A Competitive Two-Agent Scheduling Problem on Parallel Machines with Release Dates and Preemption

Abstract

We consider a competitive two-agent scheduling problem on multiple identical machines with release dates and preemption. In the scheduling model, there are two agents 𝑎 and 𝑏 each having their own job sets 𝒥_𝑎={𝐽^𝑎₁,…,𝐽^𝑎_𝑛𝑎} and 𝒥_𝑏={𝐽^𝑏₁,…,𝐽^𝑏_𝑛𝑏}, respectively. Each job 𝐽_𝑗∈𝒥_𝑎∪𝒥_𝑏 has a release date 𝑟_𝑖 and the 𝑛=𝑛_𝑎+𝑛_𝑏 jobs need to be preemptively scheduled on 𝑚 identical machines. For 𝑚=2, we show that the trade-off curve of all the Pareto optimal points can be characterized in polynomial time. When 𝑚 is input, we show that 𝑃|𝑟_𝑗,pmtn|𝐿^𝑎_max∶𝐿^𝑏_max≤𝑄 can be solved in strongly polynomial time.

1. Introduction and Problem Formulation

In recent years, multiagent scheduling problems are under extensive research. A multiagent scheduling problem means that there are multiple agents which must compete to perform their own tasks on the common processing resource. Each agent wants to optimize his/her own objective function. The objective function considered in this paper is to minimize the maximum lateness of the jobs. First, let us briefly recall the history of the classic scheduling problems of minimizing the maximum lateness of the jobs that is, there is only one agent in such a problem. Horn [1] considered the problem of scheduling the jobs preemptively on identical machines with release dates and deadlines and showed that it can be determined in strongly polynomial time if the problem has a feasible schedule by reducing it to a network flow problem which is well known to be solved in strongly polynomial time. Sahni [2] presented a faster algorithm to determine if the problem with identical release dates has a feasible schedule. Furthermore, Sahni and Cho [3] showed that it also can be determined in strongly polynomial time if the problem on related machines has a feasible schedule. Lawler and Labetoulle [4] proved that the feasibility problem on unrelated machines can be settled in weakly polynomial time by means of linear programming. Labetoulle et al. [5] studied the problems of scheduling the jobs on the parallel machines preemptively to minimize the maximum lateness of the jobs with release dates. They gave a strongly polynomial-time algorithm to solve the problem on identical machines which is based on the same network flow structure introduced by Horn [1] and a weakly polynomial-time algorithm to solve the problem on uniform machines in terms of the polymatroidal network flow model introduced by Lawler and Martel [6]. For the latter, a similar result can be found in Martel [7]. The multiagent scheduling models were initially introduced by Baker and Smith [8] and Agnetis et al. [9]. Their research focused on the problems of nonpreemptively scheduling the jobs which belong to two agents on a single machine. Agnetis et al. [10] investigated the multiagent single machine problem of finding a nonpreemptive schedule in which the cost of each agent does not exceed a given threshold value which is also studied in Cheng et al. [11]. Cheng et al. [12] considered the feasibility model of multiagent scheduling on a single machine for which each agent competes to minimize the total weighted number of his/her own tardy jobs and showed that the problem is strongly NP-hard. For more papers about the multiagent problems of scheduling the jobs nonpreemptively on a single machine, the readers are referred to Yuan et al. [13], Ng et al. [14], and Mor and Mosheiov [15]. Leung et al. [16] which initiated the preemptively multiagent scheduling problems investigated the two-agent scheduling problems of scheduling the jobs preemptively on a single machine or identical machines. Yuan et al. [17] studied a competitive two-agent scheduling problem on a single machine with release dates and preemption for which the objective of each agent is to minimize the maximum lateness and showed that all Pareto optimal points can be found in strongly polynomial time. Wan et al. [18] investigated the same two-agent scheduling model for which one agent’s objective is to minimize the maximum lateness and the other agent’s objective is to minimize the total completion time of his/her jobs. They proved that the problem is NP-hard in the ordinary sense by means of reduction from even-odd partition which is well known to be ordinarily NP-hard [19].

