1. 10/16/2010
Outline
D Waiting-Line Models
Queuing Theory
Characteristics of a Waiting-Line
System
PowerPoint presentation to accompany
p
Heizer and Render
p y Arrival Characteristics
Operations Management, 10e
Principles of Operations Management, 8e
Waiting-Line Characteristics
PowerPoint slides by Jeff Heyl
Service Characteristics
Measuring a Queue’s Performance
Queuing Costs
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Outline – Continued Outline – Continued
The Variety of Queuing Models
Model A(M/M/1): Single-Channel Other Queuing Approaches
Queuing Model with Poisson Arrivals
and Exponential Service Times
Model B(M/M/S) M lti l Ch
M d l B(M/M/S): Multiple-Channel
l
Queuing Model
Model C(M/D/1): Constant-Service-
Time Model
Little’s Law
Model D: Limited-Population Model
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Learning Objectives Learning Objectives
When you complete this module you When you complete this module you
should be able to: should be able to:
1. Describe the characteristics of 4. Apply the multiple-channel
arrivals,
arrivals waiting lines and service
lines, queuing model formulas
systems 5. Apply the constant-service-time
2. Apply the single-channel queuing model equations
model equations 6. Perform a limited-population
3. Conduct a cost analysis for a model analysis
waiting line
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2. 10/16/2010
Common Queuing
Queuing Theory Situations
Situation Arrivals in Queue Service Process
The study of waiting lines Supermarket Grocery shoppers Checkout clerks at cash
register
Waiting lines are common Highway toll booth Automobiles Collection of tolls at booth
situations Doctor’s office Patients Treatment by doctors and
nurses
Useful in both Computer system Programs to be run Computer processes jobs
manufacturing Telephone company Callers Switching equipment to
forward calls
and service Bank Customer Transactions handled by teller
areas Machine Broken machines Repair people fix machines
maintenance
Harbor Ships and barges Dock workers load and unload
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Characteristics of Waiting-
Waiting- Parts of a Waiting Line
Line Systems Population of Arrivals Queue Service Exit the system
dirty cars from the (waiting line) facility
1. Arrivals or inputs to the system general
population …
Dave’s
Population size, behavior, statistical Car Wash
distribution
2.
2 Queue discipline or the waiting line
discipline, Enter Exit
itself
Limited or unlimited in length, discipline Arrivals to the system In the system Exit the system
of people or items in it
Arrival Characteristics Waiting Line Service Characteristics
3. The service facility Size of the population Characteristics Service design
Behavior of arrivals Limited vs. Statistical distribution
Design, statistical distribution of service Statistical distribution unlimited of service
of arrivals Queue discipline
times
Figure D.1
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Arrival Characteristics Poisson Distribution
1. Size of the population
Unlimited (infinite) or limited (finite) e-λλx
P(x) = for x = 0, 1, 2, 3, 4, …
2. Pattern of arrivals x!
Scheduled or random, often a Poisson
distribution where P(x) = probability of x arrivals
3. Behavior of arrivals x = number of arrivals per unit of time
Wait in the queue and do not switch lines λ = average arrival rate
e = 2.7183 (which is the base of the
No balking or reneging natural logarithms)
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3. 10/16/2010
Poisson Distribution Waiting-
Waiting-Line Characteristics
e-λλx
Probability = P(x) =
x!
