two-machine flow-shop minimum-length scheduling with interval processing times

Two-machine flowshop minimum-length scheduling problem with

random and bounded processing times

Ali Allahverdi and aYuri Sotskov

Department of Industrial and Management Systems Engineering, College of Engineering and Petroleum, Kuwait University,

P.O. Box 5969, Safat, Kuwait and aInstitute of Engineering Cybernetics, Surganova St. 6, 220012 Minsk, Belarus and

Universite de Technologie Troges, 12 rue Marie-Curie, BP2060, 10010 Troges, France

E-mail: [email protected][Allahverdi]; [email protected][Sotskov]

Received 17 January 2001; received in revised form 12 December 2001; accepted 1 February 2002

Abstract

We address the two-machine flowshop scheduling problem to minimize makespan where jobs have random and

bounded processing times. The probability distributions of job processing times are unknown and only the lower

and upper bounds of processing times are given before scheduling. In such cases there may not exist a unique

schedule that remains optimal for all possible realizations of processing times, and therefore, a set of schedules

has to be considered which dominates all other schedules. In this paper, we find some sufficient conditions for

the considered problem.

Keywords: Scheduling; flowshop; makespan; random processing times

1. Introduction

The two-machine flowshop problem with the objective of minimizing makespan has been discussed in

the OR literature for several decades for both deterministic and stochastic environments. Johnson

(1954) provided a polynomial time solution for deterministic environments where the processing times

are known with certainty before applying a scheduling algorithm. The following two types of problems

have been defined for stochastic environments. In a stochastic job problem, job processing times are

assumed to be random variables following certain probability distributions; the exponential probability

distribution is the one most often considered in the literature. A particular case of the stochastic job

problem for the two-machine flowshop has been addressed in the literature by Ku and Niu (1986),

Kamburowski (1999), and Elmaghraby and Thoney (1999). In a stochastic machine problem, job

processing times are usually deterministic (and fixed before scheduling); however, job completion

times are random variables as a result of possible machine breakdowns. Allahverdi (1995, 1996, 1997),

Allahverdi and Mittenthal (1994, 1995), and Allahverdi and Tatari (1997) addressed the stochastic

machine problem for the case of the two-machine flowshop.

Intl. Trans. in Op. Res. 10 (2003) 65–76

# 2003 International Federation of Operational Research Societies.

Published by Blackwell Publishing Ltd.

In this paper, we address the stochastic job problem for cases where it is hard to obtain exact

probability distributions for random processing times, and where assuming a specific probability

distribution is not realistic. Commonly, solutions obtained after assuming a certain probability

distribution are not even close to the optimal solution. Fortunately, it has been observed that, although

the exact probability distribution of job processing times may not be known, upper and lower bounds

on job processing times are easy to obtain in many cases. This paper shows that information on the

bounds of job processing times is important and should be utilized in finding a solution for the

scheduling problem.

The problem considered is to minimize the length of a schedule (makespan) of n jobs j 2 J

¼ f1, 2, . . ., ng processed on k machines m 2 M ¼ f1, 2, . . ., kg when only a lower bound

tLj,m > 0 and an upper bound tU

j,m > tLj,m of the processing time t j,m of job j on machine m are

given. All n jobs have the same technological route through k given machines, namely: (1, 2, . . .,k). Using three-field notation of Lawler et al. (1993), this problem may be denoted as

FjtLj,m < t j,m < tU

j,mjCmax.

If the equality tLj,m ¼ tU

j,m holds for each job j 2 J and each machine m 2 M , then the problem


j,mjCmax becomes the deterministic flowshop problem FkCmax which is well studied

in the literature (see Lawler et al. (1993), Tanaev et al. (1994)). In the general case, the problem


j,mjCmax can be considered as a stochastic flowshop problem (see the second part of

the book by Pinedo (1995)). But under strict uncertainty when there is no prior information about

probability distributions of the random processing times, it is only known that processing times will

fall between the given lower and upper bounds with probability one.

The problem FjtLj,m < t j,m < tU

j,mjCmax represents the general case. Even if there is no information

about the possible perturbations of processing time for some job j 2 J on some machine m 2 M , one

can set tLj,m ¼ 0 and tU

j,m to be equal to the planning horizon. Moreover, two instances of a flowshop

problem with the same n and k may be distinguished only in their processing times. As a result, any

individual problem Fk�jn ¼ n�jCmax with a fixed number of machines k ¼ k� and a fixed number

of jobs n ¼ n� is a special case of the problem Fk�jn ¼ n�, tLj,m < t j,m < tU

j,mjCmax if for job j 2 J

and machine m 2 M the value tLj,m is sufficiently small and the value tU

j,m is sufficiently large.

