Queuing Theory LN 4

7/29/2019 Queuing Theory LN 4

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Queuing Theory

M S Prasad

LN 902


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Description

Each of us has spent a great deal of time waiting in lines.

In this chapter, we develop mathematical models for waiting

lines, or queues.


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8.1 Some Queuing Terminology

To describe a queuing system, an input process andan output process must be specified.

Examples of input and output processes are:

Situation Input Process Output Process

Bank Customers arrive at

bank

Tellers serve the

customers

Pizza parlor Request for pizza

delivery are received

Pizza parlor send out

truck to deliver pizzas


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The Input or Arrival Process The input process is usually called the arrival

process.

Arrivals are called customers.

We assume that no more than one arrival can occur

at a given instant. If more than one arrival can occur at a given instant,

we say that bulk arrivals are allowed.

Models in which arrivals are drawn from a small

population are called finite source models.

If a customer arrives but fails to enter the system,

we say that the customer has balked.


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The Output or Service Process To describe the output process of a queuing system,

we usually specify a probability distributiontheservice time distributionwhich governs acustomers service time.

We study two arrangements of servers: servers inparallel and servers in series.

Servers are in parallel if all servers provide the sametype of service and a customer needs only passthrough one server to complete service.

Servers are in series if a customer must pass throughseveral servers before completing service.


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Queue Discipline

The queue discipline describes the method used todetermine the order in which customers are served.

The most common queue discipline is the FCFS

discipline (first come, first served), in which

customers are served in the order of their arrival.

Under the LCFS discipline (last come, first served),

the most recent arrivals are the first to enter service.

If the next customer to enter service is randomlychosen from those customers waiting for service it

is referred to as the SIRO discipline (service in

random order).


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Finally we considerpriority queuing disciplines.

A priority discipline classifies each arrival into one

of several categories.

Each category is then given a priority level, andwithin each priority level, customers enter service

on a FCFS basis.

Another factor that has an important effect on the

behavior of a queuing system is the method that

customers use to determine which line to join.


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8.2 Modeling Arrival and Service

Processes

We define ti to be the time at which the ith customerarrives.

In modeling the arrival process we assume that the

Ts are independent, continuous random variables

described by the random variable A.

The assumption that each interarrival time is

governed by the same random variable implies that

the distribution of arrivals is independent of thetime of day or the day of the week.

This is the assumption of stationary interarrival

times.


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Stationary interarrival times is often unrealistic, butwe may often approximate reality by breaking the

time of day into segments.

A negative interarrival time is impossible. Thisallows us to write

We define1/to be the mean or average interarrival

time.

c

cdttacPdttacP

0)()(and)()( A

0

)(1

dttta


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We define to be the arrival rate, which will haveunits of arrivals per hour.

An important question is how to choose A to reflect

reality and still be computationally tractable. The most common choice forA is the exponential

distribution.

An exponential distribution with parameterhas a

density a(t) = e-t.

We can show that the average or mean interarrival

time is given by .

1)( E


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Using the fact that varA =E(A2)E(A)2, we canshow that

Lemma 1: IfA has an exponential distribution, then

for all nonnegative values oftand h,

2

1var

)()|( hPthtP A


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A density function that satisfies the equation is saidto have the no-memory property.

The no-memory property of the exponential

distribution is important because it implies that ifwe want to know the probability distribution of the

time until the next arrival, then it does not matter

how long it has been since the last arrival.


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Relations between Poisson

Distribution and Exponential

Distribution If interarrival times are exponential, the probability

distribution of the number of arrivals occurring in

any time interval of length tis given by the

following important theorem.

Theorem 1: Interarrival times are exponential with

parameterif and only if the number of arrivals to

occur in an interval of length tfollows the Poissondistribution with parametert.


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A discrete random variable N has a Poisson

distribution with parameterif, forn=0,1,2,,

What assumptions are required for interarrival times

to be exponential? Consider the following twoassumptions:

Arrivals defined on nonoverlapping time intervals are

independent.

For small t, the probability of one arrival occurring

between times tand t+tis t+o(t) refers to any

quantity satisfying

,...)2,1,0(!

)(

nn

enP

n

N

0)(

lim0

t

to

t


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Theorem 2: If assumption 1 and 2 hold, then Ntfollows a Poisson distribution with parametert,

and interarrival times are exponential with

parameter; that is, a(t) = e-t.

