Saudi Electronic University Statistics Worksheet

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I want a solution to the same problem with the Excel program, but changed the data with the QM tool, and a picture of the problem was attached, and I attached Chapter 13 .


Chapter 13
Waiting Lines and Queuing
Theory Models
To accompany
Quantitative Analysis for Management, Eleventh Edition,
by Render, Stair, and Hanna
Power Point slides created by Brian Peterson
Learning Objectives
After completing this chapter, students will be able to:
1.
2.
3.
4.
5.
Describe the trade-off curves for cost-ofwaiting time and cost of service.
Understand the three parts of a queuing
system: the calling population, the queue
itself, and the service facility.
Describe the basic queuing system
configurations.
Understand the assumptions of the common
models dealt with in this chapter.
Analyze a variety of operating characteristics
of waiting lines.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-2
13.1
13.2
13.3
13.4
13.5
Chapter Outline
Introduction
Waiting Line Costs
Characteristics of a Queuing System
Single-Channel Queuing Model with
Poisson Arrivals and Exponential Service
Times (M/M/1)
Multichannel Queuing Model with Poisson
Arrivals and Exponential Service Times
(M/M/m)
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-3
Chapter Outline
13.6
13.7
13.8
13.9
Constant Service Time Model (M/D/1)
Finite Population Model (M/M/1 with Finite
Source)
Some General Operating Characteristic
Relationships
More Complex Queuing Models and the
Use of Simulation
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13-4
Introduction
Queuing theory is the study of waiting lines.
It is one of the oldest and most widely used
quantitative analysis techniques.
The three basic components of a queuing
process are arrivals, service facilities, and the
actual waiting line.
Analytical models of waiting lines can help
managers evaluate the cost and effectiveness
of service systems.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-5
Waiting Line Costs
Most waiting line problems are focused on
finding the ideal level of service a firm should
provide.
In most cases, this service level is something
management can control.
When an organization does have control, they
often try to find the balance between two
extremes.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-6
Waiting Line Costs
There is generally a trade-off between cost of
providing service and cost of waiting time.
A large staff and many service facilities generally results
in high levels of service but have high costs.
Having the minimum number of service facilities keeps
service cost down but may result in dissatisfied
customers.
Service facilities are evaluated on their total
expected cost which is the sum of service costs
and waiting costs.
Organizations typically want to find the service
level that minimizes the total expected cost.
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13-7
Queuing Costs and Service Levels
Figure 13.1
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13-8
Three Rivers Shipping Company
Three Rivers Shipping operates a docking facility
on the Ohio River.
An average of 5 ships arrive to unload their
cargos each shift.
Idle ships are expensive.
More staff can be hired to unload the ships, but
that is expensive as well.
Three Rivers Shipping Company wants to
determine the optimal number of teams of
stevedores to employ each shift to obtain the
minimum total expected cost.
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13-9
Three Rivers Shipping Company
Waiting Line Cost Analysis
NUMBER OF TEAMS OF STEVEDORES
WORKING
1
7
(b) Average time each ship waits to
be unloaded (hours)
5
(a) Average number of ships arriving
per shift
(c) Total ship hours lost per shift
(a x b)
(d) Estimated cost per hour of idle
ship time
(e)
(f)
cost (c x d)
Stevedore team salary or service
cost
(g) Total expected cost (e + f)
2
3
5
5
3
20
$1,000
$20,000
$12,000
$32,000
4
5
4
35
$1,000
$35,000
$6,000
$41,000
2
15
$1,000
$15,000
$18,000
$33,000
10
$1,000
$10,000
$24,000
$34,000
Optimal cost
Table 13.1
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13-10
Characteristics of a Queuing System
There are three parts to a queuing system:
1. The arrivals or inputs to the system
(sometimes referred to as the calling
population).
2. The queue or waiting line itself.
3. The service facility.
These components have their own
characteristics that must be examined
before mathematical models can be
developed.
