Scheduling two agents with sum-of-processing-times-based deterioration on a single machine

0h

Applied Mathematics and Computation 219 (2013) 8848–8855

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Applied Mathe matics and Computati on

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Scheduling two agents with sum-of-processing-times-based deterioration on a single machine

096-3003/$ - see front matter � 2013 Elsevier Inc. All rights reserved.ttp://dx.doi.org/10.1016/j.amc.2013.03.040

⇑ Corresponding author.E-mail address: [email protected] (P. Liu).

Peng Liu a,⇑, Na Yi b, Xiaoye Zhou a, Hua Gong c

a School of Management, Shenyang University of Technology, Shenyang 110870, China b College of Sciences, Liaoning Shihua University, Fushun 113001, China c School of Science, Shenyang Ligong University, Shenyang 110159, China

a r t i c l e i n f o

Keywords:SchedulingTwo-agentSum-of-processing-times-basedDeteriorating jobs Single machine

a b s t r a c t

A new scheduling model with two-agent and sum-of-processi ng-times-based deterioration is proposed in this paper. Two agents comp ete to perform their respective jobs on a common single machine and each agent has his own criterion to optimize. The sum-of-processing-times-based deterioration of a job is that the actual job processing time is a function ofthe sum of the normal processing times of the already proce ssed jobs. The objective is tominimize the total completion time of the first agent with the restriction that the makespan of the second agent cannot exceed a given upper bound. The optimal properties ofthe problems are given and the optimal polynomial time algorithms are proposed to solve the sch eduling problems.

� 2013 Elsevier Inc. All rights reserved.

1. Introduction

In this paper, we introduce a new scheduling model in which both two-agen t and sum-of-p rocessing-time s-based deterioration exist simultaneou sly. The assumpti on that job processing times are known and fixed throughout the period of job processing is universally accepted in classical scheduling, but this assumption may be unrealisti c in many actual situations.The processing times of jobs may be not a constant. The sum-of-proc essing-time s-based deterioration of a job is that the actual job processing time is a function of the sum of the normal processing times of the already processed jobs. Most pre- vious researches on the deteriorating job scheduling problem assume that all jobs to be processed on a common machine belong to one agent. However, there are several agents in real production settings, each with a set of non-preempti ve jobs.The agents have to schedule their jobs on a common processin g machine, and each agent wants to minimize a cost function which depends on its own jobs’ completion times. In this paper, we address two-agent single-machine scheduling problems with sum-of-proc essing-times-bas ed deterioration, where the goal is to find a schedule that minimizes the total completion time of one agent with the restriction that the makespan of the other agent cannot exceed a given bound. The problems under considerati on are related to two well-studied scheduling problems, namely the scheduling with two-agen t and the scheduling with sum-of-p rocessing-time s-based deterioration. However, it is surprisin g that the two-agent and sum-of-processing-tim es-based deterioration have never been considered simultaneou sly in the literature.

Example involving two-agen t and deteriorati on can be found in the ingot formatio n process in the steel plant. After the molten steel is refining in the refining furnace, it can take two alternative modes accordin g to the desired ingot shape. One is

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P. Liu et al. / Applied Mathematics and Computation 219 (2013) 8848–8855 8849

the teeming mode that the molten steel in a ladle is poured into molds to solidify cylindrically-sh aped ingots. The other is the continuous-cas ting mode that the molten steel is cast and cooled to form slabs, blooms or billets in the continuous caster.Due to the differences between the teeming mode and the continuous-casti ng mode, they are regarded as two agents. Owing to the inevitabl e drop in the temperature, the processing time of the molten steel in the refining furnace deteriorates. In the processes described above, there exist the phenomena of two-agent and deterioration.

