Upper level constraints - General framework and our contributions

1.6 General framework and our contributions

2.1.1 Upper level constraints

One of the main characteristics of bilevel programming problems is the presence of two different set of

constraints, the upper level ones and the lower level ones, which impact differently on the definition of

the solutions space. In this section we highlight some theoretical results concerning the role of upper

level constraints as in the applications we describe in Chapter 5 the presence of these constraints is

shown to be not negligible. Moreover, in our opinion, in the literature the effect of the upper level

constraints does not seem to be sufficiently investigated and a large number of resolution methods,

more or less implicitly, assume they are not present in the formulation.

According to the previous definitions, it is a matter of fact that whenever there are no upper level

constraints, that is p = 0, a rational solution is also bilevel-feasible. The same result can be extended

to the case in which the upper level constraints do not involve the follower’s variables: in the previous

example x ≥ 1 is an upper level constraint in which there are no follower’s variables and, indeed, every

rational solution is also bilevel-feasible. In the same example, if the constraint x + 2y ≥ 4 is moved to

the upper level, the inducible region deeply changes, as it can be observed in Figure 2.2.

Figure 2.2. Role of the upper level constraints

Ω

(x) ⊆ [0, y

] and the reaction set R

(x) ⊆ [0, y

]. The interval [y

, y

] represents the projection

of the upper level constraint 2y ≥ 4 − x onto the y space: all the rational solutions out of this interval

are not bilevel-feasible. Hence, all the rational solutions in [0, y

) are not feasible in the leader’s

perspective. Let us consider point E, with y

= 0. Point E is a rational solution, as y

∈ R

(x

), but

it is not bilevel-feasible. The bold segment B − C is the new inducible region that is still piecewise

linear and is a supporting hyperplane of S. The optimal solution is vertex C.

From a geometrical point of view, the difference between the latter two examples occurs because in

the second there is an upper level constraint whose projection onto the y space intersects the follower’s

feasible set for a given x, and cut off some rational solutions. The shadowed area in Figure 2.2

represents the subset of solutions not satisfying the leader’s requirements but feasible for the follower.

Notice that, if in the example the upper level constraint is replaced with x + 2y ≥ 7, the only bilevel–

feasible solution is vertex C, while for higher value of the right-hand-side the BLP is an empty problem

despite S is compact and non empty. This results can be generalized as follows.

Theorem 9. Given a bilevel linear problem P

withp 6= 0, if at least one leader’s constraint is moved

to the follower problem, the new bilevel linear problemP

is a relaxation ofP

.

Proof. Let us assume that, for a given x

such that (x

, y) ∈ S, the solution (x

, y

) is rational. Let

us consider the upper level constraint c

T_j

x + d

T_j

y ≤ e

. If (x

, y

) satisfies this constraint, moving

c

T_j

x + d

T_j

y ≤ e

to the follower’s problem does not change the rationality of the solution. Conversely,

if the upper level constraint is violated, it follows that solution (x

, y

) is not bilevel–feasible for the

original problem. If constraint c

T_j

x + d

T_j

y ≤ d

is moved to the follower’s problem, (x

, y

) is not

longer a rational solution for problem P

since y

does not belong to the new set Ω

(x

). Let (x

, y

)

be the new rational solution (note that it must exist since we chose an x

such that (x

, y) ∈ S). Two

cases may happen: if (x

, y

) violates another upper level constraint it is not bilevel-feasible for P

,

otherwise P

admits a bilevel-feasible solution at x

unlike P

. The same rationale can be repeated if

the follower does not admit a finite optimal solution, i.e. R

(x

) is not a finite set. This completes the

proof.

