Midterm Exam, Fall 1999
Take Home
Due: Beginning of class, Wednesday, 3 November 1999.
This is to be all your own work. You may use any result from class,
homeworks, or the books on reserve in the library.
Do not consult anybody or anything else.
I can dispense hints to help you if you are stuck.
You can reach me by email at mitchj@rpi.edu.
My phone number is 276-6915.
The exam consists of five questions and is worth a total of 100 points.
I will have my usual office hours as follows:
| |
Friday, 29 Oct: |
1-1.50pm |
| |
Monday, 1 Nov: |
5-6pm |
- 1.
- (25 points.)
Consider the nonlinear programming problem
- (a)
- (3 points.)
Is the feasible region for this problem convex?
- (b)
- (4 points.)
Show that the gradients of the active constraints are
linearly independent at any feasible point.
- (c)
- (3 points.)
What are the Karush-Kuhn-Tucker conditions for this problem?
- (d)
- (3 points.)
Verify that the points (6,0) and (-6,0) satisfy the KKT
conditions.
- (e)
- (4 points.)
Show that there are no points which satisfy
x12+(x2+8)2>100 and also satisfy the KKT conditions.
- (f)
- (4 points.)
Are there any other points which satisfy the KKT conditions?
- (g)
- (4 points.)
Can you find the global minimizer?
- 2.
- (20 points.)
- (a)
- (10 points.)
Let
be an n x m matrix,
so yi is the ith column of Y.
Let
with
and
.
The matrix Y and the vector u are given.
Show that x=Yu solves
- (b)
- (10 points.)
Suppose you have an algorithm for minimizing a convex quadratic
function subject to linear constraints.
How could you solve the problem
where
are distinct given points in
?
(Note: a quadratic function has the form
f(w)=wTQw+cTw+d,
where Q is a matrix, c is a vector, and d is a scalar.)
- 3.
- (15 points. Each part is worth 5 points.)
Consider the nonlinear programming problem
(P)
Assume M is (n x n) and symmetric. Let the eigenvalues of
M be
.
Assume
and
.
We saw in Homework 4 that the optimal solutions to this
problem are the eigenvectors with eigenvalue
and norm
.
(Recall:
is an eigenvalue of A if
for some
nonzero vector x; a nonzero x satisfying this equation for some
is an eigenvector of A with eigenvalue
.
A is positive definite if and only if all the eigenvalues of A
are positive. A is positive semi-definite if and only if all the
eigenvalue of A are nonnegative.
If A has eigenvalues
,...,
then A+aI has eigenvalues
,...,
.)
- (a)
- What is the Lagrangian dual function
for this problem?
(Take
.)
- (b)
- Solve the Lagrangian dual problem.
- (c)
- Let
solve the Lagrangian dual problem.
Let
be the set of points in X which achieve the
infimum in the definition of
.
Show that
contains infinitely many points and that
one of them solves the primal problem.
- 4.
- (a)
- (10 points.) Consider the two sets of points
A={a1,...,ak} and
B={b1,...,bm} in
.
Show that the two sets of points can be strictly separated if
and only if there exists a vector
and a
scalar
satisfying
(Here, by strictly separate, I mean that the separating
hyperplane does not contain any points in either A or B.)
- (b)
- (10 points.)
The problem in part 4a is a linear
programming feasibility problem with variables x and
.
It can be stated as
an optimization problem if we use the objective function
,
which is zero for any feasible solution.
What is the dual of this linear program?
If the dual problem has a feasible nonzero solution,
what does that imply about the primal problem given in
part 4a? Can you give a geometric meaning for
a nonzero solution to the dual problem?
- 5.
- (20 points.)
Let
be C1 (ie, the first derivative
is continuous).
- (a)
- (8 points.) Show that the first order necessary conditions for
to be a local minimizer of f over
are
,
,
and
.
(Note:
.)
- (b)
- (6 points.) Let
be strictly monotone.
(A function G is strictly monotone if
.)
Show that F has at most one zero (an
with
)
and that the nonlinear complementarity
problem (find
with
and
)
has at most one solution.
- (c)
- (6 points.)
Using the results of parts 5a and 5b,
show that, if f is strictly convex, it has at
most one minimizer over
and at
most one minimizer over
.
(Hint: We know that f is strictly convex if
.
You may assume that
f is strictly convex if and only if
is strictly monotone.)
John E Mitchell
1999-10-27