Networking Reference
In-Depth Information
a( t )
b( t )
8
7
6
5
a(t)
4
3
N (t)
2
T 2
1
T 1
b(t)
0
t
t 1
t 2
t
Figure 4.2 Proof of Little's theorem.
arrive and packet arrival is random (inter-arrival time is not equal). Packets' service time
can also differ. Notice how the queue in this example can be empty at times and can have
up to three packets at a time.
The shaded area between α(τ) and β(τ) can be expressed as:
t
N(τ)dτ.
(4.8)
0
0, with the queue being empty, and if we select t such
that the queue is also empty at time t , the shaded are can also be given by:
If we start our analysis from τ =
α(t)
T i .
(4.9)
i =
1
Equating Equations 4.8 and 4.9 and dividing both sides by t . We get:
α(t)
i =
t
α(t)
1
t
1
t
α(t)
t
1 T i
α(t)
N(τ)dτ =
T i =
.
0
i = 1
Using Equations 4.1, 4.3, and 4.5, the above equality gives:
N t = λ t T t .
(4.10)
Taking the limits N t N , λ t λ , T t T , we reach Little's theorem in Equation 4.7
Note that this queue is assumed to eventually become empty since the service rate is greater
than the arrival rate. Asymptotically, the queue will become empty at some given time.
 
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