1. Chapter 11 Waiting Line Models
Md. Abdullah Al Mahmud
Assistant Professor
Manarat International University

2. What is the Queuing Theory?
Queue- a line of people or vehicles waiting for something
Queuing Theory- Mathematical study of waiting lines, using models to show results, and show opportunities, within arrival, service, and departure processes

3. A queue is a waiting line.
A queuing system involves customers arriving for service who sometimes have to wait.
Queuing analysis provides:
Summary measures for assessing a queuing system in terms of customers and time.
A way to balance the costs of providing service and costs of congestion.
Importance of Queuing Analysis:
Servicing customers can be costly.
Retail environments are plagued with customer congestion. Managing that has benefits.

4. Elements of Waiting Lines
A waiting line system consists of two components:
The customer population (people or objects to be processed)
The process or service system
Whenever demand exceeds available capacity, a waiting line or queue forms
There is a tradeoff between cost and service level.

5. Structure of a Waiting Line System
Four characteristics of a queuing system are:
the manner in which customers arrive
the time required for service
the priority determining the order of service
the number and configuration of servers in the system.

6. Waiting Line System
Population
Service Facility
Waiting Line

7. Waiting Line Examples Situation Arrivals Servers Service Process
Bank Customer Teller Deposit etc.
Doctor’s Patient Doctor Treatment office
Traffic Cars Light Controlled intersection passage
Assembly line Parts Workers Assembly
Tool crib Workers Clerks Check out/in tools

8. Characteristics of a Waiting Line System
Queue
Arrivals
Servers
Population of customers may be finite or infinite
Pattern of arrivals may be random
Customers may or may not be patient
Queue length may be limited
Customers are served in an FIFO sequence
There may be more than one queues
Service time may be random
There may be more than one server

9. Waiting Line Terminology
Queue: Waiting line
Arrival: 1 person, machine, part, etc. that arrives and demands service
Queue discipline: Rules for determining the order that arrivals receive service
Channel: Number of servers
Phase: Number of steps in service

10. Customer Population Characteristics
Finite versus Infinite populations:
Is the number of potential new customers materially affected by the number of customers already in queue?
Balking
When an arriving customer chooses not to enter a queue because it’s already too long.
Reneging
When a customer already in queue gives up and exits without being serviced.
Jockeying
When a customer switches between alternate queues in an effort to reduce waiting time.

11. Arrival Characteristics of a Waiting Line System
Population
Service Facility
Waiting Line
Pattern of arrivals
random
scheduled
Behavior of the arrivals
join the queue, and wait until served
No balking; refuse to join the line
No reneging; leave the line

12. Balking Service system Input source Waiting line Service facility
© 1995 Corel Corp.
Line was too long!

13. Reneging Service system Input source Waiting line Service facility
© 1995 Corel Corp.
I give up!

14. Line Characteristics of a Waiting Line System - continued
Population
Service Facility
Waiting Line
Length of the queue
limited
unlimited
Service priority
FIFO
other

15. Parts of a Waiting Line Arrival Characteristics Size of the population
Population of
dirty cars
Arrivals
from the
general
population …
Queue
(waiting line)
Service
facility
Exit the system
Dave’s
Car Wash
enter
exit
Arrivals to the system
In the system
Exit the system
Arrival Characteristics
Size of the population
Behavior of arrivals
Statistical distribution of arrivals
Waiting Line Characteristics
Limited vs. unlimited
Queue discipline
Service Characteristics
Service design
Statistical distribution of service

16. Service Facility Characteristics of a Waiting Line System - continued
Population
Waiting Line
Number of channels
single
multiple
Number of phases in service system

17. Queuing System Designs
A family dentist’s office
Queue
Service facility
Departures
after service
Arrivals
Single-channel, single-phase system
A McDonald’s dual window drive-through
Queue
Phase 1 service facility
Phase 2 service facility
Departures
after service
Arrivals
Single-channel, multiphase system

18. Queuing System Designs
Most bank and post office service windows
Service facility
Channel 1
Channel 2
Channel 3
Departures
after service
Queue
Arrivals
Multi-channel, single-phase system

19. Queuing System Designs
Some college registrations
Phase 1 service facility
Channel 1
Channel 2
Phase 2 service facility
Channel 1
Channel 2
Queue
Departures
after service
Arrivals
Multi-channel, multiphase system

20. Factors of a Queuing System
When do customers arrive?
Are customer arrivals increased during a certain time (restaurant- Denny’s: breakfast, lunch, dinner) Or is the customer traffic more randomly distributed (a café-starbucks)
Depending on what type of Queue line, How much time will customers spend
Do customers typically leave in a fixed amount of time?
Does the customer service time vary with the type of customer?

