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Information Processing Letters 115 (2015) 33–39
Contents lists available at ScienceDirect
Information Processing Letters
www.elsevier.com/locate/ipl
Scheduling a variable maintenance and linear deteriorating
jobs on a single machine
Wenchang Luo a,
Min Ji b,∗
a Faculty of Science, Ningbo University, Ningbo 315211, PR Chinab School of Computer Science and Information Engineering, Contemporary Business and Trade Research Center, Zhejiang Gongshang
University, Hangzhou 310018, PR China
a
r
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c
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e
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n
f
o a
b
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Article history:
Received 16 October 2013
Received in revised form 19 August 2014
Accepted 19 August 2014
Available online 26 August 2014
Communicated by B. Doerr
Keywords:
Scheduling
Computational complexity
Deteriorating jobs
Maintenance
We investigate a single machine scheduling problem in which the processing time of a
job is a linear function of its starting time and a variable maintenance on the machine
must
be
performed
prior
to
a
given
deadline.
The
goals
are
to
minimize
the
makespan
and the total completion time. We prove that both problems are NP-hard. Furthermore, we
show that there exists a fully polynomial time approximation scheme for the makespan
minimization problem. For the total completion time minimization problem we point out
that
there
exists
a
fully
polynomial
time
approximation
scheme
for
a
special
case.
© 2014
Elsevier
B.V.
All rights reserved.
1. Introduction
We consider the following scheduling problem: there is
a set of n linear deteriorating jobs J = { J 1, J 2, · · · , J n} to
be non-preemptively processed on a single machine, all of
which are available at time 0. For each job J j , we use p j
and C j to denote the processing time and the completion
time, respectively. The actual processing time p j of job J jis defined by p j = α j + β j s j , where α j denotes the basic
processing time, β j denotes the deteriorating rate and s j
denotes the starting time of job J j , respectively, i.e., p j is
a
(general)
linear
function
of
s j .
Moreover,
a
mandatory
maintenance must be started before a given deadline sd
on the machine and the duration of the maintenance d is
a nonnegative and nondecreasing function of its starting
time s (i.e., d = f (s) and f (s) is nonnegative and nonde-
creasing). Without loss of generality, we also assume that
all the data (the basic processing times, the deteriorating
* Corresponding author.
E-mail addresses: [email protected] (W. Luo), [email protected]
(M. Ji).
rates, the deadline and the duration of maintenance) are
integers and the function f (s) can be computed in polyno-
mial time. Furthermore, we assume that jobs cannot fit all
together before sd (i.e., αn +n
i=1(αi
n j=i+1(1 + β j)) > s)
and that every job J j can be inserted before sd (i.e.,
α j ≤ sd). The objective is to schedule all the linear deterio-
rating jobs and determine the starting time of maintenance
such that the makespan or the total completion time is
minimized.
Let C max (= max j C j) denote the maximum comple-
tion time (makespan). Following the three-field notation
introduced
by
Graham
et
al.
[4] and
the
monograph
by
Gawiejnowicz [6], we denote our problems by 1, VM| p j =
α j + β j s j |C max and 1, VM| p j = α j + β j s j|
j C j respec-
tively, where “VM” stands for a variable maintenance on
the machine.
The above problem is referred to as scheduling dete-
riorating jobs with machine unavailable constraints and
this kind of problem has received considerable attention
since 2003 (see, e.g., [14,3,5,9,7,10–12]). For the related
monograph, we refer the readers to Gawiejnowicz [6].
All of the models proposed in [14,5,9,7,3,11,12] have the
http://dx.doi.org/10.1016/j.ipl.2014.08.011
0020-0190/© 2014
Elsevier
B.V.
All rights reserved.
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34 W. Luo, M. Ji / Information Processing Letters 115 (2015) 33–39
assumptions that the machine starts at time t 0 (> 0) (to
avoid the trivial case), the processing time of a job is a
simple linear function of its starting time (i.e., p j = β j s j ),
and the machine unavailable intervals or the duration of
maintenance is prefixed, which differ from our model.
Related works Browne and Yechiali [1] first considered the
problem
of
scheduling
(general)
linear
deteriorating
jobs (without maintenance) to minimize the makespan. Using
the standard interchange principle, they showed that there
achieves the optimal makespan according to the nonde-
creasing order of α j
β j(i.e. α1
β1≤ α2
β2· · · ≤ αn
βn).
