Queueing Theory. Operations -- Prof. Juran2 Overview Basic definitions and metrics Examples of some...

Queueing Theory

Operations -- Prof. Juran 2

Overview• Basic definitions and metrics• Examples of some theoretical models


Basic Queueing TheoryA set of mathematical tools for the analysis of probabilistic systems of customers and servers.

Can be traced to the work of A. K. Erlang, a Danish mathematician who studied telephone traffic congestion in the first decade of the 20th century.

Applications:Service OperationsManufacturingSystems Analysis


Agner Krarup Erlang

(1878 - 1929)

Components of a Queuing System

Arrival Process

ServersQueue or Waiting Line

Service Process

Exit

Customer Population Sources

Population Source

Finite Infinite

Example: Number of machines needing repair when a company only has three machines.

Example: Number of machines needing repair when a company only has three machines.

Example: The number of people who could wait in a line for gasoline.

Example: The number of people who could wait in a line for gasoline.

Service Pattern

Service Pattern

Constant Variable

Example: Items coming down an automated assembly line.

Example: Items coming down an automated assembly line.

Example: People spending time shopping.

Example: People spending time shopping.

Examples of Queue Structures

Single Channel

Multichannel

SinglePhase Multiphase

One-personbarber shop

Car wash

Hospitaladmissions

Bank tellers’windows

Balking and Reneging

No Way! No Way!

Reneging: Joining the queue, then leaving

Balking: Arriving, but not joining the queue

Suggestions for Managing Queues

1. Determine an acceptable waiting time for your customers

2. Try to divert your customer’s attention when waiting

3. Inform your customers of what to expect4. Keep employees not serving the customers out

of sight5. Segment customers

Suggestions for Managing Queues

6. Train your servers to be friendly

7. Encourage customers to come during the slack periods

8. Take a long-term perspective toward getting rid of the queues


Arrival Rate refers to the average number of customers who require service within a specific period of time.

A Capacitated Queue is limited as to the number of customers who are allowed to wait in line.

Customers can be people, work-in-process inventory, raw materials, incoming digital messages, or any other entities that can be modeled as lining up to wait for some process to take place.

A Queue is a set of customers waiting for service.


Queue Discipline refers to the priority system by which the next customer to receive service is selected from a set of waiting customers. One common queue discipline is first-in-first-out, or FIFO.

A Server can be a human worker, a machine, or any other entity that can be modeled as executing some process for waiting customers.

Service Rate (or Service Capacity) refers to the overall average number of customers a system can handle in a given time period.

Stochastic Processes are systems of events in which the times between events are random variables. In queueing models, the patterns of customer arrivals and service are modeled as stochastic processes based on probability distributions.

Utilization refers to the proportion of time that a server (or system of servers) is busy handling customers.


In the literature, queueing models are described by a series of symbols and slashes, such as A/B/X/Y/Z, where

• A indicates the arrival pattern, • B indicates the service pattern, • X indicates the number of parallel servers, • Y indicates the queue’s capacity, and • Z indicates the queue discipline.

We will be concerned primarily with the M/M/1 queue, in which the letter M indicates that times between arrivals and times between services both can be modeled as being exponentially distributed. The number 1 indicates that there is one server.

We will also study some M/M/s queues, where s is some number greater than 1.


Be careful! These symbols can vary across different books, professors, etc.

Performance Measure Random Variable Expected Value Number of Customers in the System N L

Number of Customers in the Queue Nq Lq Number of Customers in Service Ns Ls Time Spent in the System T W

Time Spent in the Queue Tq Wq Time Spent in Service Ts Ws

System Parameters Number of Servers S Arrival Rate (number per unit of time) (Greek letter lambda) Service Rate (number per unit of time) (Greek letter mu) Utilization Factor (Greek letter rho)

Symbol Probability Measure

Pn Probability of n customers in the system

Pw Probability of having to wait


General(all queue models)

Single Server M/M/S

M/M/2(Model 3)

M/D/1(Model 2)

M/M/1(Model 1)

Single PhaseInfinite SourceFCFS Discipline

Infinite Queue Length


John D. C. Little

1928 —

General Formulas


The single most important formula in queueing theory is called Little’s Law:

Little’s Law applies to any subsystem as well. For example,

Average time in system L

W i

Average customers in system WL ii

Arrival Rate WL

iii

Average time in queue

qq

LW iv


L as a Function of Rho

0

20

40

60

80

100

120

140

160

180

200

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Rho

L


General Single-Server Formulas

Average time in system 1

qWW v


qq

LW vi


• There aren’t many general queueing results (see Larry Robinson’s sheet for some of them).

