Transcript Chapter 11

Chapter 16
Waiting Line Models
and Service
Improvement
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Elements of Waiting
Line Analysis
 Queue
A single waiting line
 Waiting line system consists of
Arrivals
Servers
Waiting line structures
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Components of
Queuing System
Source of
customers—
calling
population
Arrivals
Waiting Line
or
“Queue”
Server
Served
customers
Figure 16.1
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Elements of a Waiting Line
 Calling population
 Source of customers
 Infinite - large enough that one more
customer can always arrive to be served
 Finite - countable number of potential
customers
 Arrival rate ()
 Frequency of customer arrivals at waiting line
system
 Typically follows Poisson distribution
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Elements of a Waiting Line
 Service time
 Often follows negative exponential
distribution
 Average service rate = 
 Arrival rate () must be less than service
rate  or system never clears out
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Elements of a Waiting Line
 Queue discipline
Order in which customers are served
First come, first served is most
common
 Length can be infinite or finite
Infinite is most common
Finite is limited by some physical
structure
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Basic Waiting Line
Structures
 Channels are the number of
parallel servers
 Phases denote number of
sequential servers the customer
must go through
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Single-Channel Structures
Single-channel, single-phase
Waiting line
Server
Single-channel, multiple phases
Waiting line
Servers
Figure 16.2
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Multi-Channel Structures
Multiple-channel, single phase
Waiting line
Servers
Multiple-channel, multiple-phase
Waiting line
Figure 16.2
Servers
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Operating
Characteristics
 Mathematics of queuing theory does
not provide optimal or best solutions
 Operating characteristics are computed
that describe system performance
 Steady state is constant, average value
for performance characteristics that the
system will reach after a long time
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Operating
Characteristics
NOTATION
OPERATING CHARACTERISTIC
L
Average number of customers in the
system (waiting and being served)
Lq
Average number of customers in the
waiting line
W
Average time a customer spends in the
system (waiting and being served)
Wq
Average time a customer spends waiting in
line
Table 16.1
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Operating
Characteristics
NOTATION
OPERATING CHARACTERISTIC
P0
Probability of no (zero) customers in the
system
Pn
Probability of n customers in the system

Utilization rate; the proportion of time the
system is in use
Table 16.1
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Cost Relationship in
Waiting Line Analysis
Total cost
Expected costs
Service cost
Waiting Costs
Figure 16.3
Level of service
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Waiting Line Costs and
Quality Service
 Traditional view is that the level of
service should coincide with
minimum point on total cost curve
 TQM approach is that absolute
quality service will be the most costeffective in the long run
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Single-Channel, SinglePhase Models
 All assume Poisson arrival rate
 Variations
 Exponential service times
 General (or unknown) distribution of service
times
 Constant service times
 Exponential service times with finite queue
length
 Exponential service times with finite calling
population
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Basic Single-Server Model
 Assumptions:
Poisson arrival rate
Exponential service times
First-come, first-served queue discipline
Infinite queue length
Infinite calling population
  = mean arrival rate
  = mean service rate
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Formulas for SingleServer Model
Probability that no customers
are in the system (either in the
queue or being served)
Probability of exactly n
customers in the system
P0 = 1 -


Pn =


n
n
=


• P0
1-



Average number of
customers in the system
L =
Average number of
customers in the waiting line

Lq =
( - )


-
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Formulas for SingleServer Model
Average time a customer
spends in the queuing system
1
L
W =
=
 -

Average time a customer
spends waiting in line to
be served

Wq =
( - )
Probability that the server
is busy and the customer
has to wait


Probability that the server
is idle and a customer can
be served
 =
I = 1-
= 1-

= P0

To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
A Single-Server Model
Given  = 24 per hour,  = 30 customers per hour
Probability of no
customers in the
system

P0 = 1 
0.20
Average number
of customers in
the system
24

L =
=
=4
30 - 24
-
Average number
of customers
waiting in line
(24)2
2
Lq =
=
= 3.2
30(30
24)
( - )
= 1-
24
30
=
Example 16.1
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
A Single-Server Model
Given  = 24 per hour,  = 30 customers per hour
Average time in the
system per customer
1
1
W =
=
= 0.167 hour
 - 30 - 24
Average time waiting
in line per customer
24

