• No results found

Online batch scheduling on parallel machines with delivery times

N/A
N/A
Protected

Academic year: 2021

Share "Online batch scheduling on parallel machines with delivery times"

Copied!
7
0
0

Loading.... (view fulltext now)

Full text

(1)

Contents lists available atScienceDirect

Theoretical Computer Science

journal homepage:www.elsevier.com/locate/tcs

Online batch scheduling on parallel machines with delivery times

Yang Fang, Xiwen Lu

, Peihai Liu

Department of Mathematics, School of Science, East China University of Science and Technology, Shanghai 200237, China

a r t i c l e i n f o Article history:

Received 25 January 2011

Received in revised form 19 May 2011 Accepted 6 June 2011 Communicated by D.-Z. Du Keywords: Online algorithm Batch scheduling Parallel machines Delivery time

a b s t r a c t

We study the online batch scheduling problem on parallel machines with delivery times. Online algorithms are designed onmparallel batch machines to minimize the time by which all jobs have been delivered. When all jobs have identical processing times, we provide the optimal online algorithms for both bounded and unbounded versions of this problem. For the general case of processing time on unbounded batch machines, an online algorithm with a competitive ratio of 2 is given when the number of machinesm=2 or m=3, respectively. Whenm≥4, we present an online algorithm with a competitive ratio of 1.5+o(1).

©2011 Elsevier B.V. All rights reserved.

1. Introduction

In this paper, we consider the online scheduling problem onmparallel batch machines to minimize the time by which all jobs have been delivered. For each jobJj, it has a release timerj, a processing timepjand a delivery timeqj. There arem parallel batch machines and sufficient number of vehicles. Once a job finishes its processing on a batch machine, it should be delivered to its destination by a vehicle. LetCjbe the completion time on the batch machine ofJj, andLjthe time by which

Jjhas been delivered, i.e.,Lj

=

Cj

+

qj. Our goal is to minimize the timeLmax, by which all jobs have been delivered, i.e.,

Lmax

=

max

j

{

Lj

:

Lj

=

Cj

+

qj

}

. This problem can be described asPm

|

rj

,

qj

,

B

,

online

|

Lmax.

In classic machine scheduling problems, a machine can process at most one job at a time. Lee et al. [3] introduced the batch scheduling model. In our paper, the batch model is a burn-in model. A batch machine can process up toBjobs simultaneously as a batch. The processing time of a batchBiis the longest processing time of all jobs in the batch, and jobs inBihave the same beginning time and completion time. Once a batch starts to be processed, we cannot stop it. The batch machine could be boundedB

<

+∞

if the boundBof each batch size is finite, or unboundedB

= +∞

ifBis sufficiently large.

In this paper, the online setting is over-time. Each job becomes available until its release time which is not known in advance. Once a jobJjarrives, all its characteristics are known and it could be considered to be processed. Even the number of jobsnis unknown until the last job has been scheduled.

The standard measure of quality of online algorithms is competitive ratio. For minimum optimal problem, an online algorithm is called

ρ

-competitive if, for any instance, the cost output by the online algorithm is at most

ρ

times the optimal offline cost. The competitive ratio of an online algorithm is defined as the infimum of all values

ρ

. Moreover, if there are is not an online algorithm with competitive ratio less thanLfor some problem, we call the lower bound of this problem asL. If the competitive ratio

ρ

of an algorithm for this online problem matches the lower bound, i.e.

ρ

=

L, we call the algorithm optimal, or best possible.

Hoogeveen and Vestjens [2] first study the online problem 1

|

rj

,

qj

,

online

|

Lmax. They show that the lower bound is

(

5

+

1

)/

2, and give the best possible algorithm. On identical parallel machines, the lower bound could not be smaller ∗Corresponding author. Tel.: +86 21 64253262.

E-mail address:xwlu@ecust.edu.cn(X. Lu).

0304-3975/$ – see front matter©2011 Elsevier B.V. All rights reserved.

(2)

than 1.5 (see Vestjens [8]). Hall and Shmoys show that the competitive ratio of the LS algorithm is 2. Liu [5] gives a 1.618-competitive algorithm for two machines.

For the batch version 1

|

rj

,

qj

,

B

,

online

|

Lmax, it is easy to see that the lower bound could not be smaller than

(

5

+

1

)/

2 even ifpj

=

1 andqj

=

0 [1,9]. Tian et al. [6] give a 2-competitive algorithm for the unbounded case and a 3-competitive algorithm for the bounded case. When all jobs’ processing times are the same, i.e.pj

=

p, they provide optimal algorithms for both bounded and unbounded cases with competitive ratios of

(

5

+

1

)/

2.

