Seamless Channel Transition for Pyramid-based Near-VOD Services

Student: Wei-De ChienAdvisor: Prof. Ja-Shung Wang

Channel Transition

The service delay of a popular movie should be shortened to satisfy a large number of clients. If a movie is no longer popular, part of the assigned channels can be released for other movies.

Once a bottleneck occurs in the broadcasting network, the server must decrease the bandwidth usage to conform the limitation.

Seamless Property of a Channel

Transition Any proceeding service should not be

interrupted or broken off during the channel transition.

The maximum service delay should be restricted by an acceptable value during the channel transition.

Fast Broadcasting (FB)

Skyscraper Broadcasting (SB)

,34mod),1(

24mod,2)1(2

14mod),1(

04mod,1)1(2

3,2,2

1,1

)(

iif

iif

iif

iif

i

i

if

Harmonic Broadcasting (HB)

Seamless Channel Transition (SCT) Scheme

Seamless Channel Transition (SCT) Scheme (continued)

Stairway Channel Transition (SWCT)

A channel may broadcast with old video segmentation now and new segmentation later. We will refer this channel as an “old channel” or a “new channel” according to the segments it broadcasts.

Old channels should be finished gradually to satisfy the requests with old video segmentation. This is the reason why we choose “stairway” as the name of our scheme.

To coordinate with the termination of old channels, new channels should be created progressively.

The starting time of a new channel should not be earlier than the ending time of its coincident old channel.

SWCT in FB – Negative Channel Transition

Cj => Cj+m, 1 ≤ j ≤ k-m

SWCT in FB – Positive Channel Transition

Ci+m => Ci, 1 ≤ i ≤ k

SWCT in FB – Positive Channel Transition (continued)

In a positive channel transition 

d = L/(2k-1) and d = L/(2k+m-1),

d/d = (2k-1)/(2k+m-1).

Ci+m starts (2i+m-1-1)∙d later than Tb, and Ci finishes

(2i-1-1)∙d later than Tb. 

(2i+m -1-1)∙d- (2i -1-1)∙d

= d∙(2i+m-1-1)∙(2k-1)/(2k+m-1) - d∙(2i-1-1)

= d∙(2k-2i-1)∙(2m-1)/(2k+m-1) > 0,

given i ≤ k; i, k and m are positive integers.

SWCT in FB – Buffer Requirement

TR(i): received time of Si .

TP(i): played time of Si.

TR(1)=TP(1)

SWCT in FB – Buffer Requirement (continued)

Lemma 1: The maximum buffer requirement in Ck is 2k-1-1 segments for the last cycle in our channel transition scheme.

Lemma 2: The overall buffer requirement is decided by the maximum buffer requirement respecting to the last channel Ck.

Theorem 1: The maximum buffer requirement for our proposed channel transition scheme is 2k-1-1 segments, which does not exceed the maximum buffer requirement in FB scheme.

SWCT in SB – Definition of The Last Cycle

1

1

.)()(i

jaP djfTiT

TR (i)= TP(i) - (f(i)-1)∙d

SWCT in SB – Definition of The Last Cycle (continued)

SWCT in SB – Negative Channel Transition

SWCT in SB – Failure in Positive Channel Transition

SWCT in SB – Failure in Positive Channel Transition (continued)

SWCT in SB – Solution One

SWCT in SB – Solution Two

SWCT in HB – Negative Channel Transition

SWCT in SB – Positive Channel Transition

Simulations – Bandwidth Waste (SWCT in FB)

Simulations – Bandwidth Waste (SWCT vs. SCT)

Simulations – Service Delay (SWCT vs. SCT)

SCT: longer service delay due to dummy video.

SWCT: either d or d’.

Simulations – Buffer Requirement (SWCT vs. SCT)

Simulations – Bandwidth Waste (SWCT in SB)

Conclusion

In this paper, we propose a new channel transition scheme over three different broadcasting schemes.

According to the simulation results, our proposed SWCT scheme is suitable for the regularly scheduled channel re-allocation in half or full day scale.

In comparison to the existing SCT scheme, our scheme causes smaller bandwidth waste. The client waiting time and buffering space requirement are kept the same as the performances in original FB scheme, and therefore smaller than those in SCT scheme. The d and d restrict the service delay for any request issued during the channel transition.