queuing theory1
TRANSCRIPT
Queuing Theory
Introduction to Queuing Theory
• It is estimated that Americans spend a total of 37 billion
hours a year waiting in lines.
• Places we wait in line...
▪ stores ▪ hotels ▪ post offices▪ banks ▪ traffic lights ▪ restaurants▪ airports ▪ theme parks ▪ on the phone
• Waiting lines do not always contain people...• Waiting lines do not always contain people...
▪ returned videos▪ subassemblies in a manufacturing plant▪ electronic message on the Internet
• Queuing theory deals with the analysis and management of waiting lines.
Concept of loss of business due to customers’ waiting
• Cost analysis of provision of faster servicing to reduce queue
length
Waiting Lines - Queuing Theory
• Marginal cost of extra provisioning during rush hours
The Purpose of Queuing Models
• Queuing models are used to:
–describe the behavior of queuing
systems
–determine the level of service to –determine the level of service to
provide
–evaluate alternate configurations for
providing service
Queuing System Cost
• Cost of providing the service also known as
service cost
• Cost of not providing the service also known
as waiting cost
Trade-off
Total Expected Cost
Co
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of
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era
tin
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erv
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acil
ity
Service Level
Cost of providing service
Cost of waiting time
OptimalService Level
Co
st
of
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ity
• Arrival pattern
• Service pattern
• Queue discipline
• Customer’s behavior
Important factors of Queuing Situations
• Customer’s behavior
• Maximum number of customers allowed in the system
• Nature of Calling Source
Queuing System: General Structure
• Arrival Process• According to source
• According to numbers
• According to time
• Service System• Service System• Single server facility
• Multiple, parallel facilities with single queue
• Multiple, parallel facilities with multiple queues
• Service facilities in a parallel
Common Queuing System Configurations
CustomerLeavesServer 1
CustomerArrives
...
Waiting Line Server
CustomerLeaves
CustomerArrives
...
Waiting Line Server 2
CustomerLeavesServer 1
Waiting Line Server 1
Server 2
Server 3
Waiting Line
Waiting Line
CustomerLeaves
CustomerLeaves
CustomerLeaves
...
...
...
CustomerArrives
...
Waiting Line Server 2
Server 3
CustomerLeaves
CustomerLeaves
CustomerArrives
• Queue Structure• First come first served
• Last come first served
• Service in random order
• Priority service• Priority service
• Customer Behavior• Balking
• Reneging
• Jockeying
• Collusion
Characteristics of Queuing Systems:
The Arrival Process
• Arrival rate - the manner in which customers arrive at the system for service.
• Arrivals are often described by a Poisson random
variable:
L,2,1,0for λ
)(λ
==−
Xe
XcustomersP
X
where
x = no. of arrival per unit(e.g. hour)
P(x) = probability of exactly x arrivals
λλλλ =average arrival rate (e.g., calls arrive at a rate of λλλλ=5 per hour)
e = 2.7183 (known as exponential constatnt)
L,2,1,0for )( == XX!
XcustomersP
Characteristics of Queuing Systems:
The Arrival Process
• Arrivals are often described by a Poisson random
variable:
L,2,1,0for λT
)(λT
==−
Xe
XP
X
where
T = period e.g 30 minutes period
L,2,1,0for )( == XX!
XP
Characteristics of Queuing Systems:The Service Process
• Service time - the amount of time a customer
spends receiving service (not including time in the
queue).
• Service times are often described by an Exponential
random variable:
P(service time more than t) for
where µ µ µ µ is the service rate (e.g., calls can be serviced at
a rate of µµµµ=7 per hour)
• The average service time is 1/µµµµ.
P(service time more than t) = e–µµµµt, for t ≥≥≥≥ 0
P(service time less than t) = 1-e–µµµµt, for t ≥≥≥≥ 0
Exponential probability distribution used in
describing service times.
Comments
• If arrivals follow a Poisson distribution with mean λλλλ, inter
arrival times follow an Exponential distribution with
mean 1/λλλλ.
– Example
• Assume calls arrive according to a Poisson
distribution with mean λλλλ=5 per hour.distribution with mean λλλλ=5 per hour.
• Inter arrivals follow an exponential distribution
with mean 1/5 = 0.2 per hour.
• On average, calls arrive every 0.2 hours or every 12
minutes.
• The exponential distribution exhibits the Markovian
(memory less) property.
