Sub-Optimal Scheduling of a Flexible Batch Manufacturing System using an Integer Programming Solution

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1 Sub-Optimal Scheduling of a Flexible Batch Manufacturing System using an Integer Programming Solution W. Weyerman, D. West, S. Warnick Information Dynamics and Intelligent Systems Group Department of Computer Science, Brigham Young University, Provo, UT Abstract In order to determine necessary time to reach a desired quota, the manager of a flexible batch manufacturing system must know the best method of scheduling the factory. To determine a sub-optimal schedule, we first create a model of the factory. This model is used to simulate the factory. Using this simulation and an integer programming formulation, we determine a scheduling algorithm that attempts to minimize the amount of idle time of the factory. I. INTRODUCTION In a flexible batch manufacturing system (BMS), materials are processed by a chain of machines that each perform a different job. This chain of machines has the ability to process multiple types of goods. Managers of these manufacturing systems want to know not only what the plant production capacity is, but how to run more efficiently in order to increase the capacity. This information is important in making decisions such as determining how many orders a manufacturer can process and whether to make modifications or additions to existing plants in order to increase production. It is also important to know the expected capacity of a plant planned to be built or purchased. Manufacturers also have quotas that they must have the capability to meet in a certain amount of time. In order to find the highest capacity for an existing factory, it is desirable to know how to process the products in the fastest time possible; this is often posed as a scheduling problem. The manufacturing scheduling problem has been studied for non-batch flexible manufacturing systems in [4] and [5] among others. The scheduling problem for BMSs, however, has not received as much attention. When approaching the scheduling problem in a BMS with multiple workstations and a fixed quota, there is an optimal ordering of products that results in the BMS sitting idle for the least amount of time. If this idle time can be minimized, the manufacturing system can run as close to capacity as possible, thus minimizing the total time to reach quota, or make-span, and maximizing plant capacity. This is the approach taken in this paper. This paper presents a method of modeling a BMS using a discrete-time dynamic system. This model is used in a simulation to calculate a cost of switching products. The This work is supported in part by grants from ATK Thiokol, Sandia National Labs, and ORCA. Please direct comments and questions to W. Sam Weyerman at wsweyerman@gmail.com or Sean Warnick at sean@cs.byu.edu. optimization problem is then posed as a graph traversal problem and then solved using an integer programming method. The final section of this paper gives the solution to the problem of minimizing the time to produce a quota of 9 different products in a manufacturing system with 11 different types of machines. II. DESCRIPTION OF NOTATION AND PROBLEM A. Notation and Definitions Following are definitions of the terminology used in this paper. Let a machine be a general or specific-purpose resource that performs a specific job in processing a product. A machine may only be used or occupied by one product at any given time. A workstation contains one or more machines of the same type, or machines that perform the same job. A single machine requires a batch of product to be filled to capacity, the batch size of one machine may not be the same as another machine s batch size. The minimum amount of material required to run all of the material through the factory is a load; a load must be an integer amount of the largest batch in a particular route. Let an operation be represented by b,c,τ}, where b is the number of the prior machine s batches that are required to fill this machine and is limited to the positive rational numbers, c is the workstation on which this operation is to be performed, and τ is the cycle time of this machine for a certain product. Let a production line, or factory, P = F,O,Q} where F is the total number of each machine present in the factory, O is the operation matrix and defines the route of each product, and Q is the quota of each product in loads. O can be divided into three operation matrices, O b, O c, and O τ representing the three parts of an operation. The factory is said to be dominated by a product if that product has entered the final machine of its route. A product has entered the factory when it is loaded into the first machine of its route. A product is said to have left the factory when the entire load of that product has been processed. The run time, r, of a product is the time that passes from when a product dominates the factory until it leaves. The steady state run time, r ss, is the run time of a product from the empty factory. The delay, D, of a product is the time that passes from when a product enters the factory to the time it dominates the factory plus the difference of the actual run time and the steady state run time. The wait time of a product, W, is the time that a product must wait before

