IEOR 262A Mathematical Programming I
Fall 2007
Homework 6 Solution
Exercise 4.2 Solution
The dual is
max
p
0
b
s.t.
p
0
A
≤
c
0
p
≥
0
Convert it into its equivalent minimization problem:
max
(

b
0
)
p
s.t.
(

A
0
)
p
≥

c
p
≥
0
This problem is identical to the primal iff
A
0
=

A
and
b
=

c
.
For example, let
A
=
0

1
1
0
¶
,
b
=
1
2
¶
and
c
=

1

2
¶
.
Exercise 4.4 Solution
The dual is
max
p
0
c
s.t.
Ap
≤
c
p
≥
0
by the symmetry of
A
. If
Ax
*
=
c
, then
x
*
is both primal and dual feasible. Moreover, it has
the same primal and dual cost
c
0
x
*
. Therefore, it is an optimal solution of both problems.
Exercise 4.8 Solution
(a)
By the optimality of
x
*
under
c
, we have
c
0
(˜
x

x
*
)
≥
0; by the optimality of
˜
(
x
) under ˜
c
,
we have ˜
c
0
(
x
*

˜
x
)
≥
0. Add the two inequality together, we get
(
c
0

˜
c
0
)(˜
x

x
*
)
≥
0
or
(˜
c

c
)
0
(˜
x

x
*
)
≥
0
.
(b)
By strong duality,
c
0
x
*
= (
p
*
)
0
˜
x
.
By weak duality, since
p
*
is feasible under
˜
b
, we have
(
p
*
)
0
˜
b
≤
c
0
˜
x
.
(
p
*
)
0
˜
b

(
p
*
)
0
b
≥
c
0
˜
x

c
0
x
*
or
(
p
*
)
0
(
˜
b

b
)
≤
c
0
(˜
x

x
*
)
.
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 Fall '09
 DON'TRKNOW
 Operations Research, Optimization, optimal solution, max 200y

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