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quick redundant data transmission. Analytical results prove that this redundant server scheme
enables PRT to become scalable to any number of servers. Finally, we compute numerical
results to show the feasibility of the proposed architecture under real-world conditions. With
the proposed architecture, a concurrent-push-based parallel video server will be able to sus-
tain multiple simultaneous-server failure and yet, can maintain non-stop, continuous video
playback for all clients.
Appendices
A.1 Derivations of Buffer Requirement for Block Striping under FEC
Among the
N
S
servers, assume the earliest transmission for the first round starts at time
t
0
, then
the last transmission for the first round must start at the latest by time
t
0
+
τ
is the
clock jitter among servers. The time for video block group
i
to be completely filled, denoted
by
F
(
i
), is therefore bounded by
(
i
, where
τ
f
−
≤
≤
(
i
f
+
+
1)
T
F
+
t
0
+
F
(
i
)
+
1)
T
F
+
t
0
+
τ
+
(11.34)
where
T
F
is as given in equation (11.4) and
f
+
(
f
+
≥
0) and
f
−
(
f
−
≤
0) are used to model
the maximum transmission time deviation due to randomness in the system, including transmi-
ssion rate deviation, CPU scheduling, bus contention, etc.
Since the client starts playing the video after filling the first
y
groups of buffers, the playback
time for video block group 0 is simply equal to
F
(
y
−
1). Hence, the playback time for video
block group
i
, denoted by
P
(
i
), is bounded by
iN
S
T
a
v
g
+
T
E
≤
≤
iN
S
T
a
v
g
+
T
L
F
(
y
−
1)
+
P
(
i
)
F
(
y
−
1)
+
(11.35)
where
Q
R
V
T
a
v
g
=
(11.36)
is the average playback time for one video block, and
T
E
,
T
L
are the jitter bounds for video
block consumption variations.
To guarantee video playback continuity, we must ensure that a video block group arrives
before its playback deadline. In the worst-case scenario, the latest filling time must be smaller
than the earliest playback time, i.e.,
{
}≤
{
}
max
F
(
i
)
min
P
(
i
)
(11.37)
For the L.H.S., noting that
T
F
=
(
N
S
−
K
)
T
a
v
g
(cf. equations (11.4) and (11.16)) we then have
f
+
max
{
F
(
i
)
}=
(
i
+
1)(
N
S
−
K
)
T
a
v
g
+
t
0
+
τ
+
(11.38)
Similarly, the R.H.S. is
min
{
P
(
i
)
}=
iN
S
T
a
v
g
+
min
{
F
(
y
−
1)
}+
T
E
(11.39)
f
−
+
=
iN
S
T
a
v
g
+
y
(
N
S
−
K
)
T
a
v
g
+
t
0
+
T
E
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