# An Efficient Periodic Broadcasting with Small Latency and Buffer Demand for Near  Video on Demand

Authors: Ying-Nan Chen; Li-Ming Tseng
Journal: International Journal of Digital Multimedia Broadcasting
 (2012)

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## Abstract

Broadcasting Protocols can efficiently transmit videos that simultaneously shared by clients with partitioning the videos into segments. Many studies focus on decreasing clients' waiting time, such as the fixed-delay pagoda broadcasting (FDPB) and the harmonic broadcasting schemes. However, limited-capability client devices such as PDAs and set-top boxes (STBs) suffer from storing a significant fraction of each video while it is being watched. How to reduce clients' buffer demands is thus an important issue. Related works include the staircase broadcasting (SB), the reverse fast broadcasting (RFB), and the hybrid broadcasting (HyB) schemes. This work improves FDPB to save client buffering space as well as waiting time. In comparison with SB, RFB, and HyB, the improved FDPB scheme can yield the smallest waiting time under the same buffer requirements.


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## Body

## 1. Introduction

How to efficiently maintain the exhausted bandwidth with the growth in the number of clients is an important issue of VOD deployment. Dan et al. [1] presented that 80 percent of the demand is for a few number (10 or 20) of very popular videos. One way to broadcast a popular video is to partition the video into segments, which are transmitted on several channels currently and periodically. The approach (called periodic broadcasting [2]) lets multiple users share channels and thus obtains high bandwidth utilization. One of the channels only broadcasts the first segment in real time. The other channels transmit the remaining segments. When clients want to watch a video, they wait for the beginning of the first segment on the first channel. While clients start watching the video, their set-top boxes (STBs) or computers still download and buffer unreceived segments from the channels to enable them to play the video continuously.The staggered broadcasting [3] scheme treats a complete video as a single segment and then transmits it on each channel at different start times. The fast broadcasting (FB) scheme [4] improves segment partitioning and arrangement to yield shorter service latency. The harmonic broadcasting (HB) scheme [5] initially partitions a video into equally sized segments, which are further divided into smaller subsegments according to the harmonic series. HB can yield the minimum waiting time [6]; however, its implementation is difficult due to the multitude of broadcasting channels [7]. The recursive frequency-splitting (RFS) [7] scheme broadcasts a segment as close to its frequency as possible to achieve a near-minimal waiting time. The study [8] focuses on reducing the computation complexity of RFS. Paris [9] proposed a fixed-delay pagoda broadcasting (FDPB) scheme that required clients to wait for a small-fixed delay before watching the selected videos.The staircase broadcasting (SB) scheme [10] requires a client to buffer only 25% of a playing video. In modifying the FB scheme, the reverse fast broadcasting (RFB) scheme [11] also buffers 25% of video size, merely half of what is required by the FB scheme. By combining RFS and RFB the hybrid broadcasting scheme (HyB) [12] yields small client buffering space and waiting time. The study in [13] proposed a generalized reverse sequence-based model to reduce their client buffer requirements. This work aims at improving FDPB to reduce required playback latency and buffering space. We prove the applicability of the improved FDPB, and compare its performance with that of several existing approaches. Given a bandwidth of 5 channels, the new scheme reduces the broadcast latency by as much as 24% when compared to HyB and 34.5% when compared to RFB and SB. The buffer requirements for these schemes are about 25% of video size.The rest of this paper is organized as follows. Section2 introduces FDPB and proposes its improvement. Performance analysis and comparison are in Section 3. We make brief conclusions in Section 4.


## 2. Proposed Scheme

This section briefly reviews FDPB and then presents the improvement.