2 it can enter the factory. The transition time is the time it takes to switch from one product to another, specifically the time it takes for a new product to dominate the factory from when the previous product has left the factory. Transition time is used as the cost of running one load after having run a load, cost is determined not only for changing products, but also for continuing to process the same product. A strategy S tells the order to run the products and how many loads of each to run. B. Properties of the manufacturing system Several limitations, most of which are consistent with operation of a chemical plant, are put on the BMS considered in this paper. In this BMS, a general ordering of the workstations dictates the route of the batches of product travelling through the manufacturing line. In rare cases, feedback is allowed to a previous workstation, this introduces the possibility of deadlock as will be briefly addressed. The operation matrix, O, is fixed, thus not allowing modification to the production specifications of any product. Any machine type may have a workstation of multiple machines allowing for parallel processing at any stage of production. A workstation may be skipped by the route of any product: certain production steps are not necessary in the manufacture of certain products. At any given time, there may be a choice of products to load into a machine. When this choice is presented, we use a first come first served policy, the product that entered the factory first is given priority, this policy is slightly modified in the case of feedback. This policy disallows the possibility of a product passing another product in the manufacturing line. There is no queueing between machines, storage is only within the machines, so no product may proceed until there is an available machine in the next workstation of that product s route. Transport time between machines and emptying and filling time of machines is considered to be negligible. There may or may not be setup times when switching products, this time will be accounted for in the transition time. Machines will only run at capacity, so no machine will run until it has been completely filled. A machine may be partly filled with a product at which point it will wait until it has a full batch of that product before running. Yearly quota for the BMS is assumed to be known beforehand allowing for more flexibility in scheduling without forecasting demand, thus no product in the quota will have a deadline before the end of the year. In order to maintain the deterministic nature of this manufacturing system, machines are also assumed to never break down or need maintenance. C. The Problem This factory is to be used to meet given quotas in a minimum amount of time. Given a production line P, we want to minimize the make-span in when meeting quota, Q. Let C i be the completion time of product i, in a factory that makes m products, the make-span is given by C max = maxc 1,...,C m }. [2] gives a formulation for C i for a simple job-shop without parallel machines that does not hold for a factory with parallel machines. In the factory we are considering, there are no decisions once a product has entered the factory, it will flow through the factory until completion. So C i = r ssi + W i + D i. Because changing the b, c, or τ for any given operation will change the quality of the product, we must choose S such that the make-span is minimized. Thus we wish to solve: min C max S III. SIMULATION OF THE FACTORY A. Modeling the factory In order to simulate the factory, we first had to model it. Initially each machine can be viewed as a finite-state automata with a set of states Θ containing three states: loading (θ l ), running (θ r ), and unloading (θ u ). These states can be defined in terms of the portion of a full batch the machine contains, l [0,1], and the time the machine has been running, T Z. The current state, θ is defined as follows: θ = θ l l 1 T = 0 θ r l = 1 T < τ θ u T τ Assume a factory with n machines that can manufacture m different products. In order to model the factory as a dynamic system, let the state of a machine be defined by M = [l T π o], where l and T are defined as before, π N is the current product in the machine, and o N is the operation number of the product in the machine. Let R = [r 1... r m ],r i Z be the remaining supply, or raw materials, for each product given in number of batches of the first workstation for this product s route. Also, let Ψ = [ψ 1... ψ m ],ψ i Z be the amount of each product completed, given in batches of the final machine in the product s route. The state of a factory is given by x = [R M 1... M nψ ]. Allowing for workstations requires a trivial mapping from machine number to workstation number. The input to the factory, u, is the amount of raw materials added and completed product removed; the schedule is input into the factory through u. The factory can then be represented as a non-linear dynamic system: x(k + 1) = f(x,u,k) The function f(x,u,k) specifies the loading, running, and unloading of machines. When a machine is finished running, it makes the material in that machine available to the next workstation in that product s route. If a machine is empty or loading, it loads from the material that has been made available to it by a previous machine. If a loading machine is the first machine in the next product s route, then it loads from R. To determine which product to load next, the first machine checks the supply and loads the next available product, or the first non-zero entry in R. If the machine is the final machine in its product s route, then it unloads into Ψ. Loading from R and unloading to Ψ is done instantaniously.