The random processing times in the problem FjtLj,m < t j,m < tU

j,mjCmax are due to external forces in

contrast to scheduling problems with controllable processing times. In the latter problem the objective

is to choose both optimal processing times (which are under the control of a decision-maker) and an

optimal schedule with chosen processing times (see Ishii et al. (1987), Janiak (1988), Strusevich

(1995)).

The setting similar to FjtLj,m < t j,m < tU

j,mjCmax was considered by Lai et al. (1997), and Lai and

Sotskov (1999) for the jobshop scheduling problem with uncertain numerical input data. Specifically,

Lai et al. (1997) provided a formula for calculating the stability radius of the optimal schedule. The

stability radius of a schedule is defined to be the largest variation of the processing times such that this

schedule still remains optimal. In this paper, we address the same problem for a two-machine flowshop,

and obtain efficient results.

The paper is organized as follows. In Section 2, we present definitions and main notations. Section 3

presents dominance relations on the set of schedules for the problem F2jtLj,m < t j,m < tU

j,mjCmax

specifying the optimal order of two jobs. Similar dominance relations are proven in Section 4 for the

case when two jobs are adjacent in the schedule. A numerical example is discussed in Section 4.

Concluding remarks are given in Section 5.

66 A. Allahverdi and Y. Sotskov / Intl. Trans. in Op. Res. 10 (2003) 65–76

2. Definitions and notations

It is well-known that permutation schedules are dominant for the deterministic two-machine flowshop

problem of minimizing makespan. That is, one only needs to consider the same sequence of jobs on both

machines in order to find the optimal schedule. Permutation schedules are also dominant for the problem

of two-machine flowshop with random processing times (see Ku and Niu (1986)). Since we assume that

the processing times are random variables, permutation schedules are dominant for the problem under

consideration as well. Thus, there are n! sequences (permutations) S ¼ f�1, �2, . . ., �n!g for the

problem F2jtLj,m < t j,m < tU

j,mjCmax that needs to be considered in finding the optimal schedule.

For a fixed sequence �r 2 S, we use the notation t[i,m](�r) to denote the actual value of processing

time of the job which is located in position i in the sequence �r on machine m 2 M . Similar to t j,m,

the exact value t[i,m](�r) is unknown before scheduling, and the notation tL[i,m](�r) is used for the lower

bound of processing time of the job in position i on machine m, and tU[i,m](�r) is used for the upper

bound of processing time of the job in position i on machine m.

For each job j 2 J ¼ f1, 2, . . ., ng and machine m 2 M ¼ f1, 2g, any feasible realization t j,m of

processing time satisfies the following inequalities:

tLj,m < t j,m < tU

j,m: (1)

Before scheduling we do not know the exact value of t j,m. However, for our purposes we assume we

know the lower and upper bounds of processing times given by inequalities (1). These inequalities

define the polytope T of feasible vectors t ¼ (t1,1, t1,2, . . ., tn,1, tn,2) of processing times as follows:

T ¼ ft : tLj,m < t j,m < tU

j,m, j 2 f1, 2, . . ., ng, m 2 f1, 2gg:

Similar to Lai and Sotskov (1999), we use the following definition of a solution to the problem

F2jtLj,m < t j,m < tU

j,mjCmax:

Definition 1

A set of sequences S� � S is a solution to the problem F2jtLj,m < t j,m < tU

j,mjCmax if for each feasible

vector t 2 T of processing times, the set S� contains at least one optimal sequence.

Thus the whole set S of sequences is an obvious solution to the problem F2jtLj,m < t j,m < tU

j,mjCmax.

However, it is hard or practically infeasible for a decision-maker to choose the best sequence from a

large set S� of candidates as the processing of jobs evolves. Therefore, it is important to minimize the

cardinality of solution S� constructed for problem F2jtLj,m < t j,m < tU

j,mjCmax. To this end, we

introduce the following two types of dominance relations on the set of sequences S.

Definition 2

Sequence �u 2 S dominates (strongly dominates) sequence �v 2 S with respect to T if the inequality

Cmax(�u) < Cmax(�v) (inequality Cmax(�u) , Cmax(�v), respectively) holds for any vector t of proces-

sing times from polytope T.