### 2.1. Fixed-Delay Pagoda Broadcasting (FDPB) Scheme

To help explain the improved FDPB, we first review the FDPB scheme in the literature [9]. Suppose that a video of duration L is broadcast over k channels. The scheme partitions each video into m equal-size segments of duration l=L/m. These segments will be broadcast at different frequencies over the k channels. The FDPB-p scheme requires customers to wait for a fixed time interval w=pl, where p∈int,p  ≥1. Segment Si is thus transmitted at least once every p+i-1 slots. The FDPB-p scheme partitions each channel Cj into cj=p+i-1 subchannels in such a way that slot j of channel Cj belongs to its subchannel j. FDPB then maps segments into subchannels in a strict sequential fashion.Figure1 illustrates the video partition and segment arrangement of FDPB-3. Channel C1 is partitioned into two subchannels because 3+1-1 is close to 2. Let Si be the ith segment. Since p=3, segment S1 needs to be broadcast at least once every three slots and is assigned to subchannel 1. The first segment of subchannel 2 is segment S2, which appears at least once every 3 + 2 − 1 = 4 slots. We thus map two segments into subchannel 2, and channel C1 transmits three segments (i.e., S1 to S3). The first segment of channel C2 is segment S4, which is broadcast at least once every 3 + 4 −  1 = 6 slots. Since 6 is close to 4=22, channel C2 is partitioned into two subchannels. The first subchannel periodically transmits segments S4 to S6, and the second subchannel broadcasts segments S7 to S10. The first segment of channel C3 is segment S11, which is broadcast at least once every 3 + 11 − 1 = 13 slots. Similarly, since 13 approaches to 16=42, channel C3 is partitioned into four subchannels. The first subchannel sends segments S11 to S13, the second subchannel sends segments S14 to S17, the third subchannel sends segments S18 to S22, and the fourth subchannel sends S23 to S28. Hence, channel C3 broadcasts 18 segments. Figure 1(b) demonstrates how to broadcast the segments on three channels and how to download the segments colored gray. Suppose that a client starts receiving the segments at time Tp. This figure shows that the client begins to watch the video at time Tp+3L/28.Illustration of channel allocation for the FDPB-3 scheme.
(a)
The segment allocation on three channels(b)
The segment broadcasting on three channels


### 2.2. Improvement of FDPB

By integrating the RFB with FDPB, this work designs the RFB-FDPB (RFDPB) scheme, which gains small low buffer requirements as well as waiting time.


#### 2.2.1. Segment Delivery on the Server Side

Letk be the number of allocated channels. The data transmission rate of each channel is assumed to equal the playback rate b. An RFDPB server broadcasts segments according to the following steps.(1)
The largest number of segments that FDPB-p can transmit using k-2 channels is assumed to equal np. The RFDPB-p scheme then equally divides a constant-bit-rate (CBR) video into N segments, denoted by S1,S2,…,SN in sequence, where N=4np. Let L be the length of the video. The length of a segment, l, thus equals L/N.(2)
ChannelsC1 to Ck-2 periodically send segments S1 to Snp according to FDPB-p.(3)
SegmentsS2np to Snp+1 are transmitted on channel Ck-1.(4)
SegmentsS4np to S2np+1 are broadcast on channel Ck.Figure2 illustrates the segment delivery by RFDPB-3, where k=4. The server broadcasts segments S1 to S10 on channels C1 to C2 by FDPB-3 because FDPB-3 can transmit 10 segments using two channels, as indicated in Figure 1. Due to np=10, RFDPB periodically broadcasts segments S20 to S11 and S40 to S21 on channels C3 and C4.Figure 2
Segment transmission by RFDPB-3.


