Section 13 Continuous Time Queueing Systems

Continuous-Time

Queueing Systems

Professor Izhak Rubin

Electrical Engineering Department

UCLA

[email protected]

2014-2015 by Izhak Rubin

Prof. Izhak Rubin 2

M/M/1 Queueing System

Server

For M/M/1 queueing system:

1. Messages arrive in accordance with a Poisson process with intensity [ / sec]

Interarrival time distribution: 1

2. Message service time is exponentially dis

t

mess

A t e u t

tributed with rate ;

mean service time = 1/ [sec/mess].

Message service time distribution: 1

3. A single service channel is provided.

4. Unlimited storage (buffer) capacity.

tB t e u t

Prof. Izhak Rubin 3

System Size Process For M/M/1 queueing system:

The system size process, { , 0}, {0,1,2,...}

is a continuous time birth-and-death Markov chain, with:

, 0,

, 1.

The traffic intensity is

t

i

i

X X t S

i

i

= total offered arrival rate / total service rate.

Prof. Izhak Rubin 4

M/M/1 Queueing System: System Size

0

0

0

0 1 1

0

1 2

0 0

0

0,

lim

,

From the CTBD analysis: .

...We have 1, , j 0 .

...

1 .

1 .

lim

i

i

tt

i

i

j i

i

j

j j

j

j

i

i

i i

i

i

tt

a

P X j

P j a

P j a a

a a

a

a

P X

0, 1 j 0.

1 , 1jj

Prof. Izhak Rubin 5

M/M/1 Queueing System: Average

Message Waiting and Delay Times

1

For 1, the steady-state mean queue size is:

, 1.1

The steady-state mean message delay and waiting time

\(using Little's Formula, noting that = for 1) are given as:

1, 1

1

D

E X

E D E X

E W E D

1 , 1.1

Prof. Izhak Rubin 6

M/M/1 Queueing System: Waiting

Time Distribution

Steady-state waiting time distribution:

lim ( ), t 0.

We set:

( ) = P{an arriving message finds j messages ahead of itself in the queue}.

We obtain

( ) (

nn

W t P W t

j

j P j

) (1 ) , 0,

by using a result that states that Poisson Arrivals See Time Averages (PASTA).

To calculate the waiting time distribution, we assume a FCFS service policy.

An arriving

j j

message that finds upon arrival that there are j messages ahead of itself, j 1, will have to wait

until these messages are served before it is admitted into service. The services times of these j mes

sages

are statistically independent and identically distributed; this is also true for the message that is

in the midst of its service when the message for which the wait time is computed arrives into the

system. Clearly, the distribution of the sum of j independent service times (actually, the sum of the residual

service time of the message in service plus the service times of the waiting j-1 messages) is equal to the j-th order

convolution of the service time distribution. When j=0, the arriving message enters service without having to wait.

Prof. Izhak Rubin 7

M/M/1 Queueing System: Waiting

Time Distribution (Cont.)

*

1

1

Hence, we write:

(0) 1 ( ) (1 ) ( ) , t 0.

We obtain:

1 ( ), t 0 ,

Prof. Izhak Rubin 8

Example 1: M/M/1 System

Model Consider a transmission processor operating on messages queued at the output

buffer of a router to transmit messages that are served in accordance with a FIFO policy

across the single output channel to which it is attached at a rate of R=1 Mbps. Assume

messages to arrive at this output buffer in accordance with a Poisson process at a rate of

800 Kbps. Assume messages to be of random lengths so that the transmission time of

each message is exponentially distributed. The average message length is equal to 50 Kbits.

Then:

This system is modeled as an M/M/1 queueing system with the following parameters:

800 Kbps messArrival rate = = 16 [ ]

50 Kbits/mess sec

1 Mbps messService rate = = 20 [ ].

50 Kbits/mess sec

Also:

1 50 Kbits/messMean message service time = average message transmission time =

1 Mbps

50 sec,

1 messconfirming that = 20 [ ].

50 sec sec

Hence:

Traffic intensity = = 0.8 1,

so that the system is stable and reaches as it evolves steady state behavior.

m

m

Prof. Izhak Rubin 9

Example 1 (Cont.)

1

1

he steady-state mean queue size is:

0.84 [messages].

1 1 0.8

The steady-state mean message delay and waiting times are computed as:

1 40.25sec 250 sec

1 16

250 sec 50 se1

T

E X

E D E X m

E W E D m m

1

1

c 200 sec.

The waiting time distribution is give as:

1 ( ), so that at steady state the probability that the waiting time is longer than t

is expressed by: P{W > t}= , 0. Hence,

t

t

m

W t e u t

e t

1 4

the probability that the message wait time is longer

than 1 sec is computed to be:

P{W > 1}= 0.8 0.0146.