• Much of queueing theory consists of making assumptions about the specific type of queue.

• The class of models with the most analytical results is the category in which the arrival process and/or service process follows an exponential distribution.


Example: General Formula

I Average line lengthc Number of serversCi Coefficient of variation; arrival processCp Coefficient of variation; service process

Coefficient of Variation:

𝐼 ≅ 𝜌√2 (𝑐+1 )

1−𝜌×𝐶𝑖

❑❑2 +𝐶𝑝

❑❑2

2


Andrey Markov

(Андрей Андреевич Марков) 1856 — 1922


Time between Events

Pro

bab

ilit

y

tetTP

The Exponential Distribution

T is a continuous positive random number.

t is a specific value of T.


Example:

If the average time between arrivals is 45 seconds and inter-arrival times are exponentially distributed, what is the probability that two sequential arrivals will be more than 60 seconds apart?

Remember lambda is in “events per time unit”, so

02222.0451

60TP te

60*02222.0718.2

2636.0


2636.060 TP


Here’s how to do this calculation in Excel:

The EXP function raises e to the power of whatever number is in parentheses.

1234

D E F G

lambda 0.02222t 60

0.2636=EXP(-E2*E3)

=1/45


Remember that the exponential distribution has a really long tail. In probability-speak, it has strong right-skewness, and there are outliers with very large values.

In fact, the probability of any one inter-event time being longer than the mean inter-event time is:

In other words, only 37% of inter-event times will be longer than the expected value of the inter-event times.

This counter-intuitive result is because some of the 37% are really, really long.

1

TP 3679.0


1

TP 3679.0


Other Facts about the Exponential Distribution

• “Memoryless” property: The expected time until the next event is independent of how long it’s been since the previous event

• The mean is equal to the standard deviation (so the CV is always 1)

• Analogous to the discrete Geometric distribution


Siméon Denis Poisson

1781 — 1840


If the random time between events is exponentially distributed, then the random number of events in any given period of time follows a Poisson process.

A Poisson random variable is discrete. The number of events n (i.e. arrivals) in a certain space of time must be an integer.

n is a positive random integer (sometimes zero).


In English: The probability of exactly n events within t time units.

!n

etnP

tn

t


Poisson distribution; λ = 7.5

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Number of Events in 1 Time Unit

Pro

babi

lity

!n

etnP

tn

t


222324252627282930313233343536373839

E F G H Ilambda 7.5

0 0.00055311 0.00414812 0.01555553 0.03888874 0.07291645 0.10937466 0.13671827 0.14648388 0.13732869 0.114440510 0.085830411 0.058520712 0.036575413 0.021101214 0.011304215 0.0056521

=POISSON(E25,$F$22,0)

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Number of Events

Prob

abil

ity

222324252627282930313233343536373839

E F G H Ilambda 7.5

0 0.00055311 0.00470122 0.02025673 0.05914554 0.13206195 0.24143656 0.37815477 0.52463858 0.66196719 0.776407610 0.862238011 0.920758712 0.957334113 0.978435314 0.989739615 0.9953917

=POISSON(E25,$F$22,1)

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Number of Events

Prob

abil

ity


The Excel formula is good for figuring out the probability distribution for the number of events in one time unit. Here is a more general approach:

This gives the probability of exactly fifteen events in three time units, when the average number of events per time unit is 7.5.

You could adapt the Excel formula for general purposes by re-defining what “one time unit” means.

