≥
˜
g
(
˜
λ
), where ˜
g
is
the dual function for (12). This can be expressed as
p
?
≥
log ˜
g
(
˜
λ
)
.
How do these bounds compare? Are they the same, or is one better than the other?
4.12
Variable bounds and dual feasibility.
In many problems the constraints include
variable bounds
, as
in
minimize
f
0
(
x
)
subject to
f
i
(
x
)
0
,
i
= 1
, . . . , m
l
i
x
i
u
i
,
i
= 1
, . . . , n.
(13)
Let
μ
2
R
n
+
be the Lagrange multipliers associated with the constraints
x
i
u
i
, and let
⌫
2
R
n
+
be the Lagrange multipliers associated with the constraints
l
i
≥
x
i
. Thus the Lagrangian is
L
(
x,
λ
, μ,
⌫
) =
f
0
(
x
) +
m
X
i
=1
λ
i
f
i
(
x
) +
μ
T
(
x

u
) +
⌫
T
(
l

x
)
.
(a) Show that for any
x
2
R
n
and any
λ
, we can choose
μ
⌫
0 and
⌫
⌫
0 so that
x
minimizes
L
(
x,
λ
, μ,
⌫
). In particular, it is very easy to find dual feasible points.
(b) Construct a dual feasible point (
λ
, μ,
⌫
) by applying the method you found in part (a) with
x
= (
l
+
u
)
/
2 and
λ
= 0. From this dual feasible point you get a lower bound on
f
?
. Show
that this lower bound can be expressed as
f
?
≥
f
0
((
l
+
u
)
/
2)

((
u

l
)
/
2)
T

r
f
0
((
l
+
u
)
/
2)

where
 · 
means componentwise. Can you prove this bound directly?
4.13
Deducing costs from samples of optimal decision.
A system (such as a firm or an organism) chooses
a vector of values
x
as a solution of the LP
minimize
c
T
x
subject to
Ax
⌫
b,
with variable
x
2
R
n
.
You can think of
x
2
R
n
as a vector of activity levels,
b
2
R
m
as a
vector of requirements, and
c
2
R
n
as a vector of costs or prices for the activities.
With this
interpretation, the LP above finds the cheapest set of activity levels that meet all requirements.
(This interpretation is not needed to solve the problem.)
We suppose that
A
is known, along with a set of data
(
b
(1)
, x
(1)
)
,
. . . ,
(
b
(
r
)
, x
(
r
)
)
,
where
x
(
j
)
is an optimal point for the LP, with
b
=
b
(
j
)
. (The solution of an LP need not be unique;
all we say here is that
x
(
j
)
is
an
optimal solution.) Roughly speaking, we have samples of optimal
decisions, for di
↵
erent values of requirements.
You
do not
know the cost vector
c
. Your job is to compute the tightest possible bounds on the
costs
c
i
from the given data. More specifically, you are to find
c
max
i
and
c
min
i
, the maximum and
minimum possible values for
c
i
, consistent with the given data.
Note that if
x
is optimal for the LP for a given
c
, then it is also optimal if
c
is scaled by any positive
factor. To normalize
c
, then, we will assume that
c
1
= 1. Thus, we can interpret
c
i
as the relative
cost of activity
i
, compared to activity 1.
36