Optimization Problems Under One-sided (max, min)-linear Equality Constraints

WDS'12 Proceedings of Contributed Papers, Part I, 13 19, 2012. ISBN 978-80-7378-224-5 MATFYZPRESS Optimization Problems Under One-sided (max, min)-linear Equality Constraints M. Gad Charles University, Faculty of Mathematics Physics, Prague, Czech Republic. Sohag University, Faculty of Science, Sohag, Egypt. Abstract. In this article we will consider optimization problems, the obective function of which is equal to the maximum of a finite number of continuous functions of one variable. The set of feasible solutions is described by the system of (max, min)-linear equality constraints with variables on one side. Complexity of the proposed method with monotone or unimodal functions will be studied, possible generalizations extensions of the results will be discussed. 1. Introduction The algebraic structures in which (max, +) or (max, min) replace addition multiplication of the classical linear algebra have appeared in the literature approximately since the sixties of the last century (see e.g. Butkovič Hegedüs [1984], Cuninghame-Green [1979], Cuninghame-Green Zimmermann [2001], Vorobov [1967]). A systematic theory of such algebraic structures was published probably for the first time in Cuninghame-Green [1979]. In recently appeared book Butkovič [2010] the readers can find latest results concerning theory algorithms for the (max, +) linear systems of equations. Gavalec Zimmermann [2010] proposed a polynomial method for finding the maximum solution of the (max, min)-linear system. Zimmermann Gad [2011] introduced a finite algorithm for finding an optimal solution of the optimization problems under (max, +) linear constraints. Gavalec et al. [2012] provide survey of some recent results concerning the (max, min)-linear systems of equations inequalities optimization problems under the constraints descibed by such systems of equations inequalities. Gad [2012] considered the existence of the solution of the optimization problems under two-sided (max, min)-linear inequalities constraints. Let us consider the optimization problems under one-sided (max, min)-linear equality constraints of the form: max J (a i x ) = b i, i I, which can be solved by the appropriate formulation of equivalent one-sided inequality constraints using the methods presented in Gavalec et al. [2012]. Namely, we can consider inequality systems of the form: max J (a i x ) b i, i I, max J (a i x ) b i, i I. Such systems have 2m inequalities (if I = m), which have to be taken into account. It arises an idea, whether it is not more effective to solve the problems with equality constraints directly without replacing the equality constraints with the double numbers of inequalities. In this article, we are going to propose such an approach to the equality constraints. First we will study the structure of the set of all solutions of the given system of equations with finite entries a i & b i, for all i I & J. Using some of the theorems characterizing the structures of the solution set of such systems, we will propose an algorithm, which finds an optimal solution of minimization problems with obective functions of the form: f(x) max J f (x ), where f, J are continuous functions. Complexity of the proposed method of monotone or unimodal functions f, J will be studied, possible generalizations extensions of the results will be discussed. 2. One-sided (max, min)-linear Systems of Equations Let us introduce the following notations: J = {1,..., n}, I = {1,..., m}, where n m are integer numbers, R = (, ), R = [, ], R n = R R (n-times), similarly R n = R R, x = (x 1,..., x n ) R n, α β = min{α, β}, α β = max{α, β} for any α, β R, we set per definition =, 13

