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D - 1© 2011 Pearson Education, Inc. publishing as Prentice Hall
D Waiting-Line Models
PowerPoint presentation to accompany
Heizer and Render
Operations Management, 10e
Principles of Operations Management, 8e
PowerPoint slides by Jeff Heyl
D - 2© 2011 Pearson Education, Inc. publishing as Prentice Hall
Outline
 Queuing Theory
 Characteristics of a Waiting-Line
System
 Arrival Characteristics
 Waiting-Line Characteristics
 Service Characteristics
 Measuring a Queue’s Performance
 Queuing Costs
D - 3© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
 Little’s Law
 Model D: Limited-Population Model
D - 4© 2011 Pearson Education, Inc. publishing as Prentice Hall
Outline – Continued
 Other Queuing Approaches
D - 5© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
D - 6© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
D - 7© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing Theory
 The study of waiting lines
 Waiting lines are common
situations
 Useful in both
manufacturing
and service
areas
D - 8© 2011 Pearson Education, Inc. publishing as Prentice Hall
Common Queuing
Situations
Situation Arrivals in Queue Service Process
Supermarket Grocery shoppers Checkout clerks at cash
register
Highway toll booth Automobiles Collection of tolls at booth
Doctor’s office Patients Treatment by doctors and
nurses
Computer system Programs to be run Computer processes jobs
Telephone company Callers Switching equipment to
forward calls
Bank Customer Transactions handled by teller
Machine
maintenance
Broken machines Repair people fix machines
Harbor Ships and barges Dock workers load and unload
Table D.1
D - 9© 2011 Pearson Education, Inc. publishing as Prentice Hall
Characteristics of Waiting-
Line 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
D - 10© 2011 Pearson Education, Inc. publishing as Prentice Hall
Parts of a Waiting Line
Figure D.1
Dave’s
Car Wash
Enter Exit
Population of
dirty cars
Arrivals
from the
general
population …
Queue
(waiting line)
Service
facility
Exit the system
Arrivals to the system Exit the systemIn the system
Arrival Characteristics
 Size of the population
 Behavior of arrivals
 Statistical distribution
of arrivals
Waiting Line
Characteristics
 Limited vs.
unlimited
 Queue discipline
Service Characteristics
 Service design
 Statistical distribution
of service
D - 11© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
D - 12© 2011 Pearson Education, Inc. publishing as Prentice Hall
Poisson Distribution
P(x) = for x = 0, 1, 2, 3, 4, …
e-x
x!
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)
D - 13© 2011 Pearson Education, Inc. publishing as Prentice Hall
Poisson Distribution
Probability = P(x) =
e-x
x!
0.25 –
0.02 –
0.15 –
0.10 –
0.05 –
–
Probability
0 1 2 3 4 5 6 7 8 9
Distribution for  = 2
x
0.25 –
0.02 –
0.15 –
0.10 –
0.05 –
–
Probability
0 1 2 3 4 5 6 7 8 9
Distribution for  = 4
x10 11
Figure D.2
D - 14© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
D - 15© 2011 Pearson Education, Inc. publishing as Prentice Hall
Service Characteristics
 Queuing system designs
 Single-channel system, multiple-
channel system
 Single-phase system, multiphase
system
 Service time distribution
 Constant service time
 Random service times, usually a
negative exponential distribution
D - 16© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing System Designs
Figure D.3
Departures
after service
Single-channel, single-phase system
Queue
Arrivals
Single-channel, multiphase system
Arrivals
Departures
after service
Phase 1
service
facility
Phase 2
service
facility
Service
facility
Queue
A family dentist’s office
A McDonald’s dual window drive-through
D - 17© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing System Designs
Figure D.3
Multi-channel, single-phase system
Arrivals
Queue
Most bank and post office service windows
Departures
after service
Service
facility
Channel 1
Service
facility
Channel 2
Service
facility
Channel 3
D - 18© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing System Designs
Figure D.3
Multi-channel, multiphase system
Arrivals
Queue
Some college registrations
Departures
after service
Phase 2
service
facility
Channel 1
Phase 2
service
facility
Channel 2
Phase 1
service
facility
Channel 1
Phase 1
service
facility
Channel 2
D - 19© 2011 Pearson Education, Inc. publishing as Prentice Hall
Negative Exponential
Distribution
Figure D.4
1.0 –
0.9 –
0.8 –
0.7 –
0.6 –
0.5 –
0.4 –
0.3 –
0.2 –
0.1 –
0.0 –
Probabilitythatservicetime≥1
| | | | | | | | | | | | |
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
Time t (hours)
Probability that service time is greater than t = e-µt for t ≥ 1
µ = Average service rate
e = 2.7183
Average service rate (µ) =
1 customer per hour
Average service rate (µ) = 3 customers per hour
 Average service time = 20 minutes per customer
D - 20© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
D - 21© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing Costs
Figure D.5
Total expected cost
Cost of providing service
Cost
Low level
of service
High level
of service
Cost of waiting time
Minimum
Total
cost
Optimal
service level
D - 22© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing Models
The four queuing models here all assume:
 Poisson distribution arrivals
 FIFO discipline
 A single-service phase
D - 23© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing Models
Table D.2
Model Name Example
A Single-channel Information counter
system at department store
(M/M/1)
Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Exponential Unlimited FIFO
D - 24© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing Models
Table D.2
Model Name Example
B Multichannel Airline ticket
(M/M/S) counter
Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Multi- Single Poisson Exponential Unlimited FIFO
channel
D - 25© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing Models
Table D.2
Model Name Example
C Constant- Automated car
service wash
(M/D/1)
Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Constant Unlimited FIFO
D - 26© 2011 Pearson Education, Inc. publishing as Prentice Hall
Queuing Models
Table D.2
Model Name Example
D Limited Shop with only a
population dozen machines
(finite population) that might break
Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Exponential Limited FIFO
D - 27© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
D - 28© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
D - 29© 2011 Pearson Education, Inc. publishing as Prentice Hall
Model A – Single-Channel
Table D.3
 = 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)
=
Ws = Average time a unit spends in the system
(waiting time plus service time)
=