Zhang et al. [9] address the problemPm

|

rj

,

B

= +∞

,

online

|

Cmax. They give a lower bound

m+

1

2, and present an online algorithm with a competitive ratio of 1

+

α

m, where

α

m

=

(

1

α

m

)

m−1. When all jobs have identical processing times, Zhang et al. [10] show that the lower bound could not be smaller than 1

+

β

mfor the unbounded case and

(

5

+

1

)/

2 for the bounded case, where

(

1

+

β

m

)

m+1

=

2

+

β

m. They also provide the optimal algorithms for both cases. For the general case on unbounded machines, Liu et al. [4] and Tian et al. [7] prove that the lower bound is 1

+

(

m2

+

4

m

)/

2, and give

different optimal algorithms independently. Thus the problemPm

|

rj

,

B

= +∞

,

online

|

Cmaxis settled.

Here are some notations used in the paper. For any job setJ, denote byr

(

J

)

,p

(

J

)

, andq

(

J

)

, the minimum release time, the largest processing time, and the largest delivery time of jobs inJ, respectively. LetS

(

Bi

)

be the beginning time of batch

Bi. For any batchBi, denote byJ(i)

(

p(i)

,

q(i)

)

andJ(i)

(

p

(i)

,

q

(i)

)

, the longest job and the job having the largest delivery time ofBi, i.e.,p(i)

=

p

(

Bi

)

andq(i)

=

q

(

Bi

)

. We also useU

(

t

)

to denote the set of all unscheduled jobs available at timet. Let

φ

=

(

5

1

)/

2.

Inspired by [6,10], we study the problems Pm

|

rj

,

pj

=

p

,

qj

,

B

<

+∞

,

online

|

Lmax and Pm

|

rj

,

pj

=

p

,

qj

,

B

=

+∞

,

online

|

Lmax in Sections 2 and 3, and give the optimal online algorithms respectively. In Section 4, we consider

Pm

|

rj

,

qj

,

B

= +∞

,

online

|

Lmax. Whenm

=

2, a 2-competitive algorithmH2is given. Improving from the algorithmH2,

we get an algorithmHmfor the casem

3. And the competitive ratio ofHmis 1

.

5

+

o

(

1

)

, which is 1.5 whenmintends to infinity.

2. The bounded case with identical processing times

In this section, we assume that all jobs have the same processing times on bounded batch machines. This problem can be expressed asPm

|

rj

,

pj

=

p

,

qj

,

B

<

+∞

,

online

|

Lmax. A batch is calledfullif it contains exactlyBjobs. Otherwise, it is

non-full. Without loss of generality, we assume that the first job arrives at time 0.

It is well known that, for offline version of bounded batch scheduling problem on a single machine with identical release times, the FBLPT (Full-Batch Longest Processing Time) rule can be used to achieve the minimal makespan. It is also used to design an online algorithm. Similar to the FBLPT rule, we have the FBLDT (Full-Batch Largest Delivery Times) rule, and provide an algorithmHoffbased on the FBLDT rule.

Hoff:Index the jobs such thatq1

q2

≥ · · · ≥

qn. Group the firstBjobs as a batch, the nextBjobs as another batch and so on(The last batch may be non-full). Schedule the batches one by one whenever there is an idle machine.

Theorem 1. For the problem Pm

|

pj

=

p

,

qj

,

B

<

+∞|

Lmax, the algorithm Hoffis optimal.

Proof. Since all jobs have the same processing times, the job having larger delivery time should be processed earlier. It is easy to get an optimal schedule matchingHoff by pairwise interchange argument.

ByTheorem 1, it is better to schedule the jobs with larger delivery times. Now we give an online algorithm based on the FBLDT rule. By this algorithm, the schedule begins att

=

φ

p.

Algorithm HB Sett

=

φ

p. Repeat:

Apply FBLDT toU

(

t

)

. If there are more thanmbatches, then process the firstmbatches in

[

t

,

t

+

p

]

among all available batches. Otherwise, process all the available batches. Lett

=

t

+

p.

Theorem 2. For the problem Pm

|

rj

,

pj

=

p

,

qj

,

B

<

+∞

,

online

|

Lmax, Algorithm HBis optimal.

Proof. Let

σ

and

π

be the schedule obtained by AlgorithmHBand an optimal schedule. According to AlgorithmHB, we know that any batch in

σ

starts at

+

i

)

p, wherei

0 and is an integer. LetJlbe the first job such thatLl

(σ )

=

Lmax

(σ )

, which

starts at

+

k

)

p. Without changing the value obtained by AlgorithmHBand increasing the optimal value, we can remove all jobs processed at or after

+

k

)

pexceptJl. Suppose thatrl

< (φ

+

k

)

p. Otherwise, we haveLmax

(σ )

=

+

k

+

1

)

p

+

ql andLmax

(π)

+

k

+

1

)

p

+

ql, which implies that

σ

is an optimal schedule.

