Contents Telektronikk - Telenor

36
- Mean response time depends only on
mean value (not higher order moments)
of service time
- For exponential service times a job
with mean service time will have
response time equal to that of a batch
system. Shorter jobs come out better
and longer jobs worse in PS than in
batch. In this sense PS comes between
batch and SJF.
17 Queuing network
analysis
Up till now we have considered only single
nodes, whether loss systems or
queues. In this article we have no ambition
of making any thorough investigation
into networks, as it is a rather complicated
matter that would far exceed the
present aim.
A network model consists of nodes and
links. In traditional networks signal propagation
time is of no particular concern.
Radio propagation has a speed of
300,000 km/s, and for cable transmission
the assumption is roughly 200,000 km/s.
Thus the one-way delay via a geostationary
satellite is 0.25 seconds, and that of a
cable transmission halfway around the
globe 0.1 seconds. Compared to telephone
set-up and conversation times
those delays are short, whereas in a real
time dialogue the 0.25 second delay is
critical. The delay on a 500 m local area
network is 2.5 µs, corresponding to transmission
of 25 bits with 10 Mb/s transmission
speed. With a CSMA/CD access
system even that short delay is of significance.
The interesting parameter is the
Propagation time – to – Transmission
time ratio: Tp /Tt = a. Propagation time is
a fixed value irrespective of signal type,
whereas transmission time is proportional
to frame length in bits and inversely
proportional to the bit rate.
For simple one-way transmission messages
may be chained continuously at the
sending end, and a utilisation of U = 1
can be obtained. If, on the other hand,
after one transmission a far end user
wants to start sending, the utilisation will
be maximum
U = 1/(1 + a) (99)
In data communication flow control is
often performed by acknowledgement,
adding one propagation time (plus a negligible
transmission time) to one transmission
time and one propagation time in
the forward direction. Thus the gross
time used per transmitted information
frame is Tt + 2Tp , and the utilisation
(transmission efficiency) is
U = Tt /(Tt + 2Tp ) = 1/(1 + 2a) (100)
Transmission of a 1 kb frame with
acknowledgement at 64 kb/s over a 10
km cable gives a utilisation U = 0.994,
whereas the same communication over a
satellite leads to U = 0.06.
In these estimates we have assumed that
all transmitted data is useful information.
There will always be a need for some
overhead, and the real utilisation will be
reduced by a factor I/(I + O), where
I = information and O = overhead.
17.1 Basic parameters and
state analysis
The discussion above is included to give
a realistic sense of the propagation delay
importance. In the following approach
we assume that the signal delay may be
neglected, so that a utilisation U = 1 can
be obtained. In data communication networks
information messages are usually
sent as serial bitstreams. Messages may
be sent as complete entities, or they may
be subdivided in frames as sub-entities
sent separately. The time that a link is
occupied, which is the holding time in a
traffic context, depends on the transmission
capacity C. In a similar way a computer
may be considered a server with a
certain capacity C. C may be measured
arbitrarily as operations per second or
bits per second. We can define a service
time by 1/µC, being the time to complete
a job, whether it is a computer job or a
message to be transmitted. Examples are
1 C: operations/second
1/µ: operations/job
µC: jobs/second
2 C: bits/second
1/µ: bits/message
µC: messages/second
If we stick to the second example, µC
means the link capacity that is obtained
by chaining messages continuously on
the link. Assuming that the messages
arrive at a rate λ < µC, we obtain an
average load on the link A = λ/µC. Note
that the product µC here replaces µ in
previous contexts. In a data network the
dynamic aspect of a flow of messages
along some path is often emphasised.
The number of messages per second, λ,
is often termed the throughput. The link
load A expresses the number of messages
arriving per message time, and is thus a
normalised throughput.
Assume a network of N nodes and M
links. The total number of message
arrivals to (= number of departures from)
the network per second is
N N
γ = ∑ ∑ γ jk
j =1 k =1
(101)
where γ jk is the number of messages per
second originated at node j and destined
for node k. For an arbitrary node i the
total number of arrivals per second
(external and internal) is
N
λi = γ i + ∑λ j ⋅ p ji
j =1
(102)
where pji = P{a customer leaving j proceeds
to i}. Since a customer can leave
the network from any node,
P{departure from network at node i} =
N
1− ∑ pij j =1
If each node i is an M/M/ni system,
departures from each node is Poisson,
and the nodes may be analysed independently.
This leads to the important property
of product form solution for the network,
which means that the joint probability
of states ki can be expressed as the
product of the individual probabilities:
p(k1 , k2 , …, kN ) =
p1 (k1 ) ⋅ p2 (k2 ) ⋅ … pN (kN ) (103)
The solution for pi (ki ) is the one developed
for the M/M/n waiting system, equations
(76):
p(i) = (A/i) ⋅ p(i – 1) = (Ai /i!) ⋅ p(0),
0 i n
and
p(i) = (A/n) ⋅ p(i – 1) =
(An /n!) ⋅ (A/n) i–n ≤ ≤
⋅ p(0),
n≤i≤∞ 17.2 Delay calculations
Delay calculations are fairly simple for
the network of M/M/1 nodes. The
sojourn time at a node i is expressed by
Ti = 1/(µCi – λi ) (104)
The average message delay in the network
can be expressed by [14]:
M
M
∑λ
iTi ∑λ
i
i=1
i=1
T = =
γ γ µCi−λi ( )
(105)