1 introduction queuing is the study of waiting lines, or queues. the objective of queuing analysis...
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IntroductionIntroduction
• Queuing is the study of waiting lines, or queues.• The objective of queuing analysis is to design
systems that enable organizations to perform optimally according to some criterion.
• Possible Criteria– Maximum Profits.
– Desired Service Level.
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IntroductionIntroduction
• Analyzing queuing systems requires a clear understanding of the appropriate service measurement.
• Possible service measurements– 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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Elements of the Queuing ProcessElements of the Queuing Process
• A queuing system consists of three basic components:– Arrivals: Customers arrive according to some arrival
pattern.
– 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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The Arrival ProcessThe Arrival Process
• There are two possible types of arrival processes
– Deterministic arrival process.
– Random arrival process.• The random process is more common in
businesses.
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• Under three conditions the arrivals can be modeled as a Poisson process– Orderliness : one customer, at most, will arrive during any
time 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 ProcessThe Arrival Process
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P(X = k) =
Where = 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).
tke- t
k!
The Poisson Arrival ProcessThe Poisson Arrival Process
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HANK’s HARDWARE – HANK’s HARDWARE – Arrival ProcessArrival Process
• Customers arrive at Hank’s 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 arrive between 8:00 and 8:30 in the morning (k = 0, 1, 2,…)?
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k
• Input to the Poisson distribution
= 6 customers per hour.t = 0.5 hour.t = (6)(0.5) = 3.
t e- t
k !
0
0.049787
0
1!
1
0.1493612
2!0.224042
3
3! 0.224042
1 2 3 4 5 6 7
P(X = k )=
8
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HANK’s HARDWARE – HANK’s HARDWARE – An illustration of the Poisson distribution. An illustration of the Poisson distribution.
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HANK’s HARDWARE – HANK’s HARDWARE – Using Excel for the Poisson probabilitiesUsing 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, t, FALSE)• Example: P(X = 0; t = 3) = POISSON(0, 1.5, FALSE)
– Cumulative probability: P(Xk) = ?• Example: P(X3; t = 3) = Poisson(3, 1.5, TRUE)
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HANK’s HARDWARE – HANK’s HARDWARE – Excel PoissonExcel Poisson
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• Factors that influence the modeling of queues
– Line configuration
– Jockeying
– Balking
The Waiting Line CharacteristicsThe Waiting Line Characteristics
– Priority
– Tandem Queues
– Homogeneity
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• A single service queue.• Multiple service queue with single waiting line.• Multiple service queue with multiple waiting
lines.• Tandem queue (multistage service system).
Line ConfigurationLine Configuration
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• Jockeying occurs when customers switch lines once they perceived that another line is moving faster.
• Balking occurs if customers avoid joining the line when they perceive the line to be too long.
Jockeying and BalkingJockeying and Balking
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• These rules select the next customer for service.• There are several commonly used rules:
– First come first served (FCFS).– Last come first served (LCFS).– Estimated service time.– Random selection of customers for service.
Priority RulesPriority Rules
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Tandem QueuesTandem Queues
• These are multi-server systems. • A customer needs to visit several service
stations (usually in a distinct order) to complete the service process.
• Examples– Patients in an emergency room.– Passengers prepare for the next flight.
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• A homogeneous customer population is one in which customers require essentially the same type of service.
• A non-homogeneous customer population is one in which customers can be categorized according to: – Different arrival patterns– Different service treatments.
HomogeneityHomogeneity
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• In most business situations, service time varies widely among customers.
• When service time varies, it is treated as a random variable.
• The exponential probability distribution is used sometimes to model customer service time.
The Service ProcessThe Service Process
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f(t) = e-t
= the average number of customers who can be served per time period.Therefore, 1/ = the mean service time.
The probability that the service time X is less than some “t.”