Theorem 2 states that if the arrival rate is stationary,

if bulk arrives cannot occur, and if past arrivals do

not affect future arrivals, then interarrival times will

follow an exponential distribution with parameter,and the number of arrivals in any interval of length t

is Poisson with parametert.


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The Erlang Distribution

If interarrival times do not appear to be exponential

they are often modeled by an Erlang distribution.

An Erlang distribution is a continuous random

variable (call it T) whose density functionf(t) is

specified by two parameters: a rate parameterR anda shape parameterk(kmust be a positive integer).

Given values ofR and k, the Erlang density has the

following probability density function:

)0()!1(

)()(

1

tk

eRtRtf

Rtk


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Using integration by parts, we can show that ifT is an Erlan

distribution with rate parameterR and shape parameterk,

then

The Erlang can be viewed as the sum of independent andidentically distributed exponential random variable with rate

1/

2varand)(

Rk

RkE T


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Using EXCEL to Computer Poisson

and Exponential Probabilities

EXCEL contains functions that facilitate thecomputation of probabilities concerning the Poisson

and Exponential random variable.

The syntax of the Poisson EXCEL function is asfollows:

=POISSON(x,Mean,True) gives probability that a

Poisson random variable with mean = Mean is less than

or equal to x. =POISSON(x,Mean,False) gives probability that a

Poisson random variable with mean =Mean is equal to x.


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The syntax of the EXCEL EXPONDIST function isas follows:

=EXPONDIST(x,Lambda,TRUE) gives the probability

that an exponential random variable with parameter

Lambda assumes a value less than or equal to x.

=EXPONDIST(x,Lambda,FALSE) gives the probability

that an exponential random variable with parameter

Lambda assumes a value less than or equal to x.


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Modeling the Service Process

We assume that the service times of different

customers are independent random variables and

that each customers service time is governed by a

random variable S having a density functions(t).

We let 1/ be the mean service time for a customer. The variable 1/ will have units of hours per

customer, so has units of customers per hour. For

this reason, we call the service rate.

Unfortunately, actual service times may not be

consistent with the no-memory property.


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For this reason, we often assume thats(t) is anErlang distribution with shape parameters kand rate

parameterk.

In certain situations, interarrival or service times

may be modeled as having zero variance; in thiscase, interarrival or service times are considered to

be deterministic.

For example, if interarrival times are deterministic,

then each interarrival time will be exactly 1/, and ifservice times are deterministic, each customersservice time is exactly 1/.


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The Kendall-Lee Notation for

Queuing Systems

Standard notation used to describe many queuingsystems.

The notation is used to describe a queuing system in

which all arrivals wait in a single line until one ofsidentical parallel servers is free. Then the first

customer in line enters service, and so on.

To describe such a queuing system, Kendall devised

the following notation. Each queuing system is described by six characters:

1/2/3/4/5/6


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The first characteristic specifies the nature of thearrival process. The following standard

abbreviations are used:

M= Interarrival times are independent, identically

distributed (iid) and exponentially distributed

D = Interarrival times are iid and deterministic

Ek= Interarrival times are iid Erlangs with shape parameter

k.

GI= Interarrival times are iid and governed by some

general distribution


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The second characteristic specifies the nature of theservice times:

M= Service times are iid and exponentially distributed

D = Service times are iid and deterministic

Ek= Service times are iid Erlangs with shape parameter

k.

G = Service times are iid and governed by some general

distribution


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The third characteristic is the number of parallel servers.

The fourth characteristic describes the queue discipline:

FCFS = First come, first served

LCFS = Last come, first served

SIRO = Service in random order

GD = General queue discipline

The fifth characteristic specifies the maximum allowable

number of customers in the system.

The sixth characteristic gives the size of the population from

which customers are drawn.


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In many important models 4/5/6 is GD//. If thisis the case, then 4/5/6 is often omitted.

M/E2/8/FCFS/10/ might represent a health clinic

with 8 doctors, exponential interarrival times, two-phase Erlang service times, a FCFS queue

discipline, and a total capacity of 10 patients.


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The Waiting Time Paradox

Suppose the time between the arrival of buses at thestudent center is exponentially distributed with a

mean of 60 minutes.

If we arrive at the student center at a randomly

chosen instant, what is the average amount of timethat we will have to wait for a bus?

