A mixed-discrete bilevel programming problem

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1 A mixed-discrete bilevel programming problem Stephan Dempe 1 and Vyacheslav Kalashnikov 2 1 TU Bergakademie Freiberg, Freiberg, Germany 2 Instituto de Tecnologías y Educación Superior de Monterrey, Monterrey, Nuevo León, México Abstract. We investigate a special mixed-discrete bilevel programming problem resulting from the task to compute cash-out penalties due to the supply of incorrect amounts of natural gas through large gas networks. For easier solution we move the discreteness demand from the lower to the upper levels. This clearly changes the problem and it is our aim to investigate the relations between both problems. After the move, a parametric generalized transportation problem arises in the lower level for which the formulation of conditions for the existence of an optimal solution is our first task. In the second part of the paper, requirements guaranteeing that an optimal solution of the original problem is obtained by solving the modified one as well as bounds for the optimal value of the original problem are derived. 1 Introduction Bilevel programming problems are hierarchical optimization problems where the set of variables is partitioned between two variables x and y. The values of x = x(y) are determined by the follower knowing the selection of the other decision maker, the leader. The bilevel problem is the problem of the leader who has to determine the best choice of y knowing the reaction x = x(y) of the follower on his decisions. To formulate the bilevel optimization problem consider the follower s parametric problem first. Given y Y R m he has to take x(y) Ψ(y) := Argmin {f(x, y) : g(x, y) 0}, (1) x i.e. x(y) is an optimal solution of a parametric optimization problem. Now, the leader s problem (or the bilevel optimization problem) is min{f (x, y) : x Ψ(y), y Y }. (2) x,y Problems of this type have been investigated in the monographs [1,2]. The number of possible applications is large and quickly growing (cf. the annotated bibliography [3]). Other than for continuous problems, (mixed-) discrete bilevel programming has been the topic of rather a small number of papers.

2 2 Stephan Dempe and Vyacheslav Kalashnikov Here we consider a special bilevel programming problem where the lower level problem (1) is a generalized transportation problem: c ij x ij + d 1 i x i,m+1 + d 2 jx n+1,j min x ij + (1 q)x i,m+1 = a i, i = 1,..., n, u ij x ij + qu n+1,j x n+1,j = b j, j = 1,..., m, (3) x ij 0, i = 1,..., n + 1, j = 1, 2,..., m + 1, q {0, 1}. Here, the coefficients u ij (0, 1) and the 0 1 Variable q indicates that we either have inequalities in the first or in the second block of constraints. This is a generalized transportation problem since the coefficients in the constraint matrix can be different from zero and one and possible slacks are penalized in the objective function. Moreover, the objective function is not linear but the absolute value of a linear function. Denote the set of optimal solutions of this problem by Ψ(a, b), where a = (a 1,..., a n ), b = (b 1,..., b m ). The bilevel programming problem reads then as c ij x ij + d 1 i x i,m+1 + d 2 jx n+1,j min,a,b (x, q) Ψ(a, b), (a, b) Y (4) where Y is a fixed set, Y R n + R m +. Problem (3), (4) is a reformulation of a problem for minimizing the cashout penalties of a natural gas shipping company [4,7,8]. Bilevel programming problems are N P hard [5], and discrete bilevel programming problems are even harder so solve. This is especially true if the discreteness demand is located in the lower level problem [9]. Hence, we feel that it may be advantageous to shift the integrality demand from the lower to the upper levels. This, of course, changes the problem, but it has two implications. First, since there is only one 0 1 variable in the problem, this is reduced to the solution of two bilevel programming problems, the better solution of the two is taken as result of the problem. Next, to solve the resulting two linear bilevel programming problems we can e.g. use an idea in [10] reducing both problems to quadratic optimization problems involving a penalty term in the objective function.

3 2 The subproblems A mixed-discrete bilevel programming problem 3 If the discreteness demand is shifted into the upper level problem, the lower level ones (3) reduce to problems of minimizing the absolute value of a linear function on a polyhedron. If all the values on the right hand side are finite numbers, the polyhedron is bounded. Hence, this problem has an optimal solution whenever its feasible set is nonempty. We give conditions guaranteeing nonemptiness of this set. Consider the lower level problem (3) with q = 1 (the case with q = 0 can be dealt with similarly) and take an arbitrary linear objective function for convenience: c ij x ij + d 2 jx n+1,j min x ij = a i, i = 1,..., n + 1, u ij x ij + u n+1,j x n+1,j = b j, j = 1,..., m, (5) x ij 0, i = 1,..., n + 1, j = 1, 2,..., m. Note that we have added one constraint with unknown (to be fixed later) but nonnegative right-hand side a n+1 to give the model a more familiar formulation. If this problem has a feasible solution, then this is also valid for the original one. Now, we use an idea originating from [6] to transform the problem. Take any set of m + (n + 1) 1 = m + n basic variables x ij, (i, j) U. Then, if we take s 1 = 1, the system of equations s i t j u ij = 0, (i, j) U (6) has a unique solution (s, t) with s i > 0, t j > 0 for all i, j. Using this solution, the generalized transportation problem can be transformed into c ij x ij + d 2 jx n+1,j min s i x ij = s i a i, i = 1,..., n + 1, t j t j u ij s i x ij + u n+1,j s n+1 x n+1,j = t j b j, j = 1,..., m, (7) s i s n+1 x ij 0, i = 1,..., n + 1, j = 1, 2,..., m. This problem arises if each equation in the first set is multiplied by s i and each one in the second system by t j. But now, the coefficients in the second