Limited or unlimited queue length
0.25 – 0.25 –
0.02 – 0.02 –
Queue discipline - first-in, first-out
(FIFO) is most common
lity
lity
Probabil
Probabil
0.15 – 0.15 –
Other priority rules may be used in
0.10 – 0.10 –
special circumstances
0.05 – 0.05 –
– –
0 1 2 3 4 5 6 7 8 9 x 0 1 2 3 4 5 6 7 8 9 10 11 x
Distribution for λ = 2 Distribution for λ = 4
Figure D.2
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Service Characteristics Queuing System Designs
A family dentist’s office
Queuing system designs
Queue
Single-channel system, multiple- Service Departures
Arrivals
channel system facility after service
Single phase
Single-phase system, multiphase Single-channel, single phase
Single channel single-phase system
system
A McDonald’s dual window drive-through
Service time distribution Queue
Phase 1 Phase 2 Departures
Constant service time Arrivals service
facility
service
facility after service
Random service times, usually a
Single-channel, multiphase system
negative exponential distribution
Figure D.3
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Queuing System Designs Queuing System Designs
Most bank and post office service windows Some college registrations
Service
facility Phase 1 Phase 2
Channel 1 service service
Queue Queue facility facility
Channel 1 Channel 1
Service Departures Departures
Arrivals facility Arrivals after service
Channel 2
after service Phase 1 Phase 2
service service
facility facility
Service Channel 2 Channel 2
facility
Channel 3
Multi-channel, single-phase system Multi-channel, multiphase system
Figure D.3 Figure D.3
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4. 10/16/2010
Negative Exponential Measuring Queue
Distribution Performance
Probability that service time is greater than t = e-µt for t ≥ 1
1. Average time that each customer or object
µ = Average service rate
1.0 – e = 2.7183 spends in the queue
Probability that servic time ≥ 1
0.9 –
Average service rate (µ) = 3 customers per hour
2. Average queue length
0.8 –
⇒ Average service time = 20 minutes per customer 3. Average time each customer spends in the
ce
0.7
07 –
0.6 – system
0.5 –
4. Average number of customers in the system
0.4 –
Average service rate (µ) =
0.3 – 1 customer per hour 5. Probability that the service facility will be idle
0.2 –
6. Utilization factor for the system
0.1 –
0.0 |– | | | | | | | | | | | | 7. Probability of a specific number of customers
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 in the system
Time t (hours)
Figure D.4
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Queuing Costs Queuing Models
Cost
The four queuing models here all assume:
Poisson distribution arrivals
Minimum
Total Total expected cost FIFO discipline
cost
Cost of providing service
A single-service phase
Cost of waiting time
Low level Optimal High level
of service service level of service
Figure D.5
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Queuing Models Queuing Models
Model Name Example Model Name Example
A Single-channel Information counter B Multichannel Airline ticket
system at department store (M/M/S) counter
(M/M/1)
Number Number Arrival Service Number Number Arrival Service
of of Rate Time Population Queue of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Exponential Unlimited FIFO Multi- Single Poisson Exponential Unlimited FIFO
channel
Table D.2 Table D.2
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5. 10/16/2010
Queuing Models Queuing Models
Model Name Example Model Name Example
C Constant- Automated car D Limited Shop with only a
service wash population dozen machines
(M/D/1) (finite population) that might break
Number Number Arrival Service Number Number Arrival Service
of of Rate Time Population Queue of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Constant Unlimited FIFO Single Single Poisson Exponential Limited FIFO
Table D.2 Table D.2
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Model A – Single-Channel
Single- Model A – Single-Channel
Single-
1. Arrivals are served on a FIFO basis and 4. Service times vary from one customer
every arrival waits to be served to the next and are independent of one
regardless of the length of the queue another, but their average rate is
2.
