waiting line analysis

7/29/2019 Waiting line analysis

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Copyright 2006 John Wiley & Sons, Inc.Beni Asllani

University of Tennessee at Chattanooga

Waiting Line Analysis

for Service ImprovementOperations Management - 5 th Edition

Chapter 17

Roberta Russell & Bernard W. Taylor, III



Copyright 2006 John Wiley & Sons, Inc. 17-2

Lecture Outline

Elements of Waiting Line Analysis

Waiting Line Analysis and Quality

Single Server Models

Multiple Server Model




Waiting Line Analysis

Operating characteristics average values for characteristics that describe the

performance of a waiting line system Queue

A single waiting line

Waiting line system consists of Arrivals

Servers

Waiting line structures




Elements of a Waiting Line

Calling population Source of customers

Infinite - large enough that one more customer canalways arrive to be served

Finite - countable number of potential customers

Arrival rate (λ) Frequency of customer arrivals at waiting line

system

Typically follows Poisson distribution




Elements of a Waiting Line (cont.)

Service time

Often follows negative exponentialdistribution

Average service rate = μ

Arrival rate (λ) must be less than servicerate (μ) or system never clears out




Elements of a Waiting Line (cont.)

Queue discipline

Order in which customers are served First come, first served is most common

Length can be infinite or finite

Infinite is most common Finite is limited by some physical


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Basic Waiting Line Structures

Channels are the number of parallel servers

Single channel

Multiple channels

Phases denote number of sequential servers thecustomer must go through

Single phase

Multiple phases Steady state

A constant, average value for performance characteristics thatsystem will reach after a long time



Operating Characteristics

NOTATION OPERATINGCHARACTERISTIC

L Average number of customers in the system (waiting and

being served)Lq Average number of customers in the waiting line

W Average time a customer spends in the system (waiting andbeing served)

W q Average time a customer spends waiting in lineP 0 Probability of no (zero) customers in the system

P n Probability of n customers in the system

ρ Utilization rate; the proportion of time the system is in use

Table 16.1



Cost Relationship in Waiting

Line Analysis

E x p e c t e d

c o s t s

Level of service

Total cost

Service cost

Waiting Costs



Waiting Line Costs and Quality

Service Traditional view is that the level of

service should coincide with minimum

point on total cost curve TQM approach is that absolute quality

service will be the most cost-effective in

the long run



Single-server Models

All assume Poisson arrival rate

Variations

Exponential service times

General (or unknown) distribution of service times

Constant service times

Exponential service times with finite queue length Exponential service times with finite calling

population



Basic Single-Server

Model: Assumptions Poisson arrival rate

Exponential service times

First-come, first-served queue discipline Infinite queue length

Infinite calling population

= mean arrival rate

= mean service rate



Formulas for Single-

Server Model

L =-

Average number ofcustomers in the system

Probability that no customers

are in the system (either in thequeue or being served)

P 0 = 1 -

Probability of exactly n customers in the system

P n = • P 0

n

= 1 -

n

Average number ofcustomers in the waiting line

Lq =( - )



Formulas for Single-

Server Model (cont.)

=

Probability that the serveris busy and the customer

has to wait

Average time a customerspends in the queuing system

W = =1

-

L

Probability that the serveris idle and a customer canbe served

I = 1 -

= 1 - = P 0

Average time a customerspends waiting in line tobe served

W q = ( - )



A Single-Server Model

Given = 24 per hour, = 30 customers per hour

Probability of nocustomers in thesystem

P 0 = 1 - = 1 - =0.20

2430

L = = = 4Average numberof customers in

the system-

24

30 - 24

Average numberof customerswaiting in line

Lq = = = 3.2(24)2

30(30 - 24)

2

( - )



A Single-Server Model

Average time in thesystem per customer

W = = = 0.167 hour1

-

1

30 - 24

Average time waitingin line per customer

W q = = = 0.133( - )

24

30(30 - 24)

Probability that theserver will be busy andthe customer must wait = = = 0.80

24

30

Probability theserver will be idle

I = 1 - = 1 - 0.80 = 0.20



Service Improvement Analysis

Possible Alternatives

Another employee to pack up purchases

service rate will increase from 30 customers to 40customers per hour

waiting time will reduce to only 2.25 minutes

Another checkout counter arrival rate at each register will decrease from 24 to 12 per

hour customer waiting time will be 1.33 minutes

Determining whether these improvements are worththe cost to achieve them is the crux of waiting lineanalysis




Constant Service Times

Constant service timesoccur with machinery

and automatedequipment

Constant service timesare a special case of the

single-server model withundefined service times




Operating Characteristics for

Constant Service Times

P 0 = 1 -Probability that no customersare in system

Average number ofcustomers in system

L = Lq +

Average number ofcustomers in queue

Lq =2

2 ( - )




Operating Characteristics for

Constant Service Times (cont.)

