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Queuing Models
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The other line
always moves faster.
If you change lines, the one you left
will start to move faster than the oneyoure in.
Murphys Law
1995 Corel Corp.
Thank you for holding.Hello...are you there?
Youve Been There Before!
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Introduction
Queuing is the study of waiting lines, or queues.
The objective of queuing analysis is to design
systems that enable organizations to performoptimally according to some criterion.
Possible Criteria
Maximum Profits.
Desired Service Level.
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Introduction
First studied by A. K. Erlang in 1913
Analyzed telephone facilities
Body of knowledge called queuing theory
Queue is another name for waiting line
Decision problem Balance cost of providing good service with cost of
customers waiting
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Level of service
Cost
Waiting time cost
Optimal
Introduction: Waiting Line Costs
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Introduction
Analyzing queuing systems requires a clearunderstanding of the appropriate service
measurement (KPIs). Possible service measurements (KPIs)
Average time a customer spends in line.
Average length of the waiting line. The probability that an arriving customer must wait
for service.
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Introduction: The Queuing System
Customer
Server
Customers waits in queue
Service
Customer completes
service and leaves
Customers arrives
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Bank Customers Teller Deposit etc.
Doctors Patient Doctor Treatment
office
Traffic Cars Light Controlledintersection passage
Assembly line Parts Workers Assembly
Tool crib Workers Clerks Check out/in tools
Situation Arrivals Servers Service Process
Waiting Line Examples
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Three Parts of a Queuing System at Daves Car-Wash
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Elements of the Queuing Process
A queuing system consists of three basic components:
Arrivals: Customers arrive according to some arrivalpattern.
Waiting in a queue: Arriving customers may have to wait in
one or more queues for service.
Service: Customers receive service and leave the system.
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ServiceFacilityWaiting Line
Arrival rate distributionPoissonOther
Pattern of arrivalsRandomScheduled
Arrival Characteristics
Characteristics of a Waiting Line System
Size of the source populationLimitedUnlimited
Behavior of the arrivalsJoin the queue, and wait until servedBalk; refuse to join the lineRenege; leave the line
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Input Source(Population)
Size
Infinite
Input Characteristics
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Input Characteristics
Input Source(Population)
Size
1995 Corel Corp.
Infinite Finite
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Input Source(Population)
SizeArrival
Pattern
FiniteInfinite Random Non-Random
Input Characteristics
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Input Source(Population)
SizeArrivalPattern
FiniteInfinite Random Non-Random
Poisson Other
Input Characteristics
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Under three conditions the arrivals can be modeled as aPoisson process
Orderliness : one customer, at most, will arrive during anytime interval.
Stationarity: for a given time frame, the probability of arrivals
within a certain time interval is the same for all time intervals of
equal length.
Independence : the arrival of one customer has no influence
on the arrival of another.
The Arrival Process
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P(X = k) =
Wherel = mean arrival rate per time unit.
t = the length of the interval.
e = 2.7182818 (the base of the natural logarithm).k! = k (k -1) (k -2) (k -3) (3) (2) (1).
(lt)ke- lt
k!
The Poisson Arrival Process
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Number of events thatoccur in an interval of time t
Example: Number of
customers that arrive in15 min.
Mean = l (e.g., 5/hr.)
Probability:
,0
,3
,6
0 1 2 3 4 5
X
P(X)
,0,3
,6
0 2 4 6 8 10X
P(X)
l = 0.5
l = 6.0
Poisson Distribution
!
)()(
k
tekXP
kt ll-
P i Di t ib ti f A i l Ti
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Poisson Distributions for Arrival Times
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0 1 2 3 4 5 6 7 8 9 10 11 12x
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0 1 2 3 4 5 6 7 8 9 10 11 12
x
Probability
Probability
l=2 l=4
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HANKs HARDWARE Arrival Process
Customers arrive at Hanks Hardware according to a
Poisson distribution.
Between 8:00 and 9:00 A.M. an average of 6 customers
arrive at the store.