Thus, 1.46% of the messages are experiencing waiting times that are longer than 1 sec.,

e e

noting that

the average message waiting time in the queue is equal to 0.2 sec.

Prof. Izhak Rubin 10

M/M/m Queueing System

For M/M/m queueing system:


2. The message service time is exponentially distributed with rate

3. The number of se

mess

rvice channels is , 1.

4. Unlimited storage(buffer, waiting room) capacity.

The system size process { , 0}, {0,1,2,...},

is a continuous time birth-and-death Markov chain, with

,

t

i

m m

X X t S

i

0,

min , , 1

The offered traffic load is [ ],

and the traffic intensity is .

i i m i

f Erlangs

m

Server 1

Server 2

Server m


M/M/m System-Size: Steady-State

0

0

0

0

0,

lim

,

From the CTBD analysis: , j 0.

, 0! !

We have 1,

, 1! !

To reach steady state:

i

i

tt

i

i

j i

i

j j

j

j j j

j m j j m

a

P X j

P j a

P j a a

fj m

j ja a

fj m

m m m m

a

0 0

1.

0, 1lim

, 1.

j

j

j j

tt

P X jP j


M/M/m Queueing System: Steady State

Behavior

1

0

0

For 1, the queue size steady-state distribution exists and is given by:

0 , 0

where 0

The steady-state mean queue size is:

, 1

, 1.

Departing message rate [m

j

j

j

j

P j P a j

P a

j P jX

,max ,max

,max ,max

, 1ess/sec] =

, 1

, 1Departing load [Erlangs] =

, 1

Noting that = m, = , the normalized throughput rate is given by:

, 1

1,

D

DD

D D

D D

D D

m

ff

m

f m

f

f

1.


M/M/m Queueing System: Waiting

Time Analysis

1

1

Using Little's Formula, for 1, = , so that we have:

Steady-state waiting time distribution: lim ( ), t 0.

Under a FCFS policy, a message that finds upon arrival j messag

D

nn

X D D X

W D

W t P W t

es in the system,

will not have to wait if j < m. In turn, if j m, the message will have to wait until the number of

messages ahead of itself in the system falls down to m-1 (so that a server beco

mes available

to admit it into service); in this case, its waiting time is equal to the sum of j-(m-1)=j+1-m time intervals between

message departures. But the departure process during this period is a

1

0

Poisson

process with departure rate equal to , so that each inter-departure time interval is exponentially distributed

with mean interdeparture time equal to 1/ . Hence, we write:

( )m

j

m

m

W t j

* 1

1

0 0

( ) (1 ) ( ) .

Noting that ( ) ( ),we have for 1:

( ) ( ) .!

j mm t

j m

j mtmm y

j j m

j e u t

j P j

m yW t P j P j e m dy

j m


M/M/m/N Queueing System M/M/m/N queuing system:


2. Message service time is exponentially distributed with rate [sec/mess].

3. The number

mess

of service channel is , 1.

4. System message storage (buffering, waiting room) capacity (including both

messages waiting in the queue and messages in service) is limited to N messages, .

The syst

m m

N m

em size process, { , 0}, {0,1,2,..., },

is a continuous time birth-and-death (CTBD) Markov chain, with

, 0 1

0,

min , , 1.

t

i

i

X X t S N

i N

i N

i m i

N

Server 1

Server 2

Server m


M/M/m/N Queueing System

1

1

0

Since the process is irreducible and finite, steady-state distribution always exists:

0 , 0!

0 , !

where: 0 , ! !

j

j

j m

j jm N

j mj j m

fP j m

jP j

fP m j N

m m

f fP f

j m m

1 1

/ .

The call blocking probability is given by:

0!

The message departure rate (message throughout) is:

1 1 [mess/sec]

The throughput load (Erlangs) is:

1 1

N

B N m

D B

D D B

fP N P N P

m m

P P N

f P P

[Erlangs].B f


M/M/m/N Queueing System: Waiting Time

Analysis

The steady-state probability that an arriving message that is into the system

(and thus not blocked) finds messages in the system is given by:

, 0 11 1

The message steady-sta

admitted

A

B B

j

j P jj j N

P P

1 1 * 1

0

1

0 0

1

0

te waiting time distribution lim is given by :

1

Since , 0 11 1

where 1, and 1 , we have:

11

1

nn

m N i mm t

A A

i i m

A

B

N N

A

j j

m

i

W t P W t

W t i i e u t

P i P ii i N

P P N

P j j

W t P i P iP N

1 * 1

, t 0.N i m

m t

i m

e


Special Case: M/M/1/N

Queueing System

i

1

We have: ,0 ; , 1. Hence, for f= = / :

(1 )0 = , 0 , for 1;

1

1 = , for 1.

