EE263
Prof. S. Boyd
Dec. 7–8 or Dec. 8–9, 2007.
Final exam solutions
1.
Optimal initial conditions for a bioreactor.
The dynamics of a bioreactor are given by
˙
x
(
t
) =
Ax
(
t
), where
x
(
t
)
∈
R
n
is the state, with
x
i
(
t
) representing the total mass of
species or component
i
at time
t
.
Component
i
has (positive) value (or cost)
c
i
, so
the total value (or cost) of the components at time
t
is
c
T
x
(
t
). (We ignore any extra
cost that would be incurred in separating the components.)
Your job is to choose
the initial state, under a budget constraint, that maximizes the total value at time
T
.
More specifically, you are to choose
x
(0), with all entries nonnegative, that satisfies
c
T
x
(0)
≤
B
, where
B
is a given positive budget. The problem data (
i.e.
, things you
know) are
A
,
c
,
T
, and
B
.
You can assume that
A
is such that, for any
x
(0) with nonnegative components,
x
(
t
)
will also have all components nonnegative, for any
t
≥
0. (This occurs, by the way, if
and only if the offdiagonal entries of
A
are nonnegative.)
(a) Explain how to solve this problem.
(b) Carry out your method on the specific instance with data
A
=
0
.
1
0
.
1
0
.
3
0
0
0
.
2
0
.
4
0
.
3
0
.
1
0
.
3
0
.
1
0
0
0
0
.
2
0
.
1
,
c
=
3
.
5
0
.
6
1
.
1
2
.
0
,
T
= 10
,
B
= 1
.
Give the optimal
x
(0), and the associated (optimal) terminal value
c
T
x
(
T
).
Give us the terminal value obtained when the initial state has equal mass in each
component,
i.e.
,
x
(0) =
α
1
, with
α
adjusted so that the total initial cost is
B
.
Compare this with the optimal terminal value.
Also give us the terminal value obtained when the same amount,
B/n
, is spent
on each initial state component (
i.e.
,
x
(0)
i
=
B/
(
nc
i
)). Compare this with the
optimal terminal value.
Solution.
(a) We have
c
T
x
(
T
) =
c
T
e
tA
x
(0) =
b
T
x
(0), where we define
b
= (
e
TA
)
T
c
, so our
problem is to maximize
b
T
x
(0), subject to
x
(0)
≥
0 (this means all its entries are
nonnegative), and
c
T
x
(0)
≤
B
. You can think of
c
i
as the cost of investing in a
unit of component
i
, and
b
i
as the payoff received. Thus, the gain is
b
i
/c
i
. The
1
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solution to this problem is to invest everything (
i.e.
, the whole budget
B
) in any
component that has maximum gain. More formally, we choose any
k
for which
b
k
/c
k
= max
{
b
1
/c
1
, . . . , b
n
/c
n
}
, and then set
x
(0) =
Be
k
. (Recall that we assume
b
i
≥
0 and
c
i
>
0 here.)
We didn’t require a completely formal proof that this is the optimal strategy.
But here is one, just so you know what one looks like. Suppose that
x
(0) satisfies
x
(0)
≥
0,
c
T
x
(0)
≤
B
. Then we have
b
T
x
(0)
=
n
summationdisplay
i
=1
(
b
i
/c
i
)(
c
i
x
(0)
i
)
≤
parenleftbigg
max
i
=1
,...,n
(
b
i
/c
i
)
parenrightbigg
parenleftBigg
n
summationdisplay
i
=1
c
i
x
(0)
i
parenrightBigg
≤
B
max
i
=1
,...,n
(
b
i
/c
i
)
.
This shows that no feasible choice of
x
(0) can yield terminal value
c
T
x
(
T
) =
b
T
x
(0) more than
B
max
i
=1
,...,n
(
b
i
/c
i
). But the choice described above yields this
value of
b
T
x
(0), and so must be optimal.
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 Fall '08
 BOYD,S
 Microeconomics, Transaction cost, rms value, Singular value decomposition, Singular value

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