The problems in the paper are stated as follows. There are two agents 𝑎 and 𝑏 each having their own job sets 𝒥_𝑎={𝐽^𝑎₁,…,𝐽^𝑎_𝑛𝑎} and 𝒥_𝑏={𝐽^𝑏₁,…,𝐽^𝑏_𝑛𝑏}, respectively. We make the assumption that 𝒥_𝑎∩𝒥_𝑏=∅. The jobs in 𝒥_𝑎 are called 𝑎-jobs and the jobs in 𝒥_𝑏 are called 𝑏-jobs. Each job 𝐽_𝑗∈𝒥_𝑎∪𝒥_𝑏 has a release date 𝑟_𝑗, a due date 𝑑_𝑗, and a processing time 𝑝_𝑗. And the 𝑛=𝑛_𝑎+𝑛_𝑏 jobs need to be preemptively scheduled on 𝑚 identical machines. Two agents have the same objective of minimizing the maximum lateness. We use the following notation throughout this paper.(i)𝑝^𝑥_𝑗 is the processing time of job 𝐽^𝑥_𝑗, 𝑥∈{𝑎,𝑏}.(ii)𝑟^𝑥_𝑗 is the release date of job 𝐽^𝑥_𝑗, 𝑥∈{𝑎,𝑏}.(iii)𝑑^𝑥_𝑗 is the due date of job 𝐽^𝑥_𝑗, 𝑥∈{𝑎,𝑏}.(iv)𝐶^𝑥_𝑗 is the completion time of job 𝐽^𝑥_𝑗, 𝑥∈{𝑎,𝑏}.(v)𝐿^𝑥_𝑗=𝐶^𝑥_𝑗−𝑑^𝑥_𝑗 is the lateness of job 𝐽^𝑥_𝑗, 𝑥∈{𝑎,𝑏}.(vi)𝐿^𝑥_max=max{𝐿^𝑥_𝑗∶1≤𝑗≤𝑛_𝑥} is the maximum lateness of (the jobs of) agent 𝑥, 𝑥∈{𝑎,𝑏}.

A schedule 𝜎 is called Pareto optimal if there is no schedule 𝜌 such that 𝐿^𝑎_max(𝜌)≤𝐿^𝑎_max(𝜎), 𝐿^𝑏_max(𝜌)≤𝐿^𝑏_max(𝜎), and at least one inequality strictly holds, that is, a schedule is Pareto optimal for any schedule; if it is better for one agent, then it must be worse for the other agent. We say (𝐿^𝑎_max(𝜎),𝐿^𝑏_max(𝜎)) is a Pareto optimal point if schedule 𝜎 is a Pareto optimal schedule. The first problem we consider is to find all the Pareto optimal points and a corresponding schedule for each Pareto optimal point when 𝑚=2 and all the jobs are released at 0. The second problem we consider is to minimize the maximum lateness of agent 𝐴 with the maximum lateness of agent 𝐵 bounded by a given threshold 𝑄 when 𝑚 is input.

By use of the well-known three-field notation [20], the first and second problems can be formulated as follows.(i)The first problem: 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max), which is a Pareto optimization problem seeking to minimize 𝐿^𝑎_max and 𝐿^𝑏_max simultaneously.(ii)The second problem: 𝑃|𝑟_𝑗,pmtn|𝐿^𝑎_max∶𝐿^𝑏_max≤𝑄.

The rest of the paper is organized as follows. In Section 2, we show that the tradeoff curve of the Pareto optimal points can be characterized in strongly polynomial time for 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max). Section 3 gives a polynomial time algorithm to solve Problem 𝑃|𝑟_𝑗,pmtn|𝐿^𝑎_max∶𝐿^𝑏_max≤𝑄. We draw some conclusions and present some further research.

2. 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max)

First, let us state a feasibility problem of scheduling 𝑛 jobs preemptively on two identical machines with the deadlines and give a characterization of feasibility. Let 𝒥={𝐽₁,𝐽₂,…,𝐽_𝑛} be the set of the jobs and 𝑑_𝑗 and 𝑝_𝑗 are the deadline and processing time of job 𝐽_𝑗, respectively; 𝑗=1,2,…,𝑛. The problem is denoted by 𝑃2|pmtn;𝑑_𝑗|— in terms of the three-field notation. Without loss of generality, we assume that the jobs are ordered by 𝑑₁≤𝑑₂≤⋯≤𝑑_𝑛. Assume that there are total 𝑘 different deadlines and 𝑑₍₁₎,𝑑₍₂₎,…,𝑑_(𝑘) with 𝑑₍₁₎<𝑑₍₂₎<⋯<𝑑_(𝑘) are the 𝑘 different deadlines. Denote by 𝒥^(𝑖) the set of the jobs of deadline 𝑑_(𝑖), 𝑖=1,2,…,𝑘. Let 𝑟₀=0,𝑟₁,…,𝑟_𝑘=𝑛 be the 𝑘+1 numbers such that 𝒥^(𝑖)={𝐽_{𝑟𝑖−1+1},𝐽_{𝑟𝑖−1+2},…,𝐽_𝑟𝑖}, 𝑖=1,2,…,𝑘. We assume that 𝑑₀=0 and 𝐽_{𝑟𝑖−1+1} is the largest job of 𝒥^(𝑖) for convenience, 𝑖=1,2,…,𝑘.

