A projection-based reformulation and decomposition algorithm for global optimization of a class of mixed integer bilevel linear programs

Abstract

We propose an extended variant of the reformulation and decomposition algorithm for solving a special class of mixed-integer bilevel linear programs (MIBLPs) where continuous and integer variables are involved in both upper- and lower-level problems. In particular, we consider MIBLPs with upper-level constraints that involve lower-level variables. We assume that the inducible region is nonempty and all variables are bounded. By using the reformulation and decomposition scheme, an MIBLP is first converted into its equivalent single-level formulation, then computed by a column-and-constraint generation based decomposition algorithm. The solution procedure is enhanced by a projection strategy that does not require the relatively complete response property. To ensure its performance, we prove that our new method converges to the global optimal solution in a finite number of iterations. A large-scale computational study on random instances and instances of hierarchical supply chain planning are presented to demonstrate the effectiveness of the algorithm.

Appendices

Appendix A: Toy example 1

The following example is adapted from [8] and is a classical MIBLP problem. We use toy example 1 to verify the results of the proposed algorithm and demonstrate the solution procedure.

$$ \begin{aligned} & \mathop {\hbox{min} }\limits_{{y^{u} ,y^{l0} }} \, - y^{u} - 10y^{l0} \\ & \,\,\,{\text{s}} . {\text{t}} . \quad \quad y^{u} \in {\mathbf{\mathbb{Z}}}_{ + } ,y^{l0} \in {\mathbf{\mathbb{Z}}}_{ + } \\ & \qquad\quad\,\,\, y^{l0} \in \mathop {\arg \hbox{max} }\limits_{{y^{l} }} \left\{ {\begin{array}{*{20}l} { - y^{l} :} \hfill \\ { - 25y^{u} + 20y^{l} \le 30} \hfill \\ {y^{u} + 2y^{l} \le 10} \hfill \\ {2y^{u} - y^{l} \le 15} \hfill \\ { - 2y^{u} - 10y^{l} \le - 15} \hfill \\ {y^{l} \in {\mathbf{\mathbb{Z}}}_{ + } } \hfill \\ \end{array} } \right\} \\ \end{aligned} $$

This problem does not have lower-level continuous variables, so the reformulation (P4) can be simplified and the KKT conditions are not required. In addition, since there is no upper-level constraint, the second subproblem (P7) is always feasible. Thus, it is guaranteed that a bilevel feasible solution can be obtained in each iteration. The solution procedure of the proposed algorithm is presented below, and a graphical illustration is shown in Fig. 1.

Fig. 1

The solution procedure of toy example 1

In iteration $l = 0$ we solve master problem (P5) to obtain $\left( {y^{u,*} ,y^{l0,*} } \right) = \left( {2,4} \right)$ and $LB = - 42$; given $y^{u,*} = 2$, we solve subproblems (P6) and (P7) to obtain $y_{0}^{l,*} = 2$ and $UB = - 22$; at step 6, we add the following constraint to master problem(P5): $\left[ {1 \le y^{u} \le 6} \right] \Rightarrow \left[ {y^{l0} \le 2} \right]$.

In iteration $l = 1$ we solve master problem (P5) to obtain $\left( {y^{u,*} ,y^{l0,*} } \right) = \left( {6,2} \right)$ and $LB = - 26$; given $y^{u,*} = 6$, we solve subproblems (P6) and (P7) to obtain $y_{1}^{l,*} = 1$ and $UB = \hbox{min} \left\{ { - 22, - 16} \right\} = - 22$; at step 6, we add the following constraint to master problem (P5): $\left[ {2.5 \le y^{u} \le 8} \right] \Rightarrow \left[ {y^{l0} \le 1} \right]$.

In iteration $l = 2$ we solve master problem (P7) to obtain $\left( {y^{u,*} ,y^{0,*} } \right) = \left( {2,2} \right)$ and $LB = - 22$; now we have $UB = LB$ so that the algorithm terminates in step 3.