The no-memory property of the exponential

distribution implies that no matter how long it hasbeen since the last bus arrived, we would still expect

to wait an average of 60 minutes until the next bus

arrived.


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8.3 Birth-Death Processes

We subsequently use birth-death processes to

answer questions about several different types of

queuing systems.

We define the number of people present in any

queuing system at time tto be the state of thequeuing systems at time t.

We call j the steady state, or equilibrium

probability, of statej.

The behavior ofPij(t) before the steady state is

reached is called the transient behavior of the

queuing system.


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A birth-death process is a continuous-timestochastic process for which the systems state at

any time is a nonnegative integer.


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Laws of Motion for Birth-Death Law 1

With probability jt+o(t), a birth occurs between timetand time t+t. A birth increases the system state by 1,

toj+1. The variable j is called the birth rate in statej. In

most queuing systems, a birth is simply an arrival.

Law 2 With probability jt+o(t), a death occurs between time

tand time t+t. A death decreases the system state by 1,

toj-1. The variablej is the death rate in statej. In most

queuing systems, a death is a service completion. Notethat0 = 0 must hold, or a negative state could occur.

Law 3

Births and deaths are independent of each other.

R l ti f E ti l Di t ib ti


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Relation of Exponential Distribution

to Birth-Death Processes

Most queuing systems with exponential interarrivaltimes and exponential service times may be

modeled as birth-death processes.

More complicated queuing systems withexponential interarrival times and exponential

service times may often be modeled as birth-death

processes by adding the service rates for occupied

servers and adding the arrival rates for differentarrival streams.

D i ti f St d St t


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Derivation of Steady-State

Probabilities for Birth-Death

Processes We now show how the js may be determined for

an arbitrary birth-death process.

The key role is to relate (for small t) Pij(t+t) toPij(t).

The above equations are often called the flow

balance equations, orconservation of flow

equations, for a birth-death process.

,...)2,1()(1111 jjjjjjjj

0011

The Flow-Balancing Approach (Entry-Exit


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The Flow-Balancing Approach (Entry-ExitRate Balancing Approach)

In the rate diagram given below, think of thefollowing:

Each circle representing a state (i.e., number ofcustomer in the system) has an unknown probability

pj, j= 0, 1, 2, associated with it

0 1 2 3

4


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We obtain the flow balance equations for a birth-death process:

1111

3311222

2200111

1100

)()equationth(

)()2(

)()1(

)0(

jjjjjjjj

j

j

j


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,...)2,1(0 jcjj

Cj = (012j-1)/(1 2 3.. j)

S l ti f Bi th D th Fl


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Solution of Birth-Death Flow

Balance Equations

If is finite, we can solve for0:

It can be shown that if is infinite, then no

steady-state distribution exists.

The most common reason for a steady-state failing

to exist is that the arrival rate is at least as large asthe maximum rate at which customers can be

served.

j

j jc1

j

j

jc1

0

1

1

j

j jc

1

8 4 The M/M/1/GD// Q e ing


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8.4 TheM/M/1/GD// Queuing

System and the Queuing Formula

L=W We define . We callp the traffic intensity

(utilization) of the queuing system.

We now assume that 0 p < 1 thus

Ifp 1, however, the infinite sum blows up.Thus, ifp 1, no steady-state distribution exists.

p

)10()1(

)10(10

ppp

pp

j

j


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Derivation ofL Throughout the rest of this section, we assume that

p


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Derivation ofLq

In some circumstances, we are interested in the expectednumber of people waiting in line (or in the queue).

We denote this number byLq.

)(11

22

p

pp

p

pLq

j

j

jq jL1

)1(


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Derivation ofLs

Also of interest isLs, the expected number ofcustomers in service.

ppLs )1(11)(10 0210

p

pp

p

pLLL sq

11

2

0

1

j

j


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The Queuing FormulaL=W

We define Was the expected time a customerspends in the queuing system, including time in line

plus time in service, and Wq as the expected time a

customer spends waiting in line.

By using a powerful result known as Littles

queuing formula, Wand Wq may be easily

computed fromL andLq.

