchanging jobs in the ith and in the (i + 1)th positions of �. Under �,

L[i](�) = 1

u

⎛⎝1 −

m∏j=1

(1 − uaj [1])i∏

k=2

(1 − uam[k])

⎞⎠− d[i],

L[i+1](�) = 1

u

⎛⎝1 −

m∏j=1

(1 − uaj [1])i+1∏k=2

(1 − uam[k])

⎞⎠− d[i+1],

whereas under �′ it is

L[i](�′) = 1

u

⎛⎝1 −

m∏j=1

(1 − uaj [1])i−1∏k=2

(1 − uam[k])(1 − uam[i+1])

⎞⎠− d[i+1],

L[i+1](�′) = 1

u

⎛⎝1 −

m∏j=1

(1 − uaj [1])i+1∏k=2

(1 − uam[k])

⎞⎠− d[i].

It is easily verified that if d[i] > d[i+1],max{L[i](�′), L[i+1](�′)}� max{L[i](�), L[i+1](�)}.

Hence, the EDD rule is optimal. �

To solve the problems Fm|aij (1−ut), idm|Lmax, Fm|aij (1−ut), no−wait, idm|Lmax and Fm|aij (1−ut), no − idle, idm|Lmax, therefore, all jobs can be considered at the first sequence position to generate nschedules. The one with the minimum maximum lateness among these n schedules is an optimal schedule.


Table 1

Problem Complexity

Fm|aij (1 − ut), idm|Cmax O(mn)

Fm|aij (1 − ut), no − wait, idm|Cmax O(mn)

Fm|aij (1 − ut), no − idle, idm|Cmax O(mn)

Fm|aij (1 − ut), ddm|Cmax O(mn)

Fm|aij (1 − ut), no − wait, ddm|Cmax O(mn)

Fm|aij (1 − ut), no − idle, ddm|Cmax O(mn)

Fm|aij (1 − ut), idm − ddm|Cmax O(mn)

Fm|aij (1 − ut), no − wait, idm − ddm|Cmax O(mn)

Fm|aij (1 − ut), no − idle, idm − ddm|Cmax O(mn)

Fm|aij (1 − ut), no − idle, ddm − idm|Cmax O(mn)

Fm|aij (1 − ut), idm|∑wjCj O(n2 log n)

Fm|aij (1 − ut), no − wait, idm|∑wjCj O(n2 log n)

Fm|aij (1 − ut), no − idle, idm|∑wjCj O(n2 log n)

Fm|aij (1 − ut), ddm|∑wjCj O(mn log n)

Fm|aij (1 − ut), no − wait, ddm|∑wjCj O(mn log n)

Fm|aij (1 − ut), no − idle, ddm|∑wjCj O(n2 log n)

Fm|aij (1 − ut), idm − ddm|∑wjCj O(mn2 log n)

Fm|aij (1 − ut), no − wait, idm − ddm|∑wjCj O(mn2 log n)

Fm|aij (1 − ut), no − idle, idm − ddm|∑wjCj O(n3 log n)

Fm|aij (1 − ut), no − idle, ddm − idm|∑wjCj O(n3 log n)

Fm|aij (1 − ut), idm|Lmax O(n2 log n)

Fm|aij (1 − ut), no − wait, idm|Lmax O(n2 log n)

Fm|aij (1 − ut), no − idle, idm|Lmax O(n2 log n)

Fm|aij (1 − ut), no − idle, ddm|Lmax O(n2 log n)

Fm|aij (1 − ut), no − idle, idm − ddm|Lmax O(n3 log n)

Fm|aij (1 − ut), no − idle, ddm − idm|Lmax O(n3 log n)

Fm|aij (1 − ut), ddm|Lmax Open problemFm|aij (1 − ut), no − wait, ddm|Lmax Open problemFm|aij (1 − ut), idm − ddm|Lmax Open problemFm|aij (1 − ut), no − wait, idm − ddm|Lmax Open problem

Theorem 11. For the problem Fm|aij (1 − ut), no − idle, ddm|Lmax and a fixed job in the last sequenceposition, an optimal schedule can be obtained by sequencing the remaining n− 1 jobs in non-decreasingorder of dj (EDD rule).