Problem 1
• On an average 5 customers reach a barber’sshop every hour. Determine the probabilitythat exactly 2 customers will reach in a 30minutes period.
Problem 2Problem 2
� The manager of a bank observes that, on average,18customers are served by a cashier in an hour.Assuming that the service time has an exponentialdistribution, what is the probability that (a) acustomer will be free within 3 minutes,(b)a customershall be served in more than 12 minutes.
Kendall Notation• Queuing systems are described by 3 parameters:
A/B/s– Parameter A
M = Markovian interarrival times
D = Deterministic interarrival time
– Parameter B
M = Markovian service timesM = Markovian service times
G = General service times
D = Deterministic service times
– Parameter s
A number Indicating the number of servers.
• Examples,
M/M/1 D/G/4 M/G/2
Operating Characteristics
Typical operating characteristics of interest include:
ρ - Utilization factor, % of time that all servers are busy.
P0 - Prob. that there are no zero units in the system.
Lq
- Avg number of units in line waiting for service.
L - Avg number of units in the system (in line & being served).
Wq
- Avg time a unit spends in line waiting for service.Wq
- Avg time a unit spends in line waiting for service.
W - Avg time a unit spends in the system (in line & being served).
Pw
- Prob. that an arriving unit has to wait for service.
Pn
- Prob. of n units in the system.
• Model 1: Poisson-exponential single server
model – infinite populationAssumptions:
�Arrivals are Poisson with a mean arrival rate of, say λ
� Service time is exponential, rate being µ
� Source population is infinite
�Customer service on first come first served basis
� Single service station
For the system to be workable, λ ≤ µ
• Model 2: Poisson-exponential single
server model – finite population
Has same assumptions as model 1, except that
population is finitepopulation is finite
• Model 3: Poisson-exponential multiple server model – infinite population
Assumptions
� Arrival of customers follows Poisson law, mean rate λ
� Service time has exponential distribution, mean service rate µ� Service time has exponential distribution, mean service rate µ
� There are K service stations
� A single waiting line is formed
� Source population is infinite
� Service on a first-come-first-served basis
� Arrival rate is smaller than combined service rate of all service facilities
Model: 1 Operating Characteristics
a) Queue length � average number of customers in queue waiting to get service
b) System length� average number of customers in the system
c) Waiting time in queue c) Waiting time in queue � average waiting time of a customer to get service
d) Total time in system � average time a customer spends in the system
e) Server idle time� relative frequency with which system is idle
• Measurement parameters
• λ= mean number of arrivals per time period (eg. Per hour)
• μ = mean number of customers served per time period
• Probability of system being busy/traffic intensity
ρ = λ / μ
• Probability of an empty facility/system being idle
P(0) = 1– λ / μ=1- ρP(0) = 1– λ / μ=1- ρ
• Probability that exactly one customer in the system
P(1) = ρP(0)
• Probability that exactly two customer in the system
P(2) = ρP(1)= ρ2P(0)
• Probability that n customer in the system
P(n) = ρnP(0)
• Probability of having exactly n customers in the system
P(n) = ρnP(0)=ρn (1- ρ)
• Expected no. of customers in the system
Ls= ∑∞
=0n
nnP
This can be solve to obtain
• Expected number of customers in the system
Ls = λ/ (μ- λ)
=0n
• Expected number of customers in the queue(including empty queue)
= Expected no. of customer in the system-expected no. of customers being served
i.e. Lq= Ls- λ / μi.e. Lq= Ls- λ / μ
i.e. Lq= λ/ (μ- λ) - λ / μ
i.e. Lq = λ2/ μ(μ- λ)= ρ2/ 1- ρ
• Expected number of customers in the non empty
i.e. Lq ‘= μ / μ- λ= 1/ 1- ρ
• Average waiting time in queue
= product of expected queue length and expected time between arrivals
Wq= 1/ λ. Lq
Wq= 1/ λ. λ2/ μ(μ- λ)
Wq= λ/ μ(μ- λ)Wq= λ/ μ(μ- λ)
• Average waiting time in the system
= product of expected no. of customers in the system and expected time between arrivals
Ws= 1/ λ. Ls
Ws= 1/ λ. λ/ (μ- λ)
Ws= 1/ μ- λ
• Probability that a customer spends more than t units of time in the system
= Ws(t)= e–t/Ws
• Probability that a customer spends more than t units of time in the queue
= Wq(t)= ρe–t/Ws
• Problem 3. A television repairman finds that the time
spent on his job has an exponential distribution
with a mean of 30 minutes .If he repairs sets in the
order in which they came in,and if arrival of sets
follows a Poisson distribution with an average rate
of 10 per day ,what is the repairmen's expected
idle time each day? How many jobs are ahead ofidle time each day? How many jobs are ahead of
the average set just brought in? Assuming he
works for 8 hours a day.