We first define the following quantitiesL = average number of arrivals enteringthe system per

unit time


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L = average number of customers present in the queuing system

Lq = average number of customers waiting in line

Ls = average number of customers in service

W= average time a customer spends in the system

Wq = average time a customer spends in line Ws = average time a customer spends in service

Theorem 3For any queuing system in which a steady-

state distribution exists, the following relations hold:

L =WLq =Wq

Ls =Ws


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Example 4

Suppose that all car owners fill up when theirtanks are exactly half full.

At the present time, an average of 7.5 customers

per hour arrive at a single-pump gas station. It takes an average of 4 minutes to service a car.

Assume that interarrival and service times are both

exponential.

1. For the present situation, computeL and W.


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2. Suppose that a gas shortage occurs and panic buyingtakes place.

To model the phenomenon, suppose that all car owners now

purchase gas when their tank are exactly three-fourths full.

Since each car owner is now putting less gas into the tank

during each visit to the station, we assume that the average

service time has been reduced to 3 1/3 minutes.

How has panic buying affectedL and W?


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Solutions1. We have anM/M/1/GD// system with = 7.5

cars per hour and = 15 cars per hour. Thusp =7.5/15 = .50.L = .50/1-.50 = 1, and W=L/ =

1/7.5 = 0.13 hour. Hence, in this situation,

everything is under control, and long lines appear

to be unlikely.2. We now have anM/M/1/GD// system with =

2(7.5) = 15 cars per hour. Now = 60/3.333 = 18

cars per hour, andp = 15/18 = 5/6. Then

Thus, panic buying has cause long lines.

minutes20hours3

1

15

5andcars5

6

51

6

5

LWL


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Problems in which a decision maker must choosebetween alternative queuing systems are called

queuing optimization problems.


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More onL =W

The queuing formulaL =Wis very general andcan be applied to many situations that do not seem

to be queuing problems.

L = average amount of quantity present.

= Rate at which quantity arrives at system.

W = average time a unit of quantity spends in system.

ThenL =W or W = L/


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A Simple Example

Example Our local MacDonalds uses an average of 10,000 pounds

of potatoes per week.

The average number of pounds of potatoes on hand is

5000 pounds.

On the average, how long do potatoes stay in the

restaurant before being used?

Solution

We are given that L=5000 pounds and= 10,000

pounds/week. Therefore W = 5000 pounds/(10,000

pounds/week)=.5 weeks.

A Queueing Model Optimization


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Problems in which a decision maker must choose betweenalternative queueing systems

Example: An average of 10 machinists per hour arrive seekingtools. At present, the tool center is staffed by a clerk who is

paid $6 per hour and who takes an average of 5 minutes tohandle each request for tools. Since each machinist produces$10 worth of goods per hour, each hour that a machinists

spends at the tool center costs the company $10. The companyis deciding whether or not it is worthwhile to hire (at $4 perhour) a helper for the clerk. If the helper is hired the clerk willtake an average of only 4 minutes to process requirements fortools. Assume that service and arrival times are exponential.Should the helper be hired?



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Goal: Minimize the sum of the hourly service cost and expected hourly costdue to the idle times of machinists

Delay cost is the component of cost due to customers waiting in line Goal: Minimize Expected cost/hour = service cost/hour + expected delay

cost/hour

Expected delay cost/hour = (expected delay cost/customer) (expectedcustomers/hour)

Expected delay cost/customer = ($10/machinist-hour)(average hoursmachinist spends in the system) = 10W

Expected delay cost/hour = 10W

Now compute expected cost/hour if the helper is not hired and alsocompute the same if the helper is hired



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If the helper is not hired = 10 machinists per hour and = 12machinists per hour

W = 1/(-) for M/M/1/GD//. Therefore, W = 1/(12-10) = = 0.5 hour

Service cost /hour = $6/hour and expected delay cost/hour =10(0.5)(10) = $50

Without the helper, the expected hourly cost is $6 + $50 = $56

With the helper, = 15 customers/hour. Then W = 1/(-)=1/(15-10) = 0.2 hour and the expected delay cost/hour =10(0.2)(10) = $20

Service cost/hour = $6 + $4 = $10/hour

With the helper, the expected hourly cost is $10 + $20 = $30

8 5 The M/M/1/GD/c/ Queuing


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8.5 TheM/M/1/GD/c/ Queuing

System

TheM/M/1/GD/c/ queuing system is identical totheM/M/1/GD// system except for the fact that

when c customers are present, all arrivals are turned

away and are forever lost to the system.