Appendix B: Toy example 2

The following example is adapted from [25]. We use toy example 2 to demonstrate how the proposed algorithm solves an MIBLP problem with upper-level connecting constraints. Note that this toy example and following ones in Appendices C and E cannot be computed by the original reformulation-and-decomposition method.

$$ \begin{aligned} & \mathop {\hbox{min} }\limits_{y^u\!,y^{l0}} \, - y^{u} - 2y^{l0} \\ & \,\,{\text{s}} . {\text{t}} . \;\;\;{ }y^{u} \in {\mathbb{Z}}_{ + } ,y^{l0} \in {\mathbb{Z}}_{ + } \\ & \quad \quad - 2y^{u} + 3y^{l0} \le 12,y^{u} + y^{l0} \le 14 \\ &\quad \quad y^{l0} \in \mathop {\arg \hbox{max} }\limits_{{y^{l} }} \left\{ {y: - 3y^{u} + y^{l} \le - 3,3y^{u} + y^{l} \le 30,y^{l} \in {\mathbb{Z}}_{ + } } \right\} \\ \end{aligned} $$

This problem does not have lower-level continuous variables either, so the reformulation (P4) can be simplified and the KKT conditions are not required. However, there are two upper-level constraints that involve lower-level variables. Thus, the second subproblem (P7) could be infeasible. The solution procedure of the proposed algorithm is presented below, and a graphical illustration is shown in Fig. 2.

Fig. 2

The solution procedure of toy example 2

In iteration $l = 0$ we solve master problem (P5) to obtain $\left( {y^{u,*} ,y^{l0,*} } \right) = \left( {6,8} \right)$ and $LB = - 22$; given $y^{u,*} = 6$, we solve the first subproblem (P6) and obtain $\hat{y}_{0}^{l} = 12$. As the second subproblem (P7) is infeasible, at step 6 we add the following constraint to master problem (P5): $\left[ {5 \le y^{u} \le 6} \right] \Rightarrow \left[ {y^{l0} \ge 12} \right]$.

In iteration $l = 1$ we solve master problem (P5) to obtain $\left( {y^{u,*} ,y^{l0,*} } \right) = \left( {7,7} \right)$ and $LB = - 21$; given $y^{u,*} = 7$, we solve the first subproblem (P6) and obtain $\hat{y}_{1}^{l} = 9$; but we find that the second subproblem (P7) is still infeasible; at step 6, we add the following constraint to master problem (P5): $\left[ {4 \le y^{u} \le 7} \right] \Rightarrow \left[ {y^{l0} \ge 9} \right]$.

In iteration $l = 2$ we solve master problem (P5) to obtain $\left( {y^{u,*} ,y^{l0,*} } \right) = \left( {8,6} \right)$ and $LB = - 20$; given $y^{u,*} = 8$, we solve subproblems (P6) and (P7) to obtain $y_{2}^{l,*} = 6$ and $UB = - 20$; now we have $UB = LB$, so the algorithm terminates in step 7.

Appendix C: Toy example 3

The two examples above represent a special class of MIBLPs, which include merely an upper-level integer variable and a lower-level integer variable. To show all features of the proposed algorithm while ensuring simplicity for demonstration, we propose the following illustrative example.

$$ \begin{aligned} &\mathop {\hbox{min} }\limits_{{x^{u} ,y^{u} ,x^{l0} ,y^{l0} }} \, 20x^{u} - 38y^{u} + x^{l0} + 42y^{l0} \hfill \\ &\quad\,\,\,{\text{ s}} . {\text{t}} .\quad { }7y^{u} + 5x^{l0} + 7y^{l0} \le 62 \hfill \\&\quad\quad\qquad { 6}x^{u} + 9y^{u} + 10x^{l0} + 2y^{l0} \le 117 \hfill \\ &\quad\quad\qquad x^{u} \in {\mathbb{R}}_{ + } ,y^{u} \in {\mathbb{Z}}_{ + } ,x^{l0} \in {\mathbb{R}}_{ + } ,y^{l0} \in {\mathbb{Z}}_{ + } \, \hfill \\ &\quad\qquad\quad \left( {x^{l0} ,y^{l0} } \right) \in \mathop {\arg \hbox{max} }\limits_{{x^{l} ,y^{l} }} \, 39x^{l} + 27y^{l} \hfill \\&\quad\quad\qquad\quad\qquad\quad\quad {\text{ s}} . {\text{t}} . \quad{ 8}x^{u} + 2x^{l} + 8y^{l} \le 53 \hfill \\ &\quad\quad\qquad\quad\quad\qquad\quad\qquad\,\,\,{ 9}x^{u} + 2x^{l} + y^{l} \le 28 \hfill \\&\quad\quad\qquad\quad\quad\qquad\quad\qquad\,\,\, x^{l} \in {\mathbb{R}}_{ + } ,y^{l} \in {\mathbb{Z}}_{ + } \hfill \\ \end{aligned} $$