For convenience, we denote byIithe interval

[

+

i

)

p

, (φ

+

i

+

1

)

p

) (

0

i

k

)

.

LetIxbe the last interval beforeJlsuch that duringIxin

σ

at least one of the following conditions holds: 1. there is at least one idle machine,

2. there is at least one non-full batch,

(3)

IfIxexists, letG

(

l

)

denote the set which consists of jobJland all jobs betweenIxandJlin

σ

. Otherwise, letG

(

l

)

denote the set of jobJland all jobs scheduled beforeJlin

σ

. Due to the definition ofIx, all jobs inG

(

l

)

have delivery time no less thanql. We shall refer toIxas the interference interval for the schedule

σ

as it may delay the start times of the jobs inG

(

l

)

.

IfIxexists, then by the definition ofG

(

l

)

, all jobs inG

(

l

)

are released after

+

x

)

p. We have thatLmax

(π)

+

x

)

p

+

(

k

x

)

p

+

ql

=

+

k

)

p

+

qlandLmax

(σ )

=

+

k

+

1

)

p

+

ql, wherek

1. Hence, we have

Lmax

(σ )

Lmax

(π)

+

k

+

1

)

p

+

ql

+

k

)

p

+

ql

1

+

p

+

1

)

p

=

1

+

φ.

Which means thatLmax

(σ )

(

1

+

φ)

Lmax

(π)

.

IfIxdoes not exist, then due to the definition ofG

(

l

)

, we have thatLmax

(π)

(

k

+

1

)

p

+

qlandLmax

(σ )

=

+

k

+

1

)

p

+

ql. ThusLmax

(σ )

Lmax

(π)

φ

p

φ

Lmax

(π)

which means thatLmax

(σ )

(

1

+

φ)

Lmax

(π)

. Since the lower bound is 1

+

φ

, AlgorithmHBis optimal.

3. The unbounded case with identical processing times

In this section, we deal with the problemPm

|

rj

,

pj

=

p

,

qj

,

B

= +∞

,

online

|

Lmax. According to [10], the lower bound

cannot be smaller than 1

+

β

m, where

(

1

+

β

m

)

m+1

=

2

+

β

m. SinceBis infinite, we just need to decide when to start unscheduled jobs. Without loss of generality, we assume that the first job arrives at time 0.

Similar toHB, we consider to start a batch at time

[

(

1

+

β

m

)

k

1

]

p. Since

(

1

+

β

m

)

m+1

=

2

+

β

mandk

1, we have

(

1

+

β

m

)

k+m

1

=

(

1

+

β

m

)

k−1

(

2

+

β

m

)

1

=

(

1

+

β

m

)

k

+

(

1

+

β

m

)

k−1

1

(

1

+

β

m

)

k

.

Thus, the batch, which starts at

[

(

1

+

β

m

)

k

1

]

p, will complete not after

[

(

1

+

β

m

)

k+m

1

]

p. The algorithm and its competitive analysis are as follows.

Algorithm H

Step 1. Setk

=

1,t

=

β

mp.

Step 2. If

|

U

(

t

)

|

>

0, put all jobs ofU

(

t

)

in batchBkand startBkat timeton machinek(modm). Otherwise,Bk

= ∅

. Step 3. Setk

=

k

+

1, andt

= [

(

1

+

β

m

)

k

1

]

p. Return to Step 2.

According to AlgorithmH∞, batchB(i−1)m+jshould be assigned on machineMj. We have proved that the batch which starts at

[

(

1

+

β

m

)

k

1

]

pwill complete before or at the time

[

(

1

+

β

m

)

k+m

1

]

p. Therefore, the batchBk+mcan start its processing on the same machine at

[

(

1

+

β

m

)

k+m

1

]

p. The beginning time of any batchBiis

[

(

1

+

β

m

)

i

1

]

p.

Theorem 3.The competitive ratio of algorithm His1

+

β

m.

Proof. Let

σ

and

π

be the schedule produced byHand an optimal schedule. Choose any jobJ

j, and denote the batch containing it byBk. We haverj

>

S

(

Bk−1

)

(setS

(

B0

)

=

0). Note thatLj

(σ )

=

(

1

+

β

m

)

kp

+

qjandLj

(π)

(

1

+

β

m

)

k−1p

+

qj. Thus,

Lj

(σ)/

Lj

(π)

1

+

β

m

,

Jj

Therefore,Lmax

(σ)/

Lmax

(π)

1

+

β

m.

Since the lower bound cannot be smaller than 1

+

β

m, AlgorithmHis optimal. Its competitive ratio is 1

+

β

m.

4. The general case on unbounded machines

In this section, we first give a 2-competitive online algorithm forP2

|

rj

,

qj

,

B

= +∞

,

online

|

Lmax. Then, for the casem

3,

we provide an algorithm with a competitive ratio of 1

.