Before we present the characterization of feasibility of 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max), we first state a well-known result from [2].

Algorithm Sahni. Consider the following steps.

Step 1. 𝑖=1, 𝐿₁=𝐿₂=0.

Step 2. Find a unscheduled job 𝐽_𝑗 of 𝒥^(𝑖). If 𝑝_𝑗≥𝑑_(𝑖)−𝐿₁, then return infeasibility. Otherwise, we schedule job 𝐽_𝑗. If 𝑝_𝑗≤𝑑_(𝑖)−𝐿₂, then we assign time interval [𝐿₂,𝐿₂+𝑝_𝑗] to 𝐽_𝑗 on machine 2 and reset 𝐿₂∶=𝐿₂+𝑝_𝑗; else, we assign time interval [𝐿₂,𝑑_(𝑖)] on machine 2 and time interval [𝐿₁,𝐿₁+𝐿₂+𝑝_𝑗−𝑑_(𝑖)] on machine 1 to 𝐽_𝑗 and reset 𝐿₂∶=𝑑_(𝑖), 𝐿₁∶=𝐿₁+𝐿₂+𝑝_𝑗−𝑑_(𝑖).

Step 3. If all the jobs of 𝒥^(𝑖) are scheduled, then reset 𝑖∶=𝑖+1; else, go back to Step 2.

Step 4. If 𝑖=𝑛+1, then stop; else, go back to Step 2.

𝐿₁ and 𝐿₂ denote the current loads of machine 1 and machine 2 in algorithm Sahni, respectively. And algorithm Sahni first schedules the jobs of 𝒥⁽¹⁾, then schedule 𝒥⁽²⁾ and so on. The following theorem is from [2].

Theorem 1. If problem 𝑃2|𝑝𝑚𝑡𝑛;𝑑_𝑗|—is feasible, then algorithm Sahni gives a feasible schedule.

Theorem 2. Problem 𝑃2|𝑝𝑚𝑡𝑛;𝑑_𝑗|—is feasible if and only if it satisfies the following conditions:(1)𝑑_(𝑗−1)+𝑑_(𝑖)≥∑^𝑟𝑗−1_𝑡=1𝑝_𝑡+𝑝_{𝑟𝑖−1+1}, 𝑖=1,2,…,𝑘, 𝑗=1,2,…,𝑖;(2)2𝑑_(𝑖)≥∑^𝑟𝑖_𝑗=1𝑝_𝑗, 𝑖=1,2,…,𝑘.