21. Important characteristics
Arrival Process: The probability distribution that determines the customer arrivals in the system.
Service Process: determines the customer service times in the system.
Number of Servers: Amount of servers available to provide service to the customers

22. Characteristics Arrival Process Service Process Number of Servers
The probability density distribution that determines the customer arrivals in the system.
Service Process
The probability density distribution that determines the customer service times in the system.
Number of Servers
Number of servers available to service the customers.
Number of Channels
Single channel
N independent channels
Multi channels
Number of Phases/Stages
Single Queue
Series or Tandem
Cyclic -Network
Queue Discipline -Selection for Service
First com first served (FCFS or FIFO)
Last in First out (LIFO) Random Priority

23. Queuing Models Calculate:
Average number of customers in the system waiting and being served
Average number of customers waiting in the line
Average time a customer spends in the system waiting and being served
Average time a customer spends waiting in the waiting line or queue.
Probability no customers in the system
Probability n customers in the system
Utilization rate: The proportion of time the system is in use.

24. Decision Areas Arrival rates Number of service channels
Number of phases
Service time
Priority rule
Line arrangement

25. Assumptions of the Basic Queuing Model
Arrivals are served on a first come, first served basis
Arrivals are independent of preceding arrivals
Arrival rates are described by the Poisson probability distribution, and customers come from a very large population
Service times are independent of one another
Service times are described by the negative exponential probability distribution
The service rate is greater than the arrival rate

26. Poisson Distribution Poisson role in the arrival and service process:
Poisson (or random) processes: means that the distribution of both the arrival times and the service times follow the exponential distribution. Because of the mathematical nature of this exponential distribution, we can find many relationships based on performance which help us when looking at the arrival rate and service rate.
Poisson process. An arrival process where customers arrive one at a time and where the interval s between arrivals is described by independent random variables

27. Poisson Distribution e-x P(x) = for x = 0, 1, 2, 3, 4, … x!
where P(x) = probability of x arrivals
x = number of arrivals per unit of time
 = average arrival rate
e = (which is the base of the natural logarithms)

28. Queuing System Operating Characteristics
P0 = probability the service facility is idle
Pn = probability of n units in the system
Pw = probability an arriving unit must wait for service
Lq = average number of units in the queue awaiting service
L = average number of units in the system
Wq = average time a unit spends in the queue awaiting service
W = average time a unit spends in the system

29. Queuing Systems
A three part code of the form A/B/k is used to describe various queuing systems.
A identifies the arrival distribution, B the service (departure) distribution and k the number of channels for the system.
Symbols used for the arrival and service processes are: M - Markov distributions (Poisson/exponential), D - Deterministic (constant) and G - General distribution (with a known mean and variance).
For example, M/M/k refers to a system in which arrivals occur according to a Poisson distribution, service times follow an exponential distribution and there are k servers working at identical service rates.

30. Analytical Formulas
For nearly all queuing systems, there is a relationship between the average time a unit spends in the system or queue and the average number of units in the system or queue.
These relationships, known as Little's flow equations are:
L = W and Lq = Wq

31. Analytical Formulas
When the queue discipline is FCFS, analytical formulas have been derived for several different queuing models including the following:
M/M/1
M/M/k
M/G/1
M/G/k with blocked customers cleared
M/M/1 with a finite calling population
Analytical formulas are not available for all possible queuing systems. In this event, insights may be gained through a simulation of the system.