What is the probability that k customers will arrivebetween 8:00 and 8:30 in the morning (k = 0, 1, 2,)?
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k
Input to the Poissondistribution
l = 6 customers per hour.t = 0.5 hour.lt = (6)(0.5) = 3.
(lt) e- lt
k !
0
0!
0.049787
0
1!
1
0.149361
2
2!
0.2240423
3!
0.224042
1 2 3 4 5 6 7
P(X = k )=
8
0123
HANKs HARDWARE
An illustration of the Poisson distribution.
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HANKs HARDWARE
Using Excel for the Poisson probabilities Solution
We can use the POISSON function in Excel to
determine Poisson probabilities. Point probability: P(X = k) = ?
Use Poisson(k, lt, FALSE)
Example: P(X = 0; lt = 3) = POISSON(0, 3, FALSE)
Cumulative probability: P(Xk) = ?
Example: P(X3; lt = 3) = Poisson(3, 3, TRUE)
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HANKs HARDWARE Excel Poisson
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Input Source(Population)
Size BehaviorArrivalPattern
FiniteInfinite Random Non-Random
Patient Impatient
Poisson Other
Input Characteristics
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Input Source(Population)
Size BehaviorArrivalPattern
FiniteInfinite Random Non-Random
Patient Impatient
BalkPoisson Other
Input Characteristics
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Inputsource
Service
facility
Waiting
line
Service system
1995 Corel Corp.
Line was
too long!
Balking
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Input Source(Population)
Size BehaviorArrivalPattern
FiniteInfinite Random Non-Random
Patient Impatient
Balk RenegePoisson Other
Input Characteristics
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RenegingInput
source Servicefacility
Waitingline
1995 Corel Corp.
I give
up!
Service System
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Waiting Line
Length
Unlimited
1995 Corel Corp.
Waiting Line Characteristics
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Waiting Line
Length
LimitedUnlimited
1995 Corel Corp.
1995 Corel Corp.
Waiting Line Characteristics
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Waiting Line
Length QueueDiscipline
LimitedUnlimited FIFO(FCFS) Random Priority
Waiting Line Characteristics
LIFO(LCFS)
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ServiceFacility
Configuration
Multi-Channel
SingleChannel
SinglePhase
Service Characteristics
SinglePhase
MultiPhase
MultiPhase
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ArrivalsServedunits
Servicefacility
Queue
Service system
DockWaiting ship line
Ships atsea
Ship unloading system Empty
ships
Single-Channel, Single-Phase System
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Cars
& food
Single-Channel, Multi-Phase System
ArrivalsServedunits
Servicefacility
Queue
Service system
Pick-upWaiting cars
Carsin area
McDonalds drive-through
Pay
Servicefacility
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Arrivals
Servedunits
ServicefacilityQueue
Service system
Servicefacility
Example: Bank customers wait in single line for one ofseveral tellers.
Multi-Channel, Single Phase System
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Servicefacility
Arrivals
Servedunits
ServicefacilityQueue
Service system
Servicefacility
Example: At a laundromat, customers use one of severalwashers, then one of several dryers.
Servicefacility
Multi-Channel, Multi-Phase System
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In most business situations, service time varies widelyamong customers.
When service time varies, it is treated as a randomvariable.
The exponential probability distribution is used
sometimes to model customer service time.
The Service Process
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The memoryless property. No additional information about the time left for the completion of a
service, is gained by recording the time elapsed since the service
started. For Hanks, the probability of completing a service within the next 3
minutes is (0.52763) independent of how long the customer has beenserved already.
The Exponential and the Poisson distributions are related toone another. If customer arrivals follow a Poisson distribution with mean rate l,
their interarrival times are exponentially distributed with mean time
1/l.
The Exponential Distribution -
Characteristics
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f(t) = me-mt
m = the average number of customerswho can be served per time period.Therefore, 1/m = the mean service time.
The probability that the service time X is less than some t.