µ – 
1
µ – 
D - 30© 2011 Pearson Education, Inc. publishing as Prentice Hall
Model A – Single-Channel
Table D.3
Lq = Average number of units waiting in the
queue
=
Wq = Average time a unit spends waiting in the
queue
=
 = Utilization factor for the system
=
2
µ(µ – )

µ(µ – )

µ
D - 31© 2011 Pearson Education, Inc. publishing as Prentice Hall
Model A – Single-Channel
Table D.3
P0 = Probability of 0 units in the system (that is,
the service unit is idle)
= 1 –
Pn > k = Probability of more than k units in the
system, where n is the number of units in
the system
=

µ

µ
k + 1
D - 32© 2011 Pearson Education, Inc. publishing as Prentice Hall
Single-Channel Example
 = 2 cars arriving/hour µ = 3 cars serviced/hour
Ls = = = 2 cars in the system on average
Ws = = = 1 hour average waiting time in
the system
Lq = = = 1.33 cars waiting in line
2
µ(µ – )

µ – 
1
µ – 
2
3 - 2
1
3 - 2
22
3(3 - 2)
D - 33© 2011 Pearson Education, Inc. publishing as Prentice Hall
Single-Channel Example
Wq = = = 2/3 hour = 40 minute
average waiting time
 = /µ = 2/3 = 66.6% of time mechanic is busy

µ(µ – )
2
3(3 - 2)

µ
P0 = 1 - = .33 probability there are 0 cars in the
system
 = 2 cars arriving/hour µ = 3 cars serviced/hour
D - 34© 2011 Pearson Education, Inc. publishing as Prentice Hall
Single-Channel Example
Probability of more than k Cars in the System
k Pn > k = (2/3)k + 1
0 .667  Note that this is equal to 1 - P0 = 1 - .33
1 .444
2 .296
3 .198  Implies that there is a 19.8% chance that
more than 3 cars are in the system
4 .132
5 .088
6 .058
7 .039
D - 35© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
= (16) = 10 hours
2
3
2
3
Customer waiting-time cost = $10 10 = $106.67
2
3
Total expected costs = $106.67 + $56 = $162.67
D - 36© 2011 Pearson Education, Inc. publishing as Prentice Hall
Multi-Channel Model
Table D.4
M = number of channels open
 = average arrival rate
µ = average service rate at each channel
P0 = for Mµ > 
1
1
M!
1
n!
Mµ
Mµ - 
M – 1
n = 0