4 4 Stephan Dempe and Vyacheslav Kalashnikov set of equations reduce to one for (i, j) U. By simply summing up the constraints of problem (7) and using that x ij = 0 for (i, j) U we get Theorem 1. If the generalized transportation problem (5) with q = 1 has a feasible solution then there is a basic feasible solution with x ij = 0 for (i, j) U such that n+1 s i a i = t j b j (8) for the solution s, t of (6). Equation (8) shows that a n+1 has to be taken as s n+1 a n+1 = t j b j s i a i and that a necessary condition for solvability of this problem is a n+1 0. Not every set of basic variables corresponds to feasible solutions of the generalized transportation problem (5). To find a condition characterizing correct basic solutions, consider a subset I {1,..., n + 1} and the set N(I) := {j : (i, j) U for some i I}. Then, for a feasible solution of the generalized transportation problem we have s i a i = i I i I (i,j) U s i x ij j N(I) (i,j) U u ij t j s i s i x ij = j N(I) t j b j (9) due to x ij = 0 for (i, j) U. Here, the inequality is implied by {(i, j) U : i I} {(i, j) U : j N(I)} and u ij t j /s i = 1 for (i, j) U. Also, using the set M(J) := {i : (i, j) U for some j J} for J {1,..., m} we get the similar inequality i M(J) s i a i = i M(J) (i,j) U s i x ij j J (i,j) U t j u ij s i x ij = t j b j. (10) s i j J Theorem 2. Consider the generalized transportation problem (5) with q = 1. Then a set of basic variables corresponds to a feasible solution if and only if the conditions (8) together with (9) and (10) are satisfied for all subsets I {1,..., n + 1} and J {1,..., m}. To prove this theorem use a method like the north-west corner rule for computing a starting solution for the transportation problem. Such methods fix the basic variables one after the other such that in any step one of the equations in the definition of the problem (7) is satisfied. Then, in the next step, a basic variable x ij is to be determined with (i, j) {(i, j) U : j N(I)} \ {(i, j) U : i I}

5 A mixed-discrete bilevel programming problem 5 respectively (i, j) {(i, j) U : i M(J)} \ {(i, j) U : j J} for which is enough space by the assumptions of the theorem. Here, U denotes the index set of basic variables, and I and J are the index sets of the equations being already satisfied in that step. To derive another result consider the variables x n+1,j in the problem (5) as being slack variables. Then, to investigate solvability of the lower level problems we can investigate the two problems c ij x ij min x ij = a i, i = 1,..., n, u ij x ij b j, j = 1,..., m, (11) x ij 0, i = 1,..., n, j = 1, 2,..., m. and c ij x ij min x ij a i, i = 1,..., n, u ij x ij = b j, j = 1,..., m, (12) x ij 0, i = 1,..., n, j = 1, 2,..., m. Theorem 3. For any natural numbers n, m and vectors a = (a 1,..., a n ), b = (b 1,..., b m ), with a i > 0, i = 1,..., n, b j > 0, j = 1,..., m at least one of the problems (11) and (12) has a feasible solution. Proof. To prove our assertion, we use the modified North-West Corner Rule in the following form. 0: Set k = 0, i k = 1, j k = 1. 1: Put { aik, if u x ik,j k = ik,j k a ik b jk ; b jk /u ik,j k, otherwise. 2: If i k < n then set { ik + 1, if u i k+1 = ik,j k a ik b jk, i k, otherwise.

6 6 Stephan Dempe and Vyacheslav Kalashnikov If i k = n and u ik,j k a ik b jk, then set K = k and stop. If i k+1 = i k + 1 goto Step 3. Symmetrically, if j k < m, put { jk + 1, if u j k+1 = ik,j k a ik b jk, j k, otherwise. At last, if j k = m and u ik,j k a ik b jk, then set K = k and stop. 3: Now we update the parameters: a ik := a ik x ik j k ; b jk := b jk u ik j k x ik j k. If a i = 0 for all i = 1,..., n, or b j = 0 for each j = 1,..., m, then set K = k and stop. Otherwise, set k := k + 1 and return to Step 1. Let the algorithm stop in Step 3. Then by construction, either a = 0 which means that we have m x ij = a i, i = 1,..., n or b = 0 implying n u ij x ij = b j, j = 1,..., m. Since the other system of inequalities is also satisfied, we obtain non-emptiness of the feasible set of either problem (11) or (12). If the algorithm stops in Step 2 then either i K = n and m x ij = a i, i = 1,..., n or j K = m and n u ij x ij = b j, j = 1,..., m. In both cases, the other system of inequalities is also satisfied by construction. This again implies non-emptiness of the feasible set of either problem (11) or (12). The theorem is proved completely. 3 Relation between the bilevel problems As mentioned in the Introduction, we replaced problem (3), (4) by c ij x ij + d 1 i x i,m+1 + d 2 jx n+1,j min,a,b (x, q) Ψ q (a, b), (a, b) Y, q {0, 1} (13) where Ψ q (a, b) denotes the set of optimal solutions of problem (3) with q being fixed to q = 0 or q = 1. In doing so we cannot expect to obtain optimal solutions for problem (3), (4) in any case. Theorem 4. Let (x 0, a 0, b 0 ) be a locally optimal solution for problem min c x,a,b ij x ij + d 1 i x i,m+1 : (x, q) Ψ q (a, b), (a, b) Y (14)