#### 2.2.2. Segment Reception on the Client Side

Suppose that a client has enough buffers to store portions of a playing video. Letl be a basic time unit during video playing. We also let Tp be the time that the client starts downloading the video and be the origin (i.e., the first time unit) of the time axis throughout the paper. The following steps are involved in playing a video at the client.(1)
The client receives the segments on channelsC1 to Ck-2 immediately when they are available on networks. Figure 3 further demonstrates the segment downloading, where the segments downloaded by a client are gray.(2)
The segment downloading on channelCk-1 is as follows. Suppose that the client first sees segment Sj on channel Ck-1 at time Tnow and next segment Sj at time Tnext. The client is also assumed to play segments Sx and Sj at time Tnow and Tuse, respectively. The work in [11, 12] indicates that if Tnext≤Tuse, the client does not need to receive segment Sj at time Tnow and actually performs the downloading at time Tnext, without causing playing interruption. FDPB further reveals that a client has to wait for a fix time unit p to play received segments. Thus, Tuse=j+pth time unit, Tnow=x+pth time unit, and Tnext=Tnow+np. Substituting these equations into Tnext≤Tuse, we obtain
(1)x+np≤j.
If this inequality holds, a client does not receive segment Sj at time Tnow; otherwise, it immediately downloads it. Suppose that the client first sees segment S15 with only diagonal lines on channel C3 at the 6th time unit, as indicated in Figure 3. Due to p=3, x=3, j=15, and np=10, Tnow=x+p=6, Tuse=j+p=18, and Tnext=Tnow+np=16. Clearly, it is unnecessary to download segment S15 at this time unit because inequality (1), x+np=13≤j=15, holds. The result also reflects that the client does not have to receive the segment when Tnext=16≤Tuse=18. When next segment S15 colored gray in Figure 3 arrives at the 16th time unit, the inequality is no more true, 13+10>15, and the client thus downloads it.(3)
The segments on channelCk are received in the similar way, as also indicated in Figure 3. Suppose that the client first sees segment Sj′ on channel Ck at time T′now, and next segment Sj′ at time T′next. The client is also assumed to play segments Sx′ and Sj′ at time T′now and T′use. Similarly, if T′next≤T′use, the client does not receive segment Sj′. Substituting T′use=j′+pth time unit, T′now=x′+pth time unit, and T′next=T′now+2np into inequality T′next≤T′use obtains
(2)x′+2np≤j′.
If the inequality is true, the client skips the downloading of segment Sj′. Suppose that the client first sees segment S29 with only diagonal lines on channel C4 at the 12th time unit, as indicated in Figure 3. Due to p=3, x′=9, j′=29, and np=10, Tnow′=x′+p=12, Tuse′=j′+p=32, and Tnext′=T′now+2np=32. Clearly, it is unnecessary to download segment S29 at this time unit because inequality (2), x′+2np=29≤j′=29, is true. The result also reflects that the client does not have to receive the segment when Tnext′=32≤Tuse′=32. When next segment S29 colored gray in Figure 3 arrives at the 32nd time unit, the inequality is no more true, 29+20>29, and the client thus downloads it.(4)
The client plays the video in the order ofS1,S2,…,SN at time Tp+p.(5)
Stop loading data from networks when all the segments have been received.Figure 3
An illustration of segment downloading by RFDPB-3, wherek=4, np=10, and N=40.


#### 2.2.3. Workable Verification

This section describes that the RFDPB scheme guarantees continuous playing on the client side. Because the segment broadcasting on channelsC1 to Ck-2 is based on FDPB [9], RFDPB is workable for these channels. We next prove that the segment arrangements on channels Ck-1 and Ck also ensure a client to continuously play a video. FDPB further indicates that a video server must broadcast segment Sj at least once in every j+p-1 time units to keep on-time data delivery on the client side. For the RFDPB scheme, a server broadcasts segment Sj on channel Ck-1 in every np time units, where np+1≤j≤2np. Because j+p-1>np, an RFDPB client can receive video segments on channel Ck-1 on time. Similarly, the RFDPB scheme requires a server to transmit segment Sj′ on channel Ck in every 2np time units, where 2np+1≤j′≤4np. Because j′+p-1>2np, the segment delivery on this channel is also on time. Accordingly, RFDPB ensures continuous video playing at the client.


## 2.1. Fixed-Delay Pagoda Broadcasting (FDPB) Scheme

To help explain the improved FDPB, we first review the FDPB scheme in the literature [9]. Suppose that a video of duration L is broadcast over k channels. The scheme partitions each video into m equal-size segments of duration l=L/m. These segments will be broadcast at different frequencies over the k channels. The FDPB-p scheme requires customers to wait for a fixed time interval w=pl, where p∈int,p  ≥1. Segment Si is thus transmitted at least once every p+i-1 slots. The FDPB-p scheme partitions each channel Cj into cj=p+i-1 subchannels in such a way that slot j of channel Cj belongs to its subchannel j. FDPB then maps segments into subchannels in a strict sequential fashion.Figure1 illustrates the video partition and segment arrangement of FDPB-3. Channel C1 is partitioned into two subchannels because 3+1-1 is close to 2. Let Si be the ith segment. Since p=3, segment S1 needs to be broadcast at least once every three slots and is assigned to subchannel 1. The first segment of subchannel 2 is segment S2, which appears at least once every 3 + 2 − 1 = 4 slots. We thus map two segments into subchannel 2, and channel C1 transmits three segments (i.e., S1 to S3). The first segment of channel C2 is segment S4, which is broadcast at least once every 3 + 4 −  1 = 6 slots. Since 6 is close to 4=22, channel C2 is partitioned into two subchannels. The first subchannel periodically transmits segments S4 to S6, and the second subchannel broadcasts segments S7 to S10. The first segment of channel C3 is segment S11, which is broadcast at least once every 3 + 11 − 1 = 13 slots. Similarly, since 13 approaches to 16=42, channel C3 is partitioned into four subchannels. The first subchannel sends segments S11 to S13, the second subchannel sends segments S14 to S17, the third subchannel sends segments S18 to S22, and the fourth subchannel sends S23 to S28. Hence, channel C3 broadcasts 18 segments. Figure 1(b) demonstrates how to broadcast the segments on three channels and how to download the segments colored gray. Suppose that a client starts receiving the segments at time Tp. This figure shows that the client begins to watch the video at time Tp+3L/28.Illustration of channel allocation for the FDPB-3 scheme.
(a)
The segment allocation on three channels(b)
The segment broadcasting on three channels