Proof. “only if” part. Problem 𝑃2|pmtn;𝑑_𝑗|—is feasible and denoted by 𝜎 a feasible schedule for 𝑃2|pmtn;𝑑_𝑗|. For any 𝑖 and any 𝑗 with 𝑗≤𝑖, only 𝐽_{𝑟𝑖−1+1} out of {𝐽₁,𝐽₂,…,𝐽_𝑟𝑗−1}∪{𝐽_{𝑟𝑖−1+1}} can be processed in [𝑑_(𝑗−1),𝑑_(𝑖)] in 𝜎. Note that 𝐽_{𝑟𝑖−1+1} must be completed by 𝑑_(𝑖) and all the jobs of {𝐽₁,𝐽₂,…,𝐽_𝑟𝑗−1} must be completed by 𝑑_(𝑗−1). Then we have 2𝑑_(𝑗−1)+(𝑑_(𝑖)−𝑑_(𝑗−1))≥∑^𝑟𝑗−1_ℎ=1𝑝_ℎ+𝑝_{𝑟𝑖−1+1}, which means that (1) holds. Note that all the jobs of all the jobs of {𝐽₁,𝐽₂,…,𝐽_𝑟𝑖} must be completed by 𝑑_(𝑖). Then we can get that 2𝑑_(𝑖)≥∑^𝑟𝑖_𝑗=1𝑝_𝑗, which means that (2) holds.
“If” Part. When 𝑘=1, which means that all the jobs have the same deadline 𝑑. By the conditions, we have 𝑝_max=𝑝₁≤𝑑 and ∑^𝑛_𝑗=1𝑝_𝑗/2≤𝑑. According to the result of [21], we know that there exists a feasible schedule of the 𝑛 jobs meeting with the same deadline. Now assume that the conclusion holds for 𝑘=𝑡. We consider the case of 𝑘=𝑡+1. Let 𝒥={𝐽₁,𝐽₂,…,𝐽_𝑟𝑡}=∪^𝑡_𝑖=1𝒥^(𝑖). By the assumption of 𝑘=𝑡, we know that there exists a feasible schedule 𝜎 for 𝒥. Furthermore, by Theorem 1, we can assume that 𝜎 is generated by algorithm Sahni. Denote by 𝑥 and 𝑦 the loads of machine 1 and machine 2, respectively. By the assumption of 𝑘=𝑡+1 and by the algorithm Sahni, we have that 𝑥+𝑝_𝑟𝑡+1≤𝑑_(𝑡+1) and 𝑥+𝑦+∑^𝑟𝑡+1_{𝑗=𝑟𝑡+1}𝑝_𝑗≤2𝑑_(𝑡+1). Similar to McNaughton’s algorithm of [21], we can schedule the jobs of 𝒥^(𝑡+1) starting from 𝜎 to meet with the deadlines of the jobs of 𝒥^(𝑡+1). Then we can get a feasible schedule for the job set ∪^𝑡+1_𝑖=1𝒥^(𝑖).

Remark 3. Theorem 2 still holds for 𝑑₍₁₎≤𝑑₍₂₎≤⋯≤𝑑_(𝑘).

By Theorem 2, we can easily get the following lemma.

Lemma 4. Problem 𝑃2|𝑝𝑚𝑡𝑛|𝐿_max can be solved in 𝑂(𝑛²) time.

In the following, let us consider problem 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max). Let (𝑥₁,𝑦₁) and (𝑥₂,𝑦₂) be two Pareto optimal points of 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max); then 𝑥₁>𝑥₂ means that 𝑦₁<𝑦₂. So we can assume 𝑦=𝑓(𝑥) such that (𝑥,𝑦) is a Pareto optimal point of 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max), and we know that 𝑦=𝑓(𝑥) is a strictly decreasing function on 𝑥. In order to calculate the tradeoff curve of 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max), we must determine the domain 𝐷 of 𝑥 and present an efficient calculation of 𝑓(𝑥). Let 𝑡(𝑥) be the optimal value of 𝑃2|pmtn|𝐿^𝑎_max≤𝑥∶𝐿^𝑏_max then {(𝑥,𝑡(𝑥))∶0≤𝑥≤𝑥⁰} includes all the Pareto optimal points of 𝑃2|pmtn|(𝐿^𝑎_max∶𝐿^𝑏_max); that is, {(𝑥,𝑓(𝑥))∶𝑥∈𝐷} is the remaining set after {(𝑥,𝑡(𝑥))∶0≤𝑥≤𝑥⁰} deletes all the non-Pareto optimal points. We call (𝑥,𝑡(𝑥)) a transit point if the due date order of the jobs in 𝒥 with the due date of 𝐽^𝑎_𝑗 considered as 𝑑^𝑎_𝑗+𝑥−𝜖 and the due date of 𝐽^𝑏_𝑖 considered as 𝑑^𝑏_𝑖+𝑡(𝑥−𝜖) are different from the due date order of the jobs in 𝒥 with the due date of 𝐽^𝑎_𝑗 considered as 𝑑^𝑎_𝑗+𝑥+𝜖 and the due date of 𝐽^𝑏_𝑖 considered as 𝑑^𝑏_𝑖+𝑡(𝑥+𝜖) for sufficiently small 𝜖. By Lemma 4, we first get the optimal value 𝑥₀ of 𝑃2|pmtn|𝐿^𝑎_max∶𝐿^𝑏_max<+∞ and the optimal value 𝑇₀ of 𝑃2|pmtn|𝐿^𝑎_max<+∞∶𝐿^𝑏_max in polynomial time. We simply assume that 𝑥₀=𝑇₀=0. If not, we can make 𝑥₀=𝑇₀=0 by changing the due dates of the jobs of 𝒥_𝑎 and 𝒥_𝑏 to the appropriate numbers.