µ
n

µ
M
+∑
Ls = P0 +
µ(/µ)
M
(M - 1)!(Mµ - )
2

µ
D - 37© 2011 Pearson Education, Inc. publishing as Prentice Hall
Multi-Channel Model
Table D.4
Ws = P0 + =
µ(/µ)
M
(M - 1)!(Mµ - )
2
1
µ
Ls

Lq = Ls –

µ
Wq = Ws – =
1
µ
Lq

D - 38© 2011 Pearson Education, Inc. publishing as Prentice Hall
Multi-Channel Example
 = 2 µ = 3 M = 2
P0 = =
1
1
2!
1
n!
2(3)
2(3) - 2
1
n = 0
2
3
n
2
3
2
+∑
1
2
Ls = + =
(2)(3(2/3)2 2
3
1! 2(3) - 2 2
1
2
3
4
Wq = = .0415
.083
2
Ws = =
3/4
2
3
8
Lq = – =
2
3
3
4
1
12
D - 39© 2011 Pearson Education, Inc. publishing as Prentice Hall
Multi-Channel Example
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 - 40© 2011 Pearson Education, Inc. publishing as Prentice Hall
Waiting Line Tables
Table D.5
Poisson Arrivals, Exponential Service Times
Number of Service Channels, M
ρ 1 2 3 4 5
.10 .0111
.25 .0833 .0039
.50 .5000 .0333 .0030
.75 2.2500 .1227 .0147
.90 3.1000 .2285 .0300 .0041
1.0 .3333 .0454 .0067
1.6 2.8444 .3128 .0604 .0121
2.0 .8888 .1739 .0398
2.6 4.9322 .6581 .1609
3.0 1.5282 .3541
4.0 2.2164
D - 41© 2011 Pearson Education, Inc. publishing as Prentice Hall
Waiting Line Table Example
Bank tellers and customers
 = 18, µ = 20
From Table D.5
Utilization factor  = /µ = .90 Wq =
Lq

Number of
service windows M
Number
in queue Time in queue
1 window 1 8.1 .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
D - 42© 2011 Pearson Education, Inc. publishing as Prentice Hall
Constant-Service Model
Table D.6
Lq = 2
2µ(µ – )
Average length
of queue
Wq =