2 Arrivals are independent of preceding known
arrivals but the average number of 5. Service times occur according to the
arrivals does not change over time negative exponential distribution
3. Arrivals are described by a Poisson 6. The service rate is faster than the
probability distribution and come from arrival rate
an infinite population
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Model A – Single-Channel
Single- Model A – Single-Channel
Single-
λ = Mean number of arrivals per time period Lq = Average number of units waiting in the
µ = Mean number of units served per time period queue
Ls = Average number of units (customers) in the = λ2
system (waiting and being served) µ(µ – λ)
= λ Wq = Average time a unit spends waiting in the
µ–λ queue
Ws = Average time a unit spends in the system λ
=
(waiting time plus service time) µ(µ – λ)
= 1 ρ = Utilization factor for the system
µ–λ λ
=
Table D.3
µ Table D.3
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6. 10/16/2010
Model A – Single-Channel
Single- Single-
Single-Channel Example
P0 = Probability of 0 units in the system (that is, λ = 2 cars arriving/hour µ = 3 cars serviced/hour
the service unit is idle) 2
λ
λ Ls = = = 2 cars in the system on average
= 1– µ–λ 3-2
µ
Pn > k = Probability of more than k units in the 1 1
system, where n is the number of units in Ws = = = 1 hour average waiting time in
µ–λ 3-2
the system the system
k+1
λ λ2 22
= Lq = = = 1.33 cars waiting in line
µ µ(µ – λ) 3(3 - 2)
Table D.3
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Single-
Single-Channel Example Single-
Single-Channel Example
Probability of more than k Cars in the System
λ = 2 cars arriving/hour µ = 3 cars serviced/hour
k Pn > k = (2/3)k + 1
λ 2 0 .667 ← Note that this is equal to 1 - P0 = 1 - .33
Wq = = = 2/3 hour = 40 minute
µ(µ – λ) 3(3 - 2) 1 .444
average waiting time
2 .296
ρ = λ/µ = 2/3 = 66.6% of time mechanic is busy 3 .198 ← Implies that there is a 19.8% chance that
more than 3 cars are in the system
λ 4 .132
P0 = 1 - = .33 probability there are 0 cars in the
µ 5 .088
system
6 .058
7 .039
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Single-
Single-Channel Economics Multi-
Multi-Channel Model
Customer dissatisfaction M = number of channels open
and lost goodwill = $10 per hour λ = average arrival rate
Wq = 2/3 hour
µ = average service rate at each channel
Total arrivals = 16 per day
Mechanic’s salary = $56 per day 1
P0 = for Mµ > λ
Total hours M–1
1 λ
n
1 λ
M
Mµ
customers spend
waiting per day
=
2
3
(16) = 10
2
3
hours ∑ n! µ
+
M! µ Mµ - λ
n=0
2
Customer waiting-time cost = $10 10 = $106.67 M
3 λµ(λ/µ) λ
Ls = P0 +
(M - 1)!(Mµ - λ)
2 µ
Total expected costs = $106.67 + $56 = $162.67
Table D.4
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7. 10/16/2010
Multi-
Multi-Channel Model Multi-
Multi-Channel Example
λ = 2 µ = 3 M = 2
M Ls
λµ(λ/µ) 1
Ws = P0 + = 1 1
(M - 1)!(Mµ - λ)
2 µ λ P0 = =
1 n 2 2
1 2(3)
∑ n! 2
3
+
1
2!
2
3 2(3) - 2
n=0
λ
Lq = Ls –
µ
(2)(3(2/3)2 1 2 3
Ls = + =
2 3 4
1! 2(3) - 2 2
1 Lq
Wq = Ws – =
µ λ 3/4 3 3 2 1 .083
Ws = = Lq = – = Wq = = .0415
2 8 4 3 12 2
Table D.4
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Multi-
Multi-Channel Example Waiting Line Tables
Poisson Arrivals, Exponential Service Times
Number of Service Channels, M
Single Channel Two Channels ρ 1 2 3 4 5
.10 .0111
P0 .33 .5 .25 .0833 .0039
.50 .5000 .0333 .0030
Ls 2 cars .75 cars
75 .75
75 2.2500
2 2500 .1227
1227 .0147
0147
.90 3.1000 .2285 .0300 .0041
Ws 60 minutes 22.5 minutes 1.0 .3333 .0454 .0067
1.6 2.8444 .3128 .0604 .0121
Lq 1.33 cars .083 cars
2.0 .8888 .1739 .0398
Wq 40 minutes 2.5 minutes 2.6 4.9322 .6581 .1609
3.0 1.5282 .3541
4.0 2.2164
Table D.5
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Waiting Line Table Example Constant-
Constant-Service Model
Bank tellers and customers Average length λ2
λ = 18, µ = 20 Lq =
of queue 2µ(µ – λ)