Using Definition 2 we can rewrite Definition 1 as follows. A set of sequences S� � S is a solution to

the problem F2jtLj,m < t j,m < tU

j,mjCmax if for each sequence �v 2 S there exists a sequence �u 2 S�which dominates the sequence �v.

A. Allahverdi and Y. Sotskov / Intl. Trans. in Op. Res. 10 (2003) 65–76 67

Let T j,m(�r) denote the sum of processing times of the jobs in positions 1, 2, . . ., j on machine

m 2 f1, 2g for the sequence �r 2 S, i.e.,

T j,m(�r) ¼Xj

i¼1

t[i,m](�r), j ¼ 1, 2, . . ., n and m 2 f1, 2g:

Also let

� j(�r) ¼ T j,1(�r) T j1,2(�r), j ¼ 1, 2, . . ., n, (2)

and

I(�r) ¼ maxf0, �1(�r), �2, (�r), . . ., �n(�r)g (3)

where T0,2(�r) ¼ 0.

Then, the makespan Cmax(s) is given by equality

Cmax(�r) ¼ Tn,2(�r) þ I(�r): (4)

Observe that Tn,2(�r) is independent of the sequence �r 2 S. Therefore, minimizing Cmax(�r) is

equivalent to minimizing I(�r) which is total idle time on the machine 2 2 M ¼ f1, 2g until all the

jobs are completed.

3. Transposition of two jobs

Let sk denote a part (a subsequence) of a sequence �. For example, � ¼ (s1, j, s2) means a sequence in

which job j is surrounded by subsequences s1 and s2. If w 6¼ j, the notations �u ¼ (s1, w, s2, j, s3) and

�v ¼ (s1, j, s2, w, s3) mean that these sequences �u and �v are the same sequence, except that the two

jobs w and j are interchanged. If sequence �u dominates (strongly dominates, respectively) sequence

�v, then sequence �u is no worse (is better) than sequence �v, and hence, in an optimal sequence job w

may (must) precede job j.

Theorem 1. If inequalities tUw,1 < tL

j,1 and tUj,2 < tL

w,2 hold, then the sequence �1 ¼(s1, w, s2, j, s3) 2 S dominates the sequence �2 ¼ (s1, j, s2, w, s3) 2 S with respect to T.

Proof. We consider two job sequences �1 ¼ (s1, w, s2, j, s3) 2 S and �2 ¼ (s1, j, s2, w, s3) 2 S

where �1 is a sequence in which job w is in position � and job j in position �, where � , �, whereas

sequence �2 is obtained from sequence �1 by interchanging only the jobs in positions � and �. We

assume that tUw,1 < tL

j,1 and tUj,2 < tL

w,2. Next, we show that Cmax(�1) < Cmax(�2).

It is obvious that T�1,1(�1) ¼ T�1,1(�2) and T�1,2(�1) ¼ T�1,2(�2) since both sequences have the

same jobs in positions 1, 2, . . ., � 1.

Notice that

� j(�1) ¼ � j(�2) for j ¼ 1, 2, . . ., � 1 (5)

since both sequences �1 and �2 have the same jobs in these positions. It can also easily be shown that


� j(�1) ¼ � j(�2) for each j ¼ �þ 1, �þ 2, . . ., n: (6)

For j ¼ �, we have

��(�1) ¼ T�1,1(�1) þ tw,1 T�1,2(�1),

��(�2) ¼ T�1,1(�2) þ t j,1 T�1,2(�2):

From the above two equations, we obtain ��(�1) ��(�2) ¼ tw,1 t j,1. By hypothesis tUw,1 < tL

j,1, and

hence, whatever the values tw,1 and t j,1 take on, it will always satisfy the inequality

��(�1) < ��(�2): (7)

For j ¼ �,

��(�1) ¼ T�1,1(�1) þ tw,1 þX�1

r¼�þ1

t[r,1] þ t j,1 [T�1,2(�1) þ tw,2 þX�1

r¼�þ1

t[r,2]],

��(�2) ¼ T�1,1(�2) þ t j,1 þX�1

r¼�þ1

t[r,1] þ tw,1 [T�1,2(�2) þ t j,2 þX�1

r¼�þ1

t[r,2]]:

From the last two equations, we obtain ��(�1) ��(�2) ¼ t j,2 tw,2. By hypothesis we have

tUj,2 < tL

w,2. Therefore, regardless of what values of t j,2 and tw,2 we assume, it will always be true that