Machine scheduling problems with multi-agent have received increasing attention in recent years. Agnetis et al. [1] andBaker and Smith [4] are among the pioneers to introduce the concept of multi-agent into scheduling problems. The objective functions include the maximum values of regular performanc e measures, the number of late jobs, and the total weighted completion time. Cheng et al. [7] consider the multi-agent scheduling problem on a single machine, where the agent’s objective functions are of the max-form. They study the feasibility model and the minimality model. Agnetis et al. [2] developbranch-and- bound algorithms for several hard, two-agen t scheduling problems and devise a Lagrangian approach . Lee et al. [12] provide approximat ion algorithms for multi-agent scheduling to minimize total weighted completion time. Leung et al. [18] study a two-agent scheduling environment with m identical parallel machines , and generalize the results of Agne- tis et al. [1] and Baker and Smith [4] by including the total tardiness objective, allowing for preemptions , and considering jobs with different release dates. Mor and Mosheiov [23] consider the scheduling problems with two competin g agents tominimize minmax and minsum earliness measure. Lee et al. [14] consider a two-agen t scheduling problem on a two- machine flowshop setting, where the objective is to minimize the total tardiness of the first agent with the restriction that the number of tardy jobs of the second agent is zero. Lee et al. [13] study a two-machine flowshop problem with two agents where the objective is to minimize the total completion time of the first agent with no tardy jobs for the second agent. Nong et al. [25] study the scheduling problem of determining the sequence of the jobs such that the total cost of the two agents isminimized, where the cost of the first agent is the maximum weighted completion time of his jobs while the cost of the second agent is the total weighted completion time of his jobs. Mor and Mosheiov [24] consider a single machine two-agent problem in the context of batch scheduling. The objective is minimizing the total flowtime of one agent, subject to an upper bound on the flowtime of the second agent. Luo et al. [22] study approximat ion schemes for two-machine flowshop scheduling with two agents.

Recently, deteriorating job scheduling problems have received considerable attention in the literature. For the exten- sive reviews on research with deteriorating jobs, the reader is referred to the papers studied by Alidaee and Womer [3]and Cheng et al. [6]. Wang et al. [35] deal with the single-m achine scheduling problems with a time-dependent deterioration. The time-depend ent deterioration of a job is assumed to be an increasing function of total normal processing time of jobs scheduled in front of it. Lee et al. [16] develop a sum-of-proces sing-times-based deterioration model where the actual job processing time is a function of jobs already processed. They show that the single-machi ne makespan problem can be solved in polynomial time under the proposed deterioration model. Wang et al. [34] continue to explore Lee et al. [16] deteriorati on model and provide an optimal algorithm for the single-machi ne total completion time scheduling problem. Wang et al. [36] deal with the single-machi ne makespan minimization scheduling problem with a sum- of-processin g-times-based learning effect and deteriorati ng jobs. Cheng et al. [10] investigate a scheduling model with learning effects in which the actual processing time of a job is a function of the total normal processing times of the jobs already processed and of the job’s scheduled position. Lee et al. [17] consider the two-machi ne flowshop makespan problem with deteriorating jobs. An exact algorithm and three heuristic algorithms are provided. Wang and Cheng [27]deal with the machine scheduling problems with the effects of deterioration and learning. Cheng et al. [9] provide the optimal solutions for several scheduling problems where the phenomena of job deterioration and learning exist simultaneously. For other scheduling problems with deteriorating jobs, the reader is referred to Wang et al. [28–32], Wang and Wang [33], Wei and Wang [37], Yang and Wang [38].

However, scheduling research in the multi-agent setting with deteriorati on considerations is relative limited in the literature. Liu and Tang [19] are probably the first authors that consider the two-agent scheduling problems with simple linear deteriorating jobs on a single machine. Wan et al. [26] study several competitive agent scheduling problems with controllable processing times. Lee et al. [15] consider a single-machine scheduling problem with a linear deterioration assumption where the objective is to minimize the total weighted completion time of jobs from the first agent with the restriction that no tardy job is allowed for the second agent. Liu et al. [20] consider the two-agen t scheduling problems with deteriorating jobs and group technology on a single machine. Liu et al. [21] study the two-agen t single-m achine scheduling problems with position- dependent processing times. Cheng et al. [8] consider a two-agen t single-machi nescheduling problem involving deteriorating jobs and learning effects simultaneously . Cheng et al. [5] consider a two-agent scheduling problem in which the actual processing time of a job in a schedule is a function of the sum-of-p rocessing- times-based learning and a control paramete r of the leaning function. The literature mentioned above investiga tes the scheduling with multi-agent , however, they do not consider the scheduling with sum-of-proc essing-times-bas eddeterioration.