5

+

o

(

1

)

, which is not greater than 2. We partitionU

(

t

)

into two setsA

(

t

)

andB

(

t

)

.

A

(

t

)

= {

Jj

|

qj

α

pj

,

Jj

U

(

t

)

}

,

B

(

t

)

= {

Jj

|

qj

< α

pj

,

Jj

U

(

t

)

}

,

where

α

depends onm. Whenm

=

2, set

α

=

φ

.

Now we give an algorithm forP2

|

rj

,

qj

,

B

= +∞

,

online

|

Lmax. Denote two machines byM1andM2.

Algorithm H2

(1) WhenM1is idle andA

(

t

)

is not empty, start all jobs inA

(

t

)

attonM1ift

(

1

+

φ)

r

(

A

(

t

))

+

φ

p

(

A

(

t

))

. Otherwise, wait.
(4)

By AlgorithmH2, the jobs ofA

(

t

)

are assigned to be processed onM1and jobs ofB

(

t

)

onM2. For convenience, letAi and Bi be the batch processed onM1and M2, respectively. By the algorithm, jobs ofBi arrive afterS

(

Bi−1

)

. IfS

(

Bi

)

=

(

1

+

φ)

r

(

Bi

)

+

φ

p

(

Bi

)

, we callBiis a regular batch. Otherwise, it is called a non-regular batch. For a non-regular batch

Bi,S

(

Bi

)

=

S

(

Bi−1

)

+

p

(

Bi−1

)

andS

(

Bi

) > (

1

+

φ)

r

(

Bi

)

+

φ

p

(

Bi

)

. There are similar definitions and properties for batchAion

M1.

Denote by

σ

and

π

, the schedule produced byH2and an optimal schedule. We have an important property about the

schedule

σ

according to the following lemma given in [10]:

Lemma 1 ([10]). If Biis a regular batch, then Bi+1or Bi+2is also a regular batch.

This lemma also holds for batches onM1. According toLemma 1, the consecutive batchesBiandBi+1both cannot be

non-regular batches. Note that

φ

+

φ

2

=

1. IfB

iis regular andBi+1is not, we haveS

(

Bi

)

+

p(i)

=

S

(

Bi+1

) > (

1

+

φ)

S

(

Bi

)

+

φ

p(i+1)

.

Then,p(i)

> φ

S

(

Bi

)

+

φ

p(i+1)

φ

2p(i)

+

φ

p(i+1). Therefore,

(

1

φ

2

)

p(i)

> φ

p(i+1), i.e.p(i)

>

p(i+1).

LetJvbe the first job such thatLv

(σ )

=

Lmax

(σ)

. Now we check all possible cases.

Lemma 2. If Jvis processed on M1, Lmax

(σ )/

Lmax

(π)

2

.

Proof. Assume thatJvbelongs to batchAl, thenJv

=

J(l).

IfJ(l)

=

J(l)andAlis a regular batch, we haveLmax

(σ)

=

(

1

+

φ)(

r

(

Al

)

+

p(l)

)

+

q(l)andLmax

(π)

r

(

Al

)

+

p(l)

+

q(l). Therefore,Lmax

(σ)/

Lmax

(π)

1

+

φ

.

IfJ(l)

=

J(l)andAlis a non-regular batch, we haveLmax

(σ )

=

S

(

Al−1

)

+

p(l−1)

+

p(l)

+

q(l)andLmax

(π)

S

(

Al−1

)

+

p(l)

+

q(l). Therefore,Lmax

(σ)

Lmax

(π)

p(l−1)

Lmax

(π)

.

Thus, ifJ(l)

=

J(l)we can getLmax

(σ )/

Lmax

(π)

2. We continue our proof under the assumptionJ(l)is notJ(l).

Case1. IfAlis regular, thenLmax

(σ )

=

S

(

Al

)

+

p(l)

+

q(l)

=

(

1

+

φ)(

r

(

Al

)

+

p(l)

)

+

q(l). For the optimal value, we have

Lmax

(π)

r

(

Al

)

+

p(l)

+

q(l)

r

(

Al

)

+

(

1

+

φ)

p(l)andLmax

(π)

r

(

Al

)

+

q(l). Thus, 2Lmax

(π)

2r

(

Al

)

+

(

1

+

φ)

p(l)

+

q(l)

Lmax

(σ )

.

