An improved on-line algorithm for single parallel-batch machine scheduling with delivery times

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Contents lists available atSciVerse ScienceDirect

Discrete Applied Mathematics

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

An improved on-line algorithm for single parallel-batch machine

scheduling with delivery times

Ji Tian

a

, T.C.E. Cheng

b,∗

, C.T. Ng

b

, Jinjiang Yuan

c

a_{School of Sciences, China University of Mining and Technology, Xuzhou, Jiangsu 221116, People’s Republic of China} b_{Department of Logistics and Maritime Studies, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong} c_{Department of Mathematics, Zhengzhou University, Zhengzhou, Henan 450001, People’s Republic of China}

a r t i c l e i n f o

Article history: Received 4 May 2010

Received in revised form 29 November 2011

Accepted 4 December 2011 Available online 29 December 2011 Keywords: On-line scheduling Parallel-batch machine Delivery times Competitive ratio

a b s t r a c t

We study an on-line single parallel-batch machine scheduling problem where each job has a processing time and a delivery time. Jobs arrive over time and the batch capacity is unbounded. Jobs can be processed in a common batch, and each job is delivered independently and immediately at its completion time on the machine. The objective is to minimize the time by which all the jobs are delivered. We provide an on-line algorithm with a competitive ratio 2

√

2−1 ≈ 1.828, which improves on a 2-competitive on-line algorithm for this problem in the literature.

1. Introduction

We state the on-line scheduling problem on a single parallel-batch machine under study in this paper as follows: There arenjobs, say,J1

, . . . ,

Jn, that arrive over time. Each jobJjhas an arrival timerj, a processing timepj, and a delivery timeqjthat are not known in advance. We acquire the information on a jobJj, such aspjandqj, as it arrives. A job cannot be scheduled until its arrival and cannot be preempted once it is processed. A parallel-batch machine can process jobs simultaneously as a batch up to its capacity limit. We assume the batch capacity is unbounded. The processing time of a batch is the largest processing time among the jobs in it. The completion time of a batch is the time at which all the jobs in the batch are completed. Jobs in a batch have the same starting time and the same completion time. In this model, once the processing of a job is completed on the machine, we deliver it to its destination, which takes its delivery time. The objective is to minimize the time by which all the jobs are delivered. Given a schedule, letCjandDjdenote the completion time of processingJjon the machine and the time by whichJjis delivered, respectively. ThenDj

=

Cj

+

qj. Following the standard scheduling notation to describe scheduling problems introduced by Lawler et al. [4], we denote the problem by 1

|

p-batch,on-line,qj

,

b

=

n

|

Dmax,

wherebis the upper bound for the batch capacity andDmax

=

maxj

{

Dj

}

.

For on-line scheduling, the quality of an on-line algorithm is usually measured by its competitive ratio. We say an on-line algorithm is

ρ

-competitive if, for any instance, it produces a schedule with an objective value at most

ρ

times the objective value given by an off-line optimal schedule.

Parallel-batch scheduling is motivated by the burn-in operations in the manufacturing of integrated circuits [5,6,11,12]. There have been many research studies on on-line scheduling to minimize the time by which all the jobs are delivered.

∗_{Corresponding author. Tel.: +852 2766 5216; fax: +852 2364 5245.} E-mail address:[email protected](T.C.E. Cheng).

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Table 1

Summary of results.

Model Constraints on jobs Results References

b=n qj=0

√ 5+1

2 (Optimal) Deng et al. [1]

Zhang et al. [14] Poon and Yu [7] b<n qj=0 2 Zhang et al. [14] Poon and Yu [8] b<n qj=0,pj∈  p, √ 5+1 2 p  √ 5+1

2 (Optimal) Fang et al. [2]

b=2 qj=0 74 Poon and Yu [8]

b<n 3 Tian et al. [9]

b=norb<n pj=p

√ 5+1

2 (Optimal) Tian et al. [9]

b=n 2 Tian et al. [9]

b=n Agreeable(pj,qj)

√ 5+1

2 (Optimal) Yuan et al. [13]

b=n pj≥qj

√ 5+1

2 (Optimal) Yuan et al. [13]

b=n pj≤qj √ 5+1 2 Tian et al. [10] b=n pj∈  p, √ 5+1 2 p  √ 5+1

2 (Optimal) Fang et al. [2]

Hoogeveen and Vestjens [3] propose a best possible on-line algorithm with a competitive ratio √

5+1

2 for minimizing the

maximum delivery time on a single (non-batch) machine. Tian et al. [9] first study this problem on a single parallel-batch machine. If all the jobs have the same delivery time 0, the problem is equivalent to 1

|

p-batch,on-line

|

Cmax, which has been

extensively studied in the literature.Table 1summarizes the known results on on-line scheduling on a single batch machine to minimize the time by which all the jobs are delivered.

This paper is organized as follows: In Section2we introduce some notation used in this paper. In Section3we present an on-line algorithm and some useful observations and lemmas. In Section4we show that the on-line algorithm has a competitive ratio no greater than 2

√

2

−

1

≈

1

.

828 and prove that the bound is tight.

2. Preliminaries

A jobJjis calledavailableat timetif it has arrived but has not been processed at timet. Denote byU

(

t

)

the set of all the available jobs at timet. If two jobsJiandJjinU

(

t

)

satisfypi

≤

pjandqi

≤

qj, we say thatJiisdominatedbyJjat timet and we can always processJiin the same batch asJjbecause the batch capacity is unbounded. Furthermore, if we delete a dominated job at timet, it will not affect our analysis. Therefore we useU∗

(

t

)

to denote the set of non-dominated available jobs inU

(

t

)

and consider only scheduling of the jobs inU∗

(

_t

)

_{at time}_t_.

Observation 1. For any two jobsJ_iandJ_jin the non-dominated setU∗

(

_t

)

_{, we have}_p

i

̸=

pjandqi

̸=

qj, andpi

<

pjiffqi

>

qj. The following notation will be used in our discussion:

•

p

(

t

)

denotes the index of the job with the largest processing time inU∗

(

_t

)

_;

•

q

(

t

)

denotes the index of the job with the largest delivery time inU∗

(

t

)

;

•

rmin

(

J

)

denotes the minimum release time of the jobs inJ;

•

p

(

J

)

denotes the largest processing time of all the jobs in job setJ;

•

U∗ 1

(

t

)

= {

Jj

∈

U∗

(

t

)

:qj

≥

pj

}

;

•

U₂∗

(

t

)

= {

Jj

∈

U∗

(

t

)

:qj

<

pj

}

;

•

α

=

√

2

−

1;

•

A

(

t

)

= {

Jj

∈

U1∗

(

t

)

:qj

≥

(

1

+

α)

pp(t)

}

;

•

C

(

t

)

= {

Jj

∈

U1∗

(

t

)

:

α

pp(t)

≤

qj

< (

1

+

α)

pp(t)

}

;

•

D

(

t

)

= {

Jj

∈

U1∗

(

t

)

:qj

< α

pp(t)

}

;

•

E

(

t

)

= {

Jj

∈

U2∗

(

t

)

:qj

≥

1+₂αpp(t)

}

. Note that 2

α

=

2

/(

2

+

α)

=

(

1

+

α

2

)/(

1

+

α)

=

(

1

+

α)/



1+α 2

+

1



and

α <

1

−

α < (

1

+

α)/

2

< (

2

−

α)/

2

<

2

α

. We also useJ_a(t),J_c(t),J_d(t), andJ_e(t)to denote the job with the largest processing time in job setA

(

t

)

,C

(

t

)

,D

(

t

)

, andE

(

t

)

, respectively.

Observation 2. At any time instantt, we haveU∗

(

_t

)

=

_U∗

1

(

t

)

∪

U ∗ 2

(

t

)

,U ∗ 1

(

t

)

=

A

(

t

)

∪

C

(

t

)

∪

D

(

t

)

, andE

(

t

)

⊆

U ∗ 2

(

t

)

. Furthermore, ifD

(

t

)

̸= ∅

, thenp

(

U∗ 1

(

t

))

=

pd(t)

< α

pp(t).