In this paper we combine scheduling, sum-of-proc essing-times-bas ed deteriorati on and two-agent decisions at the same time. We propose the optimal properties and present the optimal polynomial time algorithms to solve the single-machine scheduling problems with two agents and sum-of-proc essing-time s-based deteriorati on. The remainder ofthis paper is organized as follows. In Section 2, we describe the proposed problems. In Section 3–5, we develop the optimal polynomial time algorithms for the two-agen t single-machi ne scheduling problems. Section 6 gives some concluding remarks.

8850 P. Liu et al. / Applied Mathematics and Computation 219 (2013) 8848–8855

2. Problem description

We now describe our problems formally. There are two agents A and B, each with a set of non-pree mptive jobs. Two agents compete for the usage of a common single machine, and each agent wants to minimize an objective function that depends on its own job’s completion times. The jobs belonging to agent A and agent B are called A-agent’s jobs and B-agent’s

jobs, respectively . All jobs are available for processin g at time zero. There are nA A-agent’s jobs JA1; J

A2; . . . ; JA

nA

n oready to be

processed on the common single machine. Associated with each A-agent’s job JAh , h = 1, 2, . . ., nA, there is a normal processing

time pAh , and pA

½r� is the normal processing time of a job if schedule d in the rth position in a sequence. Let pAhr be the processing

time of job JAh if it is scheduled in position r in a sequence. There are two sum-proces sing-time-base d deterioration models as

follows: (1) pAhr ¼ 1þ

Pr�1l¼1 pA

½l�

� �pA

h , (2) pAhr ¼ 1þ

Pr�1

l¼1pA½l�PnA

l¼1pA

l

� �pA

h , where P0

i¼1pA½i� ¼ 0. Similarly, there are nB B-agent’s jobs

JB1; J

B2; . . . ; JB

nB

n oready to be processed on the common single machine. Associated with each B-agent’s job JB

k , k = 1, 2, . . .,

nB, there is a normal processing time pBk , and pB

½v � is the normal processing time of a job if scheduled in the vth position in

a sequence. Let pBkv be the processing time of job JB

k if it is schedule d in position v in a sequence. There are two sum-pro cess-

ing-time-bas ed deteriorati on models as follows: (1) pBkv ¼ 1þ

Pv�1l¼1 pB

½l�

� �pB

k , (2) pBkv ¼ 1þ

Pv�1

l¼1pB½l�PnB

l¼1pB

l

� �pB

k , where P0

i¼1pB½i� ¼ 0 .

Let S indicate a feasible schedule of the n = nA + nB jobs. Let CAhðSÞ denote the completion time of A-agent’s job JA

h underschedule S. The objective function of agent A is to minimize the total completion time

PCA

hðSÞ. The objective function of

agent B is to minimize the makespan CBmax ¼ max

k¼1;2;...;nB

CBkðSÞ

n o. The goal is to find a schedule that minimizes the total comple-

tion time of agent A with the restrictio n that the makespan of agent B cannot exceed a given bound U. If the value U is too

small, an instance of the scheduling problem may not have feasible solutions. If there is at least one feasible solution, we say

that the instance is feasible. We adopt the three-field notation w1|w2|w3 of Graham et al. [11] to denote our problems. Under

w2, we consider two sum-proces sing-time-base d deterioration functions: (1) DA1 : pA

hr ¼ 1þPr�1

l¼1 pA½l�

� �pA

h ; and

DB1 : pB

kv ¼ 1þPv�1

l¼1 pB½l�

� �pB

k ;, (2) DA2 : pA

hr ¼ 1þPr�1

l¼1pA½l�PnA

l¼1pA

l

� �pA

h , and DB2 : pB

kv ¼ 1þPv�1

l¼1pB½l�PnB

l¼1pB

l

� �pB

k . For simplicity, we denote the

two scheduling problems as 1 DA1;D

B1

�� PCAh : CB

max 6 U and 1 DA2;D

B2

�� PCAh : CB

max 6 U, respectively.

