Open-Loop Algorithms - Scalable Continuous Media Streaming Systems - page 323

We can determine the number of channels in group j from solving for n 2 , j in

n 1 + n 2 , 0 + ... + n 2 , j − 1 n 1 + n 2 , 0 + ... + n 2 , j

B i ≤ C and B i > C (18.22)

i = h j + 1 i = h j + 1

where B i = W j for all i 's in the range g j ≤

<

g j + 1 .
To prove playback continuity for media segments broadcast in Type-II channels, we consider
an arbitrary Type-II group, say group j , comprising media segments

i

{ L i | g j ≤ i < g j + 1 } .As

the client begins receiving all channels in group j at time

t 0 + ( m + h j ) · U (18.23)

and it takes ( U W j seconds to completely receive the media segments, the time s j at which

all media segments in the group is ready for playback can be computed from

· b ) /

·

U b

s j = t 0 + ( m + h j ) · U + (18.24)

W j

Substituting W j from equation (18.19) into equation (18.24) we obtain

s j = t 0 + ( m + h j ) · U + ( g j − h j ) · U

= t 0 + ( m + g j ) · U

≤ t 0 + ( m + i ) · U , for g j ≤ i < g j + 1 (18.25)

which is equal to or earlier than the playback schedule and thus guaranteeing playback conti-
nuity.

18.6.2 Client Buffer Requirement
Compared to CB, GCB generally requires more client buffer because all but the first media
segments in a Type-II group are not transmitted in a just-in-time manner. Instead, they are
transmitted at a higher rate so that reception can be completed at the same time as the first media
segment. Consequently, thesemedia segments are received completely before the playback time
and thus occupy more client buffer.
Specifically, the client will play back media segment L i at time t 0 + U , where t 0

is the time the client entered the system. We define H i as the amount of media data received
but not yet played back at time t i as defined in (18.14). As channel switching occurs only
at the time instants t 0 +

( m + i ) ·

we only need to consider the buffer
occupancy at these instants. We can compute H h j recursively from

( m + h j ) · U for j = 0 , 1 ,...,

n 1 −

k = 0

1

m · U · B k ,

for j = 0

H h j + 1 =

n 1 + n 2 , 0 + ... + n 2 , j − 1 − 1

k = h j

for j > 0

H h j − 1 + 1 − ( h j − h j − 1 ) · U · b + ( h j − h j − 1 ) · U · B k ) ,

(18.26)