Case2. IfAlis not regular, thenLmax

(σ)

=

S

(

Al−1

)

+

p(l−1)

+

p(l)

+

q(l). Note thatq(l)

φ

p(l)andq(l−1)

φ

p(l−1). For the optimal value, we have (1)Lmax

(π)

S

(

Al−1

)

+

p(l)

+

q(l)

φ

p(l−1)

+

(

1

+

φ)

p(l); (2)Lmax

(π)

r

(

Al−1

)

+

p(l−1)

+

q(l−1)

(

1

+

φ)

p(l−1); (3)Lmax

(π)

S

(

Al−1

)

+

q(l). By

(

1

)

×

φ

+

(

2

)

×

φ

2

+

(

3

)

, we have

2Lmax

(π)

S

(

Al−1

)

+

φ

2p(l−1)

+

p(l)

+

φ

p(l−1)

+

q(l)

=

S

(

Al−1

)

+

p(l−1)

+

p(l)

+

q∗(l)

=

Lmax

(σ ).

The proof is complete.

Lemma 3. If Jvis processed on M2, Lmax

(σ )/

Lmax

(π)

2

Proof. Assume thatJvbelongs to batchBl; thenJv=J(∗l). Obviously, whenJ(l)

=

J

(l), we can getLmax

(σ)/

Lmax

(π)

2 by making similar analysis asLemma 2. The following proof is under the assumption thatJ(l)is notJ(l).

Case1.Blis regular.Lmax

(σ )

=

(

1

+

φ)(

r

(

Bl

)

+

p(l)

)

+

q(l). Note thatp

∗ (l)

(

1

+

φ)

q ∗ (l), we have (1)Lmax

(π)

r

(

Bl

)

+

p(l)

+

q∗ (l)

r

(

Bl

)

+

(

2

+

φ)

q(l)and (2)Lmax

(π)

r

(

Bl

)

+

p(l). By

(

1

)

×

φ

2

+

(

2

)

×

(

1

+

φ)

, we have 2Lmax

(π)

2r

(

Bl

)

+

(

1

+

φ)

p(l)

+

q(l)

Lmax

(σ).

Case2.Blis not regular.Lmax

(σ )

=

S

(

Bl−1

)

+

p(l−1)

+

p(l)

+

q(l). Without changing the value obtained by the algorithm and increasing the optimal value, we assume that there is only one jobJ(l−1)

(

r

(

Bl−1

),

p(l−1)

,

0

)

inBl−1. Note thatp(l)

(

1

+

φ)

q(l)

andp(l−1)

>

p(l).

1. J(l)andJ(∗l)are scheduled on the same machine in

π

. If they are scheduled in the same batch, thenLmax

(π)

S

(

Bl−1

)

+

p(l)

+

q(l). Otherwise,Lmax

(π)

S

(

Bl−1

)

+

p(l)

+

p(l)

S

(

Bl−1

)

+

p(l)

+

q(l). Therefore,Lmax

(σ )

Lmax

(π)

p(l−1)

Lmax

(π)

.

2. J(l−1)andJ(l)are scheduled on the same machine in

π

. If they are scheduled in the same batch, thenLmax

(π)

p(l−1)

+

q(l). Otherwise,Lmax

(π)

p(l−1)

+

p(l)

p(l−1)

+

q(l). Therefore,Lmax

(σ)

Lmax

(π)

S

(

Bl−1

)

+

p(l)

Lmax

(π)

.

3. J(l−1) and J(l)are scheduled on the same machine in

π

. If they are scheduled in the same batch, thenLmax

(π)

p(l−1)

+

S

(

Bl−1

)

p(l−1)

+

φ

p(l−1)

p(l−1)

+

φ

p

(

l

)

. Otherwise,Lmax

(π)

p(l−1)

+

p(l)

p(l−1)

+

φ

p(l). Since

p(l)

p(l)

(

1

+

φ)

q∗(l), we can derive thatLmax

(π)

p(l−1)

+

q(l). Therefore,Lmax

(σ)

Lmax

(π)

S

(

Bl−1

)

+

p(l)

Lmax

(π)

.

Hence we get this lemma.

(5)

Proof. ByLemmas 2and3, we know that the competitive ratio of AlgorithmH2is at most 2. Now we give an instance to

show that the bound is tight. At time 0, two jobsJ1

(

1

, φ)

andJ2

(

0

,

1

+

φ)

arrive. They are put in the same batch and started

at time

φ

. In the optimal schedule,J2is processed beforeJ1at time 0.Lmax

(σ )

=

2

(

1

+

φ)

, andLmax

(π)

=

1

+

φ

. Therefore,

the competitive ratio of AlgorithmH2is 2.

Similar to the idea of AlgorithmH2, we get the following algorithm. Ifmis odd, letm

=

2k

+

1. Otherwise, letm

=

2k

+

2,

wherek

1 and is an integer. Note that

m

/

2

⌉ =

k

+

1. Whenm

=

2k

+

1,

α

=

k2+k−1

k2+2k+1. Whenm

=

2k

+

2,

α

=

k +1

k+2.

