1. IEEE VTC 2010 Optimal Layered Video IPTV Multicast Streaming over IEEE 802.16e WiMAX Systems Po-Han Wu, Yu Hen Hu *, Jenq-Neng Hwang University of Washington Department of Electrical Engineering *University of Wisconsin – Madison Department of Electrical and Computer Engineering

2. Outline Introduction Problem Formulation Globally Optimal Solutions Suboptimal Heuristic Algorithm Simulation and Performance Comparison Conclusion

3. Introduction The IEEE 802.16e standard to provide IPTV services over WiMAX channels. Mobile subscribers (MSs) may view popular video programs at real time while roaming around metropolitan regions.

4. Introduction The high-speed multicast services can be the services of IPTV, video conferencing, and so on. Data requirement Base Layer Enhancement Layer 1 Base Layer Enhancement Layers 1 Base Layer Enhancement Layers 2

5. Introduction Multicast and broadcast services (MBS) – MBS zone is standardized in the frame structure of WiMAX spec. – MBS can occupy whole/partial of DL-subframe. – MBS service is offered in the downlink only. Preamble UL-MAP DL-MAP FCH MBS ZONE MBS MAP UL Sub-Frame Preamble UL-MAP DL-MAP FCH DL Burst 6 DL Burst 1 DL Burst 4 DL Burst 5 DL Burst 2 DL Burst 3 MBS ZONE MBS MAP UL Sub-Frame MBS MBS with Unicast Services

6. IPTV service in MBS zone – Higher transmission rate Utility – Lower transmission rate Resource consumption EL 1 BL EL2 Introduction Enhancement Layers 1 Base Layer Enhancement Layers 2 Video program Modulation QPSK Modulation16QAM 16QAM EL 1 BL EL2

7. Introduction Goal Design an efficient IPTV multicast algorithm over WiMAX network – Determine the modulation of each layer of video program such that Maximized the total utility while the resource is no larger than a bound

8. Notations and Definitions – Assumption V video streams are available for subscription by up to N MSs Each MS may subscribe one or more video streams – Matrix R = [R vl ] is specifies the sizes of video streams R vl is the size of the l th layer of the v th video per OFDMA frame Problem Formulation Enhancement Layers Base Layer Enhancement Layers Video program 2 R 21 = 5 bits/sec R 22 = 7 bits/sec R 23 = 9 bits/sec

9. Problem Formulation Notations and Definitions – Each layer is modulated at one of M different rates {c(m); 1 ≤ m ≤ M} c(1) =9.6kbps/slotQPSK ½ c(2) =14.4kbps/slotQPSK ½ c(3) =19.2kbps/slot16QAM ½ c(4) =28.8kbps/slot16QAM ¾ c(5) =38.4kbps/slot64QAM ⅔ c(6) =43.2kbps/slot64QAM ¾ ModulationTransmission Rate

10. a 21 = 2 a 22 = 3 a 23 = 3 Problem Formulation Radio Resource Allocation Process – which layer of which video stream to be transmitted – which modulation rate to be used to transmit the selected video. The decision can be summarized in a matrix A= [a vl ] – a vl = mif l th layer of v th video is modulated at c(m) – a vl = 0if l th layer of v th video is not to be transmitted Enhancement Layers Base Layer Enhancement Layers Video program 2 QPSK ½ 16QAM ½

11. The total number of OFDM symbols needed to multicast all V video programs is Problem Formulation slot consume = size / transmission rate Indicator function: I(x) = 1 if the condition x is true I(x) = 0 if the condition x is false S 11 S 12 S 13 S 14 S 21 S 22 S 23 S 24 S 31 S 32 S 33 S 34

12. Problem Formulation A matrix H = [h nv ] is represent the subscription status – h nv = 1if the n th MS subscribes v th video program – h nv = 0Otherwise A matrix P = [p vl ] denotes the prices for all video streams – Depends on how many layers video streams the MS can receive – p vl is the per subscriber price for that particular layer video stream Enhancement Layers Base Layer Enhancement Layers Video program 2 SubscriberPriceRevenue 20$ 3$60 40$ 5$200 80$10$800 $1060

13. Utility Function – The utility of delivering the v th video will be evaluated as the total revenue generated by serving that video to N users: – Aggregating for all V video programs, one has the total utility: Problem Formulation Examines the channel condition d n of the n th subscriber whether is larger than the SNR requirement d m of m th MCS Examines the MS whether subscribe the video Examines the layer of the video whether be transmitted

14. Problem Formulation The resource allocation problem can be described as follows: – Given H:the subscription matrix P : the price matrix R : the video payload matrix d : the channel condition vector c : the data modulation rate vector G 0 : the radio resource – Determine an assignment matrix A, such that the total revenue u is maximized the resource constraint S ≤ G 0 The number of potential solutions grows exponentially with respect to VL, it is obviously an NP-hard problem.

