MATP6600/DSES6780 Nonlinear Programming, Homework 6.

Due: Friday, December 2, 2011, by the end of class.

Each question is worth 20 points.

  1. The problem
    min  8x1x2 + 1(x1 - x2)4
     x ∈ IR2  4

    has a saddle point at the origin and two global minima at ±(1,-1). We investigate solving this problem using Newton’s method.

    1. Let x0 be the starting point and x1,x2, be the subsequent iterates. Show that xk is on the line x1 + x2 = 0 for k 1, for any x0, provided (x1 - x2)243.
    2. Show that if we take a steplength equal to one then the algorithm converges cubically to the saddle point if |x1 - x2| < √25-.
    3. Show that if we take a steplength equal to one then the algorithm converges to a global minimum if |x1 - x2|≥ 1.
  2. Consider using a quasi-Newton approach to the problem from question 1: if the Hessian is not positive definite, approximate the inverse of the Hessian by the matrix
        1 [  2  - 1 ]
G = --           .
    8   - 1   2

    Assume the algorithm starts from a nonzero point with an indefinite Hessian. Show that the algorithm eventually obtains a point with a positive definite Hessian. What can you then conclude from question 1?

  3. Let (NLP) denote the nonlinear programming problem
    min        - x21  -  (x2 - 9)2
subject to 4x21  +     x22     ≤   400.

    This problem has optimal solution x = (0,-20). One penalty function approach to this problem is to use a penalty parameter μ and solve

                  2            2        2
min        - x21 -  (x2 -29)  +   μs
subject to  4x1  +      x2     -    s   ≤  400.

    Use AMPL and a solver on the NEOS server (or another package) to solve this problem for various values of μ. Show empirically that increasing μ by a factor of ten results in a solution to (NLP) with about one more digit of accuracy.

    John Mitchell
    Amos Eaton 325
    x6915.
    mitchj at rpi dot edu
    Office hours: Tuesday 12.0 – 2.0, Wednesday 2.0 – 4.0.