Seamless Channel Transition for Pyramidbased Near-VOD Services
Student: Wei-De Chien
Advisor: 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)
1, i  1


2, i  2,3

 2 f (i  1)  1, i mod 4  0
f (i )  
 f (i  1), i mod 4  1
 2 f (i  1)  2, i mod 4  2

 f (i  1), i mod 4  3
,
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
i 1
TP (i)  Ta   f ( j )  d .
j 1
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
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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.