Case Study Solution Network of Queues

8/6/2019 Case Study Solution Network of Queues

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Probability and RandomProcesses

STA 3533

Module 6

Networks of Queues

Continue


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Topics

Case Study Solution

Introduction to Queuing

Open queuing networks

Closed queuing networks


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Problem 1

A data communication line delivers a block of informationevery 10 microseconds. A decoder checks each block forerrors and corrects the errors if necessary. It takes 1 s todetermine whether a block has any errors. If the block haserror, it ta es s to correct it, an i it as more t an oneerror it takes 20 s to correct the error. Blocks wait in aqueue when the decoder falls behind. Suppose that the decoderis initially empty and that the numbers of errors in the first 10

blocks are 0, 1, 3, 1, 0, 4, 0, 1, 0, 0.a.Plot the number of blocks in the decoder as a function of time.

b.Find the mean number of blocks in the decoder.

c. What percentage of the time is the decoder empty?


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Solution 1:

1. Interarrivals are constant with

interarrival times = 10 msec

2. Service time

if 0 error 1 sec

1 error 1+5 sec

>1 error1+20 sec

Arrival time 10 20 30 40 50 60 70 80 90 100

Errors 0 1 3 1 0 4 0 1 0 0

Service time 1 6 21 6 1 21 1 6 1 1

Dep. Time 11 26 51 57 58 81 82 88 91 101


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a. The plot

N(t)

b.

96.0

]1162320101123201061[100

1

)(100

1 110

10

=

+++++++++++++=

dttN

10 3020 40 6050 70 80 90 100 110

TObservation = 100 msec


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c.

Server is working during 65 msec = service times

proportion idle time = 1 65/100

= 0.35


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Problem 2:

Three queues are arranged in the loop as shown below. Assume that

the mean service time in the queue i is mi = 1/.

a. Suppose the queue has a single customer circulating in the loop. Find

the mean time E[T] it takes the customer to cycle around the loop.

De uce rom E T t e mean arriva rate at eac o t e queues. Veri y

that Littles formula holds for these two quantities.

b. If there are N customers circulating in the loop, how are the mean

arrival rate and mean cycle time related?

1 2 3


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Solution:

a. One customerno waiting

321

321

1

][ mmmmmmT ++=++=

Littles Formula

systemincustomeronemmm

mN

iqueueincustomertime

mmm

mmN

i i

ii

iii

1][

%][

3

1

3

1 321

321

=++

=

=

++

==

= =


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b.Let[T] = mean cycle time per customer, then

Total # in system = N [T] by Littles formula


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Open Queuing Networks

Consider a network of K queues in which customers

arrive from outside the network to queue k according to

independent Poisson processes of rate k. We assume

that the service time of a customer in queue k isexponentially distributed with rate kand independent

of all other service times and arrival processes. We also

suppose that queue k has ckservers.

After completion of service in queue k, a customerproceeds to queue i with probability Pki and exists the

network with probability

=

K

i

kiP

1

1


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The total arrival rate kinto queue k is the sum of the

external arrival rate and the internal arrival rates:

.,...,11 KkP

K

jjkjkk =+= =

It can be shown that above equation has a unique

solution if no customer remains in the network

indefinitely. Such networks are known as Open Queuing Networks


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The vector of the number of customers in all queues,

is a Markow process.

Now Jackson s theorem gives steady state pmf for N(t).

)),(),...,(),(),(()( 321 tNtNtNtNtN K=

k k k,

state n= (n1, n2,,nK),

Where P[ Nk= nk] is the steady state pmf of an M/M/cksystem with arrival rate k and service rate k

][]...[][])([2211 KK

nNPnNPnNPntNP =====


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Jacksons Theorem states that the numbers of

customers in the queues at time t are independent

random variables.

n addition, it states that the steady state probabilities

of the individual queues are those of an M/M/cksystem.

This is an amazing result because in general the input

process to a queue is not Poisson, as was demonstrated

in the sample queue with feedback discussed in thebeginning of this section.


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Example: Messages arrive at a concentrator according to a

Poisson process of rate . The time required to transmit a

message and receive an acknowledgment is exponentially

distributed with mean 1/m. Suppose that a message need to.

pmf for the number of messages in the concentrator.

Solution: The overall system can be represented by the

simple queue with feedback shown below.