Proof. The former part follows from the above definitions. Recall thatU∗

(

_t

)

_{is a non-dominated job set and}_D

(

_t

)

⊆

_U∗

1

(

t

)

.

IfD

(

t

)

̸= ∅

, then the definitions ofU₁∗

(

t

)

andD

(

t

)

implyp

(

U₁∗

(

t

))

=

p_d(t)andp_d(t)

≤

q_d(t)

< α

p_p(t).Observation 2

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3. An on-line algorithm

Given an instance, we use

σ

and

π

to denote the schedule produced by an on-line algorithm and an off-line optimal schedule, respectively. LetDmax

(σ )

andDmax

(π)

be the objective values generated by schedules

σ

and

π

, respectively. At

each decision timetof an on-line algorithm, we need to select jobs fromU∗

(

_t

)

_{to form a single batch to process. The rule of} selecting jobs is that the available jobs with larger delivery times always have a higher priority. In order to obtain an on-line algorithm with a competitive ratio less than 2, we observe that the jobs inU∗

(

t

)

should be divided into several groups to be scheduled separately. Consider the instance in which two jobsJ1andJ2arrive at time 0, wherep1

=

ϵ,

q1

=

1,p2

=

1, and q2

=

ϵ

with

ϵ

being a sufficiently small positive value. IfJ1andJ2are processed in a common batch in an on-line schedule, we

haveDmax

(σ)

≥

2 whileDmax

(π)

=

1

+

2

ϵ

. As

ϵ

tends to zero,Dmax

(σ)/

Dmax

(π)

tends to 2. Therefore, we divideU∗

(

t

)

into

two subsetsU∗

1

(

t

)

andU

∗

2

(

t

)

. If all the jobs inU

∗

2

(

t

)

have the same arrival time, it is easy to observe that all these jobs should

be processed in a single batch in any optimal schedule. The difficulty is to deal with the jobs inU∗

1

(

t

)

. If all the jobs inU

∗

1

(

t

)

are always processed in a single batch, we cannot obtain an on-line algorithm with a competitive ratio less than 2. Consider the instance in which two jobsJ1andJ2arrive at time 0, wherep1

=

1K

,

q1

=

K2,p2

=

1,q2

=

0, andKis a sufficiently large

positive value. Before time 1, if jobJ1is scheduled as a single batch, a copy ofJ1, i.e., one with the same processing time and

the same delivery time, will arrive exactly

ϵ

time later than the starting time of this batch. ThenDmax

(σ )

≥

1

+

1, while Dmax

(π)

=

1

+

_K1

+

2_K. AsKtends to

∞

,Dmax

(σ)/

Dmax

(π)

tends to 2.

An intuitive idea to deal with the jobs inU∗

1

(

t

)

is that the jobs with larger delivery times inU

∗

1

(

t

)

should be scheduled

in a single batch and the jobs with smaller delivery times inU∗

1

(

t

)

should be scheduled with all the jobs inU

∗

2

(

t

)

in a single

batch. So we further divideU₁∗

(

t

)

into three subsetsA

(

t

)

,C

(

t

)

, andD

(

t

)

. Specifically, we useA

(

t

)

∪

C

(

t

)

to denote the jobs with larger delivery times andD

(

t

)

to denote the jobs with smaller delivery times inU∗

1

(

t

)

. However, in order to design an

effective on-line algorithm and facilitate the analysis of the competitive ratio in Section4, we need to establish more precise procedures to determine the jobs to form a single batch to process. At each decision timet, we useR

(

t

)

to denote the set of selected jobs to process as a single batch in an on-line algorithm, whereR

(

t

)

is defined by the following sub-procedureH1

(Conditions (a) and (b) inH1are presented afterH1): Sub-procedure H1

Case1.U₁∗

(

t

)

= ∅

. ThenU∗

(

t

)

=

U₂∗

(

t

)

. SetR

(

t

)

=

U₂∗

(

t

)

.

Case2.p

(

U∗

1

(

t

)) < α

pp(t). ThenJp(t)

∈

U2∗

(

t

)

. We consider three subcases: Case2.1.A

(

t

)

̸= ∅

. SetR

(

t

)

=

A

(

t

)

.

Case2.2.A

(

t

)

= ∅

andC

(

t

)

̸= ∅

. Do the following:

If Condition (a) holds, setR

(

t

)

=

C

(

t

)

∪

D

(

t

)

; otherwise, setR

(

t

)

=

C

(

t

)

.

Case2.3.A

(

t

)

= ∅

andC

(

t

)

= ∅

. ThenU∗

1

(

t

)

=

D

(

t

)

. Do the following:

If Condition (a) holds, setR

(

t

)

=

D

(

t

)

; otherwise, setR

(

t

)

=

D

(

t

)

∪

U₂∗

(

t

)

.

Case3.p

(

U∗

1

(

t

))

≥

α

pp(t). ThenD

(

t

)

= ∅

andU

∗

1

(

t

)

=

A

(

t

)

∪

C

(

t

)

.

Case3.1.A

(

t

)

̸= ∅

. Ifp_a(t)

≥

α

p_p(t), setR

(

t

)

=

A

(

t

)

∪

C

(

t

)

; otherwise, setR

(

t

)

=

A

(

t

)

.

Case3.2.A

(

t

)

= ∅

. ThenU₁∗

(

t

)

=

C

(

t

)

. We consider two situations:

Case3.2.1.p_c(t)

≥

1+α

2 pp(t)orE

(

t

)

̸= ∅

. SetR

(

t

)

=

C

(

t

)

∪

U

∗

2

(

t

)

. Case3.2.2.p_c(t)

<

1+₂αp_p(t)andE

(

t

)

= ∅

. Do the following:

If Condition (b) holds, setR

(

t

)

=

C

(

t

)

∪

U∗

2

(

t

)

; otherwise, setR

(

t

)

=

C

(

t

)

. Condition(a):rmin

(

D

(

t

)) < α

pp(t)andt

≤

1+₂αpp(t).

Condition(b):t is the completion time of a previously formed batchBh, the starting time ofBh, say,th, is not less than

1+α

2 pp(t),Jc(t)

∈

C

(

th

)

, andpc(t)

≥

1+α

2 pp(th).

Remark.If Condition (a) holds, thenD

(

t

)

, which includes all the jobs with smaller delivery times inU∗

1

(

t

)

, should not be

delayed at timetsince there exists some job inD

(

t

)

having an earlier arrival time. If Condition (b) holds, then the arrival time of jobJ_p(t)is at least 1+₂αp_p(t)andJ_p(t)cannot be delayed at timet. So we setR

(

t

)

=

C

(

t

)

∪

U₂∗

(

t

)

.

Now we present an on-line algorithmHin the following. LetJr(t)denote the job with the largest processing time inR

(

t

)

. The main idea of algorithmHis as follows: At each decision timet, the jobs inR

(

t

)

determined by sub-procedureH1form a

waiting batch. The starting time of the waiting batch depends on the relationship betweenJ_r(t)andJ_p(t). IfJ_r(t)

̸=

J_p(t), then

U∗

(

_t

)

_{is divided into two parts}_R

(

_t

)

_and_U∗

(

_t

)

\

_R

(

_t

)

_{, and the starting time of}_R

(

_t

)

_{should be as early as possible under some} delay strategy. IfJr(t)

=

Jp(t), by sub-procedureH1(Case 1, Case 2.3, and Case 3.2), we haveR

(

t

)

=

U∗

(

t

)

. In this case, since

all the jobs inU∗

(

t

)

form the single batchR

(

t

)

, the starting time of it should be as late as possible under some delay strategy.

Algorithm H.Step0: Sett

=

0.

