13-1 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Queuing Analysis Chapter 13

13-1Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall

Queuing Analysis

Chapter 13


■ Elements of Waiting Line Analysis

■ The Single-Server Waiting Line System

■ Undefined and Constant Service Times

■ Finite Queue Length

■ Finite Calling Population

■ The Multiple-Server Waiting Line

■ Additional Types of Queuing Systems

Chapter Topics


A significant amount of time is spent in waiting lines by people, products, etc.

Providing quick service is an important aspect of quality customer service.

The basis of waiting line analysis is the trade-off between the cost of improving service and the costs associated with making customers wait.

Queuing analysis is a probabilistic form of analysis. The results are referred to as operating

characteristics. Results are used by managers of queuing

operations to make decisions.

Overview


Waiting lines form because people or things arrive at a service faster than they can be served.

Most operations have sufficient server capacity to handle customers in the long run.

Customers however, do not arrive at a constant rate nor are they served in an equal amount of time.

Elements of Waiting Line Analysis (1 of 2)


Waiting lines are continually increasing and decreasing in length and approach an average rate of customer arrivals and an average service time in the long run.

Decisions concerning the management of waiting lines are based on these averages for customer arrivals and service times.

Averages are used in formulas to compute operating characteristics of the system which in turn form the basis of decision making.

Elements of Waiting Line Analysis (2 of 2)


Components of a waiting line system include arrivals (customers), servers, (cash register/operator), customers in line form a waiting line.

Factors to consider in analysis: The queue discipline. The nature of the calling population The arrival rate The service rate.

The Single-Server Waiting Line System (1 of 2)


The Single-Server Waiting Line System (2 of 2)

Figure 13.1 The Fast Shop Market waiting line system


Queue Discipline: The order in which waiting customers are served.

Calling Population: The source of customers (infinite or finite).

Arrival Rate: The frequency at which customers arrive at a waiting line according to a probability distribution (frequently described by a Poisson distribution).

Service Rate: The average number of customers that can be served during a time period (often described by the negative exponential distribution).

Single-Server Waiting Line SystemComponent Definitions


Assumptions of the basic single-server model: An infinite calling population A first-come, first-served queue discipline Poisson arrival rate Exponential service times

Symbols:

= the arrival rate (average number of arrivals/time period)

= the service rate (average number served/time period)

Customers must be served faster than they arrive ( < ) or an infinitely large queue will build up.

Single-Server Waiting Line SystemSingle-Server Model

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Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall

Probability that no customers are in the queuing system:

Probability that n customers are in the system:

Average number of customers in system:

Average number of customer in the waiting line:

01P

01

n

n nP P

2

qL

L

Single-Server Waiting Line SystemBasic Single-Server Queuing Formulas (1 of 2)

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Average time customer spends waiting and being served:

Average time customer spends waiting in the queue:

Probability that server is busy (utilization factor):

Probability that server is idle:

1 LW

qW

U

1 1I U

Single-Server Waiting Line SystemBasic Single-Server Queuing Formulas (2 of 2)

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= 24 customers per hour arrive at checkout counter

= 30 customers per hour can be checked out

0

1 (1 - 24/30)

.20 probability of no customers in the system

P

24/(30 - 24) 4 customers on average in the systemL

2

2 (24) /[30(30 -24)] 3.2 customers on average in the waiting line

qL

Single-Server Waiting Line SystemOperating Characteristics: Fast Shop Market (1 of 2)

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Single-Server Waiting Line SystemOperating Characteristics for Fast Shop Market (2 of 2)

130 -24

1

0.167 hour (10 min) avg time in the system per customer

LW

2430(30 24)

0.133 hour (8 min) average time in the waiting line

qW

24

30

.80 probability server busy; .20 probability server will be idle

U

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Single-Server Waiting Line SystemSteady-State Operating CharacteristicsBecause of the steady-state nature of operating

characteristics: Utilization factor, U, must be less than

one:

U < 1, or / < 1 and < .

The ratio of the arrival rate to the service rate must be less than one. In other words, the service rate must be greater than the arrival rate.

The server must be able to serve customers faster than the arrival rate in the long run, or waiting line will grow to infinite size.

