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Schemes for Video on demand
Yuan-Shiang Yeh
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Outline
Introduction Previous Works Study
Buffer Requirement Channel Adjustment Bandwidth reduction in multi-layer
videos
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introduction
True VoD Batch Patch
Near VoD Fast Data Broadcasting Harmonic Boradcasting
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Comparison
True VoD Near VoD
Delay Time no yes( 依方法不同 )
Receive Type Peer to Peer Broadcast
Client number dependent independent
Suitable Video
any hot
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Objective Bandwidth
Ex: FB -> Pagoda -> New Pagoda Buffer Storage
Ex: PB -> PPB Scalable (wide range)
Ex: FB -> UD VCR Functionality
Ex: Staggered Broadcast -> SAM Channel Adjustment
Ex: FB -> Seamless Channel Transition I/O BANDWIDTH
Ex: FB -> SB
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Harmonic Broadcasting Scheme
Parameters: Movie length --- D (e.g., 120 minutes) Consumption rate of the movie --- b (e.g., 10Mb
ps) Size of the movie --- S = D*b The movie is equally divide into N segments, an
d Si is the ith segment of the movie. Viewer waiting time --- d
d = D / ND
S1 S2 S3 S4
d
bandwidth
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Harmonic Broadcasting Scheme
Parameters The ith segment of the movie Si is equally
divided into i sub-segment(s) {Si, 1, Si, 2 --- Si, i}
Let the i sub-segment(s) of Si be put on a logical channel Ci, the bandwidth of Ci is b/i.
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Harmonic Broadcasting Scheme
The total bandwidth(B) allocated for the movie is as follows:
Where HN is called the harmonic number of N
B = b + b/2 + b/3 + b/4 = 2.083b
HN = 1 + 1/2 + 1/3 + 1/4 = 2.083
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S3, 1S2, 1
S1 S3, 2
S3, 3
S2, 2
S4, 1
S4, 2
S4, 3
S4, 4
D
B
S1 S1 S1 S1
S2, 1 S2, 2 S2, 1 S2, 2
S3, 1 S3, 2 S3, 3 S3, 1
S4, 1 S4, 2 S4, 3 S4, 4
C1 B
C2 2B
C3 3B
C4 4B
d
Time
Bandwidth
S1
S2, 1
S3, 1
S4, 1
S2, 2
S3, 2
S4, 2
S2, 1
S3, 3
S4, 3
S3, 1 S3, 2
S4, 4S4, 1 S4, 2 S4, 3
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Fast broadcasting scheme(FB)
In FB scheme, we divide a movie into 2k - 1 segments, k channels is needed.
Waiting time: d = D / 7 ( D: the video length )
S1
S2
S4
S1
S3
S5
S1
S2
S6
S1
S3
S7
S1
S2
S4
S1
S3
S5
S1
S2
S6
S1
S3
S7
bChannel 1
Channel 2
Channel 3
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S1
D
d
S2 S7
C1
C2
C3
· · ·S1 S1 S1 S1 S1 S1 S1 S1
S2 S3 S2 S3 S2 S3 S2 S3
S4 S5 S6 S7 S4 S5 S6 S7
· · ·
· · ·
S2 S2 S2 S2
S1
S2
S4
S3
S5
S2
S6
S3
S7S4
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Pagoda Broadcasting Scheme
C1
C2
C3
C2k
C2k+1
…
Segment Sz needs to be transmitted at minimum frequency 1/(zd)
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Pagoda Broadcasting Scheme
Segments Stream Broadcasting Frequency
S10 to S14 4 1/10d
S15 to S19 5 1/15d
S20 to S29 4 1/20d
S30 to S49 5 1/30d
4z = 40
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Example : kou’s researchSegment number: 7×2i, i = 0, 1, … 5;Number of channels: i + 4;Client buffer size: 2i segments;
S1C0
t0
d
C0 S1
t0 + d
S2C1
S1C0 S1
t0 + 2d
S2C1
S1
S3
Playing segment
Buffered segment
V0 S1
V0 S2V1 S1 S2
V0 S3V1 S2 S3V2 S1 S3
S1
No
No
No
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S1C0 S1
t0 + 3d
S2C1
S1
S3
S1
S2
S4C2
S1C0 S1
t0 + 4d
S2C1
S1
S3
S1
S2
S4C2
S1
S5
S6
C3 S4
V0 S4V1 S3 S4V2 S2 S3V3 S1 S2
V0 S5V1 S4 S5V2 S3 S4V3 S2 S4V4 S1 S4
S6
No
S1C0 S1
t0 + 7d
S2C1
S1
S3
S1
S4
S2C2
S1
S5
S6
S4
S1
S2
S3
S7
S1
S5
S6
S4
S2
S3
S7
S1C0 S1
t0 + 6d
S2C1
S1
S3
S1
S4
S2C2
S1
S5
S6
C3 S4
S1
S2
S3
S7
S1
S5
S6
S4
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Buffer Requirement - FB
1
24
9
13510
12611
13712
12413
13514
126
8
137
124
135
126
137
124
