Contents Telektronikk - Telenor

220
Notes on a theorem of L. Takács
on single server queues with feedback
BY ELIOT J JENSEN
Abstract
L. Takács published in 1963 a paper
introducing a queuing model with
feedbacks. Such models may be used to
calculate the performance of certain
telecommunication processor systems.
This note proposes alternative formulations
of some of the results in the
paper. They may be convenient when
calculating response times for strings
of processing tasks representing e.g.
call handling functions. The impact of
task scheduling on the range of
response times has been indicated.
Background
L. Takács [1] examines the system
sojourn times of tasks offered to a single
server, with an infinite queue in front of
it. Tasks arrive to the system either from
outside, according to a Poisson process
with intensity λ, or as feedbacks. A feedback
may be generated at the termination
of the service of a task, and immediately
put at the tail of the queue. The probability
for this is p and for no feedback
q =1–p. Tasks are assumed to have
independent but identically distributed
service times. This distribution is
assumed to have the Laplace-Stieltje
transform Ψ(s). Let its mean value be α.
Theorem 3 of the Takács paper considers
the stationary distribution of the total
sojourn time θn (n = 1, 2, ...) for a random
task and its feedbacks, and reads,
cit:
If λα < q, then θn has a unique stationary
distribution P{θn ≤x},
which is
given by the following Laplace-Stieltje
transform
∞�
Φ(s) =q p (1)
k−1 Uk(s, 1) (R(s) ≥ 0)
where
k=1
U1 (s,z) = P0Ψ(s + λ(1 – z))
+ U(s + λ(1 – z),
(q + pz)Ψ(s + λ(1 – z)))(2)
for R(s) ≥0
and |z| ≤1,
P0 = 1 – λα/q,
U(s,z) is defined by (20) of [1], and
Uk+1 (s,z) = Ψ(s + λ(1 – z))
Uk (s,(q + pz)Ψ
(s + λ(1 – z))) (3)
for k = 1, 2, ...
The formula U(s,z) (eq. 20 of [1]) is defined
as the combined Laplace-Stieltje
transform and z-transform of a distribution
P j (t), giving the probability of, in
the stationary state (assuming λα < q), to
encounter exactly j (> 0) tasks in the system,
where the task in the server has
remaining service time shorter or equal
to t. It reads:
U(s, z) =
�
1 − λα
�
q
(4)
λz(1 − z)(Ψ(s) − Ψ(λ(1 − z)))
(z − (q + pz)Ψ(λ(1 − z)))(s − λ(1 − z))
We also note that Uk (s,z) is the Laplace-
Stieltje transform and z-transform of the
combined distribution of the total sojourn
time of a task and its feedbacks, and the
number of tasks left behind when the last
of its feedbacks leaves the server, provided
the number of feedbacks is exactly
k – 1 for that particular task.
From theorem 3 is obtained
∞�
Φ(s, z) =q p k−1 Uk(s, z) =
k=1
qU1(s, z)+pΨ(s + λ(1 − z))
Φ(s, (q + pz)Ψ(s + λ(1 − z)))
Then the moments
E {θ r n} =(−1) r
� �
r ∂
Φ(s, z)
∂sr for the stationary sojourn time process
may be found by solving a set of r +1
linear equations for the determination of
Φij =
� �
i+j ∂ Φ(s, z)
∂s i ∂z j
Takács has given explicit formulae for
E{θ n r } for r = 1 and 2.
(5)
(6)
An alternative formulation
In the recurrence formula for Uk (s,z) in
theorem 3, make the substitution
ξ1 (s,z) = (q + pz)Ψ(s + λ(1 – z)) (7)
and introduce the recurrence
( s=0
z=1)
( s=0
z−1)
ξk (s,z) = (q + pξk–1 (s,z))
Ψ(s + λ(1 – ξk–1 (s,z))) (8)
starting with ξ 0 (s,z) = z. Then, by rearrangement
we obtain
Uk+1(s, z) =
� k−1
U1(s, ξk(s, z))
(k = 0, 1, 2, ...)
(9)
It is easily seen that ξℓ(s, z) is the
Laplace-Stieltje transform and z-transform
of the combined distribution of system
sojourn time and the number of tasks
left in the system for a task which needs
to enter the server ℓ times, but always
with zero service time, the task joining
the system finding only one (other) task
there, which is just about to enter service.
It follows that
for |z| ≤1
and λα < q.
�
�
Ψ(s + λ(1 − ξℓ(s, z)))
ℓ=0
lim
ℓ→∞ ξℓ(s, z) = ¯ B(s)
¯B(s) is the Laplace-Stieltje transform of
the busy period distribution.
For convenience the formula for Uk (s,z)
may be slightly rewritten, using the ztransform
and the Laplace-Stieltje transform
of the number of tasks in the system
and the remaining service time of the
task in the server for the stationary process
at a random time. If the system is
empty, the remaining service time is zero.
Based on theorem 1 of [1] we may put
�
P (s, z) = 1 − λα
�
q
�
λ(1 − z)
1+z
s − λ(1 − z)
�
Ψ(s) − Ψ(λ(1 − z))
·
z − (q + pz)Ψ(λ(1 − z))
(10)
A task (which we may call a tagged task)
arriving at a random instant to the system
will see this distribution. When the
remaining service time finishes, a feedback
may, or may not be generated. At
the moment the server is about to start
processing the next task, the tagged task
will see a number i of tasks ahead of it in
the queue and a number j of tasks behind,
and at that time it will have waited a certain
time t (identical to the remaining service
time of the task initially in the
server). The Laplace-Stieltje and z-transform
of the corresponding distribution is
�
Q(s, y, z) = 1 − λα
�
(1 + (q + pz)
q
λ(1 − y)
s + λ(1 − z) − λ(1 − y) ·
�
Ψ(s + λ(1 − z)) − Ψ(λ − y))
y − (q + py)Ψ(λ(1 − y))
(11)