Since 2003, researchers have begun to consider the
problems of scheduling simple linear deteriorating jobs
with machine unavailable constraints. They assumed that
the jobs are available at time t 0 (= 1). Depending on
whether the job processing can be interrupted by the un-
available intervals or not, there are two versions: resum-
able version and non-resumable version.
For the resumable version, Wu and Lee [14] first stud-
ied
the
problem
of
scheduling
simple
linear
deteriorat-ing jobs with an unavailable constraint to minimize the
makespan. They showed the model can be solved by using
the 0–1 integer programming technique. In some special
case, the 0–1 integer programming can be solved in poly-
nomial time. For the general case, the complexity of the
problem is open. Gawiejnowicz and Kononov [7] consid-
ered the proposed model by Wu and Lee [14]. They proved
that the problem is weakly NP-hard and showed that there
exists a fully polynomial time approximation scheme. Fur-
thermore, for the problem with two or more unavailable
intervals, they showed that there does not exist a polyno-
mial time approximation algorithm with a constant worst-
case ratio,
unless
P
= NP.
Ji
and
Cheng
[10] considered a single machine scheduling problem in which the pro-
cessing time of each job is a simple linear deteriorating
function of its waiting time with the machine subject to
an unavailable constraint. The problem was similar to the
proposed problem by Gawiejnowicz and Kononov [7]. For
the makespan objective, they showed that the problem can
be modeled by 0–1 integer programming and then proved
that the problem is weakly NP-hard. Finally they showed
that there exists a fully polynomial time approximation
scheme. Fan et al. [3] considered the problem of schedul-
ing resumable deteriorating jobs on a single machine with
unavailable constraints. The goal is to minimize the total
completion
time.
They
proved
that
the
problem
with
a
sin-gle unavailable interval is NP-hard and showed that there
exists a fully polynomial time approximation scheme. Fur-
thermore, they showed that there does not exist a polyno-
mial time approximation algorithm with a constant worst-
case ratio with two or more unavailable intervals, unless
P = NP.
For the non-resumable version, Ji et al. [9] considered
the single machine scheduling problem in which the jobs
are simple linear deteriorating and the machine is subject
to an unavailable constraint. The goal is to minimize the
makespan or the total completion time. They showed that
both problems are weakly NP-hard. Furthermore, for the
makespan
objective,
they
proposed
an
optimal
on-line al-gorithm for the on-line case, and a fully polynomial time
approximation scheme for the off-line case. For the total
completion time objective, they provided a heuristic al-
gorithm with computational experiments to evaluate its
efficiency.
Our results In this paper, we show both problems 1, VM| p j
= α j + β j s j |C max and 1, VM| p j = α j + β j s j | j C j are
NP-hard.
Furthermore,
for
the
1,
VM| p j = α j + β j s j |C max
problem, we show that there exists a fully polynomial
time approximation scheme (FPTAS). For the 1, VM| p j =
α j + β j s j|
j C j problem, we point out that there also
exists an FPTAS when α j = β j . But for the general case
α j = β j whether it has an FPTAS is open.
The remainder of this paper is organized as follows:
In Section 2, we prove that the 1, VM| p j = α j + β j s j |C max
problem is weakly NP-hard and show that there exists
an FPTAS. In Section 3, we prove that the 1, VM| p j =
α j + β j s j |
j C j problem is also NP-hard and point out
that there exists an FPTAS for the special case α j = β j . The
concluding remarks are given in Section 4.
2. The 1, VM| p j =α j + β j s j |C max problem
In this section, we consider the 1, VM| p j = α j +
β j s j |C max problem. By reducing the Subset Product Prob-
lem to the considered problem, we show that the problem
is NP-hard and then show it admits an FPTAS.
2.1. Proof of NP-hardness
The NP-hardness proof of the problem is demonstrated
by performing a reduction from the NP-complete Subset
Product Problem [8].
Subset Product Problem. Given a finite set S = {1, 2,
· · · , m}, a positive integer x j ∈ Z + for each j ∈ S , and
a positive integer B , does there exist a subset S 1 ⊆ S
such that the product of the elements in S 1 satisfies j∈S 1
x j = B ?
Clearly, we can assume x j ≥ 2 for all j.
Theorem 1. The 1, VM| p j = α j + β j s j |C max problem is NP-
hard.
Proof. Given an arbitrary instance I of the Subset Product
Problem,
we
construct
an
corresponding
instance
I of
the 1, VM| p j = α j + β j s j|C max problem as follows:
• There are n
= m linear deteriorating jobs such that
α j = β j = x j − 1 for j = 1, 2, · · · , n.