��(�1) < ��(�2): (8)

For j ¼ �þ 1, �þ 2, . . ., � 1, we obtain

� j(�1) ¼ [T�1,1(�1) þ tw,1 þXj

r¼�þ1

t[r,1]] [T�1,2(�1) þ tw,2 þXj1

r¼�þ1

t[r,2]],

� j(�2) ¼ [T�1,1(�2) þ t j,1 þXj

r¼�þ1

t[r,1]] [T�1,2(�2) þ t j,2 þXj1

r¼�þ1

t[r,2]],

where we assume thatP�

r¼�þ1 t[r,2] ¼ 0. Since both sequences have the same jobs in all positions

except for position �, it follows that � j(�1) � j(�2) ¼ (tw,1 þ t j,2) (t j,1 þ tw,2). By hypothesis

tUw,1 < tL

j,1 and tUj,2 < tL

w,2. Therefore, regardless of the values of t j,1, tw,1, t j,2 and tw,2, the following

inequalities will always be satisfied for each j ¼ �þ 1, �þ 2, . . ., � 1

� j(�1) < � j(�2): (9)

It follows from (5)–(9) that � j(�1) ¼ � j(�2) for each j ¼ 1, 2, . . ., n. Hence, I(�1) < I(�2) (see

equation (3)). Equivalently, Cmax(�1) < Cmax(�2) (see equation (4)). This completes the proof. j

Notice that Theorem 1 may be used for any two sequences �1 ¼ (s1, w, s2, j, s3) 2 S and

�2 ¼ (s1, j, s2, w, s3) 2 S such that sequence �1 may be either optimal or not. The proofs of the

following three theorems (Theorems 2–4) are based on Johnson’s sequence which is optimal for the

problem F2kCmax.

Johnson’s sequence may be constructed as follows. Let us partition the set of jobs J ¼ f1, 2, . . ., nginto two subsets J1 and J2, J ¼ J1 [ J2, J1 \ J2 ¼ ˘, provided that set J1 contains job w with

tw,1 < tw,2 and set J2 contains job j with t j,1 > t j,2. In a Johnson’s sequence, the jobs from set J1 are


located before the jobs from set J2. Jobs within the set J1 are ordered in shortest processing time (SPT)

order of processing times tw,1, i.e., if tw,1 < tv,1, then job w precedes job v. Jobs within the set J2 are

ordered in longest processing time (LPT) order of processing times t j,2, i.e., if ti,2 > t j,2, then job i

precedes job j. If the above four sets of inequalities are all strict, then there exists a unique Johnson’s

sequence, otherwise the cardinality of the set ¸ of all Johnson’s sequences is more than 1.

For a single-element set T ¼ ftg (in other words, for the case of deterministic problem F2kCmax)

any Johnson’s sequence dominates any sequence from the set S. Note that it is possible to find a

sequence that is not Johnson’s sequence but is optimal for the problem F2kCmax. Therefore even if

T ¼ ftg and ¸ ¼ f�g, one cannot guarantee that sequence � strongly dominates all sequences from

the set Snf�g. On the other hand, if Johnson’s sequence is a unique optimal sequence for problem

F2kCmax, then it strongly dominates all other sequences from S with respect to ftg.

Theorem 2. If inequalities

tUw,1 < tL

w,2 (10)

and tLj,1 > tU

j,2 (11)

hold, then for each vector t 2 T of processing times, there exists a sequence �1 ¼ (s1, w, s2, j, s3) 2 S

which dominates all sequences from the set S with respect to ftg.

Proof. Let polytope T ¼ ft : tLj,m < t j,m < tU

j,m, j 2 f1, 2, . . ., ng, m 2 M ¼ f1, 2gg satisfy inequal-

ities (10) and (11). We consider any fixed vector t 2 T of processing times. Let jobs w and j in the

vector t have processing times tw,1, tw,2, t j,1 and t j,2. From (10) it follows that tw,1 < tw,2, and

therefore, there exists a Johnson’s sequence (for vector t of processing times) in which job j is included

in the set J1. From equation (11) it follows that t j,1 > t j,2, and therefore, there exists a Johnson’s

sequence for vector t of processing times in which job j is included in the set J2. Let �1 denote such a

Johnson’s sequence in which inclusions w 2 J1 and j 2 J2 simultaneously hold. Due to Johnson’s rule,

job w has to precede job j in optimal sequence �1, i.e., sequence �1 may be represented in the form

�1 ¼ (s1, w, s2, j, s3). For this fixed vector t of processing times, sequence �1 dominates any sequence

from the set S with respect to ftg. j


tUw,1 < tL

w,2, (12)

and tUw,1 < tL

j,1 (13)

hold, then for each vector t 2 T of processing times, there exists a sequence �1 ¼ (s1, w, s2, j, s3) 2 S

which dominates all sequences from the set S with respect to ftg.