Obviously, 𝑥≥0. Let 𝑥⁰ be the optimal value of 𝑃2|pmtn|𝐿^𝑎_max∶𝐿^𝑏_max≤0; then [𝑥₀,𝑥⁰] is the domain of 𝑥. We present a strongly polynomial time algorithm to achieve the optimal value of 𝑃2|pmtn|𝐿^𝑎_max∶𝐿^𝑏_max≤0. Without loss of generality, for 𝑥∈{𝑎,𝑏}, we assume that the jobs of agent 𝑥 are ordered by 𝑑^𝑥₁≤𝑑^𝑥₂≤⋯≤𝑑^𝑥_𝑛𝑥. Assume that there are total 𝑘_𝑥 different due dates and 𝑑^𝑥₍₁₎,𝑑^𝑥₍₂₎,…,𝑑^𝑥_(𝑘𝑥) with 𝑑^𝑥₍₁₎<𝑑^𝑥₍₂₎<⋯<𝑑^𝑥_(𝑘𝑥) are the 𝑘_𝑥 different due dates. Denote by 𝒥_𝑥^(𝑖) the set of the jobs of due date 𝑑^𝑥_(𝑖), 𝑖=1,2,…,𝑘_𝑥. Let 𝑟₀=0,𝑟₁,…,𝑟_𝑘𝑥=𝑛_𝑥 be the 𝑘_𝑥+1 numbers such that 𝒥_𝑥^(𝑖)={𝐽^𝑥_{𝑟𝑖−1+1},𝐽^𝑥_{𝑟𝑖−1+2},…,𝐽^𝑥_𝑟𝑖}, 𝑖=1,2,…,𝑘_𝑥. We assume that 𝑑₀=0 and 𝐽^𝑥_{𝑟𝑖−1+1} are the largest jobs of 𝒥_𝑥^(𝑖) for convenience, 𝑖=1,2,…,𝑘_𝑥. For any 𝑡≥0, we define ℎ(𝑡) as the smallest number of {𝑑^𝑏₍₁₎,𝑑^𝑏₍₂₎,…,𝑑^𝑏_(𝑘𝑏)}∪{+∞} strictly larger than 𝑡. We determine a series of 𝛿₁, 𝛿₂,…,𝛿_𝑣0 iteratively, where 𝑣₀ will be determined later. First, we determine 𝛿₁=min_{1≤𝑖≤𝑛𝑎}{ℎ(𝑑^𝑎₍₁₎)−𝑑^𝑎₍₁₎}. Then we determine 𝛿₂=min_{1≤𝑖≤𝑛𝑎}{ℎ(𝑑^𝑎₍₁₎+𝛿₁)−𝑑^𝑎₍₁₎−𝛿₁}. For general 𝑗, we determine 𝛿_𝑗=min_{1≤𝑖≤𝑛𝑎}{ℎ(𝑑^𝑎₍₁₎+∑^𝑗−1_𝑘=1𝛿_𝑘)−𝑑^𝑎₍₁₎−∑^𝑗−1_𝑘=1𝛿_𝑘}. Go on the process until the 𝑣₀−𝑡ℎ iteration such that 𝑑^𝑎₍₁₎+∑^𝑣0−1_𝑘=1𝛿_𝑘≥𝑑^𝑏_𝑘𝑏. Then we have 𝑣₀≤𝑘_𝑎𝑘_𝑏≤𝑛_𝑎𝑛_𝑏. Assume that 𝛿₀=0 and 𝛿_𝑣0+1=+∞. We use a bisection search to determine 𝑡₀ such that 𝑃2|pmtn|𝐿^𝑎_max≤∑^𝑡0−1_𝑖=1𝛿_𝑖∶𝐿^𝑏_max≤0 is infeasible and 𝑃2|pmtn|𝐿^𝑎_max≤∑^𝑡0_𝑖=1𝛿_𝑖∶𝐿^𝑏_max≤0 is feasible. And it can be completed in 𝑂(log𝑣₀) implementations of algorithm Sahni. Therefore, we have ∑^𝑡0−1_𝑖=1𝛿_𝑖<𝑥₀≤∑^𝑡0_𝑖=1𝛿_𝑖. By the definition of 𝛿_𝑡0, let 𝑑^𝑎_𝑖=𝑑^𝑎_𝑖+𝑥 and 𝑑^𝑎_(𝑗)=𝑑^𝑎_(𝑗)+𝑥, 𝑖=1,2,…,𝑛_𝑎 and 𝑗=1,2,…,𝑘_𝑎. We know that {𝑑^𝑎₍₁₎,𝑑^𝑎₍₂₎,…,𝑑^𝑎_(𝑘𝑎)}∪{𝑑^𝑏₍₁₎,𝑑^𝑏₍₂₎,…,𝑑^𝑏_(𝑛𝑏)} remains the same increasing order for any ∑^𝑡0_𝑖=1𝛿_𝑖<𝑥<∑^𝑡0−1_𝑖=1𝛿_𝑖. For ∑^𝑡0−1_𝑖=1𝛿_𝑖<𝑥<∑^𝑡0_𝑖=1𝛿_𝑖, we apply Theorem 2 to the job set 𝒥_𝑎∪𝒥_𝑏 with the deadline 𝑑^𝑎_𝑖+𝑥 of 𝐽^𝑎_𝑖 and the deadline 𝑑^𝑏_𝑗 of 𝐽^𝑏_𝑗, 𝑖=1,2,…,𝑛_𝑎 and 𝑗=1,2,…,𝑛_𝑏. Let 𝒥=𝒥_𝑎∪𝒥_𝑏={𝐽₁,𝐽₂,…,𝐽_𝑛} and 𝑑₍₁₎,𝑑₍₂₎,…,𝑑_(𝑘) with 𝑑₍₁₎<𝑑₍₂₎<⋯<𝑑_(𝑘) are the distinct deadlines of the jobs of 𝒥, where 𝑘=𝑘_𝑎+𝑘_𝑏 and 𝑛=𝑛_𝑎+𝑛_𝑏. Then 𝑑_(𝑗) is some 𝑑^𝑎_(𝑟)+𝑥 or some 𝑑^𝑏_(𝑡) for any 𝑗 with 1≤𝑗≤𝑘. And 𝑝_𝑖 is fixed as some 𝑝^𝑎_𝑟 or some 𝑝^𝑏_𝑡 for any 𝑥 with ∑^𝑡0−1_𝑖=1𝛿_𝑖<𝑥<∑^𝑡0_𝑖=1𝛿_𝑖. So by Theorem 2, we can get the exact value of 𝑥⁰ by solving at most (𝑘(𝑘+1)/2)+𝑘 linear equalities. We demonstrate the idea as two steps.