2µ(µ – )
Average waiting time
in queue

µ
Ls = Lq +
Average number of
customers in system
Ws = Wq +
1
µ
Average time
in the system
D - 43© 2011 Pearson Education, Inc. publishing as Prentice Hall
Net savings = $ 7 /trip
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
Wq = = hour
8
2(12)(12 – 8)
1
12
Waiting cost/trip
with compactor = (1/12 hr wait)($60/hr cost) = $ 5 /trip
Savings with
new equipment
= $15 (current) – $5(new) = $10 /trip
Cost of new equipment amortized = $ 3 /trip
D - 44© 2011 Pearson Education, Inc. publishing as Prentice Hall
Little’s Law
 A queuing system in steady state
L = W (which is the same as W = L/
Lq = Wq (which is the same as Wq = Lq/
 Once one of these parameters is known, the
other can be easily found
 It makes no assumptions about the probability
distribution of arrival and service times
 Applies to all queuing models except the limited
population model
D - 45© 2011 Pearson Education, Inc. publishing as Prentice Hall
Limited-Population Model
Table D.7
Service factor: X =
Average number running: J = NF(1 - X)
Average number waiting: L = N(1 - F)
Average number being serviced: H = FNX
Average waiting time: W =
Number of population: N = J + L + H
T
T + U
T(1 - F)
XF
D - 46© 2011 Pearson Education, Inc. publishing as Prentice Hall
Limited-Population Model
D = Probability that a unit
will have to wait in
queue
N = Number of potential
customers
F = Efficiency factor T = Average service time
H = Average number of units
being served
U = Average time between
unit service
requirements
J = Average number of units
not in queue or in
service bay
W = Average time a unit
waits in line
L = Average number of units
waiting for service
X = Service factor
M = Number of service
channels
D - 47© 2011 Pearson Education, Inc. publishing as Prentice Hall
Finite Queuing Table
Table D.8
X M D F
.012 1 .048 .999
.025 1 .100 .997
.050 1 .198 .989
.060 2 .020 .999
1 .237 .983
.070 2 .027 .999
1 .275 .977
.080 2 .035 .998
1 .313 .969
.090 2 .044 .998
1 .350 .960
.100 2 .054 .997
1 .386 .950
D - 48© 2011 Pearson Education, Inc. publishing as Prentice Hall
Limited-Population Example
Service factor: X = = .091 (close to .090)
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
2
2 + 20
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
D - 49© 2011 Pearson Education, Inc. publishing as Prentice Hall
Limited-Population Example
Service factor: X = = .091 (close to .090)
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
2
2 + 20
Each of 5 printers require 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
Number of
Technicians
Average
Number
Printers
Down (N - J)
Average
Cost/Hr for
Downtime
(N - J)$120
Cost/Hr for
Technicians
($25/hr)
Total
Cost/Hr
1 .64 $76.80 $25.00 $101.80
2 .46 $55.20 $50.00 $105.20
D - 50© 2011 Pearson Education, Inc. publishing as Prentice Hall
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
D - 51© 2011 Pearson Education, Inc. publishing as Prentice Hall
All rights reserved. No part of this publication may be reproduced, stored in a retrieval
system, or transmitted, in any form or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written permission of the publisher.
Printed in the United States of America.