Step1: IfU∗

(

t

)

= ∅

, go to Step 4; otherwise, determineR

(

t

)

by sub-procedureH1. Step2: IfJ_r(t)

̸=

J_p(t), do the following:

Ift

≥

(

1

−

α)

pp(t), scheduleR

(

t

)

as a single batch starting at timet. Resett

=

t

+

pr(t). Go to Step 1. Ift

< (

1

−

α)

p_p(t), then we postpone the decision onR

(

t

)

till the momentt′

=

(

₁

−

α)

_p

p(t)or till the first moment

(4)

Step3: IfJr(t)

=

Jp(t), do the following: Ift

≥

1+α

2 pp(t), scheduleR

(

t

)

as a single batch starting at timet. Resett

=

t

+

pr(t). Go to Step 1.

Ift

<

1+₂αp_p(t), then we postpone the decision onR

(

t

)

till the momentt′

=

1+α

2 pp(t)or till the momentt

′′

<

1+α

2 pp(t)if

there is a new job that arrives att′′_{. Reset}_t

=

_t′′_if_t′′_{exists; otherwise, reset}_t

=

_t′_{. Go to Step 1.}

Step4: Wait until a new job arrives and lettbe the release time of such a job. Go to Step 1.

We provide an example to illustrate how algorithmHworks. There are four jobsJ1

,

J2

,

J3, andJ4. The first three jobs arrive

at time 0 and the last one arrives at time1+₂α. The processing times and delivery times of the jobs are given by

p1

=

1

,

p2

=

α

−

α

2

,

p3

=

2

α

2

,

p4

=

α

2

;

q1

=

0

,

q2

=

1

+

α,

q3

=

2

α

2

,

q4

=

α.

According to the above definitions, we haveU∗

1

(

0

)

= {

J2

,

J3

}

,A

(

0

)

= {

J2

}

,D

(

0

)

= {

J3

}

, andU2∗

(

0

)

= {

J1

}

. By sub-procedure H1(Case 2.1),R

(

t

)

=

A

(

t

)

,

∀

t

∈ [

0

,

1

−

α

]

. By Step 2 of algorithmH,A

(

1

−

α)

= {

J2

}

acts as the first batch with starting

time 1

−

α

and completion time 1

−

α

2

=

₂

α

_{. At time 2}

α

_{, we observe that}_U∗

1

(

2

α)

= {

J3

,

J4

}

,C

(

2

α)

= {

J4

}

,D

(

2

α)

= {

J3

}

,

andU∗

2

(

2

α)

= {

J1

}

. By sub-procedureH1(Case 2.2) and Step 2 of algorithmH,R

(

2

α)

=

C

(

2

α)

= {

J4

}

acts as the second

batch with starting time 2

α

and completion time 1. At time 1, we observe thatU₁∗

(

1

)

=

D

(

1

)

= {

J3

}

andU2∗

(

1

)

= {

J1

}

. By

sub-procedureH1(Case 2.3) and Step 3 of algorithmH,R

(

1

)

= {

J1

,

J3

}

acts as the third batch starting at time 1. Hence we

haveDmax

(σ )

=

(

1

−

α)

+

p2

+

p4

+

p1

+

q3

=

(

1

−

α)

+

(α

−

α

2

)

+

α

2

+

1

+

2

α

2

=

2

(

1

+

α

2

)

.

When it does not cause confusion, we writeDonandDoptforDmax

(σ)

andDmax

(π)

, respectively. Given an instance, let

¯

J

be one of the jobs that have the maximum delivery time among the jobs assuming the valueDon. ThenD¯_J

=

Don. Denote by B0the batch containing

¯

Jin

σ

. Let

τ

be the starting time ofB0. Thenq

¯

=

q

(τ)

andJq(τ)

∈

B0. Hence we may assume that

¯

J

=

J_q(τ). Assume that there aremcontiguously processed batches immediately before time

τ

in

σ

, if they exist. Denote the

m

+

1 batches (includingB0) byB0

,

B1

,

B2

, . . . ,

Bmin decreasing order of their completion times. Denote bySithe starting time of batchBiin

σ

, for 0

≤

i

≤

m, thenSm

<

· · ·

<

S1

<

S0

=

τ

. LetJ(i)be the job with the largest processing time

inBi,

∀

i

∈ {

0

, . . . ,

m

}

. We also user(i),p(i), andq(i)to denote the arrival time, processing time, and delivery time of the job J₍i), respectively. Thenp

(

Bi

)

=

p(i). Note thatp(0)

≤

pp(τ). The equality does not always hold becauseB0may not be the last

batch in

σ

. According to algorithmHand sub-procedureH1, we have the following observations:

Observation 3. For each batchBkin

σ

,

(a) ifJ_(k)

̸=

J_p(S_k₎, thenBk

⊆

U1∗

(

Sk

)

andSk

≥

(

1

−

α)

pp(Sk)

> (

1

−

α)

p(k); (b) ifJ₍k)

=

Jp(Sk), thenBk

=

U ∗

(

_S k

)

andSk

≥

1+₂αpp(Sk)

=

1+_α 2 p(k); (c) ifJ_(k)

∈

U∗ 1

(

Sk

)

, thenBk

⊆

U1∗

(

Sk

)

.

Proof. IfJ_(k)

̸=

J_p(S_k₎, by sub-procedureH1(Cases 2 and 3),R

(

Sk

)

⊆

U1∗

(

Sk

)

. According to Step 2 of algorithmH, we have Bk

=

R

(

Sk

)

⊆

U1∗

(

Sk

)

andSk

≥

(

1

−

α)

pp(Sk)

> (

1

−

α)

p(k).

IfJ₍k)

=

Jp(Sk), by sub-procedureH1(Cases 1, 2.3 and 3),R

(

Sk

)

=

U ∗

(

_S

k

)

. By Step 3 of algorithmH, we haveBk

=

R

(

Sk

)

=

U∗

(

Sk

)

andSk

≥

1+₂αpp(Sk)

=

1+α

2 p(k).

IfJ₍k)

∈

U1∗

(

Sk

)

, thenp(k)

≤

q(k). Recall thatU∗

(

Sk

)

is a non-dominated set. From the definition ofJ(k), for each job Jx

∈

Bk

\

J(k), we havepx

<

p(k)andqx

>

q(k), soqx

>

q(k)

≥

p(k)

>

px. HenceBk

⊆

U1∗

(

Sk

)

.Observation 3follows.

Observation 4. Suppose that there exist two batchesBiandBjin

σ

such thatp(j)

≥

p(i), where 0

≤

i

<

j

≤

m. Then rmin

(

Bi

) >

Sj.

Proof. Sincep_(j)

≥

p_(i)andJ_(i)has the largest processing time inBi, we havep(j)

≥

p′

,

∀

J′

∈

Bi. Supposermin

(

Bi

)

≤

Sj. Let Jxbe a job with the minimum arrival time inBi. Thenrx

≤

Sj,Jx

∈

U∗

(

Sj

)

andpx

≤

p(j). By sub-procedureH1and algorithm H, either jobJxis processed before jobJ(j)orJxandJ(j)are processed in the same batch in

σ

. Either outcome contradicts the assumption thatJx

∈

Bi,J(j)

∈

Bj, andi

<

j.Observation 4follows.

Observation 5. For any two consecutive batchesBiandBi+1in

σ

, ifA

(

Si

)

̸= ∅

, thenA

(

Si

)

⊆

Bi

⊆

U1∗

(

Si

)

andrmin

(

A

(

Si

)) >

Si+1.

Proof. The conclusionA

(

Si

)

⊆

Bi

⊆

U1∗

(

Si

)

follows from sub-procedureH1(Cases 2.1 and 3.1). Supposermin

(

A

(

Si

))

≤

Si+1.

