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Transcript of Waiting line analysis
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Copyright 2006 John Wiley & Sons, Inc.Beni Asllani
University of Tennessee at Chattanooga
Waiting Line Analysis
for Service ImprovementOperations Management - 5 th Edition
Chapter 17
Roberta Russell & Bernard W. Taylor, III
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Lecture Outline
Elements of Waiting Line Analysis
Waiting Line Analysis and Quality
Single Server Models
Multiple Server Model
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Waiting Line Analysis
Operating characteristics average values for characteristics that describe the
performance of a waiting line system Queue
A single waiting line
Waiting line system consists of Arrivals
Servers
Waiting line structures
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Elements of a Waiting Line
Calling population Source of customers
Infinite - large enough that one more customer canalways arrive to be served
Finite - countable number of potential customers
Arrival rate (λ) Frequency of customer arrivals at waiting line
system
Typically follows Poisson distribution
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Elements of a Waiting Line (cont.)
Service time
Often follows negative exponentialdistribution
Average service rate = μ
Arrival rate (λ) must be less than servicerate (μ) or system never clears out
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Elements of a Waiting Line (cont.)
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
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Basic Waiting Line Structures
Channels are the number of parallel servers
Single channel
Multiple channels
Phases denote number of sequential servers thecustomer must go through
Single phase
Multiple phases Steady state
A constant, average value for performance characteristics thatsystem will reach after a long time
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Operating Characteristics
NOTATION OPERATINGCHARACTERISTIC
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 andbeing served)
W q Average time a customer spends waiting in lineP 0 Probability of no (zero) customers in the system
P n Probability of n customers in the system
ρ Utilization rate; the proportion of time the system is in use
Table 16.1
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Cost Relationship in Waiting
Line Analysis
E x p e c t e d
c o s t s
Level of service
Total cost
Service cost
Waiting Costs
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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 cost-effective in
the long run
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Single-server 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
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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
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Formulas for Single-
Server Model
L =-
Average number ofcustomers in the system
Probability that no customers
are in the system (either in thequeue or being served)
P 0 = 1 -
Probability of exactly n customers in the system
P n = • P 0
n
= 1 -
n
Average number ofcustomers in the waiting line
Lq =( - )
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Formulas for Single-
Server Model (cont.)
=
Probability that the serveris busy and the customer
has to wait
Average time a customerspends in the queuing system
W = =1
-
L
Probability that the serveris idle and a customer canbe served
I = 1 -
= 1 - = P 0
Average time a customerspends waiting in line tobe served
W q = ( - )
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A Single-Server Model
Given = 24 per hour, = 30 customers per hour
Probability of nocustomers in thesystem
P 0 = 1 - = 1 - =0.20
2430
L = = = 4Average numberof customers in
the system-
24
30 - 24
Average numberof customerswaiting in line
Lq = = = 3.2(24)2
30(30 - 24)
2
( - )
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A Single-Server Model
Average time in thesystem per customer
W = = = 0.167 hour1
-
1
30 - 24
Average time waitingin line per customer
W q = = = 0.133( - )
24
30(30 - 24)
Probability that theserver will be busy andthe customer must wait = = = 0.80
24
30
Probability theserver will be idle
I = 1 - = 1 - 0.80 = 0.20
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Service Improvement Analysis
Possible Alternatives
Another employee to pack up purchases
service rate will increase from 30 customers to 40customers per hour
waiting time will reduce to only 2.25 minutes
Another checkout counter arrival rate at each register will decrease from 24 to 12 per
hour customer waiting time will be 1.33 minutes
Determining whether these improvements are worththe cost to achieve them is the crux of waiting lineanalysis
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Constant Service Times
Constant service timesoccur with machinery
and automatedequipment
Constant service timesare a special case of the
single-server model withundefined service times
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Operating Characteristics for
Constant Service Times
P 0 = 1 -Probability that no customersare in system
Average number ofcustomers in system
L = Lq +
Average number ofcustomers in queue
Lq =2
2 ( - )
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Operating Characteristics for
Constant Service Times (cont.)