## 2.2. Improvement of FDPB

By integrating the RFB with FDPB, this work designs the RFB-FDPB (RFDPB) scheme, which gains small low buffer requirements as well as waiting time.


### 2.2.1. Segment Delivery on the Server Side

Letk be the number of allocated channels. The data transmission rate of each channel is assumed to equal the playback rate b. An RFDPB server broadcasts segments according to the following steps.(1)
The largest number of segments that FDPB-p can transmit using k-2 channels is assumed to equal np. The RFDPB-p scheme then equally divides a constant-bit-rate (CBR) video into N segments, denoted by S1,S2,…,SN in sequence, where N=4np. Let L be the length of the video. The length of a segment, l, thus equals L/N.(2)
ChannelsC1 to Ck-2 periodically send segments S1 to Snp according to FDPB-p.(3)
SegmentsS2np to Snp+1 are transmitted on channel Ck-1.(4)
SegmentsS4np to S2np+1 are broadcast on channel Ck.Figure2 illustrates the segment delivery by RFDPB-3, where k=4. The server broadcasts segments S1 to S10 on channels C1 to C2 by FDPB-3 because FDPB-3 can transmit 10 segments using two channels, as indicated in Figure 1. Due to np=10, RFDPB periodically broadcasts segments S20 to S11 and S40 to S21 on channels C3 and C4.Figure 2
Segment transmission by RFDPB-3.


### 2.2.2. Segment Reception on the Client Side

Suppose that a client has enough buffers to store portions of a playing video. Letl be a basic time unit during video playing. We also let Tp be the time that the client starts downloading the video and be the origin (i.e., the first time unit) of the time axis throughout the paper. The following steps are involved in playing a video at the client.(1)
The client receives the segments on channelsC1 to Ck-2 immediately when they are available on networks. Figure 3 further demonstrates the segment downloading, where the segments downloaded by a client are gray.(2)
The segment downloading on channelCk-1 is as follows. Suppose that the client first sees segment Sj on channel Ck-1 at time Tnow and next segment Sj at time Tnext. The client is also assumed to play segments Sx and Sj at time Tnow and Tuse, respectively. The work in [11, 12] indicates that if Tnext≤Tuse, the client does not need to receive segment Sj at time Tnow and actually performs the downloading at time Tnext, without causing playing interruption. FDPB further reveals that a client has to wait for a fix time unit p to play received segments. Thus, Tuse=j+pth time unit, Tnow=x+pth time unit, and Tnext=Tnow+np. Substituting these equations into Tnext≤Tuse, we obtain
(1)x+np≤j.
If this inequality holds, a client does not receive segment Sj at time Tnow; otherwise, it immediately downloads it. Suppose that the client first sees segment S15 with only diagonal lines on channel C3 at the 6th time unit, as indicated in Figure 3. Due to p=3, x=3, j=15, and np=10, Tnow=x+p=6, Tuse=j+p=18, and Tnext=Tnow+np=16. Clearly, it is unnecessary to download segment S15 at this time unit because inequality (1), x+np=13≤j=15, holds. The result also reflects that the client does not have to receive the segment when Tnext=16≤Tuse=18. When next segment S15 colored gray in Figure 3 arrives at the 16th time unit, the inequality is no more true, 13+10>15, and the client thus downloads it.(3)
The segments on channelCk are received in the similar way, as also indicated in Figure 3. Suppose that the client first sees segment Sj′ on channel Ck at time T′now, and next segment Sj′ at time T′next. The client is also assumed to play segments Sx′ and Sj′ at time T′now and T′use. Similarly, if T′next≤T′use, the client does not receive segment Sj′. Substituting T′use=j′+pth time unit, T′now=x′+pth time unit, and T′next=T′now+2np into inequality T′next≤T′use obtains
(2)x′+2np≤j′.
If the inequality is true, the client skips the downloading of segment Sj′. Suppose that the client first sees segment S29 with only diagonal lines on channel C4 at the 12th time unit, as indicated in Figure 3. Due to p=3, x′=9, j′=29, and np=10, Tnow′=x′+p=12, Tuse′=j′+p=32, and Tnext′=T′now+2np=32. Clearly, it is unnecessary to download segment S29 at this time unit because inequality (2), x′+2np=29≤j′=29, is true. The result also reflects that the client does not have to receive the segment when Tnext′=32≤Tuse′=32. When next segment S29 colored gray in Figure 3 arrives at the 32nd time unit, the inequality is no more true, 29+20>29, and the client thus downloads it.(4)
The client plays the video in the order ofS1,S2,…,SN at time Tp+p.(5)
Stop loading data from networks when all the segments have been received.Figure 3
An illustration of segment downloading by RFDPB-3, wherek=4, np=10, and N=40.