135
126
137
C1
C2
C3
C4 15 9 10 11 12 13 148 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
248
48
3
95
48
95
106
8
95
106
117
89
106
117
12
8910
117
1213
891011121314
91011121314
1011121314
11121314
121314
1314
14
Current segment
Buffer
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Buffer Requirement - FB
1
9
13510
12611
13712
12413
13514
126
13715
248
7 0 1 2 3 4 5 6
28
Buffer Requirement = 7/15
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Our permutation
1249
13611
12513
13715
1248
13610
12512
13714
1249
13611
12513
13715
1248
13610
12512
13714
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 3 48
6810
681012
781012
891012
910
1211
10
1211
13 15
1211
13 15
1213 15
13 15 15
1249
C1
C2
C3
C4
Current segment
Buffer
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Buffer Requirement - ours
1
10
13612
12514
137
9
12411
13613
125
13715
248
4 3 3 3 3 4 4 4
28
Theorem 1:Assuming that a video is equally divided into segments and allocated to n channels as Fast Data Broadcast did, our broadcasting schedule requires the minimum storage space. #
)12( n
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Result
channels FB kou oursBuffer Total
segment
Buffer Total segmen
t
Buffer Total segmen
t
4 2 7 1 7 1 7
5 4 15 2 14 2 15
6 8 31 4 28 4 31
7 16 63 8 56 8 63
8 32 128 16 112 16 128
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Bandwidth - Algorithm 1 1. Divide channels into slots. Channel i
would broadcast segment i. 2. For current segments k, find the slo
t with minimum wasted bandwidth. 3. Suppose the slot is divided into x su
bslots ,then allocate the segment k to (x+k-1) to this slot.
4. Repeat 2~3 until no free slots.
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1. Divide into slots
C1
C2
C3
S1
S3
S2
free free
Broadcasting frequency = 1
free
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2. Find suitable slot
C1
C2
C3
S1
S3
S2
free free
free
Current segment = S4
free free
1/2 * 1/2 - 1/4 = min
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3. Allocate segments
C1
C2
C3
S1
S3
S2
free free
free
Current segment = S4
free free
1/2 * 1/2 - 1/4 = min
Num of sub-slot = 2
S4 S5
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4.Repeat step 2~3
C1
C2
C3
S1
S3
S2
S6
S4 S5
S7 S8 S9
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Bandwidth - Algorithm 2
1. Divide free channel or slots with minimum wasted bandwidth for segment i.
2. Allocate segment i into current slots.
3. Repeat 1~2 until no enough free bandwidth.
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Example for Algorithm 2
freeC1
C2
C3 free
free
S1
Broadcasting frequency = 1
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Example for Algorithm 2
C1
C2
C3
S1
S3
S2 S4 S5
free freeS6 free
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Resultchannels
FB PB Algo. 1
NPB Algo. 2
HB
3 7 9 9 9 9 10
4 15 19 24 26 25 30
5 31 49 63 66 73 82
6 63 99 158 172 201 226
7 127 249 412 442 565 615
8 255 499 1056 1522 1673
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1
2
4
1
3
5
1
2
6
1
3
7
1
2
4
1
3
5
1
2
6
1
3
7
1
2
4
1
3
5
1
2
6
1
3
7
1
2
4
1
3
5
1
2
6
1
3
7
Base Layer
Enhancement Layer 1
Enhancement Layer 2
1 2
4
3
5 6 7
1 2 43 5 6 7
1 2
4
3
5 6 7
1 2 43 5 6 7 1 2
4 5 6 7 4 5 6 7
1 2 3 1 2 3 1 2
2 channels
1 channels
dummy
popular
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Delay in enhancement layer
Delay Slots = k
Frequency (Si) = 1/(i+k)
S1
1
1/2
S2
1/2
1/3
S7
1/7
1/8
Delay slots = 1
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Example
C1
C2
C3
Current segment = S1
S1
S2 free free
free freeS3
Current segment = S2Current segment = S3
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Result – delay slots
Delay slots
0 1 2 3 4 5 6
Algo.1 158 411 702 1022 1322 1722 2022
Algo.2 201 564 928 1282 1664 1987 2389
Channels = 6 Number of segment in FB scheme = 63