MthSc810 – Mathematical Programming
Fall 2011, Homework #9. Due
Tuesday
, November 8, 2011, 6PM EDT.
From the syllabus: homework will be penalized
50%
for each day they are late.
After two days, they will not be accepted.
No exception.
Problem 1.
Given the problem
min
−
x
1
−
x
2
x
1
+ 5
x
2
≤
10 +
α
x
1
+
x
2
≤
6 +
β
x
1
,
x
2
≥
0
1. Compute the set
M
=
{
(
α, β
)
∈
R
2
: the basis
B
=
{
1
,
2
}
is optimal
}
.
2. For each (
α, β
)
∈
M
, compute the change in objective function.
3. Is
M
convex? Is it a polyhedron? Is it a cone?
Solutions.
1. Let’s put the problem in standard form:
min
−
x
1
−
x
2
x
1
+ 5
x
2
+
x
3
= 10 +
α
x
1
+
x
2
+
x
4
= 6 +
β
x
1
,
x
2
,
x
3
,
x
4
≥
0
The given basis
B
is optimal if it is primal feasible and dual feasible, i.e.
(a)
x
B
=
B
−
1
b
≥
0. Now,
B
=
parenleftbigg
1 5
1 1
parenrightbigg
, hence
B
−
1
=
parenleftBigg
−
1
4
5
4
1
4
−
1
4
parenrightBigg
and
B
−
1
(10 +
α,
6 +
β
)
⊤
=
parenleftBigg
−
5
2
−
α
4
+
15
2
+
5
4
β
5
2
+
α
4
−
3
2
−
1
4
β
parenrightBigg
=
parenleftBigg
5
−
α
4
+
5
4
β
1 +
α
4
−
1
4
β
parenrightBigg
.
Therefore, necessary and sufficient conditions for primal feasibility are
as follows:
braceleftBigg
−
1
4
α
+
5
4
β
≥ −
5
1
4
α
−
1
4
β
≥ −
1
≡
braceleftBigg
α
−
5
β
≤
20
α
−
β
≥ −
4
.
(b) This does not depend on
α
or
β
, as the vector of reduced cost does
not depend on
b
. Hence
B
is dual feasible either for any
α, β
or for
none. We need to check ¯
c
⊤
=
c
⊤
N
−
c
⊤
B
B
−
1
N
≥
0
, or, since
N
=
I
2
,
c
⊤
N
−
c
⊤
B
B
−
1
≥
0
, that is,
(0
,
0)
−
(
−
1
,
−
1)
parenleftBigg
−
1
4
5
4
1
4
−
1
4
parenrightBigg
= (0
,
1)
≥
0
,
and the basis is optimal (when feasible) for any
α, β
.
To recap,
M
=
{
(
α, β
) :
α
−
5
β
≤
20
, α
−
β
≥ −
4
}
.
2. The objective function value is
c
⊤
B
x
B
=
c
⊤
B
B
−
1
b
= (
−
1
,
−
1)
parenleftBigg
5
−
α
4
+
5
4
β
1 +
α
4
−
1
4
β
parenrightBigg
=
−
6
−
β,
and it depends on
β
only, given that
β
relaxes (or restricts) the constraint
whose orthogonal is parallel to the objective function’s gradient.
3. Since
M
=
{
(
α, β
) :
α
−
5
β
≤
20
, α
−
β
≥ −
4
}
, it is not a cone as right-hand
sides are nonzero. However, it’s a polyhedron and thus convex.