Semi-Online Scheduling on Two Machines with GoS Levels and Partial Information of Processing Time

This paper investigates semi-online scheduling problems on two parallel machines under a grade of service (GoS) provision subject to minimize the makespan. We consider three different semi-online versions with knowing the total processing time of the jobs with higher GoS level, knowing the total processing time of the jobs with lower GoS level, or knowing both in advance. Respectively, for the three semi-online versions, we develop algorithms with competitive ratios of 3/2, 20/13, and 4/3 which are shown to be optimal.


Introduction
Scheduling problem under a grade of service provision was first proposed by Hwang et al. [1]. The jobs should be assigned irrevocably to machines as soon as they arrive. Each job and machine are labeled with the GoS levels. A job can be processed by a particular machine if and only if the GoS level of the job is not less than that of the machine. In practice, the scheduling with GoS eligibility constraints is widely used. For more details, please refer to [1][2][3].
For the offline scheduling on machines under a grade of service provision, Hwang et al. [1] presented an lg-lpt algorithm which has a tight bound of 5/4 for two machines and 2 − 1/( − 1) for ( ≥ 3) machines. Ji and Cheng [4] gave an fptas for this problem. However, Woeginger [5] gave two simpler fptass for the same problem. For the online version, Jiang [6] proved a lower bound of 2 for the case with two levels of GoS and presented an online algorithm with a competitive ratio of 2.52. Zhang et al. [7] improved the result and showed an algorithm with a competitive ratio of 1 + ( 2 − )/( 2 − + 2 ) ≤ 7/3. For = 2, Park et al. [8] presented an optimal algorithm with a competitive ratio of 5/3. However, there are many papers focusing on the semi-online scheduling on two machines under GoS [8][9][10][11][12][13]. As the total processing time of all jobs is known, Park et al. [8] gave an optimal algorithm with a competitive ratio of 3/2. When the largest processing time of all jobs is known, Wu et al. [12] presented an optimal algorithm with a competitive ratio of ( √ 5 + 1)/2. In the same paper, Wu et al. [12] gave an optimal algorithm with a competitive ratio of 3/2 when the optimal value of the instance is known. For the processing times bounded in an interval, Liu et al. [9] gave a competitive algorithm under some conditions and Zhang et al. [13] improved the result and gave an optimal algorithm for different intervals. In this paper, we focus on the semi-online scheduling problem where only partial information of the total processing time is known. The semi-online versions concerned in our paper are listed as follows.
sum ⋅ higher: the total processing time of all jobs with higher GoS level is known in advance.
sum ⋅ lower: the total processing time of all jobs with lower GoS level is known in advance.
We use 2 ⋅ | | max to denote the semi-online problem with information , where ∈ { ⋅ ℎ ℎ , ⋅ }. Moreover, we use 2 ⋅ | 1& 2 | max to denote the semi-online problem where both information 1 and information 2 are available in advance.
Our results indicate that competitive ratios of three algorithms are better than 5/3 of the online version [8]. When knowing sum ⋅ higher in advance, the competitive ratio is 2 The Scientific World Journal the same as the optimal algorithm [8] presented for the semionline problem where the total processing time of all jobs is known. For designing algorithm, the results indicate that knowing sum ⋅ higher&sum ⋅ lower is more useful than sum ⋅ higher or sum ⋅ lower. Moreover, knowing sum ⋅ higher is more useful than sum ⋅ lower. However, knowing sum ⋅ lower is better than knowing the largest processing time of all jobs [12].
The rest of this paper is organized as follows: in Section 2, we give some basic definitions. In Sections 3-5, we prove lower bounds and present algorithms for the three semionline problems, respectively.

Basic Definitions
We are given two machines and a series of jobs arriving online which are to be scheduled irrevocably at the time of their arrivals. The arrival of a new job occurs only after the current job is scheduled. We denote by the set = 1, . . . , the set of all job indices arranged in the order of arrival. The th arriving job is referred to as job . We denote each job by = ( , ). The GoS assigned to job is denoted by , which is 1 if the job must be processed only by the first machine and 2 if it may be processed by either of the two; is the processing time of job ; and are not known until the arrival of job . Let The scheduled can be seen as the partition of into two subsets, denoted by ( 1 , 2 ), where 1 and 2 contain job indices assigned to the first and second machine, respectively. Let ( ) = Σ ∈ for an arbitrary subset of . Then the maximum of ( 1 ) and ( 2 ), denoted by max{ ( 1 ), ( 2 )}, is the makespan of the scheduled ( 1 , 2 ). The problem is to minimize max{ ( 1 ), ( 2 )}.
The minimum makespan obtained by an optimal offline will scheduled job 1 and job 2 on the first machine and scheduled job 3 on the second machine. Hence, / opt = 3/2. The proof is completed.

Optimal Semi-Online Algorithm GoS-TH.
Since we know 1 in advance and all the jobs with = 1 must be scheduled on the first machine, we can regard them as one job; that is, 0 = ( 1 , 1). We scheduled job 0 on the first machine at first and do not need to care about the job with = 1 later.
At the arrival of each job, is updated to become a half of the total processing time of the jobs which include the jobs with GoS = 2 and job 0 ; is updated to become the maximum processing time. We define , 1 , 2 , and to be , 1 , 2 , and after we scheduled job . Then, clearly the optimum makespan opt ≥ = max( , ). Combined with the online algorithm presented by Park et al. [8], we propose Algorithm -.
(3) Suppose that the incoming job is and = 1; assign job to the first machine.