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13-11
Characteristics of a Queuing System
Arrival Characteristics have three major
characteristics: size, pattern, and behavior.
The size of the calling population can be either
unlimited (essentially infinite) or limited (finite).
The pattern of arrivals can arrive according to
a known pattern or can arrive randomly.
Random arrivals generally follow a Poisson
distribution.
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13-12
Characteristics of a Queuing System
Behavior of arrivals
Most queuing models assume customers are
patient and will wait in the queue until they are
served and do not switch lines.
Balking refers to customers who refuse to join
the queue.
Reneging customers enter the queue but
become impatient and leave without receiving
their service.
That these behaviors exist is a strong
argument for the use of queuing theory to
managing waiting lines.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-13
Characteristics of a Queuing System
Waiting Line Characteristics
Waiting lines can be either limited or unlimited.
Queue discipline refers to the rule by which
customers in the line receive service.
The most common rule is first-in, first-out (FIFO).
Other rules are possible and may be based on other
important characteristics.
Other rules can be applied to select which
customers enter which queue, but may apply
FIFO once they are in the queue.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-14
Characteristics of a Queuing System
Service Facility Characteristics
Basic queuing system configurations:
Service systems are classified in terms of
the number of channels, or servers, and the
number of phases, or service stops.
A single-channel system with one server is
quite common.
Multichannel systems exist when multiple
servers are fed by one common waiting line.
In a single-phase system, the customer
receives service form just one server.
In a multiphase system, the customer has to
go through more than one server.
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13-15
Four basic
queuing
system
configurations
Figure 13.2
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13-16
Characteristics of a Queuing System
Service time distribution
Service patterns can be either constant or
random.
Constant service times are often machine
controlled.
More often, service times are randomly
distributed according to a negative
exponential probability distribution.
Analysts should observe, collect, and plot
service time data to ensure that the
observations fit the assumed distributions
when applying these models.
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13-17
Identifying Models Using
Kendall Notation
D. G. Kendall developed a notation for queuing
models that specifies the pattern of arrival, the
service time distribution, and the number of
channels.
Notation takes the form:
Arrival
distribution
Service time
distribution
Number of service
channels open
Specific letters are used to represent probability
distributions.
M = Poisson distribution for number of occurrences
D = constant (deterministic) rate
G = general distribution with known mean and variance
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13-18
Identifying Models Using
Kendall Notation
A single-channel model with Poisson arrivals and
exponential service times would be represented
by:
M/M/1
If a second channel is added the notation would
read:
M/M/2
A three-channel system with Poisson arrivals and
constant service time would be
M/D/3
A four-channel system with Poisson arrivals and
normally distributed service times would be
M/G/4
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-19
Single-Channel Model, Poisson Arrivals,
Exponential Service Times (M/M/1)
Assumptions of the model:
Arrivals are served on a FIFO basis.
There is no balking or reneging.
Arrivals are independent of each other but the
arrival rate is constant over time.
Arrivals follow a Poisson distribution.
Service times are variable and independent but
the average is known.
Service times follow a negative exponential
distribution.
Average service rate is greater than the
average arrival rate.
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13-20
Single-Channel Model, Poisson Arrivals,
Exponential Service Times (M/M/1)
When these assumptions are met, we
can develop a series of equations that
operating
characteristics.
Queuing Equations:
Let
= mean number of arrivals per time period
= mean number of customers or units
served per time period
The arrival rate and the service rate must be
defined for the same time period.
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13-21
Single-Channel Model, Poisson Arrivals,
Exponential Service Times (M/M/1)
1. The average number of customers or units in the
system, L:
2. The average time a customer spends in the
system, W:
3. The average number of customers in the queue, Lq:
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13-22
Single-Channel Model, Poisson Arrivals,
Exponential Service Times (M/M/1)
4. The average time a customer spends waiting in
the queue, Wq:
5. The utilization factor for the system, , the
probability the service facility is being used:
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13-23
Single-Channel Model, Poisson Arrivals,
Exponential Service Times (M/M/1)
6. The percent idle time, P0, or the probability no one
is in the system:
7. The probability that the number of customers in
the system is greater than k, Pn>k:
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13-24
3 per hour.