P(X t) = 1 - e-mt
The Exponential Service Time Distribution
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Schematic illustration of the exponential
distribution
The probability that service is completed
within t time unitsP(X t) = 1 - e-mt
X = t
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HANKs HARDWARE Service time
Hanks estimates the average service time to be 1/m =4 minutes per customer.
Service time follows an exponential distribution.
What is the probability that it will take less than 3minutes to serve the next customer?
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We can use the EXPDIST function in Excel to determineexponential probabilities.
Probability density: f(t) = ? Use EXPONDIST(t, m, FALSE)
Cumulative probability: P(Xk) = ? Use EXPONDIST(t, m, TRUE)
Using Excel for the Exponential Probabilities
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The mean number of customers served perminute is = (60) = 15 customers per hour.
P(X < .05 hours) = 1 e-(15)(.05) = ? From Excel we have:
EXPONDIST(.05,15,TRUE) = .5276
HANKs HARDWARE
Using Excel for the Exponential Probabilities
3 minutes = .05 hours
HANK HARDWARE
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HANKs HARDWARE Using Excel for the Exponential Probabilities
=EXPONDIST(B4,B3,TRUE)
Exponential Distribution for Mu = 15
0.000
2.000
4.000
6.000
8.000
10.000
12.000
14.000
16.000
0.000 0.075 0.150 0.225 0.300 0.375
t
f(t)
=EXPONDIST(A10,$B$3,FALSE)Drag to B11:B26
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0.
.1
.2
.3
.4
0 2 4 6 8 10x
Probability
t>x
m=1m=2m=3m=4
Service time, &time between arrivals
Example: Service time is20 min.
Mean service rate = m
e.g., customers/hr.
Mean service time = 1/m
Equation:
Negative Exponential Distribution
f(t) = me-mt
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Performance Measures of Queuing System
Performance can be measured by focusing on:
Customers in queue. Customers in the system.
Performance is measured for a system in steadystate.
P f M f Q i S t
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Roughly, thisis a transientperiod
n
Time
Performance Measures of Queuing System
The transient periodoccurs at the initial
time of operation.
Initial transientbehavior is not
indicative of long runperformance.
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This is a
steady stateperiod..
n
Time
Performance Measures of Queuing
System The steady state
period follows the
transient period. Meaningful long run
performancemeasures can be
calculated for thesystem when insteady state.
Roughly, thisis a transientperiod
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l< kmEach with
service rate ofml< m1 +m2++mk
For k serverswith service rates mi
l< mFor one server
In order to achieve steady state, the
effective arrival rate must be less thanthe sum of the effective service rates .
Performance Measures of Queuing
System
k servers
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P0 = Probability that there are no customers in the system.
Pn = Probability that there are n customers in the system.
L = Average number of customers in the system.
Lq = Average number of customers in the queue.
W = Average time a customer spends in the system.
Wq = Average time a customer spends in the queue.
Pw = Probability that an arriving customer must wait for service.
r = Utilization rate for each server (% of time that each server is busy).
Steady State Performance Measures
Deciding on the
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Cost
Cost of providingservice
Total expected cost
Cost of waiting time
Low levelof service
High level ofservice
Optimalservice level
Minimumtotal cost
Deciding on theOptimum Level of Service
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Littles Formulas represent important relationshipsbetween L, Lq, W, and Wq.
These formulas apply to systems that meet the
following conditions: Single queue systems,
Customers arrive at a finite arrival rate l, and
The system operates under a steady state condition.
L = l W Lq = l Wq L = Lq + l/m
Littles Formulas
For the case of an infinite population W = Wq + 1/m
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Queuing system can be classified by:
Arrival process.
Service process.
Number of servers. System size (infinite/finite waiting line).
Population size.
Notation M (Markovian) = Poisson arrivals or exponential service time. D (Deterministic) = Constant arrival rate or service time.
G (General) = General probability for arrivals or service time.
Example:
M / M / 6 / 10 / 20
Classification of Queues
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M/M/1 Queuing System -Assumptions Poisson arrival process.
Exponential service time distribution.
A single server.
Potentially infinite queue.
