MD021 - Management and Operations
Service Operations Design Using Waiting Line Models
Outline

Structure of waiting line problems

Single-server model

Multiple-server model

Finite-source, single-server model

Using waiting line models to analyze and design service operations
1
Structure of Waiting Line Problems

Customer population




The service system



The source of input to the service system
Whether the input source is finite or infinite
Whether the customers are patient or impatient
Number of lines - single vs. multiple lines
Arrangement of service facilities - channels and phases
Priority rule
 First-come, first-served (FCFS) discipline
 Other rules - e.g. earliest due date (EDD), shortest processing time (SPT)
 Preemptive discipline
2
Single-Server Model
Assumptions:
Number of servers = 1
Number of phases = 1
Input source: infinite, no balking or reneging
Arrival distribution: Poisson; mean arrival rate = 
Service distribution: Exponential; mean service rate =  ; mean service time = 1 / 
Waiting line: single line; unlimited length
Priority discipline: FCFS
3
Single-Server Operating Characteristics
Average utilization:  


Probability that n customers are in the system: Pn  (1   )  n
Average number of customers in the system: L 
Average number of customers in line: Lq  L
Average time spent in the system: W 
1

Average time spent in line: Wq  W
4


Single-Server Application
Customers arrive at a checkout counter at an average of 20
per hour, according to a Poisson distribution. They are
served at an average rate of 25 per hours, with exponential
service times. Use the single-server model to estimate the
operating characteristics of this system.

L=
Lq 
W=
Wq 
What service rate is required to have customers
average only 10 minutes in the system?
5
Multiple-Server Model
Assumptions:
Number of servers = s
Number of phases = 1
Input source: infinite, no balking or reneging
Arrival distribution: Poisson; mean arrival rate = 
Service distribution: Exponential; mean service rate =  ; mean service time = 1 / 
Waiting line: single line; unlimited length
Priority discipline: FCFS
6
Multiple-Server Operating Characteristics
Average utilization:  

s
Probability that zero customers are in the system:
(  /  ) n (  /  ) s 1 1
P0  [ 

(
)]
n!
s!
1 
n0
s 1
Probability that n customers are in the system:
( /  )n
P0 0  n  s
n!
( /  )n
P0 n  s
s! sn  s
Average number of customers in line:
P0 (  /  ) s 
Lq 
s!(1   ) 2
Average time spent in line: Wq 
Lq

Average time spent in the system: W  Wq 
1

Average number of customers in the system: L  W
7
Multiple-Server Application
Suppose the manager of the checkout system decides to
add another counter. The arrival rate is still 20 customers
per hour, but now each checkout counter will be designed
to service customers at the rate of 12.5 per hour. Use the
multiple-server model to estimate the operating
characteristics of this system.

P0 
1
20 (20 / 12.5) 2 1
[1 

(
)]
12.5
2!
1 
Lq 
Wq 
How do the single-server and multiple-server models
compare?
8
Finite-Source, Single-Server Model
Assumptions:
Number of servers = 1
Number of phases = 1
Input source: finite, equals N customers
Arrival distribution: Exponential interarrival times; mean = 1 / 
Service distribution: Exponential; mean service rate =  ; mean service time = 1 / 
Waiting line: single line; no more than N - 1
Priority discipline: FCFS
9
Finite-Source, Single-Server Operating Characteristics
N!

( ) n ]1
n  0 ( N  n)! 
N
Probability that zero customers are in the system: P0  [ 
Average utilization:   1  P0
Average number of customers in line: Lq  N 

(1  P0 )


Average number of customers in the system: L  N  (1  P0 )

Average time spent in line: Wq  Lq [( N  L) ]1
Average time spent in the system: W  L[( N  L) ]1
10
Finite-Source, Single-Server Application
DBT Bank has 8 copy machines located in various offices
throughout the building. Each machine is used
continuously and has an average time between failures of
50 hours. Once failed, it takes 4 hours for the service
company to send a repair person to have it fixed. What is
the average number of copy machines in repair or waiting
to be repaired?
11
Hilltop Produce Problem
The Hilltop Produce store is staffed by one checkout clerk.
The average checkout time is exponential distributed
around an average of two minutes per customer. An
average of 20 customers arrive per hour.
a. What is the average utilization rate?
b. What is the probability that three or more customers
will be in the checkout area?
c. What is the average number of customers in the waiting
line?
d. If the customers spend an average of 10 minutes
shopping for produce, what is the average time
customers spend in the store?
12
Using Waiting Line Models to Analyze Service Operations
Balance costs against benefits of improving service system. Also, consider the costs of
not making improvements.
Line length - Long lines indicate poor customer service, inefficient service, or inadequate
capacity.
Number of customers in system - A large number causes congestion and dissatisfaction.
Waiting time in line - Long waits are associated with poor service.
Total time in system - May indicate problems with customers, server efficiency, or
capacity.
Service facility utilization - Control costs without unacceptable reduction in service.
13
Using Waiting Line Models to Design Service Operations
Arrival rates - Adjust through advertising, promotions, pricing, appointments.
Number of service facilities - Adjust service system capacity.
Number of phases - Consider splitting service tasks.
Number of servers per facility - Work force size.
Server efficiency - Training, incentives, work methods, capital investment.
Priority rule - Decide whether to allow preemption.
Line arrangement - Single or multiple lines.
14
Supplementary Material:
A. Arrival Distribution
The probability distribution of the number of arrivals (number of customers who arrive
in T time periods), n, follows Poisson distribution:
P ( n) 
(  T )  T
e
for n = 0,1,2…
n
where P(n) = probability of n arrivals in T time periods
 = average number of customer arrivals per time period
e = 2.7183
Mean E (n)  T
Variance Var (n)  T
Let T =1.Then, n, the number of arrivals per unit of time period follows a Poisson
distribution with the following probability function:
P ( n) 
 
e
n
Mean E (n) 

Variance
Var (n)  
15
B. Service Time Distribution
The service time for a customer, t, follows exponential distribution:
P(t  T )  1  e  T
 = average number of customers completing service per time period
where
t = service time of the customer
T = target service time
Mean
E (t ) 
1

1
Var
(
t
)

Variance
2
16