Step 1. We get the optimal value 𝑇⁰ of 𝑃2|pmtn|𝐿^𝑎_max≤0∶𝐿^𝑏_max, reset the due date 𝑑^𝑏_𝑗∶=𝑑^𝑏_𝑗+𝑇⁰, 𝑗=1,2,…,𝑛_𝑏, and keep the due dates 𝑑^𝑎₁,𝑑^𝑎₂,…,𝑑^𝑎_𝑛𝑎 unchanged. Let 𝜋 be the order of the jobs in 𝒥 according to the due dates from small to large. Set 𝛿₁=min_{1≤𝑖≤𝑛𝑎}{ℎ(𝑑^𝑎₁)−𝑑^𝑎₁}. Let 𝑥 and let 𝑦=𝑢(𝑥) be the numbers such that 𝒥 has the due date order 𝜋 when the due date of 𝐽^𝑎_𝑗 is regarded as 𝑑^𝑎_𝑗+𝑥 and the due date of 𝐽^𝑏_𝑖 is regarded as 𝑑^𝑏_𝑗−𝑦. Consider 𝑑^𝑎_𝑗+𝑥 as the deadline of 𝐽^𝑎_𝑗 and 𝑑^𝑏_𝑗−𝑦 as the deadline of 𝐽^𝑏_𝑖, applying the Theorem 2, subject to 𝑥+𝑦≤𝛿₁, we can find the exact representation of 𝑢(𝑥), where 𝑢(𝑥) is a curve of 𝑂(𝑘²) broken lines and the slopes of the broken lines are increasing and 𝑥₁ such that (𝑥₁,𝑡(𝑥₁)) is the transit point in polynomial time.

Step 2. We reset 𝑑^𝑏_𝑗∶=𝑑^𝑏_𝑗+𝑡(𝑥₁) and 𝑑^𝑎_𝑖∶=𝑑^𝑎_𝑖+𝑥₁, repeat the process of Step 1 to find the next transit point (𝑥₂,𝑡(𝑥₂)), go on the process until we find all the transit points and note that the number of the transit points is bounded by 𝑛_𝑎𝑛_𝑏.
In the following, we provide an instance to help the understanding of this algorithm.