Proof. Let polytope T ¼ ft : tLj,m < t j,m < tU

j,m, j 2 f1, 2, . . ., ng, m 2 M ¼ f1, 2gg satisfy inequal-

ities (12) and (13). We consider any fixed vector t 2 T of processing times. Let job w, in the vector t,

have processing times tw,1 and tw,2, and job j have processing times t j,1 and t j,2. From (12) it follows

that tw,1 < tw,2, and therefore there exists a Johnson’s sequence �1 for vector t of processing times in

which job j is included in the set J1: In sequence �1, job j may be either in the set J1 or in the set J2.

Next, we consider both cases.


Case 1. Let job j be in the set J1 in the Johnson’s sequence �1. Then it follows from inequality (13)

that tw,1 < tUw,1 < tL

j,1 < t j,1. Since tw,1 < t j,1 for jobs w and j in the set J1, then job w has to precede

job j in the Johnson’s sequence �1, i.e., sequence �1 may be represented in the form

�1 ¼ (s1, w, s2, j, s3). For this fixed vector t of processing times sequence �1 dominates any sequence

from the set S with respect to ftg.

Case 2. Let job j be in the set J2 in the Johnson’s sequence �1. By an argument similar to the proof

of Theorem 2, one can show that job w precedes job j in the Johnson’s sequence �1, and so sequence �1

dominates any sequence from the set S. Thus, we can conclude that for each vector t 2 T of processing

times in both cases, there exists a sequence �1 ¼ (s1, w, s2, j, s3) 2 S which dominates all sequences

from the set S with respect to ftg. j


tLj,1 > tU

j,2,

and tLw,2 > tU

j,2

hold, then for each vector t 2 T of processing times, there exists a sequence �1 ¼ (s1, w, s2, j, s3)

which dominates all sequences from the set S with respect to {t}.

Proof. Taking into account the rules for constructing Johnson’s sequence we can see that the content of

Theorem 4 is reversed to the content of Theorem 3. Therefore, the proof of Theorem 4 is analogous to

that of Theorem 3. j

Observe that in the statements of Theorems 2–4, subsequence s2 may be empty, i.e., jobs w and j may

be adjacent in optimal sequence �1.

4. Transposition of adjacent jobs

In this section, we consider dominance relations between sequences that differ in two adjacent jobs

only. Such a dominance relation is common in the literature and known as local dominance relation.

The dominance relations presented in Section 3 are known as global dominance relations. Both

dominance relations are usually used in implicit enumeration techniques such as the branch-and-bound

algorithm to further reduce the solution tree. These dominance relations are also used to construct

efficient heuristics. The dominance relation presented in this section is particularly useful for the

algorithms developed by Lai et al. (1999) for scheduling problems with random and bounded

processing times.

Theorem 5. The sequence �1 ¼ (s1, w, j, s2) 2 S dominates the sequence �2 ¼ (s1, j, w, s2) 2 S with

respect to T if any one of the following four conditions is satisfied.

condition 1: (i) tLj,1 > tU

w,2, and either (iia) tLw,2 > tU

w,1 or (iib) tLw,2 > tU

j,2,

condition 2: (i) tLw,2 > tU

j,1, and either (iia) tLj,1 > tU

w,1 or (iib) tLj,1 > tU

j,2,




j,2 or (iib) tLw,2 > tU

w,1,


j,2, and either (iia) tLw,2 > tU

w,1 or (iib) tLw,2 > tU

j,2:

Proof. We consider exchanging the position of two adjacent jobs in a sequence. The sequence �1 has

job w in position � and job j in position �þ 1, and the sequence �2 is obtained from the sequence �1 by

interchanging the jobs in positions � and �þ 1. We assume that one of the conditions given in the

statement of Theorem 5 is satisfied.