### 2.2.3. Workable Verification

This section describes that the RFDPB scheme guarantees continuous playing on the client side. Because the segment broadcasting on channelsC1 to Ck-2 is based on FDPB [9], RFDPB is workable for these channels. We next prove that the segment arrangements on channels Ck-1 and Ck also ensure a client to continuously play a video. FDPB further indicates that a video server must broadcast segment Sj at least once in every j+p-1 time units to keep on-time data delivery on the client side. For the RFDPB scheme, a server broadcasts segment Sj on channel Ck-1 in every np time units, where np+1≤j≤2np. Because j+p-1>np, an RFDPB client can receive video segments on channel Ck-1 on time. Similarly, the RFDPB scheme requires a server to transmit segment Sj′ on channel Ck in every 2np time units, where 2np+1≤j′≤4np. Because j′+p-1>2np, the segment delivery on this channel is also on time. Accordingly, RFDPB ensures continuous video playing at the client.


## 2.2.1. Segment Delivery on the Server Side

Letk be the number of allocated channels. The data transmission rate of each channel is assumed to equal the playback rate b. An RFDPB server broadcasts segments according to the following steps.(1)
The largest number of segments that FDPB-p can transmit using k-2 channels is assumed to equal np. The RFDPB-p scheme then equally divides a constant-bit-rate (CBR) video into N segments, denoted by S1,S2,…,SN in sequence, where N=4np. Let L be the length of the video. The length of a segment, l, thus equals L/N.(2)
ChannelsC1 to Ck-2 periodically send segments S1 to Snp according to FDPB-p.(3)
SegmentsS2np to Snp+1 are transmitted on channel Ck-1.(4)
SegmentsS4np to S2np+1 are broadcast on channel Ck.Figure2 illustrates the segment delivery by RFDPB-3, where k=4. The server broadcasts segments S1 to S10 on channels C1 to C2 by FDPB-3 because FDPB-3 can transmit 10 segments using two channels, as indicated in Figure 1. Due to np=10, RFDPB periodically broadcasts segments S20 to S11 and S40 to S21 on channels C3 and C4.Figure 2
Segment transmission by RFDPB-3.