=, a i R, b i R, i I, J are given finite numbers, consider the following system of equations: max (a i x ) = b i, i I, (1) J x x x. (2) The set of all solutions of the system (1), will be denoted M =. Before investigating properties of the set M =, we will bring an example, which shows one possible application, which leads to solving this system. Example 2.1 Let us assume that m places i I {1, 2,..., m} are connected with n places J {1, 2,..., n} by roads with given capacities. The capacity of the road connecting place i with place is equal to a i R. We have to extend for all i I, J the road between i by a road connecting with a terminal place T choose an appropriate capacity x for this road. If a capacity x is chosen, then the capacity of the road from i to T via is equal to a i x = min(a i, x ). We require that the maximum capacity of the roads connecting i to T via over = 1, 2,..., n, which is described by max J (a i x ), is equal to a given number b i R the chosen capacity x lies in a given finite interval i.e. x [x, x ], where x, x R are given finite numbers. Therefore feasible vectors of capacities x = (x 1, x 2,..., x n ) (i.e. the vectors, the components of which are capacities x having the required properties) must satisfy the system (1) (2). In what follows, we will investigate some properties of the set M = described by the system (1). Also, to simplify the formulas in what follows we will set a i (x) max J (a i x ) i I. Let us define for any fixed J the set I = {i I & a i b i }, S (x ) = {k I & a k x = b k }, define the set M = (x) = {x M = & x x}, Lemma 2.1 Let us set for all i I J T = i = {x ; (a i x ) = b i & x x } Then for any fixed i, the following equalities hold: (i) T = i = {b i} if a i > b i & b i x ; (ii) T = i = [b i, x ] if a i = b i & b i x ; (iii) T = i = if either a i < b i or b i > x. Proof: (i) If a i > b i, then a i x > b i for any x > b i, also a i x < b i for any x < b i, so that the only solution for equation a i x = b i is x = b i x. (ii) If a i = b i & b i x, then a i x < b i for arbitrary x < b i, but a i x = b i for arbitrary b i x x. (iii) If a i < b i, then either a i x = a i < b i for arbitrary x a i, or a i x = x < b i for arbitrary x < a i. Therefore there is no solution for equation a i x = b i, which means T i = =. Also if b i > x, then either x b i > x, so that T i = =, or x < b i there are two cases, the first is x a i a i x = x < b i the second case is x > a i a i x = a i < x < b i. therefore there is no solution for equation a i x = b i, which means T i = =. 14

Lemma 2.2 Let us set for all i I J let Then we have x (i) = GAD: OPTIMIZATION PROBLEMS { b i if a i > b i & b i x, x if a i = b i & b i x or T = i =, ˆx = S (ˆx ) = the following statements hold: (i) ˆx M = (x) J S (ˆx ) = I { min k I x (k) if I, x if I =, { k I ; x (k) = ˆx }, J, (ii) Let M = (x), then ˆx M = (x) for any x M = (x) x ˆx, i.e. ˆx is the maximum element of M = (x). Proof: (i) To prove the necessary condition we suppose ˆx M = (x), then max J (a i ˆx ) = b i, for all i I. So that for all i I, there exists at least one (i) J such that a i(i) ˆx (i) = max J (a i ˆx ) = b i, then either ˆx (i) = b i, if a i(i) > b i & b i x (i), or ˆx (i) > b i, if a i(i) = b i & b i x (i), so that i I (i) ˆx (i) = x (i) (i). Otherwise a ik ˆx k < b i, for all k (i) & k J, then ˆx k < b i, a ik < b i, I k =, therefore we can choose ˆx k = x k. Hence for all i I there exists at least one (i) J such that S (i) (ˆx (i) ) i S (i) (ˆx (i) ) J S (ˆx ) = I. To prove the sufficient condition, let J S (ˆx ) = I, then for all i I there exists at least one (i) J