J Comb OptimDOI 10.1007/s10878-014-9744-y

Two approximation algorithms for two-agentscheduling on parallel machines to minimize makespan

Kejun Zhao · Xiwen Lu

© Springer Science+Business Media New York 2014

Abstract A two-agent scheduling problem on parallel machines is considered. Ourobjective is to minimize the makespan for agent A, subject to an upper bound onthe makespan for agent B. When the number of machines, denoted by m, is chosenarbitrarily, we provide an O(n) algorithm with performance ratio 2 − 1

m , i.e., themakespan for agent A given by the algorithm is no more than 2 − 1

m times the optimalvalue, while the makespan for agent B is no more than 2− 1

m times the threshold value.This ratio is proved to be tight. Moreover, when m = 2, we present an O(nlogn)

algorithm with performance ratio 1+√17

4 ≈ 1.28 which is smaller than 32 . The ratio is

weakly tight.

Keywords Two-agent scheduling · Parallel machines · Makespan · Approximationalgorithm

1 Introduction

In recent years agent scheduling problems have become an important research area,due to their various applications in some applied disciplines. For example, Brewerand Plott (1996) considered a problem of scheduling trains (agents) on a shared singletrack; Schultz et al. (2002) discussed a problem in multimedia telecommunicationservices in which the agents competed for the use of a commercial satellite to transfervoice, image and data files to their clients.

K. Zhao · X. Lu (B)Department of Mathematics, School of Science East China University of Science and Technology,Shanghai 200237, Chinae-mail: xwlu@ecust.edu.cn

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J Comb Optim

The difference between classic and agent scheduling problems is that the jobs inagent scheduling problem, unlike classic one, belong to two or more owners whichwe call agents. Each agent owns a set of jobs and tries to optimize its own objectivevalue. All the agents share the same processing resource. In this paper, we study atwo-agent scheduling problem. The two agents are called agent A and B, respectively.All the jobs of agent A and agent B need to be scheduled on identical machines. Ourobjective is to minimize the makespan for agent A while keeping the makespan foragent B under a given value. The given value is indicated as threshold value for agentB. We only consider feasible threshold value in every instance of the problem, i.e.,there always exists a schedule such that the makespan for agent B is no more than thethreshold value. According to the notation introduced by Graham et al. (1979), theproblem is denoted by P||C A

max : C Bmax ≤ Q.

The relevant reference list contains the following papers. Agent scheduling prob-lems were introduced by Baker and Smith (2003) and Agnetis et al. (2004). Theformer paper focused on minimizing a linear combination of the agents’ criteria onsingle machine and gave several complexity results. In the latter paper, the authorsstudied two-agent scheduling problems. They tried to find the optimal cost value forone agent while the other agent’s cost value is constrained. They also tried to findPareto optimal solutions. Agnetis et al. (2007) extended some of those problems intomulti-agent version. Cheng et al. (2008) also provided complexity results for somespecial cases of multi-agent scheduling on single machine. Each agent holds max-formcriterion. For three NP-hard two-agent scheduling problems, Agnetis et al. (2009)developed branch and bound algorithms based on Lagrangian relaxation. An opentwo-agent scheduling problem introduced by Agnetis et al. (2004) was proved to beNP-hard under high multiplicity encoding and solved in pseudo-polynomial time byNg et al. (2006). Leung et al. (2010) confirmed that the problem is NP-hard. Theauthors also considered two-agent scheduling problems with release time and pre-emption on single machine and identical machines. Li and Yuan (2012) and Fan et al.(2013) studied unbounded and bounded parallel-batch scheduling problems with twoagents, respectively. In both papers, the compatibility of agents is considered, namelythe jobs from different agents can(not) be processed in a same batch when these agentsare (in)compatible.

Balasubramanian et al. (2009) considered two-agent scheduling problem on parallelmachines. One agent’s criterion is makespan and the other one’s is the total completiontime. They presented a heuristic algorithm as well as a genetic algorithm. Elvikis et al.(2010) and Elvikis and T’kindt (2012) paid attention to two-agent scheduling problemson uniform machines with equal-size jobs. They both developed efficient algorithmsto enumerate all strict pareto-optimal solutions.

For some NP-hard agent scheduling problems, approximation schemes were pro-vided in the following papers. Cheng et al. (2006) focused on multi-agent schedulingproblem on single machine. Each agent’s objective is to minimize the weighted num-ber of tardy jobs. An FPTAS was given in their paper. Lee et al. (2009) were interestedin a two-agent scheduling problem on single machine in which each agent’s objec-tive is to minimize the total weighted completion time. They proposed an FPTAS byreducing the problem to the restricted shortest path problem. Two-machine flow shopscheduling problems with two agents were considered by Luo et al. (2012). They

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J Comb Optim

gave two FPTASs for the problems F2||C Amax + θC B

max and F2||C Amax : C B

max ≤ Q,respectively.

Zhao and Lu (2013) proposed two FPTASs for two scheduling problems on identicalmachines with two agents. The objective is to minimize the makespan and the totalcompletion time for agent A respectively, subject to an upper bound on the makespanfor agent B. The authors proved that there is no polynomial-time algorithm withworst-case ratio of constant c > 1 for these two problems unless P = NP . Thus, therelaxation of the constraint for agent B has to be allowed.

In this paper, we study approximation algorithms for P||C Amax : C B

max ≤ Q. Thedefinition of (β1, β2, . . . , βg)−approximation algorithm for multi-agent scheduling isintroduced by Lee et al. (2009). By their approximation concept, we give the definitionof (βA, βB)−approximation algorithm for P||C A

max : C Bmax ≤ Q.

Definition 1.1 Let z∗A denote the optimal makespan for agent A and Q denote the

threshold value for agent B in an instance of P||C Amax : C B

max ≤ Q. An algorithmis called a (βA, βB)−approximation algorithm if it provides a schedule such thatzH

A ≤ βAz∗A and zH

B ≤ βB Q, where zHA and zH

B are objective values for agent A andB, respectively.

In this paper, we focus on (β, β)−approximation algorithms for P||C Amax : C B

max ≤Q. We say that a (β, β)−approximation algorithm has a performance ratio of β.Furthermore, the performance ratio β is tight if for any β ′, 0 < β ′ < β, there is aninstance satisfying zH

A > β ′z∗A and zH

B > β ′Q. The performance ratio β is weakly tightif for any β ′, 0 < β ′ < β, there is an instance satisfying zH

A > β ′z∗A or zH

B > β ′Q.

The rest of the paper is organized as follows. In Sect. 3, we consider P||C Amax :

C Bmax ≤ Q and provide an approximation algorithm with performance ratio 2 − 1

m .The ratio is proved to be tight. In Sect. 4, a better approximation algorithm is provided

with performance ratio 1+√17

4 for two-machines version. We claim that the ratio isweakly tight.

