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In-Depth Information
1
/
b
i
=
0
1
/
b
i
=
0
λ
i
β
i
=
k
1
/
b
i
=
0
β
i
=
m
i
β
i
=
and
so
1
/
b
i
=
0
β
i
/
1
/
b
i
=
0
v
(
a
,
b
)=
1
−
(
i
β
i
)
.
1
/
b
Thus, putting
g
j
=(
β
j
/
∑
i
β
i
)
,
a lower bound for
v
(
a
,
b
)
is 1
−
H
where
i
=
0
max
1
/
b
j
=
0
H
=
g
j
subject to
1
/
b
j
=
0
λ
j
g
j
=
1
/
b
j
=
0
jg
j
=
1
,
and
g
j
≥
0
(
j
=
0
,
1
,...,
1
/
b
)
.
This is a linear program with dual
min
x
+
y
subject to
λ
i
x
+
iy
≥
1
,
x
,
y
unrestricted
(
i
=
0
,
1
,...,
1
/
b
)
.
∈
Λ
+
and
q
∈
Λ
−
be such that
G
Let
s
(
s
,
q
)=
G
,
where
G
(
i
,
j
)
and
G
are given
,∈
Λ
−
by (
9.4
)and(
9.5
) respectively. It is routine to check that, for
i
)=
(
s
−
λ
s
)(
i
(
λ
q
−
λ
s
)+
λ
i
(
s
−
q
)
−
(
s
λ
q
−
q
λ
s
))
0
≤
G
(
s
,
q
)
−
G
(
s
,
i
(
−
)(
−
)
s
λ
q
λ
s
λ
i
λ
q
s
i
s
∈
Λ
+
,
and that, for
i
)=
(
λ
q
−
q
)(
i
(
λ
q
−
λ
s
)+
λ
i
(
s
−
q
)
−
(
s
λ
q
−
q
λ
s
))
0
≤
G
(
s
,
q
)
−
G
(
i
,
q
.
(
s
λ
−
q
λ
)(
s
λ
−
i
λ
)
q
s
i
s
It is now straightforward to verify by the duality theorem that
H
=
G
(
s
,
q
)
by
taking
g
q
=
G
(
s
,
q
)
−
g
s
=(
s
−
λ
s
)
/
(
s
λ
q
−
q
λ
s
)
,
g
i
=
0 otherwise and
x
=
G
(
s
,
q
)
−
y
=(
s
−
q
)
/
(
s
λ
q
−
q
λ
s
)
.
9.5 A Second Illustrative Example
In the previous section we showed that an optimal Defender strategy gives rise
to a number of chains of intervals and used this fact to obtain a lower bound for
v
(
a
,
b
)
.
However the type of chains and the number of each type that give this lower
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