Since both sequences have the same jobs in positions 1, 2, . . ., � 1, we have

� j(�1) ¼ � j(�2) for each j ¼ 1, 2, . . ., � 1: (14)

It can also easily be shown that

� j(�1) ¼ � j(�2) for each j ¼ �þ 2, �þ 3, . . ., n: (15)

Hence, it follows from (14) and (15) that in order to prove that I(�1) < I(�2) (or equivalently

Cmax(�1) < Cmax(�2)) it suffices to show that maxf��(�1), ��þ1(�1)g < maxf��(�2), ��þ1(�2)g.

It follows from (2) that for these two sequences,

��(�1) ¼ T�1,1(�1) þ tw,1 T�1,2(�1),

��(�2) ¼ T�1,1(�2) þ t j,1 T�1,2(�2),

��þ1(�1) ¼ T�1,1(�1) þ tw,1 þ t j,1 T�1,2(�1) tw,2, and

��þ1(�2) ¼ T�1,1(�2) þ t j,1 þ tw,1 T�1,2(�2) t j,2,

where T�1,1(�1) ¼ T�1,1(�2) and T�1,2(�1) ¼ T�1,2(�2) since both sequences have the same jobs in

positions 1, 2, . . ., � 1.

From these four equations, we obtain

��(�1) �(�2) ¼ tw,1 t j,1, (16)

�(�1) ��þ1(�2) ¼ t j,2 t j,1, (17)

��þ1(�1) ��(�2) ¼ tw,1 tw,2, (18)

��þ1(�1) ��(�1) ¼ t j,1 tw,2, (19)

��þ1(�1) ��þ1(�2) ¼ t j,2 tw,2: (20)

Condition 1: Assume that condition 1 is satisfied, i.e., (i) tLj,1 > tU


w,1 or (iib)

tLw,2 > tU

j,2. Then, by (19) and (i)

��(�1) < ��þ1(�1) (21)

regardless what values t j,1 and tw,2 assume.

Independent of the exact values of tw,1 and tw,2, by (18) and (iia), we have

��þ1(�1) < ��(�2): (22)

And again independent of the exact values of t j,2 and tw,2, by equation (20) and (iib), we have


��þ1(�1) < ��þ1(�2): (23)

Now if (21) and either (22) or (23) are satisfied, then,

maxf��(�1), ��þ1(�1)g < maxf��(�2), ��þ1(�2)g:

Condition 2: If condition 2 is satisfied, then it is true that (i) tLw,2 > tU

j,1, and either (iia) tLj,1 > tU

w,1 or

(iib) tLj,1 > tU

j,2. Then, regardless what values t j,1 and tw,2 take on, it follows from (19) and (i) that

��þ1(�1) < ��(�1): (24)

By equation (16) and (iia), we obtain

��(�1) < ��(�2) (25)

regardless of the exact values of tw,1 and t j,1.

Also regardless of the exact values of t j,1 and t j,2, by (17) and (iib), we have

��(�1) < ��þ1(�2): (26)

It is obvious that maxf��(�1), ��þ1(�1)g < maxf��(�2), ��þ1(�2)g if (24) and either (25) or (26) are

satisfied.

Condition 3: Assume that condition 3 is satisfied, i.e., (i) tLj,1 > tU


j,2 or (iib)

tLw,2 > tU

w,1. Then, whatever values t j,1 and tw,1 take on, it follows from (16) and (i) that

��(�1) < ��(�2): (27)

Independent of the exact values of tw,2 and t j,2, by (20) and (iia), we have

��þ1(�1) < ��þ1(�2): (28)

And also independent of the exact values of tw,1 and tw,2, by (18) and (iib), we have

��þ1(�1) < ��(�2): (29)

It is obvious that if (27) and either (28) or (29) are satisfied, then

maxf��(�1), ��þ1(�1)g < maxf��(�2), ��þ1(�2)g:

Condition 4: If condition 4 is satisfied, then it is true that (i) tLj,1 > tU

j,2, and either (iia) tLw,2 > tU

w,1 or

(iib) tLw,2 > tU

j,2. Then, it follows from (17) and (i) that

��(�1) < ��þ1(�2) (30)

whatever values t j,1 and t j,2 are.

We have from (18) and (iia),

��þ1(�1) < ��(�2) (31)

independent of the exact values of tw,2 and tw,1. And also independent of the exact values of t j,2 and

tw,2, by (20) and (iib),

��þ1(�1) < ��þ1(�2): (32)


It is obvious that maxf��(�1), ��þ1(�1)g < maxf��(�2), ��þ1(�2)g if (30) and either (31) or (32) are

satisfied. This completes the proof of Theorem 5. j

To demonstrate the obtained results we consider the following example of two-machine flowshop

problem with five jobs.