2 Preliminaries

For a given instance I , agent A has n A jobs to be processed on m identical machines.We denote the job set of agent A by J A = {

J A1 , J A

2 , . . . , J An A

}. Agent B has nB jobs

to be scheduled. We denote the job set of agent B by J B = {J B

1 , J B2 , . . . , J B

nB

}. Let

n = n A + nB . We call A-jobs the jobs of agent A and B-jobs the jobs of agent B.The processing time of J A

j is pAj for j = 1, 2, . . . , n A and the processing time of J B

j

is pBj for j = 1, 2, . . . , nB . Define PA and PB as the total processing times of all

the A-jobs and all the B-jobs, respectively. Agent B has a threshold value Q. For anarbitrary schedule σ , let C A

max (σ ) be the makespan for agent A and C Bmax (σ ) be the

makespan for agent B.

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3 Problem P||C Amax : C B

max ≤ Q

In this section, we present an approximation algorithm for P||C Amax : C B

max ≤ Q.Given an arbitrary optimal schedule π ′, Zhao and Lu (2013) has shown that ifC A

max (π′) ≤ C B

max (π′), then π ′ can be transformed into a new schedule π such that A-

jobs are all scheduled before B-jobs on each machine without increasing the makespanfor agent A or changing the makespan for agent B, vice versa. Obviously, π is also anoptimal schedule. Let π be the optimal schedule considered in the following analysisin this section.

Based on the characteristic of π mentioned above, our algorithm schedules jobsof one agent on m machines, then schedules jobs of the other agent. Since we cannotmake sure which agents’ jobs should be scheduled before the other agent’s in π , weconstruct two schedules: one with A-jobs scheduled before B-jobs and the other onewith B-jobs scheduled before A-jobs. We choose the better one. Now, we use theextension of the LS rule to determine the assignment of A-jobs and B-jobs. The LSrule is an algorithm for the problem P||Cmax : whenever a machine becomes available,the algorithm schedules job in a job list on this machine. The approximation algorithmA − L S for P||C A

max : C Bmax ≤ Q is described as follows.

Algorithm A–LS

Step 1 Schedule the jobs in J A according to the LS rule, then schedule the jobs inJ B according to the LS rule. Denote the schedule by σ L S

1 .Step 2 Schedule the jobs in J B according to the LS rule, then schedule the jobs in

J A according to the LS rule. Denote the schedule by σ L S2

.

Step 3 If C Bmax

(σ L S

1

)≤

(2 − 1

m

)Q, then choose σ L S

1 . Otherwise, choose σ L S2 .

In Step 1 and Step 2, it takes O(n) times to assign A-jobs and B-jobs. Therefore,the complexity of algorithm A − L S is O(n). In the rest of this section, we showthat algorithm A − L S has performance ratio

(2 − 1

m

)for P||C A

max : C Bmax ≤ Q. We

distinguish two cases where C Amax (π) ≤ C B

max (π) and C Amax (π) ≥ C B

max (π) to provethe result. π is the optimal schedule for P||C A

max : C Bmax ≤ Q mentioned above.

For the case where C Amax (π) ≤ C B

max (π), we propose the following lemma.

Lemma 3.1 If C Amax (π) ≤ C B

max (π), then C Amax

(σ L S

1

)/C A

max (π) ≤ 2 − 1m and

C Bmax

(σ L S

1

)/C B

max (π) ≤ 2 − 1m .

Proof By Step 1 of algorithm A − L S, A-jobs are assigned before B-jobs accordingto the LS rule in σ L S

1 . Graham (1966) proved that the LS rule has worst-case ratio2 − 1

m for P||Cmax . Thus, we have C Amax (σ

L S1 ) ≤ (

2 − 1m

)C A

max (π).We now consider the second inequality about agent B. Under condition that

C Amax (π) ≤ C B

max (π), it indicates that B-jobs are scheduled after A-jobs on eachmachine in π . Hence, we have (PA + PB)/m ≤ C B

max (π). Let J Bl be the last com-

pleted B-job in σ L S1 and its starting time is s B

l . Therefore, the above arguments yield

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J Comb Optim

that

C Bmax

(σ L S

1

)= s B

l + pBl ≤ PA + PB − pB

l

m+ pB

l = PA + PB

m+

(1 − 1

m

)pB

l

≤ C Bmax (π) +

(1 − 1

m

)C B

max (π) =(

2 − 1

m

)C B

max (π).

�By a similar way, we can deduce the following lemma for the case where C A

max (π) ≥C B

max (π).

Lemma 3.2 If C Amax (π) ≥ C B

max (π), then C Amax

(σ L S

2

)/C A

max (π) ≤ 2 − 1m and

C Bmax

(σ L S

2

)/C B

max (π) ≤ 2 − 1m .

�Theorem 3.1 For P||C A

max : C Bmax ≤ Q, the performance ratio of algorithm A − L S

is 2 − 1m . This ratio is tight.

Proof We have known that C Amax

(σ L S

1

) ≤ (2 − 1

m

)C A

max (π) by Step 1 of algo-rithm A − L S and Graham (1966). If C B

max

(σ L S

1

) ≤ (2 − 1

m

)Q, then σ L S

1 is a(2 − 1

m , 2 − 1m

) −approximation solution. If C Bmax

(σ L S

1

)>

(2 − 1

m

)Q, then we

apply Lemma 3.1 to conclude that C Amax (π) > C B

max (π). It follows from Lemma3.2 and C B

max (π) ≤ Q that σ L S2 is a

(2 − 1

m , 2 − 1m

) −approximation solution. There-fore, algorithm A − L S has performance ratio 2 − 1

m .Now we give an instance to show that the ratio is tight. In this instance,

agent A has job set J A ={

J A1 , J A

2 , . . . , J Am2−m+1

}. The processing time of

J Aj

(j = 1, 2, . . . , m2 − m

)is pA

j = 1 and the processing time of J Am2−m+1

is pAm2−m+1

= m. Agent B has job set J B ={

J B1 , J B

2 , . . . , J Bm2−m+1

}. The

processing time of J Bj

(j = 1, 2, . . . , m2 − m + 1

)is pB

j = LpAj , where L is a

number much larger than m. Let threshold value Q be (L + 1)m. Denote by σ

the schedule generated by algorithm A − L S. It is shown that σ = σ L S1 since

C Bmax

(σ L S

1

) = (L + 1)(2m − 1) − m < (L + 1)(2m − 1) = (2 − 1

m

)Q. More-

over, C Amax (π) = m. Therefore, we have that C A

max (σ )/C Amax (π) = 2m−1

m = 2 − 1m

and C Bmax (σ )/Q = ((L + 1)(2m − 1) − m)/(L + 1)m = 2 − 1

m − 1L+1 → 2 − 1

m forL → ∞. �

4 An approximation algorithm for m = 2

In this section, we consider a special case where the number of machines is two. Theproblem is denoted as P2||C A

max : C Bmax ≤ Q. We provide a new approximation

algorithm with a better performance ratio.Recall that in an instance of P||C A

max : C Bmax ≤ Q algorithm A−L S constructs two

schedules which is based on the LS rule and chooses a(2 − 1

m , 2 − 1m

) −approximation

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J Comb Optim

Fig. 1 An example to show that any algorithm only constructing σ L PT1 and σ L PT