Customers arrive at a rate of 2 per hour.
So:
= 2 cars arriving per hour
= 3 cars serviced per hour
2 cars in the system
on average
1 hour that an average car
spends in the system
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13-25
1.33 cars waiting in line
on average
40 minutes average
waiting time per car
percentage of time
mechanic is busy
probability that there
are 0 cars in the system
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13-26
Probability of more than k cars in the system
0.039
7
0.058
6
0.088
5
0.132
4
0.198
3
0.296
2
0.444
1
0.667
0
Pn>k = (2/3)k+1
k
Note that this is equal to 1
P0 = 1
0.33 = 0.667
Implies that there is a 19.8% chance that more
than 3 cars are in the system
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-27
Muffler Example
Program 13.1
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13-28
Introducing costs into the model:
Arnold wants to do an economic analysis of
the queuing system and determine the waiting
cost and service cost.
The total service cost is:
Total
(Number of channels)
=
service cost
x (Cost per channel)
Total
= mCs
service cost
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13-29
Waiting cost when the cost is based on time in the
system:
Total
(Total time spent waiting by all
=
waiting cost
arrivals) x (Cost of waiting)
(Number of arrivals) x
=
(Average wait per arrival)Cw
Total
= ( W)Cw
waiting cost
If waiting time cost is based on time in the queue:
Total
= ( Wq)Cw
waiting cost
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13-30
So the total cost of the queuing system when based
on time in the system is:
Total cost = Total service cost + Total waiting cost
Total cost = mCs + WCw
And when based on time in the queue:
Total cost = mCs + WqCw
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13-31
Arnold estimates the cost of customer waiting
time in line is $50 per hour.
Total daily
waiting cost = (8 hours per day) WqCw
= (8)(2)(2/3)($50) = $533.33
Arnold has identified the mechanics wage $7 per
hour as the service cost.
Total daily
service cost = (8 hours per day)mCs
= (8)(1)($15) = $120
So the total cost of the system is:
Total daily cost of
the queuing system = $533.33 + $120 = $653.33
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13-32
Arnold is thinking about hiring a different
mechanic who can install mufflers at a faster rate.
The new operating characteristics would be:
= 2 cars arriving per hour
= 4 cars serviced per hour
1 car in the system
on the average
1/2 hour that an average car
spends in the system
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13-33
1/2 car waiting in line
on the average
15 minutes average
waiting time per car
percentage of time
mechanic is busy
probability that there
are 0 cars in the system
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13-34
Probability of more than k cars in the system
0.004
7
0.008
6
0.016
5
0.031
4
0.062
3
0.125
2
0.250
1
0.500
0
Pn>k = (2/4)k+1
k
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13-35
The customer waiting cost is the same $50 per
hour:
Total daily
waiting cost = (8 hours per day) WqCw
= (8)(2)(1/4)($50) = $200.00
The new mechanic is more expensive at $20 per
hour:
Total daily
service cost = (8 hours per day)mCs
= (8)(1)($20) = $160
So the total cost of the system is:
Total daily cost of
the queuing system = $200 + $160 = $360
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13-36
The total time spent waiting for the 16 customers
per day was formerly:
(16 cars per day) x (2/3 hour per car) = 10.67 hours
It is now is now:
(16 cars per day) x (1/4 hour per car) = 4 hours
The total daily system costs are less with the new
mechanic resulting in significant savings:
$653.33
$360 = $293.33
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13-37
Enhancing the Queuing Environment
Reducing waiting time is not the only way
to reduce waiting cost.
Reducing the unit waiting cost (Cw) will
also reduce total waiting cost.