32. Queuing Models Model Name Example A Single-channel Information counter
system at department store
(M/M/1)
Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Exponential Unlimited FIFO

33. Queuing Models Model Name Example B Multichannel Airline ticket
(M/M/k) counter
Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Multi- Single Poisson Exponential Unlimited FIFO
channel

34. Queuing Models Model Name Example C Constant- Automated car
service wash
(M/G/1)
Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Constant Unlimited FIFO

35. Queuing Models Model Name Example D Limited Shop with only a
population dozen machines
(finite population) that might break
Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Exponential Limited FIFO

36. Model A – Single-Channel
Arrivals are served on a FIFO basis and every arrival waits to be served regardless of the length of the queue
Arrivals are independent of preceding arrivals but the average number of arrivals does not change over time
Arrivals are described by a Poisson probability distribution and come from an infinite population

37. Model A – Single-Channel
Service times vary from one customer to the next and are independent of one another, but their average rate is known
Service times occur according to the negative exponential distribution
The service rate is faster than the arrival rate

38. Single Channel – Single Phase System
Arrivals
Queue
Served units
Service facility
Assumptions
Arrivals have a Poisson distribution
Service times have an exponential distribution
This system is called an M/M/1 Queuing System.

39. Waiting Line Performance Measures
Lq = The average number of customers waiting in queue
L = The average number of customers in the system
Wq = The average waiting time in queue
W = The average time in the system
p = The system utilization rate (% of time servers are busy)

40. Formulas: Single-Server Case

41. M/M/1 Queuing System Single channel Poisson arrival-rate distribution
Exponential service-time distribution
Unlimited maximum queue length
Infinite calling population
Examples:
Single-window theatre ticket sales booth
Single-scanner airport security station

42. Formulas: Single-Server Case con’t

43. State University Computer Lab
A help desk in the computer lab serves students on a first-come, first served basis. On average, 15 students need help every hour. The help desk can serve an average of 20 students per hour.
Based on this description, we know:
µ = 20 students/hour (average service time is 3 minutes)
 = 15 students/hour (average time between student arrivals is 4 minutes)

44. Average Utilization

45. Average Number of Students in the System, and in Line

46. Average Time in the System & in Line

47. Probability of n Students in the Line

48. Example: SJJT, Inc. (A) M/M/1 Queuing System
Joe Ferris is a stock trader on the floor of the New York Stock Exchange for the firm of Smith, Jones, Johnson, and Thomas, Inc. Stock transactions arrive at a mean rate of 20 per hour. Each order received by Joe requires an average of two minutes to process.
Orders arrive at a mean rate of 20 per hour or one order every 3 minutes. Therefore, in a 15 minute interval the average number of orders arriving will be  = 15/3 = 5.

49. Example: SJJT, Inc. (A) Arrival Rate Distribution Question
What is the probability that no orders are received within a 15-minute period?
Answer
P (x = 0) = (50e -5)/0! = e -5 =

50. Example: SJJT, Inc. (A) Arrival Rate Distribution Question
What is the probability that exactly 3 orders are received within a 15-minute period?
Answer
P (x = 3) = (53e -5)/3! = 125(.0067)/6 =

51. Example: SJJT, Inc. (A) Arrival Rate Distribution Question
What is the probability that more than 6 orders arrive within a 15-minute period?
Answer
P (x > 6) = 1 - P (x = 0) - P (x = 1) - P (x = 2)
- P (x = 3) - P (x = 4) - P (x = 5)
- P (x = 6)
= =

52. Example: SJJT, Inc. (A) Service Rate Distribution Question
What is the mean service rate per hour?
Answer
Since Joe Ferris can process an order in an average time of 2 minutes (= 2/60 hr.), then the mean service rate, µ, is µ = 1/(mean service time), or 60/2.
m = 30/hr.

53. Example: SJJT, Inc. (A) Service Time Distribution Question
What percentage of the orders will take less than one minute to process?
Answer
Since the units are expressed in hours,
P (T < 1 minute) = P (T < 1/60 hour).
Using the exponential distribution, P (T < t ) = 1 - e-µt.
Hence, P (T < 1/60) = 1 - e-30(1/60)
= = = %

54. Example: SJJT, Inc. (A) Service Time Distribution Question
What percentage of the orders will require more than 3 minutes to process?
Answer
The percentage of orders requiring more than 3 minutes to process is:
P (T > 3/60) = e-30(3/60) = e = = %

55. Example: SJJT, Inc. (A) Average Time in the System Question
What is the average time an order must wait from the time Joe receives the order until it is finished being processed (i.e. its turnaround time)?
Answer
This is an M/M/1 queue with  = 20 per hour and  = 30 per hour. The average time an order waits in the system is: W = 1/(µ -  )
= 1/( )
= 1/10 hour or 6 minutes