The rate diagram for the queuing system can be

found in Figure 13 in the book.

WL

LW

LW

j

j

jc

c

q

q

c

0

)1(

)1(and

)1(


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j j j0 0

1

. .

. .

c 0

Effective Arrival Rate

1,1

1

1,1

1

0

10

0

ifc

ifc

j

j

Verification by Flow-Balancing Equations


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Verification by Flow-Balancing Equations

p0 = p1

p1+ p1 = p0 + p2

p2+ p2 = p1 + p3

p2= p3 p0 + p1 + p2 + p3 = 1

Substituting the values of = 1 and = 2, we have

2p1 = p0 p0 + 2p2 = 3p1 p1+2p3 = 2p2 p2 = 2p3 p0 + p1 + p2 + p3 = 1

p0 = 8/15, p1 = 4/15, p2 = 2/15, and p3 = 1/15

0 1 2 3

=1

=2

=1 =1

=2 =2


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For theM/M/1/GD/c/ system, a steady state willexist even if .

This is because, even if , the finite capacity of

the system prevents the number of people in the

system from blowing up.

The M/M/s/GD// Queuing


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TheM/M/s/GD// Queuing

System

We now consider theM/M/s/GD// system. We assume that interarrival times are exponential

(with rate), service times are exponential (with

rate), and there is a single line of customers

waiting to be served at one of thes parallel servers.

Ifjs customers are present, then allj customers

are in service; ifj >s customers are present, then all

s servers are occupied, andjs customers arewaiting in line.

j j j


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j j j

0 0

1

2

.

.

2

.

s s

s+1

.

.

s

.

s


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Summarizing, we find that theM/M/s/GD//

system can be modeled as a birth-death process withparameters

we definep= /s. Forp


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An M/M/s Queueing Optimization Example

1)(

/1/1//

/

)()(/,)1()(

)1(!

ss)P(j

2.....)s1,ss,(j!

s)1,2,....,(j!

)1(!!

1

0

0

j

0

)1(

0

0

s

sjP

WLLW

LL

sjs

sjPLWsjPL

s

sss

j

s

ss

is

qq

q

sj

jqqq

s

sj

j

j

j

si

i

si


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M/M/s/GD/c/ - Multiple server waiting queueproblem with exponential arrival and service times

with finite capacity



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Bank Staffing Example: The manager of a bank must determine howmany tellers should work on Fridays. For every minute a customer standsin line, the manager believes that a delay cost of 5 cents is incurred. Anaverage of 2 customers per minute arrive at the bank. On the average ittakes, a teller 2 minutes to complete a customers transaction. It costs the

bank $9 per hour to hire a teller. Inter-arrival times and service times areexponential. To minimize the sum of service costs and delay costs, howmany tellers should the bank have working on Fridays?

= 2 customer per minute and = 0.5 customer per minute, /s requires

that 4/s < 1. Thus, there must be at least 5 tellers, or the number ofcustomers present will blow up.

Now compute for s = 5, 6. Expected service cost/minute + expecteddelay cost/minute



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Q g p p

Each teller is paid $9/60 = 15 cents per minute.

Expected service cost/minute = 0.15s

Expected delay cost/minute = (expected

customers/minute) (expected delay cost/customer)

Expected delay cost/customer = 0.05Wq

Expected delay cost/minute = 2(0.05) Wq = 0.10 Wq

For s = 5, =/s = 2/.5(5) = 0.8



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P(j 5)=0.55

Wq = .55/(5(.5)-2) = 1.1 minutes For s = 5, expected delay cost/minute = 0.10(1.1) =11 cents

For s = 5, total expected cost/minute = 0.15(5) + 0.11

= 86 cents Since s = 6 has a service cost per minute of 6(0.15) =

90 cents, 6 tellers cannot have a lower total cost than5 tellers. Hence, having 5 tellers serve is optimal

TheM/M//GD// and


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e / / /G / / a d

GI/G//GD// Models

There are many examples of systems in which acustomer never has to wait for service to begin.

In such a system, the customers entire stay in the

system may be thought of as his or her service time.

Since a customer never has to wait for service, there

is, in essence, a server available for each arrival, and

we may think of such a system as an infinite-server

(or self-service).


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Using Kendall-Lee notation, an infinite serversystem in which interarrival and service times may

follow arbitrary probability distributions may be

written as GI/G//GD// queuing system.

