fast broadcasting scheme(fb) in fb scheme, we divide a movie into 2 k - 1 segments, k channels is...
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Fast broadcasting scheme(FB)• In FB scheme, we divide a movie into 2k - 1 segments,
k channels is needed.
• S = S1· S2· S3· S4· S5· S6· S7
• Waiting time: d = 70 / 7 = 10 min.
K=3
S1
S2
S4
S1
S3
S5
S1
S2
S6
S1
S3
S7
S1
S2
S4
S1
S3
S5
S1
S2
S6
S1
S3
S7
d
bChannel 0
Channel 1
Channel 2
70 min.
Fast broadcasting scheme(FB)
• Gn = the combination of segments which is numbered from 2n to 2n+1 – 1. – G0 = {S1}, G1 = {S2 , S3}, G2 = {S4 , S5 , S6 , S7}
• Channel n broadcasts Gn periodically.
• Drawback: Client has to save D/2 buffer size in average
S1
S2
S4
S1
S3
S5
S1
S2
S6
S1
S3
S7
S1
S2
S4
S1
S3
S5
S1
S2
S6
S1
S3
S7
d
bChannel 0
Channel 1
Channel 2
70 min.
My Design Issue: Minimize the client buffer size• Design Issue:
– Divide a movie into 2k - 1 channels, at most k+1 channels are needed
– Each client will save at most 2k-3 segment buffer size
– No static broadcasting schedule, we only broadcast segments which are needed during this period or next period (if free channel is available)
Design issue comparison
Movie
division
Number of
broadcasting
channels
Broadcasting
schedule
Client buffer size
FB 2k – 1 (k 3)
K Static At most 0.5D
My 2k – 1
(k 3)
K+1 Dynamic At most 2k-3 segment buffer size
( at most 0.143D)
7 Segment
• Design Issue– Divide a movie into 7 segments
– Each client will save at most 2k-3 = 23-3 =1 segment buffer size
– 4 (3+1 = 4)broadcasting channels
7 segments ( 根據上面的結果 , 開始進行 4 channel 的控
制 ) S1
U1
S1
U2
S2
S1
U3
S3
S1
U4
S4
S2
S1
U5
S5
S1
U6
S6
S4
S3
S2
S1
U7
S7
S1
U8
S6
S4
S3
S2
S5
S1
U9
S1
S7
U10
S6
S5
S4
S3
S2
S1
U11
S1
S7
U12
S6
S5
S4
S3
S2
S1
U13
S1
S7
U14
S6
S5
S4
S3
S2
Original:
(no channel control, client will save at most 1 segment)
Channel Control:
S1
U1
S1
U2
S2
S1
U3
S3
S1
U4
S4
S2
S1
U5
S5
S1
U6
S6
S4
S3
S2
S1
U7
S7
S1
U8
S6
S4
S3
S2
S5
S1
U9
S1
S7
U10
S6
S5
S4
S3
S2
S1
U11
S1
S7
U12
S6
S5
S1
U13
S1
S7
U14
S6
S5
S2 S3
S2
S4
S3
S2
S4
S3
S2
S4
S3
S2
S4
S3
S2
S4
S3
S2
15 Segment
• Design Issue– Divide a movie into 15 segments– Each client will save at most 2 segment buffer
size– 5 broadcasting channels
• We have found a broadcasting schedule applied to this design issue
Comparison
A movie is divided into 7 segments
The number of channels
Waiting time Buffer size
FB 3 One segment length
43% of a movie size
My Scheme 3 ~ 4 One segment length
14.3% of a movie size
A movie is divided into 15 segments
FB 4 One segment length
47% of a movie size
My Scheme 4 ~ 5 One segment length
13.3% of a movie size
Adaptive Fast Data Broadcasting Scheme for Video-on-Demand Service
Li-Shen Juhn and Li-Ming Tseng
IEEE Transactions on Broadcasting, VOL. 44, NO. 2, JUNE 1998
Introduction
• For a hot video, fast data broadcasting (FB) scheme substantially reduce bandwidth requirements compared with batching schemes.
• However, FB scheme has to predict which movie is hot.