2 will not have perfor-

mance ratio less than√

3+12

solution. When m = 2, the algorithm has performance ratio 32 which exactly matches

the worst-case ratio of the LS rule for P2||Cmax . Therefore, if we want to achieve abetter performance ratio, we should design an algorithm which is not based on the LSrule.

It is well known that the LPT rule is one of the best approximation algorithms forP2||Cmax with worst-case ratio 7

6 (see Graham 1969). The LPT rule is described asfollows: whenever a machine becomes available, the algorithm schedules the largestunscheduled job on the machine. In an instance of P2||C A

max : C Bmax ≤ Q, let σ L PT

1and σ L PT

2 denote the schedules constructed by a similar way to Step 1 and Step 2of algorithm A − L S, where we replace the LS rule with the LPT rule, respectively.Accordingly, there is a question whether or not we can find a

( 76 , 7

6

) −approximationsolution between σ L PT

1 and σ L PT2 for P2||C A

max : C Bmax ≤ Q. This question is even-

tually solved in the negative by a counterexample. The example is stated as follows.Agent A has two jobs with unit processing times and agent B has two jobs withprocessing times pB

1 = √3 + 1 and pB

2 = √3 − 1. The threshold value for agent B

is Q = √3 + 1. We have σ L PT

1 , σ L PT2 and optimal schedule π (see Fig. 1).

It follows from Fig. 1 that C Amax

(σ L PT

2

)/C A

max (π) = C Bmax

(σ L PT

1

)/Q =

√3+12 .

Consequently, this example implies that any algorithm which only constructs σ L PT1

and σ L PT2 will not have performance ratio less than

√3+12 for P2||C A

max : C Bmax ≤ Q.

So, we will consider another two schedules obtained by modifying σ L PT1 and σ L PT

2respectively in order to design an approximation algorithm with performance ratiowhich is closer to 7

6 .Notice that σ L PT

2 is actually close enough to the optimal schedule π in the exampleabove. We can get π by putting the two A-jobs J A

1 , J A2 before the B-job J B

2 on thesecond machine in σ L PT

2 .Based on this idea, our algorithm will construct a new schedule σ L PT

3 from σ L PT1

by the following method. First we put all the B-jobs before all the A-jobs on eachmachine in σ L PT

1 . Next we adjust the schedule if necessary. When the starting time ofthe first A-job on the first machine is later than the completion time of the last job onthe second machine, we move all the A-jobs on the first machine after all the jobs onthe second machine. When the starting time of the first A-job on the second machineis later than the completion time of the last job on the first machine, we move all theA-jobs on the second machine after all the jobs on the first machine. Similarly, thealgorithm constructs another new schedule σ L PT

4 from σ L PT2 . Finally, we choose the

best schedule from four schedules.

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J Comb Optim

To describe the algorithm formally, we will introduce some notations as follows.For a given schedule σ , let Ai (σ ) and Bi (σ ) denote the job set of A-jobs and B-jobsassigned to the i-th machine for i = 1, 2, respectively. Define S(J ) as the starting timeof the first started job and C(J ) as the completion time of the last completed job in agiven job set J . An approximation algorithm A − L PT E for P2||C A

max : C Bmax ≤ Q

is described as follows.

Algorithm A − L PT E

Step 1 Schedule the jobs in J A according to the LPT rule, then schedule the jobs in J B accordingto the LPT rule. Denote the schedule by σ L PT

1 .

Step 2 Schedule the jobs in J B according to the LPT rule, then schedule the jobs in J A accordingto the LPT rule. Denote the schedule by σ L PT

2 .

Step 3 Construct schedule σ L PT3 from σ L PT

1 as follows.

Step 3.1 Exchange A1

(σ L PT

1

)and B1

(σ L PT

1

)on the first machine, and exchange A2

(σ L PT

1

)and

B2

(σ L PT

1

)on the second machine. Denote the new schedule by σ̂ L PT

3 .

Step 3.2 If A1

(σ L PT

1

)�= φ and S

(A1

(σ L PT

1

))> C

(A2

(σ L PT

1

) ⋃ B2

(σ L PT

1

))in σ̂ L PT

3 ,

then move A1

(σ L PT

1

)after A2

(σ L PT

1

)on the second machine to get schedule σ L PT

3 . If

A2

(σ L PT

1

)�= φ and S

(A2(σ L PT

1 ))

> C(A1

(σ L PT

1

)⋃ B1

(σ L PT

1

))in σ̂ L PT

3 , then

move A2

(σ L PT

1

)after A1

(σ L PT

1

)on the first machine to get schedule σ L PT

3 .

Otherwise, let σ L PT3 = σ̂ L PT

3 .

Step 4 Construct schedule σ L PT4 from σ L PT

2 as follows.

Step 4.1 Exchange B1

(σ L PT

2

)and A1

(σ L PT

2

)on the first machine, and exchange B2

(σ L PT

2

)and

A2

(σ L PT

2

)on the second machine. Denote the new schedule by σ̂ L PT

4 .

Step 4.2 If B1

(σ L PT

2

)�= φ and S

(B1

(σ L PT

2

))> C

(A2

(σ L PT

2

)⋃ B2

(σ L PT

2

))in σ̂ L PT

4 ,

then move B1

(σ L PT

2

)after B2

(σ L PT

2

)on the second machine to get schedule σ L PT

4 . If

B2

(σ L PT

2

)�= φ and S

(B2

(σ L PT

2

))> C

(A1

(σ L PT

2

)⋃ B1

(σ L PT

2

))in σ̂ L PT

4 , then

move B2

(σ L PT

2

)after B1

(σ L PT

2

)on the first machine to get schedule σ L PT

4 . Otherwise,

let σ L PT4 = σ̂ L PT

4 .

