A Test Set of Semi-Infinite Programs Alexander Mitsos, revised by Hatim Djelassi Process Systems Engineering (AVT.SVT), RWTH Aachen University, Aachen, Germany February 26, 2016 (first published August 26, 2009) Keywords SIP ; NLP ; Slater point ; nonconvex ; global optimization 1 Test Set Test problems for semi-infinite programs (SIPs) without convexity assumptions are presented 1.1 Problems based on Watson [9] For consistency purposes the problem labels of Watson [9] are used. X = [ 1000, 1000] 2 f(x) = x 2 1/3 + x 2 2 + x 1 /2 g(x, y) = (1 x 2 1 y 2 ) 2 x 1 y 2 x 2 2 + x 2. (2) Financial support from the Deutsche Forschungsgemeinschaft (German Research Association) through grant GSC 111 is gratefully acknowledged. amitsos@alum.mit.edu hatim.djelassi@avt.rwth-aachen.de.
X = [ 1000, 1000] 3 f(x) = exp(x 1 ) + exp(x 2 ) + exp(x 3 ) g(x, y) = 1/(1 + y 2 ) x 1 x 2 y x 3 y 2. (5) X = [ 1000, 1000] 2 f(x) = (x 1 2 x 2 + 5 x 2 2 x 2 2 x 2 13) 2 + (x 1 14x 2 + x 2 2 + x 3 2 29) 2 g(x, y) = x 2 1 + 2 x 2 y 2 + exp(x 1 + x 2 ) exp(y). (6) Note that in [1] the exponent is missing in the first term of the objective function. X = [ 1000, 1000] 3 2 f(x) = x 2 1 + x 2 2 + x 2 3 g(x, y) = x 1 (y 1 + y2 2 + 1) + x 2 (y 1 y 2 y2) 2 + x 3 (y 1 y 2 + y2 2 + y 2 ) + 1. (7) X = [ 1000, 1000] 6 2 f(x) = x 1 + x 2 /2 + x 3 /2 + x 4 /3 + x 5 /4 + x 6 /3. g(x, y) = exp(y1 2 + y2) 2 x 1 x 2 y 1 x 3 y 2 x 4 y1 2 x 5 y 1 y 2 x 6 y2. 2 (8) Note that presumably in Watson s collection [9] the coefficient of x 4 in the objective function is mistyped. This is suggested by the optimal solution value reported in [9] and by the symmetry of the problem with respect to the variables x 4 and x 6.
X = [ 1000, 1000] 6 Y = [ 1, 1] 2 f(x) = 4 x 1 2/3 (x 4 + x 6 ) g(x, y) = x 1 + x 2 y 1 + x 3 y 2 + x 4 y1 2 + x 5 y 1 y 2 + x 6 y2 2 3 (y1 2 y2) 2 2. (9) Note that the last term was rewritten from (y 1 y 2 ) 2 (y 1 + y 2 ) 2 since bilinear products result in weak relaxations, and higher CPU requirements for the solution of the NLPs. X = [0, 1] [ 1000, 1000] Y = [ 1, 1] f(x) = x 2 g(x, y) = (x 1 y) 2 x 2. (H) This problem was first proposed without bounds on x 2 as Example 4.1 in [7]. The bounds were subsequently added in [2]. X = [ 1000, 1000] 2 Y = [ 1, 1] f(x) = x 2 g(x, y) = 2 x 2 1 y 2 y 4 + x 2 1 x 2. (N) 1.2 New Continuous Problems The following two problems are based on problem 4 by Watson [9]
1. X = [ 1000, 1000] 3 f(x) = x 1 + x 2 /2. + x 3 /3. g(x, y) = exp(y 1) x 1 x 2 y x 3 y 2. (4 3) 2. X = [ 1000, 1000] 6 f(x) = x 1 + x 2 /2. + x 3 /3 + x 4 /4 + x 5 /5 + x 6 /6. g(x, y) = exp(y 1) x 1 x 2 y x 3 y 2 x 4 y 3 x 5 y 4 x 6 y 5. (4 6) The following problem is based on the Shekel Function in [5] 1. X = [ 1000, 1000] 2 Y = [ 10, 10] 1 f(x) = (x 1 4) 2 + (x 2 4) 2 + 0.1 1 (x 1 1) 2 + (x 2 1) 2 + 0.2 1 (x 1 8) 2 + (x 2 8) 2 + 0.2 g(x, y) = 0.1 (x 1 y) 2 (x 2 y) 2. (S) The following problem is aimed at provoking a dense discretization in methods similar to [3]. 1. X = [0, 6] Y = [2, 6] f(x) = 10 x g(x, y) = y 2 + x y 2. (DP) 1 + exp( 40(x y)) 1.3 New Problems Involving Integer Variables The following examples involve also discrete variables
