### Module 8A Waiting Lines

```C
Waiting Lines
For Operations Management, 9e by
Krajewski/Ritzman/Malhotra
C–1
Two ways to address waiting lines
 Queuing theory
 Certain types of lines can be described mathematically
 Requires that assumptions are valid for your situation
 Systems with multiple lines that feed each other are
too complex for queuing theory
 Simulation
 Building mathematical models that attempt to act like
real operating systems
 Real-world situations can be studied without imposing
on the actual system
C-2
C–2
Why do we have to wait?
Why do services (and many manufacturing
situations) have queues?
 Arrival
time variance
 Processing
time variance
of capacity – can we afford to always
have more people/servers than customers?
 Costs
C-3
C–3
Waiting Costs and Service Costs
Cost
Total expected cost
Cost of providing
service
Minimum
total cost
Cost of waiting
time
Low level
of service
C-4
Optimal
service level
High level
of service
C–4
Costs of Queues
Cost of capacity
Annoyed customers
Wasted capacity
Lost customers
Space
Too Little Queue
Too Much Queue
Possible opportunity:
e.g. wait in the bar
for a restaurant table
C-5
C–5
Waiting Line Examples
Situation
Arrivals
Servers
Service Process
Bank
Customers
Teller
Deposit, etc.
Doctor’s
office
Patient
Doctor
Treatment
Traffic
intersection
Cars
Light
Controlled
passage
Assembly line
Parts
Workers
Assembly
1–800 software
support
User call-ins
Tech support
personnel
Technical support
C-6
C–6
Waiting Line Terminology
 Queue: Waiting line
 Arrival: 1 person, machine, part, etc. that
arrives and demands service
 Queue discipline: Priority rules for
determining the order that arrivals receive
service
 Channel: Number of waiting lines
 Phase: Number of steps in service
C-7
C–7
Assumptions
 Arrivals
 At
random (Poisson, exponential distributions)
 Fixed
(appointments, service intervals)
 Service times
 Variable
 Fixed
(exponential, normal distributions)
(constant service time)
 Other
 Size
of arrival population, priority rules, balking,
reneging
C-8
C–8
Structure of Waiting-Line Problems
Customer
population
Service system
Waiting line
Priority
rule
Service
facilities
Served
customers
Figure C.1 – Basic Elements of Waiting-Line Models
C–9
Structure of Waiting-Line Problems
1. An input, or customer population, that
generates potential customers
2. A waiting line of customers
3. The service facility, consisting of a person
(or crew), a machine (or group of
machines), or both necessary to perform
the service for the customer
4. A priority rule, which selects the next
customer to be served by the service
facility
C – 10
Customer Population
 The source of input
 Finite or infinite source

Customers from a finite source reduce the
chance of new arrivals

Customers from an infinite source do not
affect the probability of another arrival
 Customers are patient or impatient

Patient customers wait until served

Impatient customer can either balk (not even
enter a long line) or join the line and later
renege (leave before being served)
C – 11
The Service System
 Number of lines

A single-line keeps servers uniformly busy and levels
waiting times among customers

A multiple-line arrangement is favored when servers
provide a limited set of services
 Arrangement of service facilities

Single-channel, single-phase

Single-channel, multiple-phase

Multiple-channel, single-phase

Multiple-channel, multiple-phase

Mixed arrangement
C – 12
The Service System
Service facilities
Service facilities
(a) Single line
Figure C.2 – Waiting-Line
Arrangements
(b) Multiple lines
C – 13
The Service System
Service
facility
(a) Single channel, single phase
Service
facility 1
Service
facility 2
(b) Single channel, multiple phase
Figure C.3 – Examples of Service
Facility Arrangements
C – 14
The Service System
Service
facility 1
Service
facility 2
(d) Multiple channel, multiple phase
(e.g., Laundromat)
(c) Multiple channel, single phase
Figure C.3 – Examples of Service
Facility Arrangements
Service
facility 1
Service
facility 3
Service
facility 2
Service
facility 4
C – 15
The Service System
Routing for
Routing for
Routing for
(e) Mixed arrangement
Service
facility 1
: 1–2–4
: 2–4–3
: 3–2–1–4
Service
facility 2
Service
facility 3
Service
facility 4
Figure C.3 – Examples of Service
Facility Arrangements
C – 16
Priority Rule
 First-come, first-served (FCFS)—used by
most service systems
 Other rules

Earliest due date (EDD) = urgent service

Shortest processing time (SPT) = express line
 Preemptive discipline—allows a higher
priority customer to interrupt the service of
another customer or be served ahead of
another who would have been served first
C – 17
Probability
CUSTOMERS
IN THE
SYSTEM
ARRIVALS
x
(Poisson)
n
Probability that
Service Time  t
Probability
SERVICE
(Exponential)
Time t
C – 18
Probability Distributions
 The sources of variation in waiting-line problems
