Waiting Line Models

© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e

Chapter 16

Waiting Line Models for Service Improvement

Ch 16 - 2© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e

Elements Of Waiting Line Analysis

Queue a single waiting line

Waiting line system arrivals servers waiting line structures


Calling population– source of customers– an infinite population assumes such a large

number of customers that it is always possible for one more customer to arrive to be served

– a finite population consists of a countable number of potential customers

Arrival rate, – frequency of customer arrivals to system – typically follows Poisson distribution


Service time– often follows negative exponential distribution– average service rate =

Arrival rate must be less than service rate or system never clears out

(


Components Of A Queuing System

Source of

customersArrivals Server Served

customersWaiting Line

or queue


Queue Discipline And Length

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

structure


Basic Waiting Line Structures

Channels are the number of parallel servers

Phases denote number of sequential servers the customer must go through


Single-Channel Structures

ServersWaiting line

Single-channel, multiple phases

Waiting line Server

Single-channel, single-phase


Multi-Channel Structures

Servers

ServersWaiting line

Multiple-channel, single phase

Multiple-channel, multiple-phase

Waiting line


Operating Characteristics The mathematics of queuing theory does not

provide optimal or best solutions

Instead, operating characteristics are computed that describe system performance

Steady state provides the average value of performance characteristics that the system will reach after a long time


Operating Characteristics

Notation Description

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 and being served)

Wq Average time a customer spends waiting in line


P0 Probability of no (zero) customers in the system

Pn Probability of n customers in the system

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


Cost Relationship In Waiting Line Analysis

Exp

ecte

d co

sts

Level of service

Total cost

Service cost

Waiting Costs


Waiting Line Analysis And Quality

Traditional view - the level of service should coincide with the minimum point on the total cost curve

TQM view - absolute quality service is the most cost-effective in the long run


Single-Channel, Single-Phase 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

P0 =

(1 - )

Pn =

nP0

=( )

(1 - )

Lq =

L =

Probability that no customersare in system

Probability of exactly n customers in system

Average number of customers in system

Average number of customers in queue

n

)(


W = =L

Wq =

=

= = = P0

(1 - )

Average time customerspends in system

Average time customer spends in queue

Probability that serveris busy, utilization factor

Probability that server is idle & customer can be served


Single-Server Example

Given: = 24 per hour, = 30 customers per hour

P0 =

(1 - )

Lq =

L =




= 1 - (24/30) = 0.20

= 24/(30-24) = 4

= 242/30(30-24) = 3.2


W =

Wq =

=

I =




Probability that server is idle & customer can be served

= 1(30-24) = 0.167 hr = 10 min

= 1 - 0.80 = 0.20

= 24/30(30-24) = 0.133 hr = 8 min

= 24/30 = 0.80


Waiting Line Cost Analysis

Management wants to test two alternatives to reduce customer waiting time:

1. Hire another employee to pack up purchases

2. Open another checkout counter


Alternative 1 Extra employee costs $150 / week Each one-minute reduction in customer waiting time avoids $75

in lost sales Extra employee will increase service rate to 40 customers per

hour Recompute operating characteristics Wq = 0.038 hours = 2.25 minutes, originally was 8 minutes 8.00 - 2.25 = 5.75 minutes 5.75 x $75/minute/week = $431.25 per week New employee saves $431.25 - 150.00 = $281.25 / week


Alternative II New counter costs $6000 plus $200 per week for checker Customers divide themselves between two checkout lines Arrival rate is reduced from = 24 to = 12 Service rate for each checker is = 30 Recompute operating characteristics Wq = 0.022 hours = 1.33 minutes, originally was 8 minutes 8.00 - 1.33 = 6.67 minutes 6.67 x $75/minute/week = $500.00/wk - 200.00 = $300/wk Counter is paid off in 6000/300 = 20 weeks Counter saves $300/wk; choose alternative II


Constant Service Times

Constant service times occur with machinery and automated equipment

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


Operating Characteristics For Constant Service Times

Wq =Lq

(1 - )P0 =

Lq =

L = Lq +







W = Wq +Average time customerspends in system

=

When service time isconstant and = 0, formula can be simplified

Lq =

=

=

=


Constant Service Time Example

Automated car wash with service time = 4.5 min Cars arrive at rate = 10/hour (Poisson) = 60/4.5 = 13.3/hour

