1. Fast broadcasting scheme(FB) In FB scheme, we divide a movie into 2 k - 1 segments, k channels is needed. S = S 1 · S 2 · S 3 · S 4 · S 5 · S 6 · S 7 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 b Channel 0 Channel 1 Channel 2 70 min.

2. Fast broadcasting scheme(FB) G n = the combination of segments which is numbered from 2 n to 2 n+1 – 1. –G 0 = {S 1 }, G 1 = {S 2, S 3 }, G 2 = {S 4, S 5, S 6, S 7 } Channel n broadcasts G n 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 b Channel 0 Channel 1 Channel 2 70 min.

3. My Design Issue: Minimize the client buffer size Design Issue: –Divide a movie into 2 k - 1 channels, at most k+1 channels are needed –Each client will save at most 2 k-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)

4. Design issue comparison Movie division Number of broadcasting channels Broadcasting schedule Client buffer size FB 2 k – 1 (k  3) KStaticAt most 0.5D My 2 k – 1 (k  3) K+1Dynamic At most 2 k-3 segment buffer size ( at most 0.143D)

5. 7 Segment Design Issue –Divide a movie into 7 segments –Each client will save at most 2 k-3 = 2 3-3 =1 segment buffer size –4 (3+1 = 4)broadcasting channels

6. 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 U9U9 S7 U 10 S6 S5 S4 S3 S2 S1 U 11 S1 S7 U 12 S6 S5 S4 S3 S2 S1 U 13 S1 S7 U 14 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 U9U9 S7 U 10 S6 S5 S4 S3 S2 S1 U 11 S1 S7 U 12 S6 S5 S1 U 13 S1 S7 U 14 S6 S5 S2S3 S2 S4 S3 S2 S4 S3 S2 S4 S3 S2 S4 S3 S2 S4 S3 S2

7. 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

8. Comparison A movie is divided into 7 segments The number of channels Waiting timeBuffer size FB3One segment length 43% of a movie size My Scheme3 ~ 4One segment length 14.3% of a movie size A movie is divided into 15 segments FB4One segment length 47% of a movie size My Scheme4 ~ 5One segment length 13.3% of a movie size

9. 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

10. 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

11. Introduction – FB Scheme G 0 = {S 1 }, G 1 = {S 2, S 3 }, G 2 = {S 4, S 5, S 6, S 7 }, G 3 = {S 8, S 9, S 10, S 11, S 12, S 13, S 14, S 15 } K = 4

12. 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

13. 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.

14. AFB Scheme Parameter definition –K: the number of channels in FB scheme –C i : C 0, C 1, ---- C k-1 represent the possible allocated video channels for a movie –G i : the combination of broadcasting segments of C i in FB scheme –h: the highest channel number that can be assigned for a new video channel. Initially, h = k-1

15. 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 C r. 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 the next step.

16. 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 C h for the new channel. Let C a = C h Update h to the highest channel number that is still empty Put the rest data segment in sequence into C a, C a+1, --- C k.

17. 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 C 0 to C 3 –Waiting time: d –Initially, h=3 –G n G 0 = {S 1 }, G 1 = {S 2, S 3 }, G 2 = {S 4, S 5, S 6, S 7 }, G 3 = {S 8, S 9, S 10, S 11, S 12, S 13, S 14, S 15 }

18. Example explanation [t 0 ] Step3: - Assign C h to the new channel - C a = C h = C 3 - allocate rest segments to new channel C 3 by Table 1  allocate G 0, G 1, G 2, G 3 to C 3 - update h = 2

19. Example explanation [t 0 + d ] Step1: - release S 1 from C 3 Step3: - C a = C h = C 2 - allocate rest segments to all channels by Table 1  allocate S 1 to C 2 - update h = 1

20. Example explanation [t 0 + 2d ] Step1: - release S 2 from C 3 - release S 1 from C 2, and release C 2 - update h = 2 Step3: - C a = C h = C 2 - allocate rest segments to all channels by Table 1  allocate S 1, S 2 to C 2 - update h = 1

21. Example explanation [t 0 + 3d ] Step1: - release S 3 from C 3 - release S 1 from C 2 Step3: - C a = C h = C 1 - allocate rest segments to new channel C 1 by Table 1  allocate S 1, S 2 to C 1 - update h = 0

22. Example explanation AFB Scheme at time t 0 + 7d FB Scheme

23. Another example explanation Suppose there are requests until t 0 + 6d, at time interval [t 0 + 6d, t 0 + 7d], there is no request At time interval [t 0 + 7d, t 0 + 8d], new requests arrives

24. Workable verification FB Scheme –We can receive any data segments before we need to use it at client end. –The reason is that we will consume all of the 2 i – 1 segments within {C 0, --- C i-1 } before we start to consume the first data segments of C i, but there are at most data 2 i segments in C i. 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 2i2i 2 i – 1

25. 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 C a and place the rest data segments in sequence according to Table 1