Operations
Management
Module D –
Waiting-Line Models
PowerPoint presentation to accompany
Heizer/Render
Principles of Operations Management, 7e
Operations Management, 9e
© 2008 Prentice Hall, Inc.
D–1
Outline
 Characteristics of a Waiting-Line
System
 Arrival Characteristics
 Waiting-Line Characteristics
 Service Characteristics
 Measuring a Queue’s Performance
 Queuing Costs
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D–2
Outline – Continued
 The Variety of Queuing Models
 Model A(M/M/1): Single-Channel
Queuing Model with Poisson Arrivals
and Exponential Service Times
 Model B(M/M/S): Multiple-Channel
Queuing Model
 Model C(M/D/1): Constant-Service-Time
Model
 Model D: Limited-Population Model
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D–3
Outline – Continued
 Other Queuing Approaches
© 2008 Prentice Hall, Inc.
D–4
Learning Objectives
When you complete this module you
should be able to:
1. Describe the characteristics of
arrivals, waiting lines, and service
systems
2. Apply the single-channel queuing
model equations
3. Conduct a cost analysis for a
waiting line
© 2008 Prentice Hall, Inc.
D–5
Learning Objectives
When you complete this module you
should be able to:
4. Apply the multiple-channel
queuing model formulas
5. Apply the constant-service-time
model equations
6. Perform a limited-population
model analysis
© 2008 Prentice Hall, Inc.
D–6
Waiting Lines
 Often called queuing theory
 Waiting lines are common situations
 Useful in both
manufacturing
and service
areas
© 2008 Prentice Hall, Inc.
D–7
Common Queuing
Situations
Situation
Supermarket
Arrivals in Queue
Grocery shoppers
Highway toll booth
Automobiles
Doctor’s office
Patients
Computer system
Programs to be run
Telephone company
Callers
Bank
Customer
Switching equipment to
forward calls
Transactions handled by teller
Machine
maintenance
Harbor
Broken machines
Repair people fix machines
Ships and barges
Dock workers load and unload
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Service Process
Checkout clerks at cash
register
Collection of tolls at booth
Treatment by doctors and
nurses
Computer processes jobs
Table D.1
D–8
Characteristics of WaitingLine Systems
1. Arrivals or inputs to the system
 Population size, behavior, statistical
distribution
2. Queue discipline, or the waiting line
itself
 Limited or unlimited in length, discipline
of people or items in it
3. The service facility
 Design, statistical distribution of service
times
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D–9
Arrival Characteristics
1. Size of the population
 Unlimited (infinite) or limited (finite)
2. Pattern of arrivals
 Scheduled or random, often a Poisson
distribution
3. Behavior of arrivals
 Wait in the queue and do not switch
lines
 No balking or reneging
© 2008 Prentice Hall, Inc.
D – 10
Parts of a Waiting Line
Population of
dirty cars
Arrivals
from the
general
population …
Queue
(waiting line)
Service
facility
Dave’s
Car Wash
Enter
Arrivals to the system
Arrival Characteristics
 Size of the population
 Behavior of arrivals
 Statistical distribution
of arrivals
© 2008 Prentice Hall, Inc.
Exit the system
In the system
Waiting Line
Characteristics
 Limited vs.
unlimited
 Queue discipline
Exit
Exit the system
Service Characteristics
 Service design
 Statistical distribution
of service
Figure D.1
D – 11
Poisson Distribution
e-x
P(x) =
x!
for x = 0, 1, 2, 3, 4, …
where P(x) = probability of x arrivals
x = number of arrivals per unit of time
 = average arrival rate
e = 2.7183 (which is the base of the
natural logarithms)
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D – 12
Poisson Distribution
0.25 –
0.25 –
0.02 –
0.02 –
Probability
Probability
e-x
Probability = P(x) =
x!
0.15 –
0.10 –
0.05 –
–
Figure D.2
© 2008 Prentice Hall, Inc.
0.15 –
0.10 –
0.05 –
0 1 2 3 4 5 6 7 8 9 x
Distribution for  = 2
–
0 1 2 3 4 5 6 7 8 9 10 11 x
Distribution for  = 4
D – 13
Waiting-Line Characteristics
 Limited or unlimited queue length
 Queue discipline - first-in, first-out
(FIFO) is most common
 Other priority rules may be used in
special circumstances
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D – 14
Service Characteristics
 Queuing system designs
 Single-channel system, multiplechannel system
 Single-phase system, multiphase
system
 Service time distribution
 Constant service time
 Random service times, usually a
negative exponential distribution
© 2008 Prentice Hall, Inc.
D – 15
Queuing System Designs
A family dentist’s office
Queue
Service