2

2.1 Polyhedral properties 23

Figure 2.3. Upper level constraints may induce infeasibility

In the proof of Theorem 9 it is interesting to note that, in some particular cases, the upper level

constraints play a fundamental role in defining feasibility of BLP. Let us consider this simple example:

(P

)

min

x,y

y

s.t.

y ≤ 1

x ≥ 0

y ∈ argmin

−y

s.t.

x + y ≥ 1

x − y ≤ 1

y ≥ 0

(P

)

min

x,y

y

s.t.

x ≥ 0

y ∈ argmin

−y

s.t.

y ≤ 1

x + y ≥ 1

x − y ≤ 1

y ≥ 0

In Figure 2.3 we can see exactly what is stated in the proof of Theorem 9: at x = ¯x in P

the follower’s

feasible set Ω

(¯x) is unbounded and R

(¯x) is not a finite set, while in P

the set Ω

(¯x) is bounded

and a bilevel-feasible solution exists. While problem P

is infeasible, problem P

has a set of multiple

optimal solutions (x, 1) with x ∈ [0, 2].

Shi et al. [135][132][133][134] propose a new definition of bilevel linear problems and state that

the formulation they introduce is able to solve instances that the classical model fails to solve. The

model they propose is obtained placing the upper level constraints in the follower’s problem, thus

for Theorem 9 they relax the original formulation of the problem. The authors start from the wrong

observation that if S is non empty and compact and Ω

(x) 6= ∅ ∀x such that (x, y) ∈ S, a Pareto

optimal solution must exist. In fact, as we showed in the previous chapter, a bilevel–feasible solution

is not necessarily a Pareto optimal solution and the two properties are distinct concepts. Finally, as

observed by Audet et al. [11] and Mersha and Dempe [112], the authors propose a weaker formulation

of the problem because they ignore the different role played by the upper and lower level constraints

and solve a relaxation of the original BLP.

Finally, in order to understand the meaning of upper level constraints in terms of application on

real problems, let us consider the following hazardous material transportation problem. Given a graph

G = (V, E) that represents the road network of a geographical area, the leader wants to minimize the

risk associated to the hazmat transportation, while the follower has the objective of minimizing the

transportation costs for carrying hazmat materials from an origin s to a destination t. The leader can

allow or forbid the transit on a given arc (i, j) and the follower, once the network has been designed by

the leader, chooses the shortest path for realizing the transportation. The bilevel model that formulates

this problem is:

min

x,y X (i,j)∈E

ρ

y

s.t.

x

∈ {0, 1}

y ∈ argmin

y X (i,j)∈E

c

y

s.t.

X j|(i,j)∈E

y

−

X j|(j,i)∈E

y

=

      

1 i = s

0 i 6= s, t

−1

i = t

y

≤ M · x

y

∈ {0, 1}

Variable x

is equal to 1 if the leader allows the follower to transport hazmat materials on arc (i, j) and

0 otherwise. Variable y

is equal to 1 if the follower uses the arc (i, j) in the s − t path and 0 otherwise.

A risk ρ

and a cost c

are associated to every arc (i, j): the cost is associated to the arc’s length,

while the risk is due to the presence of critical infrastructures that may be damaged in case of accident.

Let us think to a road passing in the city center of a small city, it may have a small transportation cost,

but a high risk because of impact and damages that a possible accident may cause. The objective of the

follower is to find the least expensive path, while the objective of the leader is to minimize the total risk.

The follower problem is a shortest path problem in which constraints y

≤ M · x

are added: if the

transit on an arc (i, j) is forbidden, i.e. x

= 0, the follower can not choose arc (i, j) in his shortest

path problem.

Let us consider the instance in Figure 2.4 with s = 1 and t = 4.

The optimal solution for the leader has x

∗₁₄

= 0 as this is the only way to forbid follower to transit

on arc (1, 4) that represents both the shortest path and the path with the highest risk. The follower’s

shortest path is 1 − 3 − 4 of cost 22 and risk 3 and this is the optimal solution of the model. If we move

the set of constraints y

≤ M · x

from the lower to the upper level, the problem does not apparently

change very much. Actually, regardless the choice of the leader, the follower will always select path

1 − 4 of cost 1 and risk 100. Thus, if the leader sets x

= 0, the follower’s best response is not

bilevel-feasible because of the upper level constraints y

≤ M · x

. It follows that all bilevel-feasible

solutions allow transit on arc (1, 4) and the optimal solution of the problem has risk equal to 100.

This result shows that, just modifying the position of a set of constraints, their role in the formulation

changes and the solutions space may be completely different.

In document Integer Bilevel Linear Programming Problems: New Results and Applications (Page 39-43)