Contents Telektronikk - Telenor

18
Frame 5
Equilibrium calculations for some typical cases
The Poisson case
The simplest assumption is that the number of sources (N) and
the number of servers (n) are both unlimited, while each
source j contributes with an infinitesimal call rate (λj), so that
the overall call rate is state independent λi = λ = N ⋅λj . The
departure rate in state i is proportional to the number in service,
which is of course i: µ i = i ⋅ µ= i/s, where s is the mean
holding time per call.
The parameters λ and µ are constants, and equation (43) becomes
p(i) ⋅λ= p(i + 1) ⋅ µ ⋅ (i + 1) (48)
leading to
p(1) = (λ/µ) ⋅ p(0)
p(2) = (λ/2µ) ⋅ p(1) = (λ/µ) 2 /2! ⋅ p(0)
:
:
:
p(i) = (λ/µ) i /i! ⋅ p(0)
:
:
Thus, using equation (44) we get
∞
∑ p(i) = p(0)⋅ (λ / µ )
i=0
i
∞
∑ / i!
i=0
= p(0)⋅e λ / µ −λ / µ
= 1 ⇒ p(0) = e
and
p(i) = e –λ/µ ⋅ (λ/µ) i /i!
Using Little’s formula: A = λ ⋅ s = λ/µ, we obtain
p(i) = e –A ⋅ Ai /i! (49)
which we recognise as the Poisson distribution with parameter
A. The mean value of the distribution is also A, as it ought to
be. Recalling the interpretation of Little’s formula we can
state that
A = λ/µ = λ ⋅ s = average number of customers being served
= average traffic intensity.
The truncated Poisson case (the Erlang case)
In practice all server groups are limited, so the pure Poisson
case is not realistic. A truncation by limiting the number of
servers to n leads to a recursive equation set of n equations
(plus one superfluous) of the same form as that of the unlimited
Poisson case above. The normalising condition is different,
as
n
∞
∑ p(i) = 1and ∑ p(i) = 0
i=0
i=n+1
The truncated Poisson distribution thus becomes
0 ≤ i ≤ n
p(i) = 0, i > n
(50)
Any call arriving when i < n will find a free server. Since λ is
independent of i, the probability of an arbitrary call finding all
servers busy is equal to the probability that i = n:
(51)
This is the Erlang congestion (loss) formula for an unlimited
number of sources. The formula is not very practical for calculations,
so there are numerous tabulations and diagrams
available.
In principle one must distinguish between time congestion
and call congestion, time congestion being the probability of
finding all servers busy at an arbitrary point in time, and call
congestion being the probability that an arbitrary call finds all
servers busy. In the Erlang case the two probabilities are identical,
since all calls are Poisson arrivals, independent of state.
The binomial (Bernoulli) case
We assume a limited number of traffic sources (N) and
enough servers to cover any need, i.e. n ≥ N. Furthermore,
each free traffic source generates λ calls per time unit (λ is
here the individual rate, rather than the total rate, and busy
sources naturally do not generate any calls). Departure rate
per call in progress is unchanged µ = 1/s. The equilibrium
equations then take the form
p(i) ⋅ (N – i) ⋅ λ = p(i + 1) ⋅ µ ⋅ (i + 1) 0 ≤ i ≤ N (52)
By recursion from i = 0 upwards we get
p(i) = p(0)⋅
b = λ/µ = offered traffic per free source
N i
⎛ ⎞ ⎛ λ ⎞
⎝ i
⋅
⎠ ⎜ ⎟ = p(0)⋅
⎝ µ ⎠
N ⎛ ⎞
⎝ i ⎠ ⋅bi
The normalising condition is
(53)
The p(i)-expression above is seen to consist of the common
factor p(0) and the ith binomial term of (1 + b) N . Thus, by the
normalising condition, we obtain
p(0) = (1 + b) –N
and
p(i) =
P(n, A) = E(n, A) =
N
∞
∑ p(i) = 1and ∑ p(i) = 0
i=0
i= N +1
p(i) = N ⎛
⎝ i
A i
i!
A j
n
∑
j =0
j!
⎞
⎠ ⋅bi ⋅ 1+ b
A n
n!
A j
n
∑
j = 0
j!
( ) − N = N ⎛ ⎞
⎝ i ⎠ ⋅ai N −i
⋅( 1− a)
Here, a = b/(1 + b) = λ /(λ + µ) = offered traffic per source.
(54)