• The starting time of the maintenance s is prior to B
− 1
(i.e., s ≤ sd = B
− 1) and the duration is B (i.e., d
=
f (s) = B ).
• The threshold value G is defined as G = 2 X − 1, where
X =
j∈S x j .
Obviously the above construction can be completed in
polynomial time. Next, we show that instance I has a so-
lution
if
and
only
if
there
exists
a
feasible
schedule
π for instance I such that C max(π ) ≤ G .
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W. Luo, M. Ji / Information Processing Letters 115 (2015) 33–39 35
To avoid the confusion, we write S 1(S 2, S ) for instance
I to denote the job set with their indices corresponding to
S 1(S 2, S ) of instance I .
Given a solution S 1 ⊆ S to instance I such that j∈S 1
x j = B , we construct a schedule π for instance I
in the following way:
• First
we
process
the
jobs
in
S 1 according
to
the
nonde-creasing order of
α j
β j, then start the maintenance “VM”
and process the remaining jobs according to nonde-
creasing order of α j
β j.
For convenience, we denote the schedule π as π =
( S 1, VM, S 2), where S 2 (= S \S 1) denotes the remaining
jobs. With simple calculations, we have
C S 1
(π ) =
j∈ S 1
(1 + α j ) − 1 =
j∈S 1
x j − 1 = B − 1,
where C S 1
(π ) denotes the completion time of the final job
in
S 1.
Similarly
define
C S 2(π ) as
the
completion
time
of
the final job in S 2 . Then we have
C S 2
(π ) = 2B
j∈S 2
(1 + α j ) − 1 = 2B
j∈S 2
x j − 1.
Since
j∈S 2
x j = X /B , we achieve that C S 2
(π ) = 2 X − 1,
i.e., there exists a feasible schedule π for instance I such
that C max(π ) = C S 2
(π ) ≤ G .
Conversely, assume that there is a feasible schedule
π for instance I such that C max(π ) ≤ G . Again we use
( S 1, VM, S 2) to denote the schedule π , where the jobs in
ˆS 1 and
ˆS 2 can
be
processed
according
to
nondecreasing
order
of
α j
β j.
Assume that the starting time of maintenance “VM” is
B
− 1
− ∆, ∆
≥ 0. Then with simple calculations we obtain
C S 1
(π ) =
j∈S 1
(1 + α j ) − 1 = B − 1 − ∆
and
C S 2
(π ) = (2B − ∆)
j∈S 2
(1 + α j ) − 1
= (2B − ∆) X
B − ∆− 1.
If ∆ > 0, then we have
C max(π ) = C S 2
(π ) = (2B − ∆) X
B − ∆− 1
> (2B − 2∆) X
B − ∆− 1 = 2 X − 1,
which achieves a contradiction. Thus we obtain ∆ = 0,
which implies
j∈S 1(1
+ α j ) =
j∈S 1 x j = B . Then we ob-
tain a solution of instance I .
In order to solve the 1, VM| p j = α j + β j s j |C max prob-
lem,
we
first
consider
a
close
variant,
in
which
the
starting time of maintenance is fixed (i.e., equal to s). We denote
the variant by 1, nr-a| p j = α j + β j s j |C max, where “nr-a”
in the first field denotes a non-resumable availability con-
straint. The problem can be described as follows:
Given n linear deteriorating jobs J 1, J 2, · · · , J n to be
non-preemptively processed on a single machine, all of
which are available at time 0. The machine is not available
during a fixed interval [s, s + d]. The task is to schedule all
the
jobs
such
that
the
makespan
is
minimized.The 1, nr-a| p j = α j + β j s j |C max problem can be opti-
mally solved by performing the following dynamic pro-
gramming algorithm.
First, we order the jobs as 1, 2, 3, · · · , n according to
the nondecreasing order of α j
β j, i.e., α1
β1≤ α2
β2≤ · · · ≤ αn
βn.
The algorithm contains n phases. In each phase j ( j =
1, 2, · · · , n), a state space E j is generated. Each state
[ j, t , C ] in E j can be associated to a feasible partial sched-
ule for the first j jobs. The variable j denotes the first
j jobs that we have scheduled, t denotes the comple-
tion time of the last job scheduled before s and C is
the makespan of the associated partial schedule. This al-
gorithm
can
be
described
as
follows:
Dynamic programming algorithm:
Initialize E 1 := {[1, α1, α1], [1, 0, s + d + α1 + β1(s + d)]}
For j ∈ {2, 3, · · · , n}
E j = {} .