This might be less expensive to achieve
than reducing either W or Wq.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-38
Multichannel Queuing Model with Poisson
Arrivals and Exponential Service Times
(M/M/m)
Assumptions of the model:
Arrivals are served on a FIFO basis.
There is no balking or reneging.
Arrivals are independent of each other but the
arrival rate is constant over time.
Arrivals follow a Poisson distribution.
Service times are variable and independent but
the average is known.
Service times follow a negative exponential
distribution.
The average service rate is greater than the
average arrival rate.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-39
Multichannel Queuing Model with Poisson
Arrivals and Exponential Service Times (M/M/m)
Equations for the multichannel queuing model:
Let
m = number of channels open
= average arrival rate
= average service rate at each channel
1. The probability that there are zero customers in
the system is:
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13-40
Multichannel Model, Poisson Arrivals,
Exponential Service Times (M/M/m)
2. The average number of customers or units in the
system
3. The average time a unit spends in the waiting line or
being served, in the system
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13-41
Multichannel Model, Poisson Arrivals,
Exponential Service Times (M/M/m)
4. The average number of customers or units in line
waiting for service
5. The average number of customers or units in line
waiting for service
6. The average number of customers or units in line
waiting for service
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13-42
Arnold wants to investigate opening a second
garage bay.
He would hire a second worker who works at the
same rate as his first worker.
The customer arrival rate remains the same.
probability of 0 cars in the system
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13-43
Average number of cars in the system
Average time a car spends in the system
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13-44
Average number of cars in the queue
Average time a car spends in the queue
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13-45
Adding the second service bay reduces the
waiting time in line but will increase the service
cost as a second mechanic needs to be hired.
Total daily waiting cost = (8 hours per day) WqCw
= (8)(2)(0.0415)($50) = $33.20
Total daily service cost = (8 hours per day)mCs
= (8)(2)($15) = $240
So the total cost of the system is
Total system cost = $33.20 + $240 = $273.20
This is the cheapest option: open the second bay
and hire a second worker at the same $15 rate.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-46
Operating Characteristics
LEVEL OF SERVICE
OPERATING
CHARACTERISTIC
ONE
MECHANIC
=3
40 minutes
Average time spent in the
queue (Wq)
1.33 cars
Average number of cars in
the queue (Lq)
60 minutes
Average time spent in the
system (W)
2 cars
Average number of cars in
the system (L)
0.33
Probability that the system
is empty (P0)
TWO
MECHANICS
= 3 FOR BOTH
ONE FAST
MECHANIC
=4
0.50
0.50
0.75 cars
22.5 minutes
1 car
30 minutes
0.083 car
2.5 minutes
0.50 car
15 minutes
Table 13.2
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13-47
Muffler Multichannel Example
Program 13.2
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13-48
Constant Service Time Model (M/D/1)
Constant service times are used when
customers or units are processed
according to a fixed cycle.
The values for Lq, Wq, L, and W are always
less than they would be for models with
variable service time.
In fact both average queue length and
average waiting time are halved in
constant service rate models.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-49
Constant Service Time Model (M/D/1)
1. Average length of the queue
2. Average waiting time in the queue
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13-50
Constant Service Time Model (M/D/1)
3. Average number of customers in the system
4. Average time in the system
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13-51
Garcia-Golding Recycling, Inc.
The company collects and compacts
aluminum cans and glass bottles.
Trucks arrive at an average rate of 8 per hour
(Poisson distribution).
Truck drivers wait about 15 minutes before
they empty their load.
Drivers and trucks cost $60 per hour.
A new automated machine can process
truckloads at a constant rate of 12 per hour.
