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Chapter 4

Fundamental Queueing System

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Ref: Mischa Schwartz “Telecommunication Networks” Addison-Wesley publishing company 1988

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nn

nnn

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(input rate/output rate) (the probability that the system is nonempty)

The throughput (customer/see) = λ

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net arrival rate= = r

The throughput r/μ= )1()1(

111)1( 110

N

N

NP

)1()/( BPseecustomerthroughput

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Example 1 Statistical Multiplexing Compared with TDM and FDM

Assume m statistically iid Poisson packet streams each with an arrival rate of packets/sec.

The packet lengths for all streams are independent and exponentially distributed.

The average transmission time is . If the streams are merged into a single Poisson stream, with rate , the

average delay per packet is

If, the transmission capacity is divided into m equal portions, as in TDM and FDM, each portion behaves like an M/M/1 queue with arrival rate and average service rate . Therefore, the average delay per packet is

.

m

1

1T

mm

mT

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Example 2 Using One vs. Using Multiple Channels Statistical MUX(1)

A communication link serving m independent Poisson traffic streams with overall rate .

Packet transmission times on each channel are exponentially distributed with

mean . The system can be modeled by the same Markov chain as the M/M/m queue. The average delay per packet is given by

An M/M/1 system with the same arrival rate and service rate (statistical multiplexing with one channel having m times larger capacity), the average delay per packet is

and denote the queueing probability

1

m

mP

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mP

mT Q

ˆ1ˆ

QP Qp̂

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When << 1 (light load) , , and

At light load, statistical MUX with m channels produces a delay almost m timeslarger than the delay of statistical MUX with the m channels combined in one.

When , , , << , and

At heavy load, the ratio of the two delays is close to 1.

QP 0 0

QP 1 1 1 )(1 mQp̂

Qp̂

mTT

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A.

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second moment of service time and load

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Roll-call PollingStations are interrogated sequentially, one by one, by the central system, which

asks if they have any messages to transmit.

: walk time : frame transmission time

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it

time

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The scan or cycle time is given by

The average scan time

, are the ave. walk time and the ave. time to transmit pkt at station .L is the total walk time of the complete poling system.

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iw it i

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For station , let : the ave. pkt arrival rate : the ave. packet length : the number of overhead bits C : the channel capacity in bps : the ave. frame length in time

The average number of packets waiting to be transmitted when station ispolled is , the time required to transmit is

With the traffic intensity, the average scan time is given

With representing the total traffic intensity on the common channel.

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For small the average access delay should be .

Assume that each station has the same , same frame-length statistics,and the same .

The average access delay is

is the second moment of the frame length, .The access delay is the average time a packet must wait at a station from the time it first arrives until the time transmission begins. Access delay is thusthe average wait time in an M/G/1 queue.

Ref: Mischa Schwarty: “Telecommunication Networks, Protocols Modeling and Analysis”, Addison-Wesley Publishing Company, 1988, PP. 408-422

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Hub Polling� Control is passed sequentially from one station to another.

� Let the polling message be a fixed value, tp sec in length.

� The time required per station to synchronize to a polling message is ts sec.

� The total propagation delay for the entire N-station system is sec.

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Hub Polling (Cont’)� The total walk time for roll-call polling is .

� Let the stations all be equally spaced, and the round-trip propagation delay between the controller and station N be τ sec.

� The overall propagation delay is just

� The analysis of the hub-polling strategy is identical to that of roll-call polling.

� The only difference is that the walk time L is reduced through the use of hub polling.

� For hub polling, .shub NtL

sp NtNtL

)1(2

N