LetJx be a job with the minimum arrival timerxinA

(

Si

)

. Thenrx

≤

Si+1, soJx

∈

U∗

(

Si+1

)

. GivenJx

̸∈

Bi+1, we have J_(i+1)

̸=

Jp(Si+1). SinceBiandBi+1are two batches processed contiguously in

σ

, we haveJp(Si+1)

∈

U

(

Si

)

, sopp(Si+1)

≤

pp(Si). GivenJx

∈

A

(

Si

)

, we obtainqx

≥

(

1

+

α)

pp(Si)

≥

(

1

+

α)

pp(Si+1), soJx

∈

A

(

Si+1

)

. ThenJx

∈

Bi+1becauseA

(

Si+1

)

⊆

Bi+1, contradictingJx

∈

Bi.Observation 5follows.

Observation 6. Suppose there is some jobJx

∈

U∗

(

Si

)

withrx

≤

Sj, where 0

≤

i

<

j

≤

m. Then, for anyksatisfyingi

<

k

≤

j, we haveBk

⊆

U1∗

(

Sk

)

,p(k)

<

px

≤

pp(Sk)

≤

pp(Si), andq(k)

>

qx. Furthermore, ifJx

∈

C

(

Si

)

, thenBk

=

A

(

Sk

)

withp(k)

< α

pp(Sk) andJx

∈

C

(

Sk

)

.

Proof. SinceJx

∈

U∗

(

Si

)

andrx

≤

Sj, we haveJx

∈

U∗

(

Sk

)

. GivenJx

̸∈

Bk, we haveJ(k)

̸=

Jp(Sk), soJp(Sk)

∈

U

(

Sk−1

)

. Then

px

≤

pp(Sk)

≤

pp(Sk−1)

≤ · · · ≤

pp(Si). ByObservation 3(a), we haveBk

⊆

U ∗

1

(

Sk

)

. Recall that the available jobs with larger delivery times always have a higher priority to form a batch. Thenq_(k)

>

qxandp(k)

<

pxbecauseU∗

(

Sk

)

is a non-dominated set.

(5)

Furthermore, ifJx

∈

C

(

Si

)

, thenqx

≥

α

pp(Si)

≥

α

pp(Sk), soJx

∈

A

(

Sk

)

∪

C

(

Sk

)

. SinceJx

̸∈

Bkandrx

≤

Sj, byObservation 5, we haveJx

∈

C

(

Sk

)

. By sub-procedureH1 (Cases 2.1 and 3.1), we getBk

=

A

(

Sk

)

andp(k)

< α

pp(Sk).Observation 6 follows.

Observation 7. Suppose thatJ_(i)

=

J_p(S_i₎andi

>

0. Then, for anyJx

∈



i−1

k=0U

(

Sk

)

, we haverx

>

Si

≥

1+₂αp(i).

Proof. SinceJ_(i)

=

J_p(S_i₎, byObservation 3(b), we haveBi

=

U∗

(

Si

)

andSi

≥

1+₂αp(i). Hence eachJx

∈



i−1

k=0U

(

Sk

)

arrives afterSi.Observation 7follows.

Observation 8. Suppose 0

≤

k

≤

mandJx

∈

U∗

(

Sk

)

such thatrx

<

SkandSk

>

1+2αpp(Sk). Thenm

≥

k

+

1.

Proof. Supposem

=

k. Then there exists an idle time interval immediately before timeSk. By Steps 2 and 3 of algorithmH, the starting time ofJxin

σ

is no later than max

{

rx

,

1+₂αpp(Sk)

}

<

Sk. This contradicts the assumptionJx

∈

U

∗

(

_S

k

)

.Observation 8 follows.

Observation 9. SupposeBk

⊆

U1∗

(

Sk

)

andSk

>

1+2αpp(Sk)with 0

≤

k

≤

m. Then, for anyJx

∈

Bk, we haveJx

̸∈

D

(

Sk

)

,

qx

≥

α

pp(Sk), andBk

⊆

A

(

Sk

)

∪

C

(

Sk

)

.

Proof. The result follows from sub-procedureH1(Case 2.2).

To facilitate the analysis of the competitive ratio, we introduce the following useful notation:

•

S_iπdenotes the starting time ofJiin

π

for each jobJi;

•

Ji

≺

Jjdenotes the event that jobJiis scheduled before jobJjin

π

, i.e.,Siπ

<

Sπj;

•

L

(S)

denotes the lower bound onDoptdetermined by the jobs inSwith the assumption that all the jobs inSarrive at

time 0. Specifically,L

(

Ji

)

≥

pi

+

qifor each jobJi.

Observation 10.For any two jobsJiandJj, we haveL

(

Ji

,

Jj

)

≥

pj

+

min

{

pi

,

qi

}

.

Proof. In any optimal schedule, ifJi andJjare scheduled in the same batch, thenL

(

Ji

,

Jj

)

≥

pj

+

qi; ifJi andJjbelong to different batches, thenL

(

Ji

,

Jj

)

≥

pj

+

pi. HenceL

(

Ji

,

Jj

)

≥

min

{

pj

+

pi

,

pj

+

qi

} =

pj

+

min

{

pi

,

qi

}

.Observation 10 follows.

Observation 11.SupposeBk

=

C

(

Sk

)

∪

U2∗

(

Sk

)

with eitherpc(Sk)

>

1+α

2 pp(Sk)orE

(

Sk

)

̸= ∅

. Then there exists a jobJx

∈

Bk such that min

{

px

,

qx

} ≥

1+₂αpp(Sk)andL

(

Jx

,

Jp(Sk)

)

≥

(

1

+

1+α

2

)

pp(Sk).

Proof. Ifpc(Sk)

>

1+_α

2 pp(Sk), setJx

=

Jc(Sk); ifE

(

Sk

)

̸= ∅

, setJx

=

Je(Sk). By the definitions ofJc(Sk)andJe(Sk), we have min

{

px

,

qx

} ≥

1+₂αpp(Sk). ByObservation 10, we haveL

(

Jx

,

Jp(Sk)

)

≥

pp(Sk)

+

min

{

px

,

qx

} ≥

(

1

+

1+_α

2

)

pp(Sk).Observation 11 follows.

Observation 12.For eachkwith 0

≤

k

≤

m,Dopt

≥

L

(

Ja(Sk)

,

J(k)

,

Jp(Sk)

)

≥

p(k)

+

(

1

+

α)

pp(Sk)if one of the following two conditions holds:

(a)1+₂αp_p(S_k₎

≤

p_(k)

<

p_p(S_k₎; (b)A

(

Sk

)

̸= ∅

.

Proof. We assert that condition (a) implies that condition (b). Suppose 1+₂αp_p(S_k₎

≤

p_(k)

<

p_p(S_k₎andA

(

Sk

)

= ∅

. Then J_(k)

̸=

J_p(S_k₎. ByObservation 3(a), we haveBk

⊆

U1∗

(

Sk

)

. Sincep(k)

≥

1+₂αpp(Sk), by sub-procedureH1(Case 3.2.1), we have

Bk

=

C

(

Sk

)

∪

U2∗

(

Sk

)

. Sop(k)

=

pp(Sk), a contradiction. The assertion follows.

