1. March 15, 2001Mark Kalman - ee368c Analysis of Adaptive Media Playout for Stochastic Channel Models Mark Kalman Class Project EE368c

2. March 15, 2001Mark Kalman - ee368c Overview Background Previous Work New Work Experimental Results Future Work Conclusions

3. March 15, 2001Mark Kalman - ee368c Background: Streaming media Server Client Server pushes frames over the network at regular intervals Client plays out frames as they arrive Unhappy User Playout is jerky because packets don’t arrive at regular intervals. Network Network adds jitter and losses to the packet stream

4. March 15, 2001Mark Kalman - ee368c Background: Streaming media Server Client User less unhappy but still irritated at buffering delays Network Buffer at client smoothes out variance of the channel and allows retransmissions

5. March 15, 2001Mark Kalman - ee368c Background: Adaptive Playout Server Client Jubilant user! Network Adaptive Playout allows a smaller buffer at the client (and thus smaller delays)

6. March 15, 2001Mark Kalman - ee368c Past Work (1) Yuang 1998: Random packet inter-arrivals times OnOff good Off     Markov channel model p 2,1 p 2,n p 2,5 12345n Markov buffer model A state represents the number of packets in the buffer. Transitions occur each time a frame is removed to be played out

7. March 15, 2001Mark Kalman - ee368c t f1 t f2 Frame leaves buffer frame period p 2,1 p 2,n p 2,5 12345n GBGGBBBBGGGG Channel states Past work (2) Finding the transition probabilities:

8. March 15, 2001Mark Kalman - ee368c New Work Assume random inter-arrival times. In ‘Bad’ state packets arrive in error with probability 1 – P c. Simplify by considering errors as a reduction in bandwidth. Use Yuang’s Markov Buffer model.

9. March 15, 2001Mark Kalman - ee368c Scenario 1 30 second clip streamed at the maximum rate of the channel The distribution on the state of the buffer is (what you see in each frame on the left) is given by: d n = P n d 0 Where P is the transition matrix, n is frame number, d is the distribution. Probability of underflow shown by the red marker at state zero.

10. March 15, 2001Mark Kalman - ee368c Scenario 2 Long program with rebuffering after underflow. Streamed at the maximum rate of the channel. Probability of underflow is the probability on buffer state zero in the stationary distribution.

11. March 15, 2001Mark Kalman - ee368c Scenario 3 Based on Steinbach (2001). Long programs streamed at 90% of the maximum rate. Long good channel periods between bad states. Probability of underflow is:

12. March 15, 2001Mark Kalman - ee368c Results: Scenario 1 Probability of underflow vs. target buffer size for a 30 second clip.

13. March 15, 2001Mark Kalman - ee368c Results Scenario 2 Mean time between buffer underflows vs. target buffer size

14. March 15, 2001Mark Kalman - ee368c Results: Scenario 3

15. March 15, 2001Mark Kalman - ee368c Future Work In scenario 1 results not accurate for long good and bad channel durations. Time average of arrival rate over short clip does not converge to ensemble average. Solution is to run two models in parallel one with good and one with bad initial state and take weighted sum of the two paths.

16. March 15, 2001Mark Kalman - ee368c Conclusions Shown an analysis that accurately predicts underflow probabilities for stochastic channel models. More work needs to be done to apply the analysis to a more general set of channel conditions.

17. March 15, 2001Mark Kalman - ee368c Appendix: Scenario 1 Scheme not accurate when P c is too low.

18. March 15, 2001Mark Kalman - ee368c Appendix 2: Simple method works well Frequency of underflow accurately predicted by target buffer size vs. average rate of loss

19. March 15, 2001Mark Kalman - ee368c Appendix 3: Long channel states Analysis is not accurate for when channel stays in the bad and good states for long periods