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EAS6939 Homework #4
1. Consider the following design optimization problem:
22
12
1
11
3
32
1
Minimize
( )
4
4
Subject to
( )
0
()
0
(
1
) 0
fx
x
x
gx
x
=
+− +
=− ≤
=
−− ≤
x
x
x
x
(i)
Find the optimum point graphically
(ii)
Show that the optimum point does not satisfy KT condition. Explain
(i)
As shown in the figure, (1, 0) is the optimum point and
f
= 1 at the optimum point.
x
*
1.0
∇
g
3
(
x
*)
∇
g
2
(
x
*)
∇
f
(
x
*)
x
1
x
2
f
=1
f
=4
(ii) The Lagrangian function for the problem can be defined as
2
2
1
1
1
1
2
3
2
1
3
44(
)(
)[(
1)
]
Lx x x
x s
x
λλ
λ
=+− ++−+ + −+ +
−− +
The KT condition is
3
3
1
223
2
2
21
3
24
3
(
0
20
0
0
(1
)
0
0,
1,2,3
ii
xx
x
xs
s
si
−
=
−+
=
=
=
=
==
At
x
= (1, 0) since g
1
is inactive, and g
2
and g
3
are active, the slack variables should be
123
0,
0,
0
ss
===
The first equation in the KT condition can’t be satisfied by substituting into these values.
As is clear from the figure, the gradients of two active constraints are not independent: [0, 1]
and [0, 1]. In the mathematical term,
x
* is not a
regular point
of the feasible domain. The KT
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 Spring '08
 Staff

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