## 2.2.2. Segment Reception on the Client Side

Suppose that a client has enough buffers to store portions of a playing video. Letl be a basic time unit during video playing. We also let Tp be the time that the client starts downloading the video and be the origin (i.e., the first time unit) of the time axis throughout the paper. The following steps are involved in playing a video at the client.(1)
The client receives the segments on channelsC1 to Ck-2 immediately when they are available on networks. Figure 3 further demonstrates the segment downloading, where the segments downloaded by a client are gray.(2)
The segment downloading on channelCk-1 is as follows. Suppose that the client first sees segment Sj on channel Ck-1 at time Tnow and next segment Sj at time Tnext. The client is also assumed to play segments Sx and Sj at time Tnow and Tuse, respectively. The work in [11, 12] indicates that if Tnext≤Tuse, the client does not need to receive segment Sj at time Tnow and actually performs the downloading at time Tnext, without causing playing interruption. FDPB further reveals that a client has to wait for a fix time unit p to play received segments. Thus, Tuse=j+pth time unit, Tnow=x+pth time unit, and Tnext=Tnow+np. Substituting these equations into Tnext≤Tuse, we obtain
(1)x+np≤j.
If this inequality holds, a client does not receive segment Sj at time Tnow; otherwise, it immediately downloads it. Suppose that the client first sees segment S15 with only diagonal lines on channel C3 at the 6th time unit, as indicated in Figure 3. Due to p=3, x=3, j=15, and np=10, Tnow=x+p=6, Tuse=j+p=18, and Tnext=Tnow+np=16. Clearly, it is unnecessary to download segment S15 at this time unit because inequality (1), x+np=13≤j=15, holds. The result also reflects that the client does not have to receive the segment when Tnext=16≤Tuse=18. When next segment S15 colored gray in Figure 3 arrives at the 16th time unit, the inequality is no more true, 13+10>15, and the client thus downloads it.(3)
The segments on channelCk are received in the similar way, as also indicated in Figure 3. Suppose that the client first sees segment Sj′ on channel Ck at time T′now, and next segment Sj′ at time T′next. The client is also assumed to play segments Sx′ and Sj′ at time T′now and T′use. Similarly, if T′next≤T′use, the client does not receive segment Sj′. Substituting T′use=j′+pth time unit, T′now=x′+pth time unit, and T′next=T′now+2np into inequality T′next≤T′use obtains
(2)x′+2np≤j′.
If the inequality is true, the client skips the downloading of segment Sj′. Suppose that the client first sees segment S29 with only diagonal lines on channel C4 at the 12th time unit, as indicated in Figure 3. Due to p=3, x′=9, j′=29, and np=10, Tnow′=x′+p=12, Tuse′=j′+p=32, and Tnext′=T′now+2np=32. Clearly, it is unnecessary to download segment S29 at this time unit because inequality (2), x′+2np=29≤j′=29, is true. The result also reflects that the client does not have to receive the segment when Tnext′=32≤Tuse′=32. When next segment S29 colored gray in Figure 3 arrives at the 32nd time unit, the inequality is no more true, 29+20>29, and the client thus downloads it.(4)
The client plays the video in the order ofS1,S2,…,SN at time Tp+p.(5)
Stop loading data from networks when all the segments have been received.Figure 3
An illustration of segment downloading by RFDPB-3, wherek=4, np=10, and N=40.


## 2.2.3. Workable Verification

This section describes that the RFDPB scheme guarantees continuous playing on the client side. Because the segment broadcasting on channelsC1 to Ck-2 is based on FDPB [9], RFDPB is workable for these channels. We next prove that the segment arrangements on channels Ck-1 and Ck also ensure a client to continuously play a video. FDPB further indicates that a video server must broadcast segment Sj at least once in every j+p-1 time units to keep on-time data delivery on the client side. For the RFDPB scheme, a server broadcasts segment Sj on channel Ck-1 in every np time units, where np+1≤j≤2np. Because j+p-1>np, an RFDPB client can receive video segments on channel Ck-1 on time. Similarly, the RFDPB scheme requires a server to transmit segment Sj′ on channel Ck in every 2np time units, where 2np+1≤j′≤4np. Because j′+p-1>2np, the segment delivery on this channel is also on time. Accordingly, RFDPB ensures continuous video playing at the client.