SupposeA

(

Sk

)

̸= ∅

. By sub-procedureH1(Cases 2.1 and 3.1), eitherBk

=

A

(

Sk

)

orBk

=

A

(

Sk

)

∪

C

(

Sk

)

withpa(Sk)

≥

α

pp(Sk). IfBk

=

A

(

Sk

)

, we haveJ(k)

=

Ja(Sk), soL

(

Ja(Sk)

,

J(k)

,

Jp(Sk)

)

≥

L

(

Ja(Sk)

)

≥

pa(Sk)

+

qa(Sk)

≥

p(k)

+

(

1

+

α)

pp(Sk). Suppose

Bk

=

A

(

Sk

)

∪

C

(

Sk

)

withpa(Sk)

≥

α

pp(Sk). WhenJa(Sk)is not scheduled before jobsJ(k)andJp(Sk)in

π

,L

(

Ja(Sk)

,

J(k)

,

Jp(Sk)

)

≥

min

{

p₍k)

,

pp(Sk)

} +

qa(Sk)

≥

p(k)

+

(

1

+

α)

pp(Sk). WhenJa(Sk)is scheduled before both jobsJ(k)andJp(Sk)in

π

, byObservation 10,

L

(

J_a(S_k₎

,

J_(k)

,

J_p(S_k₎

)

≥

p_a(S_k₎

+

L

(

J_(k)

,

J_p(S_k₎

)

≥

p_a(S_k₎

+

p_(k)

+

p_p(S_k₎

≥

p_(k)

+

(

1

+

α)

p_p(S_k₎.Observation 12follows. In the following we provide some lemmas that will be repeatedly used in the analysis to be presented in Section4.

Lemma 3.1. Don

/

Dopt

≤

1

+

2

α

if any of the following conditions holds for some X

,

Y

,

Z

>

0:

(i) X

≤

α

Y , Don

−

Dopt

≤

X

+

Y , and Dopt

≥

(

1

+

1+₂α

)

Y ;

(ii) X

≤

1+α

2 Y , Don

−

Dopt

≤

X

+

1+α

2 Y , and Dopt

≥

X

+

Y ;

(iii) X

≤

Y , Don

−

Dopt

≤

X

+

Y , and Dopt

≥

X

+

(

1

+

α)

Y ;

(iv) X

≤

Y , Don

−

Dopt

≤

X

+

β

Y , and Dopt

≥

X

+

Y , where

β

∈ {

α,

1

−

α

}

;

(v) X

≤

Y , Don

−

Dopt

≤

X

+

1+₂αY , and Dopt

≥

α

X

+

(

1

+

1+₂α

)

Y ;

(vi) X

≤

Y , Z

≤

Y , Don

−

Dopt

≤

X

+

Z , and Dopt

≥

α

X

+

Y

+

Z .

Proof. The results can be easily verified by the fact that 2

α

=

(

1

+

α)/



1+α

2

+

1



=

2

/(

2

+

α)

and

(

1

+

α)/

2

< (

2

−

α)/

2

<

(

2

+

α)/

3

<

2

α

.

(6)

In the remaining part of this section, fromLemmas 3.2to3.5, we suppose thatm

≥

1 and there exists at least one jobJ_(k)

∈ {

J₍1)

,

J(2)

, . . . ,

J(m)

}

satisfyingJ(k)

=

Jp(Sk). LetJ(i)be the job among such jobs with the minimum index. From Observation 7, for anyJ′

∈



i−1

k=0Bk

∪ {

Jp(τ)

}

, we haver′

>

Si

≥

1+₂αp(i).

Lemma 3.2. Suppose m

≥

i

+

1and r_(i)

<

α₂p_(i). Then r_(i)

<

Sm

=

(

1

−

α)

p(i).

Proof. We establish the result by contradiction. Supposer_(i)

≥

Sm. LetBkbe the processing batch whenJ(i)arrives. Then r_(i)

∈ [

Sk

,

Ck

)

. FromObservation 3(a, b),Sk

≥

(

1

−

α)

p(k). Then

(

1

−

α)

p(k)

≤

Sk

≤

r(i)

<

α₂p(i), sop(k)

<

1+₄αp(i). Then Sk−1

=

Sk

+

p(k)

≤

r(i)

+

p(k)

<

3α₄+1p(i)

< (

1

−

α)

p(i). SinceSi

≥

1+₂αp(i), we havei

<

k

−

1

≤

m. FromObservation 6and the assumption thatJ_(i)

=

J_p(S_i₎, for anyjwithi

<

j

<

k, we haveB_j

⊆

U∗

1

(

Sj

)

andJp(Sj)

=

J(i). ByObservation 3(a), we have

Sk−1

≥

(

1

−

α)

pp(Sk−1)

=

(

1

−

α)

p(i), a contradiction. Hencer(i)

<

Sm. Then the definition ofJ(i)impliesp(i)

=

pp(Sm). Given

m

≥

i

+

1, we obtainSm

=

(

1

−

α)

p(i)from Step 2 of algorithmH.Lemma 3.2follows.

Lemma 3.3. Suppose m

≥

i

+

1, Sπ₍_i₎

<

α₂p_(i)and Don

−

Dopt

≤

min

{

(

1

+

α

2

)

p(i)

,

Si

+

α

2p(i)

}

. Then Don

/

Dopt

≤

1

+

2

α

.

Proof. Sincem

≥

i

+

1 andr_(i)

≤

S_(i)π

<

α₂p_(i), byLemma 3.2, we haver_(i)

≤

S_(i)π

<

Sm

=

(

1

−

α)

p(i). From the assumption thatJ_(i)

=

J_p(S_i₎andObservation 6, for anyjsatisfyingi

<

j

≤

m, we haveBj

⊆

U1∗

(

Sj

)

andJp(Sj)

=

J(i). If we can show that

Dopt

≥

(

1

+

α)

p(i), given the fact that

(

1

+

α

2

)/(

1

+

α)

=

2

α

and the assumption thatDon

−

Dopt

≤

(

1

+

α

2

)

p(i), we are

done. We consider three cases as follows:

Case1. There exists somekwithi

<

k

≤

msuch thatA

(

Sk

)

̸= ∅

. ThenDopt

≥

L

(

Ja(Sk)

)

≥

pa(Sk)

+

qa(Sk)

≥

(

1

+

α)

pp(Sk)

=

(

1

+

α)

p_(i).

Case2. There exists somekwithi

<

k

≤

msuch thatp_(k)

≥

α

p_(i). SinceBk

⊆

U1∗

(

Sk

)

, we haveDopt

≥

L

(

J(k)

,

J(i)

)

≥

p₍i)

+

p(k)

≥

(

1

+

α)

p(i)byObservation 10.

Case3. Eachjwithi

<

j

≤

msatisfiesA

(

Sj

)

= ∅

andp(j)

< α

p(i). GivenBj

⊆

U1∗

(

Sj

)

, by sub-procedureH1(Cases 2.2, 2.3 and

3.2), we have eitherBj

=

C

(

Sj

)

orBj

=

C

(

Sj

)

∪

D

(

Sj

)

. There are two situations:

First,m

≥

i

+

2. We assert thatBi+1

=

C

(

Si+1

)

. SupposeBi+1

=

C

(

Si+1

)

∪

D

(

Si+1

)

. By sub-procedureH1(Case 2.2),

there exists at least one jobJx

∈

D

(

Si+1

)

with arrival timerx

< α

p(i). Since

α

p(i)

< (

1

−

α)

p(i)

=

Sm

<

1+₂αp(i), by sub-procedureH1(Case 2.2), we haveJx

∈

D

(

Sm

)

⊆

Bm. This contradictsJx

∈

Bi+1. HenceBi+1

=

C

(

Si+1

)

. Furthermore, we

assert thatr₍i+1)

>

Si+2. Supposer(i+1)

≤

Si+2. Recall thatBi+2

=

C

(

Si+2

)

orBi+2

=

C

(

Si+2

)

∪

D

(

Si+2

)

. SinceJp(Sj)

=

J(i),

∀

j

∈ {

i

, . . . ,

m

}

, we haveJ_(i+1)

∈

C

(

Si+2

)

⊆

Bi+2, a contradiction. Hencer(i+1)

>

Si+2

≥

Sm

>

S_(i)π. ThenJ(i)

≺

J(i+1), so Dopt

≥

p(i)

+

p(i+1)

+

q(i+1)

≥

(

1

+

α)

p(i).

Second,m

=

i

+

1. ThenSi+1

=

Sm

=

(

1

−

α)

p(i). FromObservation 10, we haveDopt

≥

L

(

J(i)

,

J(i+1)

)

≥

p(i)

+

p(i+1).