3. Problem 1 DA1 ;D

B1

�� +CAh : CB

max <U

In this section, we derive an optimal polynomi al time algorithm to solve the problem 1 DA1;D

B1

�� PCAh : CB

max 6 U. Before presenting the algorithm, we first propose several optimal propertie s of the problem. In order to denote the objective function of all jobs, we put only one objective function in the last field of the three-field notation, e.g., Cmax in the field w3 is the makespan of all jobs.

Lemma 1. In the scheduling problem that we consider, there exists an optimal schedule without idle times between jobs on the single machine.

Proof. Since the objective function fmax andP

Cj are non-decreasing function of the completion time Cj, they are the regular measure of performance. If there are idle times between jobs on the single machine, the subsequent jobs can bemoved earlier without increasing the objective value. h

Lemma 2. For the problem 1 DA1;D

B1

�� Cmax, the makespan is sequence independent.

Proof. Assume that schedule S has two adjacent jobs Ji and Jj with job Ji immedia tely preceding job Jj, and a pairwise interchange of jobs Ji and Jj is performed to derive a new schedule S

0. That is S = (p, Ji, Jj, p0) and S0 = (p, Jj, Ji, p0), where p and p0

are partial sequence . Without loss of generality, we assume that there are r � 1 jobs in the partial sequence p. Thus, Ji and Jj

are the rth and the (r + 1)th jobs, respectively, in the schedule S. Likewise, Jj and Ji are scheduled in the rth and the (r + 1)thpositions in the schedule S0. To further simplify the notation, let T denote the completion time of the last job in the partial sequence p. Depending on whether jobs belong to agents A or B, we classify them into the following four cases.

Case 1: Both jobs Ji and Jj are from agent A in sequences S and S0 .Under S, the completion times of jobs Ji and Jj are CiðSÞ ¼ T þ 1þ

Pr�1l¼1 pA

½l�

� �pA

i and CjðSÞ ¼ CiðSÞ þ pAj 1þ

Pr�1l¼1 pA

½l� þ pAi

� �.

Under S0, the completion times of jobs Jj and Ji are CjðS0Þ ¼ T þ 1þPr�1

l¼1 pA½l�

� �pA

j and CiðS0Þ ¼ CjðS0Þ þ pAi 1þ

Pr�1l¼1 pA

½l� þ pAj

� �.

Taking the differenc e between Cj(S) and Ci(S0), one derives that


CjðSÞ � CiðS0Þ ¼ T þ 1þXr�1

l¼1

pA½l�

!pA

i þ pAj 1þ

Xr�1

l¼1

pA½l� þ pA

i

!" #

� T þ 1þXr�1

l¼1

pA½l�

!pA

j þ pAi 1þ

Xr�1

l¼1

pA½l� þ pA

j

!" #¼ 0

As a conseque nce, the makespan of S is equal to that of S0. Thus, the makespan is sequence independen t for Case 1.Case 2: Both jobs Ji and Jj are from agent B in sequences S and S0 .Similarly to Case 1, we have


l¼1

pB½l�

!pB

i þ pBj 1þ

Xr�1

l¼1

pB½l� þ pB

i

!" #

� T þ 1þXr�1

l¼1

pB½l�

!pB

j þ pBi 1þ

Xr�1

l¼1

pB½l� þ pB

j

!" #¼ 0

Thus, the makespan is sequence independen t for Case 2.Case 3: Job Ji is from agent A, but job Jj is from agent B in sequences S and S0 .

Under S, the completion times of jobs Ji and Jj are CiðSÞ ¼ T þ 1þPr�1

l¼1 pA½l�

� �pA

i and CjðSÞ ¼ CiðSÞ þ pBj 1þ

Pr�1l¼1 pB

½l�

� �.


l¼1 pB½l�

� �pB


Pr�1l¼1 pA

½l�

� �.