2Lq =

(10)2

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

Wq =Lq

=1.14/10 = .114 hour or 6.84 minutes


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 ()

P0 =

L =

M

Pn = (P0 )( )n

for n M

(M + 1() M + 1

1 - ( )M+1

Probability that no customers are in system







LW = (1 - PM)

Lq =(1- PM)

L

WWq =

Let PM = probability a customer will not join the system


Finite Queue ExampleQuick Lube has waiting space for only 3 cars = 20, = 30, M = 4 cars (1 in service + 3 waiting)

Probability that no cars are in system

Probability of exactly n cars in system

Average number of cars in system

P0 =M

1 - 20/30

20/305= = 0.38

Pm = (P0 )( )n=M (= (0.38) )420

30= 0.076

L =

(M + 1() M + 1

1 - ( )M+1

= 20/30

1 -20/30(5(20/30) 5

1 - (20/30)5= 1.24


Average number of cars in queue

Average time car spends in system

Average time carspends in queue

LW = (1 - PM)

Lq =(1- PM)

WWq =

L - 20(1-0.076)

30

= 1.24 - = 0.62

1.24

20 (1-0.076)= = 0.67 hours

= 4.03 min

30

0.067 -= = 0.033 hours

= 2.03 min


Finite Calling Population Arrivals originate from a finite (countable) population N = population size

Lq =+

N -

Pn = P0

( )nN!

(N - n)!where n = 1, 2, ..., N

(1- P0)

Probability that no customers are in system



P0 =

nN!

(N - n)!N

n = 0


W = Wq +

Lq +L = (1- P0)

Wq = (N - L) Lq





Finite Calling Pop’n Example

20 machines which operate an average of 200 hrs before breaking down = 1/200 hr = 0.005/hr

Mean repair time = 3.6 hrs = 1/3.6 hr = 0.2778/hr

Probability that no machines are in system

P0 =1

nN!

(N - n)!N

n = 0

=1

(0.005/0.2778)n20!

(20 - n)!20

n = 0

= 0.652


Lq =

0.005 + 0.2778 0.005

= 20 (1- 0.652)

Average number of machines in queue

W = Wq +1

L = Lq + (1-P0) = 0.169 + (1-0.62) = 0.520

Wq =(N - L)

Lq

Average time machinespends in system

Average time machinespends in queue

Average number of machines in system

= 0.169

+

N (1- P0)

(20 - 0.520) 0.005

0.169= = 1.74

1.74 +1

0.278= = 5.33 hrs


Multiple-Channel, Single-Phase Models

Two or more independent servers serve a single waiting line

Poisson arrivals, exponential service, infinite calling population

s>

P0 =1

n = s - 1

n = 0] +

( )n1

n!1s!

( )s s

s - ( )


L =

Pn = P0,( )n1

s! sn-sfor n > s

Pn = P0,( )n1

n!

Pw = P0

( )s1

s!

ss - ( )

( )

()s

(s - 1 ! (s - P0 +



Probability an arrivingcustomer must wait

for n <= s


/s

Lq =

L

=

Wq =

1W

=

Lq

W =L




Utilization factor


Multiple-Server Example

Customer service area = 10 customers/area = 4 customers/hour per service reps = (3)(4) = 12

P0 =1

n = s - 1

n = 0] +

( )n1

n!1s!

( )s s

s - ( )=

1

]1( )1

1!13!

( )3 3(4)

3(4)-10( )( )10!

( )1

2! 20

+ + +

= 0.045


L =( )

()s

(s - 1 ! (s - P0 +Average number of

customers in system

(10)(4) (10/4) 3

=(3-1)! [3(4)-10] 2

(0.045) + (10/4) = 6

Average time customerspends in system W =

L

= 6/10 = 0.60 hr = 36 min




Pw = P0

( )s1 s

s - ( )Probability an arrivingcustomer must wait

Wq = = 3.5/10 = 0.35 hrs = 21 min

Lq

Lq =

L = 6 - 10/4 = 3.5

s!

= (0.45) = 0.703104( )31 3(4)

3(4)-10( ) 3!


Improving Service

Add a 4th server to improve service Recompute operating characteristics

Po = 0.073 prob of no customers

L = 3.0 customersW = 0.30 hour, 18 min in serviceLq = 0.5 customers waiting

Wq = 0.05 hours, 3 min waiting, versus 21 earlier

Pw = 0.31 prob that customer must wait

Waiting Line Models

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