1. Z = {0, 1} 3 3 f(x) = z 1 + 2z 2 /2. 3z 3 /3 (IP1) g(x, z, y) = z 1 y 2 1 y 2 2 + z 2 (y 2 y 1 y 3 ) + z 3 (exp(y 3 ) 3). 2. X = [ 1000, 1000] 2 Z = {0, 1} 3 f(x) = x 2 1 + exp(x 2 ) + z 1 x 1 + 2 z 2 x 1 x 2 3 z 3 (MINLP1) g(x, z, y) = x 1 exp(y 1 ) + x 2 (y 1 y 2 2) + z 1 (y 2 1 y 2 2) + z 2 (y 2 y 1 y 3 ) + z 3 (exp(y 3 ) 3). 3. X = [ 10, 10] 2 Z = {0, 1} 3 f(x) = x 2 1 + exp(x 2 ) + z 1 x 1 + 2 z 2 x 1 x 2 + 3 z 3 (MINLP2.) g(x, z, y) = x 1 exp(y 1 ) + x 2 (y 1 y 2 2) + z 1 (y 2 1 y 2 2) + z 2 (y 2 y 1 y 3 ) + z 3 (exp(y 3 ) 3). 1.4 Design Centering The goal of design centering is to place a body B(x) into a container C. Typically, some metric of the body, e.g., volume, or area, is maximized. The constraint B(x) C, leads in general to a GSIP, but in some cases an equivalent SIP can be formulated. The following problems are based on the design centering problem by Floudas and Stein [6, Section 5.2] with C R 2 after some modifications to avoid the trigonometric functions used therein, which are not currently supported by BARON [8]. The body is a disk, parameterized by the midpoint (x 1, x 2 ) and the radius x 3, i.e., B(x) = {z R 2 : (z 1 x 1 ) 2 + (z 2 x 2 ) 2 x 2 3}. The container is described by two inequality constraints.
Since C is simply connected, Floudas and Stein [6] use B(x) C, or equivalently the constraints c i (z) 0, z B(x), i {1, 2}. However, the parameterization of B(x) leads to trigonometric functions. Here, the following formulation is used instead: c i (x 1, x 2 ) 0, i {1, 2} (y x 1 ) 2 + (w i (y) x 2 ) 2 x 2 3, y Y, i {1, 2} where Y R is a large enough set and the intermediate w i (y) is calculated by c i (y, w i ) = 0. Obviously, this calculation is not possible for any function c i, but it is for the functions of interest herein. The first constraint ensures that the center of the circle is within the container. The second constraint mandates that the boundary of the container is sufficiently far away from the center of the disk, or equivalently the radius is sufficiently small. The container C used by Floudas and Stein is described by two functions, c l (y) = 0.3 sin(π y 1 ) y 2 (for the lower side) and c u = y1 2 + 0.3 y2 2 1 (for the upper side). The former contains a trigonometric function, and therefore is replaced by (the simpler) x 2 x 3. For the latter setting c 2 = 0 directly gives y 2 = (1 y1 2)/0.3. min x 3 x s.t. x 2 (1 x 2 1 )/0.3 0 x 2 3 (y x 1 ) 2 (1 y2 )/0.3 x 2 ) 2 0, y [ 1, 1] x 2 + x 3 0 x [ 1, 1] [0, 2] [0, 1]. (D1 0 1) Note the presence of regular constraints. Note also that 1 t 2 is not differentiable at t = ±1 (the boundaries of X and Y ). Arguably, Problem (D1 0 1) is significantly simpler than the original problem by Floudas and Stein. Therefore, a second design centering problem is formulated with c 2 (z) = 0.25 (z 1 0.5) 2 z 2. The
resulting design problem is min x 3 x s.t. x 2 (1 x 2 1 )/0.3 0 x 2 3 (y x 1 ) 2 (1 y2 )/0.3 x 2 ) 2 0, y [ 1, 1] 0.25 (x 1 0.5) 2 x 2 0 x 2 3 (y x 1 ) 2 ( 0.25 (y 0.5) 2 x 2 ) 2 0, y [ 1, 1] x [ 1, 1] [0, 2] [0, 1]. (D2 0 1) Note the presence of multiple semi-infinite constraints and regular constraints. Note also that 0.25 (t 0.5) 2 is not differentiable at t = 0 (in the interior of X or Y ). To solve problems involving in GAMS [4] the keyword DNLP must be used instead of NLP. Design problems with two discs are considered, resulting in twice as many variables and parameters. Additionally it must be ensured that the discs don t overlap which results into the constraint (x 4 x 1 ) 2 + (x 5 x 4 ) 2 (x 3 + x 6 ) 2 An obvious strategy for upper bounding is to use the solution with one disc as an initial guess. However, this is not exploited here.