131415161718

H I J K L M Nn 15lambda 7.5e 2.718t 3

0.02481

=EXP(1)

=(((I14*I16)^I13)*((I15)^(-I14*I16)))/(FACT(I13))

Waiting Line Models

These four models share the following characteristics:• Single Phase• Poisson Arrivals• FCFS Discipline• Unlimited Queue Capacity

Model Layout Source Population Service Pattern 1 Single Channel Infinite Exponential 2 Single Channel Infinite Deterministic 3 Multichannel Infinite Exponential 4 Single or Multi Finite Exponential


Model 1 (M/M/1) Formulas

Utilization vii

Average customers in the system

1

L viii

L ix



Average queue length

1qL x

1

2

qL xi

2

qL xii



Average time in system

1

W xiii


qW xiv

L

Wq xv



Probability of a specific number of customers in the system

n

nP

1 xvi

nnNP xvii

Probability of no wait

10P xviii

Example: Model 1 (M/M/1)

Assume a drive-up window at a fast food restaurant.Customers arrive at the rate of 25 per hour.The employee can serve one customer every two minutes.Assume Poisson arrival and exponential service rates.

Determine:A. What is the average utilization of the employee?B. What is the average number of customers in line?C. What is the average number of customers in the

system?D. What is the average waiting time in line?E. What is the average waiting time in the system?F. What is the probability that exactly two cars will be

in the system?

Determine:A. What is the average utilization of the employee?B. What is the average number of customers in line?C. What is the average number of customers in the

system?D. What is the average waiting time in line?E. What is the average waiting time in the system?F. What is the probability that exactly two cars will be

in the system?

.8333 = cust/hr 30cust/hr 25

= =

cust/hr 30 = mins) (1hr/60 mins 2

customer 1 =

cust/hr 25 =


A) What is the average utilization of the employee?

Example: Model 1

B) What is the average number of customers in line?

4.167 = 25)-30(30

(25) =

) - ( =

22

qL

C) What is the average number of customers in the system?

5 = 25)-(30

25 =

- =

L

Example: Model 1

D) What is the average waiting time in line?

mins 10 = hrs.1667 =

=qL

qW

E) What is the average time in the system?

mins 12 = hrs .2 = =L

W

Example: Model 1F) What is the probability that exactly two cars

will be in the system (one being served and the other waiting in line)?

n

n μλ

μλ

-p

1

1157.03025

3025

12

2

-p


M/D/1 Formulas

Average queue length

2

2

qL xix

Average customers in the system

qLL xx

Example: Model 2 (M/D/1)

An automated pizza vending machine heats and dispenses a slice of pizza in 4 minutes.

Customers arrive at an average rate of one every 6 minutes, with the arrival rate exhibiting a Poisson distribution.Determine:

A) The average number of customers in line.B) The average total waiting time in the system.

Determine:

A) The average number of customers in line.B) The average total waiting time in the system.

Example: Model 2A) The average number of customers in line.

.6667 = 10)-(2)(15)(15

(10) =

) - (2 =

22

qL

B) The average total waiting time in the system.

mins 4 = hrs.06667 = 10)-51)(15(2

10 =

) - (2 =

qW

mins 8 = hrs .1333 = 15/hr

1 + hrs.06667 =

1

+ =qWW


M/M/S Formulas

Utilization S

xvii

Average customers in system qLL xviii

qL formula is nasty; best to use the Excel template or the table in the book.


Recall the Model 1 example:• Drive-up window at a fast food restaurant.• Customers arrive at the rate of 25 per hour.• The employee can serve one customer every two

minutes.• Assume Poisson arrival and exponential service

rates.

If an identical window (and an identically trained server) were added, what would the effects be on the average number of cars in the system and the total time customers wait before being served?

If an identical window (and an identically trained server) were added, what would the effects be on the average number of cars in the system and the total time customers wait before being served?

