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Queuing Theory

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Page 1: Queuing Theory1

Queuing Theory

Page 2: Queuing Theory1

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.

Page 3: Queuing Theory1

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

Page 4: Queuing Theory1

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

Page 5: Queuing Theory1

Queuing System Cost

• Cost of providing the service also known as

service cost

• Cost of not providing the service also known

as waiting cost

Page 6: Queuing Theory1

Trade-off

Total Expected Cost

Co

st

of

op

era

tin

g s

erv

ice f

acil

ity

Service Level

Cost of providing service

Cost of waiting time

OptimalService Level

Co

st

of

op

era

tin

g s

erv

ice f

acil

ity

Page 7: Queuing Theory1

• 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

Page 8: Queuing Theory1

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

Page 9: Queuing Theory1

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

Page 10: Queuing Theory1

• Queue Structure• First come first served

• Last come first served

• Service in random order

• Priority service• Priority service

• Customer Behavior• Balking

• Reneging

• Jockeying

• Collusion

Page 11: Queuing Theory1

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

Page 12: Queuing Theory1

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

Page 13: Queuing Theory1

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.

Page 14: Queuing Theory1

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.

Page 15: Queuing Theory1

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.

Page 16: Queuing Theory1

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

Page 17: Queuing Theory1

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.

Page 18: Queuing Theory1

• 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

Page 19: Queuing Theory1

• Model 2: Poisson-exponential single

server model – finite population

Has same assumptions as model 1, except that

population is finitepopulation is finite

Page 20: Queuing Theory1

• 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

Page 21: Queuing Theory1

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

Page 22: Queuing Theory1

• 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)

Page 23: Queuing Theory1

• 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

Page 24: Queuing Theory1

• 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

Page 25: Queuing Theory1

• 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

Page 26: Queuing Theory1

• 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

Page 27: Queuing Theory1

• 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.

Page 28: Queuing Theory1

• 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.

Page 29: Queuing Theory1

• 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.

Page 30: Queuing Theory1

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.

Page 31: Queuing Theory1

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.

Page 32: Queuing Theory1

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?

Page 33: Queuing Theory1

• 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

Page 34: Queuing Theory1

When no. of customers limited to x

1

)0(P1x

=

+

µ

λ

λ

)0()(P

1

Pn

n

=

µ

λ

µ

λ

Page 35: Queuing Theory1

• 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.

Page 36: Queuing Theory1

• 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?