come from the random arrivals of customers and
the variation of service times
 Arrival distribution

Customer arrivals can often be described by the
Poisson distribution with mean = T and variance
also = T

Arrival distribution is the probability of n arrivals in T
time periods

Interarrival times are the times between successive
arrivals
C – 19
Poisson Distributions for Arrival Rates
0.30
0.30
0.25
0.20
0.20
Probabilit
y
Probabilit
Probabilit
yy
0.25
0.15
0.15
0.10
0.10
0.05
0.00
0.05
0 1 2 3 4 5 6 7 8 9 10 11 12
x
=2
0.00
0
1
2
3
4
5
6 7
x
8
9 10 11 12
=4
 = average arrival rate = number/period
C-20
C – 20
Poisson Distribution
(T)n -T
Pn =
e
for n = 0, 1, 2,…
n!
where
Pn = Probability of n arrivals in T time periods
 = Average numbers of customer arrivals
per period
e = 2.7183
C – 21
Probability of Customer Arrivals
EXAMPLE C.1
Management is redesigning the customer service process in a
large department store. Accommodating four customers is
important. Customers arrive at the desk at the rate of two
customers per hour. What is the probability that four customers
will arrive during any hour?
SOLUTION
In this case using customers per hour, T = 1 hour, and n = 4
customers. The probability that four customers will arrive in
any hour is
[2(1)]4
16 –2
–2(1)
P4 =
=
e = 0.090
e
4!
24
C – 22
Negative Exponential Distribution
Probability that Service Time  t
1.0
–-t
t , for t > 0
Probability
Service
is greater
Probability
that that
Service
TimeTime
is greater
thanthan
t = et=e
for t > 0
0.9
0.8
Average Service Rate () = 3 customers per
hour
Average
Service Rate () = 3 customers per hour
Average
per
Average
ServiceTime
Time==20
20minutes
minutes
per customer
customerService
0.7
0.6
0.5
Average Service Rate
Service
() =Average
1 customer
per Rate () =
hour1 customer per hour
0.4
0.3
0.2
0.1
0.0
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
Time t in Hours
C-23
C – 23
Service Time
 Service time distribution can be described by an
exponential distribution with mean = 1/ and
variance = (1/ )2
 Service time distribution: The probability that the
service time will be no more than T time periods
can be described by the exponential distribution
P(t ≤ T) = 1 – e–T
where
μ = average number of customers completing
service per period
t = service time of the customer
T = target service time
C – 24
Service Time Probability
EXAMPLE C.2
The management of the large department store in Example C.1
must determine whether more training is needed for the
customer service clerk. The clerk at the customer service desk
can serve an average of three customers per hour. What is the
probability that a customer will require less than 10 minutes of
service?
SOLUTION
We must have all the data in the same time units. Because
 = 3 customers per hour, we convert minutes of time to
hours, or T = 10 minutes = 10/60 hour = 0.167 hour. Then
P(t ≤ T) = 1 – e–T
P(t ≤ 0.167 hour) = 1 – e–3(0.167) = 1 – 0.61 = 0.39
C – 25
Using Waiting-Line Models
 Balance costs against benefits
 Operating characteristics
 Line
length
 Number
of customers in system
 Waiting
time in line
 Total
time in system
 Service
facility utilization
C – 26
Single-Server Model
 Single-server, single line of customers, and
only one phase
 Assumptions are
1. Customer population is infinite and patient
2. Customers arrive according to a Poisson
distribution, with a mean arrival rate of 
3. Service distribution is exponential with a mean
service rate of 
4. Mean service rate exceeds mean arrival rate
5. Customers are served FCFS
6. The length of the waiting line is unlimited
C – 27
Single-Server Model
 = Average utilization of the system =


Rn = Probability that n customers are in the system
= (1 –  ) n
L = Average number of customers in the service system
=

–
Lq = Average number of customers in the waiting line
= L
W = Average time spent in the system, including service
1
=
–
Wq = Average waiting time in line
= W
C – 28
Calculating the Operating
Characteristics
EXAMPLE C.3
The manager of a grocery store in the retirement community of
Sunnyville is interested in providing good service to the senior
citizens who shop in her store. Currently, the store has a
separate checkout counter for senior citizens. On average, 30
senior citizens per hour arrive at the counter, according to a
Poisson distribution, and are served at an average rate of 35
customers per hour, with exponential service times. Find the
following operating characteristics:
a. Probability of zero customers in the system
b. Average utilization of the checkout clerk
c. Average number of customers in the system
d. Average number of customers in line
e. Average time spent in the system
f. Average waiting time in line
C – 29
Calculating the Operating
Characteristics
SOLUTION
The checkout counter can be modeled as a single-channel,
single-phase system. Figure C.4 shows the results from the
Waiting-Lines Solver from OM Explorer.
