queuing theory. introduction queuing is the study of waiting lines, or queues. the objective of...

Queuing TheoryQueuing Theory

Introduction

• Queuing is the study of waiting lines, or queues.• The objective of queuing analysis is to design

systems that enable organizations to perform optimally according to some criterion.

• Possible Criteria– Maximum Profits.

– Desired Service Level.

• Analyzing queuing systems requires a clear understanding of the appropriate service measurement.

• Possible service measurements– Average time a customer spends in line.– Average length of the waiting line.– The probability that an arriving customer must wait

for service.

• The Arrival Process

– There are two possible types of arrival processes

• Deterministic arrival process.

• Random arrival process.

– The random process is more common in businesses.

– A Poisson distribution can describe the random

arrival process.

Types of Queuing Systems

• Single Channel Single Phase: Trucks unloading shipments into a dock.


• Single Line Multiple Phase: Wendy’s Drive Thru -> Order + Pay/Pickup


• Multiple Line Single Phase: Walgreens Drive-Thru Pharmacy


• Multiple Line Multiple Phase: Hospital Outpatient Clinic, Multi-specialty

Fundamentals of Queuing Theory• Microscopic traffic flow• Arrivals

– Uniform or random• Departures

– Uniform or random• Service rate

– Departure channels• Discipline

– FIFO and LIFO are most popular– FIFO is more prevalent in traffic engineering

Poisson Distribution

• Count distribution– Uses discrete values– Different than a continuous distribution

!n

etnP

tn

P(n) = probability of exactly n vehicles arriving over time t

n = number of vehicles arriving over time t

λ = average arrival rate

t = duration of time over which vehicles are counted

Poisson Distribution

• Video

Poisson Ideas

• Probability of exactly 4 vehicles arriving– P(n=4)

• Probability of less than 4 vehicles arriving– P(n<4) = P(0) + P(1) + P(2) + P(3)

• Probability of 4 or more vehicles arriving– P(n≥4) = 1 – P(n<4) = 1 - P(0) + P(1) + P(2) + P(3)

• Amount of time between arrival of successive vehicles

tetet

P

!0

00

Poisson Distribution ExampleGroup Activity

Vehicle arrivals at the Olympic National Park main gate are assumed Poisson distributed with an average arrival rate of 1 vehicle every 5 minutes. What is the probability of the following:

1. Exactly 2 vehicles arrive in a 15 minute interval?2. Less than 2 vehicles arrive in a 15 minute

interval?3. More than 2 vehicles arrive in a 15 minute

interval?

!

minveh20.0 minveh20.0

n

etnP

tn

Example Calculations

%4.22224.0

!2

1520.02

1520.02

e

PExactly 2:

Less than 2:

More than 2:

102 PPnP

21012 PPPnP

The Poisson Arrival Distribution

P X ke

k != =

l lt) k t

( )( -

Wherel = mean arrival rate per time unit.

t = the length of the interval.

e = 2.7182818 (the base of the natural logarithm).k! = k (k -1) (k -2) (k -3) … (3) (2) (1).

The Waiting Line

– Line configuration• A Single service Queue.• Multiple service queue with single waiting line.• Multiple service queue with multiple waiting lines.• Tandem queue (multistage service system).

– Jockeying• Jockeying occurs if customers switch lines when they

perceived that another line is moving faster.

– Balking• Balking occurs if customers avoid joining the line when

they perceive the line to be too long.

– Priority rules• Priority rules define the line discipline.• These rules select the next customer for service.• There are several commonly used rules:

– First come first served (FCFS).– Last come first served (LCFS).– Estimated service time.– Random selection of customers for service.

– Homogeneity• An homogeneous customer population is one in which

customers require essentially the same type of service.• A Nonhomogeneous customer population is one in which

customers can be categorized according to: – Different arrival patterns– Different service treatments.

The Service Process

– Some service systems require a fixed service time.

– In most business situations, however, service time varies widely among customers.

