This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Department of Industrial Engineering & Operations Research IEOR160 Operations Research I Exam 1 10/13/2004 Name: Grade: Closed book, closed notes exam. No cheatsheets. Programmable calculators not allowed. 1. (15 points) Determine whether the following statements are true or false . T F A concave function cannot have a minimizer over an equality constrained feasible region. False T F If the objective function of an optimization problem is convex and the feasible region is convex, then it is a minimization problem. False T F If f is a continuously differentiable function, then all of its local maxima are among its stationary points. True T F If x is a local maximum of a concave function, then there exists a direction vector d for which the directional derivative at x is negative. False, the directional derivative is zero T F For a KKT point, if the Lagrange multiplier of a constraint is zero, then the constraint is inactive at this point. False 2. (15 points) Definition: A function f : IR n IR is strictly quasiconvex on S IR n if for each x, y S such that x 6 = y the following inequality holds: f ( x + (1 ) y ) < max { f ( x ) , f ( y ) } for all < < 1 Prove that if f is strictly quasiconvex on S , then a local minimum of f on S is also a global minimum of f on S . Suppose x S such that there exists y S with f ( y ) < f ( x ) (i.e., x is not a global minimum). Let (0 , 1). Then, from definition, we have that f ( x +(1 ) y ) < max { f ( x ) , f ( y ) } = f ( x ). Hence, given any > 0, one can find a (0 , 1) such that z = (1 ) y + x lies in the neighbourhood of x , and we know that f ( z ) < f ( x ). This implies that x cannot be a local minimum. Hence, any local minimum should be a global minimum too....
View Full
Document
 Spring '08
 Lim
 Operations Research

Click to edit the document details