Algorithm Hm:

(1) When machineMjofM1

,

M2

, . . . ,

Mm/2⌉is idle andA

(

t

)

̸= ∅

, make decision as follows. Ift

(

1

+

δ)

r

(

A

(

t

))

+

δ

p

(

A

(

t

))

, process all jobs inA

(

t

)

attonMj. Otherwise, wait.

(2) When machineMjofMm/2⌉+1

, . . . ,

Mmis idle andB

(

t

)

̸= ∅

, make decision as follows. Ift

(

1

+ ˆ

δ)

r

(

B

(

t

))

+ ˆ

δ

p

(

B

(

t

))

, process all jobs inB

(

t

)

attonMj. Otherwise, wait.

Where

δ

=

1

/

m

/

2

and

δ

ˆ

=

1

/

m

/

2

.

Denote by

σ

and

π

, the schedule produced byHmand an optimal schedule. According to AlgorithmHm, we can get this property about

σ

.

Lemma 4. In

σ

, any batch Bischeduled on machine Mj, j

=

1

, . . . ,

m

/

2

, is a regular batch, i.e. S

(

Bi

)

=

(

1

+

δ)

r

(

Bi

)

+

δ

p

(

Bi

)

. Proof. Note that

m

/

2

⌉ =

k

+

1 whethermis an odd number 2k

+

1 or an even number 2k

+

2. Suppose to the contrary that there exists at least one non-regular batch on machineMj,j

=

1

, . . . ,

k

+

1. LetBvbe the first one, which starts at

t

=

S

(

Bv

) > (

1

+

δ)

r

(

Bv

)

+

δ

p

(

Bv

)

. Thus on every machineMj,j

=

1

, . . . ,

k

+

1, there is one batch processed in

[

r

(

Bv

),

t

]

. Thesek

+

1 batches start beforer

(

Bv

)

and complete not beforet. For ease of description, we denote these batches by

B′ 1

, . . . ,

Bk+1such thatS

(

B ′ 1

) <

S

(

B ′ 2

) <

· · ·

<

S

(

B

k+1

)

. SinceBvis the first non-regular batch, the batchesB

1,

· · ·

,B

k+1are

regular.

SinceBvis not regular, we have min

1≤ik+1

{

S

(

Bi

)

+

p(i)

} ≥

S

(

Bv

) > (

1

+

δ)

r

(

Bv

)

+

δ

p

(

Bv

)

. Byr

(

Bv

) >

S

(

Bk+1

)

, we derive that min 1≤ik+1

{

S

(

Bi

)

+

p ′ (i)

)

}

> (

1

+

δ)

S

(

Bk+1

)

.

We will prove thatS

(

B

k+1

) >

1

1−jδS

(

Bk+1−j

)

,j

=

1

, . . . ,

k. (*) In fact, whenj

=

1, we haveS

(

Bk+1

)

+

p

′ (k+1)

> (

1

+

δ)

S

(

Bk+1

)

, then p(k+1)

> δ

S

(

Bk′+1

) > (

1

+

δ)

·

δ

S

(

Bk

)

+

δ

2p ′ (k+1)

.

We getp′ (k+1)

>

1−δδS

(

Bk

)

. Therefore, S

(

Bk+1

) > (

1

+

δ)

S

(

Bk

)

+

δ

p ′ (k+1)

>

1 1

δ

S

(

Bk

).

Then the inequality (*) is right forj

=

1.

Now letj

2 and the inequality (*) is true for 1

, . . . ,

j

1, thenS

(

Bk+1

) >

1(j11)δS

(

Bk+2j

)

. Thus,

S

(

Bk+2j

)

+

p(k+2j)

> (

1

+

δ)

S

(

Bk+1

) >

1

+

δ

1

(

j

1

S

(

Bk+2−j

).

We have p(k+2j)

>

j

δ

1

(

j

1

S

(

Bk+2−j

) >

j

δ

1

(

j

1

[

(

1

+

δ)

S

(

Bk+1−j

)

+

δ

p ′ (k+2−j)

]

.

This implies that

(

1

+

δ)(

1

j

δ)

1

(

j

1

p ′ (k+2−j)

>

j

δ(

1

+

δ)

1

(

j

1

S

(

Bk+1−j

).

Then,p(k+2j)

>

1jδjδS

(

Bk+1−j

)

. Therefore, S

(

Bk+2−j

) > (

1

+

δ)

S

(

Bk+1−j

)

+

δ

p ′ (k+2−j)

>

1

(

j

1

1

j

δ

S

(

Bk+1−j

).

Reminding ofS

(

Bk+1

) >

1−(j1−1)δS

(

Bk+2−j

)

, we can getS

(

Bk+1

) >

1−1jδS

(

B

k+1−j

)

. Hence the inequality (*) is true. According to the inequality (*) and

δ

=

1

/(

k

+

1

)

, we have

(

1

+

δ)

S

(

Bk+1

)

S

(

B1

)

p(1)

>

1

+

δ

1

k

δ

S

(

B ′ 1

)

S

(

B ′ 1

)

p ′ (1)
(6)

=

(

k

+

1

)

S

(

B1

)

p(1)

(

k

+

1

p(1)

p(1)

=

0

.

