#### 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 and each having their own job sets and , respectively. Each job has a release date and the jobs need to be preemptively scheduled on identical machines. For , we show that the trade-off curve of all the Pareto optimal points can be characterized in polynomial time. When is input, we show that 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 and each having their own job sets and , respectively. We make the assumption that . The jobs in are called -jobs and the jobs in are called -jobs. Each job has a release date , a due date , and a processing time . And the jobs need to be preemptively scheduled on identical machines. Two agents have the same objective of minimizing the maximum lateness. We use the following notation throughout this paper.(i) is the processing time of job , .(ii) is the release date of job , .(iii) is the due date of job , .(iv) is the completion time of job , .(v) is the lateness of job , .(vi) is the maximum lateness of (the jobs of) agent , .

A schedule is called Pareto optimal if there is no schedule such that , , 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 is a Pareto optimal point if schedule 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 and all the jobs are released at . The second problem we consider is to minimize the maximum lateness of agent with the maximum lateness of agent bounded by a given threshold when 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: , which is a Pareto optimization problem seeking to minimize and simultaneously.(ii)The second problem: .

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 . Section 3 gives a polynomial time algorithm to solve Problem . We draw some conclusions and present some further research.

#### 2.

First, let us state a feasibility problem of scheduling jobs preemptively on two identical machines with the deadlines and give a characterization of feasibility. Let be the set of the jobs and and are the deadline and processing time of job , respectively; . The problem is denoted by — in terms of the three-field notation. Without loss of generality, we assume that the jobs are ordered by . Assume that there are total different deadlines and with are the different deadlines. Denote by the set of the jobs of deadline , . Let be the numbers such that , . We assume that and is the largest job of for convenience, .

Before we present the characterization of feasibility of , we first state a well-known result from [2].

*Algorithm Sahni*. Consider the following steps.

*Step 1. *, .

*Step 2. *Find a unscheduled job of . If , then return * infeasibility*. Otherwise, we schedule job . If , then we assign time interval to on machine 2 and reset ; else, we assign time interval on machine 2 and time interval on machine 1 to and reset , .

*Step 3. *If all the jobs of are scheduled, then reset ; else, go back to Step 2.

*Step 4. *If , then stop; else, go back to Step 2.

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

Theorem 1. *If problem —is feasible, then algorithm Sahni gives a feasible schedule.*

Theorem 2. *Problem —is feasible if and only if it satisfies the following conditions:*(1)*, , ;*(2)*, .*

*Proof. *“only if” part. Problem —is feasible and denoted by a feasible schedule for . For any and any with , only out of can be processed in in . Note that must be completed by and all the jobs of must be completed by . Then we have , which means that (1) holds. Note that all the jobs of all the jobs of must be completed by . Then we can get that , which means that (2) holds.

“If” Part. When , which means that all the jobs have the same deadline . By the conditions, we have and . According to the result of [21], we know that there exists a feasible schedule of the jobs meeting with the same deadline. Now assume that the conclusion holds for . We consider the case of . Let . By the assumption of , we know that there exists a feasible schedule for . Furthermore, by Theorem 1, we can assume that is generated by algorithm Sahni. Denote by and the loads of machine 1 and machine 2, respectively. By the assumption of and by the algorithm Sahni, we have that and . Similar to McNaughton’s algorithm of [21], we can schedule the jobs of starting from to meet with the deadlines of the jobs of . Then we can get a feasible schedule for the job set .

*Remark 3. *Theorem 2 still holds for .

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

Lemma 4. *Problem can be solved in time.*

In the following, let us consider problem . Let and be two Pareto optimal points of ; then means that . So we can assume such that is a Pareto optimal point of , and we know that is a strictly decreasing function on . In order to calculate the tradeoff curve of , we must determine the domain of and present an efficient calculation of . Let be the optimal value of then includes all the Pareto optimal points of ; that is, is the remaining set after deletes all the non-Pareto optimal points. We call a transit point if the due date order of the jobs in with the due date of considered as and the due date of considered as are different from the due date order of the jobs in with the due date of considered as and the due date of considered as for sufficiently small . By Lemma 4, we first get the optimal value of and the optimal value of in polynomial time. We simply assume that . If not, we can make by changing the due dates of the jobs of and to the appropriate numbers.