Such a system operated as follows:

Interarrival times are iid with common distribution A.

DefineE(A) = 1/. Thus is the arrival rate.

When a customer arrives, he or she immediately entersservice. Each customers time in the system is governed

by a distribution S havingE(S)= 1/.


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LetL be the expected number of customers in thesystem in the steady state, and Wbe the expected

time that a customer spends in the system.

L

8.8 TheM/G/1/GD// Queuing


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Q g

System

Next we consider a single-server queuing system inwhich interarrival times are exponential, but the

service time distribution (S) need not be

exponential.

Let () be the arrival rate (assumed to be measured

in arrivals per hour).

Also define 1/ =E(S) and 2=varS.

In Kendalls notation, such a queuing system isdescribed as anM/G/1/GD// queuing system.


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Determination of the steady-state probabilities forM/G/1/GD// queuing system is a difficult matter.

Fortunately, however, utilizing the results of

Pollaczek and Khinchin, we may determineLq, L,

Ls, Wq, W, Ws.


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Pollaczek and Khinchin showed that for the

M/G/1/GD// queuing system,

It can also be shown that 0, the fraction of the time

that the server is idle, is 1-p.

The result is similar to the one for the

M/M/1/GD// system.

1

)1(2

222

q

q

q

q

q

WW

L

W

pLL

p

pL

/1/6/5)/1(22 hrhrmeanExpS


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77.2)6/51(2

)18/3(5)6/5(

18/3/)(

/1)18/3(/)(

)3,/18(~min),33.3(~,

27.2)6/51(2

)60/3(5)6/5(

)60/3(min),3min,10(~

083.2)1(2

0),6(tan~

16.4)1(

/1,/6,/5),/1(~

222

22

321

222

22

2

2

2

q

i

q

q

q

L

RKsV

RKsE

khrRErlangSmeanExpSSSSS

L

meanNS

p

pL

tConsS

p

pL

hrhrmeanExpS

8.9 Finite Source Models: The


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Machine Repair Model M/M/R/GD/K/K With the exception of theM/M/1/GD/c/ model,

all the models we have studied have displayed

arrival rates that were independent of the state of

the system.

There are two situations where the assumption ofthe state-independent arrival rate may be invalid:

1. If customers do not want to buck long lines, the arrival

rate may be a decreasing function of the number of

people present in the queuing system.2. If arrivals to a system are drawn from a small

population, the arrival rate may greatly depend on the

state of the system.


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Models in which arrivals are drawn from a small

population are called finite source models. In the machine repair problem, the system consists

ofKmachines andR repair people.

At any instant in time, a particular machine is in

either good or bad condition.

The length of time that a machine remains in good

condition follows an exponential distribution with

rate . Whenever a machine breaks down the machine is

sent to a repair center consisting ofR repair people.


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The repair center services the broken machines as if

they were arriving at anM/M/R/GD// system. Thus, ifj R machines are in bad condition, a

machine that has just broken will immediately be

assigned for repair; ifj >R machines are broken,j

R machines will be waiting in a single line for a

repair worker to become idle.

The time it takes to complete repairs on a broken

machine is assumed exponential with rate. Once a machine is repaired, it returns to good

condition and is again susceptible to breakdown.


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The machine repair model may be modeled as abirth-death process, where the statej at any time is

the number of machines in bad condition.

Note that a birth corresponds to a machine breaking

down and a death corresponds to a machine having

just been repaired.

When the state isj, there areK-j machines in good

condition. When the state isj, min (j,R) repair people will be

busy.


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Since each occupied repair worker completesrepairs at rate, the death ratej is given by

If we definep = /, an application of steady-state

probability distribution:

),...2,1(

),...,1,0(

KRRjR

Rjj

j

j

),...2,1(!

!

),...,1,0(

0

0

KRRjRR

jpj

K

Rjpj

K

Rj

j

jj


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Using the steady-state probabilities shown on theprevious slide, we can determine the following

quantities of interest:

L = expected number of broken machines

Lq = expected number of machines waiting for service

W= average time a machine spends broken (down time)

Wq = average time a machine spends waiting for service

Unfortunately, there are no simple formulas forL,Lq, W, Wq. The best we can do is express these

quantities in terms of the js:


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Kj

Rj

jq

Kj

j

j

RjL

jL

)(

0

)()(00

Lkjk

LW

L

W

j

K

j

j

K

j

j

q

q


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2 repairman, 3 machine, = 2/day, 1/ = 12 hours,

=4/day, 1/= 6 hours

j j j0 6, K 0

1 4,(K-1) 4,

2,R

.