For every state
[ j − 1, t , C ] ∈ E j−1 do:
1. Put [ j, t , C + α j + β j C ] in E j ;
2. Put [ j, t + α j + β jt , C ] in E j If t + α j + β jt ≤ s.
Endfor
Remove E j−1
Endfor
Output OPT = min[n,t ,C ]∈E n {C } and the corresponding
optimal
schedule,
which
can
be
found
by
backtracking.
Obviously, the set of E 1 consists of two partial sched-
ules, in the first of which job J 1 is processed before s, and
in the second schedule job J 1 is placed after the main-
tenance. For each state [ j − 1, t , C ] in E j−1, job J j has
two possible choices in E j . Due to the nondecreasing or-
der of α j
β j, the first possible position is the last position
in the partial schedule, and job J j completes in this case
at time C + α j + β j C . The second possible position is the
one just before the maintenance: J j starts at time t if
t + α j + β jt ≤ s.
It
can
be
easily
verified
that
the
above
dynamic
pro-gramming algorithm runs in O (ns(αn +
ni=1(1 + βi )(s +
d) +n−1
j=1
ni= j+1(1 + βi )α j )) time.
Next we transform the above dynamic programming
into an FPTAS for problem 1, nr-a| p j = α j + β j s j |C max. The
main idea is that we iteratively remove some states gen-
erated by the dynamic programming to reduce the size
of the state space down to polynomial size, which is at-
tributed to Woeginger [13].
Algorithm H 0:
Step 1. Let ∆ = 1 + ε/(2n). Partition
• the
interval
[0,
s] into
r 1 = log∆ s subintervals [0, ∆], · · · , [∆t 1 , ∆t 1+1], · · · , [∆r 1−1, s] and
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W. Luo, M. Ji / Information Processing Letters 115 (2015) 33–39 37
Step 3. Output the schedule π = arg minπ (i),i=0,1,···,l ×
{C max(π (i))} as the solution.
Note that in each time interval [si , si+1] we have si+1 ≤(1 + ε)si and f (si+1) ≤ (1 + ε) f (si ).
Theorem 5. If f (s) is a continuous and nondecreasing func-
tion,
and
can
be
computed
in
polynomial
time,
then
Algo-rithm H 1 is a fully polynomial time approximation scheme for
the 1, VM| p j = α j + β j s j|C max problem.
Proof. Let s∗ denote the starting time of maintenance in
an optimal schedule π ∗. Let S ∗1 denote the set of jobs
scheduled before s∗ in π ∗ and S ∗2(= J \ S ∗1) denote the set
of jobs scheduled after the maintenance in π ∗. Clearly we
have s∗ = s0, s1 or s∗ ∈ (si−1, si], i = 2, · · · , l. If s∗ = s0 or
s1, it is easy to obtain the optimal schedule. Thus we only
need to consider the cases that s∗ ∈ (si−1, si ], i = 2, 3, · · · , l.
For schedule π (i), with simple calculations we have
C max
π (i)
=
C max
S i1
+ f
C max
S i1 n
k=|S i1|+1
(1 + β[k])
+ α[n] +
n−1k=|S i1|+1
α[k]
n
j=k+1
(1 + β[ j])
,
where
|S i1| denote the number of jobs in S i1 and
[k] denote
the job location in schedule π (i).
Now consider the 1, nr-a| p j = α j + β j s j|C max problem,
where the starting time of maintenance is si and the du-
ration
of
maintenance
is
f (si),
we
construct
a
schedule
π∗
corresponding to schedule π ∗ and a schedule π (i) cor-
responding to schedule π (i). The subset of jobs before
the maintenance and after the maintenance are the same
in π ∗ and π ∗. The only difference in π ∗ and π ∗ is the
different starting time and the duration of maintenance.
The same relations on schedules π ∗ and π ∗ also apply to
schedules π (i) and π (i).
Clearly we have C max( π (i)) ≤ (1 + ε)C max( π ∗). Further-
more, we have
C max
π ∗
=
s∗ + f
s∗
n
k=|S ∗1|+1
(1 + β[k])
+ α[n] +
n−1k=|S ∗1|+1
α[k]
n
j=k+1
(1 + β[ j])
,
and
C max
π ∗
=
si + f (si ) n
k=|S ∗1|+1
(1 + β[k])
+ α[n] +
n−1k=|S ∗1|+1
a[k]
n
j=k+1
(1 + β[ j])
.