This problem includes continuous and integer variables in both upper- and lower-level programs. There are two upper-level constraints involving lower-level variables. Thus, the second subproblem (P7) could be infeasible. The solution procedure of the proposed algorithm is presented below.

In iteration $l = 0$ we solve master problem (P5) to obtain $\left( {x^{u,*} ,y^{u,*} ,x^{l0,*} ,y^{l0,*} } \right) = \left( {2.844,8,1.200,0} \right)$ and $LB = - 245.911$; given $\left( {x^{u,*} ,y^{u,*} } \right) = \left( {2.844,8} \right)$, we solve the first subproblem (P6) and obtain $\left( {\hat{x}_{0}^{l} ,\hat{y}_{0}^{l} } \right) = \left( {0.200,2} \right)$; the second subproblem (P7) is infeasible; at step 6 we add a set of KKT-condition-based inequalities to master problem (P5).

In iteration $l = 1$ we solve master problem (P5) to obtain $\left( {x^{u,*} ,y^{u,*} ,x^{l0,*} ,y^{l0,*} } \right) = \left( {2.889,8,1.000,0} \right)$ and $LB = - 245.222$; given $\left( {x^{u,*} ,y^{u,*} } \right) = \left( {2.889,8} \right)$, we solve the first subproblem (P6) and obtain $\left( {\hat{x}_{1}^{l} ,\hat{y}_{1}^{l} } \right) = \left( {0.500,1} \right)$; we find that the second subproblem (P7) is still infeasible; at step 6, we add another set of KKT-condition-based inequalities to master problem (P5).

In iteration $l = 2$ we solve master problem (P5) to obtain $\left( {x^{u,*} ,y^{u,*} ,x^{l0,*} ,y^{l0,*} } \right) = \left( {3.000,8,0.500,0} \right)$ and $LB = - 243.500$; given $\left( {x^{u,*} ,y^{u,*} } \right) = \left( {3.000,8} \right)$, we solve subproblems (P6) and (P7) to obtain $\left( {x_{2}^{l,*} ,y_{2}^{l,*} } \right) = \left( {0.500,0} \right)$ and $UB = - 243.500$; now we have $UB = LB$, so the algorithm terminates in step 7.

The solution procedure above takes a total of 3 iterations. If the KKT-condition-based tightening constraints (74)–(77) are not used, the algorithm takes a total of 5 iterations. Therefore, it is shown that the KKT-condition-based tightening constraints help reduce the number of iterations and computational time.

Appendix D: Inputs for generating computational examples

The following Table 4 provides the inputs to the GAMS code for generating computational instances corresponding to example 2. We note that seed is the factor used to generate random parameters, std. stands for the standard deviation used when generating $m_{R}$, $m_{Z}$, $n_{R}$, and $n_{Z}$.

Table 4 Inputs to the GAMS code for generating computational instances in example 2

In the following Table 5, we provide the inputs to GAMS for generating instances in example 3 from (hscp_6_6_1) through (hscp_12_12_5).

Table 5 Inputs to GAMS for generating instances in example 3

Appendix E: Hierarchical supply chain planning model

In this section, we present the bilevel model formulation of the hierarchical supply chain planning problem adapted from [60]. Before the model is presented, we first give the notations used in the model.