Step 5 Choose the best schedule with the makespan no more than 1+√17

4 Q for agent B.

In Step 1 and Step 2, the LPT rule needs no more than O(n log n) times to assignA-jobs and B-jobs. In Step 3 and Step 4, all operations need at most O(n) times. InStep 5, only constant time is needed to find the best schedule. Therefore, the totalrunning time of algorithm A − L PT E is no more than O(n log n).

In the following part of this section, we show the performance ratio of algorithm

A − L PT E is no more than α = 1+√17

4 ≈ 1.28. Let π ′ be an optimal solution forP2||C A

max : C Bmax ≤ Q which is also Pareto optimal, i.e., there is no optimal solution π̃

such that C Bmax (π̃) < C B

max (π′). By an analogous method in Sect. 3, we can transform

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J Comb Optim

π ′ into an optimal schedule π such that one agent’s jobs are assigned before theother agent’s jobs on each machine. Let π be the optimal schedule considered in thefollowing analysis in this section.

To study the performance ratio of algorithm A − L PT E , we again distinguishbetween two cases for π where C A

max (π) ≤ C Bmax (π) and C A

max (π) ≥ C Bmax (π).

Since the proofs for the two situations are quite similar, we mainly discuss the situationwhere C A

max (π) ≤ C Bmax (π). In this situation, it is obvious that A-jobs are assigned

before B-jobs on each machine in π , and we can easily get (PA + PB)/2 ≤ C Bmax (π).

We will show that schedule σ L PT1 satisfies C A

max

(σ L PT

1

) ≤ αC Amax (π) and

C Bmax

(σ L PT

1

) ≤ αC Bmax (π) in most cases. In the only case where the last completed

job in σ L PT1 is the largest B-job, we will prove that σ L PT

4 satisfies C Amax

(σ L PT

4

) ≤αC A

max (π) and C Bmax

(σ L PT

4

) ≤ αC Bmax (π) if C B

max

(σ L PT

1

)> αC B

max (π).

Observation 4.1 C Amax

(σ L PT

1

) ≤ 76 C A

max (π).

Proof By Step 1 of algorithm A−L PT E , A-jobs are assigned before B-jobs accordingto the LPT rule. Hence, we have C A

max

(σ L PT

1

) ≤ 76 C A

max (π) by Graham (1969). �Observation 4.1 indicates that the cost value for agent A in σ L PT

1 never exceeds α

times the optimal value. For the cost value for agent B in σ L PT1 , we have the lemmas

below.

Lemma 4.1 If the last completed job is an A-job in σ L PT1 , then C B

max

(σ L PT

1

) ≤76 C B

max (π).

Proof Since the last completed job belongs to agent A, the makespan for agent Ais no less than the makespan for agent B in σ L PT

1 . Therefore, C Bmax

(σ L PT

1

) ≤C A

max

(σ L PT

1

) ≤ 76 C A

max (π) ≤ 76 C B

max (π). �Now we assume that the last completed job is a B-job in σ L PT

1 . We initially supposethat all the B-jobs have been indexed in the decreasing order of their processing times,i.e., pB

1 ≥ pB2 ≥ · · · ≥ pB

nB. Let J B

l be the last completed job in σ L PT1 .

Lemma 4.2 If l ≥ 2, then C Bmax

(σ L PT

1

) ≤ 54 C B

max (π).

Proof When l ≥ 3, there are at least two B-jobs with processing time no less than pBl

assigned to the same machine in π . This implies C Bmax (π) ≥ 2pB

l . Hence, we get

C Bmax

(σ L PT

1

)= s B

l + pBl ≤ PA + PB − pB

l

2+ pB

l = 1

2(PA + PB) + 1

2pB

l

≤ C Bmax (π) + 1

4C B

max (π) = 5

4C B

max (π).

When l = 2, the result above remains valid in the case where J B1 and J B

l are assignedto the same machine in π . Then we should discuss the case where J B

1 and J Bl are

assigned to different machines in π . In this case, we have C Bmax (π) ≥ C A

max (π)+ pBl .

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J Comb Optim

If there is at least one A-job with processing time no less than pBl , then C B

max (π) ≥C A

max (π) + pBl ≥ 2pB

l . In a similar way, we have C Bmax

(σ L PT

1

) ≤ 54 C B

max (π).If every A-job has processing time smaller than pB

l , then in σ L PT1 the difference

between the two workloads for agent A on both machines is also smaller than pBl

since this difference is no larger than the processing time of the last completed A-job. Due to the LPT rule, we deduce that J B

1 and J Bl will be assigned to different

machines in σ L PT1 . Thus, C B

max

(σ L PT

1

) = C Amax

(σ L PT

1

) + pBl . Moreover, we have

C Bmax (π) ≥ C A

max (π) + pBl . Therefore, C B

max

(σ L PT

1

) = C Amax

(σ L PT

1

) + pBl ≤

76 C A

max (π) + pBl ≤ 7

6 C Bmax (π). �

Notice that we have proved that C Bmax

(σ L PT

1

) ≤ 54 C B

max (π) for l ≥ 2. So, we onlyhave to consider the situation where the last completed job in σ L PT

1 is the largest B-job, namely J B

1 (i.e., l = 1). If C Bmax

(σ L PT

1

) ≤ αC Bmax (π), then σ L PT

1 is exactly theschedule that we want to find by Observation 4.1. This result suggests us to considerthe left situation where l = 1 and C B

max

(σ L PT

1

)> αC B

max (π).

Lemma 4.3 If l = 1 and C Bmax

(σ L PT

1

)> αC B

max (π), then C Amax

(σ L PT

4

) ≤αC A

max (π) and C Bmax

(σ L PT

4

) ≤ αC Bmax (π).

Proof Since l = 1, J B1 is the first scheduled as well as the last completed B-job in

σ L PT1 . This implies that all the B-jobs in J B\{J B

1 } are assigned to the same machineand J B

1 is assigned to the other machine in σ L PT1 . Additionally, we will show that

each B-job in J B\{J B1 } is completed earlier than J B

1 in σ L PT1 . In fact, suppose there

is a B-job in J B\{J B1 } completed at the same time as J B

1 in σ L PT1 . It follows that

C Bmax

(σ L PT

1

) = PA+PB2 ≤ C B

max (π). This is a contradiction with C Bmax

(σ L PT

1

)>

αC Bmax (π). Let �B be the total processing times of the B-jobs in J B\{J B

1 }. Thus,�B < pB

1 .Note that σ L PT

4 is constructed from σ L PT2 by Step 4 of algorithm A − L PT E .