facility
Arrivals
Departures
after service
Single-channel, single-phase system
A McDonald’s dual window drive-through
Queue
Arrivals
Phase 1
service
facility
Phase 2
service
facility
Departures
after service
Single-channel, multiphase system
Figure D.3
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D – 16
Queuing System Designs
Most bank and post office service windows
Service
facility
Channel 1
Queue
Arrivals
Service
facility
Channel 2
Departures
after service
Service
facility
Channel 3
Multi-channel, single-phase system
Figure D.3
© 2008 Prentice Hall, Inc.
D – 17
Queuing System Designs
Some college registrations
Queue
Phase 1
service
facility
Channel 1
Phase 2
service
facility
Channel 1
Phase 1
service
facility
Channel 2
Phase 2
service
facility
Channel 2
Arrivals
Departures
after service
Multi-channel, multiphase system
Figure D.3
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D – 18
Probability that service time ≥ 1
Negative Exponential
Distribution
1.0 –
0.9 –
0.8 –
0.7 –
Probability that service time is greater than t = e-µt for t ≥ 1
µ = Average service rate
e = 2.7183
Average service rate (µ) = 3 customers per hour
 Average service time = 20 minutes per customer
0.6 –
0.5 –
0.4 –
0.3 –
Average service rate (µ) =
1 customer per hour
0.2 –
0.1 –
|
|
|
|
|
|
|
|
|
|
|
|
0.0 |–
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00
Figure D.4
© 2008 Prentice Hall, Inc.
Time t (hours)
D – 19
Measuring Queue
Performance
1. Average time that each customer or object
spends in the queue
2. Average queue length
3. Average time each customer spends in the
system
4. Average number of customers in the system
5. Probability that the service facility will be idle
6. Utilization factor for the system
7. Probability of a specific number of customers
in the system
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D – 20
Queuing Costs
Cost
Minimum
Total
cost
Total expected cost
Cost of providing service
Cost of waiting time
Low level
of service
Optimal
service level
High level
of service
Figure D.5
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D – 21
Queuing Models
The four queuing models here all assume:
 Poisson distribution arrivals
 FIFO discipline
 A single-service phase
© 2008 Prentice Hall, Inc.
D – 22
Queuing Models
Model
Name
Example
A
Single-channel
system
(M/M/1)
Information counter
at department store
Number Number
of
of
Channels Phases
Arrival
Rate
Pattern
Service
Time
Pattern
Single
Poisson
Exponential
Single
Population Queue
Size
Discipline
Unlimited
FIFO
Table D.2
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D – 23
Queuing Models
Model
Name
Example
B
Multichannel
(M/M/S)
Airline ticket
counter
Number Number
of
of
Channels Phases
Arrival
Rate
Pattern
Service
Time
Pattern
MultiSingle
channel
Poisson
Exponential
Population Queue
Size
Discipline
Unlimited
FIFO
Table D.2
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D – 24
Queuing Models
Model
Name
C
Constantservice
(M/D/1)
Example
Automated car
wash
Number Number
of
of
Channels Phases
Arrival
Rate
Pattern
Service
Time
Pattern
Population Queue
Size
Discipline
Single
Poisson
Constant
Unlimited
Single
FIFO
Table D.2
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D – 25
Queuing Models
Model
Name
Example
D
Limited
population
(finite population)
Shop with only a
dozen machines
that might break
Number Number
of
of
Channels Phases
Arrival
Rate
Pattern
Service
Time
Pattern
Single
Poisson
Exponential
Single
Population Queue
Size
Discipline
Limited
FIFO
Table D.2
© 2008 Prentice Hall, Inc.
D – 26
Model A – Single-Channel
1. Arrivals are served on a FIFO basis and
every arrival waits to be served
regardless of the length of the queue
2. Arrivals are independent of preceding
arrivals but the average number of
arrivals does not change over time
3. Arrivals are described by a Poisson
probability distribution and come from
an infinite population
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D – 27
Model A – Single-Channel
4. Service times vary from one customer
to the next and are independent of one
another, but their average rate is
known
5. Service times occur according to the
negative exponential distribution
6. The service rate is faster than the
arrival rate
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D – 28
Model A – Single-Channel
 = Mean number of arrivals per time period
µ = Mean number of units served per time period
Ls = Average number of units (customers) in the
system (waiting and being served)