Parameters

$a_{ij}$ :

Capacity consumption ratio for processing product j in plant i

$c_{i}^{U}$ :

Upper bound of production capacity in plant i

$d_{j}$ :

Customer demand of product j

$e_{ij}$ :

Resource factor for processing product j in plant i

$f_{i}$ :

Opening cost for plant i

$g_{ij}$ :

Fixed cost for opening production line j in plant i

$p_{i}$ :

Opportunity cost for unused production capacity of plant i after it is opened

$q$ :

Resource availability

$r_{ij}$ :

Transportation cost for transferring product j from plant i to the principal firm

$s_{ij}$ :

Fixed operation cost for processing product j in plant i

$w_{i}$ :

Cost to use production capacity in plant i

$n$ :

Number of product types

Continuous variables

$Cap_{i}$ :

Designated production capacity in plant i

$X_{ij}$ :

Fraction of demand of product j produced in plant i

Binary variables

$Y_{i}$ :

1 if plant I is selected and opened; 0 otherwise

$Z_{ij}$ :

1 if production line for product j in plant i is used; 0 otherwise

With the above notations, the model for the hierarchical supply chain planning problem is formulated as follows.

$$ \hbox{min} \;z_{1} = \sum\limits_{i} {f_{i} Y_{i} } + \sum\limits_{i} {\sum\limits_{{j \in IS_{i} }} {g_{ij} Z_{ij} } } + \sum\limits_{i} {p_{i} \left( {Cap_{i} - \sum\limits_{{j \in IS_{i} }} {d_{j} a_{ij} X_{ij} } } \right)} $$

(E.1)

$$ {\text{s}} . {\text{t}}.\;\sum\limits_{i} {\sum\limits_{{j \in IS_{i} }} {d_{j} e_{ij} X_{ij} } } \le q $$

(E.2)

$$\qquad\qquad Cap_{i} \le c_{i}^{U} \, \forall i $$

(E.3)

$$ \qquad\qquad Y_{i} \in \left\{ {0,1} \right\},Cap_{i} \in {\mathbb{R}}_{ + } $$

(E.4)

$$ \hbox{min} \;z_{2} = \sum\limits_{i} {w_{i} \left( {\sum\limits_{{j \in IS_{i} }} {d_{j} a_{ij} X_{ij} } } \right)} + \sum\limits_{i} {\sum\limits_{{j \in IS_{i} }} {\left( {s_{ij} Z_{ij} + d_{j} r_{ij} X_{ij} } \right)} } $$

(E.5)

$$ {\text{s}} . {\text{t}}.\;\sum\limits_{{i \in JS_{j} }} {X_{ij} } = 1 , { }\forall j $$

(E.6)

$$\qquad\qquad \sum\limits_{{j \in IS_{i} }} {d_{j} a_{ij} X_{ij} } \le Cap_{i} , { }\forall i $$

(E.7)

$$ \qquad\qquad\sum\limits_{{j \in IS_{i} }} {X_{ij} } \le nY_{i} , { }\forall i $$

(E.8)

$$ \qquad\qquad X_{ij} \le Z_{ij} , { }\forall i,j \in IS_{i} $$

(E.9)

$$ \qquad\qquad X_{ij} \in {\mathbb{R}}_{ + } ,Z_{ij} \in \left\{ {0,1} \right\} $$

(E.10)

The principal firm’s objective (E.1) is to minimize the sum of the plant opening cost, the production line opening cost, and the opportunity cost of over-setting production capacities. Constraint (E.2) enforces that the use of resources does not exceed their availabilities. Although only one type of resource is considered in this model, it can be easily extended to include multiple types of resources by adding an index for resources. Constraint (E.3) imposes a limitation on plant capacity. The lower-level objective function (E.5) is to minimize the operational costs, including the cost related to production capacity consumption, the fixed charge cost, and transportation costs for shipping products from auxiliary plants to the principal firm. Constraint (E.6) indicates that the demands must be fully satisfied. Constraint (E.7) indicates that production should not exceed capacity. Constraint (E.8) suggests that no product can be produced if the plant is not opened. Constraint (E.9) indicates that no product can be produced if the production line is not opened. Constraints (E.4) and (E.10) are non-negative and binary constraints for upper- and lower-level decision variables. In this problem setting, the principal firm first determines which plant to open ($Y_{i}$) and the capacity to install ($Cap_{i}$). Then the auxiliary plants determine which production line to use ($Z_{ij}$) and the production level of each product ($X_{ij}$).