Without loss of generality, we assume that J B1 is scheduled on the first machine in

both σ L PT2 and π . It is obvious that all the B-jobs in J B\{J B

1 } are assigned to thesecond machine in σ L PT

2 . Let xA be the total processing times of all the A-jobs onthe first machine in σ L PT

2 and x∗A be the total processing times of all the A-jobs on

the first machine in π . Assume that J Al ′ is the last completed A-job in σ L PT

2 and itsstarting time is s A

l ′ .First we present the preliminary results as follows. Since J B

1 is scheduled after theA-jobs on the first machine with total processing times of x∗

A in π , we obtain a lowerbound for C B

max (π) that

C Bmax (π) ≥ x∗

A + pB1 . (1)

Moreover, since J B1 is the first scheduled B-job in σ L PT

1 , it has to be assigned to themachine with smaller workload for agent A by Step 1 of algorithm A − L PT E . Thus,we have

1

2PA + pB

1 ≥ C Bmax

(σ L PT

1

)> αC B

max (π). (2)

123

J Comb Optim

It follows from (1) and (2) that

1

2PA + pB

1 > α(x∗A + pB

1 ). (3)

By C Bmax (π) ≥ PA+PB

2 , we have

1

2PA + pB

1 > α · PA + PB

2. (4)

Moreover, it is not hard to find that

C Amax (π) = PA − x∗

A. (5)

In fact, we only need to prove PA − x∗A ≥ x∗

A. If PA − x∗A < x∗

A, then we haveC B

max (π) ≥ x∗A+pB

1 ≥ 12 PA+pB

1 > αC Bmax (π)by (1) and (2) which is a contradiction.

Next we will show that C Amax

(σ L PT

4

) ≤ αC Amax (π) and C B

max

(σ L PT

4

) ≤αC B

max (π) by analyzing the schedule σ L PT2 . To prove this result, we distinguish two

cases based on whether or not there is A-jobs assigned to the first machine in σ L PT2 .

Case 1 xA = 0, i.e., there is no A-jobs assigned to the first machine in σ L PT2 .

Case 1.1 C Amax

(σ L PT

2

) ≤ C Bmax

(σ L PT

2

).(see Fig. 2)

Figure 2 indicates that pB1 ≥ PA. By (3), we have 1

2 PA > αx∗A + (α − 1)pB

1 .Combining the two inequalities yields 1

2 PA > αx∗A + (α − 1)PA. So, x∗

A < 3−2α2α

PA.Moreover, C A

max

(σ L PT

4

) = PA by Step 4 of algorithm A − L PT E and C Amax (π) =

PA − x∗A by (5). Therefore, we have

C Amax

(σ L PT

4

)

C Amax (π)

= PA

PA − x∗A

≤ PA

PA − 3−2α2α

PA= 2α

4α − 3≈ 1.21

and C Bmax

(σ L PT

4

) = pB1 ≤ C B

max (π).Case 1.2. C A

max

(σ L PT

2

)> C B

max

(σ L PT

2

)(see Fig. 3).

We see that the last completed A-job J Al ′ has the minimum processing time of all

the A-jobs by the LPT rule. We consider two possible cases on the first machine in π .Case 1.2.1. There is at least one A-job on the first machine in π , i.e., x∗

A > 0.

Fig. 2 xA = 0 and

C Amax

(σ L PT

2

)≤

C Bmax

(σ L PT

2

)in σ L PT

2

Fig. 3 xA = 0 and

C Amax

(σ L PT

2

)>

C Bmax

(σ L PT

2

)in σ L PT

2

123

J Comb Optim

It is obvious that x∗A ≥ pA

l ′ . By Fig. 3, we have �B + PA − pAl ′ ≤ pB

1 . Combiningthe two inequalities yields

�B + PA − x∗A ≤ pB

1 . (6)

By (3) and (6), we get that 12 PA > αx∗

A + (α − 1)pB1 ≥ αx∗

A + (α − 1)(�B +PA − x∗

A) ≥ αx∗A + (α − 1)(PA − x∗

A). It follows that x∗A < ( 3

2 − α)PA. Moreover,C A

max

(σ L PT

4

) = PA by Step 4 of algorithm A − L PT E and C Amax (π) = PA − x∗

A by(5). Consequently, we have

C Amax

(σ L PT

4

)

C Amax (π)

= PA

PA − x∗A

≤ PA

PA − ( 32 − α)PA

= α.

By (1), (6) and Step 4 of algorithm A − L PT E , we immediately get

C Bmax

(σ L PT

4

)≤ PA + �B ≤ pB

1 + x∗A ≤ C B

max (π).

Case 1.2.2. There is no A-job on the first machine in π , i.e., x∗A = 0.

It is obvious that C Amax

(σ L PT

4

) = PA = C Amax (π).

Then we analyze the makespan for agent B in σ L PT4 . If pB

1 ≥ PA, then the jobs inJ B\{J B

1 } are still scheduled on the second machine in σ L PT4 by Step 4.2 of algorithm

A − L PT E . This implies that C Bmax

(σ L PT

4

) = PA + �B . By (2) and the assumptionof pB

1 ≥ PA, we have αC Bmax (π) < 1

2 PA + pB1 ≤ 3

2 pB1 . It turns out obviously that

pB1 > 2

3αC Bmax (π). Therefore, we obtain

C Bmax

(σ L PT

4

)= PA + �B = 2 · PA + PB

2− pB

1 ≤ 2C Bmax (π)

− 2

3αC B

max (π) ≈ 1.15 C Bmax (π).

If pB1 < PA, then the jobs in J B\{J B

1 } will be moved after J B1 on the first machine

in σ L PT4 by algorithm A − L PT E . This means C B

max

(σ L PT

4

) = PB . By (2) and theassumption of pB

1 < PA, αC Bmax (π) < 1

2 PA + pB1 < 3

2 PA. It turns out obviously thatPA > 2

3αC Bmax (π). Therefore, we have

C Bmax

(σ L PT

4

)= PB = 2 · PA + PB

2− PA ≤

(2 − 2

3α

)C B

max (π) ≈ 1.15 C Bmax (π).

In Case 1, we have proved that the lemma is true when there is no A-jobs assignedto the first machine in σ L PT

2 . Now we move to the other case.Case 2 xA > 0, i.e., there exists at least one A-job assigned to the first machine inσ L PT

2 .Case 2.1 The last completed A-job J A

l ′ is assigned to the first machine in σ L PT2 (see

Fig. 4).

123

J Comb Optim

Fig. 4 xA > 0 and the lastcompleted A-job J A

l′ is assigned

to the first machine in σ L PT2

In Fig. 4, the job set J A3 contains the A-jobs which are scheduled on the first machine

in σ L PT2 . Let J A

s be the first scheduled A-job in J A3 . We denote J A

1 as the job set thatincludes the A-jobs such that their start times are earlier than the start time of J A

s inσ L PT

2 . J A2 = J A\(J A

1

⋃ J A3 ). Let �A

1 = ∑J A

j ∈J A1

pAj and �A

2 = ∑J A

j ∈J A2

pAj .