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Heizer om10 mod_d

  • 1. D - 1© 2011 Pearson Education, Inc. publishing as Prentice Hall D Waiting-Line Models PowerPoint presentation to accompany Heizer and Render Operations Management, 10e Principles of Operations Management, 8e PowerPoint slides by Jeff Heyl
  • 2. D - 2© 2011 Pearson Education, Inc. publishing as Prentice Hall Outline  Queuing Theory  Characteristics of a Waiting-Line System  Arrival Characteristics  Waiting-Line Characteristics  Service Characteristics  Measuring a Queue’s Performance  Queuing Costs
  • 3. D - 3© 2011 Pearson Education, Inc. publishing as Prentice Hall 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  Little’s Law  Model D: Limited-Population Model
  • 4. D - 4© 2011 Pearson Education, Inc. publishing as Prentice Hall Outline – Continued  Other Queuing Approaches
  • 5. D - 5© 2011 Pearson Education, Inc. publishing as Prentice Hall 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
  • 6. D - 6© 2011 Pearson Education, Inc. publishing as Prentice Hall 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
  • 7. D - 7© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Theory  The study of waiting lines  Waiting lines are common situations  Useful in both manufacturing and service areas
  • 8. D - 8© 2011 Pearson Education, Inc. publishing as Prentice Hall Common Queuing Situations Situation Arrivals in Queue Service Process Supermarket Grocery shoppers Checkout clerks at cash register Highway toll booth Automobiles Collection of tolls at booth Doctor’s office Patients Treatment by doctors and nurses Computer system Programs to be run Computer processes jobs Telephone company Callers Switching equipment to forward calls Bank Customer Transactions handled by teller Machine maintenance Broken machines Repair people fix machines Harbor Ships and barges Dock workers load and unload Table D.1
  • 9. D - 9© 2011 Pearson Education, Inc. publishing as Prentice Hall Characteristics of Waiting- Line 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
  • 10. D - 10© 2011 Pearson Education, Inc. publishing as Prentice Hall Parts of a Waiting Line Figure D.1 Dave’s Car Wash Enter Exit Population of dirty cars Arrivals from the general population … Queue (waiting line) Service facility Exit the system Arrivals to the system Exit the systemIn the system Arrival Characteristics  Size of the population  Behavior of arrivals  Statistical distribution of arrivals Waiting Line Characteristics  Limited vs. unlimited  Queue discipline Service Characteristics  Service design  Statistical distribution of service
  • 11. D - 11© 2011 Pearson Education, Inc. publishing as Prentice Hall 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
  • 12. D - 12© 2011 Pearson Education, Inc. publishing as Prentice Hall Poisson Distribution P(x) = for x = 0, 1, 2, 3, 4, … e-x x! 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)
  • 13. D - 13© 2011 Pearson Education, Inc. publishing as Prentice Hall Poisson Distribution Probability = P(x) = e-x x! 0.25 – 0.02 – 0.15 – 0.10 – 0.05 – – Probability 0 1 2 3 4 5 6 7 8 9 Distribution for  = 2 x 0.25 – 0.02 – 0.15 – 0.10 – 0.05 – – Probability 0 1 2 3 4 5 6 7 8 9 Distribution for  = 4 x10 11 Figure D.2
  • 14. D - 14© 2011 Pearson Education, Inc. publishing as Prentice Hall 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
  • 15. D - 15© 2011 Pearson Education, Inc. publishing as Prentice Hall Service Characteristics  Queuing system designs  Single-channel system, multiple- channel system  Single-phase system, multiphase system  Service time distribution  Constant service time  Random service times, usually a negative exponential distribution
  • 16. D - 16© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing System Designs Figure D.3 Departures after service Single-channel, single-phase system Queue Arrivals Single-channel, multiphase system Arrivals Departures after service Phase 1 service facility Phase 2 service facility Service facility Queue A family dentist’s office A McDonald’s dual window drive-through
  • 17. D - 17© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing System Designs Figure D.3 Multi-channel, single-phase system Arrivals Queue Most bank and post office service windows Departures after service Service facility Channel 1 Service facility Channel 2 Service facility Channel 3
  • 18. D - 18© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing System Designs Figure D.3 Multi-channel, multiphase system Arrivals Queue Some college registrations Departures after service Phase 2 service facility Channel 1 Phase 2 service facility Channel 2 Phase 1 service facility Channel 1 Phase 1 service facility Channel 2
