UC Berkeley
Department of Electrical Engineering and Computer Science
EECS 227A
Nonlinear and Convex Optimization
Solutions 1
Fall 2009
Review 1.1
For future reference, the function
f
has the following first and second derivatives:
∂f
∂x
(
x, y
)
=
4
x
3
−
4
yx
−
20
x
∂f
∂y
(
x, y
)
=
−
2
x
2
+ 2
y
∂
2
f
∂y
2
(
x, y
)
=
−
4
x
∂
2
f
∂x∂y
(
x, y
)
=
2
(a) Note that
f
(1
,
0) =
f
(
−
1
,
0) =
−
9, so that
1
2
f
(1
,
0) +
1
2
f
(
−
1
,
0) =
−
9. On the other
hand, we have
f
(0
,
0) = 0
>
−
9, showing that
f
is not convex.
(b) We need to solve
∇
f
(
x, y
) = 0 for (
x, y
)
∈
R
2
. Taking derivatives yields
∇
f
(
x, y
)
=
bracketleftbig
4
x
(
x
2
−
y
)
−
20
x
−
2(
x
2
−
y
)
bracketrightbig
T
By inspection, (
x, y
) = (0
,
0) is the only stationary point. It is neither a local maximum
nor a local minimum, since
f
(
δ,
0) =
δ
4
−
10
δ
2
<
0 for
δ
small enough, and
f
(0
, ǫ
) =
ǫ
2
>
0 for
ǫ
negationslash
= 0. Alternatively, we may compute the Hessian matrix
∇
2
f
(0
,
0) =
parenleftbigg
−
20
0
0
2
parenrightbigg
which is indefinite, showing (0
,
0) is neither a local minimum nor a local maximum.
(c) The function
f
is convex. The set
C
is bounded in the
y
direction, and the function
value tends to infinity as

x

increases, so that a global minimum exists and is attained.
By necessary optimality condition for constrained optimization to solve this problem, if
(
x
∗
, y
∗
) is global minimum it needs to satisfy
∇
f
(
x
∗
, y
∗
)
T
bracketleftbiggparenleftbigg
x
y
parenrightbigg
−
parenleftbigg
x
∗
y
∗
parenrightbiggbracketrightbigg
≥
0
,
∀
(
x, y
)
∈
X
In part (b), we showed that
f
has only one stationary point in the interior of the given set,
and it is not a minimum. Hence, the minimum must occur on the boundary of the set—
i.e., either at some point (
x
∗
,
1) or some point (
x
∗
,
0), and we must have
∂f
∂x
(
x
∗
,
1) = 0 or
∂f
∂x
(
x
∗
,
0) = 0. For
y
= 0, we have
f
(
x,
0) =
x
4
−
10
x
2
. Since
∂f
∂x
(
x,
0) = 4
x
3
−
20
x
2
, the
relevant stationary points are at
x
=
±
√
5, with
f
(
±
√
5
,
0) = 25
−
50 =
−
25. Otherwise,
for
y
= 1, we have
f
(
x,
1) =
x
4
−
12
x
2
+ 1. We have
∂f
∂x
(
x,
1) = 4
x
3
−
24
x
, so stationary
points are found at
x
=
±
√
6 with
f
(
±
√
6
,
0) =
−
35. Checking second derivatives shows
that this is the global optimum.
1
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Review 1.2
Using KKT conditions, we have
g
j
(
x
∗
)
≤
0
, j
= 1
, . . . , m
λ
∗
j
≥
0
, j
= 1
, . . . , m
λ
∗
j
g
j
(
x
∗
) = 0
, j
= 1
, . . . , m
∇
f
(
x
∗
) +
m
summationdisplay
j
=1
λ
∗
j
∇
g
j
(
x
∗
) = 0
So
∇
f
(
x
∗
)
T
(
x
−
x
∗
) =
−
m
summationdisplay
j
=1
λ
∗
j
∇
g
j
(
x
∗
)
T
(
x
−
x
∗
)
Since
g
j
, j
= 1
, . . . , m
are convex functions:
∇
g
j
(
x
∗
)
T
(
x
−
x
∗
)
≤
g
j
(
x
)
−
g
j
(
x
∗
), therefore
∇
f
(
x
∗
)
T
(
x
−
x
∗
)
≥ −
m
summationdisplay
j
=1
λ
∗
j
(
g
j
(
x
)
−
g
j
(
x
∗
))
=
−
m
summationdisplay
j
=1
λ
∗
j
g
j
(
x
) +
m
summationdisplay
j
=1
λ
∗
j
g
j
(
x
∗
)
=
−
m
summationdisplay
j
=1
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 Spring '11
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 Electrical Engineering, Derivative, Optimization, Convex set, Convex function, x∗

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