=Probability that theserver is busy

Average time customerspends in the system W = W q +

1

Average time customerspends in queue W q =

Lq




Constant Service Times:

ExampleAutomated car wash with service time = 4.5 min

Cars arrive at rate = 10/hour (Poisson)

= 60/4.5 = 13.3/hour

W q = = 1.14/10 = .114 hour or 6.84 minutesLq

(10)2

2(13.3)(13.3 - 10)Lq = = = 1.14 cars waiting

2

2 ( - )




Finite Queue Length

A physical limit exists on length of queue M = maximum number in queue

Service rate does not have to exceed arrival rate ( )to obtain steady-state conditions

P 0 =Probability that nocustomers are in system

1 - /

1 - ( / )M + 1

Probability of exactly n customers in system P n = (P 0) for n ≤ M

n

L = -Average number ofcustomers in system

/

1 - /

(M + 1)( / )M + 1

1 - ( / )M + 1




Finite Queue Length (cont.)

Let P M = probability a customer will not join system

Average time customerspends in system

W =L

(1 - P M )

Lq = L -(1- P M )Average number of

customers in queue

Average time customerspends in queue

W q = W -1




Finite Queue: ExampleFirst National Bank has waiting space for only 3 drive in windowcars. = 20, = 30, M = 4 cars (1 in service + 3 waiting)

Probability thatno cars are inthe system P 0 = = = 0.38

1 - 20/30

1 - (20/30)5

1 - /

1 - ( / )M + 1

P n = (P 0) = (0.38) = 0.076

Probability ofexactly 4 cars in

the system

20

30

4

n =M

L = - = 1.24

Average numberof cars in thesystem

/

1 - /

(M + 1)( / )M + 1

1 - ( / )M + 1




Finite Queue: Example (cont.)

Average time a carspends in the system

W = = 0.067 hrL

(1 - P M )

Lq = L - = 0.62(1- P M )Average number of

cars in the queue

Average time a carspends in the queue

W q = W - = 0.033 hr1




Finite Calling Population

Arrivals originate from a finite (countable) population

N = population size

Probability of exactly n

customers in systemP n

= P 0 where n = 1, 2, ..., N

n

N !

(N - n )!

Average number ofcustomers in queue

Lq = N - (1- P 0)+

Probability that nocustomers are in system

P 0 =

n = 0

N !

(N - n )!

N n

1




Finite Calling Population (cont.)

W q =Lq

(N - L)

Average time customerspends in queue

L

= Lq

+ (1 - P 0)

Average number of

customers in system

W = W q +Average time customerspends in system

1




20 trucks which operate an average of 200 days before breakingdown ( = 1/200 day = 0.005/day)

Mean repair time = 3.6 days ( = 1/3.6 day = 0.2778/day)

Probability that notrucks are in the system P 0 = 0.652

Average number oftrucks in the queue

Lq = 0.169

Average number of

trucks in system L = 0.169 + (1 - 0.652) = .520

Finite Calling Population:

Example

Average time truckspends in queue

W q = 1.74 days

Average time truckspends in system

W = 5.33 days




Two or more independent servers serve a single waiting line

Poisson arrivals, exponential service, infinite calling

population

s >

P 0 =1

1

s!

s s

s -

n =s -1

n =0

1

n !

n

+

Basic Multiple-server Model

Computing P 0 can be time-consuming.

Tables can used to find P 0 for selected values of and s .




Probability of exactlyn customers in the

system

P n =

P 0, for n > s 1

s ! s n-s

n

P 0, for n > s 1n !

n

Probability an arrivingcustomer must wait P w = P 0

1

s !

s

s -

s

Average number ofcustomers in system

L = P 0 +( / )s

(s - 1)!(s - )2

Basic Multiple-server

Model (cont.)




W =L

Average time customerspends in system

= /s Utilization factor

Average time customerspends in queue W q = W - =1Lq

Lq = L -Average number ofcustomers in queue

Basic Multiple-server Model

(cont.)




Multiple-Server System:Example

Student Health Service Waiting Room

= 10 students per hour

= 4 students per hour per service representative

s = 3 representativess = (3)(4) = 12

P 0 = 0.045Probability no studentsare in the system

Number of students inthe service area

L = 6




Multiple-Server System:Example (cont.)

Lq = L - / = 3.5Number of studentswaiting to be served

Average time students

will wait in line W q = Lq / = 0.35 hours

Probability that astudent must wait

P w = 0.703

Waiting time in theservice area

W = L / = 0.60




Add a 4th server to improve service

Recompute operating characteristics

P 0 = 0.073 prob of no students L = 3.0 students

W = 0.30 hour, 18 min in service

Lq = 0.5 students waiting

W q = 0.05 hours, 3 min waiting, versus 21 earlier

P w = 0.31 prob that a student must wait

Multiple-Server System:

Example (cont.)



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