Contents Telektronikk - Telenor

where q(i) = 0 for i < 0 and i > M.
The finite buffer model may be analysed
using an iterative approach by solving
the governing equations (2.37) numerically.
However, we take advantage of the
analysis in section 2.1 and we apply the
same method as for the infinite buffer case.
We introduce the generating functions
Q M (z) =
and by using (2.37) we get:
where
and
M�
q(i)z i ,
i=0
Q M z − 1
(z) =
zn − A(z)
⎡
n−1 �
⎣ P M j z j − z M+n
∞� 1 − zs 1 − z gM ⎤
(s) ⎦
j=0
P M j =
j�
i=0
p M i
p M i = �
i1+i2=i
q(i1)a(i2)
(2.38)
is the probability that there are i cells in
the system at the end of a slot, and is the
probability that exactly s cells are lost
during a slot.
By examining (2.38) we see that for the
major part of the queue-length distribution
is of the same form as for the infinite
buffer case:
Q (2.39)
M min[i,n−1] �
(i) = P M j Θ(i − j)
j=0
The coefficients P j M’s are determined by
the fact that (2.39) gives the queue length
distribution also for i = M,...,M+n-1, and
whence:
n−1 �
P M j Θ(M + k − j) =1; k =0, 1, ..., n − 1
j=0
s=1
(2.40)
The n equations (2.40) are non-singular
and determine the P j M’s uniquely. For
small buffers (2.40) is well suited to calculate
the coefficients. However, to avoid
numerical instability for larger M we
expand (2.40) by taking partial expansion
of the function Φ. After some algebra we
get:
�
n−1
P
j=0
M j
= δ 1,l (n – A) (2.41)
where
A(rl,rs) =
�
r j
�
l + r
s=n+1
j � � �
M
rl
sA(rl,rs)
rs
�� n
�
nA(rl) − rlA ′ (rl)
nA(rs) − rsA ′ �
(rs)
k=1,k�=s
� n
k=1,k�=l
�
−1 rs − r
(2.42)
−1�
�
k
� �
−1
rl − r−1
k
The generating function for the P M’s j in
this case may be expressed as:
n−1 �
P M j x j
j=0
= (n – A)E 1 (I + W) -1 V(x) (2.43)
where E1 is the n dimensional row vector
E1 = (1,0,....,0), and W =(W(i,l)) is the
n × n matrix:
W (i, l) = �
� �M+n−1 rl
s=n+1
(2.44)
and V(x) is the n dimension column vector
V(x) = (V l (x)):
(2.45)
The form of (2.41) is useful to determine
the overall cell loss probability. Inserting
z = 1 in (2.38) we obtain the mean volume
of cells lost in a slot;
∞�
E[VL] = sg M (s) :
(2.46)
The volume lost can also be determined
by direct arguments since we have that the
rs
� nA(rl) − rlA ′ (rl)
nA(rs) − rsA ′ (rs)
�n k=1,k�=l
�
Vl(rs)Vi(rs)
(x − rk)
Vl(x) = �n k=1,k�=l (rl − rk)
s=1
n−1 �
E[VL] = P M j − (n − A)
j=0
lost volume of cells in a slot equals the
offered load minus the carried load i.e.
⎛
n−1 �
E[VL] =A − ⎝ jp M ⎞
M�
j + ⎠
which equals (2.46).
From (2.41) taking l = 1 we get the cell
loss probability
pL = E[VL]
A
evaluating we get
j=0
j=0 s=n+1
(2.47)
(2.48)
For large M we expand (I + W) -1 in
(2.48), and by taking only the first term
in the expression for W we get the following
simple approximate formula for
the cell loss probability:
where P is the product
(2.49)
(2.50)
By using (2.14) and (2.15) we obtain the
following integral expression for P:
1
P =
(2.51)
(n − A) (rn+1 − 1) exp(I)
where I is the integral
j=n
np M j
as:
pL = −1
n−1 �
A
�
P M j r j �
1
sA(1,rs)
(n − A)
pL = −
A E1(I + W ) −1 WE T 1
(n − A) 2
rs
� M
pL ≈
A[rn+1A ′ (rn+1) − nA(rn+1)]
� �M+n−1 1
P
rn+1
2
P =
� n
k=2 (rn+1 − rk)
� n
k=2
(1 − rk)
(rn+1) n
I = 1
�
(rn+1 − 1)
2πi |ζ|=r (rn+1 − ζ)(ζ − 1)
�
log 1 − A(ζ)
ζn �
dz
(2.52)
211