MATH 164/2 Optimization, Winter 2006, Midterm Exam Solutions
Instructor: Luminita Vese
Teaching Assistant: Ricardo Salazar
[1]
(10 points)
(a)
Consider the following feasible set
S
=
n
x
∈
R
n
:
Ax
≥
b,
x
≥
~
0
o
. Show
that if the vector
d
satisfies
d
6
=
~
0,
d
≥
~
0 and
Ad
≥
~
0, then
d
is a direction of
unboundedness for the set
S
.
(b)
Consider the following linear programming problem:
Minimize
z
=
x
1

5
x
2
+
x
3

3
x
4
subject to
3
x
1

x
2
+
2
x
4
≥
2

2
x
1
+
3
x
4
≥

1
x
2

x
3
≥
2
5
x
1

3
x
3
≥
2
x
1
,
x
2
,
x
3
,
x
4
≥
0
(b1)
Show that
x
= (2
,
4
,
2
,
1)
T
is a feasible point to the problem and label each
of the constraints as active or inactive.
(b2)
Show that the vector
d
= (1
,
2
,
1
,
1)
T
is a direction of unboundedness (note
that this problem, as given, is not in standard form!)
Solution:
(a)
We know by definition that
d
is a direction of unboundedness for
S
if
d
6
=
~
0
and if
x
+
γd
∈
S
, for any
x
∈
S
and any
γ
≥
0.
Let
x
∈
S
and
γ
≥
0 arbitrary. Then
Ax
≥
b
, and using the assumptions on
d
,
we have:
A
(
x
+
γd
) =
Ax
+
A
(
γd
) =
Ax
+
γAd
≥
b
+
γ
~
0 =
b
.
Therefore
A
(
x
+
γd
)
≥
b
(1)
Also, because
x
≥
~
0,
γ
≥
0 and
d
≥
~
0, we have
x
+
γd
≥
~
0 +
γ
~
0 =
~
0
(2)
From (1) and (2), we deduce that
x
+
γd
∈
S
for any
x
∈
S
and any
γ
≥
0,
therefore
d
6
= 0 is a direction of unboundedness for the set
S
.
(b1)
We verify that
x
= (2
,
4
,
2
,
1)
T
satisfies all eight
constraints given.
3
·
2

4 + 2
·
1 = 4
>
2 (1st inactive)

2
·
2 + 3
·
1 =

1 =

1 (2nd active)
4

2 = 2 = 2 (3rd active)
5
·
2

3
·
2 = 4
>
2 (4th inactive)
x
1
= 2
>
0 (inactive),
x
2
= 4
>
0 (inactive),
x
3
= 2
>
0 (inactive),
x
4
= 1
>
0
(inactive).
Therefore the point
x
= (2
,
4
,
2
,
1)
T
satisfies all eight constraints of the problem,
so it is a feasible point,
x
∈
S
.
(b2)
Notice that
d
= (1
,
2
,
1
,
1)
T
6
=
~
0 and
d
≥
~
0. Therefore, using the proof for
(a), it is sufficient to show that
Ad
≥
~
0, where
A
is the matrix corresponding to the