19. Knapsack, Bin Packing and Plant Location problems

19.1. Introduction

One of the “simplest” integer linear optimization models is the so-called knapsack problem. Assume we are given a set of \(n\) “objects” and that to each of these both a “value” and a “weight” are associated. Assume also that a finite capacity, a non negative real number, is given. The problem then becomes to select a subset of objects whose total weight does not exceed capacity. Among these sets of objects, it is required to find one with the highest possible total value.

Denoting by \(\mathpar{v}_{j}\) the value of the \(j\)-th object, with \(\mathpar{w}_{j}\) its weight and with \(\mathpar{b}\) the capacity, introducing a binary variable associated with each object, the following model is obtained:

\begin{align*} \max \sum_{j = 1}^{n} \mathpar{v}_{j} \mathvar{\delta}_{j} & \\ \sum_{j = 1}^{n} \mathpar{w}_{j} \mathvar{\delta}_{j} & \leq \mathpar{b} \\ \mathvar{\delta}_{j} & \in \{0,1 \} & j \in 1,n \end{align*}

This model is very basic, but applications of some practical interest exist, although the main interest in this model consists in the fact that its constraints are often found in other, more complex, models.

To describe possible applications, we start from the name of the model. The idea is that a knapsack is available, which can carry at most a certain capacity (which might be weight, volume, number of standard pallets, …). A set of objects is available, the choice of each of which consumes part of the available capacity. The problem consists then in filling the knapsack with the most valuable collection of objects. It must be said that this application is just used for teaching purposes, and nobody would ever prepare a knapsack this way. However, if we slightly change the context, some direct application comes to mind. As an example, the knapsack might be a truck, or a container; the objects might be parcels to be delivered. The value is a measure of the priority or of the urgency of delivery.

In other contexts, objects might be projects to be chosen. A company might have a number of possible projects to perform, each of which requires some budget in order to be completed. The budget is the weight associated to each project and the capacity is the overall budget available. The value of each project is the revenue associated to project delivery. Slightly changing the application, instead of a monetary budget, we might have an overall time budget, and different projects will have different time consumption and different gains associated. A similar application arises in advertising: there are a number of available slots for placing a TV advertisement. Each slot is sold from the TV company, and the price is the “weight” of the object. The available budget for this marketing operation is the capacity. The value associated to each slot might be the expected audience, i.e., the expected number of people who will see the advertisement.

This model is called binary knapsack, as, for any object, the decision to be taken is whether to choose or not each single object. It would be trivial to extend the model to cover the case in which some objects are available in multiple copies, changing binary variables into non negative integer ones, with an upper bound equal to the number of available objects.

The basic knapsack model is only apparently a simple one, with a single constraint (apart from binary constraints). In integer optimization, the number of constraints or variables is not, in general, an indicator of the complexity of a problem. Indeed, often, the reverse is true, with more efficient models characterized by formulations with a huge number of constraints.

There exist many heuristics for the approximate solution of knapsack problems; moreover, the simple structure of the constraints makes the solution of the linear relaxation particularly simple. In fact, the optimal solution of the linear relaxation is obtained by first sorting the objects in non-increasing order with respect to the ratio between value and weight. Scanning this list, if an object fits the knapsack, it is chosen (its binary variable is set to 1). As soon as a first object is found which cannot be included in the solution, the corresponding variable is set to the fraction of the object necessary to “saturate” the capacity constraint.

\begin{align*} \bar{\jmath} &:= \arg \max \{i: \sum_{j=1}^{i} \mathpar{w}_{i} \leq b \} \\ \mathvar{\delta}_j & = 1 & \forall \, j \leq \bar{\jmath} \\ \mathvar{\delta}_{\bar{\jmath}+1} & = \frac{b - \sum_{i=1}^{\bar{\jmath}} \mathpar{w}_j}{\mathpar{w}_{\bar{\jmath}+1}} \\ \mathvar{\delta}_j & = 0 & \textrm{ otherwise} \end{align*}

A simple implementation of the basic model is given:

Knapsack.mod
set OBJECTS;
param weight{OBJECTS};
param value{OBJECTS};
param cap;

var delta{OBJECTS}, binary;

maximize total_value: sum {j in OBJECTS} value[j] * delta[j];

s.t. knapsack:
   sum{j in OBJECTS} weight[j] * delta[j] <= cap; 

while the following is a tiny numerical example:

Knapsack.dat
set OBJECTS := A B C D E F G H I J K L M N O;

param: 
  value weight:=
A 135     70    
B 139     73    
C 149     77    
D 150     80    
E 156     82    
F 163     87    
G 173     90    
H 184     94    
I 192     98    
J 201    106    
K 210    110    
L 214    113    
M 221    115    
N 229    118    
O 240    120    
;

param cap := 730;

Solving the model, given its very small size, is indeed very easy. A solver like, e.g., CBC, in a fraction of a second, and 24 Branch and Bound iterations, returns the optimal solution consisting in choosing objects A, B, C, D, H, I, N, O, with total weight 730 ( full capacity) and total profit 1418. The relaxed solution has objects A, C, G, H, I, N, O fully in the knapsack, while object M is chosen at 54.78% level. The optimal value of the relaxed problem is 1423.07, from which we see that the gap between the optimal and the relaxed solution is relatively small (0.36%).

We can already see from this tiny example that the integer solution is not a rounding of the solution of the linear relaxation.

19.1.1. Advanced modeling: cover inequalities

Although the purpose of this part is to introduce modeling techniques, some algorithmic aspects need to be addressed too, as, for difficult integer optimization problems, the possibility of finding an optimal solution is connected with the goodness of the formulation, something which goes beyond the purely formal correctness. That is, there might be different correct formulations of the same problem, all of which are exact from the point of view of modeling, but different from the point of view of their closedness to the ideal representation. Within integer optimization, the boundary between models and algorithms is quite vague.

Many of the most recent integer linear optimization approaches require a reformulation of the problem using additional constraints whose purpose is, mainly, to improve the value of the linear relaxation, decreasing the gap between the optimal solution and the relaxed one. Inserting additional constraints without eliminating any feasible solution leads to a smaller polyhedron in the relaxation and, as a consequence, to a possible improvement in the value of the relaxed problem. There exist no general and efficient technique to define these additional constraints, and specific classes need to be devised for each specific class of problems. Finding classes of these inequalities is, thus, in a certain sense a modeling problem, albeit different from those encountered so far.

For the binary backpack problem, a few simple arguments allow us to deduce some valid constraints for the problem which are in general not valid for the linear relaxation. The family of cover inequalities can be deduced from the trivial observation that if \(k\) items have a total weight greater than the capacity, then no more than \(k-1\) of them can belong to a feasible solution. In formulae, if \(S\) is a subset of objects such that

\[ \sum_{j \in S} \mathpar{w}_{j}> \mathpar{b} \]

then the following inequality is valid

\[ \sum_{j \in S} \mathvar{\delta}_{j} \leq |S| -1 \]

where \(| S |\) denotes the cardinality of the set \(S\).

An inequality like this one is not necessarily satisfied in the points of linear relaxation polyhedron. Adding all possible cover inequalities to the formulation is not a feasible approach, in general, as their number can be huge; moreover, many of these inequalities are redundant, as they are dominated by others of the same family. An improvement on this family of inequalities corresponds to considering only minimal sets, with respect to the property of being a cover (that is, to contain objects that exceed the capacity of the container).

Let us consider a simpler numerical example, to outline the basic idea of adding cover inequalities. Consider a knapsack problem with the following objects:

\begin{align*} \begin{array}{ccccccc|c} A & B & C & D & E & F & G & b \\ 11 & 6 & 6 & 5 & 5 & 4 & 1 & 19 \end{array} \end{align*}

We do not write the objective function as our purpose here is just to show how to improve the formulation, i.e., how to add valid inequalities. In a real situation, however, a procedure, called separation should be defined which, given the solution of a relaxation of the problem, finds at least one cut inequality, i.e., a valid inequalities which makes the current solution of the relaxed problem infeasible. Adding such a cut would lead to an improvement in the relaxation and to finding a different solution of the relaxed problem.