K-1

2,(K-2)

.

(K-(K-1))

8,R

.

R3,K 0 8,R

2909.0


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0545.

2182.

4364.

3

2

1

0

Find the expected number of machines working

=3-L=3-(1(.4364)+2(.2182)+3(.0545))=3-1.0363=1.9637

Find the utilization of repairman

=.4909

Ls= 1(.4364)+2(.2182)+2(.0545)=.9818

Find the expected wait time for the repairman

0545.)0545(.1

4)0363.13(2)(

63.4/0545.

q

q

q

L

Lk

hoursL

W

8.10 Exponential Queues in Series


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and Open Queuing Networks

In the queuing models that we have studied so far, acustomers entire service time is spent with a single

server.

In many situations the customers service is not

complete until the customer has been served by

more than one server.

A system like the one shown in Figure 19 in the

book is called a k-stage series queuing system.


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Theorem 4If (1)interarrival times for a series

queuing system are exponential with rate , (2)

service times for each stageIserver are exponential,

and (3) each stage has an infinite-capacity waiting

room, then interarrival times for arrivals to eachstage of the queuing system are exponential with

rate .

For this result to be valid, each stage must have

sufficient capacity to service a stream of arrivalsthat arrives at rate ; otherwise, the queue will

blow up at the stage with insufficient capacity.

Open Queuing Networks


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Open Queuing Networks Open queuing networks are a generalization of

queues in series. Assume that stationj consists ofsjexponential servers, each operating at ratej.

Customers are assumed to arrive at station j from

outside the queuing system at rate rj.

These interarrival times are assumed to be

exponentially distributed.

Once completing service at stationI, a customer

joins the queue at stationj with probabilitypij andcompletes service with probability

kj

j

ijp1

1


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Define j, the rate at which customers arrive atstationj.

1, 2, kcan be found by solving the following

systems of linear equations:

This follows, because a fractionpij of the i arrivals

to station i will next go to stationj. Supposesjj > j holds for all stations.

),...,2,1(1

kjprki

i

iijjj


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Then it can be shown that the probabilitydistribution of the number of customers present at

stationj and the expected number of customers

present at stationj can be found by treating stationj

as anM/M/sj/GD// system with arrival rate jand service ratej.

If for somej,sjj j, then no steady-state

distribution of customers exists. The number of customers present at each station are

independent random variables.


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That is, knowledge of the number of people at allstations other than stationj tells us nothing about

the distribution of the number of people at stationj! This result does not hold, however, if either

interarrival or service times are not exponential.

To findL, the expected number of customers in thequeuing system, simply add up the expected numberof customers present at each station.

To find W, the average time a customer spends inthe system, simply apply the formulaL=Wto theentire system.

p


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p11

p22

p33

332211 ppp

ii

=8,s=1

=15,s=1

=12,s=1

=3,s=2

.5

.5

=5

=10

=5+.5(10)=10

=.5(10)=5

W=1/(12-10)=.5W=1/(8-5)=.333

W=1/(15-10)=.2

Network Models of Data


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Communication Networks Queuing networks are commonly used to model

data communication networks.

The queuing models enable us to determine the

typical delay faced by transmitted data and also to

design the network. We are interested, of course, in the expected delay

for a packet.

Also, if total network capacity is limited, a natural

question is to determine the capacity on each arcthat will minimize the expected delay for a packet.


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The usual way to treat this problem is to treat each

arc as if it is an independentM/M/1 queue anddetermine the expected time spent by each packet

transmitted through that arc by the formula

We are assuming a static routing in which arrival

rates to each node do not vary with the state of the

network. In reality, many sophisticated dynamic routing

schemes have been developed.

1W

8.11 TheM/G/s/GD/s/ System


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(Blocked Customers Cleared)

In many queuing systems, an arrival who finds allservers occupied is, for all practical purposes, lost to

the system.

If arrivals who find all servers occupied leave the

system, we call the system a blocked customerscleared, or BCC, system.