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A manager wishes to test several alternatives for reducing customer waiting time:

1. Addition of another employee to pack up purchases

2. Addition of another checkout counter.

Alternative 1: Addition of an employee

(raises service rate from = 30 to = 40 customers per hour). Cost $150 per week, avoids loss of $75 per week

for each minute of reduced customer waiting time.

System operating characteristics with new parameters:

Po = .40 probability of no customers in the system

L = 1.5 customers on average in the queuing system

Single-Server Waiting Line SystemEffect of Operating Characteristics (1 of 6)

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System operating characteristics with new parameters (continued):

Lq = 0.90 customer on the average in the waiting line

W = 0.063 hour average time in the system per customer

Wq = 0.038 hour average time in the waiting line per customer

U = .60 probability that server is busy and customer must wait

I = .40 probability that server is available

Average customer waiting time reduced from 8 to 2.25 minutes worth $431.25 per week.


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Alternative 2: Addition of a new checkout counter ($6,000 plus $200 per week for additional cashier). = 24/2 = 12 customers per hour per checkout

counter = 30 customers per hour at each counter System operating characteristics with new

parameters:

Po = .60 probability of no customers in the system

L = 0.67 customer in the queuing systemLq = 0.27 customer in the waiting line

W = 0.055 hour per customer in the systemWq = 0.022 hour per customer in the

waiting lineU = .40 probability that a customer must

waitI = .60 probability that server is idle


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Savings from the reduced waiting time worth:

$500 per week - $200 = $300 net savings per week.

After $6,000 is recovered, alternative 2 would provide:

$300 -281.25 = $18.75 more savings per week.


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Table 13.1 Operating characteristics for each alternative system


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Figure 13.2 Cost trade-off for service levels


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Exhibit 13.1

Single-Server Waiting Line SystemSolution with Excel and Excel QM (1 of 2)

Formula for Lq, average number in queue

=(1/(D4-D3))*60

=(D3/(D4*(D4-D3)))*60

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Exhibit 13.2

Single-Server Waiting Line SystemSolution with Excel and Excel QM (2 of 2) Click on “Add-Ins” to

access the “Excel QM” menu

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Exhibit 13.3

Single-Server Waiting Line SystemSolution with QM for Windows

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Constant, rather than exponentially distributed service times, occur with machinery and automated equipment.

Constant service times are a special case of the single-server model with undefined service times.

Queuing formulas for the undefined service time model:

01P

22 2 /2 1 /q

L

qL L

qq

LW

1q

W W

U

Single-Server Waiting Line SystemUndefined and Constant Service Times

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0201 1 .33 probability that machine not in use30

2 2 2 22 2 / 20 1/15 20/302 1 / 2 1 20/30

3.33 employees waiting in line

3.33 (20/30)

4.0 employees

q

q

P

L

L L

in line and using the machine

Data: Single fax machine; arrival rate of 20 users per hour, Poisson distributed; undefined service time with mean of 2 minutes, standard deviation of 4 minutes.

Operating characteristics:

Single-Server Waiting Line SystemUndefined Service Times Example (1 of 2)

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3.33 0.1665 hour 10 minutes waiting time20

1 10.1665 0.1998 hour30

12 minutes in the system

20 67% machine utilization30

qq

q

LW

W W

U

Operating characteristics (continued):

Single-Server Waiting Line SystemUndefined Service Times Example (2 of 2)

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In the constant service time model there is no variability in service times; = 0.

Substituting = 0 into equations:

All of the remaining formulas are the same as the single-server formulas.

22 2 22 2 2 2/ 0 / /

2 1 / 2 1 / 2 1 / 2qL

Single-Server Waiting Line SystemConstant Service Times Formulas

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22 (10) 1.14 cars waiting2 ( ) 2(13.3)(13.3 10)

1.14 0.114 hour or 6.84 minutes waiting time10

q

qq

L

LW

Car wash servicing one car at a time; constant service time of 4.5 minutes; arrival rate of customers of 10 per hour (Poisson distributed).

Determine average length of waiting line and average waiting time.

= 10 cars per hour, = 60/4.5 = 13.3 cars per hour

Single-Server Waiting Line SystemConstant Service Times Example

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Exhibit 13.4

Undefined and Constant Service TimesSolution with Excel

Average number in the queue, Lq

=D8+(1/D4)*60

=(D6/D3)*60

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Exhibit 13.5

Undefined and Constant Service TimesSolution with QM for Windows

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0 01 / ( ) for n M

11 ( / )

1 (1 )( 1)( / )/ 1 / 11 ( / )

1 (1 )

n

Mq

qM

nP P P

M

M PML L LM

LW W WP

In a finite queue, the length of the queue is limited.

