Two-agent Scheduling in Open Shops Subject to Machine Availability and Eligibility Constraints

Purpose: The aims of this article are to develop a new mathematical formulation and a new heuristic for the problem of preemptive two-agent scheduling in open shops subject to machine maintenance and eligibility constraints. Design/methodology: Using the ideas of minimum cost flow network and constraint programming, a heuristic and a network based linear programming are proposed to solve the problem. Findings: Computational experiments show that the heuristic generates a good quality schedule with a deviation of 0.25% on average from the optimum and the network based linear programming model can solve problems up to 110 jobs combined with 10 machines without considering the constraint that each operation can be processed on at most one machine at a time. In order to satisfy this constraint, a time consuming Constraint Programming is proposed. Fo r n = 80 and m = 10, the average execution time for the combined models (linear programming model combined with Constraint programming) exceeds two hours. Therefore, the heuristic algorithm we developed is very efficient and is in need. Practical implications: Its practical implication occurs in TFT-LCD and E-paper manufacturing wherein units go through a series of diagnostic tests that do not have to be performed in any specified order.


Introduction
Consider two agents who have to schedule two sets of jobs on an m-machine open shop.Each job has k operations, k ≤ m, and each operation must be performed on the corresponding specialized machine.The order in which the operations of each job are performed is irrelevant.
Operation preemption is allowed and the machine availability and eligibility constraints are considered.The machine availability arises when machines are subject to breakdowns, maintenance, or perhaps high priority tasks are prescheduled in certain time intervals.The machine eligibility constraints are imposed when the number of operations of each job i can be less than m.Each machine can handle at most one operation at a time and each operation can be processed on at most one machine at a time.Two agents are called Agent A and B. signifies job preemption, implies that each job has a release date, denotes the specific subset of machines to process job j, and denotes agent B will accept a schedule of time up to Q.
The multi-agent scheduling problems have received increasing attention recently.Baker and Smith (2003) perhaps the first to consider the problem in which two agents compete on the use of a single machine.They demonstrated that although determining a minimum cost schedule according to any of three criteria: makespan, minimizing maximum lateness, and minimizing total weighted completion time for a single machine, is polynomial, the problem of minimizing a mix of these criteria is NP-hard.Agnetis, Mirchandani, Pacciarelli and Pacifici  (2009), Leung, Pinedo and Wan (2010) and Lee, Chung and Huang (2013).When release times are further considered, see (Lee, Chung & Hu, 2012;Yin, Wu, Cheng and Wu, 2012;Yin, Wu, Cheng, Wu & Wu, 2014;Wu, Wu, Chen, Yin & Wu, 2013).The multi-agent problems are extended by considering variations in job processing time such as controllable processing time (Wan, Vakati, Leung & Pinedo, 2010), deteriorating job processing time (Cheng, Wu, Cheng and Wu, 2011;Liu, Yi & Zhou, 2011), and learning effect (Cheng, Cheng, Wu, Hsu and Wu, 2011;Lee & Hsu, 2012;Wu, Huang & Lee, 2011;Yin, Cheng and Wu, 2012).Yu, Zhang, Xu and Yin (2013) considered a two agents problem to minimize an aggregate increasing objection function of two agents' objective function and a piece-rate maintenance which is implemented once a fixed number of jobs is completed.Mor and Mosheiov (2010) considered minimizing the maximum earliness cost or total weighted earliness cost of one agent, subject to an upper bound on the maximum earliness cost of the other agent.They showed that both minimax and minsum cases are polynomially solvable while the weighted minsum case is NPhard.
As for the preemptive open shop problem, Gonzalez and Sahni (1976) proposed a polynomial time algorithm to obtain the minimum makespan.Breit, Schmidt and Strusevich (2001) studied a two-machine open shop where one machine is not available for processing during a given time interval.The objective is to minimize the makespan.They showed that the problem is NP-hard and presented a heuristic with a worst-case ratio of 4/3.When time-windows is considered for each job on an open shop and the objective is to minimize makespan, Sedeno-Noda, Alcaide and Gonzalez-Martin (2006) introduced a network flow procedure to check feasibility and a max-flow parametrical algorithm to minimize the makespan.
Sedeno-Noda, Pablo and Gonzalez-Martin (2009) extended the same problem by considering performance costs including resource and personnel involvement.
This two-agent problem under consideration arises in TFT-LCD and E-Paper manufacturing wherein units go through a series of diagnostic tests that do not have to be performed in any specified order.
This paper is organized as follows: Section 2 formulates the problem and provides a heuristic.
Section 3 is devoted to describing the minimum cost flow network and its corresponding linear programming model.Section 4 illustrates the effectiveness of the heuristic algorithm and computational efficiency of the minimum cost flow network, and finally in Section 5, we provide conclusions.Numerical Example 1.There are two jobs for each agent to be processed on three machines.