Instance. There are two different deadlines 𝑑₍₁₎=3 and 𝑑₍₂₎=6 and six jobs partitioned into two subsets: 𝒥⁽¹⁾={𝐽₁,𝐽₂,𝐽₃} and 𝒥⁽²⁾={𝐽₄,𝐽₅,𝐽₆}, 𝑝₁=1, 𝑝₂=1, 𝑝₃=3, 𝑝₄=2, 𝑝₅=2, and 𝑝₆=6.

In the instance, we can calculate all the transit points by the algorithm. There are four transit points: (𝑥₀,𝑦₀)=(0,2), (𝑥₁,𝑦₁)=(1/2,3/2), (𝑥₂,𝑦₂)=(3/2,3/2), and (𝑥₃,𝑦₃)=(3/2,0). So the tradeoff curve is demonstrated as Figure 1.

Figure 1 

The solid line and the solid point show all the Pareto optimal points and the dotted line and hollow point are the dominated points.

3. 𝑃|pmtn|𝐿^𝑎_max∶𝐿^𝑏_max≤𝑄

We now consider problem 𝑃|𝑟_𝑗,pmtn|𝐿^𝑎_max∶𝐿^𝑏_max≤𝑄. Without loss of generality, we can assume 𝑄=0. Otherwise, we can shift the due date of 𝐽^𝑏_𝑖 to some appropriate number, 𝑖=1,2,…,𝑛_𝑏. Similarly, we first deal with the relaxed version of this problem which is to test a trial value of 𝐿^𝑎_max for feasibility. That is, for a given value 𝑦, one has to determine whether or not there exists a schedule for which 𝐿^𝑎_max≤𝑦. This condition is equivalent to the requirement that no job 𝐽^𝑎_𝑖 is completed after an induced deadline 𝑑^𝑎_𝑖+𝑦 and no job 𝐽^𝑏_𝑗 is completed after 𝑑^𝑏_𝑗. For the feasibility problem 𝑃|𝑟_𝑗,pmtn|𝐿^𝑎_max≤𝑦∶𝐿^𝑏_max≤0, Horn [1] proposed a network flow algorithm to solve it.

Horn’ approach is as follows. Suppose 𝑦 is a trial value for 𝐿^𝑎_max. Let {𝑒₁,𝑒₂,…,𝑒_2𝑛} with 𝑒₁≤𝑒₂≤⋯≤𝑒_2𝑛 be the ordered collection of release dates 𝑟^𝑎_𝑗, 𝑟^𝑏_𝑖 and deadlines 𝑑^𝑏_𝑖, induced deadlines 𝑑^𝑎_𝑖+𝑦. We arbitrarily break the ties. Further, Define the time interval 𝐸_𝑘=[𝑒_𝑘,𝑒_𝑘+1] for 𝑘=1,2,…,2𝑛−1.

We construct the flow network as follows. There are 𝑛=𝑛_𝑎+𝑛_𝑏 job vertices 𝐽₁,𝐽₂,…,𝐽_𝑛 (𝐽_𝑗 is some 𝐽^𝑎_𝑠 or some 𝐽^𝑏_𝑟) interval vertices 𝐸₁,𝐸₂,…,𝐸_2𝑛−1, a source vertex 𝑆, and a sink vertex 𝑇. There ia an arc (𝐽_𝑗,𝐸_𝑘) of capacity 𝑒_𝑘+1−𝑒_𝑘 if and only if the release date of 𝐽_𝑗 is no more that 𝑒_𝑘 and 𝑒_𝑘+1 are no more than the (induced) deadlines of 𝐽_𝑗. In addition, there is an arc (𝑆,𝐽_𝑗) of capacity 𝑝_𝑗 which is the processing time of 𝐽_𝑗 for 𝑗=1,2,…,𝑛 and an arc (𝐸_𝑘,𝑇) of capacity 𝑚(𝑒_𝑘+1−𝑒_𝑘) for 𝑘=1,2,…,2𝑛−1. Now, a maximum value flow is found in 𝑂(𝑛³) time [22, 23]. It should be evident that the trial value 𝑦 is feasible if and only if the maximum flow value is 𝑃=∑^𝑛_𝑗=1𝑝_𝑗, where 𝑝_𝑗 is the processing time of 𝐽_𝑗. If the maximum flow value is indeed 𝑃, a feasible schedule is easily constructed: for each interval 𝐸_𝑘, the corresponding schedule part can be achieved by applying the McNaughton’s algorithm [21]. Notice that there are certain critical trial values of 𝐿^𝑎_max. There are at most 𝑛_𝑎(𝑛+𝑛_𝑏) values 𝑦 such that 𝑑^𝑎_𝑗+𝑦=𝑑^𝑏_𝑖 or 𝑑^𝑎_𝑗+𝑦=𝑟^𝑎_𝑘 or 𝑑^𝑎_𝑗+𝑦=𝑟^𝑏_ℎ. The vertex-arc structure of the network remains unchanged for all trial values between two successive critical values.

