2001/10/25Sheng-Feng Ho1 Efficient and Scalable On- Demand Data Streaming Using UEP Codes Lihao Xu Washington University in St. Louis ACM Multimedia 2001 Sept. 30 – Oct. 5, 2001

2001/10/25Sheng-Feng Ho2 Outline Introduction Unequal Error Protection Codes UEP Codes Example Theorem The Scheme Resource Consumption Network Bandwidth Client’s Buffer Space Client’s Network Bandwidth Additional Initial Playout Delay Conclusion

2001/10/25Sheng-Feng Ho3 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

2001/10/25Sheng-Feng Ho4 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)

2001/10/25Sheng-Feng Ho5 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 L i (i ≦ 1 ≦ n), and it is encoded with a UEP code C of N symbols, then any L i symbols of its codeword are sufficient to retrieve the i-th symbol in the original message m.

2001/10/25Sheng-Feng Ho6 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

2001/10/25Sheng-Feng Ho7 UEP Codes Example (2) Apply the (6,2) B-Code on a to get a codeword of a ： A=A1..A6 (A i 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

2001/10/25Sheng-Feng Ho8 UEP Codes Example (3) Apply a modified (6,3) RS-Code on b to get a codeword of b ： B=B1..B6 (B i 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 C i = c i for i = 1 to 6, thus each C i is 1/6 size of c.

2001/10/25Sheng-Feng Ho9 UEP Codes Example (4) Construct a UEP codeword of the original message m ： U = U1..U6, where U i = A i B i C i for i = 1 to 6. The protection degrees of the original data symbols a, b and c are L a =2, L b =3 and L c =6 respectively. 1/L a +1/L b +1/L c =1

2001/10/25Sheng-Feng Ho10 UEP Codes Example (5) Original Message m=abc UEP Codeword U=U1U2U3 U1=A1B1C1,U2=.. Server MulticastClient Receive Network Transmit UEP Codeword U=U1U2U3 a=a1..a6,b=b1.. Original Message m=abc

2001/10/25Sheng-Feng Ho11 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 L i (i ≦ 1 ≦ n), then

2001/10/25Sheng-Feng Ho12 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 L i, the user retrieves the i-th symbol of the original data stream and plays it out.

2001/10/25Sheng-Feng Ho13 The Scheme (2) B ： normal playout unicast network bandwidth d ： the initial playout number of original data symbols The peak network bandwidth needed ： r peak B, where r peak = max(L i /(i+d)) The average network bandwidth is r ave B, where r ave B = L n /(n+d)

2001/10/25Sheng-Feng Ho14 The Scheme (3) d R C36.788%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

2001/10/25Sheng-Feng Ho15 Resource Consumption (1) The peak network bandwidth needed ： r peak B, where r peak = max(L i /(i+d)) Minimize r peak, set L i /(i+d) = R for all i’s R=H n+d -H d ≒ αlog 2 (1+n/d), where ½ ≦ α ≦ 1 R ≒ αlog 2 (1+n/d), where ½ ≦ α ≦ 1

2001/10/25Sheng-Feng Ho16 Resource Consumption (2) Let S i 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 ：

2001/10/25Sheng-Feng Ho17 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

2001/10/25Sheng-Feng Ho18 Additional Initial Playout Delay d ： additional initial playout delay ： normalized backbone network bandwidth W ： client’s normalized incoming network badnwidth

2001/10/25Sheng-Feng Ho19 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