Obviously, . Let be the optimal value of ; then is the domain of . We present a strongly polynomial time algorithm to achieve the optimal value of . Without loss of generality, for , we assume that the jobs of agent are ordered by . Assume that there are total different due dates and with are the different due dates. Denote by the set of the jobs of due date , . Let be the numbers such that , . We assume that and are the largest jobs of for convenience, . For any , we define as the smallest number of strictly larger than . We determine a series of , iteratively, where will be determined later. First, we determine . Then we determine . For general , we determine . Go on the process until the iteration such that . Then we have . Assume that and . We use a bisection search to determine such that is infeasible and is feasible. And it can be completed in implementations of algorithm Sahni. Therefore, we have . By the definition of , let and , and . We know that remains the same increasing order for any . For , we apply Theorem 2 to the job set with the deadline of and the deadline of , and . Let and with are the distinct deadlines of the jobs of , where and . Then is some or some for any with . And is fixed as some or some for any with . So by Theorem 2, we can get the exact value of by solving at most linear equalities. We demonstrate the idea as two steps.

*Step 1. *We get the optimal value of , reset the due date , , and keep the due dates unchanged. Let be the order of the jobs in according to the due dates from small to large. Set . Let and let be the numbers such that has the due date order when the due date of is regarded as and the due date of is regarded as . Consider as the deadline of and as the deadline of , applying the Theorem 2, subject to , we can find the exact representation of , where is a curve of broken lines and the slopes of the broken lines are increasing and such that is the transit point in polynomial time.

*Step 2. *We reset and , repeat the process of Step 1 to find the next transit point , go on the process until we find all the transit points and note that the number of the transit points is bounded by .

In the following, we provide an instance to help the understanding of this algorithm.

*Instance*. There are two different deadlines and and six jobs partitioned into two subsets: and , , , , , , and .

In the instance, we can calculate all the transit points by the algorithm. There are four transit points: , , , and . So the tradeoff curve is demonstrated as Figure 1.

#### 3.

We now consider problem . Without loss of generality, we can assume . Otherwise, we can shift the due date of to some appropriate number, . Similarly, we first deal with the relaxed version of this problem which is to test a * trial value* of for feasibility. That is, for a given value , one has to determine whether or not there exists a schedule for which . This condition is equivalent to the requirement that no job is completed after an * induced deadline * and no job is completed after . For the feasibility problem , Horn [1] proposed a network flow algorithm to solve it.

Horn’ approach is as follows. Suppose is a trial value for . Let with be the ordered collection of release dates , and deadlines , induced deadlines . We arbitrarily break the ties. Further, Define the time interval for .

We construct the flow network as follows. There are job vertices ( is some or some ) interval vertices , a source vertex , and a sink vertex . There ia an arc of capacity if and only if the release date of is no more that and are no more than the (induced) deadlines of . In addition, there is an arc of capacity which is the processing time of for and an arc of capacity for . Now, a maximum value flow is found in time [22, 23]. It should be evident that the trial value is feasible if and only if the maximum flow value is , where is the processing time of . If the maximum flow value is indeed , a feasible schedule is easily constructed: for each interval , the corresponding schedule part can be achieved by applying the McNaughton’s algorithm [21]. Notice that there are certain * critical* trial values of . There are at most values such that or or . 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 in two phases. In the first phase, the largest infeasible critical value is determined. A bisection search for requires the testing of trial values or 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 . Next, a value is determined in such a way that the capacity of this cut is increased to exactly . The procedure is then repeated in the network induced by . This process yields a sequence of increasing trial values . It terminates when the minimum cut capacity is exactly , that is, at an iteration where is the first feasible trial value and therefore the optimum value of . We will show that .

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

A similar situation holds for the capacities of the arcs , except that they change by or rather than by or . All arcs whose capacities are increased are incident with a vertex of type of which there are at most . If is in the cut, then no can be forward arc in the cut, so that the cut capacity increases in all arcs incident with is at most . It follows that eh capacity of the cut is increased by , where is an integer * multiplier* with . We assert that , and let be the next critical value after . Note that The vertex-arc structure of the network remains unchanged for and . Since is a feasible critical value, by the max-flow min-cut theorem, we have that , which means that . Accordingly, we set , and repeat.

Each cut in the network can be characterized by a pair , where is its multiplier and its capacity. When is increased to , the multipliers of cuts do not change, although their capacities indeed do. Suppose that the minimum cut found at iteration has multiplier and capacity and consider the replacement of by . Each cut with multiplier will have its capacity increased to at lease . Hence, . Note that for all and . It follows that there can be at most iterations.

#### Conflict of Interests

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