Similarly, we have


l¼1

pA½l�

!pA

i þ pBj 1þ

Xr�1

l¼1

pB½l�

!" #

� T þ 1þXr�1

l¼1

pB½l�

!pB

j þ pAi 1þ

Xr�1

l¼1

pA½l�

!" #¼ 0

As a conseque nce, the makespan of S is equal to that of S0. Thus, the makespan is sequence independen t for Case 3.Case 4: Job Ji is from agent B, but job Jj is from agent A in sequences S and S0 .Similarly to Case 3, we have


l¼1

pB½l�

!pB

i þ pAj 1þ

Xr�1

l¼1

pA½l�

!" #� T þ 1þ

Xr�1

l¼1

pA½l�

!pA

j þ pBi 1þ

Xr�1

l¼1

pB½l�

!" #¼ 0

Thus, the makespan is sequence independen t for Case 4.From Case 1 to Case 4, this completes the proof of Lemma 2. h

Lemma 3. An optimal schedule exists in which the A-agent’s jobs are processed in the non-decrea sing order of pAh for minimizing

total completion time.Proof. By contradiction. Depending on whether jobs Ji and Jj are adjacent, we classify them into the following two cases.Case 1. Jobs Ji and Jj are adjacent. Here, we still use the same notation as in the proof of Lemma 2. Suppose schedule S is an

optimal schedule in which the A-agent’s jobs are not ordered accordin g to the non-decreasing order of pAh , and let Ji and Jj be

the first pair of A-agent’s jobs such that pAi > pA

j . Schedule S0 is created from S by swapping jobs Ji and Jj, that is S = (p, Ji, Jj, p0)and S0 = (p, Jj, Ji, p0).

CiðSÞ þ CjðSÞ � Cj S0� �� Ci S0

� �¼ CiðSÞ � Cj S0

� � þ CjðSÞ � Ci S0

� � ¼ T þ 1þ

Xr�1

l¼1

pA½l�

!pA

i

" #� T þ 1þ

Xr�1

l¼1

pA½l�

!pA

j

" #þ 0

¼ pAi � pA

j

� �1þ

Xr�1

l¼1

pA½l�

!> 0

It follows that S0 has a strictly smaller total completion times of jobs Ji and Jj. This contradicts the optimality of schedule S.Case 2. Jobs Ji and Jj are not adjacent. There are some B-jobs between Ji and Jj, that is S = (p, Ji, B, Jj, p0) and S0 = (p, Jj, B, Ji, p0).

The proof is similar to that of Case 1 and is omitted.From Case 1 and Case 2, this complete s the proof of Lemma 3. h

Lemma 4. An optimal schedule exists in which the B-agent’s jobs are consecutively processed.Proof. By Lemma 2, the makespan is sequence independent. Using a pairwise job interchange argument, we can

consecutive ly process B-agent’s jobs and consolidate all B-agent’s jobs into a block. h


Based on these Lemmas 1–4, we introduce the following efficient solution algorithm.Algorithm 1.Step 1: Set i = 1 .Step 2: Arrange the A-agent’s jobs as JA

1; JA2; . . . ; JA

nA

n oaccording to the non-decreas ing order of pA

h and denote all B-agent’s jobs as a dummy job B1.

Step 3: Define sequence S ¼ JA1; J

A2; . . . ; JA

nA�1; JAnA;B1

n oand calculate the maknspan CB

max for agent B. If CBmax 6 U, then the

sequence S is an optimal schedule .Step 4: Select the dummy job B1 to the (nA � i)th position in the sequence JA

1; JA2; . . . ; JA

nA�i; JAnA�iþ1; . . . ; JA

nA

n oand calculate the

maknspan CBmax for agent B. If CB

max 6 U, then the sequence Si ¼ JA1; J

A2; . . . ; JA

nA�i;B1; JAnA�iþ1; . . . ; JA

nA�1; JAnA

n ois an optimal

schedule.Step 5: If i < nA and CB

max > U, then i = i + 1 and go to Step 4. Otherwise, stop.