• This paper presents an adaptive fast data broadcasting scheme (based on FB) that the bandwidth allocation for a movie is always efficient whether the movie is popular or not
Introduction – FB Scheme
G0 = {S1},
G1 = {S2 , S3},
G2 = {S4 , S5 , S6 , S7},
G3 = {S8 , S9 , S10 , S11 , S12 , S13 , S14 , S15}
K = 4
Adaptive Fast Broadcasting (AFB) Scheme
• The bandwidth requirement for a movie depends on the users requests.– If the movie is popular, the bandwidth
requirement of AFB scheme is the same as FB scheme
– If there is no request for a movie, in AFB scheme, no new bandwidth will be allocated for the movie
Adaptive Fast Broadcasting (AFB) Scheme
• We do not have to predict the popularity of a movie, but the bandwidth allocation is always efficient whether or not the movie is popular.
AFB Scheme
• Parameter definition – K: the number of channels in FB scheme
– Ci: C0 , C1 , ---- Ck-1 represent the possible allocated video channels for a movie
– Gi: the combination of broadcasting segments of Ci in FB scheme
– h: the highest channel number that can be assigned for a new video channel.
• Initially, h = k-1
AFB Scheme
• On the server side, the AFB scheme serves the movie every d minutes in the following way (initially, h=k-1)– Step1:
• Release the data segments that have been sent in the previous time interval from each broadcasting channels. If there is no more data to be sent within the channel, release the channel, suppose it’s Cr . If the released channel number is is greater than h, then, h = r
– Step2:• If there is no request for the movie, goto step1. Otherwise, do t
he next step.
AFB Scheme
• Server side service– Step3: if there is no free channel, reject the request.
Otherwise, according to the admission control policy, allocate a new channel and put the rest data segments in the new channel in the following way
• Assign Ch for the new channel.
• Let Ca = Ch
• Update h to the highest channel number that is still empty
• Put the rest data segment in sequence into Ca , Ca+1 , --- Ck .
Example explanation
• Parameters – K = 4– Suppose that there are request during every time interval– We divide a movie into 15 segments– Video channels are numbered from C0 to C3
– Waiting time: d– Initially, h=3– Gn
G0 = {S1}, G1 = {S2 , S3}, G2 = {S4 , S5 , S6 , S7}, G3 = {S8 , S9 , S10 , S11 , S12 , S13 , S14 , S15}
Example explanation
[t0 ]
Step3:- Assign Ch to the new channel- Ca = Ch = C3
- allocate rest segments to new channel C3 by Table 1
allocate G0 , G1 , G2 , G3 to C3
- update h = 2
Example explanation
[t0 + d ]Step1:
- release S1 from C3
Step3:- Ca = Ch = C2
- allocate rest segments to all channels by Table 1 allocate S1 to C2
- update h = 1
Example explanation[t0 + 2d ]Step1:
- release S2 from C3
- release S1 from C2 , and release C2
- update h = 2
Step3:- Ca = Ch = C2
- allocate rest segments to all channels by Table 1 allocate S1 , S2 to C2
- update h = 1
Example explanation[t0 + 3d ]Step1:
- release S3 from C3
- release S1 from C2
Step3:- Ca = Ch = C1
- allocate rest segments to new channel C1 by Table 1 allocate S1 , S2 to C1
- update h = 0
Example explanation
AFB Scheme at time t0 + 7d FB Scheme
Another example explanation• Suppose there are requests until t0 + 6d, at time interval [t
0 + 6d , t0 + 7d], there is no request
• At time interval [t0 + 7d , t0 + 8d], new requests arrives
Workable verification
• FB Scheme– We can receive any data segments before we need to us
e it at client end.
– The reason is that we will consume all of the 2i – 1 segments within {C0, --- Ci-1} before we start to consume the first data segments of Ci , but there are at most data 2i segments in Ci .
S1
S2
S4
S1
S3
S5
S1
S2
S6
S1
S3
S7
S1
S2
S4
S1
S3
S5
S1
S2
S6
S1
S3
S7
Channel 0
Channel 1
Channel 2
2i
2i – 1
Workable verification
• AFB Scheme– We can also receive any data segments before we need
to use it at client end.
– The reason is that at any step, we will allocate a video channel Ca and place the rest data segments in sequence according to Table 1
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