Problem Formulation and the Heuristic Algorithm
The job and machine data are listed in Tables 1 and 2 and agent B will accept a schedule of time up to Q = 17.

Job (j)
Release time 1 3 2 6 Processing time Availability interval   Step 2. Since there is no available operation in [e 1 , e 2 ], set l = 2.In [e 2 , e 3 ], no operation for agent B is available and thus is selected to be processed on m 1 due to the maximum value Step 2. In [e 8 , e 9 ], is selected to be processed on m 2 , however, is not in set E, therefore go to Step 3.
Proceed in the same manner, the heuristic solution with C max = 18 is obtained and shown in Figure 2.  Upon obtaining E, T l and m(l), we formulate the problem with U as a minimum cost flow network problem on a tripartite network G(U).
The network consists of a set of nodes and a set of arcs connecting certain pairs of the nodes.
The node set in G(U) includes a source node s, a sink node t, and the following three node sets.
(i) Job nodes, , j = 1, 2, …, n x , x = a, b. (iii) Combination nodes (i, T l ) consisting of machine i and time interval T l , where T l satisfies e l+1 ≤ U and i Î m(l).
The arc set consists of directed arcs that are generated as follows.
(ii) An arc from node to node with capacity (e l+1 -e l ).
(iii) An arc from node to (i, T l ) with capacity (e l+1 -e l ) if and only if i Î  m(l).
(v) Finally, there is an arc (s, t) from node s to node t with capacity ∞.
The source node s consists of n + 1 emanating arcs.Let f(x, y) denote the flow in the arc from node x to node y, i.e. node x and node y are a pair of nodes connecting this arc.A positive f(s, t) indicates that an amount of flow equal to f(s, t) cannot be assigned to the machines under the upper bound U, i.e. there exists no feasible schedule in which the completion time of each job is no greater than U.This happens if U is a trial value which is incorrectly set or the heuristic algorithm is incorrectly calculated.Denote the latest arrival times of agent A and B and the time epochs Q and U as e u , e w , e z and e v , respectively.
The corresponding directed tripartite network G(U) using the data of Numerical Example 1 is shown in Figure 4.  Table 3 illustrates that the deviation of the heuristic from the optimal solution appears in descending trend as the value of m increases.The reason is that the heuristic selects the maximum total remaining processing time of each job, instead of that of each machine, to be processed on the corresponding machine.For the rate of machine availability θ, the value of 1 showing that all machines are available at any times.Since none of , i = 1, ..., m, k = 1, ..., N(i) incurred, the number of time intervals decreases and the span of time interval increases, therefore more jobs are competing to be scheduled in each time interval and hence the error increases.As to the number of jobs, the higher value that n is, the better the performance of the heuristic.One reason is that the higher value for n implies more time epochs incurred and a smaller time span for each T l making the heuristic easier to assigning the operations correctly.Another reason is that the denominator increases, whereas the deviation of the heuristic solution from the optimal solution may not increase in proportion to the denominator.There is no significant difference in the performance of the heuristic on the value of n a /n b , but n a /n b = 0.5 gives the best performance.When n a /n b increases, the deviation increases due to the fact that the heuristic gives priority to agent B and thus more operations of agent A should compete to schedule in each time interval.
Based on this analysis, we found that decreasing machine number m and the value of n a /n b reduced the deviation of the heuristic, while decreasing job number n increased the deviation of the heuristic.The heuristic generates a good quality schedule with a deviation from the optimum of 0.25% on average.
As to the execution time of the heuristic, Table 4 shows the average execution time of the network based linear programming and the heuristic.The average execution time of the heuristic is small compared to that of the linear programming.are shown in columns lingo, OPL, and Total, respectively.For n = 80 and m = 10, the average execution time for the combined model exceeds two hours.Therefore, an efficient heuristic algorithm is in great need.

