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are
n
active video sessions and
m
fully-occupied groups, the number of ways to distribute
the remaining (
n
m
) groups with none of
those groups fully occupied can be obtained from equation (10.29) as
N
(
n
−
m
/
g
) video sessions among the remaining (
G
−
−
m
/
g
,
G
−
m
,
−
(
/
g
)
1). Hence the total number of ways for exactly
m
of the groups fully occupied is
given by
G
m
N
(
n
m
g
,
,
g
−
N
full
(
n
,
m
)
=
−
G
−
m
1)
(10.30)
The probability of having
m
fully-occupied groups given
n
active video sessions can then be
obtained from
N
full
(
n
,
m
)
P
full
(
n
,
m
)
=
(10.31)
N
(
n
,
G
,
)
Knowing this, we can derive the average scheduling delay in the following way. Given
m
out
of
G
groups are fully occupied, the probability for the assigned group to be available (not fully
occupied) is given by
G
−
m
V
0
=
(10.32)
G
Hence
P
0
=
V
0
) will be the probability of the assigned group being fully occupied. It can
be shown that the probability for a client to wait
k
additional groups provided that the first
k
assigned groups are all fully occupied is
(1
−
G
−
m
V
k
=
Pr
{
(
k
+
1)th group available
|
P
k
}=
k
,
1
≤
k
≤
m
(10.33)
G
−
and the probability for the first
k
groups all being fully occupied is
m
k
−
1
−
k
m
!(
G
−
k
)!
P
k
=
=
k
)!
,
1
≤
k
≤
m
.
(10.34)
G
−
i
G
!(
m
−
i
=
0
Hence, we can solve for the probability of a client having to wait
k
additional groups, denoted
by
W
k
, from
(
G
−
m
)
m
!(
G
−
k
−
1)!
.
(10.35)
Therefore, given the number of groups that are fully occupied
m
, the average number of groups
a client has to wait can be obtained from
W
k
=
Pr
{
(
k
+
1)th group free
|
P
k
}
P
k
=
,
1
≤
k
≤
m
G
!(
m
−
k
)!
τ
m
G
T
F
W
avg
(
m
)
=
kW
k
+
+
1
(10.36)
k
=
1
Similarly, given the number of active video sessions
n
, the average number of groups a client
has to wait can be obtained from equations (10.31) and (10.36) as follows:
G
−
1
M
avg
(
n
)
=
W
avg
(
j
)
P
full
(
n
,
j
)
(10.37)
j
=
1
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