56. Example: SJJT, Inc. (A) Average Length of Queue Question
What is the average number of orders Joe has waiting to be processed?
Answer
Average number of orders waiting in the queue is:
Lq = 2/[µ(µ - )]
= (20)2/[(30)(30-20)]
= 400/300
= 4/3

57. Example: SJJT, Inc. (A) Utilization Factor Question
What percentage of the time is Joe processing orders?
Answer
The percentage of time Joe is processing orders is equivalent to the utilization factor, /. Thus, the percentage of time he is processing orders is:
/ = 20/30
= 2/3 or %

58. M/M/k Queuing System Multiple channels (with one central waiting line)
Poisson arrival-rate distribution
Exponential service-time distribution
Unlimited maximum queue length
Infinite calling population
Examples:
Four-teller transaction counter in bank
Two-clerk returns counter in retail store

59. Multichannel – Single Phase System
Service facility
Queue
Arrivals
Served units
Service facility
Assumptions
Arrivals have a Poisson distribution
Service times have an exponential distribution
This system is called an M/M/skQueuing System where s is the number of servers.

60. Example: SJJT, Inc. (B) M/M/2 Queuing System
Smith, Jones, Johnson, and Thomas, Inc. has begun a major advertising campaign which it believes will increase its business 50%. To handle the increased volume, the company has hired an additional floor trader, Fred Hanson, who works at the same speed as Joe Ferris.
Note that the new arrival rate of orders,  , is 50% higher than that of problem (A). Thus,  = 1.5(20) = 30 per hour.

61. Example: SJJT, Inc. (B) Sufficient Service Rate Question
Why will Joe Ferris alone not be able to handle the increase in orders?
Answer
Since Joe Ferris processes orders at a mean rate of µ = 30 per hour, then  = µ = 30 and the utilization factor is 1.
This implies the queue of orders will grow infinitely large. Hence, Joe alone cannot handle this increase in demand.

62. Example: SJJT, Inc. (B) Probability of n Units in System Question
What is the probability that neither Joe nor Fred will be working on an order at any point in time?

63. Example: SJJT, Inc. (B) Probability of n Units in System (continued)
Answer
Given that  = 30, µ = 30, k = 2 and ( /µ) = 1, the probability that neither Joe nor Fred will be working is:
= 1/[(1 + (1/1!)(30/30)1] + [(1/2!)(1)2][2(30)/(2(30)-30)]
= 1/( ) = 1/3 =

64. Example: SJJT, Inc. (B) Average Time in System Question
What is the average turnaround time for an order with both Joe and Fred working?

65. Example: SJJT, Inc. (B) Average Time in System (continued) Answer
The average turnaround time is the average waiting time in the system, W.
µ( /µ)k (30)(30)(30/30)2
Lq = P0 = (1/3) = 1/3
(k-1)!(kµ -  )2 (1!)((2)(30)-30))2
L = Lq + ( /µ) = 1/3 + (30/30) = 4/3
W = L/(4/3)/30 = 4/90 hr. = min.

66. Example: SJJT, Inc. (B) Average Length of Queue Question
What is the average number of orders waiting to be filled with both Joe and Fred working?
Answer
The average number of orders waiting to be filled is Lq. This was calculated earlier as 1/3 .

67. Example: SJJT, Inc. (C) Economic Analysis of Queuing Systems
The advertising campaign of Smith, Jones, Johnson and Thomas, Inc. (see problems (A) and (B)) was so successful that business actually doubled. The mean rate of stock orders arriving at the exchange is now 40 per hour and the company must decide how many floor traders to employ. Each floor trader hired can process an order in an average time of 2 minutes.

68. Example: SJJT, Inc. (C) Economic Analysis of Queuing Systems
Based on a number of factors the brokerage firm has determined the average waiting cost per minute for an order to be $.50. Floor traders hired will earn $20 per hour in wages and benefits. Using this information compare the total hourly cost of hiring 2 traders with that of hiring 3 traders.

69. Example: SJJT, Inc. (C) Economic Analysis of Waiting Lines
Total Hourly Cost
= (Total salary cost per hour)
+ (Total hourly cost for orders in the system)
= ($20 per trader per hour) x (Number of traders)
+ ($30 waiting cost per hour) x (Average number of orders in the system)
= 20k + 30L.
Thus, L must be determined for k = 2 traders and for k = 3 traders with  = 40/hr. and  = 30/hr. (since the average service time is 2 minutes (1/30 hr.).

