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or
T
E
(11.57)
Substituting the lower bound of equation (11.52) into equation (11.57) and noting
T
F
=
(
N
S
−
(
i
D
F
≤
i
(
N
S
−
f
+
+
+
1)
T
F
+
t
0
+
τ
+
K
)
T
a
v
g
+
min
{
F
F
(
y
−
1)
} +
K
)
T
a
v
g
we get
(
i
D
F
≤
i
(
N
S
−
K
)
T
a
v
g
+
y
(
N
S
−
f
+
+
+
1)(
N
S
−
K
)
T
a
v
g
+
t
0
+
τ
+
K
)
T
a
v
g
+
t
0
D
F
+
T
E
f
−
+
+
(11.58)
Rearranging, we can obtain
y
from
1
f
+
−
f
−
−
+
τ
+
T
E
y
Before
=
(11.59)
(
N
S
−
K
)
T
a
v
g
Similarly, if the failure occurs after playback has begun, then the continuity condition
becomes
min
P
After
(
i
)
max
{
F
(
i
)
} ≤
(11.60)
or
(
i
D
F
≤
i
(
N
S
−
T
E
(11.61)
f
+
+
+
1)
T
F
+
t
0
+
τ
+
K
)
T
a
v
g
+
min
{
F
N
(
y
−
1)
} +
Solving, we can obtain
y
from
1
f
+
−
f
−
−
+
τ
+
T
E
+
D
F
y
After
=
(11.62)
(
N
S
−
K
)
T
a
v
g
Similarly for
Z
, we also need to consider the two cases. First, for the case where failure
occurs before playback begins, we have
min
{
F
(
i
+
l
−
2)
}≥
max
{
P
Before
(
i
)
}
(11.63)
or
(
i
f
−
>
i
(
N
S
−
T
L
(11.64)
+
l
−
1)(
N
S
−
K
)
T
a
v
g
+
t
0
+
K
)
T
a
v
g
+
max
{
F
F
(
y
−
1)
} +
Substituting the upper bound of equation (11.52) into equation (11.64), we have
(
i
f
−
>
iN
S
T
a
v
g
+
y
(
N
S
−
+
l
−
1)(
N
S
−
K
)
T
a
v
g
+
t
0
+
K
)
T
a
v
g
+
t
0
D
F
+
T
L
f
+
+
+
τ
+
(11.65)
=
−
Rearranging we can obtain
z
(
l
y
)as
1
f
+
−
f
−
+
+
τ
+
T
L
+
D
F
z
Before
=
(11.66)
(
N
S
−
K
)
T
a
v
g
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