A new compactor would be amortized at $3
per truck unloaded.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-52
Constant Service Time Model (M/D/1)
Analysis of cost versus benefit of the purchase
Current waiting cost/trip = (1/4 hour waiting time)($60/hour cost)
= $15/trip
New system: = 8 trucks/hour arriving
= 12 trucks/hour served
Average waiting
time in queue = Wq = 1/12 hour
Waiting cost/trip
with new compactor = (1/12 hour wait)($60/hour cost) = $5/trip
Savings with
new equipment = $15 (current system) $5 (new system)
= $10 per trip
Cost of new equipment
amortized = $3/trip
Net savings = $7/trip
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-53
Excel QM Solution for Constant Service
Time Model with Garcia-Golding Recycling
Example
Program 13.3
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13-54
Finite Population Model
(M/M/1 with Finite Source)
When the population of potential customers is
limited, the models are different.
There is now a dependent relationship between
the length of the queue and the arrival rate.
The model has the following assumptions:
1. There is only one server.
2. The population of units seeking service is
finite.
3. Arrivals follow a Poisson distribution and
service times are exponentially distributed.
4. Customers are served on a first-come, firstserved basis.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-55
Finite Population Model
(M/M/1 with Finite Source)
Equations for the finite population model:
Using = mean arrival rate, = mean service
rate, and N = size of the population, the
operating characteristics are:
1. Probability that the system is empty:
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-56
Finite Population Model
(M/M/1 with Finite Source)
2. Average length of the queue:
3. Average number of customers (units) in the system:
4. Average waiting time in the queue:
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13-57
Finite Population Model
(M/M/1 with Finite Source)
5. Average time in the system:
6. Probability of n units in the system:
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13-58
Department of Commerce
The Department of Commerce has five printers
that each need repair after about 20 hours of work.
Breakdowns will follow a Poisson distribution.
The technician can service a printer in an average
of about 2 hours, following an exponential
distribution.
Therefore:
= 1/20 = 0.05 printer/hour
= 1/2 = 0.50 printer/hour
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13-59
Department of Commerce Example
1.
2.
3.
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13-60
Department of Commerce Example
4.
5.
If printer downtime costs $120 per hour and the
technician is paid $25 per hour, the total cost is:
Total
hourly
cost
(Average number of printers down)
= (Cost per downtime hour)
+ Cost per technician hour
= (0.64)($120) + $25 = $101.80
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13-61
Excel QM For Finite Population Model with
Department of Commerce Example
Program 13.4
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13-62
Some General Operating
Characteristic Relationships
Certain relationships exist among specific
operating characteristics for any queuing system
in a steady state.
A steady state condition exists when a system is
in its normal stabilized condition, usually after an
initial transient state.
Equations:
L= W
Lq = Wq
(or W = L/ )
(or Wq = Lq/ )
And
W = Wq + 1/
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13-63
More Complex Queuing Models and
the Use of Simulation
In the real world there are often variations from
basic queuing models.
Computer simulation can be used to solve these
more complex problems.
Simulation allows the analysis of controllable
factors.
Simulation should be used when standard
queuing models provide only a poor
approximation of the actual service system.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-64
Copyright
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.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
13-65
Chapter 7
Linear Programming Models:
Graphical and Computer
Methods
To accompany
Quantitative Analysis for Management, Eleventh Edition,
by Render, Stair, and Hanna
Power Point slides created by Brian Peterson
A
B C
1 Garcia-Golding Recycling
12345
Excel QM Solution for Constant Service
Time Model with Garcia-Golding Recycling
3 Waiting Lines
9
10
11
12
6 Data
7 Arrival rate (2)
8
Service rate (μ)
M/D/1 (Constant Service Times)
The arrival RATE and service RATE both must be rates and use the same time unit. Given a time
Program 13.3
Example
▬▬▬▬▬▬▬▬▬▬▬ ٧٩٤ / ۳۲۲-۳۲۱
Results
8 Average server utilization (p)
12 Average number of customers in the queue(La)
Average number of customers in the system(L)
Average waiting time in the queue(W₁)
Average time in the system(W)
Probability (% of time) system is empty (Po)
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
E
0.66667
0.66667
1.33333
0.08333
0.16667
0.33333
●●●
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