Now we will show that C Amax

(σ L PT

4

) = PA − xA. We will prove it by a contra-diction. Suppose that PA − xA < xA. Since �B < pB

1 , the largest A-job has to beassigned to the second machine in σ L PT

2 by Step 2 of algorithm A − L PT E . Thismeans that PA − xA ≥ pA

j for every J Aj ∈ J A. By the assumption of PA − xA < xA,

there are at least two A-jobs in J A3 . Moreover, J A

l ′ is the smallest job in J A3 by the LPT

rule. Hence, there are at least three A-jobs with processing time no less than pAl ′ . It

indicates that C Amax (π) ≥ 2pA

l ′ . In addition, pB1 + xA − pA

l ′ = s Al ′ ≤ �B + PA − xA by

the LPT rule. Therefore, pAl ′ ≥ pB

1 −�B +2xA − PA > pB1 −�B by the assumption of

PA − xA < xA. Then we turn to schedule σ L PT1 . The construction of σ L PT

1 figures outthat the difference between its workloads on both machines is no greater than pB

1 −�B .Denote by δ the difference. Hence, δ ≤ pB

1 − �B < pAl ′ . Combining the two inequal-

ities of C Amax (π) ≥ 2pA

l ′ and δ < pAl ′ , we have C B

max

(σ L PT

1

) = PA+PB−δ2 + δ <

12 (PA + PB) + 1

2 pAl ′ ≤ C B

max (π) + 14 C A

max (π) ≤ 54 C B

max (π) < αC Bmax (π). This is a

contradiction with C Bmax

(σ L PT

1

)> αC B

max (π). Thus, PA − xA ≥ xA. It follows theresult.

Moreover, we have by (3) that x∗A < 1

α( 1

2 PA − (α − 1)pB1 ). Since J A

l ′ is scheduledon the first machine in σ L PT

2 , we obtain pB1 + xA ≥ PA − xA. From (5), we get

C Amax (π) = PA − x∗

A. Therefore, we deduce that

C Amax

(σ L PT

4

)

C Amax (π)

= PA − xA

PA − x∗A

≤ PA − xA

PA − 1α

( 12 PA − (α − 1)pB

1

) = PA − xA(1 − 1

2α

)PA + (

1 − 1α

)pB

1

≤ PA − xA(1 − 1

2α

)PA + (

1 − 1α

)(PA − 2xA)

= PA − xA(2 − 3

2α

)PA − 2

(1 − 1

α

)xA

= 1 − xAPA(

2 − 32α

) − 2(1 − 1

α

) xAPA

≤ 1 − 0(2 − 3

2α

) − 2(1 − 1

α

) · 0= 2α

4α − 3≈ 1.21.

Next we discuss agent B. We have shown that pB1 + xA ≥ PA − xA. By Step 4.2 of

algorithm A− L PT E , the jobs in J B\{J B1 } are still scheduled on the second machine

in σ L PT4 . It is obvious that

123

J Comb Optim

C Bmax

(σ L PT

4

)≤ xA + pB

1 . (7)

When xA ≤ x∗A, it follows from (1) and (7) that C B

max

(σ L PT

4

) ≤ xA + pB1 ≤

x∗A + pB

1 ≤ C Bmax (π). When xA > x∗

A, we distinguish two subcases based on x∗A.

(i). 2x∗A > �B + PA − pB

1 .

Firstly, we will prove that there are at least two A-jobs on the first machine inσ L PT

2 . In fact, suppose that there is only one A-job on the first machine in σ L PT2 .

So, pAl ′ = xA > x∗

A. By the definition of J A1 and the LPT rule, each job in J A

1 hashigher priority than the jobs in J A

3 . Furthermore, J Al ′ is the smallest job in J A

3 . Itturns out that pA

j > x∗A for any J A

j ∈ J A1

⋃ J A3 . This implies that only the jobs in

J A2 may be scheduled on the first machine in π . Hence, �A

2 ≥ x∗A. Thus, we have

pB1 ≤ �B + �A

1 = �B + PA − xA − �A2 ≤ �B + PA − pA

l ′ − x∗A. It follows

that pAl ′ + x∗

A ≤ �B + PA − pB1 . Combining this with the assumption of 2x∗

A >

�B + PA − pB1 yields pA

l ′ < x∗A. This is a contradiction with pA

l ′ > x∗A.

The above result implies that pAl ′ ≤ 1

2 xA. So, we have �B + PA − xA ≥ s Al ′ =

pB1 + xA − pA

l ′ ≥ pB1 + xA − 1

2 xA = pB1 + 1

2 xA. It follows that �B + PA − pB1 ≥ 3

2 xA.Combining the inequality with the assumption of 2x∗

A > �B + PA − pB1 , we deduce

that

xA <4

3x∗

A. (8)

Moreover, it follows from (3) that x∗A < 1

α( 1

2 PA + pB1 ) − pB

1 . Since 2x∗A > �B +

PA − pB1 ≥ PA − pB

1 , we have

PA <2 − α

α − 1pB

1 . (9)

By (1), (7), (8), (9) and x∗A < 1

α

( 12 PA + pB

1

) − pB1 , we obtain

C Bmax

(σ L PT

4

)

C Bmax (π)

≤ xA + pB1

x∗A + pB

1

≤43 x∗

A + pB1

x∗A + pB

1

= 4

3−

13 pB

1

x∗A + pB

1

≤ 4

3−

13 pB

11α

( 12 PA + pB

1

) − pB1 + pB

1

= 4

3−

13αpB

112 PA + pB

1

≤ 4

3−

13αpB

112 · 2−α

α−1 pB1 + pB

1

= 2 − 2

3α ≈ 1.15.