SinceDon

−

Dopt

≤

Si

+

α

2p(i)

=

Si+1

+

p(i+1)

+

α

2p(i)andp(i+1)

< α

p(i), we have

(

Don

−

Dopt

)/

Dopt

≤

(

Si+1

+

p(i+1)

+

α

2_p

(i)

)/(

p_(i)

+

p_(i+1)

)

≤

(

1

−

α

+

α

+

α

2

)/(

1

+

α)

=

2

α

.Lemma 3.3follows.

Lemma 3.4. Don

/

Dopt

≤

1

+

2

α

if Don

−

Dopt

≤

p(i)

+

z and any one of the following two conditions holds:

(i) z

≤

α

pp(τ)and S(i)π

≥

Si;

(ii) z

≤

p_p(τ)and Dopt

≥

Si

+

max

{

p(i)

,

pp(τ)

} +

z.

Proof. Recall that

τ

=

S0. By Observation 7, we haverp(τ)

>

Si

≥

1+₂αp(i). If condition (i) occurs, we haveDopt

≥

Si

+

max

{

L

(

J(i)

),

L

(

Jp(τ)

)

} ≥

Si

+

max

{

p(i)

,

pp(τ)

}

. Hence

Don

−

Dopt Dopt

≤

p(i)

+

z Si

+

max

{

p(i)

,

pp(τ)

}

≤

(

1

+

α)

max

{

p(i)

,

pp(τ)

}



₁+α 2

+

1



max

{

p_(i)

,

p_p(τ)

}

=

2

α.

If condition (ii) occurs, then

Don

−

Dopt Dopt

≤

p(i)

+

z Si

+

max

{

p(i)

,

pp(τ)

} +

z

≤



2 max

{

p(i)

,

pp(τ)

}

1+α 2

+

2



max

{

p_(i)

,

p_p(τ)

}

<

2 2

+

α

=

2

α,

as

α < (

1

+

α)/

2.Lemma 3.4follows.

Lemma 3.5. Don

/

Dopt

≤

1

+

2

α

if S₍πi)

<

Si, Don

−

Dopt

≤

min

{

p(i)

+

z

,

Si

+

z

}

, and one of the following three conditions holds. (i) z

≤

α

p_p(τ)and Dopt

≥

S₍πi)

+

p(i)

+

pp(τ);

(ii) z

≤

1+α

2 pp(τ)and Dopt

≥

S

π

(i)

+

p(i)

+

p_p(τ)

+

z;

(7)

Proof. We distinguish the following three conditions:

Case1. Condition (i) holds. IfSi

≤

1+2αp(i), thenDon

−

Dopt

≤

Si

+

z

≤

1+2αp(i)

+

α

pp(τ)

≤

1+α

2

(

p(i)

+

pp(τ)

)

≤

1+α

2 Dopt. Suppose

thatSi

>

1+₂αp(i). Givenri

≤

S_(i)π

<

Siand byObservation 8, we getm

≥

i

+

1. We consider two situations: First,S_(i)π

≥

α₂p(i). Then

(

Don

−

Dopt

)/

Dopt

≤

(

p(i)

+

z

)/(

S₍πi)

+

p(i)

+

pp(τ)

)

≤

(

p(i)

+

α

pp(τ)

)/((

1

+

α2

)

p(i)

+

pp(τ)

)

≤

max

{

1

/(

1

+

α

2

), α

} =

2

α

.

Second,S_(i)π

<

α₂p_(i). Whenp_p(τ)

≥

α

max

{

p_(i)

,

p_p(τ)

}

, we have

Don

−

Dopt Dopt

≤

p(i)

+

z S_(i)π

+

p_(i)

+

p_p(τ)

≤

max

{

p₍i)

,

pp(τ)

} +

α

pp(τ) max

{

p_(i)

,

p_p(τ)

} +

p_p(τ)

≤



1

+

α

2



max

{

p_(i)

,

p_p(τ)

}

(

1

+

α)

max

{

p_(i)

,

p_p(τ)

}

=

2

α.

The second last inequality follows from the fact that the functionf

(

x

)

=

1+_αx

1+x decreases monotonically in

[

α,

1

]

. When p_p(τ)

< α

max

{

p_(i)

,

p_p(τ)

}

, we havep_p(τ)

< α

p_(i). SinceDon

−

Dopt

≤

min

{

p(i)

+

z

,

Si

+

z

}

andz

≤

α

pp(τ), we have Don

−

Dopt

≤

min

{

(

1

+

α

2

)

p(i)

,

Si

+

α

2p(i)

}

. ByLemma 3.3, we are done.

Case2. Condition (ii) holds. IfSi

≤

1+₂αp(i), then Don

−

Dopt

≤

Si

+

z

≤

1+₂α

(

p(i)

+

pp(τ)

)

≤

1+₂αDopt

<

2

α

Dopt.

IfSi

>

1+₂αp(i), from the fact thatri

≤

S_(i)π

<

Si and byObservation 8,m

≥

i

+

1. There are two possibilities: First, S_(i)π

≥

α

2p(i). Then

(

Don

−

Dopt

)/

Dopt

≤

(

p(i)

+

z

)/(

S

π

(i)

+

p(i)

+

p_p(τ)

+

z

)

≤

(

p_(i)

+

1+α

2 pp(τ)

)/((

1

+

α 2

)

p(i)

+

(

1

+

1+α 2

)

pp(τ)

)

≤

max

{

1

/(

1

+

α 2

),

1+α 2

/(

1

+

1+α 2

)

} =

2

α

. Second,S π

(i)

<

α2p(i). Whenpp(τ)

≥

(α

−

α

2

)

_max

{

_p (i)

,

p_p(τ)

}

, we have Don

−

Dopt Dopt

≤

p(i)

+

z S_(i)π

+

p_(i)

+

p_p(τ)

+

z

≤

max

{

p_(i)

,

p_p(τ)

} +

1+α 2 pp(τ) max

{

p_(i)

,

p_p(τ)

} +



1

+

1+α 2



p_p(τ)

≤



1

+

(α

−

α

2

)

1+α 2



max

{

p_(i)

,

p_p(τ)

}



1

+

(α

−

α

2

)



1

+

1+α 2



max

{

p₍i)

,

pp(τ)

}

=

1

+

α

2 1

+

α

=

2

α.

The last inequality follows from the fact that the functionf

(

x

)

=

1+

1+α 2 x 1+(1+1+₂α)xdecreases monotonically in

[

α

−

α

2

,

₁

]

_{. When} p_p(τ)

< (α

−

α

2

)

_max

{

_p (i)

,

p_p(τ)

}

, we getp_p(τ)

< (α

−

α

2

)

_p

(i). GivenDon

−

Dopt

≤

min

{

p(i)

+

z

,

Si

+

z

}

andz

≤

1+₂αpp(τ), we haveDon

−

Dopt

≤

min

{

(

1

+

α

2

)

p(i)

,

Si

+

α

2p(i)

}

. Then we are done byLemma 3.3.

Case3. Condition (iii) holds. IfSi

≤

1+₂αp(i), we get

(

Don

−

Dopt

)/

Dopt

≤

(

Si

+

z

)/(

S_(i)π

+

p(i)

+

(

1

+

α)

pp(τ)

+

z

)

≤

(

1+α

2 p(i)

+

pp(τ)

)/(

p(i)

+

(

2

+

α)

pp(τ)

)

≤

max

{

(

1

+

α)/

2

,

1

/(

2

+

α)

} =

1+α

2

<

2

α

. Assume thatSi

>

1+α

2 p(i). The assumption

thatS_(i)π

<

S_iandObservation 8implym

≥

i

+

1. We consider two situations: First,Sπ_(i)

≥

α

2p(i). Then

(

Don

−

Dopt

)/

Dopt

≤

(

p_(i)

+

z

)/(

S_(i)π

+

p_(i)

+

(

1

+

α)

p_p(τ)

+

z

)

≤

(

p_(i)

+

p_p(τ)

)/((

1

+

α

2

)

p(i)

+

(

2

+

α)

pp(τ)

)

≤

max

{

1

/(

1

+

α

2

),

1

/(

2

+

α)

} =

2

α

.