Wq =
=
= 0.133
30(30 - 24)
(-)
Probability that the
server will be busy and
the customer must wait
 =
Probability the
server will be idle
I = 1 -  = 1 - 0.80 = 0.20
24

=
= 0.80
30

Example 16.1
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Waiting Line Cost Analysis
To improve customer services
management wants to test two
alternatives to reduce customer
waiting time:
1. Another employee to pack up
purchases
2. Another checkout counter
Example 16.2
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Waiting Line Cost Analysis
 Add extra employee to increase service rate
from 30 to 40 customers per hour
 Extra employee costs $150/week
 Each one-minute reduction in customer
waiting time avoids $75 in lost sales
 Waiting time with one employee = 8 minutes
Wq = 0.038 hours = 2.25 minutes
8.00 - 2.25 = 5.75 minutes reduction
5.75 x $75/minute/week = $431.25 per week
New employee saves $431.25 - $150.00 = $281.25/wk
Example 16.2
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Waiting Line Cost Analysis
 New counter costs $6000 plus $200 per week
for checker
 Customers divide themselves between two
checkout lines
 Arrival rate is reduced from = 24 to = 12
 Service rate for each checker is  = 30
Wq = 0.022 hours = 1.33 minutes
8.00 - 1.33 = 6.67 minutes
6.67 x $75/minute/week = $500.00/wk - $200 = $300/wk
Counter is paid off in 6000/300 = 20 weeks
Example 16.2
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Waiting Line Cost Analysis
 Adding an employee results in savings and
improved customer service
 Adding a new counter results in slightly
greater savings and improved customer
service, but only after the initial investment
has been recovered
 A new counter results in more idle time for
employees
 A new counter would take up potentially
valuable floor space
Example 16.2
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Constant Service Times
 Constant service times occur with
machinery and automated
equipment
 Constant service times
are a special case
of the single-server
model with general
or undefined service times
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Operating Characteristics
for Constant Service Times


Probability that no customers
are in system
P0 = 1 -
Average number of
customers in queue
2
Lq =
2( - )
Average number of
customers in system
L = Lq +


To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Operating Characteristics
for Constant Service Times
Average time customer
spends in queue
Average time customer
spends in the system
Probability that the
server is busy
Wq =
Lq

1
W = Wq +
 =



To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Constant Service Times
Automated car wash with service time = 4.5 min
Cars arrive at rate  = 10/hour (Poisson)
 = 60/4.5 = 13.3/hour
2
(10)2
Lq =
=
= 1.14 cars waiting
2(13.3)(13.3 - 10)
2( - )
Wq =
Lq

= 1.14/10 = .114 hour or 6.84 minutes
Example 16.3
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Finite Queue Length
 A physical limit exists on length of queue
 M = maximum number in queue
 Service rate does not have to exceed arrival
rate () to obtain steady-state conditions
Probability that no
customers are in system
P0 =
1 - /
1 - (/)M + 1


Probability of exactly n
customers in system
Pn = (P0)
Average number of
customers in system
/
L=
1 - /
n
for n ≤ M
(M + 1)(/)M + 1
1 - (/)M + 1
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Finite Queue Length
Let PM = probability a customer will not join system
Average number of
customers in queue
Lq = L -
Average time customer
spends in system
Average time customer
spends in queue
W =
(1- PM)

L
(1 - PM)
Wq = W -
1

To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Finite Queue
Quick Lube has waiting space for only 3 cars.
 = 20,  = 30, M = 4 cars (1 in service + 3 waiting)
Probability that
no cars are in
the system
P0 =
1 - /
1-
=
(/)M + 1
Probability of
exactly 4 cars in
the system

Pn = (P0)

Average number
of cars in the
system
/
L=
1 - /
1 - 20/30
1-
(20/30)5
n=M
20
= (0.38)
30
(M + 1)(/)M + 1
1-
(/)M + 1
= 0.38
4
= 0.076
= 1.24
Example 16.4
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Finite Queue
Quick Lube has waiting space for only 3 cars.
 = 20,  = 30, M = 4 cars (1 in service + 3 waiting)
Average number of
cars in the queue
Average time a car
spends in the system
Average time a car
spends in the queue
Lq = L -
W =
(1- PM)

= 0.62
(1 - PM)
= 0.067 hr
L
Wq = W -
1

= 0.033 hr
Example 16.4
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Finite Calling Population
 Arrivals originate from a finite (countable) population
 N = population size
Probability that no
customers are in system
P0 =
1
N

n=0
N!