Conclusion
In this paper, we have analyzed the preemptive open-shop with machine availability and eligibility constraints for two-agent scheduling problem.The objective is to minimize makespan, given that one agent will accept a schedule of time up to Q.This problem arises in TFT-LCD and E-Paper manufacturing wherein units go through a series of diagnostic tests that do not have to be performed in any specified order.We proposed an effective heuristic to find a nearly optimal solution and a linear programming model based on minimum cost flow network to optimally solve the problem.Computational experiments show that the heuristic generates a good quality schedule with a deviation of 0.25% on average from the optimum.
Agents A and B have n a (n b ) jobs.Let n denote the total number of jobs, i.e., n = n a + n b , and each job j of agent A (B) is denoted by .The processing time of a job j of agent A (B) on machine i is denoted by .The release time of a job j of agent A (B) is denoted by .Each machine i is available for processing in the given N(i) intervals, which are , i = 1, ..., m, k = 1, ..., N(i) and , where and are the start time and end time of the kth availability interval of machine i, respectively.The operations of job j of agent A (B) are processed on a specified subset of the machines in an arbitrary order.We use to denote the completion time of J j for agent A (B) and to denote the makespan.The objective is to minimize makespan, given that agent B will accept a schedule of cost up to Q.According to the notation for machine scheduling, the problem is denoted as , where O indicates open machines, NC win means that the machines are not available in certain time intervals, pmtn a(b)

(
2004) studied a two-agent setting for a single machine, two-machine flowshop and two-machine open shop environments.The objective function value of the primary customer is minimized subject to the requirement that the objective function value of the second customer cannot exceed a given number.The objective functions are the maximum of regular functions, the number of late jobs, and the total weighted completion times.The problem in a similar two-agent single machine was further studied by Cheng, Ng and Yuan (2006, 2008), Ng, Cheng and Yuan (2006), Agnetis, Pacciarelli and Pacifici (2007), Agnetis, Pascale and Pacciarelli

First
, we rank all , , , , and Q in nondecreasing order and put them into the time epoch set E. Let e l be the lth time epoch in E, i.e., E = {e 1 , e 2 , e 3 , ..., e v }, where e l < e l+1 and .Let T l be the lth time interval between two adjacent time epochs e l and e l +1 , where l = 1, …, |E|-1.L e t m(l) be the set of machines that are available in T l , i.e., m(l) = {i|integer k, 1 ≤ k ≤ N(i), such that and }.

Figure 1 .
Figure 1.Distinct time epochs in set E of and the maximum machine load of m 1 .Since y = 1, update the values of , and a s 3, 6 and 2, respectively.Since there is no more available operation in the time interval [e 2 , e 3 ], we set l = 3.Using the same manner, the time interval [e 3 , e 4 ], [e 4 , e 5 ], [e 5 , e 6 ], [e 6 , e 7 ] and [e 7 ,e 8 ] are scheduled.In the time interval [e 8 , e 9 ], the available jobs for agent B are and is selected to be processed on m 3 .Since y = 3, update the values of , and as 0, 3 and 13, respectively.Since  E, therefore return to Step 3. Step 3. Let e 9 equal to = 13 and re-index time epochs in E. That is [e 10 , e 9 ], [e 11 , e 10 ], [e 12 , e 11 ],Iteration 2

Figure 3 .
Figure 3. Distinct time epochs in set E for Numerical Example 1

(
ii) Combination nodes consisting of job ， j = 1, 2, …, n x , x = a, b and time interval T l , where T l satisfies e l+1 ≤ U and ≤ e l for agent A and e l+1 ≤ Q and ≤ e l for agent B.

Figure 4 .
The optimal solution for Numerical Example 1 Using the convention that summations are taken only over existing arcs, we formulate O, NC win |pmtn a , , : pmtn b , , |C max : ≤ Q as a linear programming problem, called LP, as follows:

Table 1 .
Job data

Table 2 .
Machine data All values of , , and Q are ranked in ascending order.The corresponding time epoch set

Table 4 .
Execution times of Lingo and Heuristic for small-size instances -1119-Journal of Industrial Engineering and Management -http://dx.doi.org/10.3926/jiem.1352In Table 5, the average execution times in seconds for Linear Programming model, Constraint Programming model, and Combined model (Linear programming and Constraint Programming)

Table 5 .
Execution times of LP model, OPL model and those of both models -1120-Journal of Industrial Engineering and Management -http://dx.doi.org/10.3926/jiem.1352