Consider objects A, B, C: their total weight is 23, thus they can be used to form a cover inequality:

\begin{align*} \mathvar{\delta}_A + \mathvar{\delta}_B + \mathvar{\delta}_C & \leq 2 \end{align*}

If we consider, e.g., C, D, E, F, G, their total weight is 21, so we might define a cover inequality form these objects, with right hand side 4 (no more than 4 of these 5 can be in any knapsack). However, the objects still form a cover if we cancel object G, the lightest one. So the inequality

\begin{align*} \mathvar{\delta}_C + \mathvar{\delta}_D + \mathvar{\delta}_E + \mathvar{\delta}_F & \leq 3 \end{align*}

is a valid cover inequality and it dominates the one obtained including object G.

Another observation is worth at this point: if at most 3 out of the set of objects \(C,D,E,F\) can be inserted in a feasible knapsack, it is clear that, including in the set also objects \(A,B\), whose weight is even larger, will not lead to the possibility of inserting more than 3 objects. It is thus true that the following inequality

\begin{align*} \mathvar{\delta}_A + \mathvar{\delta}_B + \mathvar{\delta}_C + \mathvar{\delta}_D + \mathvar{\delta}_E + \mathvar{\delta}_F & \leq 3 \end{align*}

is also valid. Moreover, this inequality is stronger that the cover inequality associated to the whole set \(A,B,C,D,E,F\), since this would need a right hand side equal to 5. This kind of inequality is called an extension of the cover inequality. In general, given a cover associated to a subset of objects \(S\), and extended cover inequality is given by

\begin{align*} \sum_{j \in S} \mathvar{\delta}_j + \sum_{j \not\in S : \mathpar{w}_j \geq \mathpar{w}_k \forall\,k \in S} \mathvar{\delta}_j & \leq |S| -1 \end{align*}

It can be recalled here that these inequalities, i.e., non-dominated cover inequalities and extended cover inequalities, are already a potentially large family of valid inequalities; however, although large, this set of inequalities is not sufficient to fully describe the ideal polyhedron for the binary knapsack problem. Other inequalities, in particular those known as lifted inequalities can be defined and added to the set. We refer the interested reader to the vast literature on this subject, starting, e.g., by [Nemhauser and Wolsey, 1999]. In any case, for moderate to large size problem instances, generating and adding all these inequalities to the formulation is usually out of question. What can be done is to solve a relaxation of the problem, to design a procedure to separate the solution of the relaxation, if not feasible; and then to add this inequality to the formulation, solve again and repeat the loop. Recall however that separation, i.e., finding a violated inequality within the family of cover and lifted inequalities, is as hard as solving the problem from scratch and, in general, heuristic separation methods are used. These methods can often generate a valid inequality which cuts off the current solution, but, sometimes, cannot find one even if it exists.

The emphasis we have given to this advanced topic here is connected to the fact that the importance of knapsack valid inequalities goes far beyond the problem itself. In fact any inequality with binary constraints can be seen as a knapsack inequality, possibly after a few variable substitutions. Consider, as an example the generic inequality

\begin{align*} 3 \mathvar{\delta}_1 - 5 \mathvar{\delta}_2 +2 \mathvar{\delta}_3 -2\mathvar{\delta}_4 & \leq 1 \end{align*}

If we replace the variables with a negative coefficient with their complement to 1, we obtain

\begin{align*} 3 \mathvar{\delta}_1 - 5 (1- \mathvar{\bar{\delta}}_ 2) +2 \mathvar{\delta}_3 -2 (1- \mathvar{\bar{\delta}}_ 4) & \leq 1 \end{align*}

which simplifies to

\begin{align*} 3 \mathvar{\delta}_1 + 5 \mathvar{\bar{\delta}}_ 2 +2 \mathvar{\delta}_3 +2 \mathvar{\bar{\delta}}_ 4 & \leq 8 \end{align*}

and this is exactly a binary knapsack constraint. We can, thus, add cover inequalities and their generalizations in order to strengthen the formulation. A similar technique can be applied in the case of \(\geq\) constraints; thus any inequality constraint in binary variables is equivalent to a knapsack inequality and, therefore, can be strengthened through the techniques seen in this paragraph. This is indeed done by many of the advanced exact mixed integer optimization solvers available.