(N - n)! 
n
n
Probability of exactly n
customers in system
N!

Pn =
(N - n)! 
Average number of
customers in queue
+
Lq = N (1- P0)

P0
where n = 1, 2, ..., N
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Finite Calling Population
 Arrivals originate from a finite (countable) population
 N = population size
Average number of
customers in system
L = Lq + (1 - P0)
Lq
Average time customer
spends in queue
Wq =
Average time customer
spends in system
1
W = Wq +

(N - L) 
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Finite Calling Population
20 machines which operate an average of 200 hrs
before breaking down ( = 1/200 hr = 0.005/hr)
Mean repair time = 3.6 hrs ( = 1/3.6 hr = 0.2778/hr)
Probability that no
machines are in the system
P0 = 0.652
Average number of
machines in the queue
Lq = 0.169
Average number of
machines in system
L = 0.169 + (1 - 0.652) = .520
Example 16.5
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Finite Calling Population
20 machines which operate an average of 200 hrs
before breaking down ( = 1/200 hr = 0.005/hr)
Mean repair time = 3.6 hrs ( = 1/3.6 hr = 0.2778/hr)
Average time machine
spends in queue
Wq = 1.74 hrs
Average time machine
spends in system
W = 5.33 hrs
Example 16.5
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Multiple-Channel,
Single-Phase Models
 Two or more independent servers serve a
single waiting line
 Poisson arrivals, exponential service,
infinite calling population
 s>
P0 =
1
n=s-1

n=0
1
n!


n
1
+ s!


s
s
s - 
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Multiple-Channel,
Single-Phase Models
 Two or more independent
servers
serve
Computing
P0 can
be a
single waiting line time-consuming.
Tables canservice,
used to
 Poisson arrivals, exponential
find P0 for selected
infinite calling population
values of  and s.
 s>
P0 =
1
n=s-1

n=0
1
n!


n
1
+ s!


s
s
s - 
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Multiple-Channel,
Single-Phase Models
Probability of exactly
n customers in the
system
Probability an arriving
customer must wait
Average number of
customers in system


1
s! sn-s
Pn =
Pw
1
=
s!


s
P0,
P0,
for n > s
s
P
s -  0
(/)s
L =
for n > s
n


1
n!
n
(s - 1)!(s - )2

P0 +

To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Multiple-Channel,
Single-Phase Models
L

Average time customer
spends in system
W =
Average number of
customers in queue
Lq = L -
Average time customer
spends in queue
Lq
1
Wq = W =


Utilization factor
 = /s


To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Multiple-Server System
Customer service area
 = 10 customers/area
 = 4 customers/hour per service rep
s = (3)(4) = 12
Probability no customers
are in the system
P0 = 0.045
Number of customers in
the service department
L = 6
Waiting time in the
service department
W = L /  = 0.60
Example 16.6
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Multiple-Server System
Customer service area
 = 10 customers/area
 = 4 customers/hour per service rep
s = (3)(4) = 12
Number of customers
waiting to be served
Lq = L - / = 3.5
Average time customers
will wait in line
Wq = Lq/ = 0.35 hours
Probability that
customers must wait
Pw = 0.703
Example 16.6
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.
Improving Service
 Add a 4th server to improve service
 Recompute operating characteristics
 P0 = 0.073 prob of no customers
 L = 3.0 customers
 W = 0.30 hour, 18 min in service
 Lq = 0.5 customers waiting
 Wq = 0.05 hours, 3 min waiting, versus 21 earlier
 Pw = 0.31 prob that customer must wait
To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.