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next micro-round is available is
N S
m
V 1 =
Pr
next round available
|
P 0 }=
(12.45)
{
N S
1
This is also the probability for a client to wait one additional micro-round provided the
assigned micro-round is already fully occupied. It can be shown that the probability for a client
to wait k additional micro-rounds provided that the first k assigned micro-rounds are all fully
occupied is
N S
m
V k =
Pr
{
( k
+
1)th round available
|
P k }=
1
k
m
(12.46)
N S
k
We already know P 0 , and it can be shown that the probability for the first k micro-rounds
all being fully occupied is given by
m
k
1
i
m !( N S
k )!
P k =
=
k )! ,
1
k
m
(12.47)
N S
i
N S !( m
i = 0
Hence, we can solve for the probability of a client having to wait k additional micro-rounds
from
W k =
Pr
{
( k
+
1)th round free
|
P k }
P k
(12.48)
( N S
m ) m !( N S
k
1)!
=
,
1
k
m
N S !( m
k )!
Therefore, given m - the number of micro-rounds that are fully occupied - the average
number of micro-rounds a client has to wait can be obtained from
T F
1
N S
W avg ( m )
=
kW k +
+
(12.49)
k =
1
where the second term accounts for the additional delay as described in Theorem 12.1. Sim-
ilarly, given n - the number of active video sessions - the average number of micro-rounds a
client has to wait can be obtained from
N S 1
M avg ( n )
=
W avg ( j ) P full ( n
,
j )
(12.50)
j
=
1
And the corresponding average scheduling delay given a system utilization of n is
M avg ( n ) Q
R V
D S =
(12.51)
Substituting equations (12.43), (12.48), (12.49), and (12.50) into equation (12.51) gives the
desired result.
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