Operating characteristics, where M is the maximum number in the system:

Finite Queue Length

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Metro Quick Lube single bay service; space for one vehicle in service and three waiting for service; mean time between arrivals of customers is 3 minutes; mean service time is 2 minutes; both inter-arrival times and service times are exponentially distributed; maximum number of vehicles in the system equals 4.

Operating characteristics for = 20, = 30, M = 4:0

0

1 / 1 20/30 .38 probability that system is empty51 1 (20/30)1 ( / )

420( ) (.38) .076 probability that system is full30M

PM

n MP P

Finite Queue Length Example (1 of 2)

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Average queue lengths and waiting times:1( 1)( / )/

1 / 11 ( / )

5(5)(20/30)20/30 1.24 cars in the system51 20/30 1 (20/30)

(1 ) 20(1 .076)1.24 0.62 cars waiting30

1.24 0.067 hours waiting in the s(1 ) 20(1 .076)

Mq

M

MMLM

L

PL L

LWP

ystem

1 10.067 0.033 hour waiting in line30q

W W

Finite Queue Length Example (2 of 2)

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Exhibit 13.6

Finite Queue Model ExampleSolution with Excel

Formula for P0 in cell D7

+((D3/D4)/(1-(D3/D4))) - ((D5+1)*(D3/D4) ^(D5+1))/(1-(D3/D4)^(D5+1))

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Exhibit 13.7

Finite Queue Model ExampleSolution with QM for Windows

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0

0 0

0

1 !

( )!0

where N population size, and n 1, 2,...N

! (1 )( )!

1(1 ) ( )

n q

qq q q

P nN NN nn

nNP P L N PN n

LL L P W W W

N L

In a finite calling population there is a limited number of potential customers that can call on the system.

Operating characteristics for a system with Poisson arrival and exponential service times:

Finite Calling Population

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Wheelco Manufacturing Company; 20 machines; each machine operates an average of 200 hours before breaking down; average time to repair is 3.6 hours; breakdown rate is Poisson distributed, service time is exponentially distributed.

Is repair staff sufficient?

= 1/200 hour = .005 per hour

= 1/3.6 hour = .2778 per hour

N = 20 machines

Finite Calling Population Example (1 of 2)

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…The system seems woefully inadequate.

01 .652

20 20! .005(20 )! .27780

.005 .277820 1 .652 .169 machines waiting.005

.169 (1 .652) .520 machines in the system

.169 1.74 hours waiting for repair(20 .520)(.005)

11.74

q

q

P n

nn

L

L

W

W

5.33 hours in the system.2778

Finite Calling Population Example (2 of 2)

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Exhibit 13.8

Finite Calling Population ExampleSolution with Excel and Excel QM (1 of 2)

P0 = 1/G26

Summation component for n=1 in cell G6

Array with summation components for P0 formula

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Exhibit 13.9

Finite Calling Population ExampleSolution with Excel and Excel QM (2 of 2) Click on “Add-Ins” to access

the macro for the finite population model

Enter problem data in cells B7:B9

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Exhibit 13.10

Finite Calling Population ExampleSolution with QM for Windows

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Multiple-Server Waiting Line (1 of 3)

Figure 13.3 A multiple-server waiting line

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In multiple-server models, two or more independent servers in parallel serve a single waiting line.

Biggs Department Store service department; first-come, first-served basis.


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Customer Service System at Biggs

Department Store


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Multiple-Server Waiting LineQueuing Formulas (1 of 3)

Assumptions: First-come first-served queue discipline Poisson arrivals, exponential service times

Infinite calling population. Parameter definitions:

= arrival rate (average number of arrivals per time period)

= the service rate (average number served per time period) per server (channel)

c = number of servers c = mean effective service rate for the system

(must exceed arrival rate)

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0

0

0

1 probability no customers in system1 1 1

! !0

1 for n c!