19.2. Multiple knapsacks, Cutting Stock

The knapsack model can be easily generalized to the case of multiple knapsacks (or multiple containers). The problem may still be that of choosing the maximum “value” of the selected objects, with constraints imposing not to exceed the capacity of each of the available containers. The “natural” formulation of the problem is quite immediate, after the presentation of the binary knapsack problem. However an important difference characterizes these models. Now the decision to be taken is not just whether an object is to be chosen or not: if an object is chosen, we need to decide also to which container will the object be allocated. We can obtain this result by using two indices for each binary variable, one associated to each object, the other one associated to each container. Let \(\mathvar{\delta}_{ij}\) be a binary variable with value 1 if and only if the object \(j\) is placed in container \(i\). With this choice of variables, the formulation closely follows that of the standard knapsack; the main variants are the presence of a capacity constraint for each container and a constraint which forbids the same object to be included into more than one container.

Let us denote by \(\mathpar{n}\), as before, the number of objects and by \(\mathpar{C}\) the number of containers, each of which characterized by a finite capacity \(\mathpar{Cap}_i\). A model for this problem might be written as follows:

\begin{align*} \max_{\mathvar{\delta}} \sum_{j=1}^n \sum_{i=1}^C \mathpar{v}_j \mathvar{\delta}_{ij} & \\ \sum_{i=1}^C \mathvar{\delta}_{ij} & \leq 1 & \forall\,j=1,\ldots,n \\ \sum_{j=1}^n \mathpar{w}_j \mathvar{\delta}_{ij} & \leq \mathpar{Cap}_i & \forall\,i=1,\ldots,C \\ \mathvar{\delta}_{ij} & \in \{0,1\} & \forall\,i,j \end{align*}

The first group of inequalities state that each object can be placed at most once in a single container. The second group is composed of knapsack constraints, one for each container, each of which characterized by a maximum capacity. It is worth observing that these knapsack inequalities might also be generalized to cope with multiple, different, capacity measures for each container. Consider, as an example, loading a group of trucks, taking into account both the weight as well as the volume required and available. Then, for each truck, two knapsack inequalities can be imposed, one for the total weight, and another one for the total volume. Of course volume computation is especially hard, if this depends on how the parcels are loaded in the three-dimensional available space. However, in many cases, the actual loading pattern might be disregarded, and a total volume occupancy can be simply computed. This is the case, e.g., for goods delivered in pallets: the volume occupied by a pallet depends only on the base surface, as all of the vertical space above the pallet is unavailable for further loading. It remains the problem of deciding a two-dimensional arrangement, but pallet measures are standard and efficient loading patterns can be decided a priori.

A classical application of this model, apart from the natural one in transport logistics, is related to one dimensional cutting stock problems: given a set of one-dimensional objects (for example cables) and a set of purchase orders, defined as lengths of cable to be cut for a customer, a problem might be that of deciding which order to satisfy and from which of the available cables to cut it, in order to maximize a total value, which, in some cases, is a measure of the priority of each order.