Example: Model 3Average number of cars in the system

0.1874) and 0.1524between --235 p. 10.9,(Exhibit

1760= .qL

1.009 = 3025 + .176 = + =

qLL

Total time customers wait before being served

minute) 1 ( = mincustomers/ 25

customers .176 = = hours .007

qL

qW


M/M/s Calculator (Mms.xls)

1234

5

6789101112131415161718192021222324252627

A B C D E F G Hmms.xls M/M/s Queueing Formula Spreadsheet

Inputs: Definitions of terms:

lambda 25 lambda = arrival rate

mu 30 mu = service rates = number of serversLq = average number in the queueL = average number in the systemWq = average wait in the queueW = average wait in the systemP(0) = probability of zero customers in the systemP(delay) = probability that an arriving customer has to wait

Outputs:s Lq L Wq W P(0) P(delay) Utilization

01 4.1667 5.0000 0.1667 0.2000 0.1667 0.8333 0.83332 0.1751 1.0084 0.0070 0.0403 0.4118 0.2451 0.41673 0.0222 0.8555 0.0009 0.0342 0.4321 0.0577 0.27784 0.0029 0.8362 0.0001 0.0334 0.4343 0.0110 0.20835 0.0003 0.8337 0.0000 0.0333 0.4346 0.0017 0.16676 0.0000 0.8334 0.0000 0.0333 0.4346 0.0002 0.13897 0.0000 0.8333 0.0000 0.0333 0.4346 0.0000 0.11908 0.0000 0.8333 0.0000 0.0333 0.4346 0.0000 0.10429 0.0000 0.8333 0.0000 0.0333 0.4346 0.0000 0.0926

10 0.0000 0.8333 0.0000 0.0333 0.4346 0.0000 0.0833


Finite Queuing: Model 4

Finite Notation:

D = Probability that an arrival must wait in line

F = Efficiency factor, a measure of the effect of having to wait in line

Ls = Average number of units being served

J = Source population less those being served

Lq = Average number of units in line

S = Number of servers


n = Number of customers in the system

N = Number of customers in the source population

P = Probability of exactly n customers in the system

T = Average time to perform service

U = Average time between service (per customer)

W = Average waiting time in line

X = Service factor = UTT

(the proportion of time a customer is in service)

The copy center of an electronics firm has four copy machines that are all serviced by a single technician.

Every two hours, on average, the machines require adjustment. The technician spends an average of 10 minutes per machine when adjustment is required.

Assuming Poisson arrivals and exponential service, how many machines are “down” (on average)?

N, the number of machines in the population = 4M, the number of repair people = 1T, the time required to service a machine = 10 minutesU, the average time between service = 2 hours

.077 =min 120+min 10

min 10U+T

T=X

From Exhibit 10.10, p. 237, F = .980 (Interpolation)From Exhibit 10.10, p. 237, F = .980 (Interpolation)


Lq = N(1 - F) = 4(1 - 0.980) = 0.08 machines

(From Table TN6.11: F = 0.980 , using interpolation)

Ls = FNX = 0.980(4)(0.077) = 0.302

Average number of machines down = Lq + Ls = 0.382 machines

Note: book uses L instead of Lq, and H instead of Ls


Example: Airport SecurityEach airline passenger and his or her luggage must be checked to determine whether he or she is carrying weapons onto the airplane. Suppose that at Gotham City Airport, an average of 10 passengers per minute arrive, where interarrival times are exponentially distributed. To check passengers for weapons, the airport must have a checkpoint consisting of a metal detector and baggage X-ray machine.

Whenever a checkpoint is in operation, two employees are required. These two employees work simultaneously to check a single passenger. A checkpoint can check an average of 12 passengers per minute, where the time to check a passenger is also exponentially distributed.


Why is an M/M/l, not an M/M/2, model relevant here?


What is the probability that a passenger will have to wait before being checked for weapons?


On average, how many passengers are waiting in line to enter the checkpoint?


On average, how long will a passenger spend at the checkpoint (including waiting time in line)?


Difficulties with Analytical Queueing Models

• Using expected values, we can get some results

• Easy to set up in a spreadsheet• It is dangerous to replace a random

variable with its expected value• Analytical methods (beyond expected

values) require difficult mathematics, and must be based on strict (perhaps unreasonable) assumptions


Summary• Basic definitions and metrics• Examples of some theoretical models

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