15. Service Profile – A service profile is the modulation rate assigned to each layer of a given video stream. Service profile Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(6) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(6) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(5) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(5) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(4) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(4) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(2) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(2) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(1) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(1) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(6) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(6) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(5) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(5) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(4) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(4) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(2) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(2) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(1) Video 2 Layer 1 : c(1) Layer 2 : c(1) Layer 3 : c(1) Globally Optimal Solutions Enhancement Layers Base Layer Enhancement Layers Video program 2 Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(6) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(6) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(5) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(5) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(4) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(4) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(2) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(2) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(1) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(1) …… 6 3 service profiles

16. Globally Optimal Solutions Proposition 1. – Denote the modulation rate of the l th layer video as M l. – M l +1 ≥ M l or M l +1 = 0. Enhancement Layers Base Layer Enhancement Layers Video program 2 M1M1 M2M2 M3M3 = c(4) = {c(4), c(5), c(6)} = {c(5), c(6)} = c(5) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(1) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(1) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(2) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(2) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(4) Video 2 Layer 1 : c(1) Layer 2 : c(3) Layer 3 : c(4)

17. Globally Optimal Solutions A Pricing Scheme – Each service profile can be associated with a utility-resource pair (u i, S i ) evaluated based on the previous formulas. Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(3) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(3) (u,S) = (500,70) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(4) Video 2 Layer 1 : c(1) Layer 2 : c(2) Layer 3 : c(4) (u,S) = (400,65) Video 2 Layer 1 : c(2) Layer 2 : c(3) Layer 3 : c(6) Video 2 Layer 1 : c(2) Layer 2 : c(3) Layer 3 : c(6) (u,S) = (350,50) Video 2 Layer 1 : c(3) Layer 2 : c(4) Layer 3 : c(6) Video 2 Layer 1 : c(3) Layer 2 : c(4) Layer 3 : c(6) (u,S) = (240,20) ……

18. Globally Optimal Solutions Given a scatter plot of u i versus S i : 1000 900 800 700 600 500 400 020406080100120 cost : Consumption of Slots Total Utility Value

19. Globally Optimal Solutions Observation – If u A > u B, the solution A is better than solution B – The best solution of is called Plausible service profile A(u A,S i ) B(u B,S i ) SiSi uAuA uBuB S u Plausible service profile

20. Globally Optimal Solutions Given a scatter plot of u i versus S i : 1000 900 800 700 600 500 400 020406080100120 cost : Consumption of Slots Total Utility Value

21. Globally Optimal Solutions Given a scatter plot of u i versus S i : 1000 900 800 700 600 500 400 020406080100120 cost : Consumption of Slots Total Utility Value

22. Globally Optimal Solutions Optimal solution: Select one plausible service profile such that – the total utility u is maximized – the total resource S is no larger than the preset bound G 0 The exponentially growing solution space makes it unrealistic to use exhaustive search in practical operation. An efficient heuristic search algorithm that yields reasonably good solution is desirable.

23. Heuristic Algorithm for Multiple Video Programs Our approach to developing a heuristic algorithm – to develop an efficient search method to arrive at the vicinity of an optimal solution more quickly Let us consider an example – 3 video programs with 100, 80, and 40 subscribers each – The resource constraint G 0 = 165 slots – Each video stream has 4 layers

24. Heuristic Algorithm for Multiple Video Programs Given a scatter plot of E(i) = u(i)/S(i) versus S(i) : Observation: – the averaged utility per slot values are approximately the same Good movieBad moviemovie with wrong modulation

25. Heuristic Algorithm for Multiple Video Programs Given a scatter plot of E(i) = u(i)/S(i) versus S(i) : Observation: – the averaged utility per slot values are approximately the same Conclusion: – Search for a solution in which the efficiency of every video program are approximately the same

26. Heuristic Algorithm for Multiple Video Programs Step 1: Let the range of the union of u/S values of all V video programs be divided equally into B sections. B=3 I II III 43 23 15

27. Heuristic Algorithm for Multiple Video Programs Step 2: Within each section, identify a plausible service profile such that the efficient value is closest to the lower boundary of that section. B=3 I II III For section I: mini E 1 =44 mini E 2 =43 mini E 3 =null 43 23 15 For section II: mini E 1 =23 mini E 2 =28 mini E 3 =null For section III: mini E 1 =17 mini E 2 =16 mini E 3 =15

28. Heuristic Algorithm for Multiple Video Programs Step 3: Exhaustively examine each of these B candidate solutions and select the one with maximum utility value while satisfying the resource constraint Select one plausible service profile in the section such that – the total utility u is maximized – the total resource S is no larger than the preset bound G 0 B=3 I II III

29. Simulation Simulation Set up – V = 3 video programs are multicast via WiMAX MBS services. – Each video program is SVC coded using L = 4 layers – Each video stream has an identical size R vl = 250 kbps. – The price of each layer is 5, 2.5, 1.5, 1 OBEE Algorithm – IEEE WCNC 2009 – maximize the number of users who can receive the basic video service – For each enhancement layer, the modulation will be selected for transmission until out of resource or all layers are allocated.

30. Simulation Our simulations show that no matter which scenario or resource budget is conducted, – the proposed method indeed creates higher utility value – close to the global optimum. G 0 =165 – the highest total utility is 11.8% higher than OBEE – the proposed algorithm is 7.46% higher than OBEE G 0 =128 – the proposed algorithm is 5.5% higher than OBEE

31. Conclusion We introduced an advanced algorithm for layered video multi-casting over mobile WiMAX. Our simulation shows that – Searching for plausible solutions quickly – the system indeed obtains higher utility value. It is also applicable for all OFDMA real time system.