Theorem 1. Algorithm 1 generates an optimal schedule for the problem 1 DA1;D

B1

�� PCAh : CB

max 6 U in O(nA log nA + nB) time.Proof. The proof of optimalit y is straightforw ard from the results of Lemmas 1-4. Let us now turn to time complexity. We

can sequence the jobs of set JA in O(nA log nA) time according to the non-decreasing order of the normal processing time pAh .

Creating dummy job B1 incurs O(nB) operations. Therefore, the overall time complexi ty of Algorithm 1 is O(nA log nA + nB).This completes the proof. h

Furthermore, we consider the following numerical example.Example 1. For the problem 1 DA

1;DB1

�� PCAh : CB

max 6 U, we assume that U = 12. Set JA ¼ JA1; J

A2; J

A3

n o: pA

1 ¼ 1; pA2 ¼ 3; pA

3 ¼ 2;

Set JB ¼ JB1; J

B2

n o: pB

1 ¼ 1; pB2 ¼ 4.

Solution. According to Step 1, set i = 1. By Step 2, arrange the A-agent’s jobs as JA1; J

A3; J

A2

n oaccording to the non-decreas ing

order of pAh and denote all B-agent’s jobs as a dummy job B1 ¼ JB

1JB2

n o. By Step 3, define sequence S ¼ JA

1; JA3; J

A2;B1

n o. Calculate

pA11 ¼ 1, pA

32 ¼ ð1þ 1Þ � 2 ¼ 4, pA23 ¼ ð1þ 1þ 2Þ � 3 ¼ 12, pB

11 ¼ 1, pB22 ¼ ð1þ 1Þ � 4 ¼ 8. So the makespan for agent B is

CBmax ¼ 1þ 4þ 12þ 1þ 8 ¼ 26.

Since CBmax ¼ 26 > U ¼ 12, select the dummy job B1 to the 2th position in the sequence JA

1; JA3; J

A2

n oby Step 4, i.e.,

S1 ¼ JA1; J

A3;B1; JA

2

n o. By calculating, we obtain that the makespan for agent B is CB

max ¼ 1þ 4þ 1þ 8 ¼ 14. Since

CBmax ¼ 14 > U ¼ 12, we derive i = 2 and go to Step 4. So select the dummy job B1 to the 1th position in the sequence

JA1; J

A3; J

A2

n o, i.e., S2 ¼ JA

1;B1; JA3; J

A2

n o. Similarly, the makespan for agent B is CB

max ¼ 1þ 1þ 8 ¼ 10 .Since CB

max ¼ 10 < U ¼ 12, the sequence S2 ¼ JA1;B1; JA

3; JA2

n ois an optimal schedule by Step 4. And by Step 5, stop. Conse-

quently, we obtain an optimal schedule JA1 ! JB

1 ! JB2 ! JA

3 ! JA2, and

PCA

h ¼ 1þ 14þ 26 ¼ 41 .

4. Problem 1 DA2 ;D

B2

�� +CAh : CB

max <U

For the problem 1 DA2;D

B2

�� PCAh : CB

max 6 U, we have the following lemmas that are similar to Lemmas 2, 3 and 4,respectively .

Lemma 5. For the problem 1 DA2;D

B2

�� Cmax, the makespan is sequence independent.Proof. Here, we still use the same notation as in the proof of Lemma 2. Depending on whether jobs belong to agents A or

B, we classify them into the following four cases.Case 1. Both jobs Ji and Jj are from agent A in sequences S and S0.


l¼1pA½l�PnA

l¼1pA

l

� �pA

i and CjðSÞ ¼ CiðSÞ þ pAj 1þ

Pr�1

l¼1pA½l�þpA

iPnAl¼1

pAl

� �.


l¼1pA½l�PnA

l¼1pA

l

� �pA


Pr�1

l¼1pA½l�þpA

jPnAl¼1

pAl

� �.