Second,S_(i)π

<

α₂p_(i). Whenp_p(τ)

≥

α

2max

{

p_(i)

,

p_p(τ)

}

, we have

Don

−

Dopt Dopt

≤

p(i)

+

z S_(i)π

+

p_(i)

+

(

1

+

α)

p_p(τ)

+

z

≤

max

{

p_(i)

,

p_p(τ)

} +

p_p(τ) max

{

p_(i)

,

p_p(τ)

} +

(

2

+

α)

p_p(τ)

≤

(

1

+

α

2

)

_max

{

_p (i)

,

p_p(τ)

}

(

1

+

α)

max

{

p_(i)

,

p_p(τ)

}

=

2

α.

The second last inequality follows from the fact that the functionf

(

x

)

=

1+x

1+(2+α)xdecreases monotonically in

[

α

2

,

₁

]

_{. When} p_p(τ)

< α

2_max

{

_p

(i)

,

p_p(τ)

}

, we havep_p(τ)

< α

2_p

(i). SinceDon

−

Dopt

≤

min

{

p(i)

+

z

,

Si

+

z

}

andz

≤

pp(τ), we have Don

−

Dopt

≤

min

{

(

1

+

α

2

)

p(i)

,

Si

+

α

2p(i)

}

. ByLemma 3.3, we are done.Lemma 3.5follows.

4. Analysis of algorithmH

In this section we show the main result of this paper in the following theorem.

Theorem 4.1. The on-line algorithm H has a competitive ratio no greater than1

+

2

α

, where

α

=

√

2

−

1, and the bound is tight.

We construct an instance to show that the bound is tight as follows: There are four jobsJ1

,

J2

,

J3, andJ4. JobsJ1andJ2

arrive at time 0, andJ3andJ4arrive at time 1

+

ϵ

. The processing times and delivery times of the jobs are given by p1

=

1

,

p2

=

α,

p3

=

α,

p4

=

ϵ

;

q1

=

0

,

q2

=

α,

q3

=

0

,

q4

=

α

2

−

ϵ.

The on-line algorithmHgenerates a schedule

σ

: The first batch

{

J2

}

starts at time 1

−

α

, the second batch

{

J1

}

starts at time

1, and the third batch

{

J3

,

J4

}

starts at time 2. There exists an optimal schedule

π

in which the first batch

{

J1

,

J2

}

starts at

time 0, the second batch

{

J4

}

starts at time 1

+

ϵ

, and the third batch

{

J3

}

starts at time 1

+

2

ϵ

. The schedules

σ

and

π

are

demonstrated inFig. 1.

Then we haveDon

=

1

−

α

+

α

+

1

+

α

+

α

2

−

ϵ

=

2

+

α

+

α

2

−

ϵ

andDopt

=

1

+

2

ϵ

+

α

. As

ϵ

tends to zero, the ratio Don

/

Dopttends to

(

2

+

α

+

α

2

)/(

1

+

α)

=

1

+

2

α

.

(8)

Fig. 1.

We prove that algorithmHhas a competitive ratio no greater than 1

+

2

α

. The proof consists of considering all the possible situations. There are two ‘‘global’’ situations: (1)m

=

0 and (2)m

>

0, wheremis the number of contiguously processed batches immediately before time

τ

in

σ

. In turn, in situation (2), the following subcases are distinguished: (2a) J₍1)

=

Jp(S1);

(2b)J₍0)

∈

U1∗

(τ)

;

(2c) B0

=

C

(τ)

∪

U2∗

(τ)

withC

(τ)

̸= ∅

;

(2d)B0

=

D

(τ)

∪

U2∗

(τ)

withD

(τ)

̸= ∅

orB0

=

U2∗

(τ)

withqq(τ)

< α

pp(τ);

(2e) B0

=

U2∗

(τ)

withqq(τ)

≥

α

pp(τ).

We mention that (2b)–(2e) are under the situationJ₍1)

̸=

Jp(S1)and (2c)–(2e) are also under the situationJ(0)

∈

U2∗

(τ)

. By the

definition ofR

(τ)

in sub-procedureH1(Cases 1–3), we haveA

(τ)

= ∅

andB0

=

U1∗

(τ)

∪

U

∗

2

(τ)

in subcases (2c)–(2e). There

are three possibilities:B0

=

C

(τ)

∪

U2∗

(τ)

withC

(τ)

̸= ∅

,B0

=

D

(τ)

∪

U2∗

(τ)

withD

(τ)

̸= ∅

, andB0

=

U2∗

(τ)

. Then we

know that (2c)–(2e) cover all the possible situations underJ₍0)

∈

U2∗

(τ)

and (2b)–(2e) cover all the possible situations under J₍1)

̸=

Jp(S1). Therefore (2a)–(2e) cover all the possible situations underm

>

0. We consider all the above listed situations

withinTheorems 4.2–4.7and show thatDon

/

Dopt

≤

1

+

2

α

for each of them.

Note that batchesB0

,

B1

, . . . ,

Bmare processed contiguously in

σ

andSm

<

Sm−1

<

· · ·

<

S1

<

S0

=

τ

. Then Don

=

Sl

+

l



j=0

p_(j)

+

q_q(τ)

,

for any l

,

0

≤

l

≤

m

.

(1)

Theorem 4.2. Suppose that m

=

0. Then Don

/

Dopt

≤

1

+

2

α

.

Proof. Sincem

=

0, there exists an idle time interval immediately before time

τ

. If

τ >

1+₂αp_p(τ), givenm

=

0 and

Observation 8, we havermin

(

B0

∪

Jp(τ)

)

=

τ

. ThenDopt

≥

τ

+

max

{

L

(

Jq(τ)

),

L

(

Jp(τ)

)

} ≥

max

{

τ

+

qq(τ)

, (

3+₂α

)

pp(τ)

}

. So Don

−

Dopt

≤

p(0)

≤

pp(τ)

≤

3+2αDopt

<

2

α

Doptby(1). In what follows, we suppose that

τ

≤

1+_α

2 pp(τ). Case1. J₍0)

∈

U1∗

(τ)

. SinceDopt

≥

qq(τ), by(1), we have

Don

−

Dopt

≤

τ

+

p(0)

≤

1

+

α

2 pp(τ)

+

p(0)

.

(2)

There are three possibilities to consider:

First,p₍0)

<

1+2αpp(τ). ByObservation 10,Dopt

≥

L

(

J(0)

,

Jp(τ)

)

≥

p(0)

+

pp(τ). From(2)andLemma 3.1(ii), we are done.

Second, 1+₂αp_p(τ)

≤

p₍0)

<

pp(τ) orA

(τ)

̸= ∅

. ByObservation 12(a, b),Dopt

≥

p(0)

+

(

1

+

α)

pp(τ). From(2)and

Lemma 3.1(iii), we are done.

Third,p₍0)

=

pp(τ)andA

(τ)

= ∅

. SinceJ(0)

∈

U1∗

(τ)

, byObservation 3(c), we haveB0

⊆

U1∗

(τ)

andU

∗

2

(τ)

= ∅

. By

sub-procedureH1(Case 3.2.1), we haveB0

=

C

(τ)

andqq(τ)

< (

1

+

α)

pp(τ). SinceDopt

≥

L

(

J(0)

)

≥

2p(0)

=

2pp(τ), by(1), we

haveDon

−

Dopt

≤

τ

+

qq(τ)

−

pp(τ)

< (

1+₂α

+

α)

pp(τ)

<

3

α

pp(τ)

<

2

α

Dopt.