  • 19. D - 19© 2011 Pearson Education, Inc. publishing as Prentice Hall Negative Exponential Distribution Figure D.4 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – Probabilitythatservicetime≥1 | | | | | | | | | | | | | 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 Time t (hours) Probability that service time is greater than t = e-µt for t ≥ 1 µ = Average service rate e = 2.7183 Average service rate (µ) = 1 customer per hour Average service rate (µ) = 3 customers per hour  Average service time = 20 minutes per customer
  • 20. D - 20© 2011 Pearson Education, Inc. publishing as Prentice Hall 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
  • 21. D - 21© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Costs Figure D.5 Total expected cost Cost of providing service Cost Low level of service High level of service Cost of waiting time Minimum Total cost Optimal service level
  • 22. D - 22© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models The four queuing models here all assume:  Poisson distribution arrivals  FIFO discipline  A single-service phase
  • 23. D - 23© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models Table D.2 Model Name Example A Single-channel Information counter system at department store (M/M/1) Number Number Arrival Service of of Rate Time Population Queue Channels Phases Pattern Pattern Size Discipline Single Single Poisson Exponential Unlimited FIFO
  • 24. D - 24© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models Table D.2 Model Name Example B Multichannel Airline ticket (M/M/S) counter Number Number Arrival Service of of Rate Time Population Queue Channels Phases Pattern Pattern Size Discipline Multi- Single Poisson Exponential Unlimited FIFO channel
  • 25. D - 25© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models Table D.2 Model Name Example C Constant- Automated car service wash (M/D/1) Number Number Arrival Service of of Rate Time Population Queue Channels Phases Pattern Pattern Size Discipline Single Single Poisson Constant Unlimited FIFO
  • 26. D - 26© 2011 Pearson Education, Inc. publishing as Prentice Hall Queuing Models Table D.2 Model Name Example D Limited Shop with only a population dozen machines (finite population) that might break Number Number Arrival Service of of Rate Time Population Queue Channels Phases Pattern Pattern Size Discipline Single Single Poisson Exponential Limited FIFO
  • 27. D - 27© 2011 Pearson Education, Inc. publishing as Prentice Hall 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
  • 28. D - 28© 2011 Pearson Education, Inc. publishing as Prentice Hall 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
  • 29. D - 29© 2011 Pearson Education, Inc. publishing as Prentice Hall Model A – Single-Channel Table D.3  = 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) = Ws = Average time a unit spends in the system (waiting time plus service time) =  µ –  1 µ – 
  • 30. D - 30© 2011 Pearson Education, Inc. publishing as Prentice Hall Model A – Single-Channel Table D.3 Lq = Average number of units waiting in the queue = Wq = Average time a unit spends waiting in the queue =  = Utilization factor for the system = 2 µ(µ – )  µ(µ – )  µ
  • 31. D - 31© 2011 Pearson Education, Inc. publishing as Prentice Hall Model A – Single-Channel Table D.3 P0 = Probability of 0 units in the system (that is, the service unit is idle) = 1 – Pn > k = Probability of more than k units in the system, where n is the number of units in the system =  µ  µ k + 1
  • 32. D - 32© 2011 Pearson Education, Inc. publishing as Prentice Hall Single-Channel Example  = 2 cars arriving/hour µ = 3 cars serviced/hour Ls = = = 2 cars in the system on average Ws = = = 1 hour average waiting time in the system Lq = = = 1.33 cars waiting in line 2 µ(µ – )  µ –  1 µ –  2 3 - 2 1 3 - 2 22 3(3 - 2)
  • 33. D - 33© 2011 Pearson Education, Inc. publishing as Prentice Hall Single-Channel Example Wq = = = 2/3 hour = 40 minute average waiting time  = /µ = 2/3 = 66.6% of time mechanic is busy  µ(µ – ) 2 3(3 - 2)  µ P0 = 1 - = .33 probability there are 0 cars in the system  = 2 cars arriving/hour µ = 3 cars serviced/hour
  • 34. D - 34© 2011 Pearson Education, Inc. publishing as Prentice Hall Single-Channel Example Probability of more than k Cars in the System k Pn > k = (2/3)k + 1 0 .667  Note that this is equal to 1 - P0 = 1 - .33 1 .444 2 .296 3 .198  Implies that there is a 19.8% chance that more than 3 cars are in the system 4 .132 5 .088 6 .058 7 .039
  • 35. D - 35© 2011 Pearson Education, Inc. publishing as Prentice Hall 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 = (16) = 10 hours 2 3 2 3 Customer waiting-time cost = $10 10 = $106.67 2 3 Total expected costs = $106.67 + $56 = $162.67