(i i). 2x∗A ≤ �B + PA − pB

1 .If there is only one A-job on the first machine in σ L PT

2 , then pAl ′ = xA. We

have proved that �A2 ≥ x∗

A above. By the definition of π , we have C Amax (π) ≤

C Bmax (π) ≤ min{x∗

A + PB, PA − x∗A + PB}. In addition, C A

max (π) = PA − x∗A by

123

J Comb Optim

(5). Thus, one can see that PA − x∗A ≤ x∗

A + PB . Combining the two inequalities�A

2 ≥ x∗A and PA − x∗

A ≤ x∗A + PB , we obtain PB ≥ PA − 2�A

2 . Hence, PA + PB =(pA

l ′ + �A1 + �A

2 ) + PB ≥ (pAl ′ + �A

1 ) + PA − �A2 = 2(pA

l ′ + �A1 ). Let PA + PB =

2(pAl ′ + �A

1 + θ) where θ ≥ 0. By (4), we have 12 PA + pB

1 > α(pAl ′ + �A

1 + θ). Thisleads to pB

1 > (α − 12 )(pA

l ′ +�A1 )− 1

2�A2 +αθ by PA = pA

l ′ +�A1 +�A

2 . Therefore,we get

�B = PB − pB1 = PA + PB − (pA

l ′ + �A1 + �A

2 + pB1 )

= 2(pAl ′ + �A

1 + θ) − (pAl ′ + �A

1 + �A2 + pB

1 )

= pAl ′ + �A

1 + 2θ − �A2 − pB

1 ≤ pAl ′ + �A

1 + 2θ − �A2

−((

α − 1

2

)(pA

l ′ + �A1 ) − 1

2�A

2 + αθ

)

=(

3

2− α

)(pA

l ′ + �A1 ) − 1

2�A

2 + (2 − α) θ. (10)

By (7), (10) and PA + PB = 2(pAl ′ + �A

1 + θ), we have

C Bmax

(σ L PT

4

)

C Bmax (π)

≤ pAl ′ + pB

1

(PA + PB)/2≤ pA

l ′ + �B + �A1

(PA + PB)/2

≤ pAl ′ + �A

1 + ( 32 − α

)(pA

l ′ + �A1 ) − 1

2�A2 + (2 − α) θ

pAl ′ + �A

1 + θ

= 5

2− α −

12 (�A

2 + θ)

pAl ′ + �A

1 + θ≤ 5

2− α ≈ 1.22.

If there are at least two A-jobs on the first machine in σ L PT2 , then pA

l ′ ≤ 12 xA. Since

pB1 + xA − pA

l ′ = s Al ′ ≤ �B + PA − xA = PB − pB

1 + PA − xA, we have

2pB1 ≤ PA + PB − 2xA + pA

l ′ ≤ PA + PB − 3

2xA. (11)

In addition, 2pB1 > (α − 1)PA + αPB by (4). Thus, we have (α − 1)PA + αPB <

PA + PB − 32 xA. This leads to

xA <2

3

((2 − α)PA − (α − 1)PB

). (12)

By (4) and pB1 ≤ PB , we have

PA <2 − α

α − 1PB . (13)

123

J Comb Optim

Fig. 5 xA > 0 and the lastcompleted A-job J A

l′ is assigned

to the second machine in σ L PT2

It follows from (7), (11), (12) and (13) that

C Bmax

(σ L PT

4

)

C Bmax (π)

≤ xA + pB1

(PA + PB)/2≤ 2xA + PA + PB − 3

2 xA

PA + PB= 1 +

12 xA

PA + PB

≤ 1 +12 · 2

3

((2 − α) PA − (α − 1) PB

)

PA + PB= 1 + 2 − α

3−

13 PB

PA + PB

≤ 5 − α

3−

13 PB

2−αα−1 PB + PB

= 2 − 2

3α ≈ 1.15.

Case 2.2 The last completed A-job J Al ′ is assigned to the second machine in σ L PT

2(see Fig. 5).

In Fig. 5, job set F A contains the A-jobs scheduled on the first machine in σ L PT2 .

First we analyze the makespan for agent A. It is clear that xA ≤ PA −xA. Therefore,

C Amax

(σ L PT

4

)= PA − xA. (14)

To prove that C Amax

(σ L PT

4

) ≤ αC Amax (π), we distinguish two cases where s A

l ′ ≤ pB1

and s Al ′ > pB

1 .If s A

l ′ ≤ pB1 , then pA

l ′ ≥ xA and pB1 ≥ s A

l ′ = �B + PA − xA − pAl ′ . When xA ≥ x∗

A,it is clear that C A

max

(σ L PT

4

) = PA − xA ≤ PA − x∗A = C A

max (π) by (5) and (14).When xA < x∗

A, there is at least one A-job in J A\F A which is assigned to the firstmachine in π . By the LPT rule, we know that J A

l ′ is the smallest job in J A\F A. Hence,x∗

A ≥ pAl ′ . This implies that pB

1 ≥ �B + PA − xA − pAl ′ ≥ PA − xA − x∗

A. Moreover,we have pB

1 < 1α−1

( 12 PA − αx∗

A

)by (3). Combining the two inequalities, we obtain

x∗A <

(3

2− α

)PA + (α − 1)xA. (15)

It follows from (5), (14) and (15) that

C Amax

(σ L PT

4

)

C Amax (π)

= PA − xA

PA − x∗A

≤ PA − xA

PA − (( 32 − α

)PA + (α − 1)xA

)

= 1 − xAPA

α − 12 − (α − 1) xA

PA

≤ 1 − 0

α − 12 − (α − 1) · 0

= α.

123

J Comb Optim

If s Al ′ > pB

1 , then there exists an A-job J Aj ∈ J A

3 with higher priority than J Al ′ by

the LPT rule. In this case, we derive that pAl ′ ≤ pA

j ≤ xA. This leads to pB1 + xA ≥

s Al ′ = �B + PA − xA − pA

l ′ ≥ �B + PA − 2xA. Hence, pB1 ≥ PA − 3xA. Moreover,

we have pB1 < 1

α−1

( 12 PA − αx∗

A

)by (3). Combining the two inequalities, we obtain

x∗A <

(3

2α− 1

)PA + 3

(1 − 1

α

)xA. (16)

It follows from (5), (14) and (16) that

C Amax

(σ L PT

4

)

C Amax (π)

= PA − xA

PA − x∗A

≤ PA − xA

PA − (( 32α

− 1)

PA + 3(1 − 1

α

)xA

)

= 1 − xAPA(

2 − 32α

) − 3(1 − 1

α

) xAPA

≤ 1 − 0(2 − 3

2α

) − 3(1 − 1

α

) · 0

= 2α

4α − 3≈ 1.21.

Next we consider agent B. We discuss two cases where xA + pB1 < PA − xA and

xA + pB1 ≥ PA − xA.

If xA + pB1 < PA − xA, then the jobs in J B\{J B

1 } will be moved after J B1 on

the first machine in σ L PT4 by Step 4.2 of algorithm A − L PT E . This indicates that

C Bmax

(σ L PT

4

) = xA + PB . It follows from (4) that pB1 > α−1

2 PA + α2 PB . By the

assumption of xA + pB1 ≤ PA − xA, we obtain pB

1 ≤ PA − 2xA. Combining thetwo inequalities yields xA < 3−α

4 PA − α4 PB . Moreover, we have PB

PA≤ 2 since

12 PB ≤ pB

1 ≤ PA − 2xA ≤ PA. Thus, we obtain that

C Bmax

(σ L PT

4

)

C Bmax (π)

≤ xA + PB

(PA + PB)/2≤

3−α4 PA − α

4 PB + PB

(PA + PB)/2

=3−α

2 + 4−α2

PBPA

1 + PBPA

≤3−α

2 + 4−α2 · 2

1 + 2= 11 − 3α

6≈ 1.19.