An infinite population.
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The probability thata customer waits inthe system more thant is P(X>t) = e-(m - l)t
P0 = 1 (l/m)Pn = [1 (l/m)](l/m)nL = l /(ml)Lq = l
2 /[m(ml)]W = 1 /(ml)Wq = l /[m(ml)]Pw = l / mr = l / m
M / M /1 Queue - Performance Measures
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MARYs SHOES
Customers arrive at Marys Shoes every 12 minutes on
the average, according to a Poisson process.
Service time is exponentially distributed with anaverage of 8 minutes per customer.
Management is interested in determining theperformance measures for this service system.
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MARYs SHOES - Solution
Inputl = 1/12 customers per minute = 60/12 = 5 per hour.
m = 1/8 customers per minute = 60/8 = 7.5 per hour.
Performance Calculations
P0 = 1 - (l/m) = 1 - (5/7.5) = 0.3333Pn = [1 - (l/m)](l/m)n = (0.3333)(0.6667)nL = l/(m - l) = 2Lq = l
2/[m(m - l)] = 1.3333W = 1/(m - l) = 0.4 hours = 24 minutesWq = l/[m(m - l)] = 0.26667 hours = 16 minutes
Pw = l/m =0.6667r = l/m =0.6667
P(X
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MARY s SHOES - Spreadsheet solution
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M/M/k Queuing Systems Characteristics
Customers arrive according to a Poisson process at a mean
rate l.
Service times follow an exponential distribution.
There are k servers, each of who works at a rate ofmcustomers (with km> l).
Infinite population, and possibly infinite line.
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P
n k
k
k
n k
n
k0
0
1
1
1 1
+
-
-
! !
lm
lm
m
m l
Pn
P
k kP
n
n
n
n k
-
lm
lm
!
!
0
0
for n k.
P for n > k.n
M / M /k Queue - Performance Measures
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( ) ( )
Wk k
P
k
- -+
lm m
m l m1
12 0
!
Performance measurements L, Lq, Wq,, can be obtained from Littles formulas.
Pk
k
kP
w
k
-
10
!
lm
mm l
r
l
m k
M / M /k Queue - Performance Measures
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LITTLE TOWN POST OFFICE
Little Town post office is open on Saturdaysbetween 9:00 a.m. and 1:00 p.m.
Data
On the average 100 customers per hour visit the officeduring that period. Three clerks are on duty.
Each service takes 1.5 minutes on the average.
Poisson and Exponential distributions describe thearrival and the service processes respectively.
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LITTLE TOWN POST OFFICE
The Postmaster needs to know the relevant servicemeasures in order to:
Evaluate the current service level.
Study the effects of reducing the staff by one clerk.
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This is an M / M / 3 queuing system.
Input
l 100 customers per hour.m 40 customers per hour (60/1.5).
Does steady state exist (l < km )?
l 100 < km 3(40) = 120.
LITTLE TOWN POST OFFICE - Solution
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LITTLE TOWN POST OFFICE solutioncontinued
First P0 is found by
045.