In the following we report an example of implementation of a tiny example:

cutstock.mod
set OBJECTS;
set CONTAINERS;

param weight{OBJECTS};
param value{OBJECTS};
param capacity{CONTAINERS};

var delta{CONTAINERS, OBJECTS}, binary;

maximize total_value:
   sum {i in CONTAINERS, j in OBJECTS} value[j] * delta[i,j];
   
s.t. OneAtMost{j in OBJECTS}:
  sum{i in CONTAINERS} delta[i,j] <= 1;

s.t. CapacityConstraints{i in CONTAINERS}:
   sum{j in OBJECTS} weight[j] * delta[i,j] <= capacity[i];
   
cutstock.dat
param: OBJECTS: weight value:=
A 20  10
B 18   5
C 17   7
D 15   6
E 14   9
F 12  10
G 10  12
H 5    4
I 4    4
J 2    1
;

param: CONTAINERS : capacity :=
C1 35
C2 30
C3 28;

The example is referred to a cutting stock elementary case. In the example, we assume to have three cables, whose lengths are 35, 30, 28, respectively, and we wish to cut them in order to satisfy orders, given in the data file, according to their required quantity and to their value (priority). When solved with a good quality branch and bound method, we obtain the optimal solution consisting in allocating orders A and D to cable 1 (using its full capacity), orders E, F, I to cable 2 (again, used at full capacity) and orders C and G to cable 3, using 27 out of 28 units. Orders B, H , J could not be satisfied.

19.3. Bin packing - container optimization

An important variant of the multiple knapsack model consists in the optimization of the total number or total cost of the containers needed to host all of the objects. The model becomes more complex in this case. In fact it becomes necessary to add a variable for each container, whose value is 1 if and only if the container is used. We define a binary variable \(\var{y}_i, i =1,C\) and, through suitable constraints, we impose that this variable has value 1 if and only if container \(i\) contains at least an object. There are several ways to formulate this constraint, some of which related to the formulation of logical constraints as we have seen in chapter Using binary variables in logical constraints. In fact the link between object allocation and container usage can be formulated as:

\[ \exists \, j: \mathvar{\delta}_{ij} = 1 \implies \var{y}_i = 1 \]

or, equivalently,

\[ \var{y}_i = 0 \implies \mathvar{\delta}_{ij} = 0 \quad \forall \, j \]

The opposite implication is not needed as, in general, to each container a positive cost is associated, so that there is no incentive to have the binary variable associated to a container have value 1 when no object is placed in that container.

Let \(\mathpar{c}_i\) denote the cost of using container \(i\). If all costs are equal to one, the problem becomes that of finding the minimum number of containers which are sufficient to carry all of the required load. Otherwise the problem consists in minimizing the cost of used containers. The generic bin-packing problem can thus be formulated as

\begin{align*} \min_{\var{y}, \mathvar{\delta}} \sum_{i=1}^C \mathpar{c}_i \var{y}_i \\ \sum_{i=1}^C \mathvar{\delta}_{ij} & = 1 & \forall\,j=1,\ldots,n \\ \exists \, j: \mathvar{\delta}_{ij} = 1 & \implies \var{y}_i = 1 & \forall\,i=1,C \\ \sum_{j=1}^n \mathpar{w}_j \mathvar{\delta}_{ij} & \leq \mathpar{Cap}_i & \forall\,i=1,\ldots,C \\ \mathvar{\delta}_{ij} & \in \{0,1\} & \forall\,i,j \\ \var{y}_i & \in \{0,1\} & \forall\, i \end{align*}

A possible formulation of the implication is the following: the logical statement \(\mathvar{\delta} _ {ij} = 0 \, \forall \, j\) can be written as

\[ \sum_{j=1}^n \mathvar{\delta}_{ij} \leq 0 \quad \forall\,i =1,C \]

so that the implication becomes

\begin{align*} \var{y}_i = 0 & \implies \sum_{j =1}^n \mathvar{\delta}_{ij} \leq 0 \end{align*}

Recalling modeling techniques with logical variables, a redundant constraint needs to be imposed for the case in which the implicant is false. Here if \(\var{y}_i = 1\) an upper bound needs to be chosen for the right hand side. A simple upper bound is given by the cardinality of the set of objects: in fact, if a container is used, the total number of objects which might be assigned to it is bounded by the total number of available objects.