If xA + pB1 ≥ PA − xA, then the jobs in J B\{J B

1 } are still scheduled on the secondmachine in σ L PT

4 by Step 4.2 of algorithm A − L PT E . In this case, we derive thatC B

max

(σ L PT

4

) = PA − xA + �B . The assumption of xA + pB1 ≥ PA − xA leads to

2xA ≥ PA − pB1 ≥ PA − PB . It follows that 2xA ≥ 1

2 (PA − PB). Moreover, we have2pB

1 > 14 PA + 5

4 PB by (4) and α > 54 . Thus,

C Bmax

(σ L PT

4

)

C Bmax (π)

≤ PA − xA + �B

(PA + PB)/2= PA − xA + PB − pB

1

(PA + PB)/2= 2 − 2pB

1 + 2xA

PA + PB

≤ 2 −14 PA + 5

4 PB + 12 (PA − PB)

PA + PB= 2 −

34 PA + 3

4 PB

PA + PB= 5

4.

123

J Comb Optim

Therefore, we obtain the conclusion as desired. �By Observation 4.1 and Lemmas 4.1, 4.2 and 4.3, we conclude the following the-

orem.

Theorem 4.1 If C Amax (π) ≤ C B

max (π), then C Amax

(σ L PT

1

) ≤ αC Amax (π) and

C Bmax

(σ L PT

1

) ≤ αC Bmax (π), or C A

max

(σ L PT

4

) ≤ αC Amax (π) and C B

max

(σ L PT

4

) ≤αC B

max (π).

�For the situation where C A

max (π) ≥ C Bmax (π), we have the following result by the

similar method.

Theorem 4.2 If C Amax (π) ≥ C B

max (π), then C Amax

(σ L PT

2

) ≤ αC Amax (π) and

C Bmax

(σ L PT

2

) ≤ αC Bmax (π), or C A

max

(σ L PT

3

) ≤ αC Amax (π) and C B

max

(σ L PT

3

) ≤αC B

max (π).

�By Theorem 4.1 and 4.2, we present the main result in this section as follows.

Theorem 4.3 For P2||C Amax : C B

max ≤ Q, the performance ratio of algorithm A −L PT E is α = 1+√

174 . The ratio is weakly tight.

Proof By Theorems 4.1, 4.2 and C Bmax (π) ≤ Q, there always exists a ( 1+√

174 ,

1+√17

4 )−approximation solution among schedules σ L PT1 , σ L PT

2 , σ L PT3 and σ L PT

4 .

Therefore, the algorithm A − L PT E has performance ratio 1+√17

4 .Notice that the performance ratio is not tight. In fact, let σ be the schedule generated

by algorithm A − L PT E . By the analysis of Theorems 4.1 and 4.2, at least one of the

results that C Amax (σ ) ≤ 5

4 C Amax (π) < 1+√

174 C A

max (π) and C Bmax (σ ) ≤ 5

4 C Bmax (π) <

1+√17

4 Q is true.

Now we give two instances to prove that the performance ratio is weakly tight. In

the first instance, agent A has four jobs with processing times pA1 = 1+√

174 , pA

2 =5+√

178 , pA

3 = 5+√17

8 − ε and pA4 = 1 − ε for a small ε > 0. Agent B has one

job with processing time pB1 = 3+√

172 . The threshold value Q = 5+√

172 − ε. By

algorithm A−L PT E, σ L PT1 will not be chosen since C B

max

(σ L PT

1

) = 11+3√

174 −ε >

1+√17

4 Q, whereas σ L PT2 , σ L PT

3 and σ L PT4 have the same cost value for agent A and

B, respectively (see Fig. 6).Therefore, σ = σ L PT

2

(σ L PT

3 , σ L PT4

)by algorithm A − L PT E . We have that

C Amax (σ )

C Amax (π)

=5+√

172 − 2ε

3+√17

2 − ε

ε→0−−→ 1 + √17

4

and C Bmax (σ ) = 3+√

172 < Q.

123

J Comb Optim

Fig. 6 A weakly tight instance for algorithm A − L PT E

In the second instance, agent A has one job with processing time pA1 = 3+√

172

and agent B has four jobs with processing times pB1 = 1+√

174 , pB

2 = 5+√17

8 , pB3 =

5+√17

8 − ε and pB4 = 1 − ε, where ε > 0 is a small enough constant. The threshold

value is 3+√17

2 − ε. Because C Amax

(σ L PT

1

) = C Amax

(σ L PT

3

) = C Amax

(σ L PT

4

)<

C Amax

(σ L PT

2

)and C B

max

(σ L PT

1

) = C Bmax

(σ L PT

3

) = C Bmax

(σ L PT

4

)< 1+√

174 Q, we

obtain that σ = σ L PT1

(σ L PT

3 , σ L PT4

)by algorithm A − L PT E . Therefore, we have

that C Amax (σ ) ≤ C A

max (π) and C Bmax (σ )/Q → 1+√

174 for ε → 0. �

5 Conclusions

In this paper, we have presented two approximation algorithms A−L S and A−L PT Efor P||C A

max : C Bmax ≤ Q and P2||C A

max : C Bmax ≤ Q, respectively. For P||C A

max :C B

max ≤ Q, algorithm A − L S schedules two agents’ jobs based on the LS ruleand takes O(n) times to produce a

(2 − 1

m , 2 − 1m

)−approximation solution. Theperformance ratio 2 − 1

m is proved to be tight. For two-machines version, algorithmA − L PT E schedules two agent’s jobs based on the LPT rule rather than the LS

rule. The algorithm has running time O(nlogn) and performance ratio 1+√17

4 whichis weakly tight.

We only consider (β, β)−approximation algorithms for P||C Amax : C B

max ≤ Q inthis paper. In the future, (βA, βB)−approximation algorithms for P||C A

max : C Bmax ≤

Q would be very interesting to investigate. Furthermore, Lee et al. (2009) and Sauleand Trystram (2009) introduced several approximation algorithms for multi-agentscheduling. Approximation algorithms for multi-agent scheduling problems may beconsidered as well.

Acknowledgments The authors would like to thank anonymous referees for their helpful commentsand suggestions, which significantly improve the results and presentation of this paper. This research issupported by NSFC (11371137) and the Fund for the Doctoral Program of China (20120074110021).

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