Therefore, we get

(

1

+

δ)

S

(

Bk+1

) >

S

(

B1

)

+

p(1), which contradicts

min

1≤ik+1

{

S

(

Bi

)

+

p(i)

}

> (

1

+

δ)

S

(

Bk+1

).

This means thatBvis regular. Therefore, any batchBischeduled on machineMjis a regular batch, forj

=

1

, . . . ,

k

+

1. Note that the proof ofLemma 4has no connection withqj. Hence, by making similar analysis, we can derive that any batch onMj,j

= ⌈

m

/

2

⌉ +

1

, . . . ,

m, is regular. Therefore, we have the following lemma:

Lemma 5. In

σ

which is produced by Algorithm Hm, any batch is regular.

LetJvbe the first job such thatLv

(σ)

=

Lmax

(σ )

. Assuming thatJvbelongs to batchBl, thenJv

=

J(l). We make competitive

analysis by checking all possible cases whereBlis scheduled.

Lemma 6. If Blis scheduled on Mj, j

=

1

, . . . ,

m

/

2

, then Lmax

(σ)

(

1

+

R1

)

Lmax

(π)

, where R1

=

(

1

+

δ)/(

1

+

α)

.

Proof. SinceBlis regular, we haveLmax

(σ )

=

(

1

+

δ)(

r

(

Bl

)

+

p(l)

)

+

q(l). Since

α

1 and

δ

1, we have

αδ

+

δ

1

+

δ

. Then

δ

(

1

+

δ)/(

1

+

α)

=

R1.

For the optimal value, we have (1)Lmax

(π)

r

(

Bl

)

+

p(l)

+

q(l)

r

(

Bl

)

+

(

1

+

α)

p(l)and (2)Lmax

(π)

r

(

Bl

)

+

q(l). Then by

(

1

)

×

R1

+

(

2

)

, we get

(

1

+

R1

)

Lmax

(π)

(

1

+

R1

)

r

(

Bl

)

+

(

1

+

δ)

p(l)

+

q(l)

(

1

+

δ)

r

(

Bl

)

+

(

1

+

δ)

p(l)

+

q(l)

=

Lmax

(σ).

The result holds.

Lemma 7. If Blis scheduled on Mj, j

= ⌈

m

/

2

⌉ +

1

, . . . ,

m, then Lmax

(σ )

(

1

+

R2

)

Lmax

(π)

, where R2

= ˆ

δ

+

α/(

1

+

α)

.

Proof. SinceBlis regular, we haveLmax

(σ )

=

(

1

+ ˆ

δ)(

r

(

Bl

)

+

p(l)

)

+

q(l). For the optimal value, we have (1)Lmax

(π)

r

(

Bl

)

+

p(l)and (2)Lmax

(π)

r

(

Bl

)

+

p(l)

+

q∗(l)

1+ααq∗(l). Then by

(

1

)

×

(

1

+ ˆ

δ)

+

(

2

)

×

α/(

1

+

α)

, we can derive that

(

1

+

R2

)

Lmax

(π)

(

1

+ ˆ

δ)(

r

(

Bl

)

+

p(l)

)

+

q(l)

=

Lmax

(σ ).

The result follows.

Theorem 5. The competitive ratio of Algorithm Hmis

ρ

=

ρ

1

=

1

+

(k +1)(k+2) k(2k+3)

,

if m

=

2k

+

1

ρ

2

=

1

+

(k+2) 2 (k+1)(2k+3)

,

if m

=

2k

+

2,

where k

1and is an integer.

Proof. Whenm

=

2k

+

1,

δ

=

1

/(

k

+

1

)

,

δ

ˆ

=

1

/

k, the competitive ratio of AlgorithmHmis at most

ρ

=

1

+

max

{

R1

,

R2

} =

1

+

max

1 1

+

α

×

k

+

2 k

+

1

,

1 k

+

α

(

1

+

α)

.

When

α

=

k2+k−1

k2+2k+1,

ρ

achieves its minimal value

ρ

1

=

1

+

(

k

+

1

)(

k

+

2

)

k

(

2k

+

3

)

.

Whenm

→ +∞

,

ρ

1

1

.

5.

Whenm

=

2k

+

2,

δ

=

1

/(

k

+

1

)

,

δ

ˆ

=

1

/(

k

+

1

)

, the competitive ratio of AlgorithmHmis at most

ρ

=

1

+

max

{

R1

,

R2

} =

1

+

max

1 1

+

α

×

k

+

2 k

+

1

,

1 k

+

1

+

α

(

1

+

α)

.