70. Example: SJJT, Inc. (C) Cost of Two Servers P0 = 1 /
[1+(1/1!)(40/30)]+[(1/2!)(40/30)2(60/(60-40))]
= 1 / [1 + (4/3) + (8/3)]
= 1/5

71. Example: SJJT, Inc. (C) Cost of Two Servers (continued) Thus,
µ( /µ)k (40)(30)(40/30)2
Lq = P0 = (1/5) = 16/15
(k-1)!(kµ - ) !(60-40)2
L = Lq + ( /µ) = 16/15 + 4/3 = 12/5
Total Cost = (20)(2) + 30(12/5) = $ per hour

72. Example: SJJT, Inc. (C) Cost of Three Servers
[(1/3!)(40/30)3(90/(90-40))] ]
= 1 / [1 + 4/3 + 8/9 + 32/45]
= 15/59

73. Example: SJJT, Inc. (C) Cost of Three Servers (continued)
(30)(40)(40/30)3
Hence, Lq = (15/59) =128/885
(2!)(3(30)-40)2 =
Thus, L = 128/ /30 = 1308/885 (= )
Total Cost = (20)(3) + 30(1308/885) = $ per hour

74. Example: SJJT, Inc. (C) System Cost Comparison Wage Waiting Total
Cost/Hr Cost/Hr Cost/Hr
2 Traders $ $ $112.00
3 Traders
Thus, the cost of having 3 traders is less than that of
2 traders.

75. M/D/1 Queuing System Single channel Poisson arrival-rate distribution
Constant service time
Unlimited maximum queue length
Infinite calling population
Examples:
Single-booth automatic car wash
Coffee vending machine

76. Example: Ride ‘Em Cowboy!
M/D/1 Queuing System
The mechanical pony ride machine at the entrance to a very popular J-Mart store provides 2 minutes of riding for $.50. Children (accompanied of course!) wanting to ride the pony arrive according to a Poisson distribution with a mean rate of 15 per hour.
a) What fraction of the time is the pony idle?
b) What is the average number of children waiting to ride the pony?
c) What is the average time a child waits for a ride?

77. Example: Ride ‘Em Cowboy!
Fraction of Time Pony is Idle
l = 15 per hour
m = 60/2 = 30 per hour
Utilization = l/m = 15/30 = .5
Idle fraction = 1 – Utilization = = .5

78. Example: Ride ‘Em Cowboy!
Average Number of Children Waiting for a Ride
Average Time a Child Waits for a Ride
(or 1 minute)

79. M/G/k Queuing System Multiple channels
Poisson arrival-rate distribution
Arbitrary service times
No waiting line
Infinite calling population
Example:
Telephone system with k lines. (When all k lines are being used, additional callers get a busy signal.)

80. Example: Allen-Booth M/G/k Queuing System
Allen-Booth (A-B) is an OTC market maker. A broker wishing to trade a particular stock for a client will call on a firm like Allen-Booth to execute the order. If the market maker's phone line is busy, a broker will immediately try calling another market maker to transact the order.
A-B estimates that on the average, a broker will try to call to execute a stock transaction every two minutes. The time required to complete the transaction averages 75 seconds. A-B has four traders staffing its phones. Assume calls arrive according to a Poisson distribution.

81. Example: Allen-Booth
This problem can be modeled as an M/G/k system with block customers cleared with:
1/l = 2 minutes = 2/60 hour
l = 60/2 = 30 per hour
1/µ = 75 sec. = 75/60 min. = 75/3600 hr.
µ = 3600/75 = 48 per hour

82. Example: Allen-Booth
% of A-B’s Potential Customers Lost Due to Busy Line First, we must solve for P0
where k = 4 1
P0 =
1 + (30/48) + (30/48)2/2! + (30/48)3/3! + (30/48)4/4!
P0 = .536
continued

83. Example: Allen-Booth
% of A-B’s Potential Customers Lost Due to Busy Line
(l/µ) (30/48)4
Now, P4 = P0 = (.536) 4!
  =.003
Thus, with four traders 0.3% of the potential customers are lost.

84. End of Chapter 11