2001/10/25Sheng-Feng Ho1 Efficient and Scalable On- Demand Data Streaming Using UEP Codes Lihao Xu...

2001/10/25 Sheng-Feng Ho 1

Efficient and Scalable On-Demand Data Streaming Using UEP Codes

Lihao XuWashington University in St. Louis

ACM Multimedia 2001 Sept. 30 – Oct. 5, 2001


OutlineIntroductionUnequal Error Protection Codes UEP Codes Example Theorem

The SchemeResource Consumption Network Bandwidth Client’s Buffer Space Client’s Network Bandwidth

Additional Initial Playout DelayConclusion


Introduction

Streaming Method Unicast ： Point-to-point

Resource consumption is proportional to the number of requests.

Multicast ： Pyramid or Skyscraper The lowest resource consumption with no

initial data playout delay is proportional to logarithm of the average request arrival rate.

The reliability ： Automatic-Repeat-Request 、 Forward Error Correction


Unequal Error Protection Codes

(N,K) block code encodes an original message of K symbols into a codeword of N data symbols of the same size.A data symbol is a general data unit of certain size ： a bit, a byte, a packet or a frame.Original K data symbols can be recovered from any M data symbols of its codeword. (M≧K)


Unequal Error Protection Codes

For a UEP code, certain data symbols of its codeword are protected against a greater number of errors than others.For a message m with n data symbols, if the error protection degree of its i-th symbol is Li (i≦1 ≦n), and it is encoded with a UEP code C of N symbols, then any Li symbols of its codeword are sufficient to retrieve the i-th symbol in the original message m.


UEP Codes Example (1)

Message m has 3 symbols of equal size ： a, b and c. (m=abc) Partition symbol a into 6 sub-symbols of equal size ：

a=a1..a6, b into 9 sub-symbols ： b=b1..b9 c into 6 sub-symbols ： c1..c6


UEP Codes Example (2)Apply the (6,2) B-Code on a to get a codeword of a ： A=A1..A6 (Ai is ½ size of a) A1 = a1, a2+a3, a4+a6, A2 = a2, a3+a4, a5+a1, A3 = a3, a4+a5, a6+a2, A4 = a4, a5+a6, a1+a3, A5 = a5, a6+a1, a2+a4, A6 = a6, a1+a2, a3+a5, and + is the simple bit-wise binary exclusive

or (XOR) operations



Apply a modified (6,3) RS-Code on b to get a codeword of b ： B=B1..B6 (Bi is 1/3 size of b) B1 = b1, b2, b3, B2 = b1+b3+b4, b2+b4+b5, b2+b3+b6, B3 = b6+b7, b4+b6+b8, b5+b9, B4 = b3+b6+b7, b1+b4+b7+b8, b2+b5+b9 B5 = b4+b6+b9, b4+b7, ？ B6 = b7, b8, b9 “+” is the XOR operation

Let Ci = ci for i = 1 to 6, thus each Ci is 1/6 size of c.



Construct a UEP codeword of the original message m ： U = U1..U6, where Ui = AiBiCi for i = 1 to 6.

The protection degrees of the original data symbols a, b and c are La=2, Lb=3 and Lc=6 respectively.

1/La+1/Lb+1/Lc=1



Original Messagem=abc

UEP CodewordU=U1U2U3

U1=A1B1C1,U2=..

Server Multicast Client ReceiveNetwork Transmit

UEP CodewordU=U1U2U3

a=a1..a6,b=b1..

Original Messagem=abc


Theorem

For a message m with n symbols, if there exists a UEP code such that the error protection degree of the i-th symbol in the original message m is Li (i≦1 ≦n), then

1/11

≦

n

iLi


The Scheme (1)

Encoding the original data of n symbols into a UEP codeword of N symbols.Multicast the UEP codeword in a cyclic fashion.Once the number of data symbols in user’s buffer space reaches Li, the user retrieves the i-th symbol of the original data stream and plays it out.


The Scheme (2)

B ： normal playout unicast network bandwidthd ： the initial playout number of original data symbolsThe peak network bandwidth needed ： rpeakB, where rpeak = max(Li/(i+d))The average network bandwidth is raveB, where raveB = Ln/(n+d)


The Scheme (3)

d 0 1 30 60R 12.859 8.87 5.484 4.796C 36.788% 36.790% 36.941% 37.095%

Resource Consumption vs. Initial Playout Delay for Multicasting a 2-hour Video.

d ： initial playout delay in seconds

R ： normalized backbone network bandwidth

C ： normalized client buffer space needed


Resource Consumption (1)

The peak network bandwidth needed ：rpeakB, where rpeak = max(Li/(i+d))

Minimize rpeak, set Li/(i+d) = R for all i’s

R=Hn+d-Hd≒αlog2(1+n/d), where ½≦α≦1

12

1,log1/1

1/1

21

1

whereliH

Li

m

im

n

i≦

R≒αlog2(1+n/d), where ½≦α≦1


Resource Consumption (2)

Let Si be the buffer space needed between the retrieval of the i-th and the (i+1)th original data symbols, then

Client’s Buffer Space ：n

diHHS didn

ni

)1)((max

10

)(,1,)1

1(1

11 diRlniil

lnS i

i

ji

ji

n

diHHS didn

ni

)1)((max

10


Resource Consumption (3)The normalized network bandwidth a client needs to consume between the retrieval of the i-th and (i+1)th original data symbols is

10, niHHC didni

)(),1

1(1

diRll

RC i

i

jj

i

10, niHHC didni


Additional Initial Playout Delay

d ： additional initial playout delay ： normalized backbone network bandwidthW ： client’s normalized incoming network badnwidth

ddn HHdR )(

)(,

)(),(,/)()(),(

dRWd

dRWdRWnWdRdndWD

))((,))(

11(),( dndRll

dRwddWD nn


Conclusion

Our scheme utilizes nice properties of error control codes, particularly UEP codes.The scheme also tolerates packet loss during transmission, thus further reduces multicast cost.Problem ： Fast Seek

2001/10/25Sheng-Feng Ho1 Efficient and Scalable On- Demand Data Streaming Using UEP Codes Lihao Xu...

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