625.152
25.65.21
1
100)40(3
)40(3
40
100
!3
1
40
100
!n
1
1
P 2
0n
3n0
+++
-
+
P0 is used now to determine all the other performance measures.(ver slide seguinte: usando o Template Queue.xls)
LITTLE TOWN POST OFFICE spreadsheet Solution
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M/M/k Queuing Model
INPUTS Value INPUTS Value
Lambda = 100
Mu = 40
OUTPUTS
# Servers L Lq W Wq Pw Rho Cost 0 1 2
12
3 6.011236 3.511236 0.060112 0.035112 0.702247 0.833333 0 0.044944 0.11236 0.140449
4 3.033095 0.533095 0.030331 0.005331 0.319857 0.625 0 0.073695 0.184237 0.230297
5 2.630371 0.130371 0.026304 0.001304 0.130371 0.5 0 0.0801 0.20025 0.250313
6 2.533889 0.033889 0.025339 0.000339 0.047445 0.416667 0 0.08162 0.204051 0.255063
7 2.50858 0.00858 0.025086 8.58E-05 0.015443 0.357143 0 0.08198 0.204951 0.256189
8 2.502053 0.002053 0.025021 2.05E-05 0.004517 0.3125 0 0.082063 0.205157 0.2564469 2.50046 0.00046 0.025005 4.6E-06 0.001195 0.277778 0 0.082081 0.205201 0.256502
10 2.500096 9.59E-05 0.025001 9.59E-07 0.000288 0.25 0 0.082084 0.20521 0.256513
11 2.500019 1.87E-05 0.025 1.87E-07 6.34E-05 0.227273 0 0.082085 0.205212 0.256515
12 2.500003 3.4E-06 0.025 3.4E-08 1.29E-05 0.208333 0 0.082085 0.205212 0.256516
13 2.500001 5.79E-07 0.025 5.79E-09 2.43E-06 0.192308 0 0.082085 0.205212 0.256516
14 2.5 9.28E-08 0.025 9.28E-10 4.27E-07 0.178571 0 0.082085 0.205212 0.256516
15 2.5 1.4E-08 0.025 1.4E-10 7.02E-08 0.166667 0 0.082085 0.205212 0.256516
in the System where n =
Probabiility of n Customers
Server Cost =
Goodwill Cost When Waiting =
Goodwill Cost While Being Served =
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M/G/1 Queuing System Assumptions
Customers arrive according to a Poisson process with a
mean rate l.
Service time has a general distribution with mean rate m.
One server.
Infinite population, and possibly infinite line.
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( )L
+
-
+
2
2
2 1
l lm
lm
l
m
Note: It is not necessary to know the particular service time distribution.Only the mean and standard deviation of the distribution are needed.
M/G/1 Queuing SystemPollaczek - Khintchine Formula for L
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Teds repairs television sets and VCRs.
Data
It takes an average of 2.25 hours to repair a set.
Standard deviation of the repair time is 45 minutes. Customers arrive at the shop once every 2.5 hours on the average,
according to a Poisson process.
Ted works 9 hours a day, and has no help.
He considers purchasing a new piece of equipment. New average repair time is expected to be 2 hours.
New standard deviation is expected to be 40 minutes.
TEDS TV REPAIR SHOP
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Ted wants to know the effects of using the newequipment on
1. The average number of sets waiting for repair;
2. The average time a customer has to waitto get his repaired set.
TEDS TV REPAIR SHOP
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This is an M/G/1 system (service time is notexponential - note that 1/m).
Input The current system (without the new equipment)l = 1/2.5 = 0.4 customers per hour.
m = 1/2.25 = 0.4444 costumers per hour.
= 45/60 = 0.75 hours. The new system (with the new equipment)
m = 1/2 = 0.5 customers per hour.
= 40/60 = 0.6667 hours.
TEDS TV REPAIR SHOP - Solution
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TEDS TV REPAIR SHOP - Solution
Medida de Desempenho Atual NovoL 5.40 2.58
Lq 4.50 1.78
W 13.50 6.44Wq 11.25 4.44
Pw .9 .8
r .9 .8P0 .1 .2
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Poisson arrival process at mean rate l.
k servers, each having an exponential service time with mean
rate m.
Maximum number of customers that can be present in the
system at any one time is F.
Customers are blocked (and never return) if the system is full.
Characteristics of M/M/k/F Queuing System
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A customer is blocked if the system is full.
Probability that the system is full is PF (that is, 100PF%of the arriving customers dont enter the system).
The effective arrival rate = the rate of arrivals that makeit through into the system (le).
le = l(1 - PF)
M/M/k/F Queuing SystemEffective Arrival Rate
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RYAN ROOFING COMPANY
Ryan gets most of its business from customers whocall and order service.
When a telephone line is available but the secretaryis busy serving a customer, a new calling customer is willingto wait until the secretary becomes available.
When all the lines are busy, a new calling customer gets abusy signal and calls a competitor.
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Data
Arrival process is Poisson, and service process is Exponential.