Figure C.4 – Waiting-Lines Solver for Single-Channel, Single-Phase System
C – 30
Calculating the Operating
Characteristics
Both the average waiting time in the system (W) and the
average time spent waiting in line (Wq) are expressed in
hours. To convert the results to minutes, simply multiply
by 60 minutes/ hour. For example, W = 0.20(60) minutes,
and Wq = 0.1714(60) = 10.28 minutes.
C – 31
Analyzing Service Rates
EXAMPLE C.4
The manager of the Sunnyville grocery in Example C.3 wants
a. What service rate would be required so that customers
average only 8 minutes in the system?
b. For that service rate, what is the probability of having more
than four customers in the system?
c. What service rate would be required to have only a 10
percent chance of exceeding four customers in the system?
C – 32
Analyzing Service Rates
SOLUTION
The Waiting-Lines Solver from OM Explorer could be used
iteratively to answer the questions. Here we show how to solve
the problem manually.
a. We use the equation for the average time in the system
and solve for 
W=
1
–
8 minutes = 0.133 hour =
1
 – 30
0.133 – 0.133(30) = 1
 = 37.52 customers/hour
C – 33
Analyzing Service Rates
b. The probability of more than four customers in the system
equals 1 minus the probability of four or fewer customers in
the system.
4
P = 1 – n
P
=0 n
4
and
n
= 1 – n
(1
–

)

=0
=
30
= 0.80
37.52
Then,
P = 1 – 0.21(1 + 0.8 + 0.82 + 0.83 + 0.84)
= 1 – 0.672 = 0.328
Therefore, there is a nearly 33 percent chance that more
than four customers will be in the system.
C – 34
Analyzing Service Rates
c. We use the same logic as in part (b), except that  is now a
decision variable. The easiest way to proceed is to find the
correct average utilization first, and then solve for the
service rate.
P (n>k) = 1 – (1 –  )(1 +  +  2 +  3 +  4)
= 1 – (1 –  )(1 +  +  2 +  3 +  4) +  (1 +  +  2 +  3 +  4)
= 1 – 1 –  –  2 –  3 –  4 +  +  2 +  3 +  4 +  5 =  5 =  k+1
or
 = P1/5
If
P = 0.10
 = (0.10)1/5 = 0.63
C – 35
Analyzing Service Rates
Therefore, for a utilization rate of 63 percent, the probability of
more than four customers in the system is 10 percent.
For  = 30, the mean service rate must be
30
 = 0.63
 = 47.62 customers/hour
C – 36
Multiple-Server Model
 Service system has only one phase,
multiple-channels
 Assumptions (in addition to single-server
model)

There are s identical servers

The service distribution for each server is
exponential

The mean service time is 1/

s should always exceed 
C – 37
Multiple-Server Model
 = Average utilization of the system =

s
P0 = Probability that zero customers are in the system
= 
s1

 n0
 /  
n
 /    1  
s

n!
s!
1


 1  
Pn = Probability that n customers are in the system
  /  
P0 0  n  s
= 

n!