• Problem 3. A television repairman finds that the time
spent on his job has an exponential distribution
with a mean of 30 minutes .If he repairs sets in the
order in which they came in,and if arrival of sets
follows a Poisson distribution with an average rate
of 10 per hour day ,what is the repairmen's
expected idle time each day? How many jobs areexpected idle time each day? How many jobs are
ahead of the average set just brought in? Assuming
he works for 8 hours a day.
• Problem 3. A tailor specializes in ladies dresses. The number of customers
approaching the Tailor appear to be Poisson distributed with a mean of 6
customers per hour. The tailor attends the customer on a first come first
serve basis. The tailor can attend the customers at an average rate of 10
customers per hour with a service time exponentially distributed.
Required
1. Find the probability of no. of arrivals(0 through 5) (i) a 15 minutes interval (ii) a 30
minutes interval.
2. (i)The utilization parameter. (ii)probability that system remains idle.
3. Average time that the tailor is free on a 10 hour working day
4. probability of no. of arrivals(0 through 5) in the system
5. What is the expected no. of customers (i)in the tailor shop. (ii)waiting for service
6. What is the expected length of the queue that have at least one customer.
7. How much time should a customer expect to spend (i)in the queue.(ii)in the tailor
shop.
8. What is the probability that the waiting time of customer shall be more than 10
minutes.
9. What is the probability that the customer shall be in the shop for more than 15
minutes.
Problem 4:Arrival at a telephone booth are considered to be Poisson with an average time of 10 minutes between one arrival and next. The length of the phone call is assumed to be distributed exponentially with mean 3 minutes.distributed exponentially with mean 3 minutes.
Find.
1. The probability that an arrival finds that four persons are waiting for their turn.
2. The average no. of customer waiting and making telephone call.
3. The average length of the queue.
Cost analysis
Suppose that the service mechanics are
paid at Rs. 8 per hour and store room
attendants are paid Rs. 5 per hour. In a
typical 8 hour day the total arrival would typical 8 hour day the total arrival would
be 6 X 8=48.Arriver has to wait ½ an hour
before he obtains parts of his
requirements.
Problem 5:A repairmen is to be hired by a company
to repair machines that breaks down following a
Poisson process, with an average rate of 4 per hour.
The cost of non productive machine time is Rs. 90
per hour. The company has the option of choosingper hour. The company has the option of choosing
either a slow or fast repairmen .The fast repairmen
charges Rs.70 per hour and will repair machines at
the rate of 7 per hour. The slow repairmen charges
Rs.50 per hour and will repair machines at the rate of
6 per hour. Which repairmen should be hired?
• Problem 6:A tool company's quality controldepartment is managed by a single clerk, whotakes an average of 5 minutes in checkingparts of each of the machines coming forinspection. The machines arrive once in everyinspection. The machines arrive once in every8 minutes on the average.One hour ofmachine valued at Rs.15 and a clerk’s time isvalued at Rs. 4 per hour.What are the averagehourly queuing system cost associated withthe Quality Control Department
When no. of customers limited to x
1
)0(P1x
−
=
+
µ
λ
λ
)0()(P
1
Pn
n
=
−
µ
λ
µ
λ
• Problem: 7:A hospital emergency room can accommodate at most X=5 patients. The patients arrive at a rate of 4 per hour. The single staff physician can only treat 5 patients per hour. Any patient overflow is directed to another hospital.patient overflow is directed to another hospital.
(a) Determine the probability distribution for the number of patients in the hospital.
(b) Determine the mean values for the number patients in the emergency room, and number of patients waiting to see the doctor.
• Problem 8:Customers arrive at a one window drive in bank according to Poisson distribution with mean 10 per hour. Service time of each customer is exponential with mean 5 minutes. The space in front of the window,including that for the serviced car can accommodate a maximum of 3 cars.Other cars can wait outside space.cars.Other cars can wait outside space.
a) What is the probability that an arriving customer can drive directly to the space in front of the window?
b) What is the probability that an arriving customer will have to wait outside the indicated space?
c) How long is an arriving customer expected to wait before starting the service?