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f
i
,
j
=
e
j
Aq
i
t
)=
e
j
A
Ω
(
a
)
T
t
.
(
a
)
T
q
i
−
Ω
g
i
,
j
(
a
,
p
∗
∞
<
δ
−
This means that the constraint
p
can be compactly expressed by defin-
ing the polynomials
1)
k
(
f
i
,
3
g
i
,
j
(
a
h
i
,
j
,
k
(
a
,
t
)=(
−
,
t
)
−
f
i
,
j
g
i
,
3
(
a
,
t
)) +
δ
f
i
,
3
g
i
,
3
(
a
,
t
)
.
(9.17)
p
∗
∞
≤
δ
Indeed,
p
−
if and only if
h
i
,
j
,
k
(
a
,
t
)
≥
0
∀
(
i
,
j
,
k
)
∈I
(9.18)
where
I
is the set
I
=
{
(
i
,
j
,
k
) :
i
= 1
,...,
N
,
j
= 1
,
2
,
k
= 1
,
2
}
{
(9.19)
,
,
,...,
,
,
}.
(
i
j
k
) :
i
= 1
N
j
= 3
k
= 1
In fact, let us observe that the first two constraints in (9.16) are recovered by the
first set on the right hand side of (9.19), while the third constraint is recovered
by the second set on the right hand side of (9.19) since
h
i
,
3
,
1
(
a
,
t
)=
δ
f
i
,
3
g
i
,
3
(
a
,
t
).
Therefore,
s
r
(
δ
) and
s
t
(
δ
) can be rewritten as
s
r
(
δ
)=sup
a
,
t
θ
s.t. (9.18)
(9.20)
s
t
(
δ
)=sup
a
,
t
t
s.t. (9.18)
.
Let us observe that
g
i
,
j
(
a
,
t
) is a cubic polynomial in the variables
a
and
t
due to the
(
a
)
T
t
. In order to lower the degree of the polynomials in the optimization
problems in (9.20), let us define the new variable
z
term
Ω
3
as
∈
R
(
a
)
T
t
z
=
Ω
.
(
a
)
T
Let us observe that
t
can be recovered from
z
since
Ω
is always invertible, in
particular
Ω
(
a
)
T
−
1
=
Γ (
a
)
1 +
2
.
a
Hence, let us define
z
)=
e
j
A
Ω
z
(
a
)
T
q
i
−
g
i
,
j
(
a
,
h
i
,
j
,
k
(
a
1)
k
(
f
i
,
3
g
i
,
j
(
a
,
z
)=(
−
,
z
)
−
f
i
,
j
g
i
,
3
(
a
,
z
)) +
δ
f
i
,
3
g
i
,
3
(
a
,
z
)
.
It follows that (9.18) holds if and only if
h
i
,
j
,
k
(
a
,
≥
∀
,
,
∈I .
z
)
0
(
i
j
k
)
h
i
,
j
,
k
(
a
Moreover, the degree of
,
z
) is 2, which is lower than that of
h
i
,
j
,
k
(
a
,
t
) which
is equal to 3.
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