such that S (i) (ˆx (i) ) i S (i) (ˆx (i) ). Therefore ˆx (i) = b i if a i(i) > b i & b i x (i), then a i(i) ˆx (i) = ˆx (i) = b i. Otherwise ˆx (i) = x (i) if either a i(i) = b i & b i x (i), then a i(i) ˆx (i) = a i(i) = b i or T i(i) = =. Then for all i I there exists at least one (i) J such that max J (a i ˆx ) = a i(i) ˆx (i) = b i. Then ˆx M = (x). (ii) Let M = (x), then for each i I, there exists at least one (i) J such that T i(i) =, b i x (i) & a i(i) b i. Therefore there exists at least one (i) J such that either a i(i) > b i & b i x (i), then x (i) (i) = b i. Or a i(i) = b i & b i x (i), then x (i) (i) = x (i) so that ˆx (i) = b i if i I (i) a i(i) ˆx (i) = b i is satisfied. Otherwise if I =, we set ˆx = x. Then ˆx M = (x) for any x M = (x) we have x ˆx, i.e. ˆx is the maximum element of M = (x). It is appropriate now to define M = (x, x) = {x M = (x) & x x}, which is the set of all solutions of the system described by (1) (2). Theorem 2.1 Let ˆx S (ˆx ) be defined as in Lemma 2.2 then: (i) M = (x, x) if only if ˆx M = (x) & x ˆx, (ii) If M = (x, x), then ˆx is the maximum element of M = (x, x), (iii) Let M = (x, x) J J. Let us set x = ˆx if J, otherwise x = x. Then x M = (x, x) J S ( x ) = I Proof: 15

(i) If ˆx M = (x) & x ˆx, from definition M = (x, x) we have ˆx M = (x, x), then M = (x, x). If M = (x, x) so that M = (x) from lemma 2.2, it is verified that ˆx M = (x) ˆx is the maximum element of M = (x), therefore ˆx x. (ii) If M = (x, x), so that M = (x) M = (x, x) M = (x), since ˆx is the maximum element of M = (x), then ˆx is the maximum element of M = (x, x). (iii) Let x M = (x, x) x M = (x), also from definition x we have x = ˆx for all J, so that S ( x ) J. And x < ˆx for all J \ J, so that S ( x ) = J \ J. Hence x M = (x), by lemma 2.2 we have J S ( x ) = I. Let J S ( x ) = I, since J J we have J S ( x ) = I. By lemma 2.2, also we have x x therefore x M = (x, x). 3. Optimization Problems Under One-sided (max, min)-linear Equality Constraints In this section we will solve the following optimization problem: subect to x M = (x) f(x) max J f (x ) min (3) x M = (x, x) (4) We assume further that f : R R are continuous monotone functions (i.e. increasing or decreasing), M = (x, x) denotes the set of all feasible solutions of the system described by (1) (2) assuming that M = (x, x) (note that the emptiness of the set M = (x, x) can be verified using the considerations of the preceding section). Let J { f decreasing function} so that min (x ) = f (ˆx ), x [x,ˆx ] J. Then we can propose an algorithm for solving problem (3) (4) under the assumption that M = (x, x) which means J S (ˆx ) = I, i.e. for finding an optimal solution x opt of problem (3) (4). Algorithm 3.1 We will provide algorithm, which summarizes the above discussion finds an optimal solution x opt of problem (3) (4), where f (x ) are continuous monotone functions. 0 Input I, J, x, x, a i b i for all i I J; 1 Find ˆx, set x = ˆx, 2 Find J { f decreasing function} 3 F = {p max J f ( x ) = f p ( x p )} 4 If F J, then x opt = x, Stop. 5 Set y p = x p p F, & y = x, otherwise 6 If J S (y ) = I, set x = y go to 3 7 x opt = x, Stop. We will illustrate the performance of this algorithm by the following numerical example. 16