Each phone call takes 3 minutes on the average.
10 customers per hour call the company on the average.
One appointment secretary takes phone calls from3 telephone lines.
RYAN ROOFING COMPANY
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Management would like to design the following system:
The fewest lines necessary.
At most 2% of all callers get a busy signal.
Management is interested in the following information:
The percentage of time the secretary is busy.
The average number of customers kept on hold. The average time a customer is kept on hold.
The actual percentage of callers who encounter a busy signal.
RYAN ROOFING COMPANY
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This is an M/M/1/3 system Input
l = 10 per hour.m = 20 per hour (1/3 per minute).
Excel spreadsheet gives:P0 = 0.533, P1 = 0.133, P3 = 0.06
6.7% of the customers get a busy signal.
This is above the goal of 2%.
P0 = 0.516, P1 = 0.258, P2 = 0.129, P3 = 0.065, P4 = 0.032
3.2% of the customers get the busy signalStill above the goal of 2%
RYAN ROOFING COMPANY - Solution
M/M/1/4 systemM/M/1/5 system
P0 = 0.508, P1 = 0.254, P2 = 0.127, P3 = 0.063, P4 = 0.032
P5 = 0.016So 1.6% of the customers get the busy signalThe goal of 2% has been achieved.
See spreadsheet next
RYAN ROOFING COMPANY - Spreadsheet Solution
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M / M / 1 / / m Queuing Systems
In this system the number of potential customers is finite andrelatively small.
As a result, the number of customers already in the systemaffects the rate of arrivals of the remaining customers.
Characteristics
A single server.
Exponential service time, Poisson arrival process. A population size of a (finite) m customers.
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PACESETTER HOMES
Pacesetter Homes runs four different development projects.
Data
At each site running a project is interrupted once every 20 workingdays on the average.
The V.P. for construction handles each stoppage.
How long on the average a site is non-operational?
If it takes 2 days on the average to restart a projects progress (theV.P. is using the current car).
If it takes 1.875 days on the average to restart a projects progress(the V.P. is using a new car)
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PACESETTER HOMES Solution
This is an M/M/1//4 system, where: The four sites are the four customers.
The V.P. for construction is the server. Input
l = 0.05 (1/20)m = 0.5 m = 0.533
(1/2 days, using the current car) (1/1.875 days, using a new car).
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Performance Current New
Measures Car Car
Average number of customers in the system L 0.467 0.435
Average number of customers in the queue Lq 0.113 0.100
Average number of days a customer is in the system W 2.641 2.437
Average number of days a customer is in the queue Wq 0.641 0.562The probability that an arriving customer will wait Pw 0.353 0.334
Oveall system effective utilization-factor r 0.353 0.334The probability that all servers are idle Po 0.647 0.666
Summary of results
PACESETTER HOMES Solutioncontinued
PACESETTER HOMES Spreadsheet Solution
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M/M/1//m Queuing Model
INPUTS Value
0.05
Mu = 0.53333
4
Probabiility of n Customers
OUTPUTS in the System where n =
Server L Lq W Wq Pw Rho 0 1 2 3 41 0.43451 0.10025 2.43734 0.56233 0.33427 0.33427 0.66573 0.24965 0.07021 0.01317 0.00123
Lambda =
m =
E i A l i f Q i
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Economic Analysis of QueuingSystems
The performance measures previously developed areused next to determine a minimal cost queuing
system. The procedure requires estimated costs such as:
Hourly cost per server .
Customer goodwill cost while waiting in line. Customer goodwill cost while being served.
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WILSON FOODSTALKING TURKEY HOT LINE
Wilson Foods has an 800 number to answer customersquestions.
If all the customer representatives are busy when a newcustomer calls, he/she is asked to stay on the line.
A customer stays on the line if the waiting time is notlonger than 3 minutes.
H t iWILSON FOODS
- TALKING TURKEY HOT LINE
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Data
On the average 225 calls per hour are received.