Assuming that interarrival times are exponential,

such a system may be modeled as anM/G/s/GD/s/system.


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In most BCC systems, primary interest is focused

on the fraction of all arrivals who are turned away. Since arrivals are turned away only whens

customers are present, a fraction s of all arrivals

will be turned away.

Hence, an average ofsarrivals per unit time will

be lost to the system.

Since an average of(1-s) arrivals per unit time

will actually enter the system, we may conclude that

)1( ssLL

8.12 How to Tell Whether


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Interarrival Times and Service

Times are Exponential How can we determine whether the actual data are

consistent with the assumption of exponential

interarrival times and service times? Suppose for example, that interarrival times oft1, t2,

tn have been observed.

It can be shown that a reasonable estimate of the

arrival rate is given by

ni

i

it

n

1

8.13 Closed Queuing Networks


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Q g For manufacturing units attempting to implement

just-in-time manufacturing, it makes sense to

maintain a constant level of work in progress.

For a busy computer network it may be convenient

to assume that as soon as a job leaves the system

another job arrives to replace the job. Systems where there is constant number of jobs

present may be modeled as closed queuing

networks.

Since the number of jobs is always constant the

distribution of jobs at different servers cannot be

independent.

8.15 Priority Queuing Models


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y Q g There are many situations in which customers are

not served on a first come, first served (FCFS)

basis.

Let WFCFS, WSIRO, and WLCFS be the random

variables representing a customers waiting time in

queuing systems under the disciplines FCFS, SIRO,LCFS, respectively.

It can be shown that

E(WFCFS) =E(WSIRO) =E(WLCFS)

Thus, the average time (steady-state) that a

customer spends in the system does not depend on

which of these three queue disciplines is chosen.


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It can also be shown thatvarWFCFS < varWSIRO < var(WLCFS)

Since a large variance is usually associated with a

random variable that has a relatively large chance of

assuming extreme values, the above equationindicates that relatively large waiting times are most

likely to occur with an LCFS discipline and least

likely to occur with an FCFS discipline.


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In many organizations, the order in which

customers are served depends on the customerstype.

For example, hospital emergency rooms usually

serve seriously ill patients before they servenonemergency patients.

Models in which a customers type determines the

order in which customers undergo service are call

priority queuing models. The interarrival times of type i customers are

exponentially distributed with rate i.


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Interarrival times of different customer types areassumed to be independent.

The service time of a typeIcustomer is described

by a random variable Si.

Nonpreemptive Priority Models


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Nonpreemptive Priority Models

In a nonpreemptive model, a customers servicecannot be interrupted.

After each service completion, the next customer to

enter service is chosen by given priority to lower-

numbered customer types (Lower numberedhigher priority).

In the Kendall-Lee notation, a nonpreemptive

priority model is indicated by labeling the fourthcharacteristic as NPRP.

Preemptive Priorities


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Preemptive Priorities In a preemptive queuing system, a lower priority

customer can be bumped from service whenever ahigher-priority customer arrives.

Once no higher-priority customers are present, the

bumped type i customer reenters service.

In a preemptive resume model, a customers

service continues from the point at which it was

interrupted.


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In a preemptive repeat model, a customer beginsservice anew each time he or she reenters service.

Of course, if service times are exponentially

distributed, the resume and repeat disciplines are

identical.

In the Kendall-Lee notation, we denote a

preemptive queuing system by labeling the fourth

characteristic PRP.


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For obvious reasons, preemptive discipline arerarely used if the customers are people.

8.15 Transient Behavior of QueuingS t


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Systems We have assumed the arrival rate, service rate and

number of servers has stayed constant over time.This allows us to talk reasonably about the existence

of a steady state.

In many situations the arrival rate, service rate, and

number of servers may vary over time.

An example is a fast food restaurant.

It is likely to experience a much larger arrival rate during

the time noon-1:30 pm than during other hours of theday.

Also the number of servers will vary during the day with

more servers available during the busier periods.


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When the parameters defining the queuing system

vary over time we say the queuing system is non-stationary.

Consider the fast food restaurant. We call these

probability distributions transient probabilities.

We now assume that at time tinterarrival times are

exponential with rate(t) and thats(t) servers are

available at time twith service times being

exponential with rate(t). We assume the maximum number of customers

present at any time is given by N.

Queuing Theory LN 4

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