ORIE 3310/5310
Optimization II
Spring 2012
Homework # 1
Solutions posted Monday 2/13 at 4:00p.m.
Homework quiz in recitation Tuesday, 2/14 – Friday, 2/17.
1. Three representations from a (max) LP solved in the LP notes for ORIE 3310:
(
A
)
(

z
)
+
5
3
x
2

4
3
x
3
=

12
1
3
x
2
+
1
3
x
3
+
x
1
=
3
x
2

x
3
+
x
4
=
1
2
3
x
2

1
3
x
3
+
x
5
=
1
x
2
,
x
3
,
x
1
,
x
4
,
x
5
≥
0
(
B
)
(

z
)
+
4
x
1
+
3
x
2
=
0
3
x
1
+
x
2
+
x
3
=
9
3
x
1
+
2
x
2
+
x
4
=
10
x
1
+
x
2
+
x
5
=
4
x
1
,
x
2
,
x
3
,
x
4
,
x
5
≥
0
(
C
)
(

z
)
+
1
3
x
3

5
3
x
4
=

41
3
2
3
x
3

1
3
x
4
+
x
1
=
8
3

x
3
+
x
4
+
x
2
=
1
1
3
x
3

2
3
x
4
+
x
5
=
1
3
x
3
,
x
4
,
x
1
,
x
2
,
x
5
≥
0
(a) Which representation has the highest objective value? Is it optimal?
(b) Which variables are basic in representation (C)? What are the values taken by these (basic)
variables in the corresponding basic solution?
(c) Which variables are nonbasic in representation (A)? What are the values taken by these (nonbasic)
variables in the corresponding basic solution?
(d) Do variables
x
2
, x
4
, x
5
define a basis for this problem?
Why/why not?
Is this basis feasible?
Why/why not?
(e) Determine by inspection from the above data inverses of the following matrices:
3
1
0
3
2
0
1
1
1
1
1
3
0
0
1
0
0
2
3
1
Explain how you arrive at your answers.
(f) If the problem’s objective function is changed to
max
5
x
1
+ 3
x
2
, which representation is optimal?
Why? Are there alternative optimal solutions for the new objective? Explain.
(g) If the righthand side coefficients in representation (B) are changed from (9
,
10
,
4) to (3
,
2
,
5) ,
does representation (A) remain feasible? Explain.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2. We would like to prove the following result, known as the
Farkas Theorem
.
For any
m
×
n
real matrix
A
and
m
vector
b
exactly one of the two following systems is feasible:
(I)
Ax
=
b
,
x
≥
0
(II)
yA
≥
0 ,
yb <
0 .
(a) Show that
(I)
and
(II)
cannot both be feasible.
(b) If system
(I)
is infeasible then at the end of Phase I (maximizing
z
=

∑
m
i
=1
w
i
, the negative
sum of artificials) of the simplex algorithm we have a tableau of the following form:
x
1
· · ·
x
n
w
1
· · ·
w
m
RHS
(

z
)
≤
0
· · ·
≤
0
≤
0
· · ·
≤
0
>
0
≥
0
.
.
.
≥
0
Show how to determine a vector
y
which is feasible in system
(II)
directly from the data in the
above tableau. (Thus when
(I)
fails,
(II)
must be feasible, completing the proof of the theorem.)
3. The LP feasible region
F
=
{
x
:
Ax
=
b, x
≥
0
} 6
=
∅
is
bounded
provided
F
⊆ {
x
:

δ
≤
x
j
≤
δ
∀
j
}
for some positive constant
δ
; i.e., we can place the LP feasible region in a
hypercube
of side length 2
δ
.
Suppose that
F
is such a bounded feasible region.
(a) Show that the LP
max
x
∈
F
cx
has a finite optimum solution value for any objective function
c
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 BLAND
 Operations Research, Linear Programming, Optimization, x3, shadow prices, restricted master program

Click to edit the document details