Since
f (s) is
continuous
and
nondecreasing,
according to the choice of si , we have si ≤ (1
+ ε )s∗ and f (si ) ≤
(1
+ ε) f (s∗). Then we have C max( π ∗) ≤ (1
+ ε)C max(π ∗).
Then we obtain
C max
π (i)
≤ C max
π (i)
≤
1 + ε
C max
π ∗
≤
1 + ε
2
C max
π ∗
.
Let
ε = ε2 + 2ε ,
we
achieve
C max(π )
≤ C max(π (i))
≤ (1
+ε)C max(π ∗). Computing each si , i = 0, 1, 2 · · · , l needs at
most logsd
2 time by using binary search because the func-
tion f (s) is continuous and nondecreasing. Thus, with the
assumption that f (s) can be computed in polynomial time,
Algorithm H 1 runs in polynomial time.
3. The 1, VM| p j =α j + β j s j |
j C j problem
In this section, we consider the 1, VM| p j = α j +
β j s j |
j C j problem. Again we use the Subset Product
Problem for the reduction to show that the problem is
NP-hard.
Then,
we
point
out
that
there
exists
an
FPTAS
for the special case α j = β j .
3.1. Proof of NP-hardness
Again the NP-hardness proof is constructed by per-
forming a reduction from the NP-complete Subset Product
Problem.
Theorem 6. The 1, VM| p j = α j + β j s j |
j C j problem is NP-
hard.
Proof. For
a
given
instance
I of
the
Subset
Product
Prob-lem, we construct an corresponding instance I of the
1, VM| p j = α j + β j s j |
j C j problem as follows:
Let A
= X /B , where X =
j∈S x j .
• There are totally n = m + 3 linear deteriorating jobs
such that α j = β j = x j − 1 for j = 1, 2, · · · , m and
3 large jobs αm+1 = βm+1 = X 2 A − 1, αm+2 = βm+2 =
X 2 B − 1, αm+3 = βm+3 = X 4 − 1.
• The starting time of maintenance s is prior to X 3 − 1
(i.e., s ≤ sd = X 3 − 1) and the duration of maintenance
d is X 3 (i.e., d
= f (s) = X 3).
• The threshold value G is defined as G = X 3 − 1 + 2(m +
1)( X 6 − 1/2)
+ 2 X 10 − 1.
It can be verified that the above construction can be
completed in polynomial time. We show that the in-
stance I has a solution if and only if there exists a fea-
sible schedule π for the scheduling instance I such that
Z (π ) ≤ G , where Z (π ) denote the total completion time
of schedule π .
Before we present our main result, we state some aux-
iliary lemmas.
Lemma 7. (See [2,9].) For any subset S 1 ⊆ S, we have
B
j∈S \S 1 x j + A
j∈S 1 x j ≥ 2 X ,
and
the
equality
holds
if
and
only if
j∈S 1 x j = B and
j∈S \S 1
x j = A .
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38 W. Luo, M. Ji / Information Processing Letters 115 (2015) 33–39
Lemma 8. Let Z (π ) denote the total completion time of sched-
ule π . If the schedule π for the 1, VM| p j = α j + β j s j |
j C j problem satisfies Z (π ) ≤ G , then the following two claims hold.
(1) One of the jobs J m+1 and J m+2 is scheduled before the
maintenance, and the other after the maintenance;
(2) the job J m+3 is scheduled after the maintenance.
Proof. The proof is straight. We omit it.
Next we prove Theorem 6.
Suppose that there exists a subset S 1 ⊆ S satisfying j∈S 1
x j = B (i.e.,
j∈S 2 x j = A, where S 2 = S \S 1) for in-
stance I . We construct a schedule π for instance I in the
following way:
• First we process the jobs in S 1 according to the nonde-
creasing order of α j
β jstarting at time 0, then we pro-
cess J m+1, and start the maintenance right after the
completion
time
of
job
J m+1,
after
the
maintenance
we process the jobs in S 2 according to nondecreas-
ing order of α j
β j, and finally we process jobs J m+2 and
J m+3.
For convenience, we denote the schedule π as π =
(S 1, J m+1, VM, S 2, J m+2, J m+3).
With simple calculations we have
C m+1 =
j∈S 1
(1 + α j)(1 + αm+1) − 1 = X 3 − 1,
C m+2 = j∈S 1
(1 + α j )(1 + αm+1) + X 3×
j∈S 2
(1 + α j )(1 + αm+2) − 1
= 2 X 6 − 1,
C m+3 = (1 + C m+2)(1 + αm+3) − 1 = 2 X 10 − 1.