Taking the differenc e between Cj(S) and Ci(S0), one derives that

CjðSÞ � CiðS0Þ ¼ T þ 1þPr�1

l¼1 pA½l�PnA

l¼1pAl

!pA

i þ pAj 1þ

Pr�1l¼1 pA

½l� þ pAiPnA

l¼1pAl

!" #� T þ 1þ

Pr�1l¼1 pA

½l�PnAl¼1pA

l

!pA

j þ pAi 1þ

Pr�1l¼1 pA

½l� þ pAjPnA

l¼1pAl

!" #

¼ 0

Conseque ntly, the makespan of S is equal to that of S0. Thus, the makespan is sequence independen t for Case 1.Case 2. Both jobs Ji and Jj are from agent B in sequences S and S0 .Similarly to Case 1, we have



l¼1 pB½l�PnB

l¼1pBl

!pB

i þ pBj 1þ

Pr�1l¼1 pB

½l� þ pBiPnB

l¼1pBl

!" #� T þ 1þ

Pr�1l¼1 pB

½l�PnBl¼1pB

l

!pB

j þ pBi 1þ

Pr�1l¼1 pB

½l� þ pBjPnB

l¼1pBl

!" #

¼ 0

Thus, the makespan is sequence independen t for Case 2.Case 3. Job Ji is from agent A, but job Jj is from agent B in sequence s S and S0 .


l¼1pA½l�PnA

l¼1pA

l

� �pA

i and CjðSÞ ¼ CiðSÞ þ pBj 1þ

Pr�1

l¼1pB½l�PnB

l¼1pB

l

� �.


l¼1pB½l�PnB

l¼1pB

l

� �pB


Pr�1

l¼1pA½l�PnA

l¼1pA

l

� �.

Taking the difference between Cj(S) and Ci(S0), one derives that


l¼1 pA½l�PnA

l¼1pAl

!pA

i þ pBj 1þ

Pr�1l¼1 pB

½l�PnBl¼1pB

l

!" #� T þ 1þ

Pr�1l¼1 pB

½l�PnBl¼1pB

l

!pB

j þ pAi 1þ

Pr�1l¼1 pA

½l�PnAl¼1pA

l

!" #¼ 0

Consequentl y, the makespan of S is equal to that of S0. Thus, the makespan is sequence independent for Case 3.Case 4. Job Ji is from agent B, but job Jj is from agent A in sequence s S and S0 .Similarly to Case 3, we have


l¼1 pB½l�PnB

l¼1pBl

!pB

i þ pAj 1þ

Pr�1l¼1 pA

½l�PnAl¼1pA

l

!" #� T þ 1þ

Pr�1l¼1 pA

½l�PnAl¼1pA

l

!pA

j þ pBi 1þ

Pr�1l¼1 pB

½l�PnBl¼1pB

l

!" #¼ 0

Thus, the makespan is sequence independen t for Case 4.From Case 1 to Case 4, this completes the proof of Lemma 5. h

Lemma 6. An optimal schedule exists in which the A-agent’s jobs are processed in the non-decrea sing order of pAh for minimizing

total completion time.Proof. By contradiction. Depending on whether jobs Ji and Jj are adjacent, we classify them into the following two cases.Case 1. Jobs Ji and Jj are adjacent. It is similar to the proof of Lemma 3 except that:

CiðSÞ þ CjðSÞ � CjðS0Þ � CiðS0Þ ¼ CiðSÞ � CjðS0Þ

þ CjðSÞ � CiðS0Þ

¼ T þ 1þPr�1

l¼1 pA½l�PnA

l¼1pAl

!pA

i

" #� T þ 1þ

Pr�1l¼1 pA

½l�PnAl¼1pA

l

!pA

j

" #þ 0 ¼ pA

i � pAj

� �1þ

Pr�1l¼1 pA

½l�PnAl¼1pA

l

!> 0

It follows that S0 has a strictly smaller total completion times of jobs Ji and Jj. This contradicts the optimality of schedule S.Case 2. Jobs Ji and Jj are not adjacent. There are some B-agent’s jobs between Ji and Jj, that is S = (p, Ji, B, Jj, p0) and

S0 = (p, Jj, B, Ji, p0). The proof is similar to that of Case 1 and is omitted.From Case 1 and Case 2, this complete s the proof of Lemma 6. h

Lemma 7. An optimal schedule exists in which the B-agent’s jobs are consecutively processed.Proof. By Lemma 5, the makespan is sequence independent. Using a pairwise job interchange argument, we can

consecutive ly process B-agent’s jobs and consolidate all B-agent’s jobs into a block. h

Using Lemmas 1, 5, 6 and 7, we have an algorithm to determine an optimal schedule for the problem

1 DA2;D

B2

�� PCAh : CB

max 6 U. The algorithm is same to Algorithm 1, and we omit the details.