Proof. It is the same as Theorem 10. �

To solve the problem Fm|aij (1 − ut), no − idle, ddm|Lmax, therefore, all jobs can be considered at thelast sequence position to generate n schedules. The one with the minimum maximum lateness amongthese n schedules is an optimal schedule.

Theorem 12. For the problems Fm|aij (1 − ut), no − idle, idm − ddm|Lmax and Fm|aij (1 − ut), no −idle, ddm − idm|Lmax, which a fixed job in the first sequence position and a fixed job in the last sequence


position, an optimal schedule can be obtained by sequencing the remaining n− 2 jobs in non-decreasingorder of dj (EDD rule).

Proof. It is the same as Theorem 10. �

To solve the problems Fm|aij (1−ut), no− idle, idm−ddm|Lmax and Fm|aij (1−ut), no− idle, ddm−idm|Lmax, therefore, all jobs can be considered at the first sequence position and for the fixed first job theremaining n − 1 jobs can be considered at the last sequence position. Hence, there are n(n − 1) possibleschedules. The one with the minimum maximum lateness among these n(n − 1) schedules is an optimalschedule.

6. Conclusions

We consider the general, no-wait and no-idle flow shop scheduling problems with dominant machinesunder the assumption that processing times are not constants over time, a phenomenon that appears to existin various environments. We assume a linear deterioration, i.e., pij =aij (1−ut) for t �0. All problems tominimize the makespan or the weighted sum of completion time studied here are known to have polynomialsolutions in classical scheduling and are shown to remain polynomially solvable (see Table 1). However,this does not appear to be the case of minimizing the maximum lateness, i.e., Fm|aij (1 − ut), ddm|Lmax(Fm|aij (1 − ut), no − wait, ddm|Lmax) and Fm|aij (1 − ut), idm − ddm|Lmax (Fm|aij (1 − ut), no −wait, idm − ddm|Lmax) (see Table 1). Their computational complexity status remain open. Whereasthe corresponding classical scheduling problems Fm|ddm|Lmax [18] and Fm|idm − ddm|Lmax [21] aresolvable in polynomial time. These questions may be a subject of a future research.

Acknowledgements

The author is grateful for two anonymous referees for their helpful comments on an earlier version ofthis paper.

Appendix A.

In the following a formal proof of Observation 1 is given:

Proof (by induction). Without loss of generality, we consider schedule � = [J[1], J[2], . . . , J[n]]. Underthe case of idm, we have

C1[1] = a1[1] = 1

u(1 − (1 − ua1[1])),

C2[1] = C1[1] + a2[1](1 − uC1[1]) = a1[1] + a2[1](1 − ua1[1]) = 1

u

(1 −

2∏i=1

(1 − uai[1]))

,

...


C[1] = Cm[1] = 1

u

(1 −

m∏i=1

(1 − uai[1]))

,

C[2] = C[1] + am[2](1 − uC[1]) = 1

u

(1 −

m∏i=1

(1 − uai[1])(1 − uam[2]))

.

Suppose Observation 1 holds for job J[j ], i.e.,

C[j ] = 1

u

⎛⎝1 −

m∏i=1

(1 − uai[1])j∏

k=2

(1 − uam[k])

⎞⎠ .

Consider job Jj+1.

C[j+1] = C[j ] + am[j+1](1 − uC[j ]) = 1

u

⎛⎝1 −

m∏i=1

(1 − uai[1])j+1∏k=2

(1 − uam[k])

⎞⎠ .

Hence, Observation 1 holds for J[j+1]. This completes the proof of Observation 1. �

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Flow shop scheduling problems with decreasing linear deterioration under dominant machines