\begin{align*} \sum_{j=1}^n \mathvar{\delta}_{ij} \leq \mathpar{M}_i && \forall \, i =1,C \end{align*}

where \(\mathpar{M}_i\) can be chosen equal to \(n\). As an alternative, this constant might be chosen equal to an upper bound on the number of objects which can be assigned to the container taking into account also their weight. If objects are sorted in non decreasing order with respect to the weight, then we might choose

\begin{align*} M_i &= \max \{k : \sum_{j=1}^k \mathpar{w}_j \leq \mathpar{Cap}_i\} \end{align*}

Given a correct upper bound, the logical constraint can be transformed into a linear one:

\begin{align*} \sum_{j=1}^n \mathvar{\delta}_{ij} &\leq \mathpar{M}_i \var{y}_i & \forall \, i = 1,C \end{align*}

There exist other correct formulations for the same constraint. The logical implication might also be seen as equivalent to

\begin{align*} \var{y}_i = 0 & \implies \mathvar{\delta}_{ij} = 0 & \forall \, i = 1, C, j = 1, n \end{align*}

This corresponds to the linear model

\begin{align*} \mathvar{\delta}_{ij} &\leq \var{y}_i & \forall \, i =1,C, j=1,n \end{align*}

Finally another modeling possibility consists in considering the knapsack inequality and using the binary variable associated to the container to switch the actual capacity between zero and the real capacity, depending on the value of the binary variable.

\begin{align*} \sum_{j=1}^n \mathpar{w}_{j} \mathvar{\delta}_{ij} &\leq \mathpar{Cap}_{i} \var{y}_i & \forall \, i = 1,C \end{align*}

19.3.1. Comparison between the formulations

The first formulation given above, \(\sum_{j=1}^n \mathvar{\delta}_{ij} \leq \mathpar{M}_i \var{y}_i\), is weaker than the second one, \(\mathvar{\delta}_{ij} \leq \var{y}_i\), at least when the upper bound chosen for the logical implication is the total number of objects. In fact, to see this, observe that if we sum, with respect to the objects, all of the inequalities of the second formulation we obtain the first one:

\begin{align*} \mathvar{\delta}_{ij} & \leq \var{y}_i & \implies \\ \sum_{j=1}^n \mathvar{\delta}_{ij} &\leq \sum_{j =1}^n \var{y}_i \\ & = n \var{y}_i \end{align*}

This means that the polyhedron obtained by relaxing the binary constraint on all variables in the second formulation is contained in the polyhedron obtained from the first one. Thus the second formulation is at least as strong as the first one. In order to check whether it is strictly stronger, and, thus, preferable from the point of view of the quality of the relaxation, we need to exclude that the two polyhedra are identical. But this is true as we can always find a solution in the first polyhedron which does not belong to the second one, provided that the number of objects is greater than one. As an example, let

\begin{align*} \mathvar{\delta}_{ij} &= 0 & \forall \, j \ne 1 \\ \mathvar{\delta}_{i1} &= 1 \\ \var{y}_i & = 1 / n \end{align*}

This is an example of a solution which satisfies the constraints of the first formulation, but violates some of the constraints in the second one. Observe that, in order to show an example like this one, we had to search among non binary solutions. In fact, both formulations are correct, which means that they describe exactly the same set of binary feasible solutions. If a difference exists between the two formulation, this needs to be searched in the relaxed, non binary, set of solutions.

For what concerns the third formulation, \(\sum_{j=1}^n \mathpar{w}_{j} \mathvar{\delta}_{ij} \leq \mathpar{Cap}_{i} \var{y}_i\), this is not directly comparable with the others: depending on the values of the parameters, it could be stronger, weaker, or even neither stronger nor weaker than the other ones. This means that the relaxation of this formulation might neither be contained nor containing any of the other two. In this case, a possibility might be to use both formulations, in order to strengthen the resulting model.