– When service time varies, it is treated as a random variable.

– The exponential probability distribution is used sometimes to model customer service time.

• The Exponential Service Time Distribution

f(X) = me-mX

where m = is the average number of customers who can be served per time period.

The probability that the service time X is less than some “t.”

P(X t) = 1 - e-mt

Schematic illustration of the exponential distribution

The probability that service is completed within t time units

X = t

f(X)

Measures of Queuing System Performance

• Performance can be measured by focusing on:

– Customers in queue.

– Customers in the system.

• Transient and steady state periods complicate the service time analysis

• The transient period occurs at the initial time of operation.– Initial transient behavior is not indicative of long

run performance.• The steady state period follows the transient

period.• In steady state, long run probabilities of having “n”

customers in the system do not change as time goes on.• In order to achieve steady state, the effective arrival rate

must be less than the sum of effective service rates .

< < l m l m1 +m2+…+mk < l km

For one server For k servers For k servers each with service

rate m

• Steady State Performance MeasuresP0 = Probability that there are no customers in the system.

Pn = Probability that there are “n” customers in the system.

L = Average number of customers in the system.Lq = Average number of customers in the queue.

W = Average time a customer spends in the system.Wq = Average time a customer spends in the queue.

Pw = Probability that an arriving customer must wait for service.

r = Utilization rate for each server (the percentage of time that each server is busy).

• Little’s Formulas– Little’s Formulas represent important relationships between

L, Lq, W, and Wq.– These formulas apply to systems that meet the following

conditions:• Single queue systems,• Customers arrive at a finite arrival rate , l and• The system operates under steady state condition.

L = l W Lq = l Wq L = Lq + l / m

For the infinite population case

Queue Notation

• Popular notations:– D/D/1, M/D/1, M/M/1, M/M/N– D = deterministic distribution– M = exponential distribution

NYX //

Arrival rate nature

Departure rate nature

Number ofservice channels

Queuing Theory Applications• D/D/1

– Use only when absolutely sure that both arrivals and departures are deterministic

• M/D/1 – Controls unaffected by neighboring controls

• M/M/1 or M/M/N– General case

• Factors that could affect your analysis:– Neighboring system (system of signals)– Time-dependent variations in arrivals and departures

• Peak hour effects in traffic volumes, human service rate changes– Breakdown in discipline

• People jumping queues! More than one vehicle in a lane!– Time-dependent service channel variations

• Grocery store counter lines

Queue Analysis – Graphical

ArrivalRate

DepartureRate

Time

Veh

icle

s

t1

Queue at time, t1

Maximum delay

Maximum queue

D/D/1 Queue

Delay of nth arriving vehicle

Total vehicle delay

Queue Analysis – Numerical

• M/D/1– Average length of queue

– Average time waiting in queue

– Average time spent in system

0.1

12

2

Q

12

1w

1

2

2

1t

λ = arrival rate μ = departure rate


• M/M/1– Average length of queue



0.1

1

2

Q

1

w

1t


• Performance Measures for the M / M /1 Queue

P0 = 1- (l / m)

Pn = [1 - (l / m)] (l/ m)n

L = l / (m - l) Lq = l 2 / [m(m - l)]

W = 1 / (m - l)Wq = l / [m(m - l)]

Pw = l / m

r = l / m

The probability thata customer waits in the system more than“t” is P(X>t)= e-(m - l)t


• M/M/N– Average length of queue



2

10

1

1

! NNN

PQ

N

1

Q

w

Q

t

0.1N


M/M/N – More Stuff– Probability of having no vehicles

– Probability of having n vehicles

– Probability of being in a queue

1

0

0

1!!

1N

n

N

c

n

c

c

NNn

P

Nnfor !

0 n

PP

n

n

Nnfor !

0 NN

PP

Nn

n

n

NNN

PP

N

Nn

1!