An average phone call takes 1.5 minutes. A customer will stay on the line waiting at most 3 minutes.
A customer service representative is paid $16 per hour.
Wilson pays the telephone company $0.18 per minute when the
customer is on hold or when being served.
Customer goodwill cost is $20 per minute while on hold.
Customer goodwill cost while in service is $0.05.
How many customer servicerepresentatives should be used
to minimize the hourly cost ofoperation?
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TC(K) =Cwk + (Ct + gs)L + (gw - gs)Lq
WILSON FOODS Solution
The total hourly cost model
TC(K) = Cwk + CtL + gwLq + gs(L - Lq)
Total hourly wages
Total averagehourly Telephone charge
Average hourly goodwill
cost for customers on hold
Average hourly goodwillcost for customers in service
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InputCw= $16
Ct = $10.80 per hour [0.18(60)]
gw= $12 per hour [0.20(60)]gs = $3 per hour [0.05(60)]
The Total Average Hourly Cost =TC(K) = 16K + (10.8+3)L + (12 - 3)Lq = 16K + 13.8L + 9Lq
WILSON FOODS Solution continued
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Assuming a Poisson arrival process and an Exponential servicetime, we have an M/M/K system.
l = 225 calls per hour.
m = 40 per hour (60/1.5). The minimal possible value for K is 6 to ensure that steady
state exists (l
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Summary of results of the runs for k=6,7,8,9,10
K L Lq Wq TC(K)
6 18.1255 12.5 0.05556 458.64
7 7.6437 2.0187 0.00897 235.65
8 6.2777 0.6527 0.0029 220.51
9 5.8661 0.2411 0.00107 227.12
10 5.7166 0.916 0.00041 239.70
Conclusion: employ 8 customer service representatives.
WILSON FOODS Solution continued
WILSON FOODS Spreadsheet Solution
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M/M/k Queuing Model
INPUTS Value INPUTS Value
Lambda = 225 16Mu = 40 22.8
13.8
OUTPUTS
# Servers L Lq W Wq Pw Rho Cost 0 1 2
1
2
3
4
5
6 18.1255 12.5005 0.080558 0.055558 0.833367 0.9375 458.6364 0.001184 0.006659 0.01873
7 7.643727 2.018727 0.033972 0.008972 0.493467 0.803571 235.652 0.002742 0.015423 0.043376
8 6.277703 0.652703 0.027901 0.002901 0.275586 0.703125 220.5066 0.003291 0.018514 0.05207
9 5.866105 0.241105 0.026072 0.001072 0.144663 0.625 227.1222 0.003492 0.019641 0.055241
10 5.716569 0.091569 0.025407 0.000407 0.07122 0.5625 239.7128 0.003565 0.020056 0.056407
11 5.659381 0.034381 0.025153 0.000153 0.032853 0.511364 254.4089 0.003592 0.020206 0.056831
12 5.637531 0.012531 0.025056 5.57E-05 0.014202 0.46875 269.9107 0.003602 0.02026 0.056981
13 5.629393 0.004393 0.02502 1.95E-05 0.00576 0.432692 285.7252 0.003605 0.020278 0.057032
in the System where n =
Probabiility of n Customers
Server Cost =Goodwill Cost When Waiting =
Goodwill Cost While Being Served =
HARGROVE HOSPITAL MATERNITY
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HARGROVE HOSPITAL MATERNITYWARD
Hargrove Hospital is experiencing cutbacks, and is trying toreorganize operations to reduce operating costs.
There is a trade off between the costs of operating more birthing stations and
the costs of rescheduling surgeries in the surgery room when womengive birth there, if all the birthing stations are occupied.
The hospital wants to determine the optimal number of birthingstations that will minimize operating costs.
HARGROVE HOSPITAL MATERNITY
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HARGROVE HOSPITAL MATERNITYWARD
Data
Cutting one birthing station saves $25,000 per year.
Building a birthing station costs $30,000. Maternity in the surgery room costs $400 per hour.
Six women on the average need a birthing station a day.The arrival process is Poisson.
