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.
1
Introduction
• 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.
2
Elements of the Queuing Process
• A queuing system consists of three basic
components:
– Arrivals: Customers arrive according to some arrival
pattern.
– Waiting in a queue: Arriving customers may have to wait
in one or more queues for service.
– Service: Customers receive service and leave the system.
3
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.
4
The Arrival Process
• Under three conditions the arrivals can be modeled as a
Poisson process
– Orderliness : one customer, at most, will arrive during any
time interval.
– Stationarity : for a given time frame, the probability of arrivals
within a certain time interval is the same for all time intervals of
equal length.
– Independence : the arrival of one customer has no influence
on the arrival of another.
5
The Poisson Arrival Process
ke- lt
(lt)
P(X = k) =
k!
Where
l = 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).
6
HANK’s HARDWARE – Arrival Process
• Customers arrive at Hank’s Hardware according to a
Poisson distribution.
• Between 8:00 and 9:00 A.M. an average of 6
customers arrive at the store.
• What is the probability that k customers will arrive
between 8:00 and 8:30 in the morning (k = 0, 1, 2,…)?
7
HANK’s HARDWARE –
An illustration of the Poisson distribution.
• Input to the Poisson
distribution
l = 6 customers per hour.
t = 0.5 hour.
lt = (6)(0.5) = 3.
0 1 2 3 4 5 6 7 8
10k23
(lt)
e
P(X = 01k23 )=
k2!
1!
0!
3!!
lt
= 0.224042
0.149361
0.049787
0.224042
8
HANK’s HARDWARE –
Using Excel for the Poisson probabilities
• Solution
– We can use the POISSON function in Excel to
determine Poisson probabilities.
– Point probability: P(X = k) = ?
• Use Poisson(k, lt, FALSE)
• Example: P(X = 0; lt = 3) = POISSON(0, 1.5, FALSE)
– Cumulative probability: P(Xk) = ?
• Example: P(X3; lt = 3) = Poisson(3, 1.5, TRUE)
9
HANK’s HARDWARE –
Excel Poisson
10
The Waiting Line Characteristics
• Factors that influence the modeling of queues
– Line configuration
– Priority
– Jockeying
– Tandem Queues
– Balking
– Homogeneity
11
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).
12
Jockeying and Balking
• Jockeying occurs when customers switch lines
once they perceived that another line is moving
faster.
• Balking occurs if customers avoid joining the line
when they perceive the line to be too long.
13
Priority Rules
• 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.
14
Tandem Queues
• These are multi-server systems.
• A customer needs to visit several service
stations (usually in a distinct order) to complete
the service process.
• Examples
– Patients in an emergency room.
– Passengers prepare for the next flight.
15
Homogeneity
• A homogeneous customer population is one in
which customers require essentially the same
type of service.
• A non-homogeneous customer population is one
in which customers can be categorized according
to:
– Different arrival patterns
– Different service treatments.
16
The Service Process
• In most business situations, 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.
17
The Exponential Service Time Distribution
f(t) = me-mt
m = the average number of customers
who can be served per time period.
Therefore, 1/m = the mean service time.
The probability that the service time X is less than some “t.”
P(X  t) = 1 - e-mt
18
Schematic illustration of the exponential
distribution
The probability that service is completed
within t time units
P(X  t) = 1 - e-mt
X=t
19
HANK’s HARDWARE – Service time
• Hank’s estimates the average service time to be
1/m = 4 minutes per customer.
• Service time follows an exponential distribution.
• What is the probability that it will take less than 3
minutes to serve the next customer?
20
Using Excel for the Exponential Probabilities
• We can use the EXPDIST function in Excel to
determine exponential probabilities.
• Probability density: f(t) = ?
– Use EXPONDIST(t, m, FALSE)
• Cumulative probability: P(Xk) = ?
– Use EXPONDIST(t, m, TRUE)
21
HANK’s HARDWARE –
Using Excel for the Exponential Probabilities
• The mean number of customers served per
minute is ¼ = ¼(60) = 15 customers per hour.
• P(X < .05 hours) = 1 – e-(15)(.05) = ?
3 minutes = .05 hours
• From Excel we have:
– EXPONDIST(.05,15,TRUE) = .5276
22
HANK’s HARDWARE –
Using Excel for the Exponential Probabilities
=EXPONDIST(B4,B3,TRUE)
f(t)
Exponential Distribution for Mu = 15
16.000
14.000
12.000
10.000
8.000
6.000
4.000
2.000
0.000
0.000
0.075
0.150
0.225
0.300
0.375
t
=EXPONDIST(A10,$B$3,FALSE)
Drag to B11:B26
23
The Exponential Distribution Characteristics
• The memoryless property.
– No additional information about the time left for the completion of a
service, is gained by recording the time elapsed since the service
started.
– For Hank’s, the probability of completing a service within the next 3
minutes is (0.52763) independent of how long the customer has been
served already.
• The Exponential and the Poisson distributions are related to
one another.
– If customer arrivals follow a Poisson distribution with mean rate l,
their interarrival times are exponentially distributed with mean time
24
1/l.
9.3 Performance Measures of
Queuing System
• Performance can be measured by focusing on:
– Customers in queue.
– Customers in the system.