3 The amount unloaded and loaded are given by the following equations: l i (k + 1) = l i(k) min l i (k),(1 l next (k))(o b,pnext )} l + i (k + 1) = l i(k) + min 1 l i (k), l } prev(k) O b,pprev The load of machine i changes according to: l + i (k + 1) θ i = θ l and θ prev = θ u l i (k + 1) θ i = θ u and θ next = θ l l i (k + 1) = 1 o i = 1 0 o i = o final l(k) otherwise When a machine (i) is running, T is incremented at each time step. If the machine is empty, T is reset to 0. The update function for T is given by: T i (k + 1) = T i (k) + 1 θ i = θ r 0 l i = 0 T i (k) otherwise The values for π and o are set when loading a product from machine i to machine j, the update functions for π and o are: πi (k) θ π j (k + 1) = i = θ u and l j = 0 (5) π j (k) otherwise oi (k) + 1 θ o j (k + 1) = i = θ u and l j = 0 (6) o j (k) otherwise R and Ψ only change when the first machine of a product s route is loaded, the last machine is unloaded, or an input to the system changes them. Let i be the first machine in a product s route and j be the last, R and Ψ are updated as: ri (k) + u r i (k + 1) = ri (k) 1 θ i = θ l (7) r i (k) + u ri (k) otherwise ψi (k) + u ψ i (k + 1) = ψi (k) + 1 θ j = θ u (8) ψ i (k) + u ψi (k) otherwise These functions combine to form f(x,u,k). B. Deadlock Due to Feedback When feedback is introduced into a BMS, deadlock becomes a possibility. In [1], four conditions for deadlock are given; each of these will be considered. 1) Mutual Exclusion: Mutual exclusion is a condition that requires that any resource may not be used by more than one task at any time. Since each machine may only be occupied by a single batch of a single product at any given time, mutual exclusion must be preserved. Due to this, this condition for deadlock must be kept. 2) Wait for: Wait for means that a task can hold a resource while waiting assignment of other resources. Since there is no queueing, each batch of product will hold the machine it is currently in while it waits for the next machine in its route to be available. If finite sized queues were allowed each batch would still have to hold a spot in the queue while it waits for the next machine to become available. Thus the wait for condition of deadlock must be kept. (1) (2) (3) (4) M1 M2 M3 M4 Fig. 1. Example of deadlock. Two products, green and red, have different routes. M1 and M2 are occupied by green while M3 and M4 are occupied by red. The route of red feeds back to M2. In this example circular wait is present: green is waiting for M3 and red is waiting for M2, both are held by the other product. M1 M2 M3 M4 Fig. 2. A possible solution. M1 is occupied by green while M3 and M4 are occupied by red, M2 is held by red. Since red holds M2 while it has any product in M3 or M4, both of these batches of product will be able to continue to completion. After red is completed, M2 will be released and green is free to proceed. 3) No preemption: Preemption is removing a resource from a task currently holding it. In a manufacturing environment preemption is an unreasonable thing to do. In order to remove a machine from a batch in process, the batch must be removed from the manufacturing line, thus destroying it. Preempting a batch in process would require wasting all of the material and time that has been used in that batch up to this point. Preemption is not an option. 4) Circular wait: When circular wait is present, there is a circular chain of tasks where each task holds a resource that the next task is requesting. In an open loop manufacturing chain, there is no chance for circular wait to happen since there cannot be a circular chain of products. Once feedback is introduced circular wait becomes a possibility. Figure 1 shows and example of the four conditions for deadlock being met. When creating a simulator, the issue of deadlock must be addressed to ensure that deadlock never happens during a simulation. In [6] it is shown that the first come, first served (FCFS) policy is unstable. When queue length restrictions are made, [3] shows that this leads to deadlock and gives a petri net scheduling solution to avoid deadlock in a flexible manufacturing system. To guarantee that the system did not enter a deadlocked state in our simulation, the circular wait condition was removed. This was accomplished by maintaining an empty machine if there is a product in process that will feedback into that machine. When the product reaches a machine with workstation number greater than or equal to the workstation that it feeds back into, the simulator will ensure that there is at least one machine within that workstation that is always available until the product has passed through the feedback loop. An example of using this method to avoid feedback is shown in figure 2. This collapses the feedback chain into a single wait making the whole route resemble an open loop

4 manufacturing chain. This method will cause unnecessary delays in a BMS with multiple feedback loops on multiple products. IV. METHOD OF SOLUTION In order to solve the time minimization problem it has been formulated as an integer programming (IP) problem. The objective is to minimize the running time required to satisfy the given quota. This will give us the order to run the products until we have reached quota. We will view scheduling as traversing a graph. If product a is run and then product b, it will be represented by a transition from node a to node b. The cost of transitioning from node a to node b will be given by the calculated transition time from product a to product b. The total minimum transition time will be the lowest cost of traversing the graph and visiting each node the required amount of times to reach quota. The resulting formulation is based on the IP formulation of the travelling salesperson problem (TSP) given in [7]. A. The IP Formulation Let the matrix C be the cost matrix for a directed graph representing the transition costs of every product that we desire to make. The cost matrix only represents the firstorder transition times, meaning the time to transition from a b is starting with an empty factory, running product a and then running product b. This graph must contain a dummy node (0) which represents the empty factory. It is not necessary to compute the time to transition to an empty factory since this is given by the time to run the final product and is not a transition, so the transition time to the empty factory is always 0. There must be exactly one transition from node 0 to any other node and one transition going to node 0. C is thus a [m + 1 m + 1] matrix. The matrix X specifies which transitions to make to reach quota and is the same size as C. If we let x ij represent the number of times the transition from node i to node j is traversed and c ij the cost of that traversal, the objective function for the IP problem is thus: minimize j=0 c ij x ij Because we eventually want a tour, each node must have the same number of incoming arcs as outgoing arcs. This adds the constraint: x ij x jl = 0j = 0,1,...,m l=0 If the vector q is the quota vector then there must be enough arcs entering each node to meet quota; we will set the quota of node 0 to be 1. This will constrain at least one arc coming in and one going out. The following constraint is added: x ij q j j = 0,1,...,m However if going over quota is allowed, the objective function must include the time the factory must run to make a load of any product (R), thus appropriately adding time for going over quota, so the new objective function is: minimize (c ij + r ssj )x ij j=0 The values of X are limited to be integer values because they are the number of times each transition is to be taken. The full IP problem is: minimize subject to j=0 (c ij + r ssj )x ij x ij x jl = 0 l=0 j = 0,1,...,m x ij q j j = 0,1,...,m x ij 0,1,...} This formulation allows for disjoint sets. Let ς be the smallest subtour or disjoint set, ς is a set containing each transition in the subtour. Since we want to find a single tour, this problem may be iteratively solved adding a constraint each iteration to break the smallest subtour, ς, of length ς length : (9) x ij < ς length (10) i,j ς Once there is a single tour, the matrix X may be used to create S by following the transitions starting with node 0 and ending with node 0 traversing every transition defined in X. B. Calculating the Cost Matrix The cost matrix is calculated by running every combination of two products (including the same product repeated) through the simulation. The time from when the first product leaves the factory until the second product dominates the factory is recorded as the cost from the first product to the second. C. Future Work This method of modeling the transition costs assumes that there is a high correlation between the first order transitions and higher order transitions. This model will not capture the difference of the transition time from product a to product b when starting from an empty factory, processing a then b and starting from an empty factory, processing b then a followed by b. We want to find a relation between the transition time and the delay of a product given a schedule. Such a result may present a method of determining bounds of goodness on the first order transition cost.