When

α

=

k+1

k+2,

ρ

achieves its minimal value

ρ

2

=

1

+

(

k

+

2

)

2

(

k

+

1

)(

2k

+

3

)

.

Whenm

→ +∞

,

ρ

2

1

.

5.
(7)

It is easy to see that

ρ <

2 whenm

4. Whenm

=

3,

ρ

=

2

.

2

>

2. We can easily make a small modification for AlgorithmHmform

=

3 : (1) set

α

=

1; (2) process jobs inA

(

t

)

onM1ift

(

1

+

δ)

r

(

A

(

t

))

+

p

(

A

(

t

))

; (3) process jobs in

B

(

t

)

onM2orM3ift

(

1

+ ˆ

δ)

r

(

B

(

t

))

+ ˆ

δ(

B

(

t

))

, where

δ

=

1 and

δ

ˆ

=

1

/

2. Then, we can get its competitive ratio as 2 by

making a similar proof.

Our results in this section can be described as follows: 1. Whenm

=

2, upper bound is 2.

2. Whenm

=

3, upper bound is 2.

3. Whenm

=

2k

+

1,k

2, upper bound is

ρ

1

=

1

+

(k

+1)(k+2)

k(2k+3) . 4. Whenm

=

2k

+

2,k

1, upper bound is

ρ

2

=

1

+

(k

+2)2

(k+1)(2k+3). Whenmtends to

+∞

, both

ρ

1and

ρ

2tend to 1.5.

Acknowledgements

This research was supported by the grant 09ZR1407200 of the Science Foundation of Shanghai and the National Natural Science Foundation of China (No. 11071072).

References

[1] X. Deng, C.K. Poon, Y. Zhang, Approximation algorithms in batch processing, Jounal of Combinatorial Optimizations 7 (2003) 247–257.

[2] J.A. Hoogeveen, A.P.A. Vestjean, A best possible deterministic online algorithm for minimizing maximum delivery times on a single machine, SIAM Journal on Discrete Mathematics 13 (2000) 56–63.

[3] C.Y. Lee, R. Uzsoy, L.A. MartinVega, Efficient algorithms for scheduling semiconductor burn-in operations, Operations Research 40 (1992) 764–775. [4] P. Liu, X. Lu, Y. Fang, A best possible deterministic online algorithm for minimizing makespan on parallel batch machines. Journal of Scheduling,

doi:10.1007/s10951-009-0154-4. [5] P. Liu, Personal Communication.

[6] J. Tian, R. Fu, J. Yuan, Online scheduling with delivery time on a single batch machine, Theoretical Computer Science 374 (2007) 49–57.

[7] J. Tian, T.C.E. Cheng, C.T. Ng, J. Yuan, Online scheduling on unbounded parallel-batch machines to minimize the makespan, Information Processing Letters 109 (2009) 1211–1215.

[8] A.P.A. vestjens, online machine scheduling. Ph.D. Thesis, Department of mathematics and Computing Science, Eindhoven University of Techology, Eindhoven, The Netherlands, 1997.

[9] G. Zhang, X. Cai, C.K. Wong, online algorithms for minimizing makespan on batch processing machines, Naval Research Logistics 48 (2001) 241–258. [10] G. Zhang, X. Cai, C.K. Wong, Optimal online algorithms for scheduling on parallel batch processing machines, IIE Transactions 35 (2003) 175–181.

Theoretical Computer Science 412 (2011) 5333–5339 ScienceDirect doi:10.1007/s10951-009-0154-4.

References

Related documents

We reviewed a total of 47 prospective studies examining short-term abstinence effects from six different potential behavioral addictions (i.e., exercise, gaming, gambling,

The aim of this research, is the investigation of anti- inflammatory effect of silymarin in the treatment of colon ulcer induced acetic acid in mice Balb/C

Metabolic syndrome (MetS) is considered as a major health problem in recent years and recognized by a cluster of risk factors related to diabetes and amplified risk of

In black-box test context, XSS vulnerabilities arise when a tainted data (re- sulting of a partial copy of an input parameter value, a string of characters) appears in at least

7.4 mOVemenT in Berlin CiTy ZOneS 7.4.1 A German unit may not enter a Berlin city zone from a zone outside the city if the Berlin zone is contested (see glossary) or

However, the patients receiving intaperitoneal chemotherapy had more serve toxicity (grade 3 – 4, fatigue, hematologic, gastrointestinal, metabolic, and neurologic). Only 42 % of

Charlotteville, VA 22908, USA; 26 Department of Gynecology, University Medical Center Hamburg-Eppendorf, Martinistrasse 52, 20246 Hamburg, Germany; 27 Department of Medicine,

Asia In Asia, Safilo is present with Safilo Far East, with headquarters in Hong Kong, which coordinates sales activities in the Asian markets through the following companies: