A Disjunctive Model for Scheduling in a Manufacturing and Packing Facility with Intermediate Storage

Optimization and Engineering, 1, 215–231, 2000c© 2000 Kluwer Academic Publishers. Manufactured in The Netherlands.

A Disjunctive Model for Scheduling in a Manufacturingand Packing Facility with Intermediate Storage

VIPUL JAIN, IGNACIO E. GROSSMANN∗Department of Chemical Engineering, Carnegie Mellon University, Pittsburgh, PA 15213, [email protected]; [email protected]

Received August 6, 1999; Revised April 3, 2000

Abstract. The problem considered in this paper deals with the short term scheduling of a two stage continuousprocess with intermediate storage tanks. The major scheduling decisions in this problem are: a) the assignmentof orders to various storage tanks; b) the sequence of orders in each unit; c) the timing of various operations indifferent stages. The problem is highly combinatorial in nature. The major challenge is to develop strong integerprogramming formulations and to devise efficient solution techniques. An initial model is presented in the formof a disjunctive program which is later transformed to a Mixed Integer Linear Programming (MILP) problem. Anumber of example problems are solved which highlight the limitations of this model as the number of ordersincreases. A heuristic based on partial preordering is considered which solves industrial sized problems veryquickly. The objective function values for the heuristic solutions are within 7% of the optimal values.

Keywords: scheduling, disjunctive programming, mixed integer linear programming, preordering heuristics

1. Introduction

Many chemical processing plants produce different products using a variety of raw materialswith the same set of processing units. As a result, there is a need for tools and systematicmethods for scheduling the manufacturing of a portfolio of products, in these systems. Inthe past a great deal of attention was placed in chemical plants that operate in the batchmode. Recently, there has been an increase in the research for the scheduling of continuousprocess plants (Pinto and Grossmann, 1995; Sahinidis and Grossmann, 1991; Schillinget al., 1994). The major concerns in continuous scheduling may differ from the ones arisingin batch scheduling. This was highlighted in the work of Pinto and Grossmann (1995) andSahinidis and Grossmann (1991) where the selection of cycle times and the handling ofintermediate storage and inventory were the main issues. In this paper we investigate aspecial class of continuous scheduling problems. It deals with the short term scheduling ofa two stage continuous system with intermediate storage.

Most of the work regarding short term scheduling has focused on batch scheduling.Extensive reviews on batch scheduling have been reported in the literature (Shah, 1998;Zenter and Pekny, 1994). A detailed review on mixed integer based modeling techniquesfor scheduling has been presented by Pinto and Grossmann (1998). One of the main issues

∗Corresponding author.

216 JAIN AND GROSSMANN

in modeling scheduling problems is the time domain representation (Pinto and Grossmann,1995). Kondili et al. (1993) presented a general formulation based on a discrete timerepresentation, the state task network (STN), to model short term scheduling problemsthat arise in batch plants. The model, however, proved to be computationally very ex-pensive. Following this work, Shah et al. (1993) introduced a reformulation and severalsolution techniques with the objective of reducing the required computational effort. TheSTN model was extended to an even more general framework by Pantelides (1994), namelythe resource task network (RTN). Using the RTN framework a number of models based ona continuous time representation were proposed by different authors (Zhang and Sargent,1996a, 1996b; Reklaitis and Mockus, 1995; Schilling and Pantelides, 1996; Ierapetritouand Floudas, 1998a, 1998b).

The models proposed by Zhang and Sargent (1996a, 1996b), Schilling and Pantelides(1996), and Ierapetritou and Floudas (1998a, 1998b) can handle both continuous and batchprocesses. All these models have been limited to a single objective function i.e. maximiz-ing profit. In practice, depending on the problem, a different objective function might bedesirable e.g. minimizing makespan. A general model based on a continuous time repre-sentation for the short term scheduling of flowshop batch plants was presented by Pintoand Grossmann (1995). In the subsequent paper, these authors presented an alternativemodel with preordering which was computationally very fast for large problems (Pinto andGrossmann, 1996). For continuous processes, the continuous time domain representation ismore relevant as no approximation is then required for the processing times. The heuristicalgorithms that exploit the specific nature of the problems to solve them quickly have provedto be quite useful (Adams et al., 1988; Baker, 1994; Pinto and Grossmann, 1996). However,the disadvantage is that the optimality of the schedule cannot be guaranteed.

The major goal of this paper is to develop an MILP model for the scheduling of a two stagecontinuous system with intermediate storage. Based on some simplifying assumptions, thisproblem is initially modeled as a disjunctive program using a continuous time representation.Applying the equations of convex hull for each disjunction, this model is then transformedto a Mixed Integer Linear Programming (MILP) model, and a number of numerical exampleproblems are solved using a commercial MILP solver CPLEX 6.5 (1999). The computationalresults for these examples indicate that computational time increases rapidly as the problemsize increases, prohibiting the solution of industrial sized problems. A heuristic based onthe partial preordering of orders is presented to solve the industrial sized problems withmodest computational effort. The objective values for the heuristic solutions are within 7%of the optimal values.

2. Problem description

This problem is motivated by a manufacturing and packing facility of a fast moving con-sumer goods company. The process consists of two main stages (See figure 1). A product ismanufactured using a continuous processor (manufacturing stage), and is simultaneouslyfed into an intermediate storage tank. The product is then packed into cartons and bottlesof different sizes on continuous parallel lines (packing stage). It should be noted that thenumber and capacity of storage tanks is limited. Furthermore, a storage tank can be con-nected only to one manufacturing and one packing unit at a time and vice versa. The demand

A DISJUNCTIVE MODEL FOR SCHEDULING 217

Figure 1. Two stage process with storage.

(in tons) of each product in a particular type of packing defines an “order”, e.g. 15 tons ofliquid detergent in 2 L bottles. Each order requires the use of a particular manufacturing andpacking unit, and the processing rates for each order in these units is fixed. Furthermore,there is a cleanup time between the orders in all the stages of process which is independentof the sequence of manufacturing. At a given point in time there are a set of orders to bemanufactured and packed. The goal is to find a schedule to complete all the orders in theminimum possible time, i.e. minimize makespan.

3. Assumptions

The problem as stated is complex, and therefore we make some simplifying assumptionsregarding the operational policy in the manufacturing process. These assumptions are basedon engineering insights to reduce the problem size and make it solvable. The assumptionis that each order is manufactured in one lot, as a result of which processing time formanufacturing and packing of each order can be determined a priori. The assumption isreasonable because cleanup is required for every changeover. Also, it is assumed that thecapacity of storage tanks is more than the size of each order. At first sight, it might appearthat this assumption will impose a major limitation on the proposed model. However, thisassumption is not violated for most of the orders and larger orders can always be dividedinto smaller orders to satisfy the capacity constraint. Furthermore, it is assumed that orderscannot be mixed. It should be noted that in theory the proposed assumptions may leadto a suboptimal solution, but our experience shows that the solutions obtained using theproposed methods are within 7% of the best possible solution because of certain problemcharacteristics that will be highlighted later on.

4. Modeling concept

In this section, we present the underlying modeling concept that is used in the next sectionto develop a model that uses disjunctions and logic propositions to solve the scheduling


Figure 2. Timing constraints.

problem at hand, given the simplifying assumptions. Since, the manufacturing and packingstages are continuous operations, the output from the manufacturing unit has to be sent toa storage tank simultaneously. The manufacturing of a product cannot begin unless thereis a clean and empty storage tank available. By the same token, packing of a particularproduct cannot begin unless it has already been manufactured or it is being manufactured.The storage tank can be used for any other order only if packing and manufacturing of anearlier order is complete. The utilization of the manufacturing and the packing unit, and astorage tank for each order is as shown in figure 2. Note that while the processing times inthe manufacturing (dik) and packing (dik ′′ ) units are known, the time for the intermediatestorage is unknown (dik ′ ).

To model this problem the concept of production paths is used. Each path is an orderedset of units used for the production of each order. The units required to manufacture andpack a particular order are unique, and therefore the paths for each order differ only becausedifferent storage tanks can be used. Hence, the number of paths for each order is equal tonumber of storage tanks. Paths for an order that uses Manufacturing Unit 1 and PackingLine 1 are shown in figure 1. The idea of paths is useful because at any given time onlyone storage tank can be connected to a manufacturing and a packing unit. It should benoted that this idea is fairly general and can be easily extended for the case when there is achoice between parallel manufacturing or/and packing units to complete each order. Usingthis idea we can now formulate a model in terms of selection of paths using disjunctionsand logic propositions. The objective of the model is to find a schedule for manufacturing,storing and packing various orders, so as to minimize the makespan.

5. Disjunctive model

Before presenting the details of the model we will first describe various indices, sets,parameters, and variables used in the model. An order is denoted by an indexi and theset of all the orders to be completed is denoted byI. For simplicity all the manufacturingunits, packing lines, and storage tanks will be referred as units. A unit is denoted by indexk. Units are divided into three sets. The set of manufacturing units is denoted byK1, the setof storage tanks is denoted byK2, and the set of packing units is denoted byK3. Indexh isused to denote a possible manufacturing path for an order. The set of possible paths for a


particular orderi is denoted byHi . Note that an elementh of setHi is an ordered set of theform (k, k′, k′′) such thatk ∈ K1, k′ ∈ K2, k′′ ∈ K3. Let dik denote the processing time oforderi in unit k (notek ∈ K1 ∪ K3) andcik denote the cleanup time after processing orderi in unit k (notek ∈ K1 ∪ K2 ∪ K3).

There are both continuous and discrete decision variables associated with the problem.The continuous variables are start times of an orderi on any unitk, duration of an orderi in astorage tankk (k ∈ K2), and overall makespan of the schedule. They are denoted byTik , dik ,andMs, respectively. The discrete variables associated with the problem are the decisionsregarding the path to be used to manufacture a particular order (xih), and the sequence ofany two orders on any particular unit to resolve resource limitations (yii ′k). The booleanvariablexih is true if orderi is manufactured using pathh and false, otherwise. Similarly,the boolean variableyii ′k is true if orderi is completed beforei ′ on unitk. It should be notedthat these variables are defined only for a subset of combinations ofi andk.

The optimization problem corresponding to the scheduling problem at hand can be writtenas a minimization problem subject to a set of constraints. The objective of the optimizationproblem is to minimize the makespan (Ms). The constraints for the optimization problemcan be written in the form of disjunctions and logic propositions. They can be broadlydivided into four categories and are as follows.

5.1. Timing constraints

These constraints ensure that for a specific path chosen to complete a particular order,the start times and end times of that order on various units involved in that path have thefollowing relationship:

• the start time in manufacturing unit and storage tank is the same.• the end time in manufacturing unit is less than or equal to end time in storage tank• the start time in storage tank is less than or equal to start time in packing unit• the end time in storage tank and packing unit is the same

These relationships are clearly shown in figure 2. Since only one pathh can be selected(h ∈ Hi ) for a given orderi, these constraints can be expressed in the form of disjunctionsover all paths.

∨h∈Hi

xih

Tik = Tik ′

Tik + dik ≤ Tik ′ + dik ′

Tik ′ ≤ Tik ′′

Tik ′ + dik ′ = Tik ′′ + dik ′′

∀i ∈ I (1)

Recall thath is an ordered set of units of the form(k, k′, k′′), wherek ∈ K1, k′ ∈ K2,k′′ ∈ K3. A boolean variablexih is associated with each term of the disjunction and ifit is true then the corresponding disjunct is active. We can simplify Disjunction (1) by


eliminatingTik andTik ′ from the second inequality,

∨h∈Hi

xih

Tik = Tik ′

dik ≤ dik ′

Tik ′ ≤ Tik ′′

Tik ′ + dik ′ = Tik ′′ + dik ′′

∀i ∈ I (2)

5.2. Sequencing constraints

These are typical sequencing constraints and they resolve the resource limitations by se-quencing two different orders that use the same unit. If a particular orderi ′ is processedafter some other orderi in a particular unitk then the start time of orderi ′ is greater than orequal to the end time of orderi. For units belonging to manufacturing and packing stagesthese constraints can be written as,(

yii ′k

Ti ′k ≥ Tik + dik + cik

)∨(¬yii ′k

Ti ′k ≥ 0

)∀i, i ′ ∈ I , i ′ 6= i,

k ∈ K i ∩ K i ′ ∩ (K1 ∪ K3)(3)

For the storage units these constraints can be written as,(yii ′k

Ti ′k ≥ Tik + dik + cik

)∨(¬yii ′k

Ti ′k ≥ 0

)∀i, i ′ ∈ I , i ′ 6= i,

k ∈ K i ∩ K i ′ ∩ K2(4)

Here,K i andK i ′ denote the set of units that are involved in any of the paths correspondingto ordersi andi ′, respectively. The difference between the two set of constraints is that theduration of each order is a constant in manufacturing and packing unit (dik), and is variablein a storage tank (dik).

5.3. Logic constraints

There are two categories of logic constraints, essential constraints and non-essential con-straints. Essential constraints are the ones without which the model is incomplete. Con-straints relating the assignment decisions to sequencing decisions fall into this category.Non-essential constraints are redundant constraints and including them in the model doesnot change the optimal solution. However, these constraints sometimes can speed up thesolution procedure. Constraints relating sequencing decisions in different stages fall intothis category. All these logic constraints can be represented as logic propositions.

5.3.1. Essential constraints.These constraints enforce the relationship between the as-signment variables and the sequencing variables. The logical relationship can be stated asfollows.


“If a path corresponding to an order i that requires unit k is selected and a pathcorresponding to an order i′ that also requires unit k is selected then there is a sequencingconstraint between order i and i′ on unit k”

This relationship can be enforced using the boolean variablesxih, xi ′h′ , yii ′k, andyi ′ik as,( ∨h∈Hk

i

xih

)∧( ∨

h′∈Hki ′

xi ′h′

)⇒ yii ′k ∨ yi ′ik ∀i, i ′ ∈ I , i ′ > i, k ∈ K i ∩ K i ′ (5)

Here, setsHki andHk

i ′ denote the set of paths corresponding to ordersi andi ′ that use unitk. This constraint is symmetric fori andi ′ and duplication of constraints has been avoidedby enforcingi ′> i . Note that this relationship can be simplified for the case when a specificunit is always used in a particular stage for an order. The simplification is based on theobservation that all the paths corresponding to that order will always use that particularunit. For these units, the logic propositions∨h∈Hk

ixih and∨h∈Hk

i ′xi ′h′ are always true. Hence,

the logical relationship can be simplified as follows,

yii ′k ∨ yi ′ik ∀i, i ′ ∈ I , i ′ > i, k ∈ K i ∩ K i ′ ∩ K j (6)

Here,K j is the set of units belonging to stagej in which a specific unit is used to complete anorder. This simplified relationship can be used for manufacturing and packing stage becausethe manufacturing unit and the packing line to be used for each order is fixed beforehand.The set of essential logical constraints for this problem can be summarized as follows.( ∨

h∈Hki

xih

)∧( ∨

h∈Hki ′

xi ′h′

)⇒ yii ′k ∨ yi ′ik ∀i, i ′ ∈ I , i ′ > i, k ∈ K i ∩ K i ′ ∩ K2 (7)

yii ′k ∨ yi ′ik ∀i, i ′ ∈ I , i ′ > i, k ∈ K i ∩ K i ′ ∩ (K1 ∪ K3) (8)

5.3.2. Non-essential constraints.These constraints are redundant. However, they canspeed up the solution procedure by propagating the logic relationships existing between thesequence of tasks in storage tanks and their sequence in manufacturing or packing units.It should be noted that these constraints are useful only when orders use a specific unit inmanufacturing or packing stage.

The first set of constraints enforces the fact that if two ordersi and i ′ that need thesame manufacturing unit use the same storage tank then the sequence of these orders in themanufacturing unit is the same as that in the storage tank. Note that this set of constraintsis valid only for the pair of orders that share the same manufacturing unit.

yii ′k′ ⇒ yii ′k ∀i, i ′ ∈ I , i ′ 6= i, k ∈ K1 ∩ K i ∩ K i ′ , k′ ∈ K2 ∩ K i ∩ K i ′ (9)

The second set of constraints enforces the fact that if two ordersi and i ′ that need thesame packing unit use the same storage tank then the sequence in the packing unit is sameas that in storage tank. This set of constraints is valid for the pair of orders that share thesame packing unit.

yii ′k′ ⇒ yii ′k′′ ∀i, i ′ ∈ I , i ′ 6= i, k′ ∈ K2 ∩ K i ∩ K i ′ , k′′ ∈ K3 ∩ K i ∩ K i ′ (10)


5.4. Makespan constraint

This constraint relates the end time of all the orders to the makespan of the overall schedule.It enforces the fact that the makespan of the schedule is greater than or equal to the endtime of all the orders in the units belonging to the packing stage.

Ms ≥ Tik + dik ∀i ∈ I , k ∈ (K3 ∩ K i ) (11)

The minimization problem corresponding to the scheduling problem at hand can be statedas follows,

(M1) min Ms

s.t. (2), (3), (4), (7), (8), (9), (10), and (11)

Tik ≥ 0 ∀i ∈ I , k ∈ K i

dik ≥ 0 ∀i ∈ I , k ∈ K i ∩ K2

xih ∈ {True, False} ∀i ∈ I , h ∈ Hi

yii ′k ∈ {True, False} ∀i, i ′ ∈ I , i ′ 6= i, k ∈ K i ∩ K i ′

6. MILP model

The constraints in the model (M1) are in the form of disjunctions and logic propositions.Currently, there are no algorithms available to solve the problems represented using disjunc-tions and logic propositions. However, such problems can be easily converted to an MILPproblem using known transformations. Note that for every boolean variable in model (M1)there is a corresponding binary variable in the MILP problem. For the ease of understandingthe same symbols have been used for the equivalent binary variables.

6.1. Timing constraints

Using the convex hull formulation for each disjunction (Balas, 1985; T¨urkay and Grossmann,1996), the set of timing constraints (2) can be written as:∑

h∈Hi

xih = 1

Tik =∑h∈Hk

i

thik ∀k ∈ K i

dik ′ =∑

h∈Hk′i

δhik ′ ∀k′ ∈ K i ∩ K2

thik = th

ik ′

xihdik ≤ δhik ′

thik ′ ≤ th

ik ′′

thik ′ + δh

ik ′ = thik ′′ + xihdik ′′

thik ≤ U xih, th

ik ′ ≤ U xih

thik ′′ ≤ U xih, δh

ik ′ ≤ U xih

∀h = (k, k′, k′′) ∈ Hi

k ∈ K1, k′ ∈ K2,

k′′ ∈ K3

∀i ∈ I (12)


Here thik andδh

ik ′ are disaggregated variables required for the convex hull formulationsandU is a valid upper bound.

6.2. Sequencing constraints

Using the big-M formulation, constraint (3) and (4) can be written as,

Ti ′k ≥ Tik + dik + cik −U (1− yii ′k) ∀ i, i ′ ∈ I , i ′ 6= i,k ∈ K i ∩ K i ′ ∩ (K1 ∪ K3)

(13)

Ti ′k ≥ Tik + dik + cik −U (1− yii ′k) ∀ i, i ′ ∈ I , i ′ 6= i,k ∈ K i ∩ K i ′ ∩ K2

(14)

respectively. It should be noted that for the sequencing constraints we did not use the convexhull formulation. Usually, this formulation is recommended to reformulate disjunctions eventhough it requires extra variables and constraints. This is because the convex hull formulationgives a tighter linear programming relaxation, and therefore usually fewer branch and boundnodes are needed to solve the problem. However, aside from the fact that a larger numberof variables and constraints are required for the convex hull formulation, in the case ofconstraints (3) and (4) the tightening that is achieved is rather poor due to the large sizeof the convex hull implied by these constraints. Therefore, for these constraints it is moreeffective to instead use the big-M formulation.

6.3. Logic constraints

The logic constraints (7) to (10) which were written as logic propositions can be tranformedinto linear inequalities using the methodologies presented by Raman and Grossmann (1991).

6.3.1. Essential constraints.The following inequalities are equivalent to the essentialconstraints (7) and (8), respectively.

yii ′k + yi ′ik ≥∑h∈Hk

i

xih +∑h∈Hk

i ′

xi ′h − 1 ∀i, i ′ ∈ I , i ′ > i, k ∈ K i ∩ K i ′ ∩ K2 (15)

yii ′k + yi ′ik = 1 ∀i, i ′ ∈ I , i ′ > i, k ∈ K i ∩ K i ′ ∩ (K1 ∪ K3) (16)

The derivation of Eq. (16) is straight forward. However, the derivation of inequality (15) issomewhat involved and therefore it is presented in Appendix A.

6.3.2. Non-essential constraints.The non-essential constraints (9) and (10) can be writtenas the following inequalities.

yii ′k′ ≤ yii ′k ∀i, i ′ ∈ I , i ′ 6= i, k ∈ K1 ∩ K i ∩ K i ′ , k′ ∈ K2 ∩ K i ∩ K i ′ (17)

yii ′k′ ≤ yii ′k′′ ∀i, i ′ ∈ I , i ′ 6= i, k′ ∈ K2 ∩ K i ∩ K i ′ , k′′ ∈ K3 ∩ K i ∩ K i ′ (18)


The MILP model equivalent to the disjunctive formulation (M1) can then be written as,

(M2) min Ms

s.t. (11), (12) to (18)

Tik ≥ 0 ∀i ∈ I , k ∈ K i

thik ≥ 0 ∀i ∈ I , k ∈ K i , h ∈ Hk

i

dik ≥ 0 ∀i ∈ I , k ∈ K i ∩ K2

δhik ≥ 0 ∀i ∈ I , k ∈ K i ∩ K2, h ∈ Hk

i

xih ∈ {0, 1} ∀i ∈ I , h ∈ Hi

yii ′k ∈ {0, 1} ∀i, i ′ ∈ I , i ′ 6= i, k ∈ K i ∩ K i ′

It is clear that the size of the of the MILP will increase with the cardinality of the set ofpaths (H).

7. Computational results

The MILP model M2 can be used to solve the problem at hand. Let us first consider asmall example problem to demonstrate the methodology. Consider a plant consisting of 1manufacturing unit, 2 packing lines, and 2 storage tanks. The aim of the problem is to derivea schedule to complete 5 orders in minimum possible time. The time required to completemanufacturing and packing steps for all the 5 orders is as shown in Table 1. Orders 1 and2 require Packing Line 1, and orders 3, 4, and 5 require Packing Line 2. A cleanup time of0.5 hour is required between two consecutive orders in all the three stages. This problemcan be solved using the MILP model M2 and the optimal schedule is as shown in figure 3.The MILP model for this problem has 72 continuous variables, 78 binary variables, and279 constraints. There are a number of different modeling tools (AMPL, 1993; GAMS,1992) and commercial solvers available (CPLEX 6.5, 1995; XPRESS-MP, 1997) that canbe used to model and solve the MILP model M2. In our case, the problem was modeledusing the GAMS modeling language (1992) and CPLEX 6.5 was used to solve the MILPproblem. It took 27 branch and bound nodes and 0.68 seconds in computational time on

Table 1. Problem data.

Processing time (hours)

Order Manufacturing Packing

1 0.85 3.05

2 3.65 15.94

3 0.3 2.21

4 0.75 2.57

5 0.35 2.58


Figure 3. Optimal schedule for the example problem.

a HP9000/C110 workstation to solve the problem to optimality. It is interesting to notethat the Packing Line 1 is the bottleneck unit and the makespan of the schedule is equal tothe sum of total processing time of orders 1 and 2 and the cleanup time between orders 1and 2. Also, note that the manufacturing of Order 2 has to wait till acleanstorage tank isavailable.

While the MILP model M2 provides a way for solving the scheduling problems con-sidered in this paper, an important aspect that has to be investigated is the computationaltime required to solve this MILP model. To study the computational performance of thescheduling methodology a number of example problems of different sizes should be consid-ered. An industrial case study provided to us had 2 manufacturing units, 5 storage tanks, 5packing lines, and 28 orders. The data is confidential and cannot be presented in this paper.A number of examples of varying complexity can be generated by considering the sameplant topology, but with different number of orders to be scheduled. The model size (numberof variables and constraints) increases as the number of orders to be scheduled increases.Seven different example problems were considered with the number of orders ranging from7 to 28. Computational results for all the seven examples are summarized in Table 2. It canbe seen that the size of the problem grows rapidly with the increase in number of orders. Thesmallest problem considered has 226 continuous variables, 267 binary variables, and 837constraints and the largest problem considered has 898 continuous variables, 4518 binaryvariables, and 11,070 constraints.

Computational results indicate that problems with more than 15 orders could not besolved within acceptable optimality tolerance (5%) using a reasonable number of branch andbound nodes (50,000) or computational resources (CPU time<5 hours on a HP9000/C110workstation). From an industrial perspective, for such a methodology to be useful on theshop floor, it is very important that it should be able to obtain reasonably good solutionsin few minutes of computational time. In the next section, we present a heuristic that usesMILP model M2 and achieves this goal.


Table 2. Computational results for model M2.

Variables Makespana CPU timeb OptimalityOrders (bin., cont.) Constraints (hours) (seconds) Nodesc gap (%)

7 267, 226 837 71.67 2.54 36 4.9

9 445, 290 1287 68.22 9.30 113 0.0

10 552, 322 1554 84.66 111.81 232 3.6

15 1257, 482 3244 87.73 3,798.27 31,173 3.5

17 1623, 546 4116 90.59 6,278.59 50,000 6.9

20 2254, 642 5582 102.19 63,68.62 50,000 20.4

28 4518, 898 11070 – 18,000.00 21900 –

aTermination criteria: Optimality tolerance= 5%.bOn a HP9000/C110 workstation, upper limit= 18,000.00 seconds.cUpper limit= 50,000 nodes.

8. Partial preordering

One of the limitations of the above mentioned solution strategy is that it does not exploitspecial problem characteristics to speed up the search. One such characteristic that can beexploited is the fact that the processing time of an order in the packing line is much longerthan the processing time in the manufacturing unit. Even though more orders are processedin each manufacturing unit as compared to any packing line, the makespan of the scheduleis determined by one of the packing lines (bottleneck), as it takes more time to process anorder in a packing line. For instance, reconsider the small example problem presented in theprevious section. In that example the sum of processing and cleanup times for all the ordersin the manufacturing unit is 7.9 hrs, and in the two packing lines it is 19.49 and 8.36 hrs.Note that while calculating these numbers the cleanup time after the last order has not beenincluded. Clearly, in this example the optimal makespan of the schedule is determined bythe total processing and cleanup times in the packing lines and it cannot be less than 19.49hrs. Furthermore, any schedule which has a makespan of 19.49 hrs is an optimal schedule.We can generalize this observation, and say that a schedule that utilizes (order processingor cleanup) the bottleneck unit at all times is an optimal schedule. Furthermore, utilizationof the bottleneck unit can be used as a measure for the quality of a schedule.

One of the main reasons for the poor performance of the MILP model is the poor LPrelaxation of the problem. When we solve the relaxed problem, the solution suggests thatall orders should be processed in parallel to obtain a schedule with minimum makespan.Obviously, this is not a feasible schedule because integrality constraints on most of thesequencing variables (yii ′k) are violated. To enforce sequencing constraints, the branch andbound method considers a pair of orders and creates two nodes to evaluate the effect ofsequencing those two orders on a particular unit i.e.yii ′k = 1 or yii ′k = 0. This processcontinues until it finds an integer solution. Usually, a feasible schedule is obtained deepin the branch and bound tree because the relaxed problem always tries to schedule ordersin parallel. One other problem is that there is a significant difference between the value ofthe makespan obtained before and after fixing sequencing decisions (poor LP relaxation).A number of cases (large number of nodes) have to be considered before the search can


be terminated. If we can fix some of the sequencing decisions beforehand we may be ableto reduce the solution time significantly. Based on this idea and in the spirit of the workof Pinto and Grossmann (1995, 1996) and Jain and Grossmann (1998), we propose thefollowing heuristic to solve the problem at hand,

1. Sequence all the orders on a packing unit in the order of increasing processing time (fixyii ′k ∀i, i ′ ∈ I , i 6= i ′, k ∈ K i ∩ K i ′ ∩ K3).

2. For all the orders that use the same packing line use manufacturing paths that use thesame storage tank (fixxik ∀i ∈ I , h ∈ Hi ). This rule is reasonable because the numberof storage tanks is the same as the number of packing lines.

3. Solve the MILP model M2 with these additional constraints.

It is acknowledged that the heuristic presented here is very specific to the industrialexample considered in this paper and may not be useful for other scheduling problems.Nevertheless, results for the particular example, which are shown in Table 3, are interestingand worth investigating. It can be seen that for all the examples, the makespan of theschedule obtained using the proposed strategy is within 7% of the total processing andcleanup time for all the orders on the bottleneck packing line. In other words, for all theschedules obtained using this strategy, the bottleneck unit was being used more than 93%of the time and a schedule obtained using this heuristic is in fact very close to an optimalsolution. Furthermore, it took only 14.68 seconds to solve the full industrial sized problem.The schedule obtained in this case is shown in figure 4. It should be noted that one ofthe main reasons that the MILP problems are solvable now is that the CPLEX presolveis able to reduce the problem size significantly. For example, for the full sized industrialproblem, presolve reduced the problem size from 5114 variables and 11070 constraintsto 393 variables and 940 constraints. If the MILP model is solved without preorderingheuristics, presolve can reduce the problem size from 5416 variables and 11070 constraintsonly to 4995 variables and 10509 constraints. Problem sizes before and after presolve forall examples considered are listed in Table 4. Based on these results, we can say that thisheuristic can infact be used to obtain very good schedules quickly.

Table 3. Computational results for model M2 with preordering constraints.

Optimality gap Total processing &Variables Makespana CPU timeb (%) for MILP at cleanup time for

Orders (bin., cont.) Constraints (hours) (seconds) Nodesc termination bottleneck unit

7 228, 226 837 72.91 0.24 8 1.4 68.22

9 392, 290 1287 71.91 0.39 15 0 68.22

10 490, 322 1554 84.66 0.33 0 0 84.66

15 1148, 482 3244 88.39 1.02 1 0.7 87.73

17 1492, 546 4116 87.73 1.6 20 0 87.73

20 2088, 642 5582 87.73 3.25 70 0 87.73

28 4216, 898 11070 90.59 14.68 221 0.9 87.73

aTermination criteria: Optimality tolerance= 5%.bOn a HP9000/C110 workstation, upper limit= 18,000.00 seconds.cUpper limit= 50,000 nodes.


Table 4. Number of variables and constraints before and after persolve with CPLEX 6.5.

Without preordering With preordering

Before presolve After presolve Before presolve After presolve

Orders Variables Constraints Variables Constraints Variables Constraints Variables Constraints

7 493 837 387 697 454 837 40 63

9 735 1287 599 1106 682 1287 56 98

10 874 1554 723 1343 812 1554 65 121

15 1739 3244 1513 2943 1630 3244 120 257

17 2169 4116 1913 3775 2038 4116 148 328

20 2896 5582 2595 5181 2730 5582 199 448

28 5416 11070 4995 10509 5114 11070 393 940

Figure 4. Schedule for the full problem.

9. Conclusions

This paper has considered a class of problems that deals with the short term schedulingfor a two stage continuous process with intermediate storage tanks. The concept of pathswas used to model the flow of product through different stages and storage tanks. A model


was developed using the disjunctions and logic propositions after making some simplifyingassumptions. The model was transformed to an MILP model by using the convex hullformulation for each disjunction and a set of linear inequalities for logic propositions. Anumber of different example problems were solved. It was observed that the size of theproblem grows rapidly with the number of orders and larger problems can not be solvedusing the current method. A heuristic based on partial preordering in one of the stages waspresented to solve the larger problems quickly. Although the heuristic is problem specific,and hence may not be useful for some other problems in this class, it has led to reductionsin computational time by several orders of magnitude for obtaining optimal or near optimalsolutions. Ideas used in the heuristic cannot be generalized in the mathematical sense, butit is very likely that they can be exploited to solve scheduling problems that involve onebottleneck stage.

Appendix A. Transformation of Logic Proposition (7)

As a first step, let us simplify the logic proposition (7) by replacing∨h∈Hkixih by a new

boolean variablezik .

zik ∧ zi ′k ⇒ yii ′k ∨ yi ′ik ∀i, i ′ ∈ I , i ′ > i, k ∈ K i ∩ K i ′ ∩ K2 (19)

where,

zik ⇔∨

h∈Hki

xih ∀i ∈ I , k ∈ K i ∩ K2 (20)

Logic proposition (19) is equivalent to

yii ′k + yi ′ik ≥ zik + zi ′k − 1 ∀i, i ′ ∈ I , i ′ 6= i, k ∈ K i ∩ K i ′ ∩ K2 (21)

Logic proposition (20) is equivalent to the two following logic propositions,

¬zik ∨( ∨

h∈Hki

xih

)∀i ∈ I , k ∈ K i ∩ K2 (22)

¬( ∨

h∈Hki

xih

)∨ zik ∀i ∈ I , k ∈ K i ∩ K2 (23)

These two logic propositions are equivalent to the following inequalities,

zik ≤∑h∈Hk

i

xih ∀i ∈ I , k ∈ K i (24)

xih ≤ zik ∀i ∈ I , k ∈ K i , h ∈ Hki (25)


Summing inequality (25) for allh ∈ Hki ,∑

h∈Hki

xih ≤ |Hki |zik ∀i ∈ I , k ∈ K i (26)

This inequality can be strengthened by using the first equation in (12) and the fact thatHk

i ⊂ Hi . Hence,∑h∈Hk

i

xih ≤ zik ∀i ∈ I , k ∈ K i (27)

From (24) and (27) we obtain,∑h∈Hk

i

xih = zik ∀i ∈ I , k ∈ K i (28)

We can eliminate Eq. (28) by substituting forzik andzi ′k in (21). The equivalent transfor-mation for logic proposition (7) is

yii ′k + yi ′ik ≥∑h∈Hk

i

xih +∑h∈Hk

i ′

xi ′h− 1 ∀i, i ′ ∈ I , i ′ 6= i, k ∈ K i ∩ K i ′ ∩ K2 (29)

Acknowledgments

The authors are most grateful for funding by the National Science Foundation under grantCTS-9810182. They are also grateful to Dr. Turaj Tahmassebi from Unilever Research forsupplying the industrial test problem.

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A Disjunctive Model for Scheduling in a Manufacturing and Packing Facility with Intermediate Storage

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