Theorem 2. The competitive ratio of Algorithm GoS-TH is
Proof. Suppose that Theorem 2 is false. There must exist an instance with the least number of jobs to make -> (3/2) opt . The makespan is not determined until the arrival of job . Therefore, 2 )} ≤ (3/2) opt , we just need to prove Theorem 2 is true when = 2.

Lower Bound of Competitive Ratio
Theorem 3. Any semi-online algorithm for 2 ⋅ | ⋅ | max has a competitive ratio of at least 20/13.
Proof. We will construct a job sequence with 2 = 26 to make an arbitrary algorithm behave poorly. We begin with jobs / opt = 24/14 > 20/13. Otherwise, if job 3 or job 4 is scheduled on the first machine or both of them are scheduled on the first machine, we further generate job 5 = (26, 1). We will have ( 1 ) ≥ 40. Since optimal algorithm will scheduled jobs 1 , 2 , 3 , and 4 on the second machine and scheduled job 5 on the first machine, we have / opt ≥ 20/13. Case 2 ( 1 or 2 is scheduled on the first machine). We continue to generate job 3 = (2, 2). Then we discuss the following two subcases.

Optimal Semi-Online Algorithm GoS-TL.
In this subsection, we design an optimal algorithm with a competitive ratio of 20/13. Let 1 and 2 be the jobs with = 2 assigned 4 The Scientific World Journal to the first and second machine, respectively, where ( 1 ) + ( 2 ) = 2 . We define 1 and 2 to be 1 and 2 after we scheduled job . Then, we propose Algorithm -.

Algorithm GoS-TL
(1) Suppose that the incoming job is and = 1; assign job to the first machine.
Based on Lemma 4, we straightforwardly have Corollary 5.
Proof. Since Algorithm -will only scheduled a job with = 2 to the first machine only when ( −1 1 ) + ≤ (7/13) 2 , the lemma can directly be got from the Algorithm -. The proof is completed.

Lemma 8. If job is scheduled on the second machine by
Algorithm -where ( 2 ) ≤ (7/13) 2 , we have > (1/2) 2 . Then we have opt ≥ > (1/2) 2 . If job is scheduled on the second machine, Algorithm -will scheduled the remaining jobs on the first machine. Thus, Since ( ( −1 2 ) + )/ is decreasing function of the variation of and increasing function of variation of ( −1 2 ), we have The proof is completed.
Since there is no job with = 2 that will scheduled on the first machine between job V and by Algorithm -, combined with job being scheduled on the second machine, we have V + > (7/13) 2 . Since > (4/13) 2 , V , , and must satisfy which implies that opt > (7/13) 2 , hence, The proof is completed.
Based on Lemma 4 to Lemma 10 and Corollary 5 to Corollary 7, we have the following theorem naturally. Theorem 11. The competitive ratio of Algorithm GoS-TL is 20/13 for 2 ⋅ | ⋅ | max .

Optimal Algorithm for 2 ⋅ | ⋅ ℎ ℎ& ⋅ |
In this section, we show a lower bound of competitive ratio and present an optimal algorithm for the semi-online version as 1 and 2 are known in advance.

Lower Bound of Competitive Ratio
Theorem 12. Any semi-online algorithm for 2 ⋅ | ⋅ ℎ ℎ& ⋅ | max has a competitive ratio of at least 4/3.
Proof. The theorem will be proved by adversary method. Let 1 = 1/3 and 2 = 5/3 be known in advance. The first job is 1 = (1/3, 1). Job 1 must be scheduled on the first machine. Then job 2 = (1/3, 2) arrives. If job 2 is scheduled on the second machine, we further generate jobs 3 = (1, 2) and 4 = (1/3, 2). In this situation, we have ≥ 4/3 and opt = 1 since the optimal algorithm will scheduled jobs 1 , 2 , and 4 on the first machine and scheduled job 3 on the second machine. Thus, we have / opt ≥ 4/3. Otherwise, if job 2 is scheduled on the first machine, then we further generate jobs 3 = (2/3, 2) and 4 = (2/3, 2). In this situation, we also have ≥ 4/3 and opt = 1 since the optimal algorithm will scheduled jobs 1 and 3 on the first machine and scheduled jobs 2 and 3 on the second machine. The proof is completed.

Optimal Semi-Online Algorithm GoS-TB.
In this subsection, we design an optimal algorithm with a competitive ratio of 4/3. Since we know 1 in advance and all the jobs with = 1 must be scheduled on the first machine, therefore, we can regard them as one job, that is, 0 = ( 1 , 1). We scheduled job 0 on the first machine at first and do not need to care about the job with = 1 later. We present Algorithm as follows.
(3) Suppose that the incoming job is and = 1; assign job to the first machine. (4) Suppose that the incoming job is and = 2.    Proof. Since the job with = 1 is scheduled at first and we do not need to care about them after that. We focus on the jobs with = 2.

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The Scientific World Journal Assume job is the first job with = 2 to make ( 1 ) + > (2/3)( 1 + 2 ). We prove it by the following two cases.