Every birthing process occupies a birthing station for twohours on the average.
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Solution
Analysis of the current situation
Currently there are two birthing stations The current problem can be modeled as a M/G/2/2 queuing system. Using the MGkk Excel worksheet with l = 6 and m = 12/day we have:
r = .23077 W = .0833 days Pw = .076923
L = .46154
P0 = .6154
HARGROVE HOSPITAL
7.7% of the arriving women are sent tothe surgery room to give birth.
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Solution continued The birthing stations problem can be modeled as a
M/G/k/k queuing model. Input
l = 6 women per day;m = 12 women per day (24/2);
k = the number of birthing stations used
The total cost for the hospital is
TC(k) = Cost of using the surgery room for maternity
+ Additional cost of operating k stations
HARGROVE HOSPITAL
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Solution continued
Average daily cost of using the surgery room for maternity:
Pk(l)(average time in the surgery room)(hourly cost) Average additional daily cost of operating k stations
25,000/365 = $68.49 per day.
Average daily total costTC(k) = Pk(l)(24/m)(Hourly cost) + 68.49k
= Pk(6)(24/12)(400) + 68.49k
= 4800Pk + 68.49k
HARGROVE HOSPITAL
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k Pk $4800PkAdditional
Cost ofstations
Total net
average
Daily cost
12
3
4
.333333.076923
.012658
.001580
$1,600369.23
60.76
7.58
$ 68.490
82.19
163.38
$1,531.51364.23
142.95
171.96
Optimal
From repeated runs of the MGkk worksheet to determine Pk wegot the following results:
HARGROVE HOSPITAL - Solution
Current
HARGROVE HOSPITALSpreadsheet Solution
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M/G/k/k Queuing Model
INPUTS Value
6
12
2
OUTPUTS# Servers L Lq W Wq Pw Rho 0 1 2 3
2 0.461538 0 0.083333 0 0.076923 0.230769 0.615385 0.307692 0.076923
in the System where n =
Probabiility of n Customers
Lambda =
Mu =
k =
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Tandem Queuing Systems
In a Tandem Queuing System a customer must visitseveral different servers before service is completed.
BeverageMeats
Examples
All-You-Can-Eat restaurant
Tandem Queuing Systems
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BeverageMeats
In a Tandem Queuing System a customer must visitseveral different servers before service is completed.
g y
Examples
All-You-Can-Eat restaurant
Tandem Queuing Systems
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BeverageMeats
Tandem Queuing Systems
A drive-in restaurant, where first you place your order, then pay andreceive it in the next window.
A multiple stage assembly line.
Examples All-You-Can-Eat restaurant
In a Tandem Queuing System a customer must visitseveral different servers before service is completed.
Cashiers
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Tandem Queuing Systems
For cases in which customers arrive according to aPoisson process and service time in each station isexponential, .
Total Average Time in the System =Sum of all average times at the individual stations
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BIG BOYS SOUND, INC.
Big Boys sells audio merchandise.
The sale process is as follows:
A customer places an order with a sales person.
The customer goes to the cashier station to pay for the
order.
After paying, the customer is sent to the pickup desk to
obtain the good.
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Data for a regular Saturday Personnel.
8 sales persons are on the job.
3 cashiers.
2 workers in the merchandise pickup area.
Average service times.
Average time a sales person waits on a customer is 10 minutes.
Average time required for the payment process is 3 minutes.
Average time in the pickup area is 2 minutes.
Distributions.
Exponential service time at all the service stations.
Poisson arrival with a rate of 40 customers an hour.
BIG BOYS SOUND, INC.
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What is the average amount of time, a customerwho makes a purchase spends in the store?
Only 75% of the arriving customers make a purchase!
BIG BOYS SOUND, INC.
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BIG BOYS SOUND, INC. Solution
This is a Three Station Tandem Queuing System
Sales ClerksM / M / 8
CashiersM / M / 3
Pickup deskM / M / 2
W = 3 47 minutes
W3 = 2.67 minutes
(.75)(40)=30
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