=
µ–
Ws = Average time a unit spends in the system
(waiting time plus service time)
1
=
µ–
Table D.3
© 2008 Prentice Hall, Inc.
D – 29
Model A – Single-Channel
Lq = Average number of units waiting in the
queue
2

=
µ(µ – )
Wq = Average time a unit spends waiting in the
queue

=
µ(µ – )
p = Utilization factor for the system

=
µ
Table D.3
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D – 30
Model A – Single-Channel
P0 = Probability of 0 units in the system (that is,
the service unit is idle)

µ
Pn > k = Probability of more than k units in the
system, where n is the number of units in
the system
= 1–
=

µ
k+1
Table D.3
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D – 31
Single-Channel Example
 = 2 cars arriving/hour
µ = 3 cars serviced/hour
2

Ls =
=
= 2 cars in the system on average
3
2
µ–
1
1
Ws =
=
= 1 hour average waiting time in
µ–
3-2
the system
22
2
Lq =
=
= 1.33 cars waiting in line
3(3
2)
µ(µ – )
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D – 32
Single-Channel Example
 = 2 cars arriving/hour
µ = 3 cars serviced/hour
2

Wq =
=
= 2/3 hour = 40 minute
3(3
2)
µ(µ – )
average waiting time
p = /µ = 2/3 = 66.6% of time mechanic is busy

P0 = 1 = .33 probability there are 0 cars in the
µ
system
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D – 33
Single-Channel Example
Probability of more than k Cars in the System
k
0
1
2
3
4
5
6
7
© 2008 Prentice Hall, Inc.
Pn > k = (2/3)k + 1
.667  Note that this is equal to 1 - P0 = 1 - .33
.444
.296
.198  Implies that there is a 19.8% chance that
more than 3 cars are in the system
.132
.088
.058
.039
D – 34
Single-Channel Economics
Customer dissatisfaction
and lost goodwill = $10 per hour
Wq = 2/3 hour
Total arrivals = 16 per day
Mechanic’s salary = $56 per day
Total hours
customers spend
waiting per day
=
2
2
(16) = 10
hours
3
3
Customer waiting-time cost = $10 10
2
3
= $106.67
Total expected costs = $106.67 + $56 = $162.67
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D – 35
Multi-Channel Model
M = number of channels open
 = average arrival rate
µ = average service rate at each channel
P0 =
1
M–1
∑
n=0
1
n!

µ
µ(/µ)
n
1 
+
M! µ
M
for Mµ > 
Mµ
Mµ - 
M

Ls =
P +
2 0
µ
(M - 1)!(Mµ - )
Table D.4
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D – 36
Multi-Channel Model
µ(/µ)
M
1
Ws =
P +
=
2 0
µ
(M - 1)!(Mµ - )
Ls


Lq = Ls –
µ
Lq
1
Wq = Ws –
=

µ
Table D.4
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D – 37
Multi-Channel Example
 = 2
µ = 3
M = 2
1
P0 =
=
1
∑
1
n!
n
2
3
1
+
2!
2
2
3
2(3)
1
2
2(3) - 2
n=0
Ls =
Ws =
3/4
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2
(2)(3(2/3)2
1! 2(3) - 2 2
=
3
8
1
2
2
+
3
=
3
4
1
2
3
Lq =
–
=
12
3
4
Wq =
.083
2
= .0415
D – 38
Multi-Channel Example
© 2008 Prentice Hall, Inc.
Single Channel
Two Channels
P0
.33
.5
Ls
2 cars
.75 cars
Ws
60 minutes
22.5 minutes
Lq
1.33 cars
.083 cars
Wq
40 minutes
2.5 minutes
D – 39
Waiting Line Tables
Poisson Arrivals, Exponential Service Times
Number of Service Channels, M
ρ
1
.10
.0111
.25
.0833
.0039
.50
.5000
.0333
.0030
.75
2.2500
.1227
.0147
1.0
.3333
.0454
.0067
1.6
2.8444
.3128
.0604
.0121
2.0
.8888
.1739
.0398
2.6
4.9322
.6581
.1609
1.5282
.3541
3.0
4.0
2
3
4
5
2.2164
Table D.5
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D – 40
Waiting Line Table Example
Bank tellers and customers
 = 18, µ = 20
Utilization factor ρ = /µ = .90
Lq
Wq = 
From Table D.5
Number of
service windows
1 window
M
1
Number
in queue
8.1
Time in queue
.45 hrs, 27 minutes
2 windows
2
.2285
.0127 hrs, ¾ minute
3 windows
3
.03
.0017 hrs, 6 seconds
4 windows
4
.0041
.0003 hrs, 1 second
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D – 41
Constant-Service Model
Average length
of queue
2