7 A mixed-discrete bilevel programming problem 7 with q = 0 and objective function value z 0 = and let x 1 Ψ 1 (a 0, b 0 ) with z 1 = c ij x 0 ij + d 1 i x 0 i,m+1 c ij x 1 ij + d 2 jx 1 n+1,j. If 0 < z 0 < z 1, then (x 0, a 0, b 0 ) is a local optimal solution of problem (3), (4). The proof of this theorem mainly uses continuity of the optimal function value of right hand side perturbed linear optimization problems guaranteeing that the function value of feasible solutions for the subproblem with q = 1 for parameter values being near to (a 0, b 0 ) can never be lower then z 0. Hence the sequence of inequalities 0 < z 0 < z 1 implies that x 0 Ψ(a 0, b 0 ) is feasible for the problem (3), (4) and this persists in a neighborhood of (a 0, b 0 ). There are some more cases which can be treated similarly: Remark (x 0, a 0, b 0 ) is locally optimal for problem (14) with q = 0 and 0 > z 0 > z 1, 2. (x 1, a 1, b 1 ) is locally optimal for problem (14) with q = 1 and 0 > z 1 > z 0, and 3. (x 1, a 1, b 1 ) is again locally optimal for problem (14) with q = 1 and 0 < z 1 < z 0. It is not too difficult to see that an optimal solution of problem (3) is either in Ψ 0 (a, b) or in Ψ 1 (a, b). Hence, and thus Ψ(a, b) Ψ 0 (a, b) Ψ 1 (a, b) {(x, a, b) : x Ψ(a, b)} {(x, a, b) : x Ψ 0 (a, b)} {(x, a, b) : x Ψ 1 (a, b)}. Now, since the feasible set in problem (3), (4) is equal to the set on the left hand side and the feasible set of the really solved one is equal to the larger set on the right hand side of this inclusion we get Theorem 5. min q {0,1} zq z max q {0,1} zq where z q denotes the optimal function value of (14) and z is the optimal value of problem (3), (4).

8 8 Stephan Dempe and Vyacheslav Kalashnikov References 1. Bard, J. F. (1998) Practical Bilevel Optimization: Algorithms and Applications. Kluwer Academic Publishers, Dordrecht 2. Dempe, S. (2002) Foundations of Bilevel programming. Kluwer Academic Publishers, Dordrecht 3. Dempe, S. (2003) Annotated Bibliography on Bilevel Programming and Mathematical Programs with Equilibrium Constraints. Optimization 52, Dempe, S., Kalashnikov, V. (2002) Discrete bilevel programming: application to a gas shipper s problem. Preprint No , TU Bergakademie Freiberg, (available at: 5. Deng, X. (1998) Complexity issues in bilevel linear programming. In: Migdalas, A., Pardalos, P., and Värbrand, P. (Eds.): Multilevel Optimization: Algorithms and Applications,, Kluwer Academic Publishers, Dordrecht et al., Gabasov, R.F., Kirillova, F.M. (1978) Linear Programming Methods. Part 2. Transportation Problems (Metody linejnogo programmirovaniya. Chast 2. Transportnye zadachi). Izdatel stvo Belorusskogo Universiteta, Minsk, (in Russian) 7. Kalashnikov, V.V., Ríos-Mercado, R.Z. (2001) A penalty-function approach to a mixed-integer bilevel programming problem. In: Zozaya, C. e.a. (Eds.), Proceedings of the 3rd International Meeting on Computer Science, Aguascalientes, Mexico, Vol. 2, pp Kalashnikov, V.V., Ríos-Mercado, R.Z. (2002) An algorithm to solve a gas cashout problem. In: R. Clute (Ed.), Proceedings of the International Business and Economic Research Conference (IBERC 2002), Puerto Vallarta, Mexico 9. Vicente, L.N., Savard, G., Judice, J.J. (1996) The discrete linear bilevel programming problem. Journal of Optimization Theory and Applications 89, White, D.J., Anandalingam, G. (1993) A penalty function approach for solving bilevel linear programs. Journal of Global Optimization 3,

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