1 for n c probability of n customers in

n

n

P n cn c cn c cn

nP Pn cc c

nP Pn

0

system

( / ) average customers in the system2( 1)!( )

average time customer spends in the system

cL P

c c

LW


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0

average number of customers in the queue

1 average time customer is in the queue

1 probability customer must wait for service!

q

qq

w

L L

LW W

c cP Pc c


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01

0 1 2 3 3(4)1 10 1 10 1 10 1 100! 4 1! 4 2! 4 3! 4 3(4) 10

.045 probability of no customers

3(10)(4)(10/4) 10(.045)2 4(3 1)![3(4) 10]

6 customers on average in servi

P

L

ce department

6 0.60 hour average customer time in the service department10

W

Multiple-Server Waiting LineBiggs Department Store Example (1 of 2) = 10, = 4, c = 3

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1064

3.5 customers on the average waiting to be served

3.510

0.35 hour average waiting time in line per customer

3 3(4)1 10 (.045)3! 4 3(4) 10

.703 probability customer must wait for servi

q

q

w

L

W

P

ce

Multiple-Server Waiting LineBiggs Department Store Example (2 of 2)

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Exhibit 13.11

Multiple-Server Waiting LineSolution with Excel

Formula for P0

=((((D3)*(D4)*((D3/D4)^D5)*(D7))/(FACT(D5-1)*(((D5*D4)-D3)^2))))+(D3/D4)

=(1/FACT(D5))*((D3/D4)^D5)*((D5*D4*D7)/((D5)*(D4)-(D3)))

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Exhibit 13.12

Multiple-Server Waiting LineSolution with Excel QM

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Exhibit 13.13

Multiple-Server Waiting LineSolution with QM for Windows

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Figure 13.4 Single queues with single & multiple servers in sequence

Additional Types of Queuing Systems (1 of 2)

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Other items contributing to queuing systems: Systems in which customers balk from

entering system, or leave the line (renege). Servers who provide service in other than a

first-come, first-served manner Service times that are not exponentially

distributed or are undefined or constant Arrival rates that are not Poisson

distributed Jockeying (i.e., moving between queues)

Additional Types of Queuing Systems (2 of 2)

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Problem Statement: Citizens Northern Savings Bank loan officer customer interviews.

Customer arrival rate of four per hour, Poisson distributed; officer interview service time of 12 minutes per customer.

1. Determine operating characteristics for this system.

2. Add an additional officer creating a multiple-server queuing system with two channels. Determine operating characteristics for this system.

Example Problem Solution (1 of 7)

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Solution:

Step 1: Determine Operating Characteristics for the Single-Server System

= 4 customers per hour arrive, = 5 customers per hour are served

Po = (1 - / ) = ( 1 – 4 / 5) = .20 probability of no customers in the system

L = / ( - ) = 4 / (5 - 4) = 4 customers on average in the queuing system

Lq = 2 / ( - ) = 42 / 5(5 - 4) = 3.2 customers on average in the waiting line


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Step 1 (continued):

W = 1 / ( - ) = 1 / (5 - 4) = 1 hour on average in the system

Wq = / (u - ) = 4 / 5(5 - 4) = 0.80 hour (48 minutes) average time in the waiting line

Pw = / = 4 / 5 = .80 probability the new accounts officer is busy and a customer must wait


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0

0

11 1 1

! !0

10 1 21 4 1 4 1 4 2 5

5 5 50! 1! 2! 2 5 4

.429 probability no customers in system

P n cn c cn c cn

P

Step 2: Determine the Operating Characteristics for the Multiple-Server System.

= 4 customers per hour arrive; = 5 customers per hour served; c = 2 servers


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0

2

2

45 4

5(2)(5) (4)

( / )2( 1)!( )

4 5(.429)

1!

0.952 average number of customers in the system

cL P

c c

L

Step 2: Determine the Operating Characteristics for the Multiple-Server System.

= 4 customers per hour arrive; = 5 customers per hour served; c = 2 servers


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4.9525

.9524

0.152 average number of customers in the queue

0.238 hour average time customer is in the system

qL L

LW

Step 2 (continued):


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2

.1524

0.038 hour average time customer is in the queue

1!

1 4 2 4 .42952! 2 4

.229 probability customer must wait for service

qq

w

w

LW

c cP Poc c

P

Step 2 (continued):


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13-1 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Queuing Analysis Chapter 13