n
  /   P
n s
0
 s! s n  s
n
C – 38
Multiple-Server Model
Lq = Average number of customers in the waiting line
P0   /   
s
=
s! 1   
2
Wq = Average waiting time of customers in line =
Lq

W = Average time spent in the system, including service
= Wq 
1

L = Average number of customers in the service system
= W
C – 39
Estimating Idle Time and Costs
EXAMPLE C.5
The management of the American Parcel Service terminal in
Verona, Wisconsin, is concerned about the amount of time the
company’s trucks are idle (not delivering on the road), which
the company defines as waiting to be unloaded and being
unloaded at the terminal. The terminal operates with four
and each crew costs \$30 per hour. The estimated cost of an idle
truck is \$50 per hour. Trucks arrive at an average rate of three
per hour, according to a Poisson distribution. On average, a
crew can unload a semitrailer rig in one hour, with exponential
service times. What is the total hourly cost of operating the
system?
C – 40
Estimating Idle Time and Costs
SOLUTION
The multiple-server model is appropriate. To find the total cost
of labor and idle trucks, we must calculate the average number
of trucks in the system.
Figure C.5 shows the results for the American Parcel Service
problem using the Waiting-Lines Solver from OM Explorer.
Manual calculations using the equations for the multiple-server
model are demonstrated in Solved Problem 2 at the end of this
supplement. The results show that the four-bay design will be
utilized 75 percent of the time and that the average number of
trucks either being serviced or waiting in line is 4.53 trucks.
That is, on average at any point in time, we have 4.53 idle
trucks. We can now calculate the hourly costs of labor and idle
trucks:
C – 41
Estimating Idle Time and Costs
Labor cost:
\$30(s) = \$30(4)
=
\$120.00
Idle truck cost:
\$50(L) = \$50(4.53)
=
226.50
Total hourly cost
=
\$346.50
Figure C.5 – Waiting-Lines Solver for Multiple-Server Model
C – 42
Little’s Law
 Relates the number of customers in a waiting line
system to the waiting time of customers
 Using the notation from the single-server and
multiple-server models it is expressed as L = W
or Lq = Wq
 Holds for a wide variety of arrival processes,
service time distributions, and numbers of
servers
 Only need to know two of the parameters
C – 43
Little’s Law
Ls = Ws
DMV
Count arrivals ()
Periodically count #
customers in line (Lq)
Lq = Wq
Estimate average
time in line (Wq)
Widget manufacturing
Determine average
process time (Ws)
Determine average
arrival rate ()
C-44
Estimate average
work-in-process (Ls)
C – 44
Little’s Law
 Service

Estimate W
L customers
Average time
=
W
=
in the facility
 customer/hour
30
=
= 0.75 hours or 45 minutes
40
 Manufacturing

Estimate the average work-in-process L
Work-in-process = L = W
= 5 gear cases/hour (3 hours) = 15 gear cases
C – 45
Finite-Source Model
 Assumptions

Follows the assumption of the single-server,
except that the customer population is finite

Having only N potential customers

If N > 30, then the single-server model with the
assumption of infinite customer population
can be used for a rough approximation
C – 46
Finite-Source Model
P0 = Probability that zero customers are in the system
= 
N

 n  0
  
  
 N  n !    
N!
n
1
 = Average utilization of the server = 1 – P0
Lq = Average number of customers in the waiting line
= N     1  P 
0

C – 47
Finite-Source Model
L = Average number of customers in the service system
=
N 


1 
P0 
Wq = Average waiting time in line = L q  N  L  
1
Ws = Average time spent in the system, including service
=
L  N  L  
1
C – 48
Analyzing Maintenance Costs
EXAMPLE C.6
The Worthington Gear Company installed a bank of 10 robots
about 3 years ago. The robots greatly increased the firm’s labor
productivity, but recently attention has focused on
maintenance. The firm does no preventive maintenance on the
robots because of the variability in the breakdown distribution.
Each machine has an exponential breakdown (or interarrival)
distribution with an average time between failures of 200 hours.
Each machine hour lost to downtime costs \$30, which means
that the firm has to react quickly to machine failure. The firm
employs one maintenance person, who needs 10 hours on
average to fix a robot. Actual maintenance times are
exponentially distributed. The wage rate is \$10 per hour for the
maintenance person, who can be put to work productively
elsewhere when not fixing robots. Determine the daily cost of
labor and robot downtime.