• Performance is measured for a system in steady
state.
25
9.3 Performance Measures of
Queuing System
• The transient period
occurs at the initial
time of operation.
• Initial transient
behavior is not
indicative of long run
performance.
n
Roughly, this
is a transient
period…
Time
26
9.3 Performance Measures of
Queuing System
• The steady state
period follows the
transient period.
• Meaningful long run
performance
measures can be
calculated for the
system when in
steady state.
n
Roughly, this
is a transient
period…
This is a
steady state
period………..
Time
27
9.3 Performance Measures of
Queuing System
In order to achieve steady state, the
effective arrival rate must be less than
the sum of the effective service rates .
k servers
l< m
For one server
l< m1 +m2+…+mk
For k servers
with service rates mi
l< km
Each with
service rate of m
28
Steady State Performance Measures
P0 = 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).
29
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 a steady state condition.
L=lW
Lq = l Wq
For the case of an infinite population
L = Lq + l/m
30
Classification of Queues
• Queuing system can be classified by:
–
–
–
–
–
Arrival process.
Service process.
Number of servers.
System size (infinite/finite waiting line).
Population size.
Example:
M / M / 6 / 10 / 20
• Notation
– M (Markovian) = Poisson arrivals or exponential service time.
– D (Deterministic) = Constant arrival rate or service time.
– G (General) = General probability for arrivals or service time.
31
M/M/1 Queuing System - Assumptions
– Poisson arrival process.
– Exponential service time distribution.
– A single server.
– Potentially infinite queue.
– An infinite population.
32
M / M /1 Queue - Performance Measures
P0 = 1 – (l/m)
Pn = [1 – (l/m)](l/m)n
L = l /(m – l)
Lq = l2 /[m(m – l)]
W = 1 /(m – l)
Wq = l /[m(m – l)]
Pw = l / m
r =l/m
The probability that
a customer waits in
the system more than
“t” is P(X>t) = e-(m - l)t
33
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.
34
MARY’s SHOES - 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 Calculations
m –l = 7.5 – 5 = 2.5 per hr.
P0 = 1 - (l/m) = 1 - (5/7.5) = 0.3333
P(X<10min) = 1 – e-2.5(10/60)
n
n
Pn = [1 - (l/m)](l/m) = (0.3333)(0.6667)
= .565
L = l/(m - l) = 2
Pw = l/m = 0.6667
2
Lq = l /[m(m - l)] = 1.3333
r = l/m = 0.6667
W = 1/(m - l) = 0.4 hours = 24 minutes
Wq = l/[m(m - l)] = 0.26667 hours = 16 minutes
35
MARY’s SHOES Spreadsheet solution
=B4/B5
=A11-B4/B5
=B4/(B5-B4)
=A11/B
4
=C11-1/B5
=1B4/B5
=1E11
=H11*($B$4/$B$5
)
Drag to Cell AL11
36
Economic Analysis of Queuing
Systems
• The performance measures previously developed
are used next to determine a minimal cost queuing
system.
• The procedure requires estimated costs such as:
– Hourly cost per server .
– Customer goodwill cost while waiting in line.
– Customer goodwill cost while being served.
37
Tandem Queuing Systems
• In a Tandem Queuing System a customer must visit
several different servers before service is completed.
Meats
Beverage
• Examples
– All-You-Can-Eat restaurant
38
Tandem Queuing Systems
• In a Tandem Queuing System a customer must visit
several different servers before service is completed.
Meats
Beverage
• Examples
– All-You-Can-Eat restaurant
39
Tandem Queuing Systems
• In a Tandem Queuing System a customer must visit
several different servers before service is completed.
Meats
Beverage
• Examples
– All-You-Can-Eat restaurant
– A drive-in restaurant, where first you place your order, then
pay and receive it in the next window.
– A multiple stage assembly line.
40
Tandem Queuing Systems
• For cases in which customers arrive according to
a Poisson process and service time in each
station is exponential, ….
Total Average Time in the System =
Sum of all average times at the individual stations
41
BIG BOYS SOUND, INC.
• Big Boys sells audio merchandise.
• The sale process is as follows:
– A customer places an order with a sales person.
– The customer goes to the cashier station to pay for
the order.
– After paying, the customer is sent to the pickup desk
to obtain the good.
42
BIG BOYS SOUND, INC.
• Data for a regular Saturday
– Personnel.
• 8 sales persons are on the job.
• 3 cashiers.
• 2 workers in the merchandise pickup area.
– Average service times.
• Average time a sales person waits on a customer is 10 minutes.
• Average time required for the payment process is 3 minutes.
• Average time in the pickup area is 2 minutes.
– Distributions.
• Exponential service time at all the service stations.
• Poisson arrival with a rate of 40 customers an hour.
43
BIG BOYS SOUND, INC.
Only 75% of the arriving customers make a purchase!
What is the average amount of time, a
customer who makes a purchase spends
in the store?
44
BIG BOYS SOUND, INC. – Solution
• This is a Three Station Tandem Queuing System
(.75)(40)=30
Sales Clerks
M/M/8
Pickup desk
M/M/2
Cashiers
M/M/3
W3 = 2.67 minutes
W1 = 14 minutes
W2 = 3.47 minutes
Total = 20.14 minutes.
45