Example. Let the bounds of processing times for problem FjtLj,m < t j,m < tU

j,mjCmax be given in Table

1. These bounds define the polytope T of feasible vectors t of job processing times.

It is easy to see that Theorem 1 implies the following dominance relations between jobs 1, 3, 4, and

5. Since tU1,1 ¼ 6 < tL

3,1 ¼ 7, tU1,1 ¼ 6 < tL

4,1 ¼ 6, tU1,1 ¼ 6 < tL

5,1 ¼ 9, tU3,2 ¼ 4 < tL

1,2 ¼ 6, tU4,2 ¼ 5 <

tL1,2 ¼ 6, tU

5,2 ¼ 4 < tL1,2 ¼ 6, job 1 precedes jobs 3, 4 and 5. Similarly, the theorem implies that job 2

precedes jobs 3 and 5 since tU2,1 ¼ 7 < tL

3,1 ¼ 7, tU2,1 ¼ 7 < tL

5,1 ¼ 9, tU3,2 ¼ 4 < tL

2,2 ¼ 8, tU5,2 ¼ 4 <

tL2,2 ¼ 8. Due to Theorem 2, for each feasible vector t 2 T of processing times, there exists an optimal

sequence in which job 2 precedes job 4 since tU2,1 ¼ 7 < tL

2,2 ¼ 8 and tU4,2 ¼ 5 < tL

4,1 ¼ 6. Therefore,

instead of considering 5! ¼ 120 sequences S of five jobs, it is sufficient to consider 2! 3! ¼ 12

Table 1

Upper and lower bounds of job processing times

Job j j ¼ 1 j ¼ 2 j ¼ 3 j ¼ 4 j ¼ 5

tLj,1 4 5 7 6 9

tUj,1 6 7 11 8 11

tLj,2 6 8 2 3 2

tUj,2 10 12 4 5 4

Table 2.

Sequences representing

the solution set

Sequence

1,2,3,4,5

1,2,3,5,4

1,2,4,3,5

1,2,4,5,3

1,2,5,3,4

1,2,5,4,3

2,1,3,4,5

2,1,3,5,4

2,1,4,3,5

2,1,4,5,3

2,1,5,3,4

2,1,5,4,3


sequences of five jobs (given in Table 2) which presents the solution S� to problem F|tLj,m <

t j,m < tUj,m|Cmax with numerical input data given by Table 1.

5. Conclusions

Job processing times cannot be precisely obtained in many practical flowshop environments. This

means that a solution obtained by only considering the average processing times and/or assuming a

certain probability distribution may not be even close to an optimal schedule for the realization of the

process. It has also been observed that even if exact distribution of processing times are not known,

upper and lower bounds on the processing time of a job on each machine can easily be obtained. This

information is important and it needs to be utilized in finding a solution for the scheduling problem.

We have addressed this problem for a two-machine flowshop and obtained some useful dominance

relations that minimize the makespan.

There are different possible extensions to this research. One is to investigate strong dominance

relations. Another is to consider flowshops with more than two machines with the same criterion.

Minimizing makespan is important in situations where a simultaneously received batch of jobs is

required to be completed as soon as possible. An example is a multi-item order submitted by a single

customer that needs to be delivered within the shortest possible time. The makespan criterion also

helps to increase the utilization of resources.

There are other real-life situations in which each completed job is needed as soon as it is processed.

In such situations, one is interested in minimizing the mean or sum of completion times of all jobs

rather than minimizing makespan. This objective is particularly important in real-life situations in

which reducing inventory or holding cost is of primary concern. An extension would be to consider the

same problem with mean flow time criterion for two and more machines.

An assumption that is made in this research is that set-up times are included in processing times.

This assumption is valid for some flowshops and not valid for others. For the latter, set-up times should

be considered as separate from processing times. See Allahverdi et al. (1999) for a recent survey paper

on deterministic flowshops with separate set-up times. This case provides another possible extension.

Acknowledgments

The research of the second author was partially supported by INTAS (project 00-217), ISTC (project

B-104-98), and Fundamental Research Fund of Belarus (project FFI 99-119). The authors would like to

thank anonymous referees for useful comments on the early version of the paper.

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