C – 49
Analyzing Maintenance Costs
SOLUTION
The finite-source model is appropriate for this analysis because
the customer population consists of only 10 machines and the
other assumptions are satisfied. Here,  = 1/200, or 0.005 breakdown per hour, and  = 1/10 = 0.10 robot per hour. To calculate
the cost of labor and robot downtime, we need to estimate the
average utilization of the maintenance person and L, the
average number of robots in the maintenance system at any
time. Figure C.6 shows the results for the Worthington Gear
Problem using the Waiting-Lines Solver from OM Explorer.
C – 50
Analyzing Maintenance Costs
Manual computations using the equations for the finite-source
model are demonstrated in Solved Problem 3 at the end of this
supplement. The results show that the maintenance person is
utilized only 46.2 percent of the time, and the average number
of robots waiting in line or being repaired is 0.76 robot.
However, a failed robot will spend an average of 16.43 hours in
the repair system, of which 6.43 hours of that time is spent
waiting for service. While an individual robot may spend more
than two days with the maintenance person, the maintenance
person has a lot of idle time with a utilization rate of only 42.6
percent. That is why there is only an average of 0.76 robot
being maintained at any point of time.
C – 51
Analyzing Maintenance Costs
Figure C.5 – Waiting-Lines Solver for Finite-Source Model
The daily cost of labor and robot downtime is
Labor cost:
(\$19/hour)(8 hours/day)(0.462 utilization)
=
\$36.96
Idle robot cost:
(0.76 robot)(\$30/robot hour)(8 hours/day)
=
182.40
Total daily cost
=
\$219.36
C – 52
Decision Areas for Management
1. Arrival rates
2. Number of service facilities
3. Number of phases
4. Number of servers per facility
5. Server efficiency
6. Priority rule
7. Line arrangement
C – 53
Using Waiting Line Methods to Analyze
Operations
 Line Length
 Number of Customers in System
– customers in line and being served
 Waiting Time in Line
 Total Time in System
– time in line and being served
 Service Facility Utilization
– percentage of time servers are busy or idle
C-54
C – 54
Remember:
 and  are Rates
  = Mean number of
arrivals per time
period

If average service time is
15 minutes, then μ is 4
customers/hour
e.g., 3 units/hour
  = Mean number of
people or items
served per time period

e.g., 4 units/hour

1/ = 15 minutes/unit
C – 55
Multi-Channel, Single Phase System
(Classic Setup)
Service system
Queue
Arrivals
Service
facility
Served
units
Service
facility
Example: Bank customers pick shortest line
for one of several tellers.
C-56
C – 56
Multi-Channel, Single Phase System
(Modern Approach)
Service system
Queue
Arrivals
Service
facility
Served
units
Service
facility
Example: Bank customers wait in single line
for one of several tellers.
C-57
C – 57
Multi-Channel, Single Phase System
(Classic versus Modern Approach)
Performance measures are the same if customers in
Classic approach are allowed to switch lines if their
line is moving slower. So why the modern approach?
 More uniform customer experience (variance
around average is smaller)
 Customers are kept moving more often
(psychological benefit)
 More privacy at service window
 Easier to manage lines
C-58
C – 58
Solved Problem 1
A photographer takes passport pictures at an average rate of 20
pictures per hour. The photographer must wait until the
customer smiles, so the time to take a picture is exponentially
distributed. Customers arrive at a Poisson-distributed average
rate of 19 customers per hour.
a. What is the utilization of the photographer?
b. How much time will the average customer spend with the
photographer?
SOLUTION
a. The assumptions in the problem statement are consistent
with a single-server model. Utilization is

 =  = 19 = 0.95
20
C – 59
Solved Problem 1
b. The average customer time spent with the photographer is
W =
1
1
=
= 1 hour
–
20 – 19
C – 60
Solved Problem 2
The Mega Multiplex Movie Theater has three concession clerks
serving customers on a first come, first-served basis. The
service time per customer is exponentially distributed with an
average of 2 minutes per customer. Concession customers wait
in a single line in a large lobby, and arrivals are Poisson
distributed with an average of 81 customers per hour. Previews
run for 10 minutes before the start of each show. If the average
time in the concession area exceeds 10 minutes, customers
become dissatisfied.
a. What is the average utilization of the concession clerks?
b. What is the average time spent in the concession area?
C – 61
Solved Problem 2
SOLUTION
a. The problem statement is consistent with the multiple-server
model, and the average utilization rate is
=

s
=
81 customers/
hour
60 minutes/se rver hour 
 3 servers 

2
minutes/cu
stomer


= 0.90
The concession clerks are busy 90 percent of the time.
