1. Waiting Line Models And Service Improvement
Chapter 16
Waiting Line Models And Service Improvement
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e

2. Why do waiting lines form at a service facility, even though there may be more than enough service capacity to meet normal demand in the long run?
2000 by Prentice-Hall, Inc

3. Elements Of Waiting Line Analysis
Queue
a single waiting line
Waiting line system consists of
arrivals
servers
waiting line structures
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

4. Components Of Queuing System
Source of
customers
Arrivals
Waiting Line
or queue
Server
Served
customers
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

5. Russell/Taylor Oper Mgt 3/e
Calling population
source of customers
infinite - large enough that one more customer can always arrive to be served
finite - countable number of potential customers – examples?
Arrival rate ()
frequency of customer arrivals at waiting line system
typically follows Poisson distribution
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

6. Russell/Taylor Oper Mgt 3/e
Service time
often follows negative exponential distribution
average service rate = 
Arrival rate () must be less than service rate or system never clears out
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

7. Queue Discipline And Length
order in which customers are served
first come, first served is most common
When would FCFS not be appropriate?
Length can be infinite or finite
infinite is most common
finite is limited by some physical structure – examples?
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

8. Basic Waiting Line Structures
Channels are the number of parallel servers
Phases denote number of sequential servers the customer must go through
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

9. Single-Channel Structures
Single-channel, single-phase
Waiting line Server
Single-channel, multiple phases
Waiting line
Servers
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

10. Multi-Channel Structures
Multiple-channel, single phase
Servers
Multiple-channel, multiple-phase
Waiting line
Servers
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

11. Operating Characteristics
Mathematics of queuing theory does not provide optimal or best solutions
Operating characteristics are computed that describe system performance
Steady state is constant, average value for performance characteristics that the system will reach after a long time
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

12. Operating Characteristics
Notation Description
L Average number of customers in the system (waiting to be 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
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

13. Russell/Taylor Oper Mgt 3/e
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
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

14. Cost Relationship In Waiting Line Analysis
Total cost
Expected costs
Service cost
Waiting Costs
Level of service
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

15. Waiting Line Analysis And Quality
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
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

16. 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
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

17. 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
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

18. Formulas For Single-Server Model
P0 =  
(1 - )
Probability that no customers
are in system   ( ) n
Probability of exactly n
customers in system
Pn = P0 =   ( ) n  
(1 - ) 

Average number of
customers in system
L = 

Average number of
customers in queue
Lq =
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

19. Russell/Taylor Oper Mgt 3/e
Average time customer
spends in system 
 L

W = =
Wq = 

Average time customer
spends in queue 

Probability that server
is busy, utilization factor
=  
(1 - )
 =
 =
Probability that server is
idle & customer can be served
= P0
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

20. Single-Server Example
Given  = 24 per hour,  = 30 customers per hour, calculate  
(1 - ) =
1 - (24/30) = 0.20
Probability that no
customers
are in system
P0 = 

Average number of
customers in system
L =
= 24/(30-24) = 4 

Average number of
customers in queue
Lq =
= 242/30(30-24) = 3.2
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

21. Russell/Taylor Oper Mgt 3/e
Average time customer
spends in system 

W =
= 1(30-24) = hr = 10 min
Average time
customer
spends in queue 

Wq=
= 24/30(30-24) = hr = 8 min 

Probability that server
is busy, utilization factor
=
= 24/30 = 0.80
Probability that server is
idle & customer can be
served
I =

= = 0.20
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

22. Waiting Line Cost Analysis
To improve customer services management wants to test two alternatives to reduce customer waiting time:
1. Another employee to pack up purchases
2. Another checkout counter
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

23. Russell/Taylor Oper Mgt 3/e
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 = hours = 2.25 minutes, originally was 8 minutes
= 5.75 minutes
5.75 x $75/minute/week = $ per week
New employee saves $ = $281.25/wk
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

24. Russell/Taylor Oper Mgt 3/e
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 = hours = 1.33 minutes, originally was 8 minutes
= 6.67 minutes
6.67 x $75/minute/week = $500.00/wk = $300/wk
Counter is paid off in 6000/300 = 20 weeks
Counter saves $300/wk; choose alternative II
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

25. 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
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

26. Operating Characteristics For Constant Service Times
Probability that no customers
are in system  
(1 - )
P0 =
Average number of
customers in queue
Lq =


Average number of
customers in system
L =
Lq +  
Average time customer
spends in queue
Wq = Lq

© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

27. Russell/Taylor Oper Mgt 3/e 
W =
Wq +
Average time customer
spends in system 

=
Probability that server
is busy, utilization factor


When service time is
constant,  = 0
Formula can be simplified
Lq = =



 = =


© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

28. 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

2
(10)2
Lq = =
= 1.14 cars waiting
2(13.3)( )
Wq = Lq

=1.14/10 = .114 hour or 6.84 minutes
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

29. Russell/Taylor Oper Mgt 3/e
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

M
Probability that no
customers are in system
P0 =  
Probability of exactly n
customers in system ( ) n
Pn =
(P0 )
for n M


(M + 1() M + 1
1 - ( )M+1
Average number of
customers in system
L =
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

30. Russell/Taylor Oper Mgt 3/e
Let PM = probability a customer will not join system
Lq =
(1- PM)
 L
Average number of
customers in queue L
Average time customer
spends in system
W =
(1 - PM)
Average time customer
spends in queue   W
Wq =
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