P(X t) = 1 - e-t
The Exponential Service Time DistributionThe Exponential Service Time Distribution
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Schematic illustration of the exponential Schematic illustration of the exponential distributiondistribution
The probability that service is completed within t time units
P(X t) = 1 - e-t
X = t
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HANK’s HARDWARE – HANK’s HARDWARE – Service timeService time
• Hank’s estimates the average service time to be 1/ = 4 minutes per customer.
• Service time follows an exponential distribution.• What is the probability that it will take less than 3
minutes to serve the next customer?
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• We can use the EXPDIST function in Excel to determine exponential probabilities.
• Probability density: f(t) = ?– Use EXPONDIST(t, , FALSE)
• Cumulative probability: P(Xk) = ?– Use EXPONDIST(t, , TRUE)
Using Excel for the Exponential ProbabilitiesUsing Excel for the Exponential Probabilities
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• The mean number of customers served per minute is ¼ = ¼(60) = 15 customers per hour.
• P(X < .05 hours) = 1 – e-(15)(.05) = ?• From Excel we have:
– EXPONDIST(.05,15,TRUE) = .5276
HANK’s HARDWARE – HANK’s HARDWARE – Using Excel for the Exponential ProbabilitiesUsing Excel for the Exponential Probabilities
3 minutes = .05 hours
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HANK’s HARDWARE – HANK’s HARDWARE – Using Excel for the Exponential ProbabilitiesUsing Excel for the Exponential Probabilities
=EXPONDIST(B4,B3,TRUE)
Exponential Distribution for Mu = 15
0.0002.0004.0006.0008.00010.00012.00014.00016.000
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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• 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 Hank’s, the probability of completing a service within the next 3 minutes is (0.52763) independent of how long the customer has been served already.
• The Exponential and the Poisson distributions are related to one another.
– If customer arrivals follow a Poisson distribution with mean rate , their interarrival times are exponentially distributed with mean time 1/
The Exponential Distribution -The Exponential Distribution - CharacteristicsCharacteristics
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9.3 Performance Measures of 9.3 Performance Measures of Queuing System Queuing System
• Performance can be measured by focusing on:– Customers in queue.
– Customers in the system.
• Performance is measured for a system in steady state.
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Roughly, this is a transient period…
n
Time
9.3 Performance Measures of 9.3 Performance Measures of Queuing System Queuing System
• The transient period occurs at the initial time of operation.
• Initial transient behavior is not indicative of long run performance.
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This is a steady state period………..
n
Time
9.3 Performance Measures of 9.3 Performance Measures of Queuing System Queuing System
• The steady state period follows the transient period.
• Meaningful long run performance measures can be calculated for the system when in steady state.
Roughly, this is a transient period…
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kEach with
service rate of
kEach with
service rate of
…
For k serverswith service rates
…
For k serverswith service rates
For one server
For one server
In order to achieve steady state, theeffective arrival rate must be less than the sum of the effective service rates .
In order to achieve steady state, theeffective arrival rate must be less than the sum of the effective service rates .
9.3 Performance Measures of 9.3 Performance Measures of Queuing System Queuing System
k servers
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P0 = Probability that there are no customers in the system.P0 = Probability that there are no customers in the system.
Pn = Probability that there are “n” customers in the system.Pn = Probability that there are “n” customers in the system.
L = Average number of customers in the system.L = Average number of customers in the system.
Lq = Average number of customers in the queue.Lq = Average number of customers in the queue.
W = Average time a customer spends in the system.W = Average time a customer spends in the system.
Wq = Average time a customer spends in the queue.Wq = Average time a customer spends in the queue.
Pw = Probability that an arriving customer must wait for service.
Pw = Probability that an arriving customer must wait for service.
= Utilization rate for each server (the percentage of time that each server is busy).
= Utilization rate for each server (the percentage of time that each server is busy).
Steady State Performance MeasuresSteady State Performance Measures
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• Little’s Formulas represent important relationships between 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 and– The system operates under a steady state condition.
