Download WI4087TU sheets week..

Karush-Kuhn-Tucker conditions (Ch. 12.7)
If the derivative in a point is zero and its second
derivative is positive-definit, then the poiny is locally
optimal.
With constraints these conditions will be a lot more
complicated because an optimum can lie on the
boundary of the feasible region, without the derivative
being zero.
Consider the following on-linear optimization problem
with non-linear constraints:
Max
s.t.
and
f(x)
gi(x)  bi
xi  0
for all i = 1, …, m
for all i = 1, …, n
x* can be a locally optimal solution if and only if there
are numbers, u1, …, um such that:
m
1.
2.
3.
4.
5.
6.
f ( x )   u i g i ( x * )  0
*
i 1
m
 f

g
*
x 
( x )   ui i ( x * )   0
 x

x j
i 1
 j

*
j
gi(x*)  bi
ui(gi(x*)-bi) = 0
x*  0.
ui  0.
for all j = 1, …, n
for all i = 1, …, m
for all i = 1, ..., m
The numbers ui, are associated toe ach constraints are
called Lagrange multipliers. They are non-negative and
can be positief only (4.) of the corresponding constraint
is active, i.e., if gi(x*) = bi. The solution x* is the non the
boundary of the feasibility region determined by this
constraint.
3. and 5. are just requirements for feasibility:
The above Karush-Kuhn-Tucker (KKT) conditions
(1939/51) characterize locally optimal solutions to an
optimization problem with constraints.
They can be used the check (prove) whether a certain
solution is optimal. In principe the optimal solutions
could be determined with it, but this is usually difficult.
What are the KKT conditions in the one-dimensional
case?
Max
s.t.
and
f(x)
xb
x0
x* is a locally optimal solution, if and only if there are
numbers u such that:
1.
2.
3.
4.
5.
6.
f’(x*) – u  0
x*(f’(x*)-u) = 0
x*  b
u(x*-b) = 0
x*  0.
u  0.
It follows that one of the following situations hold:
1.
x* = 0 and f’(0)  0
2.
x* = b and f’(b)  0
3.
0  x*  b and f’(x*) = 0
This means: an optimal solution on the left boundary,
on the right boundary, or an interior optimal solution.
Example:
Max
s.t.
and
f(x1, x2) = log(x1+1) + x2
2x1 + x2  3
x1, x2  0
The KKT conditions are now:
1a
1
 2u1
x1  1
1b u1  1
 1



2a x1  x  1  2u1   0
 1

2b
3
4
5
6
x2 1  u1   0
2x1  x2  3
u1 2x1  x2  3  0
x1 , x2  0
u1 , u2  0
From 1b we have u1  1 and because x1 is non-negative
1
(5), the following holds: x  1  2u1  1  2  0 .
1
From 2a it follows that x1 = 0.
From 4 it follows that x2 = 3
From 2b it follows that u1 = 1.
It is easy to see that x1 = 0, x2 = 3, u1 = 1 satisfies all
KKT conditions, so (0,3) is the optimale solution of the
problem and the objective value is f(0, 3) = 3.