.1

.9


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The net arrival rate into the queue is = + p, that

is,

p=

1

Thus, the pmf for the number of messages in the

concentrator is

)1/(/

,...1,0)1(][

==

===

where

nnNP n


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Example:New programs arrive at a CPU according to a Poisson

process of rate as shown in below fig. A program spends as

exponentially distributed execution time of mean 1/1 in the CPU.

At the end of this service time, the program execution is completewith robabilit or it re uires retrievin additional in ormation

from secondary storage with probability 1-p. Suppose that the

retrieval of information from secondary storage requires an

exponentially distributed amount of time with mean 1/2.Find the

mean time that each program spends in the system.

1

2

1p /CPU

p


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Solution: the net arrival rates into the two queues are

Thus

1221 )1( pand =+=

p

p

andp

)1(

21

==

Each queue behaves like an M/M/1 system so,

Where1= 1/1and2= 2/2. Littles formula then gives the

mean for total time spent in the system:

2

22

1

11

1][and

1][

=

= NENE

.11

1][][

2

2

1

121

+

=

+=

NNETE


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Close Queuing Networks

In some problems, a fixed number of customers, say I,

circulate endlessly in a closed queuing network.

For Example, some computer system models assume that at

any time a fixed number of programs use the CPU and input/.

1 2

1p

/CPU

p


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We now consider queuing networks that are identical to the

previously discussed Open Network except that the external

arrivals rates are zero and the networks always contain a fixed

number of customers I. We show that the steady state pmf for such

s stem is roduct rom but that the states o the ueues are no

longer independent.

The net arrival rate into the queue k is now given by

(1)

Note that these equations have the same from as the set of

equations that define the stationary pmf for a discrete-time

Markow chain with transition probabilities Pjk.

= ==K

jjkjk KkP1 ,...,1


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The only difference is that the sum of the ks need notbe one. Thus the solution vector to eq. (1) must be

proportional to the stationary pmf{j} corresponding to{Pjk} :

K

(2)

And where (I) is a constant that depends on I, the

number of customers in the queuing network. If we sumboth sides of eq. (2) over k, we see that(I) is the sum ofthe arrival rates in all the queues un the network, andk= k/(I) is the fraction of the total arrivals to queue k.

=

==j

jkjkkk whereI1

,)(


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Theorem:Let lk= (I) kbe a solution to

eq.(1), and let n = (n1, n2, ,nk) by any state of

the network for which n1, n

2, ,n

k 0 and

Then

where P[ Nk= nk] is the steady state pmf of an M/M/cksystem with arrival rate kand service rate k, and

where S(I) is the normalization constant given by

........21 K

)4.....(,

)(

][]...[][])([ 2211

IS

nNPnNPnNPntNP KK

=====


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Equation 4 states that P[ N(t) = n] has a product form.

)5.....(,][]...[][)(1...:

2211

1

=++

====Knnn

KK nNPnNPnNPIS

of the marginal pmf s because of the normalizationconstant S(I).

This constant arises because the fact that there arealways I customers in the network implies that theallowable states n must satisfy equation (3).

The theorem can be proved by taking the approachused to prove Jackson s theorem.


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Example: Suppose the computer system example that we haveseen in open queuing network is operated so that there are always

I programs in the system. The resulting network of queue is shown

below. Note that the feedback loop around the CPU signifies the

another one. Find the steady state pmf of the system. Find the rate

at which programs are completed.

1 2

1p

/CPU

p


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Solution: The stationary probabilities associated with equation

(1) are found by solving

1)1( 11,12,211 =+=+= andpp

e stationary pro a i ities are t en

And the arrival rates are

Solution Continues

p

p

p

=

=

2

1and

2

121

p

Ip

p

II

=

==

2

)()1(and

2

)()( 211


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The stationary pmf for the network is then

221121 0

)(

)1()1(],[

===

IiIS

iINiNP

I

iIi

Solution Continues

2

22

1

11

1

2

12

21

2

1

121

0

2121

,)1(

0

1

1],[

)1)(1()(

==

===

===

=

+

=

andp

where

IiiINiNPSo

ISand

i

I

i

iIi


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The rate at which programs are completed is pl1 . We find l1from

the relation between server utilization and probability of an

empty system:

1 1 ===

111

11

1

1)1(

1

+

+

=

I

I

I

pp

thatimplieswhich

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Case Study Solution Network of Queues