31. Russell/Taylor Oper Mgt 3/e
Finite Queue Example
Quick 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
P0 =

M
1 - 20/30
20/305 =
= 0.38
Pm =
(P0 )   ( )
n=M
= (0.38) 4 20 30
= 0.076
Probability of exactly n
cars in system
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 system
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

32. Russell/Taylor Oper Mgt 3/e
Average number of
cars in queue
(1- PM)

 20( )
30
Lq=
L - =
= 0.62 L
1.24
Average time car
spends in system =
= 0.67 hours
= 4.03 min
W =
(1 - PM)
20 ( )   W 
= hours
= 2.03 min
Average time car
spends in queue
Wq = = 30
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

33. Finite Calling Population
Arrivals originate from a finite (countable) population
N = population size
P0 = 
n N!
(N - n)!  N
n = 0
Probability that no
customers are in system
Probability of exactly n
customers in system N!   ( ) n
Pn = P0
where n = 1, 2, ..., N
(N - n)!
Average number of
customers in queue
+
Lq = N
(1- P0)

© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

34. Russell/Taylor Oper Mgt 3/e
Average number of
customers in system
Lq +
L =
(1- P0)
Wq =
(N - L)  Lq
Average time customer
spends in queue
Average time customer
spends in system
W =
Wq +  
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

35. Finite Calling Population 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 = /hr
P0 = 1
n N!
(N - n)!  N
n = 0
Probability that no
machines are in system 1 =
= 0.652 20
20! 
(0.005/0.2778)n
(20 - n)!
n = 0
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

36. Russell/Taylor Oper Mgt 3/e
+
Average number of
machines in queue
Lq = N
(1- P0)


= 20
( )
= 0.169
0.005
Average number of
machines in system
L = Lq + (1-P0) = (1-0.62) = 0.520 Lq
0.169
Average time machine
spends in queue
Wq = =
= 1.74
(N - L) 
( ) 0.005 1
Average time machine
spends in system 1  =
W =
Wq +
1.74 +
= 5.33 hrs
0.278
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

37. 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
] +   ( ) n n! s! s s
s - 
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

38. Russell/Taylor Oper Mgt 3/e1   ( ) n
Probability of exactly n
customers in system
Pn =
P0,
for n > s
s! sn-s 1   ( ) n
Pn =
P0,
for n <= s n!
Probability an arriving
customer must wait 1   ( ) s ( s )
Pw = P0
s -  s!  
( ) s
(s - 1 ! (s - 
Average number of
customers in system   ( )
L =
P0 +
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

39. Russell/Taylor Oper Mgt 3/e
Average time customer
spends in system
W = L  
Average number of
customers in queue
Lq = L  1  Lq
Average time customer
spends in queue
Wq = W = 
Utilization factor
 =
/s
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

40. Multiple-Server Example
Customer service area
 = 10 customers/area
 = 4 customers/hour per service rep
s = (3)(4) = 12
P0 = 1

n = s - 1
n = 0
] +   ( ) n n! s! s s
s -  1 =
= 0.045 ]  ( ) 1 0!    ( ) 1 1! 1  ( ) 1 2!  2 1 3!   ( ) 3
3(4)
3(4)-10 + + +
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

41. Russell/Taylor Oper Mgt 3/e 
( ) s
(s - 1 ! (s -    ( )
Average number of
customers in system
L =
P0 +
(10)(4) (10/4) 3 =
(3-1)! [3(4)-10] 2
(0.045) + (10/4) = 6
Average time customer
spends in system
W = L 
= 6/10 = 0.60 hr = 36 min
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

42. Russell/Taylor Oper Mgt 3/e
Lq =  L
= /4 = 3.5
Average number of
customers in queue
Wq = 
= 3.5/10 = 0.35 hrs = 21 min Lq
Average time customer
spends in queue 1   ( ) s ( s ) s!
Pw = P0
s - 
Probability an arriving
customer must wait 3 1 10
3(4) ( ) ( ) =
(0.45) = 0.703
3(4)-10 3! 4
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

43. Russell/Taylor Oper Mgt 3/e
Improving Service
Add a 4th server to improve service
Recompute operating characteristics
Po = prob of no customers
L = 3.0 customers
W = 0.30 hour, 18 min in service
Lq = 0.5 customers waiting
Wq = 0.05 hours, 3 min waiting, versus 21 earlier
Pw = 0.31 prob that customer must wait
© 2000 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 3/e
2000 by Prentice-Hall, Inc Ch

44. We can have some control over the system using a variety of methods:
1. Do not overlook the effects of perception management.
2. Determine the acceptable waiting time for your customers.
3. Install distractions that entertain and physically involve the customer.
4. Get customers out of the line.
5. Only make people conscious of the time if they grossly overestimate waiting times.
6. Modify customer arrival behavior.
7. Keep resources not serving customers out of sight.
8. Segment customers by personality types.
9. Adopt a long-term perspective.
10. Never underestimate the power of a friendly server.
2000 by Prentice-Hall, Inc

45. Question: What changes can be made to improve service?
alter arrival patterns or rates
change number of service channels
change number of sequential phases of service
change number of service facilities (add new stores)
increase server efficiency through improved methods or incentives
priority rules
line arrangements
2000 by Prentice-Hall, Inc