C – 62
Solved Problem 2
1
b. The average time spent in the system, W, is W = W q 

Here,
Wq =
Lq

P0   /   
s
Lq =
s! 1   
2
 s1  /  
 /    1



n!
s!
 1 
 n0
n
and P0 =
s



1
We must solve for P0, Lq, and Wq, in that order, before we can
solve for W:
 s1  /  
 /    1


P0 =  
n!
s!
 1 
 n0
n



1
1
=
1
=
s
 81 / 30   2 . 7  2
1

2
 2 .7  
1 



6
1

0
.
9



1
1  2 . 7  3 . 645  32 . 805
3
=
1
40 . 15
= 0.0249
C – 63
Solved Problem 2
P0   /   
Lq =
Wq =
s! 1   
Lq

=
3! 1  0 . 9 
7.352 customers
=
W = Wq 
2
0 . 0249  81 / 30   0 . 9 
3
s
81 customers/
1

hour
= 0.0908 hours +
2
=
0 . 4411
6  0 . 01 
= 7.352 customers
= 0.0908 hour
1
30

hour =  0.1241 hour 
60 minutes 

hour


= 7.45 minutes
With three concession clerks, customers will spend an
average of 7.45 minutes in the concession area.
C – 64
Solved Problem 3
The Severance Coal Mine serves six trains having exponentially
distributed interarrival times averaging 30 hours. The time
required to fill a train with coal varies with the number of cars,
weather-related delays, and equipment breakdowns. The time to
fill a train can be approximated by an exponential distribution
with a mean of 6 hours 40 minutes. The railroad requires the
coal mine to pay large demurrage charges in the event that a
train spends more than 24 hours at the mine. What is the
average time a train will spend at the mine?
SOLUTION
The problem statement describes a finite-source model, with
N = 6. The average time spent at the mine is W = L[(N – L)]–1,
with 1/ = 30 hours/train,  = 0.8 train/day, and  = 3.6
trains/day. In this case,
C – 65
Solved Problem 3
 N
N!    
  



N

n
!
   
 n  0
n
P0 =
=
=
1
=
1
6
 0 .8 
  6  n !  3 . 6 
n0
6!
n
1
 6!  0 . 8    6!  0 . 8    6!  0 . 8    6!  0 . 8    6!  0 . 8    6!  0 . 8    6!  0 . 8  
  
  
  
  
  
  
 
 
6
!
3
.
6
5
!
3
.
6
4
!
3
.
6
3
!
3
.
6
2
!
3
.
6
1
!
3
.
6
0
!
3
.
6













 

 
 
 
 
 
 
0
1
2
3
1
=
1  1 . 33  1 . 48  1 . 32  0 . 88  0 . 39  0 . 09
L= N 


1 
P0 
W = L  N  L  
1
 3 .6
= 6
 0 .8
=
4
1
6 . 49
5
6
= 0.1541
1  0 . 1541   = 2.193 trains
2 . 193
 3 . 807 0 . 8

= 0.72 day
Arriving trains will
spend an average of
0.72 day at the coal
mine.
C – 66
Izzy’s Ice Cream Stand
 You are opening an ice cream stand that has a single employee (you).
You expect to see about 25 customers an hour. It takes you an average
of 2 minutes to serve each customer. Customers are served in a FCFS
manner.
 Your research suggests that if there is a line of more than 4 people that
customers have to wait more than 6 minutes to get their order filled
they are not likely to come back.
 How well will this system do at satisfying customers?
C-67
C – 67
Izzy’s Ice Cream Stand (2)
 = 25 customers/hr
ρ =  /  = .833
 = 30 customers/hr
L
=  / ( - ) = 5 customers
Pn = probability that n customers are in the system = (1- ρ) ρn
P0
P1
P2
P3
P4
=
=
=
=
=
(1 – .833) x .833
(1 – .833) x .833
(1 – .833) x .833
(1 – .833) x .833
(1 – .833) x .833
=
0
=
1
=
=
2
=
3
.167
.139
.116
.097
.080
.167
cumulative
.306
.422
.519
.599
Pmore than 4 = 1 – .599 4= .401
Customers in the
system
C-68
C – 68
Izzy’s Ice Cream Stand (3)
 = 25 customers/hr
 = 30 customers/hr
ρ =  /  = .833
L =  / ( - )
Lq = ρL
= 25 / (30 – 25) = 5 customers
= .833 x 5
= 4.17 customers
W = 1 / ( - )
Wq = ρW
= 1 / (30 – 25) = .2 hr
= .833 x 12 min
C-69
= 12 min
= 10 min
Why not
L–1?