N+1

The call blocking probability is given by:

.

The message

i

jj

N

B

i N N i

P j P j N

P N P N

1 1

departure rate (message throughout) is:

1 1 [mess/sec]

The throughput Erlang load rate is:

1 1 [Erlangs].

E(X)Mean message delay (system time) = E(D)= .

D B

D D B B

D

P P N

f P P f


Example 2: M/M/1/N System

Model Consider the same transmisison processor modeled by Example 1, except that

it is now assumed that the total storage capacity of the processor (including the

message in service and waiting messages) is equal to N messages.

Then:

This system is modeled as an M/M/1/N queueing system with the following parameters:

800 Kbps messArrival rate = = 16 [ ]

50 Kbits/mess sec

1 MbpsService rate = =

50 Kbits/mess

mess

20 [ ].sec

Traffic intensity = = 0.8 1.

Since the system's storage capacity N is finite, the system is stable and reaches as it

evolves steady state behavior.


Example 2: M/M/1/N System

Model (Cont.)

1

The message blocking probability is given by:

(1 )= , for 1.

1

Hence we compute the following results.

For N = 4 messages, we obtain the message blocking probability (blocking ratio) to

N

B NP N P N

be

equal to 0.1218, so that 12.18% of arriving messages are blocked (not admitted into the buffer).

Then, the message departure rate (throughput) is equal to:

1 1 =16(1-0.1218)= 14.05 [mess/D BP P N sec] = 702.5 Kbps.

In turn, if we double the stirage capacity level, so that we set N=8,

we obtain the message blocking probability (blocking ratio) to be

equal to 0.0387, so that only 3.87% of arriving

messages are blocked (not admitted into the buffer).

Then, the message departure rate (throughput) is equal to:

1 1 =16(1-0.0387)= 15.38 [mess/sec] = 769 Kbps..D BP P N


M/M/m/m Queueing System For the , no queueing space is provided.

An arriving call (message) is blocked and discarded

M/M/m/m QS

if it finds upon arrival that

all service channels are busy, serving other messages.

S

m

ystem Size Process = X = CTBD process; S = {0,1,2,...N} with N = m.

for N i 0; , for N i 1. Hence:

, 0; [Erlangs].! !

The steady state system-size distribution P={P

i i

i i

i i

i

fa i f

i i

0

0

(j), j S} always exists

and is given as follows:

!, 0 .

!

j

m

j k m kk

k

fj

P j a a j mf

k

Server 1

Server 2

Server m


Erlangs Loss Formula

0

: (for the call blocking probability):

.! !

where: m = number of service channels;

f = offered load rate =

Erlang's Loss For

[Erlangs].

Note: This blocking proba

mula

m km

B

k

f fP P m

m k

bility formula also holds for the M/G/m/m system model.

Messages incur no waiting time, W = 0.

Message mean delay time (as confirmed by using Little's

Little's formula) is thus equal to the message m

ean service time:

1.

1 B

XD

P


Example 3: M/M/N/N System

Model Consider a central exchange switch in a telephone system of a small organization

that does not provide queueing facility for storing waiting calls. Arriving

calls that find the system to the busy are blocked and assumed lost. Assume the switch to

be served by N=m service channels (transmission trunks).

Assume the switching system to be loaded by calls that arrive in accordance with a

Poisson procecalls

ss at a rate of = 12 [ ]. Assume the duration of each call to be hour

an exponentially distributed random variable whose average value is equal to

14 minutes.

This system is modeled as an M/M/m

/m queueing system with the following

parameters (noting the message to now be represented as a call):

callsArrival rate = = 0.2 [ ]

min

callsService rate = = 0.25 [ ].

min

Traffic intensity and traffic

loading is = = f = 0.8 [ ].Erlangs


Example 3: M/M/N/N System

Model (Cont.)

0

: (for the call blocking probability), we have:

,! !

where: m = number of service channels;

Using Er

lang's Loss Form

f = offered load rate = [Erlang

ula

m km

B

k

f fP P m

m k

s]= 0.8 Erlangs.

Hence,we obtain the following results for the blocking probabilities for different values of

selected system storage levels:

For N=m= 1, we obtain: 44.44%, so that 44.44% of arriviBP ng calls are blocked;

For N=m= 2, we obtain: 15.09%, so that 15.09% of arriving calls are blocked;

For N=m= 3, we obtain: 3.86%, so that only 3.86% of arriving calls are blocked, while

B

B

P

P

the remainder 96.64% of arriving messages are admitted into the system and are

immediately served without incurring any waiting time.

Section 13 Continuous Time Queueing Systems

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