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CO250/CM340 INTRODUCTION TO OPTIMIZATION  HW 10 SOLUTIONS
Exercise 1
(20 marks)
.
(a) The cone generated by
{
(1
,
0
,
0)
,
(0
,
1
,
0)
,
(0
,
0
,
1)
}
is
C
=
{
x
:
x
=
λ
1
(1
,
0
,
0) +
λ
2
(0
,
1
,
0) +
λ
3
(0
,
0
,
1)
, λ
i
≥
0
∀
i
}
=
{
(
λ
1
,λ
2
,λ
3
) :
λ
i
≥
0
∀
i
}
.
Le
x
1
and
x
2
denote two points in
C
, where
x
i
= (
λ
i
1
,λ
i
2
,λ
i
3
)
for
i
= 1
and
2
. Then for any
0
≤
λ
≤
1
,
λx
1
+ (1

λ
)
x
2
=
λ
(
λ
1
1
,λ
1
2
,λ
1
3
) + (1

λ
)(
λ
2
1
,λ
2
2
,λ
2
3
)
= (
λλ
1
1
+ (1

λ
)
λ
2
1
,λλ
1
2
+ (1

λ
)
λ
2
2
)
,λλ
1
3
+ (1

λ
)
λ
2
3
)
= (
μ
1
,μ
2
,μ
3
)
where each
μ
i
≥
0
because of the conditions on
λ
and the
λ
i
j
. Hence
(
μ
1
,μ
2
,μ
3
)
∈
C
. Since this was
true for any two points
x
1
and
x
2
∈
C
, it follows that
C
is convex.
To ﬁnd extreme points, consider
x
∈
C
. Then clearly
x
1
:=
1
2
x
and
x
2
:=
3
2
x
are both in
C
.
Moreover,
x
=
1
2
x
1
+
1
2
x
2
. Hence,
x
1
cannot be extreme if
x
1
6
=
x
2
. This occurs whenever
x
6
=
0
. So
the only possible extreme point of
C
is
0
.
To show that
x
=
0
is an extreme point, we assume that it is properly contained in the interval
joining distinct points
x
1
,
x
2
∈
C
, and reach a contradiction. Proper containment implies that
x
=
λx
1
+ (1

λ
)
x
2
for some
0
< λ <
1
, and also
x
1
6
=
x
=
0
, so one of the coordinates of
x
1
is nonzero.
Without loss of generality, let it be the ﬁrst coordinate,
x
1
1
. Since
0 =
x
1
=
λx
1
1
+ (1

λ
)
x
2
1
, it follows
that
x
2
1
<
0
, which contradicts
x
2
∈
C
.
(b) Yes it is convex, and the only extreme point is
0
. (The proof, not required here, is similar to (a)’s proof.)
(c)
S
is the intersection of the sets of feasible solutions of the two LP’s. We know that the set of feasible
solutions of an LP is convex, and that the intersection of two convex sets is convex. Thus
S
is convex.
(d) We know that the set of feasible solutions of an LP is convex. Also
c
T
x
= 1
is a hyperplane, which is
the intersection of two halfspaces,
c
T
x
=
≤
1
and
c
T
x
≥
1
, and we know that all halfspaces are convex.
Note that
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 Spring '10
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