Case2. J₍0)

∈

U2∗

(τ)

. ThenJ(0)

=

Jp(τ). Recall that

τ

≤

1+₂αpp(τ). ByObservation 3(b),

τ

=

1+₂αpp(τ). We consider two

subcases:

Case2.1.U∗

1

(τ)

= ∅

. By sub-procedureH1(Case 1), we haveB0

=

U2∗

(τ)

, soJq(τ)

∈

U

∗

2

(τ)

. FromObservation 10, we have Dopt

≥

L

(

Jq(τ)

,

Jp(τ)

)

≥

pp(τ)

+

qq(τ). By(1),Don

−

Dopt

≤

τ

=

1+₂αpp(τ)

≤

1+₂αDopt

<

2

α

Dopt.

Case2.2.U∗

1

(τ)

̸= ∅

. SinceJ(0)

∈

U2∗

(τ)

, by sub-procedureH1(Cases 2.3 and 3.2),A

(τ)

= ∅

and eitherB0

=

C

(τ)

∪

U2∗

(τ)

or B0

=

D

(τ)

∪

U2∗

(τ)

. We consider these two possibilities individually:

First,B0

=

C

(τ)

∪

U2∗

(τ)

. Thenqq(τ)

< (

1

+

α)

pp(τ). Sincem

=

0, according to sub-procedureH1(Case 3.2), we have either E

(τ)

̸= ∅

orp_c(τ)

>

1+₂αp_p(τ). FromObservation 11, there exists a jobJx

∈

B0such thatDopt

≥

L

(

Jx

,

Jp(τ)

)

≥

(

1+₂α

+

1

)

pp(τ). Given(1)and

τ

=

1+α

2 pp(τ), we obtainDon

−

Dopt

≤

qq(τ)

< (

1

+

α)

pp(τ)

≤

2

α

Dopt.

Second,B0

=

D

(τ)

∪

U2∗

(τ)

. Thenqq(τ)

< α

pp(τ). Sincem

=

0 andt

=

1+α

2 pp(τ), by sub-procedureH1(Case 2.3), we have r_q(τ)

≥

rmin

(

D

(τ))

≥

α

pp(τ)

>

qq(τ). IfS_p(τ)π

≥

rq(τ), thenDopt

≥

rq(τ)

+

pp(τ). IfS_p(τ)π

<

rq(τ), thenDopt

≥

pp(τ)

+

qq(τ). Hence Dopt

≥

min

{

rq(τ)

,

qq(τ)

} +

pp(τ)

≥

pp(τ)

+

qq(τ). From(1),Don

−

Dopt

≤

τ

=

1+₂αpp(τ)

<

2

α

Dopt.Theorem 4.2follows.

(9)

In the rest of Section4, we assume thatm

>

0. Then the discussions proceed under conditions (2a)–(2e) individually.

Discussion under condition(2a):J₍1)

=

Jp(S1).

Condition (2a) implies thatJ₍1)is the jobJ(i)from the preamble toLemmas 3.2–3.5.

Theorem 4.3. Suppose that J₍1)

=

Jp(S1). Then Don

/

Dopt

≤

1

+

2

α

.

Proof. SinceJ₍1)

=

Jp(S1), byObservation 7, we have rx

>

S1

≥

1

+

α

2 p(1)

,

for any Jx

∈

U ∗

(τ).

(3) We consider two situations:

Case1. J₍0)

∈

U1∗

(τ)

. From(3),Dopt

>

S1

+

pq(τ)

+

qq(τ). By(1), we have

Don

−

Dopt

<

p(1)

+

p(0)

.

(4)

Case1.1.S₍π₁₎

≥

S1. By(3)andObservation 10,Dopt

≥

S1

+

max

{

L

(

J(0)

,

J(1)

),

L

(

J(0)

,

Jp(τ)

)

} ≥

S1

+

max

{

p(1)

,

pp(τ)

} +

p(0). Set z

=

p₍0). From(4)andLemma 3.4(ii), we are done.

Case1.2. S₍π₁₎

<

S1. From (3),Dopt

≥

Sπ₍1)

+

p(1)

+

qq(τ). From (1),Don

−

Dopt

≤

S1

+

p(0). ThenDon

−

Dopt

≤

min

{

p₍1)

+

p(0)

,

S1

+

p(0)

}

by(4). There are three situations:

First,p₍0)

<

1+2αpp(τ). From(3),Dopt

≥

S₍π1)

+

p(1)

+

L

(

J(0)

,

Jp(τ)

)

≥

S₍π1)

+

p(1)

+

pp(τ)

+

p(0). Setz

=

p(0). ByLemma 3.5(ii),

we are done.

Second,1+₂αp_p(τ)

≤

p₍0)

<

pp(τ)orA

(τ)

̸= ∅

. From(3)andObservation 12(a, b),Dopt

≥

S₍π1)

+

p(1)

+

L

(

Ja(τ)

,

J(0)

,

Jp(τ)

)

≥

S₍π₁₎

+

p₍1)

+

p(0)

+

(

1

+

α)

pp(τ). Setz

=

p(0). ByLemma 3.5(iii), we are done.

Third,p₍0)

=

pp(τ)andA

(τ)

= ∅

. SinceJ(0)

∈

U1∗

(τ)

, byObservation 3(c), we haveB0

⊆

U1∗

(τ)

andU

∗

2

(τ)

= ∅

.

ThenB0

=

C

(τ)

and qq(τ)

< (

1

+

α)

pp(τ). From (3),Dopt

≥

S₍π1)

+

p(1)

+

p(0)

+

q0

>

S₍π1)

+

p(1)

+

2pp(τ) and Dopt

≥

S1

+

p(0)

+

q(0)

>

S1

+

2pp(τ). By(1),Don

−

Dopt

≤

min

{

p(1)

+

α

pp(τ)

,

S1

+

α

pp(τ)

}

. Setz

=

α

pp(τ). ByLemma 3.5(i),

we are done.

Case2. J₍0)

∈

U2∗

(τ)

. ThenJ(0)

=

Jp(τ). Case2.1.U∗

1

(τ)

= ∅

. By sub-procedureH1(Case 1), we haveB0

=

U2∗

(τ)

, soJq(τ)

∈

U2∗

(τ)

. From(3)andObservation 10, Dopt

≥

S1

+

L

(

Jq(τ)

,

Jp(τ)

)

≥

S1

+

pp(τ)

+

qq(τ). By(1), we getDon

−

Dopt

≤

p(1). Setz

=

0.

WhenS₍π₁₎

≥

S1, we are done byLemma 3.4(i). WhenS₍π1)

<

S1, By(3)and Observation 10,Dopt

≥

S₍π1)

+

p(1)

+

L

(

J_q(τ)

,

J_p(τ)

)

≥

S₍π₁₎

+

p₍1)

+

pp(τ)

+

qq(τ). From(1), we haveDon

−

Dopt

≤

S1, soDon

−

Dopt

≤

min

{

p(1)

,

S1

}

. ByLemma 3.5(i),

we are done.

Case2.2.U∗

1

(τ)

̸= ∅

. SinceJ(0)

∈

U2∗

(τ)

, by sub-procedureH1(Cases 2.3 and 3.2),A

(τ)

= ∅

and eitherB0

=

C

(τ)

∪

U2∗

(τ)

or B0

=

D

(τ)

∪

U2∗

(τ)

. We consider these two possibilities individually:

First,B0

=

C

(τ)

∪

U2∗

(τ)

. Thenqq(τ)

< (

1

+

α)

pp(τ). Recall thatJ(1)

=

Jp(S1). By sub-procedureH1(Case 3.2), we have

eitherp_c(τ)

>

1+₂αp_p(τ)orE

(τ)

̸= ∅

. ByObservation 11, the batchB0contains a jobJxsuch thatL

(

Jx

,

Jp(τ)

)

≥

(

1+₂α

+

1

)

pp(τ). From(3),Dopt

≥

S1

+

L

(

Jx