We propose to find the optimum value of 𝐿^𝑎_max in two phases. In the first phase, the largest infeasible critical value 𝑦₀ is determined. A bisection search for 𝑦₀ requires the testing of log(𝑛_𝑎(𝑛+𝑛_𝑏)) trial values or 𝑂(𝑛_𝑎⋅𝑛²log𝑛) time overall.

In the second phase, a maximum value flow and a minimum capacity cut are found in the network with capacities induced by the value 𝑦₀. Next, a value 𝑦₁>𝑦₀ is determined in such a way that the capacity of this cut is increased to exactly 𝑃. The procedure is then repeated in the network induced by 𝑦₁. This process yields a sequence of increasing trial values 𝑦_𝑖. It terminates when the minimum cut capacity is exactly 𝑃, that is, at an iteration 𝑧 where 𝑦_𝑧 is the first feasible trial value and therefore the optimum value of 𝐿^𝑎_max. We will show that 𝑧=𝑂(𝑛_𝑎max{𝑛,𝑚}).

Suppose a minimum cut with capacity 𝑃₀<𝑃 is found in the network for 𝑦₀. Consider how the capacity of this cut is changed when 𝑦₀ is increased by some positive amount 𝛿. The capacity 𝑒_𝑘+1−𝑒_𝑘 of an arc (𝐽_𝑗,𝐸_𝑘)(a)stays the same if both of 𝑒_𝑘 and 𝑒_𝑘+1 are induced deadlines or not induced deadlines,(b)increases by 𝛿 if 𝑒_𝑘 is not induced deadline and 𝑒_𝑘+1 is induced deadline, and(c)decreases by 𝛿 if 𝑒_𝑘 is induced deadline and 𝑒_𝑘+1 is not induced deadline.

A similar situation holds for the capacities of the arcs (𝐸_𝑘,𝑇), except that they change by 𝑚𝛿 or −𝑚𝛿 rather than by 𝛿 or −𝛿. All arcs whose capacities are increased are incident with a vertex 𝐸_𝑘 of type (𝑏) of which there are at most 𝑛. If (𝐸_𝑘,𝑇) is in the cut, then no (𝐽_𝑗,𝐸_𝑘) can be forward arc in the cut, so that the cut capacity increases in all arcs incident with 𝐸_𝑘 is at most 𝑚𝛿. It follows that eh capacity of the cut is increased by 𝜇₀𝛿, where 𝜇₀ is an integer multiplier with 𝜇₀≤𝑛_𝑎max{𝑛,𝑚}. We assert that 𝜇₀≥1, and let 𝑦⁰ be the next critical value after 𝑦₀. Note that The vertex-arc structure of the network remains unchanged for 𝑦₀ and 𝑦⁰. Since 𝑦⁰ is a feasible critical value, by the max-flow min-cut theorem, we have that 𝑃₀+𝜇₀(𝑦⁰−𝑦₀)≥𝑃, which means that 𝜇₀≥1. Accordingly, we set 𝛿=(𝑃−𝑃₀)/𝜇₀, 𝑦₁=𝑦₀+𝛿 and repeat.

Each cut in the network can be characterized by a pair (𝜇,𝑃^′), where 𝜇 is its multiplier and 𝑃^′ its capacity. When 𝑦_𝑙 is increased to 𝑦_𝑙+1, the multipliers of cuts do not change, although their capacities indeed do. Suppose that the minimum cut found at iteration 𝑖 has multiplier 𝜇_𝑖≥1 and capacity 𝑃_𝑖 and consider the replacement of 𝑦_𝑖 by 𝑦_𝑖+1. Each cut with multiplier 𝜇≥𝜇_𝑖 will have its capacity increased to at lease 𝑃. Hence, 𝜇_𝑖+1<𝜇_𝑖. Note that 𝜇_𝑖≥1 for all 𝑖 and 𝜇₀≤𝑛_𝑎max{𝑛,𝑚}. It follows that there can be at most 𝜇₀≤𝑛_𝑎max{𝑛,𝑚} iterations.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.