5. Extension to general cost functions

In this section, let f Bk ð�Þ be a non-decr easing function of the completion time of job JB

k . The objective function of agent B is

to minimize the maximum cost f Bmax ¼ max

k¼1;2;...;nB

ff Bk CB

kðSÞ� �

g. Note that CBmax ¼ max

k¼1;2;...;nB

fCBkðSÞg is a special case of the maximum

cost. We denote the two scheduling problems as 1 DA1;D

B1

�� PCAh : f B

max 6 U and 1 DA2;D

B2

�� PCAh : f B

max 6 U, respectively .

It is clear that Lemmas 1–4 are still valid for the problem 1 DA1;D

B1

�� PCAh : f B

max 6 U. The maximum permitted makespan UBf

for agent B is obtained by solving the equation f Bmax CB

k

� �¼ U. We have the following Algorithm 2 that are similar to Algorithm

1 for the problem 1 DA1;D

B1

�� PCAh : f B

max 6 U .


Algorithm 2.Step 1: Set i = 1 .Step 2: Arrange the A-agent’s jobs as JA

1; JA2; . . . ; JA

nA

n oaccording to the non-decreas ing order of pA

h and denote all B-agent’s jobs as a dummy job B1.

Step 3: Define sequence S ¼ JA1; J

A2; . . . ; JA

nA�1; JAnA;B1

n o, and calculate the makespan CB

max for agent B and the maximum per-

mitted makespan UBf from f B

maxðCBkÞ ¼ U. If CB

max 6 UBf , then the sequence S is an optimal schedule.

Step 4: Select the dummy job B1 to the (nA � i)th position in the sequence JA1; J

A2; . . . ; JA

nA�i; JAnA�iþ1; . . . ; JA

nA

n o, and calculate

the makespan CBmax for agent B and the maximum permitted makespan UB

f from f BmaxðC

BkÞ ¼ U. If CB

max 6 UBf , then the sequence

Si ¼ JA1; J

A2; . . . ; JA

nA�i;B1; JAnA�iþ1; . . . ; JA

nA�1; JAnA

n ois an optimal schedule.

Step 5: If i < nA and CBmax > UB

f , then i = i + 1 and go to Step 4. Otherwise, stop.

Theorem 2. Algorithm 2 generates an optimal schedule for the problem 1 DA1;D

B1

�� PCAh : f B

max 6 U in O(nA log nA + nB) time.

Proof. The proof of Theorem 2 is similar to Theorem 1, and we omit the details. h

It is clear that Lemmas 5–7 are still valid for the problem 1 DA2;D

B2

�� PCAh : f B

max 6 U. For the problem

1 DA2;D

B2

�� PCAh : f B

max 6 U, the solution algorithm is same to Algorithm 2, and we omit the details.

6. Conclusions

In this paper, we study the two-agent single-machi ne scheduling problems with sum-of-p rocessing-time s-based deterioration. The objective is to minimize the total completion time of A-agent with a constraint that the makespan of B-agent cannot exceed a given upper bound. We propose the optimal properties and present the optimal polynomi al time algorithms to solve the scheduling problems.

The consideration of other time-dependent deterioration might be interesting issues for future research, for example, the monotonic deterioration function or adding a power to the deterioration function. The other interesting future research direction is to analyze the problems with other objective functions such as minimizing the number of late jobs and the total weighted completion time.

Acknowled gements

We are grateful to the editor and the anonymous referees for their helpful comments and suggestions . This research issupported by National Natural Science Foundati on of China (Grant No. 71001074, 71101097), Science Research Foundation of Department of Education of Liaoning Province of China (Grant No. W2010302).

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Scheduling two agents with sum-of-processing-times-based deterioration on a single machine