C – 69
Application C.1
Customers arrive at a checkout counter at an average 20 per
hour, according to a Poisson distribution. They are served at an
average rate of 25 per hour, with exponential service times. Use
the single-server model to estimate the operating
characteristics of this system.
 = 20 customer arrival rate per hour
 = 25 customer service rate per hour
SOLUTION
1. Average utilization of system

 =  = 20 = 0.8
25
C – 70
Application C.1

20
=4
25 – 20
2. Average number of customers
in the service system
L=
3. Average number of customers
in the waiting line
Lq = L = 0.8(4) = 3.2
4. Average time spent in the
system, including service
W=
5. Average waiting time in line
Wq = W = 0.8(0.2) = 0.16
–
=
1
1
=
= 0.2
–
25 – 20
C – 71
Application C.2
In the checkout counter example, what service rate is required
to have customers average only 10 minutes in the system?
SOLUTION
1
W=
= 0.17 hr (or 10 minutes)
–
0.17( – ) = 1, where  = 20 customers arrival rate per hour
1 + 0.17(20)
=
= 25.88 or about 26 customers per hour
0.17
C – 72
Application C.3
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. What is the waiting time
in line of the new system?
s = 2,  = 12.5 customers per hour,  = 20 customers per hour
SOLUTION
1. Average utilization of the system
=

s
=
20
2 12 . 5 
= 0.8
C – 73
Application C.3
2. Probability that zero
customers are in the
system
P0 =
=




1
s





 

   1
 1    s!  1  








20
1 

12 . 5













1

2

 20 



1
12
.
5


 


2!
 1  0 . 8  
= 0.11
C – 74
Application C.3
3. Average number of
customers in the
waiting line
P0   /   
s
Lq =
s! 1   
2
2
=
4. Average waiting time
of customers in line
Wq =
 20 
0 . 11 
 0 .8
 12 . 5 
2! 1  0 . 8 
Lq

=
1 . 408
20
2
= 1.408
= 0.0704 hrs
(or 4.224 minutes
C – 75
Application C.4
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?
 = 1/50 = 0.02 copiers per hour
 = 1/4 = 0.25 copiers per hour
C – 76
Application C.4
SOLUTION
 N
N!    
  

 n  0  N  n !    
n
1. Probability that zero
customers are in the system
=
P0 =
1
1
8!
8!
 8!
0
1
8

0 . 08    0 . 08      0 . 08 
 8!

7!
0!
= 0.44
2. Average utilization of the
server
 = 1 – P0 = 1 – 0.44 = 0.56
3. Average number of
customers in the service
system
L= N 
= 8


1 
0 . 25
0 . 02
P0 
1  0 . 44  = 1
C – 77
Application C.5
The Hilltop Produce store is staffed by one checkout clerk. The
average checkout time is exponentially distributed around an
average of two minutes per customer. An average of 20
customers arrive per hour.
What is the average utilization rate?
SOLUTION

 =  = 20 = 0.667
30
C – 78
Application C.5
What is the probability that three or more customers will be in
the checkout area?
First calculate 0, 1, and 2 customers will be in the checkout
area:
Rn = (1 –  ) 0 = (0.333)(0.667)0 = 0.333
Rn = (1 –  ) 1 = (0.333)(0.667)1 = 0.222
Rn = (1 –  ) 2 = (0.333)(0.667)2 = 0.111
Then calculate 3 or more customers will be in the checkout area:
1 – P0 – P1 – P2 = 0.333 – 0.222 – 0.111 = 0.334
C – 79
Application C.5
What is the average number of customers in the waiting line?
Lq = L = 

–
 20

=  0 . 667 
 = 1.333
30

20


What is the average time customers spend in the store?
W=
1
1
=
= 0.1 hr  60 min/hr = 6 minutes
 –  30  20