Lq
Utilization factor ρ = λ/µ = .90 Wq = λ
Average waiting time λ
in queue Wq =
From Table D.5
D5 2µ(µ – λ)
Number of Number
service windows M in queue Time in queue Average number of λ
Ls = Lq +
1 window 1 8.1 .45 hrs, 27 minutes customers in system µ
2 windows 2 .2285 .0127 hrs, ¾ minute
Average time 1
3 windows 3 .03 .0017 hrs, 6 seconds Ws = Wq +
in the system µ
4 windows 4 .0041 .0003 hrs, 1 second
Table D.6
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8. 10/16/2010
Constant-
Constant-Service Example Little’s Law
Trucks currently wait 15 minutes on average
Truck and driver cost $60 per hour A queuing system in steady state
Automated compactor service rate (µ) = 12 trucks per hour
Arrival rate (λ) = 8 per hour
Compactor costs $3 per truck
L = λW (which is the same as W = L/λ
Current waiting cost per trip = (1/4 hr)($60) = $15 /trip
g p p ( )( ) p
Lq = λWq (which is the same as Wq = Lq/λ
8 1
Wq = = hour Once one of these parameters is known, the
2(12)(12 – 8) 12
other can be easily found
Waiting cost/trip It makes no assumptions about the probability
with compactor = (1/12 hr wait)($60/hr cost) = $ 5 /trip
Savings with
distribution of arrival and service times
= $15 (current) – $5(new) = $10 /trip
new equipment Applies to all queuing models except the limited
Cost of new equipment amortized = $ 3 /trip population model
Net savings = $ 7 /trip
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Limited-
Limited-Population Model Limited-
Limited-Population Model
D = Probability that a unit N = Number of potential
T will have to wait in customers
Service factor: X =
T+U queue
Average number running: J = NF(1 - X) F = Efficiency factor T = Average service time
H = Average number of units U = Average time between
Average number waiting: L = N(1 - F) being
b i servedd unit service
it i
requirements
Average number being serviced: H = FNX
J = Average number of units W = Average time a unit
T(1 - F) not in queue or in waits in line
Average waiting time: W = XF service bay
L = Average number of units X = Service factor
Number of population: N = J + L + H waiting for service
M = Number of service
channels
Table D.7
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Finite Queuing Table Limited-
Limited-Population Example
X M D F
.012 1 .048 .999 Each of 5 printers requires repair after 20 hours (U) of use
.025 1 .100 .997 One technician can service a printer in 2 hours (T)
.050 1 .198 .989 Printer downtime costs $120/hour
Technician costs $25/hour
.060 2 .020 .999
1 .237 .983 2
Service factor: X = = .091 (close to .090)
091 090)
.070 2 .027 .999 2 + 20
1 .275 .977 For M = 1, D = .350 and F = .960
.080 2 .035 .998 For M = 2, D = .044 and F = .998
1 .313 .969
.090 2 .044 .998
Average number of printers working:
1 .350 .960 For M = 1, J = (5)(.960)(1 - .091) = 4.36
.100 2 .054 .997 For M = 2, J = (5)(.998)(1 - .091) = 4.54
Table D.8 1 .386 .950
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9. 10/16/2010
Limited-
Limited-Population Example Other Queuing Approaches
Average Average
Number
Each of 5 printers require Cost/Hrafter 20 Cost/Hr(U) of use
repair for hours for
Number of Printers Downtime Technicians
One technician can service a printer in 2 hours (T) Total The single-phase models cover many
Technicians Down (N - J) (N - J)$120
Printer downtime costs $120/hour ($25/hr) Cost/Hr queuing situations
Technician costs $25/hour $76.80
1 .64 $25.00 $101.80
2 Variations of the four single-phase
Service factor: X =
2 .46
46 $55.20 091 $50 00 090)
$55 = .091 (close to .090) 20
2 + 20 20 $50.00 $105.20
$105 systems are possible
For M = 1, D = .350 and F = .960 Multiphase models
For M = 2, D = .044 and F = .998 exist for more
Average number of printers working:
complex situations
For M = 1, J = (5)(.960)(1 - .091) = 4.36
For M = 2, J = (5)(.998)(1 - .091) = 4.54
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