Lq =
2µ(µ – )
Average waiting time
in queue
Wq =
Average number of
customers in system
Ls = Lq +
Average time
in the system

2µ(µ – )

µ
Ws = W q +
1
µ
Table D.6
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D – 42
Constant-Service Example
Trucks currently wait 15 minutes on average
Truck and driver cost $60 per hour
Automated compactor service rate (µ) = 12 trucks per hour
Arrival rate () = 8 per hour
Compactor costs $3 per truck
Current waiting cost per trip = (1/4 hr)($60) = $15 /trip
1
8
Wq =
=
hour
12
2(12)(12 – 8)
Waiting cost/trip
with compactor = (1/12 hr wait)($60/hr cost)
Savings with
= $15 (current) – $5(new)
new equipment
Cost of new equipment amortized
Net savings
© 2008 Prentice Hall, Inc.
= $ 5 /trip
= $10 /trip
= $ 3 /trip
= $ 7 /trip
D – 43
Limited-Population Model
T
T+U
Average number running: J = NF(1 - X)
Service factor: X =
Average number waiting: L = N(1 - F)
Average number being serviced: H = FNX
T(1 - F)
Average waiting time: W =
XF
Number of population: N = J + L + H
Table D.7
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D – 44
Limited-Population Model
D = Probability that a unit
N = Number of potential
T
will havefactor:
to wait inX =
customers
Service
T+U
queue
F = Average
Efficiencynumber
factor
T = JAverage
service
running:
= NF(1
- X) time
H = Average
Average number
of waiting:
units U =LAverage
number
= N(1 time
- F) between
being served
unit service
Average number being serviced:
H = FNX
requirements
T(1 - F)time a unit
J = Average number of units W = Average
Average
waiting
not in queue
or in time: W = waitsXF
in line
service bay
Number of population: N = J + L + H
L = Average number of units X = Service factor
waiting for service
M = Number of service
channels
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D – 45
Finite Queuing Table
X
.012
.025
.050
.060
.070
.080
.090
Table D.8
© 2008 Prentice Hall, Inc.
.100
M
1
1
1
2
1
2
1
2
1
2
1
2
1
D
.048
.100
.198
.020
.237
.027
.275
.035
.313
.044
.350
.054
.386
F
.999
.997
.989
.999
.983
.999
.977
.998
.969
.998
.960
.997
.950
D – 46
Limited-Population Example
Each of 5 printers requires repair after 20 hours (U) of use
One technician can service a printer in 2 hours (T)
Printer downtime costs $120/hour
Technician costs $25/hour
2
Service factor: X =
= .091 (close to .090)
2 + 20
For M = 1, D = .350 and F = .960
For M = 2, D = .044 and F = .998
Average number of printers working:
For M = 1, J = (5)(.960)(1 - .091) = 4.36
For M = 2, J = (5)(.998)(1 - .091) = 4.54
© 2008 Prentice Hall, Inc.
D – 47
Limited-Population Example
Average
Average
Numberrequire Cost/Hr
for 20 Cost/Hr
for of use
Each of 5 printers
repair after
hours (U)
Number
Printers
Total
One of
technician
can serviceDowntime
a printer in 2Technicians
hours (T)
Technicians
Down (N - J)
(N - J)$120
($25/hr)
Cost/Hr
Printer downtime costs $120/hour
1
.64 $25/hour $76.80
Technician
costs
$25.00
$101.80
2 $55.20
$50.00to .090)
$105.20
= .091 (close
2 + 20
For M = 1, D = .350 and F = .960
2
.46 X =
Service
factor:
For M = 2, D = .044 and F = .998
Average number of printers working:
For M = 1, J = (5)(.960)(1 - .091) = 4.36
For M = 2, J = (5)(.998)(1 - .091) = 4.54
© 2008 Prentice Hall, Inc.
D – 48
Other Queuing Approaches
 The single-phase models cover many
queuing situations
 Variations of the four single-phase
systems are possible
 Multiphase models
exist for more
complex situations
© 2008 Prentice Hall, Inc.
D – 49