L =W Lq = Wq L = Lq +
Little’s FormulasLittle’s Formulas
For the case of an infinite population
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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
Example:
M / M / 6 / 10 / 20
Classification of QueuesClassification of Queues
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MMMM1 Queuing System - 1 Queuing System - AssumptionsAssumptions
– 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 in the system more than “t” is P(X>t) = e-( - )t
The probability thata customer waits in the system more than “t” is P(X>t) = e-( - )t
P0 = 1 – ()
Pn = [1 – ()]()n
L = ( – )Lq = 2 [( – )]
W = 1 ( – )Wq = [( – )]
Pw =
=
M / M /1 Queue - Performance MeasuresM / M /1 Queue - Performance Measures
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MARY’s SHOESMARY’s SHOES
• Customers arrive at Mary’s Shoes every 12 minutes on the average, according to a Poisson process.
• Service time is exponentially distributed with an average of 8 minutes per customer.
• Management is interested in determining the performance measures for this service system.
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MARY’s SHOESMARY’s SHOES - - SolutionSolution– Input = 1/12 customers per minute = 60/12 = 5 per hour. = 1/ 8 customers per minute = 60/ 8 = 7.5 per hour.
– Performance Calculations
P0 = 1 - () = 1 - (57.5) = 0.3333
Pn = [1 - ()]()n = (0.3333)(0.6667)n L = ( - ) = 2Lq = 2[( - )] = 1.3333
W = 1( - ) = 0.4 hours = 24 minutesWq = ( - )] = 0.26667 hours = 16 minutes
P0 = 1 - () = 1 - (57.5) = 0.3333
Pn = [1 - ()]()n = (0.3333)(0.6667)n L = ( - ) = 2Lq = 2[( - )] = 1.3333
W = 1( - ) = 0.4 hours = 24 minutesWq = ( - )] = 0.26667 hours = 16 minutes
Pw = =0.6667= =0.6667
P(X<10min) = 1 – e-2.5(10/60)
= .565
– = 7.5 – 5 = 2.5 per hr.
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=A11-B4/B5
=1-B4/B5
=A11/B4
=C11-1/B5
=B4/B5
=H11*($B$4/$B$5)Drag to Cell AL11
=1-E11
MARY’s SHOESMARY’s SHOES - -
Spreadsheet solutionSpreadsheet solution
=B4/(B5-B4)
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Economic Analysis of Queuing Economic Analysis of Queuing SystemsSystems
• The performance measures previously developed are used 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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Tandem Queuing SystemsTandem Queuing Systems
• In a Tandem Queuing System a customer must visit several different servers before service is completed.
Beverage Meats
• Examples– All-You-Can-Eat restaurant
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Beverage Meats
• In a Tandem Queuing System a customer must visit several different servers before service is completed.
Tandem Queuing Systems
• Examples– All-You-Can-Eat restaurant
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Beverage Meats
Tandem Queuing Systems
– A drive-in restaurant, where first you place your order, then pay and receive it in the next window.
– A multiple stage assembly line.
• Examples– All-You-Can-Eat restaurant
• In a Tandem Queuing System a customer must visit several different servers before service is completed.
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Tandem Queuing SystemsTandem Queuing Systems
• For cases in which customers arrive according to a Poisson process and service time in each station is exponential, ….
Total Average Time in the System = Sum of all average times at the individual stations
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 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.BIG BOYS SOUND, INC.
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What is the average amount of time, a customer who makes a purchase spends in the store?
Only 75% of the arriving customers make a purchase!
BIG BOYS SOUND, INC.BIG BOYS SOUND, INC.
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BIG BOYS SOUND, INC. – SolutionBIG BOYS SOUND, INC. – Solution
• This is a Three Station Tandem Queuing System
Sales ClerksM / M / 8
CashiersM / M / 3
Pickup deskM / M / 2
= 40
= 30
= 30
W1 = 14 minutesW2 = 3.47 minutes
W3 = 2.67 minutes
Total = 20.14 minutes.
(.75)(40)=30