Transcript Module D: Waiting

Operations
Management
Module D –
Waiting-Line Models
PowerPoint presentation to accompany
Heizer/Render
Principles of Operations Management, 6e
Operations Management, 8e
© 2006
Prentice
Hall, Inc. Hall, Inc.
©
2006
Prentice
D–1
Outline
 Characteristics Of A Waiting-Line
System
 Arrival Characteristics
 Waiting-Line Characteristics
 Service Facility Characteristics
 Measuring the 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
 Other Queuing Approaches
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D–3
Learning Objectives
When you complete this module, you
should be able to:
Identify or Define:
 The assumptions of the four basic
waiting-line models
Describe or Explain:
 How to apply waiting-line models
 How to conduct an economic
analysis of queues
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D–4
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–5
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–6
Arrival Characteristics
1. Size of the population
 Unlimited (infinite) or limited (finite)
2. Behavior of arrivals
 Scheduled or random, often a Poisson
distribution
3. Behavior of arrivals
 Wait in the queue and do not switch
lines
 Balking or reneging
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D–7
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
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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–8
Poisson Distribution
e-x
P(x) =
x!
where P(x)
x

e
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=
=
=
=
for x = 0, 1, 2, 3, 4, …
probability of x arrivals
number of arrivals per unit of time
average arrival rate
2.7183 (which is the base of the
natural logarithms)
D–9
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
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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 – 10
Waiting-Line Characteristics
 Limited or unlimited queue length
 Queue discipline - first-in, first-out
is most common
 Other priority rules may be used in
special circumstances
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D – 11
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
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D – 12
Queuing System Designs
Your family dentist’s office
Queue
Service
facility
Arrivals
Departures
after service
Single-channel, single-phase system
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 – 13
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
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D – 14
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 – 15
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
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Time t in hours
D – 16
Measuring Queue
Performance
1. Average time that each customer or object
spends in the queue
2. Average queue length
3. Average time in the system
4. Average number of customers in the system
5. Probability the service facility will be idle
6. Utilization factor for the system
7. Probability of a specified number of customers
in the system
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D – 17
Queuing Costs
Cost
Total expected cost
Minimum
Total
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 – 18
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 – 19
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 – 20
Queuing Models
Model
Name
C
Constant
service
(M/D/1)
Example
Automated car
wash
Number Number
of
of
Channels Phases
Arrival
Rate
Pattern
Service
Time
Pattern
Single
Poisson
Constant
Single
Population Queue
Size
Discipline
Unlimited
FIFO
Table D.2
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D – 21
Queuing Models
Model
Name
D
Limited
population
(finite)
Example
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
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D – 22
Model A - Single Channel
1. Arrivals are FIFO 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 – 23
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 – 24
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
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D – 25
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 – 26
Model A - Single Channel
P0 = Probability of 0 units in the system (that is,
the service unit is idle)

µ
= Probability of more than k units in the
system, where n is the number of units in
the system
= 1–
Pn > k
=

µ
k+1
Table D.3
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D – 27
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 – 28
Single Channel Example
 = 2 cars arriving/hour
Wq
µ = 3 cars serviced/hour
2

=
=
= 40 minute average waiting
3(3
2)
µ(µ – )
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 – 29
Single Channel Example
Probability of More Than k Cars in the System
k
0
1
2
3
4
5
6
7
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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 – 30
Single Channel Economics
Customer dissatisfaction
and lost goodwill
Wq
Total arrivals
Mechanic’s salary
Total hours
customers spend
waiting per day
=
= $10 per hour
= 2/3 hour
= 16 per day
= $56 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 – 31
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 – 32
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 =
(2)(3(2/3)2
1! 2(3) - 2
Ws =
3/4
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2
=
3
8
2
1
2
2
+
3
=
3
4
1
2
3
Lq =
–
=
12
3
4
Wq =
.083
2
= .0415
D – 33
Multi-Channel Example
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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 – 34
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 waiting
time in system

2µ(µ – )

µ
Ws = W q +
1
µ
Table D.5
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D – 35
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
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= $ 5 /trip
= $10 /trip
= $ 3 /trip
= $ 7 /trip
D – 36
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
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D – 37
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 – 38
Finite Queuing Table
X
.012
.025
.050
.060
.070
.080
.090
Table D.7
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.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 – 39
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
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D – 40
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
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D – 41