10

0.1N


Example 1

You are entering a Club to watch a basketball game. There is only one ticket line to purchase tickets. Each ticket purchase takes an average of 18 seconds. The average arrival rate is 3 persons/minute.

Find the average length of queue and average waiting time in queue assuming M/M/1 queuing.

Solution • Departure rate: μ = 18 seconds/person or 3.33

persons/minute• Arrival rate: λ = 3 persons/minute• ρ = 3/3.33 = 0.90

• Q-bar = 0.902/(1-0.90) = 8.1 people• W-bar = 3/3.33(3.33-3) = 2.73 minutes

• T-bar = 1/(3.33 – 3) = 3.03 minutes

Example 2

You are now in line to get into the Arena. There are 3 operating turnstiles with one ticket-taker each. On average it takes 3 seconds for a ticket-taker to process your ticket and allow entry. The average arrival rate is 40 persons/minute.

Find the average length of queue, average waiting time in queue assuming M/M/N queuing.

What is the probability of having exactly 5 people in the system?

Solution • N = 3• Departure rate: μ = 3 seconds/person or 20 persons/minute• Arrival rate: λ = 40 persons/minute• ρ = 40/20 = 2.0• ρ/N = 2.0/3 = 0.667 < 1 so we can use the other equations

• P0 = 1/(20/0! + 21/1! + 22/2! + 23/3!(1-2/3)) = 0.1111• Q-bar = (0.1111)(24)/(3!*3)*(1/(1 – 2/3)2) = 0.88 people• T-bar = (2 + 0.88)/40 = 0.072 minutes = 4.32 seconds• W-bar = 0.072 – 1/20 = 0.022 minutes = 1.32 seconds

Since n > N (5 > 3)• Pn = 25(0.1111)/(35-3*3!) = 0.0658 = 6.58%

Example 3

You are now inside the Arena. They are passing out Harry the Husky doggy bags as a free giveaway. There is only one person passing these out and a line has formed behind her. It takes her exactly 6 seconds to hand out a doggy bag and the arrival rate averages 9 people/minute.

Find the average length of queue, average waiting time in queue, and average time spent in the system assuming M/D/1 queuing.

Solution

• N = 1• Departure rate: μ = 6 seconds/person or 10 persons/minute• Arrival rate: λ = 9 persons/minute• ρ = 9/10 = 0.9

• Q-bar = (0.9)2/(2(1 – 0.9)) = 4.05 people• W-bar = 0.9/(2(10)(1 – 0.9)) = 0.45 minutes = 27 seconds• T-bar = (2 – 0.9)/((2(10)(1 – 0.9) = 0.55 minutes = 33 seconds

MARY’s SHOES• Customers arrive at Mary’s Shoes every 12

minutes on the average, according to a Poisson process.

• Service time is exponentially distributed with an average of 8 minutes per customer.

• Management is interested in determining the performance measures for this service system.

SOLUTION– Input

l = 1/ 12 customers per minute = 60/ 12 = 5 per hour.

m = 1/ 8 customers per minute = 60/ 8 = 7.5 per hour.

– Performance CalculationsP0 = 1- (l / m) = 1 - (5 / 7.5) = 0.3333

Pn = [1 - (l / m)] (l/ m) = (0.3333)(0.6667)n L = l / (m - l) = 2Lq = l2/ [m(m - l)] = 1.3333

W = 1 / (m - l) = 0.4 hours = 24 minutesWq = l / [m(m - l)] = 0.26667 hours = 16 minutes

P0 = 1- (l / m) = 1 - (5 / 7.5) = 0.3333

Pn = [1 - (l / m)] (l/ m) = (0.3333)(0.6667)n L = l / (m - l) = 2Lq = l2/ [m(m - l)] = 1.3333

W = 1 / (m - l) = 0.4 hours = 24 minutesWq = l / [m(m - l)] = 0.26667 hours = 16 minutes

Pw = l / = 0.6667m

r = l / = m0.6667

queuing theory. introduction queuing is the study of waiting lines, or queues. the objective of...

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