Example 3.1 Consider the optimization problem (3), where f (x ) J are continuous monotone functions in the form f (x ) c x + d, C = [ 0.2057 4.8742 2.8848 0.9861 1.7238 1.1737 3.3199 ] D = [ 1.4510 1.5346 3.6121 0.9143 2.0145 1.9373 4.8467 ] subect to x M = (x, x), where the set M = (x, x) is given by the system (1) (2) where J = {1, 2,..., 7}, I = {1, 2,..., 6}, x = 0 J x = 10 J consider the system (1) of equations where a i & b i i I J are given by the matrix A vector B as follows: 6.1221 9.0983 9.5032 6.0123 6.1112 4.1221 5.5776 8.2984 3.3920 2.5185 1.1925 8.9742 6.7594 8.6777 A = 2.0115 6.3539 4.4317 7.7452 0.6465 9.4098 1.3576 6.4355 1.6404 3.1850 3.7361 7.2605 3.0201 5.3808 8.5668 5.8310 2.5146 8.7804 3.7709 4.4770 2.3007 5.2690 9.6900 5.1598 9.2889 6.1585 1.0786 7.0121 B T = [ 6.1221 7.0955 6.3539 6.4355 6.5712 7.0121 ] By the method in section 2 we get ˆx, which is the maximum element of M = (x, x), as follows: ˆx = (6.5712, 6.1221, 6.1221, 6.3539, 6.4355, 6.3539, 7.0955) By using algorithm 3.1 after five iterations we find that if we set x 4 = 0 the third equation of the system (1) does not satisfy, therefore ALGORITHM 1 go to step 8 take x opt = x = (6.5712, 0, 0, 6.3539, 0, 0, 7.0955) stop. We obtained the optimal value for the obective function f(x opt ) = 5.3510. We can easily verify that x opt is a feasible solution. Remark 3.1 By reference to the lemma 2.2 theorem 2.1 it is not difficult to note that the maximum number of arithmetic or logic operations in any step to get ˆx can not exceed n m operations. This will happen when we calculate x (i), i I & J. Also from the above example we can remark that the maximum number of operations in each step in any iterations from algorithm 3.1 is less than or equal to the number of variables n the maximum number of iterations from step 3 to step 6 of this algorithm can not exceed n. Therefore the computational complexity of the algorithm 3.1 is O(max(n 2, n m)). In what follows let us modify algorithm 3.1, to be suitable to find an optimal solution for any general continuous functions f (x ) as follows: Algorithm 3.2 We will provide algorithm, which summarizes the above discussion find an optimal solution x opt of problem (3) (4), where f (x ) are general continuous functions. 0 Input I, J, x, x, a i b i for all i I J; 1 Find ˆx, set x = ˆx, 2 Find min x [x,ˆx ] f (x ) = f (x ), J. 3 Set J { f (ˆx ) = f (x )} 4 F = {p max J f ( x ) = f p ( x p )} 5 If F J, then x opt = x, Stop. 17

6 Set y p = x p p F, & y = x, otherwise 7 If J S (y ) = I, set x = y go to 3 8 x opt = x, Stop. We will illustrate the performance of this algorithm by the following numerical examples. Example 3.2 Consider the optimization problem (3), where f (x ) J are continuous functions given in the following form f (x ) (x ξ ) 2, ξ = (3.3529, 1.4656, 5.6084, 5.6532, 6.1536, 6.5893) subect to x M = (x, x), where the set M = (x, x) is given by the system (1) (2) where J = {1, 2,..., 6}, I = {1, 2,..., 6}, x = 0 J x = 10 J consider the system (1) of equations where a i & b i i I J are given by the matrix A vector B as follows: 3.6940 0.8740 0.5518 4.6963 2.1230 1.4673 1.9585 8.3470 5.8150 8.5545 8.9532 8.7031 A = 1.3207 8.9610 1.5718 3.7155 0.1555 4.3611 8.4664 9.1324 6.6594 2.5637 6.0204 6.0846 2.4219 9.6081 1.9312 2.5218 1.3976 4.1969 1.1172 3.6992 7.5108 4.7686 4.4845 4.3301 B T = [ 4.0195 7.2296 4.2766 6.6594 4.1969 6.9874 ] By the method in section 2 we get ˆx, which is the maximum element of M = (x, x), as follows: ˆx = (6.6594, 4.1969, 6.9874, 4.0195, 7.2296, 4.2766) By using algorithm 3.2 after three iterations we find that x opt = (3.3297, 1.4689, 6.9874, 4.0195, 7.2296, 4.2766 Here we find the algorithm 3.2 stop in step 5 since the active variable in iteration 3 is x 6, at the same time the obective function has the minimum value in this value of x 6, so that the algorithm 3.2 stop. Then we obtained the optimal value of the obective function, f(x opt ) = 5.3488. It is easy to verify that x opt is a feasible solution. Example 3.3 Consider the optimization problem (3), where f (x ) J are continuous functions