Here we have shown that the second formulation is stronger than the first one. However, it contains many more constraints and thus, for large problems, the overhead due to the need to solve large linear optimization problems might mitigate the advantages of a stronger formulation. A possibility, for large problems, could be that of using the strongest formulation in an implicit, dynamic, way. As we have already seen, the idea is to avoid the explicit insertion of all of the constraints in the initial formulation, but adding some of them, as needed and when needed. We might solve the problem without any constraint linking logical variables associate to objects with those associated to containers. Then we might insert only those constraints of the type \(\mathvar{\delta}_{ij} \leq \var{y}_i\) that are violated in the current solution. Solving the problem obtained in this way either a feasible solution is obtained, in which case it will be also optimal for the original problem. Otherwise, there must exist some violated constraint among those not yet included in the formulation. These constraints are added to the formulation and the whole process is iterated.

In the general formulation of bin packing problems, a requirement prescribes to choose exactly one copy of each item to be placed in one of the containers (and only one). If we change the constraint allowing an integer number of copies, \(\mathpar{N}_j\) of each object to be chosen, the problem remains quite similar in formulation, except that the constraints now becomes

\begin{align*} \min_{\var{y}, \mathvar{\delta}} \sum_{i=1}^C \mathpar{c}_i \var{y}_i \\ \sum_{i=1}^C \mathvar{\delta}_{ij} & = \mathpar{N}_j & \forall\,j=1,\ldots,n \\ \exists \, j: \mathvar{\delta}_{ij} = 1 & \implies \var{y}_i = 1 & \forall\,i=1,C \\ \sum_{j=1}^n \mathpar{w}_j \mathvar{\delta}_{ij} & \leq \mathpar{Cap}_i & \forall\,i=1,\ldots,C \\ \mathvar{\delta}_{ij} & \in \{0,1\} & \forall\,i,j \\ \var{y}_i & \in \{0,1\} & \forall\, i \end{align*}

This model generalizes the cutting stock one to the case in which all orders have to be satisfied.

An application of bin-packing to a context different from that of loading containers frequently found in logistics, or optimally cutting parts from a minimum number of cables, is in the context of production scheduling. Here the containers are production machines, or plants, with their maximum capacity (e.g., hours in a day); objects are jobs, with a duration. If each machine has a cost, the problem becomes that of scheduling all of the jobs so that the minimum (number or cost) of necessary machines is found. A related problem is that of minimizing the makespan, i.e. the completion time of the last finishing job. This problem can be easily modeled introducing an additional variable, the makespan, which is greater or equal to the total sum of used capacity in each machine (it is indeed an instance of Minimax problems).

19.4. Plant location

These problems arise in a quite different context, but share some interesting modeling aspect with the models just presented. Assume that a set of \(n\) customers, with a known demand for a product are to be served by a set of plants, or inventories, whose location has to be chosen based on cost considerations. A set of \(C\) possible locations are known, and building a facility at location \(i\) has a cost \(\mathpar{c}_i\). The (expected) demand of each customer is \(\mathpar{w}_j\) and the maximum capacity of each location is \(\mathpar{Cap}_i\). Every customer chooses one of the available facilities in order to satisfy all of the demand. Another relevant information is the cost of connecting a customer with a location: a quantity \(\mathpar{TC}_{ij}\) is given which represents the cost of serving customer \(j\) from location \(i\) - this might be thought of as a transportation cost. The overall model thus becomes:

\begin{align*} \min_{\var{y}, \mathvar{\delta}} \sum_{i=1}^C \mathpar{c}_i \var{y}_i + \sum_{i=1}^C \sum_{j=1}^n \mathpar{TC}_{ij} \mathvar{\delta}_{ij} \\ \sum_{i=1}^C \mathvar{\delta}_{ij} & = 1 & \forall\,j=1,\ldots,n \\ \exists \, j: \mathvar{\delta}_{ij} = 1 & \implies \var{y}_i = 1 & \forall\,i=1,C \\ \sum_{j=1}^n \mathpar{w}_j \mathvar{\delta}_{ij} & \leq \mathpar{Cap}_i & \forall\,i=1,\ldots,C \\ \mathvar{\delta}_{ij} & \in \{0,1\} & \forall\,i,j \\ \var{y}_i & \in \{0,1\} & \forall\, i \end{align*}

and the similarity with the bin-packing model is evident. Of course, the logical implication should be modeled as a regular constraint, as we have already seen.

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