Because C j < C m+2 for j = 1, 2, · · · , m, we achieve
Z (π ) ≤
m j=1
C j +
3k=1
C m+k
< X 3 − 1 + 2(m + 1) X 6 − 1/2
+ 2 X 10 − 1 = G.
Now suppose that there is a feasible schedule π such
that Z (π ) ≤ G . According to Lemma 8, we only need to
consider two cases.
Case 1. Job J m+1 is scheduled before the maintenance.
Case 2. Job J m+2 is scheduled before the maintenance.
Consider Case 1. Let S 2 ⊆ {1, 2, · · · , n} denote the in-
dices of jobs scheduled after the maintenance except the 3
large jobs and S 1 = {1, 2, · · · , n}\{S 2}. Then the maximum
completion
time
of
the
jobs
scheduled
before
the
mainte-nance is
C Bmax =
j∈ S 1
(1 + α j )(1 + αm+1) − 1
= X 2
A
j∈S 1
(1 + α j )
− 1,
and the maximum completion time of the jobs processed
after
the
maintenance
is
C Amax =
X 2
A
j∈S 1
(1 + α j )
+ X 3
×
j∈S 2
(1 + α j )(1 + αm+2)(1 + αm+3) − 1
= X 10 + X 9
B
j∈S 2
(1 + α j )
− 1.
Hence, we have
Z (π ) > C
B
max + C
A
max
= X 2
A
j∈S 1
(1 + α j )
− 1 + X 10
+ X 9
B
j∈S 2
(1 + α j )
− 1
= X 10 + X 9
A
j∈ S 1
(1 + α j ) + B
j∈ S 2
(1 + α j )
− X 9 − X 2 A j∈
ˆS 1
(1 + α j )− 2.
Clearly C Bmax ≤ sd , i.e., X 2( A
j∈S 1
(1 + α j)) ≤ X 3 − 1.
We have A
j∈S 1
(1 + α j ) ≤ X , which implies
j∈S 1(1 +
α j ) ≤ B . Suppose there is no solution for instance I , then j∈S 1
(1 + α j ) < B . By Lemma 7, we have B
j∈S 2
x j +
A
j∈S 1
x j > 2 X , and thus B
j∈S 2
x j + A
j∈S 1
x j ≥
2 X + 1. With simple calculations we have
Z (π ) > X 10 + 2 X 10 + X 9
− X 9 − X 2
A
j∈S 1
(1 + α j )
− 2
> X 10 + 2 X 10 + X 9 − X 9 − X 2
X − 2
> 2 X 10 + X 9 + X 3 − 2
> G,
which achieves a contradiction. Thus
j∈S 1(1
+ α j ) = j∈S 1
x j = B , i.e., there exists a solution for instance I .
For Case 2, the proof is similar to Case 1, and we omit
it.
For the 1, VM| p j = α j + β j s j |
j C j problem with the
special case α j = β j , it is easily verified that there exist
an optimal solution such that the jobs before the main-
tenance
are
sequenced
according
to
the
nondecreasing
α j
order, and so are the jobs scheduled after the maintenance.
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W. Luo, M. Ji / Information Processing Letters 115 (2015) 33–39 39
Using the similar idea for the 1, VM| p j = α j + β j s j |C max
problem, we can also derive an FPTAS. But for the general
case α j = β j , whether it admits an FPTAS is open.
4. Concluding remarks
In
this
paper,
we
study
the
problem
of
scheduling a variable maintenance and (general) linear deteriorating
jobs on a single machine. For the makespan and total com-
pletion time minimization objectives, we show that both
problems are NP-hard and there exists an FPTAS for the
makespan minimization problem. But for the total comple-
tion time minimization problem, in general case whether
it admits an FPTAS is open.
Acknowledgements
We thank three anonymous referees for their help-
ful
comments
on
an
earlier
version
of
our
paper.
Luo was supported by Ningbo Natural Science Foundation
(2013A610099), Hu Lan outstanding doctoral foundation
of Ningbo University, Science Foundation of Ningbo Uni-
versity (xkl1320) and the K.C. Wong Magna Foundation of
Ningbo University. Ji was supported in part by the National
Social Science Foundation of China (Grant No. 14CGL071),
the Humanities and Social Sciences Planning Foundation of
the Ministry of Education (Grant No. 13YJA630034), and
the Contemporary Business and Trade Research Center of
Zhejiang Gongshang University, which is the key Research
Institute of Social Sciences and Humanities of the Ministry
of Education.
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