min x 3 x 6 x s.t. x 2 (1 x 2 1 )/0.3 0 2 x 2 3 (y 1 x 1 ) 2 (1 y1 2) 2)/0.3 x 0, y 1 [ 1, 1] x 2 6 (y 2 x 4 ) 2 x 3 x 2 x 5 (1 x 2 4 )/0.3 0 2 (1 y2 5) 2)/0.3 x 0, y 2 [ 1, 1] x 6 x 5 (x 4 x 1 ) 2 + (x 5 x 4 ) 2 (x 3 + x 6 ) 2 x 3 x 6 x 3 + x 6 1.5 x [ 1, 1] [0, 2] [0, 1] [ 1, 1] [0, 2] [0, 1]. (D1 0 2)
min x 3 x 6 x s.t. x 2 (1 x 2 1 )/0.3 0 2 x 2 3 (y 1 x 1 ) 2 (1 y1 2) 2)/0.3 x 0, y 1 [ 1, 1] 0.25 (x 1 0.5) 2 x 2 0 x 2 3 (y 1 x 1 ) 2 ( 0.25 (y 1 0.5) 2 ) 2 x 2 0, y1 [ 1, 1] x 5 (1 x 2 4 )/0.3 0 2 x 2 6 (y 2 x 4 ) 2 (1 y2 5) 2)/0.3 x 0, y 2 [ 1, 1] 0.25 (x 4 0.5) 2 x 5 0 x 2 6 (y 2 x 4 ) 2 ( 0.25 (y 2 0.5) 2 x 5 ) 2 0, y2 [ 1, 1] (x 4 x 1 ) 2 + (x 5 x 2 ) 2 (x 3 + x 6 ) 2 x 3 x 6 x 3 + x 6 1 x [ 1, 1] [0, 2] [0, 1] [ 1, 1] [0, 2] [0, 1]. (D2 0 2) Finally, another variant of the design problems is considered where the sum of the areas of the discs is maximized. For the case of one disc the problems are mathematically equivalent, but the new objective is nonconvex and therefore potentially numerically more challenging. As expected the algorithm gives very similar results for the two different objective functions. On the other hand, for the case of two discs the objectives are different; the maximal area results in a slightly bigger difference between the radii of the two discs. min x x 2 3 Constraints same as (D1 0 1) (D1 1 1) min x x 2 3 Constraints same as (D2 0 1) (D2 1 1)
min x x 2 3 x 2 6 Constraints same as (D1 0 2) (D1 1 2) min x x 2 3 x 2 6 Constraints same as (D2 0 2) (D2 1 2) References [1] B. Bhattacharjee, W. H. Green Jr., and P. I. Barton. Interval methods for semi-infinite programs. Computational Optimization and Applications, 30(1):63 93, 2005. [2] B. Bhattacharjee, P. Lemonidis, W. H. Green Jr., and P. I. Barton. Global solution of semi-infinite programs. Mathematical Programming, Series B, 103(2):283 307, 2005. [3] J. W. Blankenship and J. E. Falk. Infinitely constrained optimization problems. Journal of Optimization Theory and Applications, 19(2):261 281, 1976. [4] A. Brooke, D. Kendrick, and A. Meeraus. GAMS: A User s Guide. The Scientific Press, Redwood City, California, 1988. [5] C. A. Floudas, P. M. Pardalos, C. S. Adjiman, W. R. Esposito, Z. H. Gümüs, S. T. Harding, J. L. Klepeis, C. A. Meyer, and C. A. Schweiger. Handbook of Test Problems in Local and Global Optimization. Nonconvex Optimization and Its Applications. Kluwer Academic Publishers, Dordrecht, 1999. [6] C. A. Floudas and O. Stein. The adaptive convexification algorithm: A feasible point method for semiinfinite programming. SIAM Journal on Optimization, 18(4):1187 1208, 2007. [7] R. Hettich and K. O. Kortanek. Semi-infinite programming - Theory, methods, and applications. SIAM Review, 35(3):380 429, 1993. [8] N. V. Sahinidis. BARON: A general purpose global optimization software package. Journal of Global Optimization, 8:201 205, 1996. [9] G. Watson. Numerical experiments with globally convergent methods for semi-infinite programming problems. Lecture Notes in Economics and Mathematical Systems, 215:193 205, 1983.