given in the following form f (x ) (x ξ )(x ), where ξ = (3.3529, 1.4656, 5.6084, 5.6532, 6.1536, 6.5893) = (0.7399, 0.1385, 4.1585, 1.1625, 2.1088, 1.2852) subect to x M = (x, x), where the set M = (x, x) is given in the same way as in example 3.2 By the method in section 2 we get ˆx, which is the maximum element of M = (x, x), as follows: ˆx = (6.6594, 4.1969, 6.9874, 4.0195, 7.2296, 4.2766) By using algorithm 3.2 we find: Iteration 1: 1 x = (6.6594, 4.1969, 6.9874, 4.0195, 7.2296, 4.2766); 2 x = 0.7325, 1.4689, 5.5899, 1.1656, 6.1452, 1.2830; 3 J = ; 4 F = {1}; f ( x) = 19.5728; 6 y = (0.7325, 4.1969, 6.9874, 4.0195, 7.2296, 4.2766); 7 J S (y ) = I; 18

x = (0.7325, 4.1969, 6.9874, 4.0195, 7.2296, 4.2766). Iteration 2: 3 J = {1}; 4 F = {3}; f ( x) = 15.3692; 6 y = (0.7325, 4.1969, 5.5899, 4.0195, 7.2296, 4.2766); 7 J S (y ) = {1, 2, 3, 5} I; 8 x opt = (0.7325, 4.1969, 6.9874, 4.0195, 7.2296, 4.2766), STOP. In Iteration 2 algorithm 3.2 stops since the fourth sixth equations in the system of equations (1) can not be verify if we change the value of the variable y 3 to be y 3 = (5.5899). If it is necessary to change this value of the variable y 3 to minimize the obective function so that our exhortation to decision-maker, it must be changed both the capacities of the ways a 45 a 65 to be a 45 = 6.6594 a 65 = 6.9874 in order to maintain the verification of the system of equations (1). In this case we can complete the operation to minimize the obective function, after Iteration 4 we obtained x opt = (0.7325, 1.4689, 5.5899, 4.0195, 7.2296, 4.2766); so that the optimal value f(x opt ) = 10.0481, we can easily verify that x opt is a feasible solution. Remark 3.2 Step 2 in algorithm 3.2 depends on the method, which finds the minimum for each function f (x ) in the interval [x, ˆx ], but this appears in the first iteration only only once. Also from the above examples we can remark that the maximum number of operations in each step in any iterations from algorithm 3.2 is less than or equal to the number of variables n the maximum number of iterations from step 3 to step 7 of this algorithm{ can not exceed n. Therefore } the computational complexity of the algorithm 3.2 is given by max O(max(n 2, n m)), Õ, where Õ is complexity of the method, which finds the minimum for each function f (x ) in the interval [x, ˆx ]. Acknowledgments. The author offers sincere thanks to Prof. Karel Zimmermann for his continuing support the great advice forever. References Butkovič, P., Hegedüs, G.: An Elimination Method for Finding All Solutions of the System of Linear Equations over an Extremal Algebra, Ekonomicko matematický obzor 20, 203 215, 1984. Butkovič, P.: Max-linear Systems: Theory Algorithms, Springer Monographs in Mathematics, 267 p, Springer-Verlag, London 2010. Cuninghame-Green, R. A.: Minimax Algebra. Lecture Notes in Economics Mathematical Systems 166, Springer-Verlag, Berlin 1979. Cuninghame-Green, R. A., Zimmermann, K.: Equation with Residual Functions, Comment. Math. Univ. Carolinae 42, 729 740, 2001. Gad, M.: optimization problems under two-sided (max, min)-linear inequalities constraints(to appear). Gavalec, M., Zimmermann, K.: Solving Systems of Two-Sided (max, min)-linear Equations, Kybernetika 46, 405 414, 2010. Gavalec, M., Gad, M., Zimmermann, K.:Optimization problems under (max, min)-linear equation /or inequality constraints (to appear). Vorobov, N. N.: Extremal Algebra of positive Matrices, Datenverarbeitung und Kybernetik 3, 39 71, 1967 (in Russian). Zimmermann, K., Gad, M.: Optimization Problems under (max, +) linear Constraints,International conference presentation of mathematics 11 (ICPC), 11, 159-165, 2011. 19