  • 36. D - 36© 2011 Pearson Education, Inc. publishing as Prentice Hall Multi-Channel Model Table D.4 M = number of channels open  = average arrival rate µ = average service rate at each channel P0 = for Mµ >  1 1 M! 1 n! Mµ Mµ -  M – 1 n = 0  µ n  µ M +∑ Ls = P0 + µ(/µ) M (M - 1)!(Mµ - ) 2  µ
  • 37. D - 37© 2011 Pearson Education, Inc. publishing as Prentice Hall Multi-Channel Model Table D.4 Ws = P0 + = µ(/µ) M (M - 1)!(Mµ - ) 2 1 µ Ls  Lq = Ls –  µ Wq = Ws – = 1 µ Lq 
  • 38. D - 38© 2011 Pearson Education, Inc. publishing as Prentice Hall Multi-Channel Example  = 2 µ = 3 M = 2 P0 = = 1 1 2! 1 n! 2(3) 2(3) - 2 1 n = 0 2 3 n 2 3 2 +∑ 1 2 Ls = + = (2)(3(2/3)2 2 3 1! 2(3) - 2 2 1 2 3 4 Wq = = .0415 .083 2 Ws = = 3/4 2 3 8 Lq = – = 2 3 3 4 1 12
  • 39. D - 39© 2011 Pearson Education, Inc. publishing as Prentice Hall Multi-Channel Example 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
  • 40. D - 40© 2011 Pearson Education, Inc. publishing as Prentice Hall Waiting Line Tables Table D.5 Poisson Arrivals, Exponential Service Times Number of Service Channels, M ρ 1 2 3 4 5 .10 .0111 .25 .0833 .0039 .50 .5000 .0333 .0030 .75 2.2500 .1227 .0147 .90 3.1000 .2285 .0300 .0041 1.0 .3333 .0454 .0067 1.6 2.8444 .3128 .0604 .0121 2.0 .8888 .1739 .0398 2.6 4.9322 .6581 .1609 3.0 1.5282 .3541 4.0 2.2164
  • 41. D - 41© 2011 Pearson Education, Inc. publishing as Prentice Hall Waiting Line Table Example Bank tellers and customers  = 18, µ = 20 From Table D.5 Utilization factor  = /µ = .90 Wq = Lq  Number of service windows M Number in queue Time in queue 1 window 1 8.1 .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
  • 42. D - 42© 2011 Pearson Education, Inc. publishing as Prentice Hall Constant-Service Model Table D.6 Lq = 2 2µ(µ – ) Average length of queue Wq =  2µ(µ – ) Average waiting time in queue  µ Ls = Lq + Average number of customers in system Ws = Wq + 1 µ Average time in the system
  • 43. D - 43© 2011 Pearson Education, Inc. publishing as Prentice Hall Net savings = $ 7 /trip 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 Wq = = hour 8 2(12)(12 – 8) 1 12 Waiting cost/trip with compactor = (1/12 hr wait)($60/hr cost) = $ 5 /trip Savings with new equipment = $15 (current) – $5(new) = $10 /trip Cost of new equipment amortized = $ 3 /trip
  • 44. D - 44© 2011 Pearson Education, Inc. publishing as Prentice Hall Little’s Law  A queuing system in steady state L = W (which is the same as W = L/ Lq = Wq (which is the same as Wq = Lq/  Once one of these parameters is known, the other can be easily found  It makes no assumptions about the probability distribution of arrival and service times  Applies to all queuing models except the limited population model
  • 45. D - 45© 2011 Pearson Education, Inc. publishing as Prentice Hall Limited-Population Model Table D.7 Service factor: X = Average number running: J = NF(1 - X) Average number waiting: L = N(1 - F) Average number being serviced: H = FNX Average waiting time: W = Number of population: N = J + L + H T T + U T(1 - F) XF
  • 46. D - 46© 2011 Pearson Education, Inc. publishing as Prentice Hall Limited-Population Model D = Probability that a unit will have to wait in queue N = Number of potential customers F = Efficiency factor T = Average service time H = Average number of units being served U = Average time between unit service requirements J = Average number of units not in queue or in service bay W = Average time a unit waits in line L = Average number of units waiting for service X = Service factor M = Number of service channels
  • 47. D - 47© 2011 Pearson Education, Inc. publishing as Prentice Hall Finite Queuing Table Table D.8 X M D F .012 1 .048 .999 .025 1 .100 .997 .050 1 .198 .989 .060 2 .020 .999 1 .237 .983 .070 2 .027 .999 1 .275 .977 .080 2 .035 .998 1 .313 .969 .090 2 .044 .998 1 .350 .960 .100 2 .054 .997 1 .386 .950
  • 48. D - 48© 2011 Pearson Education, Inc. publishing as Prentice Hall Limited-Population Example Service factor: X = = .091 (close to .090) 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 2 2 + 20 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
  • 49. D - 49© 2011 Pearson Education, Inc. publishing as Prentice Hall Limited-Population Example Service factor: X = = .091 (close to .090) 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 2 2 + 20 Each of 5 printers require 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 Number of Technicians Average Number Printers Down (N - J) Average Cost/Hr for Downtime (N - J)$120 Cost/Hr for Technicians ($25/hr) Total Cost/Hr 1 .64 $76.80 $25.00 $101.80 2 .46 $55.20 $50.00 $105.20
  • 50. D - 50© 2011 Pearson Education, Inc. publishing as Prentice Hall 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
  • 51. D - 51© 2011 Pearson Education, Inc. publishing as Prentice Hall All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.