=Probability that theserver is busy
Average time customerspends in the system W = W q +
1
Average time customerspends in queue W q =
Lq
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Constant Service Times:
ExampleAutomated car wash with service time = 4.5 min
Cars arrive at rate = 10/hour (Poisson)
= 60/4.5 = 13.3/hour
W q = = 1.14/10 = .114 hour or 6.84 minutesLq
(10)2
2(13.3)(13.3 - 10)Lq = = = 1.14 cars waiting
2
2 ( - )
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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
P 0 =Probability that nocustomers are in system
1 - /
1 - ( / )M + 1
Probability of exactly n customers in system P n = (P 0) for n ≤ M
n
L = -Average number ofcustomers in system
/
1 - /
(M + 1)( / )M + 1
1 - ( / )M + 1
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Finite Queue Length (cont.)
Let P M = probability a customer will not join system
Average time customerspends in system
W =L
(1 - P M )
Lq = L -(1- P M )Average number of
customers in queue
Average time customerspends in queue
W q = W -1
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Finite Queue: ExampleFirst National Bank has waiting space for only 3 drive in windowcars. = 20, = 30, M = 4 cars (1 in service + 3 waiting)
Probability thatno cars are inthe system P 0 = = = 0.38
1 - 20/30
1 - (20/30)5
1 - /
1 - ( / )M + 1
P n = (P 0) = (0.38) = 0.076
Probability ofexactly 4 cars in
the system
20
30
4
n =M
L = - = 1.24
Average numberof cars in thesystem
/
1 - /
(M + 1)( / )M + 1
1 - ( / )M + 1
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Finite Queue: Example (cont.)
Average time a carspends in the system
W = = 0.067 hrL
(1 - P M )
Lq = L - = 0.62(1- P M )Average number of
cars in the queue
Average time a carspends in the queue
W q = W - = 0.033 hr1
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Finite Calling Population
Arrivals originate from a finite (countable) population
N = population size
Probability of exactly n
customers in systemP n
= P 0 where n = 1, 2, ..., N
n
N !
(N - n )!
Average number ofcustomers in queue
Lq = N - (1- P 0)+
Probability that nocustomers are in system
P 0 =
n = 0
N !
(N - n )!
N n
1
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Finite Calling Population (cont.)
W q =Lq
(N - L)
Average time customerspends in queue
L
= Lq
+ (1 - P 0)
Average number of
customers in system
W = W q +Average time customerspends in system
1
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20 trucks which operate an average of 200 days before breakingdown ( = 1/200 day = 0.005/day)
Mean repair time = 3.6 days ( = 1/3.6 day = 0.2778/day)
Probability that notrucks are in the system P 0 = 0.652
Average number oftrucks in the queue
Lq = 0.169
Average number of
trucks in system L = 0.169 + (1 - 0.652) = .520
Finite Calling Population:
Example
Average time truckspends in queue
W q = 1.74 days
Average time truckspends in system
W = 5.33 days
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Two or more independent servers serve a single waiting line
Poisson arrivals, exponential service, infinite calling
population
s >
P 0 =1
1
s!
s s
s -
n =s -1
n =0
1
n !
n
+
Basic Multiple-server Model
Computing P 0 can be time-consuming.
Tables can used to find P 0 for selected values of and s .
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Probability of exactlyn customers in the
system
P n =
P 0, for n > s 1
s ! s n-s
n
P 0, for n > s 1n !
n
Probability an arrivingcustomer must wait P w = P 0
1
s !
s
s -
s
Average number ofcustomers in system
L = P 0 +( / )s
(s - 1)!(s - )2
Basic Multiple-server
Model (cont.)
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W =L
Average time customerspends in system
= /s Utilization factor
Average time customerspends in queue W q = W - =1Lq
Lq = L -Average number ofcustomers in queue
Basic Multiple-server Model
(cont.)
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Multiple-Server System:Example
Student Health Service Waiting Room
= 10 students per hour
= 4 students per hour per service representative
s = 3 representativess = (3)(4) = 12
P 0 = 0.045Probability no studentsare in the system
Number of students inthe service area
L = 6
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Multiple-Server System:Example (cont.)
Lq = L - / = 3.5Number of studentswaiting to be served
Average time students
will wait in line W q = Lq / = 0.35 hours
Probability that astudent must wait
P w = 0.703
Waiting time in theservice area
W = L / = 0.60
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Add a 4th server to improve service
Recompute operating characteristics
P 0 = 0.073 prob of no students L = 3.0 students
W = 0.30 hour, 18 min in service
Lq = 0.5 students waiting
W q = 0.05 hours, 3 min waiting, versus 21 earlier
P w = 0.31 prob that a student must wait
Multiple-Server System:
Example (cont.)
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Copyright 2006 John Wiley & Sons, Inc.
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