5 Further generalizations to the factory, such as allowing products in process to be passed by products with shorter manufacturing times may not result in a sensible transition time using the current heuristic. In order to allow for further generalization, we need to determine a different method of calculating costs. V. EXAMPLE Let P be a factory with 11 different workstations and nine products defined as follows (all times are in minutes): O b : F : ( ) O c : O τ : Q : ( ) The transition cost matrix (C) for these products was calculated to be: The solution to the LP using this cost matrix and allowing disjoint sets is:

6 Strategy IP Cost Sim. Time S OPTT SP S OPTtour S BLOCKT SP S OPTtour S BLOCKnaive S BLOCKrandom S RANDOM TABLE I TOTAL TIMES FOR EACH STRATEGY BASED ON C, R, AND SIMULATION TIMES By manually combining this solution and the corresponding solution to the TSP ( ), the optimal strategy with respect to the TSP (S OPTT SP ) was derived to be: [2] S. French. Sequencing and Scheduling: An Introduction to the Mathematics of the Job-Shop. Mathematics and its Applications. Ellis Horwood Limited, [3] A. Gürel, S. Bogdan, and F. L. Lewis. Matrix approach to deadlock-free dispatching in multi-class finite buffer flowlines. IEEE Transactions on Automatic Control, 45(11): , November [4] F. Martinelli and P. Valigi. The impact of finite buffers on the optimal scheduling of a single-machine two-part-type manufacturing system. IEEE Transactions on Automatic Control, 47(10): , October [5] J. F. O Kane, D. K. Harrison, and V. I. Vitanov. An AI approach to scheduling in flexible manufacturing systems. Factory 2000, Competitive Performance Through Advanced Technology., Third International Conference on, pages 24 28, July [6] T. I. Seidman. First come, first served can be unstable! IEEE Transactions on Automatic Control, 39(10): , October [7] R. J. Vanderbei. Linear Programming: Foundations and Extensions. Kluwer Academic Publishers, 2 nd edition, ,9,9,3,3,2,2,1,4,1,4,1,5,5,8,8,8,7 By iteratively adding subtour breaking constraints to the IP, it was possible to find two optimal tours (S OPTtour1,S OPTtour2 ) which respectively are: 6,9,3,8,9,3,2,1,4,1,4,1,5,5,2,8,8,7 6,9,3,8,9,3,2,2,1,4,1,4,1,5,5,8,8,7 Four more strategies were created for testing purposes, S BLOCKT SP, S BLOCKnaive, S BLOCKrandom, and S RANDOM. These strategies are: S BLOCKT SP :6,9,9,3,3,2,2,4,4,1,1,1,5,5,8,8,8,7 S BLOCKnaive :1,1,1,2,2,3,3,4,4,5,5,6,7,8,8,8,9,9 S BLOCKrandom :8,8,8,2,2,7,4,4,3,3,6,9,9,5,5,1,1,1 S RANDOM :8,6,8,3,9,9,7,3,5,1,1,2,1,8,5,4,2,4 The results of running each of these strategies are found in Table I. Although OPT tour1 and OPT tour2 had the best upper bound, OPT TSP had the best overall running time. There is a discrepancy between best upper bound and best actual running time, but it is relatively small when considering smart strategies. The discrepancy between the smart strategies derived using the IP and the other strategies is quite large. There is certainly an advantage to formulating the IP and finding the solution. VI. CONCLUSION We have derived a somewhat general model for a BMS. This model has been used to build a simulation. Using the simulation we determined the cost of transitioning from one product type to another. This cost is used to solve a minimization problem posed as an IP problem. The solution to this problem gives a sub-optimal ordering of product types in order to produce the most product in a minimum amount of time. REFERENCES [1] E. G. Coffman, Jr., M. J. Elphick, and A. Shoshani. System deadlocks. Computing Surveys, 3(2):67 78, June 1971.