## 3. Analysis and Comparison

Before investigating entire buffer requirements for RFDPB, this paper analyzes the segment downloading on channelCk-1 first. For RFDPB-p, a client plays the segments received from this channel during time units np+1+p to 2np+p. The possible time to download the segments is during the pth to 2np+pth time units. Suppose that a client sees segment Sj at the np+1+pth time unit. Inequality (1) makes a complete segment-downloading diagram for channel Ck-1, as indicated in Figure 4, where the segments downloaded by the client are colored gray. The figure shows that the client does not continuously download segments, and some segments are skipped. The segments are not downloaded during the first to ⌊(j+(np+1))/2⌋-np+pth time units, during the j+1-np+pth to ⌊(j+1)/2⌋+pth time units, during the ⌊(j+(np+1))/2⌋+1+pth to j+pth time units, and during the ⌊(j+1)/2⌋+np+1+pth to 2np+pth time units, respectively. Such results reflect that the client can delay downloading these segments by np time units, due to (1). Similarly, (2) also makes a complete segment-downloading diagram for channel Ck, as indicated in Figure 6, where a client is assumed to see segment Sj at the 2np+1+pth time unit.Figure 4
The segment downloading on channelCk-1 under RFDPB-p.The RFDPB scheme uses the same methodology as HyB, combining a nonharmonic scheme and the RFB to reduce clients’ waiting time and buffering spaces (Figure5). For segment broadcasting on channels C1 to Ck-2 using the nonharmonic scheme, channels Ck-1 and Ck broadcast the other segments according RFB. Yu’s work [12] has proved HyB only buffer 25% of video size for k≥3, where k is the number of broadcasting channels. The proof of the maximum number of segments buffered by a client for RFDPB is similar to that of HyB, so this paper neglects it. Therefore, let B(k) be the maximum number of segments buffered by a client, where k is the number of broadcasting channels. For RFDPB,
(3)B(1)=0B(2)=1B(k)=np,k≥3.
The previous analysis indicates that RFDPB buffers at most nr/N=25% of video size. Table 1 lists the comparison of maximum buffering spaces required by RFDPB-3, FB, SB, RFB, FDPB-3, and HyB. RFDPB-3 reduces the buffering space up to 32.4% when compared to FDPB-3 and up to 50% when compared to FB and has the same good result with the other schemes. Given a video of 120 minutes, Figure 6 shows the buffer requirements at different channels. To understand how well the RFDPB scheme performs on clients waiting time, this work calculates the values of N offered by RFDPB-6, SB, FB, RFB, FDPB-6, and HyB given different numbers of server channels, as listed in Table 2. The inverse of N offered by each scheme reflects the waiting time for a new client to start his/her VOD service. Figure 7, which is drawn in a logarithmic scale, shows that RFDPB-6 performs close to FDPB-6 stably. For k≥4, the RFDPB-6 outperforms all the schemes, except the FDPB-6. To understand the relationship between RFDPB and HyB schemes, this work calculates clients’ waiting time of RFDPB-9, RFDPB-6, HyB, and RFDPB-3, as listed in Table 3. Figure 8 shows that, for p ≥ 6, RFDPB-p outperforms HyB. The previous comparisons clearly indicate that RFDPB exhibits a good tradeoff between client buffering spaces and waiting time.Table 1
The maximum buffering space required by different schemes in the percentage of video size usingk channels.
k12345678910RFDPB-30332525252525252525HyB0332525252525252525RFB0332927262525252525FB0334347484950505050SB0172123242525252525FDPB-367604642413938383737Table 2
The maximum numbers of segments,N, offered by different schemes.
k12345678910FDPB-6725711864851286342591952479067054RFDPB-61328100284744194051441370036780HyB134123610029280422606088FB (RFB)1371531631272555111023SB1371531631272555111023Table 3
The waiting time (seconds), offered by RFDPB-9, RFDPB-6, HyB, and RFDPB-3 schemes.
k12345678910RFDPB-9720024001350385.71139.6652.619.97.472.791.04RFDPB-6720024001542.86432152.1158.0622.278.43.151.17HyB7200240018006002007224.668.963.191.18RFDPB-3720024001800540192.8672.9728.5711.074.191.58Figure 5
The segment downloading on channelCk under RFDPB-p.Figure 6
Comparison of the maximum buffer requirements in the number of minutes of a 120-minute video.Figure 7
The maximum waiting time incurred on new clients at different numbers of channels (L=120 minutes).Figure 8
The maximum waiting time (RFDPB and HyB) incurred on new clients at different numbers of channels (L=120 minutes).


## 4. Conclusions

This paper presents an improved version of FDPB for efficient periodic broadcasting of popular videos. The proposed scheme takes advantage of the FDPB and the RFB schemes to obtain small waiting time and low buffer demand. Through mathematical analysis, we prove the applicability of this scheme by demonstrating that client playback continuity is guaranteed. Given a bandwidth of 5 channels, the new scheme reduces the broadcast latency by as much as 24% when compared to HyB and 34.5% when compared to RFB and SB. The buffer requirements for these schemes are about 25% of video size.



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*Source: 717538-2012-04-09.xml*