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